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http://www.nutrition-nutritionists.com/nutritional-status/body-composition/body-composition-updates/ | 1,519,005,295,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812306.16/warc/CC-MAIN-20180219012716-20180219032716-00414.warc.gz | 525,897,888 | 12,651 | # Body composition UPDATES
## p and Animal size —
In five minutes a human will burn around 350 joules (J) of energy per kilogram of body mass. A kilogram of mice will burn 3,000 J in the same time, and a 4,000- kg African elephant will burn just 200 J per kilogram. On a gram-for-gram basis, large animals burn less energy and require less food than small ones. The exact relationship between metabolic rate and body mass, and the mechanistic basis for it, remains an open question. Elephants are some 200,000 times greater in mass than mice, but require only 15,000 times more energy each day.
In Nature April 1 2010 p 753 Kolokotrones et at. show that the standard power equation that has been used for decades to describe the relationship may not be appropriate for mammals.
They use the deviation from this equation to test one of the most prominent explanations for the relationship.
In Paris in the 1830s, Sarrus and Rameaux suggested that, because the heat produced by an animal as a by-product of metabolism must be lost through the body surface, the rate at which it produces heat (met¬abolic rate) should be proportional not to its body mass, but to the surface area over which the heat is lost. Rubner then demonstrated empirically in 1883 that the basal metabolic rate of dogs is proportional to their body sur¬face area.
In 1916, Krogh suggested that the relationship between body mass and metabolic rate is best described by a power function, so that metabolic rate is not proportional to mass, but to mass raised to some power p. When p is equal to one, the relationship is a straight line and metabolic rate is said to scale isometrically (in proportion) with mass. When p does not equal one, the relationship is curved, and meta¬bolic rate is said to scale allometrically.
Krogh also suggested that for endotherms (birds and mammals) p is close to the value of 2/3 suggested by Rubner’s ‘surface law’, whereas for ectotherms, such as insects, fish, amphibians and reptiles, the p value is closer to 1.
However, in 1932 Kleiber, Brody and Proctor showed that metabolic rate was on a scale with a p value close to ¾. ( Kleiber’s law).
Kolokotrones et al. however demonstrate that a stand¬ard power equation may not be appropriate for describing the relationship between basal metabolic rate and body mass in mammals. They show that the value of p increases with body size: metabolic rate increases more rapidly with mass for large mammals than for small ones. They show that the relationship between mass and metabolic rate has convex curvature on a logarithmic scale, and is therefore not a pure power law, even after accounting for body temperature.
These results extend our understanding of a very complicated topic .
Size, surface are , heat production and loss, rate of movement, mass, types of muscles and more.
White 2010 There is no single p Nature vol 464 p 691
Kolokotrones et al 2010 Curvature in metabolic scaling . Nature vol 464 pp 753-756
## Physical Measurements
Pengelly and Morris (Scottish Medical Journal 2009) have reviewed the relative merits of measurements of body mass index and weight distribution.
It has been accepted for many years that being overweight or obese, as indicated by a body mass index (BM I) of 25 or over for the former and 30 or over for the latter, is associated with impairment of long term health and prognosis. The World Health Organisation (WHO) has indicated that, in Caucasians, waist measurements of 94cm or more in men, and 80cm or more in women have similar adverse effects on health, with increased risks at 102 cm or more in men and 88 cm in women.
The role of waist-hip ratio (W/H) and whether it represents a better index than waist (W) measurement alone is being debated; many papers favour waist measurement alone. But two papers in 2005 discussing 27,098 subjects, 12,461 of whom had myocardial infarction and 14,637 controls, come down firmly in favour of W/H and were followed by a Lancet Editorial entitled Farewell to Body Mass Index?
Life assurance companies at medical examination usually request height and weight measurements (and therefore BMI). Most ask for waist measurements and a few hip measurements in addition (and therefore W/H).
The authors have reviewed the data in 816 consecutive subjects for life assurance examination in whom BMls, Ws and W/Hs were all recorded. In these the evidence supports the use of W as the best indicator of risk in men (634 cases), but not in the relatively small number of women (182 cases) in whom H appeared better.
We believe that BMI, W and W/H should be recorded in every subject at life assurance examination so that the insurance companies in the long term will be able to reach valid conclusions about their individual and collective value.
Pengelly and Morris Body mass index and weight distribution . Scottish Medical Journal 2009 vol 54 p 17-21
## Waist Circumference Review
In December 2008, the World Health Organization (WHO) convened a consultation to discuss cut-points for waist circumference (WC).
There are large differences in body composition in men and women, with women having more body fat. Fat distribution also differs with gender, with men having a relatively more central distribution of fat. These differences begin early in life and become more apparent in puberty due to changes in sex hormone levels. In both, men and women, waist and waist-to-hip ratio increase with age. A large portion of this increase is due to gains in body weight, but the increases observed are larger than those that would be predicted from increases in the body mass index alone, and increases in WC are seen with aging in the absence of weight gain. The current practice of using separate waist cut-points by gender is appropriate. Although WC increases with age, so does the risk of many chronic diseases. An evaluation of the need for age-specific waist cut-points in adults would need to consider disease risk.
Current waist circumference (WC) and waist-to-hip ratio (WHR) cutoffs have been identified from studies of predominantly European-derived populations. However, these cutoffs may not be appropriate for other ethnic groups. Studies investigating ethnic-specific cutoffs were identified among Aboriginal, Asian, African (Sub-Saharan), African-American, Hispanic, Middle Eastern, Pacific Islander and South American populations. The majority of studies recommending ethnic-specific cutoffs was for Asian populations. Few studies recommended cutoffs in Aboriginal, African (Sub-Saharan), Pacific Islanders and South American populations. All studies were cross-sectional, and the overwhelming majority of studies used receiver operating characteristic curves. The studies used a number of methods for assessing WC and WHR, and a variety of outcome measures, making cross-study comparison difficult. There is possible evidence that Asians should have a lower WC cutoff than Europeans. The evidence is insufficient for specific cutoffs for African-American, Hispanic and Middle Eastern populations but some studies indicate current cutoffs for Europeans may be appropriate, whereas there is insufficient evidence for the other ethnic groups. Future studies are needed to address the methodological limitations of the current literature.
J Stevens, E G Kat and Huxley (2010) Associations between gender, age and waist circumference European Journal of Clinical Nutrition 64, 6–15;
S A Lear et al (2010) Appropriateness of waist circumference and waist-to-hip ratio cutoffs for different ethnic groups European Journal of Clinical Nutrition 64, 42–61 | 1,599 | 7,615 | {"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} | 2.671875 | 3 | CC-MAIN-2018-09 | longest | en | 0.945724 |
https://www.sanfoundry.com/analog-communications-questions-answers-switching-circuit/ | 1,670,010,063,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00284.warc.gz | 1,030,672,814 | 21,117 | # Analog Communications Questions and Answers – Switching Circuit
«
»
This set of Analog Communications Multiple Choice Questions & Answers (MCQs) focuses on “Switching Circuit”.
1. If we compared to the USB, the power in the LSB is ________
a) one-fourth of USB
b) twice of USB
c) same as USB
d) half of USB
Explanation: Generally, USB (upper sideband) and LSB (lower sideband) contain same information and are identical to one another. Thus, both entail same power.
2. What will be the effect on power in amplitude modulated system if we increase the modulation index?
a) increases
b) becomes twice
c) remains same
d) decreases
Explanation: If we increase or decrease the modulation index, accordingly the power of amplitude modulated system will also increase or decrease. This is because the power is directly proportional to the modulation index.
3. What is the modulation index for getting maximum power in amplitude modulated system?
a) 1
b) 12
c) 0.3
d) 0.8
Explanation: For getting maximum power, it is required to set the maximum and minimum voltage of the modulated wave such that we get modulation index equal to unity. This implies 100% modulation.
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4. How can we derive FM from PM?
a) by integrating the modulating signal before applying to the PM oscillator
b) by integrating signal out of the PM oscillator
c) by differentiating the modulating signal before applying to the PM oscillator
d) by differentiating signal out of the PM oscillator
Explanation: Frequency modulation can be derived from phase modulated system, if we integrate our modulating signal before applying our wave to PM oscillator. Likewise, Phase modulation can be obtained from frequency modulated system, if we differentiate our modulating signal before applying our wave to FM oscillator.
5. FM bandwidth can be approximated by ________
a) Carson’s rule
b) Armstrong’s rule
c) Bessel’s rule
d) Pythagoras rule
Explanation: According to Carson’s rule, the required bandwidth is twice the sum of the maximum frequency deviation and the maximum modulating signal frequency.
B = 2(fd+fm)Hz.
6. Which effect is observed when FM reception deteriorates abruptly due to noise?
a) Threshold effect
b) Capture effect
c) Noise effect
d) Limit effect
Explanation: Signal to Noise ratio is the measure of signal present with respect to the surrounding noise. Threshold effect has low output signal to noise ratio. It contains large noise as compared to input signal in envelope detector. In threshold effect, detection of message signal is difficult.
7. FM stereo _________
a) is implemented using an SCA signal
b) has a higher signal to noise ratio than mono FM
c) is not compatible with mono FM
d) uses DSBSC AM modulation
Explanation: A stereo FM contains three major sections :- monomode, stereo mode and section common to both mono and stereo mode. The stereo section is much complicated than others. It uses DSBSC AM modulation, in which the carrier is suppressed and only the two sidebands are transmitted, for its process.
8. Signal to noise ratio of a frequency modulated system is better than amplitude modulated system.
a) True
b) False
Explanation: Signal to noise ratio is the ratio of signal power to the noise power. It is often described in decibels. Signal to noise ratio is the ratio of signal power to the noise power. It is the measure of signal present with respect to the surrounding noise. It is directly proportional to the modulation index and thus it increases with modulation index. It is better for frequency modulated systems.
9. What will be the effect on power if the distance between the sidebands and center frequency increases?
a) power decreases
b) power increases
c) power remains same
d) power becomes twice of its previous value
Explanation: If the distance between the central frequency and sidebands increases, the effective power decreases as bandwidth increases. | 862 | 3,986 | {"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} | 3.734375 | 4 | CC-MAIN-2022-49 | longest | en | 0.863009 |
https://oli.cmu.edu/courses/graphical-causal-models-open-free/ | 1,721,024,618,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00768.warc.gz | 402,064,500 | 31,064 | # Graphical Causal Models — Open & Free
An introduction to essential terminology and ways of using causal graphs to represent causal systems. Learn about Open & Free OLI courses by visiting the “Open & Free features” tab below.
### Description
In making the causal graph modules, we’ve taken a very spare approach and cover only the essential ideas in terminology on causal graphs. They include the basic concepts of causal graphs as a way to represent causal systems, but they don’t go into nuance or extended case studies.
In the modules, we present graph theoretic ideas of directed paths, undirected paths, and treks. We go all the way through D- Separation, which is a fundamental notion developed by Judea Pearl and colleagues in the late 1980s. We present the key ideas in just a 2- to 4-minute video followed immediately by several Learn By Doing exercises to see if you’ve got the ideas presented in the video. The activities contain feedback and may include several layers of hints to help you if you get confused. The entire unit through Bayes Nets should take no more than three hours.
We hope you enjoy the material, and we are confident that learning this content will help with any more extensive investigations into graphical causal models.
## Open & Free features
### Open & Free Courses
Freeforever
• Open & Free OLI courses enable independent learners to study a subject on their own terms, at their leisure. Courses are:
• Self-guided.
• Self-paced.
• Self-supported.
• Open & Free courses include only the learning materials:
• No teacher.
• No tests.
• No college credit.
• No certificate of completion.
• do not use an Open & Free course because your teacher will never see your work.
## What students will learn
By the time they finish this course, students will learn or be able to:
• Represent direct causes and effects via causal graphs.
• Represent direct and indirect causation with causal graphs.
• Represent common causes and effects with causal graphs.
• Represent feedback with causal graph
• Identify the effects of hard and soft interventions on causal graphs
• Represent and compute undirected paths.
• Recognize colliders and the number of them in a path.
• Recognize Treks
• Categorize nodes on a path as active / inactive
• Categorize paths as active / inactive
• Categorize d-separation / d-connection for any X, Y | { Z } in a causal graph.
• Connect Conditional Probability Table (CPT) Structure to a Graph.
• Encode parametric form into Conditional Probability Table (CPT).
## Learning objectives by module
UNIT 2: Graphical Causal Models
Module 2: Graphical Causal Models: Basics
• Represent direct causes and effects via causal graphs
• Represent direct and indirect causation with causal graphs
• Represent common causes and effects with causal graphs
• Represent feedback with causal graph
• Identify the effects of hard and soft interventions on causal graphs
Module 3: Graphical Causal Models: Undirected Paths, Treks, and D-Separation
• Represent and compute undirected paths.
• Recognize colliders and the number of them in a path.
• Recognize Treks
• Categorize nodes on a path as active / inactive
• Categorize paths as active / inactive
• Categorize d-separation / d-connection for any X, Y | { Z } in a causal graph
Module 4: Bayes Nets
• Connect Conditional Probability Table (CPT) Structure to a Graph
• Encode parametric form into Conditional Probability Table (CPT)
## Course assessments, activities, and outline
UNIT 1: Introduction to Center for Causal Discovery Modules
Module 1: Introduction and Acknowledgements
UNIT 2: Graphical Causal Models
Module 2: Graphical Causal Models: Basics
Module 3: Graphical Causal Models: Undirected Paths, Treks, and D-Separation
Module 4: Bayes Nets
UNIT 3: Appendix
Module 5: Appendix
## Other course details
A few hours.
August 2018
#### Author and other credits
Funders:
• National Institutes of Health
• Center for Causal Discovery
• University of Pittsburgh
• Carnegie Mellon University
• Open Learning Initiative/Simon
• Eberly Center
Authors and contributors
• Richard Scheines
Dean, Dietrich College of Humanities & Social Sciences
Professor of Philosophy, Machine Learning, and HCII
Carnegie Mellon University
• Michael Ringenberg
Senior Research Programmer
Dietrich College of Humanities & Social Sciences
Carnegie Mellon University
• Ruben Sanchez-Romero, MS
PhD student, Department of Philosophy
Carnegie Mellon University
• Kim Larson
Learning Engineer
Open Learning Initiative, Simon Initiative, and Eberly Center
Carnegie Mellon University
• Daniel Sneider
B.A., Mechanical and Biomedical Engineering, 2020
Carnegie Mellon University
• Jared Jory-Geiger
B.A., Psychology and Philosophy, 2019.
Carnegie Mellon University
## System requirements
• internet access
• an operating system that supports the latest browser update
• the latest browser update (Chrome recommended; Firefox, Safari supported; Edge and Internet Explorer are supported but not recommended)
• pop-ups enabled
Some courses include exercises with exceptions to these requirements, such as technology that cannot be used on mobile devices.
This course’s system requirements:
• None listed (subject to change)
OLI Website:
New look and
New student registration process
OLI’s website has undergone a refresh, and so has the student registration process. Watch the video to see how easily students can register with a Course Key.
Go to Top | 1,204 | 5,464 | {"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} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.931372 |
https://nbviewer.jupyter.org/urls/bitbucket.org/vitotap/notebooks/raw/master/tools/timeseries/Trend%20Analysis.ipynb | 1,575,861,645,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00229.warc.gz | 473,952,566 | 740,391 | # Timeseries Trend Analysis¶
This notebook runs a simple trend analysis on the timeseries shown in the time series viewer: https://proba-v-mep.esa.int/applications/time-series-viewer/app/app.html It demonstrates how Spark can be used on the MEP Hadoop cluster, from within a notebook.
## Disclaimer¶
This notebook is not scientifically validated, only meant as a technical demonstration on a real world use case.
## Context¶
So some context first about the data we are going to manipulate.
• The dataset contains FAPAR data (Fraction of Absorbed Photosynthetically Active Radiation)
• The data was computed using raster data taken by a satelite
• Normally, there should be one measurement every 10 days for every zone and landcover.
• Due to external conditions (e.g. clouds), data can be missing.
## Goal¶
Our goal is pretty simple: we want to detect if there are any trends for each zone.
## General approach¶
We will proceed as follows:
1. Clean missing measurements from the original dataset
2. Group measurements by zone and landcover
3. For every zone:
1. Do some cleaning for every landcover
2. Combine the different landcovers data
3. Remove seasonality by taking yearly averages
4. Compute the trend
4. Plot the trends on a world map.
We will detail each step along the way.
In [1]:
import pyspark
import numpy as np
import pandas as pd
import matplotlib
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import datetime, time
%matplotlib inline
Let's read the parquet file containing all our data, as follows:
In [2]:
from pyspark.conf import SparkConf
conf = SparkConf()
conf.set('spark.executor.memory', '4g')
Out[2]:
<pyspark.conf.SparkConf at 0x30b96d0>
The SparkContext 'sc' is our entry point to the Hadoop cluster. The sparkConf defines some parameters.
In [3]:
sc = pyspark.SparkContext(conf=conf)
In [4]:
sqlCtx = pyspark.SQLContext(sc)
In [5]:
!hdfs dfs -ls /tapdata/tsviewer/combined-stats/stats.parquet
Found 169 items
-rw-r--r-- 3 mep_tsviewer vito 0 2017-03-22 00:20 /tapdata/tsviewer/combined-stats/stats.parquet/_SUCCESS
-rw-r--r-- 3 mep_tsviewer vito 25357414 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00000.parquet
-rw-r--r-- 3 mep_tsviewer vito 25341813 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00001.parquet
-rw-r--r-- 3 mep_tsviewer vito 25344785 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00002.parquet
-rw-r--r-- 3 mep_tsviewer vito 25325074 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00003.parquet
-rw-r--r-- 3 mep_tsviewer vito 25382644 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00004.parquet
-rw-r--r-- 3 mep_tsviewer vito 25368452 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00005.parquet
-rw-r--r-- 3 mep_tsviewer vito 25338382 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00006.parquet
-rw-r--r-- 3 mep_tsviewer vito 25364026 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00007.parquet
-rw-r--r-- 3 mep_tsviewer vito 25343301 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00008.parquet
-rw-r--r-- 3 mep_tsviewer vito 25337037 2017-03-22 00:07 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00009.parquet
-rw-r--r-- 3 mep_tsviewer vito 25361096 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00010.parquet
-rw-r--r-- 3 mep_tsviewer vito 25355121 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00011.parquet
-rw-r--r-- 3 mep_tsviewer vito 25374168 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00012.parquet
-rw-r--r-- 3 mep_tsviewer vito 25373881 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00013.parquet
-rw-r--r-- 3 mep_tsviewer vito 25343740 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00014.parquet
-rw-r--r-- 3 mep_tsviewer vito 25375775 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00015.parquet
-rw-r--r-- 3 mep_tsviewer vito 25376607 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00016.parquet
-rw-r--r-- 3 mep_tsviewer vito 25332092 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00017.parquet
-rw-r--r-- 3 mep_tsviewer vito 25363406 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00018.parquet
-rw-r--r-- 3 mep_tsviewer vito 25364663 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00019.parquet
-rw-r--r-- 3 mep_tsviewer vito 25345982 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00020.parquet
-rw-r--r-- 3 mep_tsviewer vito 25364568 2017-03-22 00:08 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00021.parquet
-rw-r--r-- 3 mep_tsviewer vito 25345751 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00022.parquet
-rw-r--r-- 3 mep_tsviewer vito 25359454 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00023.parquet
-rw-r--r-- 3 mep_tsviewer vito 25362565 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00024.parquet
-rw-r--r-- 3 mep_tsviewer vito 25342732 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00025.parquet
-rw-r--r-- 3 mep_tsviewer vito 25371436 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00026.parquet
-rw-r--r-- 3 mep_tsviewer vito 25364058 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00027.parquet
-rw-r--r-- 3 mep_tsviewer vito 25314740 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00028.parquet
-rw-r--r-- 3 mep_tsviewer vito 25362923 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00029.parquet
-rw-r--r-- 3 mep_tsviewer vito 25376056 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00030.parquet
-rw-r--r-- 3 mep_tsviewer vito 25339711 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00031.parquet
-rw-r--r-- 3 mep_tsviewer vito 25397722 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00032.parquet
-rw-r--r-- 3 mep_tsviewer vito 25379909 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00033.parquet
-rw-r--r-- 3 mep_tsviewer vito 25335473 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00034.parquet
-rw-r--r-- 3 mep_tsviewer vito 25364742 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00035.parquet
-rw-r--r-- 3 mep_tsviewer vito 25383452 2017-03-22 00:09 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00036.parquet
-rw-r--r-- 3 mep_tsviewer vito 25353719 2017-03-22 00:10 /tapdata/tsviewer/combined-stats/stats.parquet/part-r-00037.parquet
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In [6]:
data = sqlCtx.read.parquet("hdfs:/tapdata/tsviewer/combined-stats/stats.parquet")
data
Out[6]:
DataFrame[landcover: string, zone: int, date: string, BIOPAR_ALB_BHV_V1: double, BIOPAR_ALB_BHV_V1_area: bigint, BIOPAR_ALB_BHV_V1_validArea: bigint, BIOPAR_ALB_DHV_V1: double, BIOPAR_ALB_DHV_V1_area: bigint, BIOPAR_ALB_DHV_V1_validArea: bigint, BIOPAR_BA_V1: double, BIOPAR_BA_V1_area: bigint, BIOPAR_BA_V1_validArea: bigint, BIOPAR_DMP: double, BIOPAR_DMP_area: bigint, BIOPAR_DMP_validArea: bigint, BIOPAR_FAPAR_V1: double, BIOPAR_FAPAR_V1_area: bigint, BIOPAR_FAPAR_V1_validArea: bigint, BIOPAR_FCOVER_V1: double, BIOPAR_FCOVER_V1_area: bigint, BIOPAR_FCOVER_V1_validArea: bigint, BIOPAR_LAI_V1: double, BIOPAR_LAI_V1_area: bigint, BIOPAR_LAI_V1_validArea: bigint, BIOPAR_NDVI_V1: double, BIOPAR_NDVI_V1_area: bigint, BIOPAR_NDVI_V1_validArea: bigint, BIOPAR_NDVI_V2: double, BIOPAR_NDVI_V2_area: bigint, BIOPAR_NDVI_V2_validArea: bigint, BIOPAR_SWI10_V3: double, BIOPAR_SWI10_V3_area: bigint, BIOPAR_SWI10_V3_validArea: bigint, BIOPAR_TOCR_B0: double, BIOPAR_TOCR_B0_area: bigint, BIOPAR_TOCR_B0_validArea: bigint, BIOPAR_TOCR_B2: double, BIOPAR_TOCR_B2_area: bigint, BIOPAR_TOCR_B2_validArea: bigint, BIOPAR_TOCR_B3: double, BIOPAR_TOCR_B3_area: bigint, BIOPAR_TOCR_B3_validArea: bigint, BIOPAR_TOCR_MIR: double, BIOPAR_TOCR_MIR_area: bigint, BIOPAR_TOCR_MIR_validArea: bigint, BIOPAR_VCI: double, BIOPAR_VCI_area: bigint, BIOPAR_VCI_validArea: bigint, BIOPAR_VPI: double, BIOPAR_VPI_area: bigint, BIOPAR_VPI_validArea: bigint, BIOPAR_WB_V1: double, BIOPAR_WB_V1_area: bigint, BIOPAR_WB_V1_validArea: bigint, BIOPAR_WB_V2: double, BIOPAR_WB_V2_area: bigint, BIOPAR_WB_V2_validArea: bigint, PROBAV_L3_S10_TOC_RED_333M: double, PROBAV_L3_S10_TOC_RED_333M_area: bigint, PROBAV_L3_S10_TOC_RED_333M_validArea: bigint, PROBAV_L3_S10_TOC_NIR_333M: double, PROBAV_L3_S10_TOC_NIR_333M_area: bigint, PROBAV_L3_S10_TOC_NIR_333M_validArea: bigint, PROBAV_L3_S10_TOC_BLUE_333M: double, PROBAV_L3_S10_TOC_BLUE_333M_area: bigint, PROBAV_L3_S10_TOC_BLUE_333M_validArea: bigint, PROBAV_L3_S10_TOC_SWIR_333M: double, PROBAV_L3_S10_TOC_SWIR_333M_area: bigint, PROBAV_L3_S10_TOC_SWIR_333M_validArea: bigint, PROBAV_L3_S10_TOC_NDVI_333M: double, PROBAV_L3_S10_TOC_NDVI_333M_area: bigint, PROBAV_L3_S10_TOC_NDVI_333M_validArea: bigint, RAINFALL: double, RAINFALL_area: bigint, RAINFALL_validArea: bigint]
Next, we want to register this data in a temporary table so we can query it using Spark's SQLContext.
In [7]:
data.registerTempTable("data")
Now although we instructed Spark where the data was and to register it, it did not actually read anything. This is because Spark is lazy: it will postpone doing any task until it the last moment. Requesting data will force it to load it, as follows:
In [8]:
data.take(1)
Out[8]:
[Row(landcover=u'Water_bodies', zone=3219, date=u'2007-11-01', BIOPAR_ALB_BHV_V1=0.14052462320017708, BIOPAR_ALB_BHV_V1_area=11163, BIOPAR_ALB_BHV_V1_validArea=11076, BIOPAR_ALB_DHV_V1=0.14695226555861848, BIOPAR_ALB_DHV_V1_area=11163, BIOPAR_ALB_DHV_V1_validArea=11076, BIOPAR_BA_V1=0.0009853981904505958, BIOPAR_BA_V1_area=11163, BIOPAR_BA_V1_validArea=11, BIOPAR_DMP=None, BIOPAR_DMP_area=None, BIOPAR_DMP_validArea=None, BIOPAR_FAPAR_V1=0.21222769551344417, BIOPAR_FAPAR_V1_area=11163, BIOPAR_FAPAR_V1_validArea=10995, BIOPAR_FCOVER_V1=0.11557305349856734, BIOPAR_FCOVER_V1_area=11163, BIOPAR_FCOVER_V1_validArea=10995, BIOPAR_LAI_V1=0.28491411937776656, BIOPAR_LAI_V1_area=11163, BIOPAR_LAI_V1_validArea=10995, BIOPAR_NDVI_V1=0.27314899691044103, BIOPAR_NDVI_V1_area=11163, BIOPAR_NDVI_V1_validArea=10995, BIOPAR_NDVI_V2=None, BIOPAR_NDVI_V2_area=None, BIOPAR_NDVI_V2_validArea=None, BIOPAR_SWI10_V3=None, BIOPAR_SWI10_V3_area=None, BIOPAR_SWI10_V3_validArea=None, BIOPAR_TOCR_B0=None, BIOPAR_TOCR_B0_area=None, BIOPAR_TOCR_B0_validArea=None, BIOPAR_TOCR_B2=None, BIOPAR_TOCR_B2_area=None, BIOPAR_TOCR_B2_validArea=None, BIOPAR_TOCR_B3=None, BIOPAR_TOCR_B3_area=None, BIOPAR_TOCR_B3_validArea=None, BIOPAR_TOCR_MIR=None, BIOPAR_TOCR_MIR_area=None, BIOPAR_TOCR_MIR_validArea=None, BIOPAR_VCI=None, BIOPAR_VCI_area=None, BIOPAR_VCI_validArea=None, BIOPAR_VPI=None, BIOPAR_VPI_area=None, BIOPAR_VPI_validArea=None, BIOPAR_WB_V1=None, BIOPAR_WB_V1_area=None, BIOPAR_WB_V1_validArea=None, BIOPAR_WB_V2=None, BIOPAR_WB_V2_area=None, BIOPAR_WB_V2_validArea=None, PROBAV_L3_S10_TOC_RED_333M=None, PROBAV_L3_S10_TOC_RED_333M_area=None, PROBAV_L3_S10_TOC_RED_333M_validArea=None, PROBAV_L3_S10_TOC_NIR_333M=None, PROBAV_L3_S10_TOC_NIR_333M_area=None, PROBAV_L3_S10_TOC_NIR_333M_validArea=None, PROBAV_L3_S10_TOC_BLUE_333M=None, PROBAV_L3_S10_TOC_BLUE_333M_area=None, PROBAV_L3_S10_TOC_BLUE_333M_validArea=None, PROBAV_L3_S10_TOC_SWIR_333M=None, PROBAV_L3_S10_TOC_SWIR_333M_area=None, PROBAV_L3_S10_TOC_SWIR_333M_validArea=None, PROBAV_L3_S10_TOC_NDVI_333M=None, PROBAV_L3_S10_TOC_NDVI_333M_area=None, PROBAV_L3_S10_TOC_NDVI_333M_validArea=None, RAINFALL=2.1919554834854207, RAINFALL_area=1406, RAINFALL_validArea=1406)]
Using Spark's SQLContext, we can query our data with SQL-like expressions. Here we are interested in the zone, landcover, date, FAPAR and area columns.
We will also filter out unwanted rows by specifying that we do not want rows for which we do not have a FAPAR value.
In [11]:
data = sqlCtx.sql("SELECT zone, landcover, date date, BIOPAR_FAPAR_V1 fapar, BIOPAR_FAPAR_V1 " +
"FROM data " +
"WHERE BIOPAR_FAPAR_V1 is not null").cache()
data.take(1)
Out[11]:
[Row(zone=3219, landcover=u'Water_bodies', date=u'2006-08-11', fapar=0.29711740277250515, BIOPAR_FAPAR_V1=0.29711740277250515)]
As you can see, although null values have been filtered out, some FAPAR values still remain nan. We can filter them out using the filter method from Spark. For that, we can either use the DataFrame's method filter, or we can use the DataFrame as a regular RDD.
The difference is that the DataFrame contains Rows, for which we can refer to the columns with their names, whereas the RDD contains tuples, where we have to select columns based on their index. However, with a regular RDD, we can pass any method we want to filter, map, reduce, ..., whereas with a DataFrame, we have to use pre-defined functions.
In [12]:
def fapar_not_nan(row):
return row[3] is not None and not np.isnan(row[3])
withoutNaNs = data.rdd.filter(fapar_not_nan).cache()
withoutNaNs.take(1)
Out[12]:
[Row(zone=3219, landcover=u'Water_bodies', date=u'2006-06-21', fapar=0.24140815792410703, BIOPAR_FAPAR_V1=0.24140815792410703)]
For grouping, we could use spark's groupBy method, or we could first map our data to a (Key, Value) pair and use the groupByKey method. We will use the latter method as we have more fined control over what goes into Value.
In [13]:
def by_zone(row):
return ((row[0]), (row[1], row[2], row[3], row[4]))
by_zone = withoutNaNs.map(by_zone).groupByKey().cache()
by_zone.take(1)
Out[13]:
[(672, <pyspark.resultiterable.ResultIterable at 0x601efd0>)]
The result is a list of pairs (Zone, Values) where Values is a list containing all the values for the specific Zone we grouped by.
Spark will combine the different values into a ResultIterable object, not really suited for the analysis we will be doing. Instead, we are going to use Pandas and Numpy. So first thing we want to do is to convert our ResultIterable into an actual Pandas DataFrame.
In [14]:
def iterable_to_pd(row):
return (row[0], pd.DataFrame(list(row[1]), columns=["landcover", "date", "fapar", "area"]))
by_zone = by_zone.map(iterable_to_pd).cache()
by_zone.take(1)
Out[14]:
[(672, landcover date \
4 Mosaic_herbaceous_cover__tree_and_shrub 2000-06-01
5 Mosaic_herbaceous_cover__tree_and_shrub 2004-06-21
6 Mosaic_herbaceous_cover__tree_and_shrub 2010-10-11
7 Mosaic_herbaceous_cover__tree_and_shrub 2011-01-01
8 Mosaic_herbaceous_cover__tree_and_shrub 2015-01-21
9 Sparse_vegetation_tree_shrub_herbaceous_cover 2000-04-21
10 Sparse_vegetation_tree_shrub_herbaceous_cover 2006-08-11
11 Sparse_vegetation_tree_shrub_herbaceous_cover 2012-12-01
12 Sparse_vegetation_tree_shrub_herbaceous_cover 2016-12-21
13 Shrubland 2002-10-11
14 Shrubland 2003-01-01
15 Shrubland 2007-01-21
16 Shrubland 2013-05-11
17 Cropland_rainfed 1999-05-01
18 Cropland_rainfed 2003-05-21
19 Cropland_rainfed 2009-09-11
20 Mosaic_natural_vegetation 2008-04-11
21 Mosaic_natural_vegetation 2014-08-01
22 Urban_areas 2002-04-01
23 Urban_areas 2006-04-21
24 Urban_areas 2012-08-11
28 Water_bodies 2000-05-11
29 Water_bodies 2006-09-01
... ... ...
11685 Mosaic_tree_and_shrub_herbaceous_cover 2002-01-01
11686 Mosaic_tree_and_shrub_herbaceous_cover 2006-01-21
11687 Mosaic_tree_and_shrub_herbaceous_cover 2012-05-11
11688 Shrub_or_herbaceous_cover_flooded_fresh_saline... 2000-06-11
11689 Shrub_or_herbaceous_cover_flooded_fresh_saline... 2006-10-01
11690 Shrub_or_herbaceous_cover_flooded_fresh_saline... 2010-10-21
11691 Shrub_or_herbaceous_cover_flooded_fresh_saline... 2011-01-11
11696 Mosaic_cropland 1999-02-21
11697 Mosaic_cropland 2005-06-11
11698 Mosaic_cropland 2011-10-01
11699 Mosaic_cropland 2015-10-21
11700 Mosaic_cropland 2016-01-11
11706 Grassland 2004-12-11
11707 Grassland 2005-03-01
11708 Grassland 2009-03-21
11709 Grassland 2015-07-11
11710 Tree_cover_flooded_saline_water 2000-04-11
11711 Tree_cover_flooded_saline_water 2006-08-01
11712 Tree_cover_flooded_saline_water 2010-08-21
11713 Tree_cover_flooded_saline_water 2016-12-11
11714 Tree_cover_flooded_saline_water 2017-03-01
fapar area
0 0.781023 0.781023
1 0.739872 0.739872
2 0.714261 0.714261
3 0.795091 0.795091
4 0.682756 0.682756
5 0.617121 0.617121
6 0.384000 0.384000
7 0.568926 0.568926
8 0.498292 0.498292
9 0.512471 0.512471
10 0.376571 0.376571
11 0.415657 0.415657
12 0.412000 0.412000
13 0.553395 0.553395
14 0.598780 0.598780
15 0.606932 0.606932
16 0.673394 0.673394
17 0.674976 0.674976
18 0.636602 0.636602
19 0.527672 0.527672
20 0.731046 0.731046
21 0.628583 0.628583
22 0.549303 0.549303
23 0.660212 0.660212
24 0.509302 0.509302
25 0.627513 0.627513
26 0.692247 0.692247
27 0.549743 0.549743
28 0.559624 0.559624
29 0.394943 0.394943
... ... ...
11685 0.469723 0.469723
11686 0.665650 0.665650
11687 0.653948 0.653948
11688 0.638316 0.638316
11689 0.405949 0.405949
11690 0.503323 0.503323
11691 0.501129 0.501129
11692 0.602000 0.602000
11693 0.296000 0.296000
11694 0.546000 0.546000
11695 0.572000 0.572000
11696 0.630002 0.630002
11697 0.814023 0.814023
11698 0.605181 0.605181
11699 0.597991 0.597991
11700 0.645325 0.645325
11701 0.576503 0.576503
11702 0.611666 0.611666
11703 0.615693 0.615693
11704 0.598025 0.598025
11705 0.624073 0.624073
11706 0.503609 0.503609
11707 0.602599 0.602599
11708 0.483708 0.483708
11709 0.498409 0.498409
11710 0.722473 0.722473
11711 0.675400 0.675400
11712 0.662609 0.662609
11713 0.610634 0.610634
11714 0.619562 0.619562
[11715 rows x 4 columns])]
Now that we are ready to write our Combine task, let's detail the algorithm we are going to write to process each zone:
1. Group the dataset by landcover, and for each:
1. Resample the data monthly (instead of 10-daily) with padding in order to fill some void if there are any
2. Reweight the FAPAR values by the area the landcover takes over the total area
3. Regroup the dataset by date and take the (monthly) sum
4. Resample the data yearly and remove the 1st and last year of data (to only include full years)
In [15]:
def combine_fapar(zone):
series = zone[1]
series["date"] = pd.to_datetime(series["date"])
series = series.set_index("date")
# Group the dataset by landcover, and for each:
by_landcover = series.groupby("landcover")
# Resample the data monthly (instead of 10-daily) with padding in order to fill some void if there are any
series = by_landcover.reset_index()
# Reweight the FAPAR values by the area the landcover takes over the total area
total_area = by_landcover["area"].mean().sum()
series["fapar"] *= series["area"] / total_area
series = series.drop("area", axis=1)
# Regroup the dataset by date and take the (monthly) sum
series = series.groupby("date").sum()
# Resample the data yearly and remove the 1st and last year of data (to only include full years)
yearly_sum = pd.DataFrame(series["fapar"].resample("12m", "mean")[1:-1])
return (zone[0], series, yearly_sum)
combined = by_zone.map(combine_fapar).cache()
combined.take(1)
Out[15]:
[(672, fapar
date
1999-01-31 9.717959
1999-02-28 8.868936
1999-03-31 10.370095
1999-04-30 12.420800
1999-05-31 12.032194
1999-06-30 9.968254
1999-07-31 9.284237
1999-08-31 8.657068
1999-09-30 6.982681
1999-10-31 5.440120
1999-11-30 5.537638
1999-12-31 8.929892
2000-01-31 10.512340
2000-02-29 11.302637
2000-03-31 13.014589
2000-04-30 14.539168
2000-05-31 14.625471
2000-06-30 13.036348
2000-07-31 10.837346
2000-08-31 9.071030
2000-09-30 6.932734
2000-10-31 8.162503
2000-11-30 7.231002
2000-12-31 10.725195
2001-01-31 11.366307
2001-02-28 10.276144
2001-03-31 10.517193
2001-04-30 10.990723
2001-05-31 10.960789
2001-06-30 11.189507
... ...
2014-10-31 8.205650
2014-11-30 8.801627
2014-12-31 11.202606
2015-01-31 10.501179
2015-02-28 9.387215
2015-03-31 10.382771
2015-04-30 11.646911
2015-05-31 11.264060
2015-06-30 12.828613
2015-07-31 11.889360
2015-08-31 10.821384
2015-09-30 8.733411
2015-10-31 7.264628
2015-11-30 6.536714
2015-12-31 7.512584
2016-01-31 7.997100
2016-02-29 10.642866
2016-03-31 9.514068
2016-04-30 9.566114
2016-05-31 8.566643
2016-06-30 7.837038
2016-07-31 8.002204
2016-08-31 6.676644
2016-09-30 5.125340
2016-10-31 5.245660
2016-11-30 6.372412
2016-12-31 9.893308
2017-01-31 10.128927
2017-02-28 9.886582
2017-03-31 10.078414
[219 rows x 1 columns], fapar
date
2000-01-31 9.083688
2001-01-31 10.903694
2002-01-31 9.291626
2003-01-31 10.591492
2004-01-31 9.401098
2005-01-31 11.984670
2006-01-31 12.621007
2007-01-31 10.920098
2008-01-31 10.022505
2009-01-31 10.516419
2010-01-31 11.414949
2011-01-31 11.175359
2012-01-31 11.270612
2013-01-31 11.411795
2014-01-31 11.138742
2015-01-31 10.768098
2016-01-31 9.688729
2017-01-31 8.130935)]
Finally, we can compute the trend using a linear regression (OLS).
In [16]:
def min_years(datas):
return len(datas[2]) >=5
def ols(datas):
yearly_series = datas[2]
x, y = yearly_series.index.astype(int) // (10**9 * 60 * 24 * 30), yearly_series["fapar"]
X = np.c_[np.ones(len(x)), x]
return (datas[0], datas[1], datas[2], np.linalg.pinv(X).dot(y))
trends = combined.filter(min_years).map(ols).cache()
Next we will plot the trends. We are interested in the highest and lowest trends in our data.
Let's define our plotting function and plot the trends by descending order.
In [21]:
#copy auxiliary data from hdfs
!hdfs dfs -copyToLocal /tapdata/GAUL2013/zonecodes.csv ./
17/03/22 13:25:52 WARN hdfs.DFSClient: DFSInputStream has been closed already
In [23]:
zonecodes = pd.read_csv("./zonecodes.csv")
def plotTopTrends(trendsCollected):
for zone, series, yearly_series, trend in trendsCollected:
if not np.isnan(trend[0]) and not np.isnan(trend[1]):
codes["Trend"] = trend[1]
print codes
pdf = series.copy()
pdf["ols"] = pdf.index.astype(int) / (10**9 * 60 * 24 * 30) * trend[1] + trend[0]
pdf[["fapar", "ols"]].plot(figsize=(16,4))
#sns.regplot(x="unix_date", y="fapar", data=series, scatter=False)
plt.show()
In [24]:
plotTopTrends(trends.sortBy(lambda x: x[3][1], ascending=False).take(10))
ADM0_NAME ADM1_NAME Trend
117 American Samoa Administrative unit not available 0.001467
/opt/rh/python27/root/usr/lib64/python2.7/site-packages/matplotlib/font_manager.py:1297: UserWarning: findfont: Font family [u'sans-serif'] not found. Falling back to DejaVu Sans
(prop.get_family(), self.defaultFamily[fontext]))
ADM0_NAME ADM1_NAME Trend
793 Egypt Al Bahr/al Ahmar (redsea) 0.000554
ADM0_NAME ADM1_NAME Trend
1311 Kiribati Administrative unit not available 0.000551
ADM0_NAME ADM1_NAME Trend
1662 Montserrat Plymouth 0.000495
ADM0_NAME ADM1_NAME Trend
1827 Pakistan Punjab 0.000426
ADM0_NAME ADM1_NAME Trend
1101 India Rajasthan 0.000421
ADM0_NAME ADM1_NAME Trend
805 Egypt As Ismailiyah (ismailia) 0.000421
ADM0_NAME ADM1_NAME Trend
104 Algeria Saida 0.000418
ADM0_NAME ADM1_NAME Trend
112 Algeria Tiaret 0.000405
ADM0_NAME ADM1_NAME Trend
2561 Turkey Denizli 0.0004
In [25]:
plotTopTrends(trends.sortBy(lambda x: x[3][1], ascending=True).take(10))
ADM0_NAME ADM1_NAME Trend
574 Christmas Island Administrative unit not available -0.001041
ADM0_NAME ADM1_NAME Trend
904 French Polynesia Administrative unit not available -0.000726
ADM0_NAME ADM1_NAME Trend
3314 Martinique La Trinite -0.000519
ADM0_NAME ADM1_NAME Trend
731 Djibouti Obock -0.000301
ADM0_NAME ADM1_NAME Trend
3102 Guinea Conakry -0.000285
ADM0_NAME ADM1_NAME Trend
ADM0_NAME ADM1_NAME Trend
ADM0_NAME ADM1_NAME Trend | 15,698 | 42,573 | {"found_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} | 2.6875 | 3 | CC-MAIN-2019-51 | latest | en | 0.742659 |
http://umgmagazine.com/books/analysis-on-manifolds | 1,553,623,481,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205600.75/warc/CC-MAIN-20190326180238-20190326202238-00423.warc.gz | 211,781,600 | 8,827 | # Analysis On Manifolds by James R. Munkres PDF
By James R. Munkres
A readable advent to the topic of calculus on arbitrary surfaces or manifolds. obtainable to readers with wisdom of uncomplicated calculus and linear algebra. Sections contain sequence of difficulties to augment concepts.
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Therefore, the space E is sometimes called a skew product of the base space B with the fiber F. An example of a non-trivial skew product is a certain skew product of the circle by a segment, viz the Mllbius strip. The existence of a local direct product structure of is clearly seen in Fig. 26. Figure 26 Let M be a smooth manifold. The space TM of the tangent bundle p:TM -+ M consists of all pairs (a, -r), where a E M and -r is a tangent vector to the manifold M starting at the point a. If M c RN ,the concept of a tangent vector is defined in an Chapter 1.
E. if its pre-image contains no critical points. g. Ref [2]). If N =Lx R4 , the Sard theorem should be applied to the composition pf, where 26 Computer Topology and 3-Manifolds p:L x Rk ~ Rk is the projection of the direct product, and the isotopy lP, in the general case is now easily constructed using local triviality of normal bundle of the manifold LeN. Transversality of two submanifolds M,L e N is understood as transversality of the embedding i: M ~ N to L. ill other words, M and N are transversal if at each point of their intersection x E M r.
Recall that the second homotopy group 1t'z (X) of a space X is trivial if and only if any map of the two-dimensional sphere into X is homotopic to a constant map, or, which is the same, if any two maps (coinciding on the boundary) of a two-dimensional disc into X are homotopic to each other under a homotopy fIxed on the boundary. 1 If a surface F is different from the sphere and from the projective plane, then 1r2 (F) = O. Proof. For the reader acquainted with the fundamentals of homotopic topology, this fact is trivial. | 818 | 3,563 | {"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} | 2.59375 | 3 | CC-MAIN-2019-13 | latest | en | 0.909227 |
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Jan 16, 2013 06:26 AM
Discussion
Survey: How many meals do you eat out or take out in a week?
Yep, I am curious how often do you eat out or take out from restaurants on an average week. To clarify the few definitions in hoping to keep the answers consistent:
We will be counting the numbers of meals, not numbers of days. If I were to eat out breakfast, lunch and dinner every day of the week, that will be counted as 21 times, not 7 times.
Getting a Big Mac for lunch counts as a take out, getting a donut and coffee for breakfast from Dunkin Donut counts as a take out, buying foods from the company caffertia counts as a eat out, but buying frozen meals from Lean Cuisine does not, pouring milk into a bowl of cereal does not.
I know, I know, it gets a bit confusing because many options appear to be in the gray area, but we need to draw the line somewhere, right? So please answer as best as you can in term of how often you eat restaurant foods. Thanks.
A) 0-2 meals a week
B) 3-5 meals
C) 6-11 meals
D) >11 meals
My answer is (B). On an average week, I eat out about 1 times (sometime 2 times), and do 1 take out from restaurants. However, I often split my single takeout into 2-3 meals, so my answer is closer to 3-4. Thanks.
P.S.: I mean your current lifestyle (think of your last 6 months to last year), not 5 or 20 years ago -- just to be clear.
1. Click to Upload a photo (10 MB limit)
Delete
1. I'd have to say A. There is the occasional work week when I have a couple of work lunches at restaurants, so that might occasionally push me up to B. Or for example, when I travel for work (e.g. when I am out of the country for a week or two), I am closer to D. But in a normal week when I am working in town, then A. (There are some weeks when the answer is zero.)
I love to eat out and take out, but my husband does a lot of cooking, so we default to A.
1. re: Cachetes
:) Thanks. You got my point. On occasional travel, I definitely eat out 100%, but I don't travel very often. Thanks for your answer. I used to be more like (A), but I actually force myself to eat out more just to get better idea of restaurant cookings.
2. I shoot for eating out once every other week. Sometimes the convenience factor of moving around gets me once more. So A.
1. A)
On a tight budget now, but still when my fiancee and I have the same day off we will generally eat 1 lunch out as its cheaper, sometimes we will eat out at dinner instead though.
1. Until the new year a D, but have resolved to cook more of my meals myself, so I'm moving to a B, last week an A.
In an ideal world I'd like to eat about 6 meals a week at home.
If I buy some pastries at the market and bring them home I take it this would count as a home meal, but if I drive-thru Mc Donalds and get an egg McMuffin and bring it home, that is a meal out? | 746 | 2,892 | {"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} | 2.71875 | 3 | CC-MAIN-2015-35 | latest | en | 0.970514 |
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VALLIAMMAI ENGINEERING COLLEGE DEPARTMENT OF CIVIL ENGINEERING CE6704: ESTIMATION AND QUANTITY SURVEYING SEMESTER: 07 th REGULATION-2013 ACADEMIC YEAR: 2016-2017 UNIT I ESTIMATE OF BUILDINGS Load bearing and framed structures Calculation of quantities of brick work, RCC, PCC, Plastering, white washing, colour washing and painting / varnishing for shops, rooms, residential building with flat and pitched roof Various types of arches Calculation of brick work and RCC works in arches Estimate of joineries for panelled and glazed doors, windows, ventilators, handrails etc. 1 List the methods of estimating quantities? BT-1 2 Define estimate. BT-1 3 What is a detailed estimate? BT-1 4 Name the types of estimate. BT-1 5 Show the contents of abstract estimate? BT-1 6 Define preliminary estimate BT-1 7 Discuss Long wall and short wall method? BT-2 8 Summarize the advantages of centre line method over long wall and short wall method? BT-2 9 Classify the type of arches? BT-2 10 Describe centre line method. BT-2 11 Write the recommendations for degree of accuracy on measurements. BT-3 12 Demonstrate the importance of estimation BT-3 13 Illustrate what is prime cost BT-3 14 State plinth area rate BT-4 15 An arch of 2.5m span subtends an angle of 80 0 at the center. The thickness of arch is 30cm and the breadth of the wall is 40cm. calculate the quantity of arch work. BT-4 16 Explain schedule of rate BT-4 17 Generalize the duties of quantity surveyor. BT-5 18 Formulate methods to be adopted for volume calculating? BT-5 19 Write the unit of measurement for earth work, D.P.C and brick work. BT-6 20 Write the units of measurements for plastering, flooring and painting BT-6 www.studentsfocus.com
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VALLIAMMAI ENGINEERING COLLEGE
DEPARTMENT OF CIVIL ENGINEERING CE6704: ESTIMATION AND QUANTITY SURVEYING
SEMESTER: 07th REGULATION-2013 ACADEMIC YEAR: 2016-2017
UNIT I ESTIMATE OF BUILDINGS
Load bearing and framed structures – Calculation of quantities of brick work, RCC, PCC, Plastering, white washing, colour washing and painting / varnishing for shops, rooms, residential building with flat and pitched roof – Various types of arches – Calculation of brick work and RCC works in arches – Estimate of joineries for panelled and glazed doors, windows, ventilators, handrails etc.
1 List the methods of estimating quantities? BT-1 2 Define estimate. BT-1 3 What is a detailed estimate? BT-1 4 Name the types of estimate. BT-1 5 Show the contents of abstract estimate? BT-1 6 Define preliminary estimate BT-1 7 Discuss Long wall and short wall method? BT-2
8 Summarize the advantages of centre line method over long wall and short wall method? BT-2
9 Classify the type of arches? BT-2 10 Describe centre line method. BT-2 11 Write the recommendations for degree of accuracy on measurements. BT-3 12 Demonstrate the importance of estimation BT-3 13 Illustrate what is prime cost BT-3 14 State plinth area rate BT-4
15 An arch of 2.5m span subtends an angle of 800 at the center. The thickness of arch is 30cm and the breadth of the wall is 40cm. calculate the quantity of arch work. BT-4
16 Explain schedule of rate BT-4 17 Generalize the duties of quantity surveyor. BT-5 18 Formulate methods to be adopted for volume calculating? BT-5 19 Write the unit of measurement for earth work, D.P.C and brick work. BT-6 20 Write the units of measurements for plastering, flooring and painting BT-6
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Fig 1
Fig 2
Fig 3 www.studentsfocus.com
S.S MANU S.SURESHBABU HANUSH PRAVEEN C
PART –B
1
Explain the following measurements of a building? (i) Carpet area (ii) Floor Area (iii) Circulation area (iv)Carpet area
BT-1
2
The Plan and sectional elevation of the building are given in Fig-1 Estimate the quantities for the following items of works. (i) RCC slabs, lintels & sunshades. (ii) Doors and windows (iii)Plastering internal
BT-1
3
The Plan and sectional elevation of the building are given in Fig-1 Estimate the quantities for the following items of works. (i) 1st class brickwork in Super structure CM1:6 (ii) PCC in foundations (iii)Ceiling plastering
BT-1
4
Prepare the Detailed Estimate for the following items of works are given in Fig-2 (i) Earth work excavation in foundation (ii) Cement plastering CM 1:6 (iii)White washing
BT-1
5
Prepare the Detailed Estimate for the following items of works are given in Fig-2 (i) RCC slabs, lintels & sunshades. (ii) Steel reinforcements bars in RCC at 1% (iii)Plastering internal
BT-2
6
Explain in detail about (i) Joineries (ii) Arch types (iii) Abstract estimate (iv) Supplementary estimate
BT-2
7 (i) What are methods to be adopted for volume calculating? (ii) Differentiate abstract and detailed estimate
BT-2
8 (i) Explain any four types of approximate estimate? ii) Differentiate Cube rate estimate and item rate estimate
BT-3
9 Briefly explain various types of estimates in detail? BT-3
Fig 3
Fig 4
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
10
The Plan and sectional elevation of the building are given in Fig-3 Estimate the quantities for the following items of works. (i) RCC slabs, lintels & sunshades. (ii) Doors and windows (iii)Plastering internal
BT-4
11
The Plan and sectional elevation of the building are given in Fig-3 Estimate the quantities for the following items of works. (i) 1st class brickwork in Super structure CM1:6 (ii) PCC in foundations (iii)Ceiling plastering
BT-4
12
Prepare the Detailed Estimate for the following items of works are given in Fig-4 (i) Earth work excavation in foundation (ii) Cement plastering CM 1:6 (iii)White washing
BT-4
13
Prepare the Detailed Estimate for the following items of works are given in Fig-4 (i) RCC slabs, lintels & sunshades. (ii) Steel reinforcements bars in RCC at 1% (iii)Plastering internal
BT-5
14 (i)Explain the following measurements of a building? (a) Plinth area (b) Plot area (ii)Write the formula for Mid ordinate rule and Prismoidal formula Rule.
BT-6
UNIT II ESTIMATE OF OTHER STRUCTURES 10 Estimating of septic tank, soak pit – sanitary and water supply installations – water supply pipe line – sewer line – tube well – open well – estimate of bituminous and cement concrete roads – estimate of retaining walls – culverts – estimating of irrigation works – aqueduct, syphon, fall.
PART-A
1 What are the different types of roads? BT-1 2 Explain the role of soak pit BT-2 3 What are factors to be considered in design of septic tank? BT-1 4 Define lead. BT-1 5 Define lift. BT-1 6 Describe the methods to determine the area of roads in excavation BT-2 7 Rewrite the mid sectional area formula. BT-5 8 Explain prismoidal rule. BT-4 9 Write about sewer line. BT-6 10 Workout the quantity of stone metals required for 2km length of a 4m wide road. BT-2 11 A cement concrete road is to be constructed over the existing water bound
macadam road. Calculate the material requirement for a slab thickness of 10cm. BT-3
12 What are all sanitary fittings? Illustrate BT-3 13 Sketch and discuss about different parts of culvert. BT-2 14 What is the importance of soak pit? Explain BT-4 15 Classify the various types of arches BT-4 16 What is the role of baffle wall in septic tank? BT-1 17 Mention and explain few irrigation structures. BT-5
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
18 What are the needs for retaining wall? BT-1 19 Explain aqueduct. BT-6 20 Calculate the size of septic tank for 25 users BT-3
PART-B
1 Describe in detail the different types of roads. BT-1 2 Describe briefly the different types of irrigation structures. BT-1 3 Prepare an estimate for the following,
i) Excavation ii) Sub Grade iii) Quantity of concrete
‘cement concrete road’ given in fig
BT-2
4 Prepare the estimate quantity for the following constrains of ‘bituminous road’ shown in fig
i) Quantity of road thar and asphalt coat for 2m length ii) Detailed estimate of sub grade
BT-2
5 Prepare the estimate for the ‘septic tank’ shown in fig BT-2
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
6 Prepare the estimate for the ‘soak pit’ in fig.5 (see the above sketch) i) Brick work ii) Quantity of brick bats
BT-2
7 Prepare an estimate for the ‘box culvert’ shown in fig
BT-5
8 Describe in detail the methods of estimation of roads. BT-1 9 What are the components of a culvert? Illustrate. BT-3 10 Illustrate about any two irrigation structures. BT-3 11 Calculate the quantities and estimate the retaining wall shown in figure below
i) Excavation
BT-3
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
ii) Reinforcement activities iii) Concreting quantities
12 Estimate the cost of earthwork for a portion of a road from the following data. Road width at the formation surface is 8m.Side slopes 2:1 in banking and 1.5:1 in cutting. Length of chain is 30m.
BT-2
13 Explain a procedure for estimation of the following i) Open well ii) Tube well.
BT-6
14 Explain a ‘cantilever retaining wall’ and prepare the estimate for the given one of 100m length.
BT-4
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
UNIT III SPECIFICATION AND TENDERS 8
Data – Schedule of rates – Analysis of rates – Specifications – sources – Preparation of detailed and general specifications – Tenders – TTT Act – e-tender – Preparation of Tender Notice and Document – Contracts – Types of contracts – Drafting of contract documents – Arbitration and legal requirements.
PART-A
1 What is analysis of rates? BT-1 2 Define specification. BT-1 3 Briefly illustrate the schedule of rates. BT-3 4 Define contract. BT-1 5 Discuss, what a tender is. BT-2 6 Explain the termstandard specification. BT-4 7 Illustrate the term arbitration. BT-3 8 The actual expenditure incurred in the construction of school building which have a
total length of main walls140m is Rs. 14.97 lakhs. Estimate the approximate cost of a school building which will have 180m length of main walls.
BT-2
9 Classify and explain the types of penalties that are imposed on a contract and why are they imposed?
BT-4
10 Discuss the different types of specification? BT-2 11 Illustrate, the information that a contract document contain. BT-3 12 What is a tender notice? BT-1 13 Explain the types of contract? BT-4 14 Rewrite the requirements of a contract. BT-5 15 Define lump sum contract. BT-1 16 Explain the objects of specification BT-6 17 Discuss the types of termination of contract BT-2 18 Define the term Contractor BT-1 19 Rewrite the Essentials of contract BT-5 20 Summarize the important legal implications of a contract BT-6
PART-B (16 Marks) 1 Illustrate the general specifications of a residential building. BT-3 2 Mention and describe the general specifications of a bituminous road. BT-1 3 Explain detailed specification of earth work. BT-2 4 Explain a detailed specification of super structure. BT-2 5 Explain and compare the different types of contract in detail. BT-4 6 Describe the legal requirements of a contract? BT-1 7 Discuss about schedule of rates. BT-2 8 Illustrate the specifications for a septic tank? BT-3 9 Illustrate the following in brief
i) General or brief specification ii) Detailed specification iii) Standard specification
BT-3
10 Explain about the contents of Tender notice? BT-2 11 Describe the general specification for first class buildings BT-1 12 Explain and rewrite the detailed specifications for any four items of work? BT-5
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
13 Explain and compare the different types of estimates & differentiate detailed estimate from cube rate estimate?
BT-4
14 Summarize the contents of contract document. BT-6
UNIT – IV VALUATION
Necessity – Basics of value engineering – Capitalised value – Depreciation – Escalation – Value of building – Calculation of Standard rent – Mortgage – Lease.
PART –A 1 List the different methods of depreciation? BT-1 2 Define valuation? BT-1 3 What is obsolescence? BT-1
4 Find the plinth area required for the residential accommodation for an assistant Engineer in the pay scale of Rs.36100 to 47500 per month. BT-1
5 Define the Gross income: BT-1 6 What is scrap value? BT-1 7 Summarize why we calculate standard rent of building? BT-2 8 Explain Gross income BT-2 9 Describe Net come BT-2
10 A property fetches a net income of Rs.900.00 deducting all outgoings. Workout the capitalized value of the property if the rate of interest is 6% per annum. BT-2
11 Demonstrate the meaning of salvage value? BT-3 12 Illustrate about Annuity BT-3 13 Illustrate about book valve BT-3 14 Differentiate between market valve and book valve. BT-4 15 Point out factors influencing compaction? BT-4
16
A pumping set with a motor has been installed in a building at a cost Rs.2500.00.Assuming the life of the pump as 15 years, workout the amount of annual installment of sinking fund to be deposited to accumulate the whole amount of 4% compound interest.
BT-4
17
An old building has been purchased by a person at a cost of Rs.30, 000/‐ excluding the cost of the land. Evaluate the amount of annual sinking fund at 4% interest assuming the future life of the building as 20 years and scarp value of the building as 10% of the cost of purchase.
BT-5
18 Sinking fund method of depreciation is more reliable” - Justify BT-5 19 Write the necessity of valuation. BT-6 20 Write short note on Escalation? BT-6
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PART –B
1
Define the following : (i) Type of lease (ii) mortage (iii) Escalation
BT-1
2
State the following terms : (1) Scrap valve (2) Salvage valve (3) Book Valve (4) Market valve
BT-1
3
In a plot of land costing Rs 20,00,000 a building has been newly constructed at a local cost of Rs80,00,000 including sanitary and water supply works , electrical insllations,etc . The building consist of four flats of four tenants. The owner expects 8 % return on the cost of construction and 5 % of return on the land. Calculate the standard rent for each flat of the building assuming. (i) The life of the building is 60 yrs and the sinking fund will be created on 4% interest basis (ii)Annual repair cost 1% of the cost of construction (iii) Other outgoings including taxes at 30% of the net return on the building?
BT-1
4 (i) Explain differ forms of valve (ii) Discuss about a freehold property
BT-1
5
Discuss the following terms : (i) Types of Lease (ii) Mortage (iii) Escalation (iv) Methods of Depreciation
BT-2
6
Explain the terms clearly: (i) Annuity Head rent (ii) Deferred income (iii) Deferred annuities (iv) Single rate Y.P.
BT-2
7
Calculate the annual rent of a building with the following data. Cost of land = Rs.20,000/- Cost of building = Rs.80,000/- Estimate life = 80years Return expected = 5% on land 6% on building Annual repairs are expected to be 0.7% of the cost construction and other out goings will be 25% of the gross rent. There is no proposal to set up a sinking fund
BT-2
8 The capitalized cost of a building is Rs.one lac, including all fittings of first class construction. if the rate of interest is 6%, Calculate net return from the property. Assume out goings as 15% on gross income.
BT-3
9
A plot measure 800sq.m.the built up area rate of this 1st class building is Rs.600/-per sq.m this rates includes cost of water supply, sanitary and electric installations. The age of the building is 50 years. The cost of the land is Rs.1800/- per sq.m Calculate the standard rent for a building located in CMA assuming the required parameters
BT-3
10 A Owner occupied property is required to be valued for the wealth tax purpose of BT-4
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S.S MANU S.SURESHBABU HANUSH PRAVEEN C
land and building.The following particulars are available. Evaluate the present valve of the property Valve of the land = Rs4,00,000.00 Cost of the building to put up such a building present =Rs10,00,000 Age of the building = 40 year Estimate cost of repair =Rs.50,000.00 Depreciation to be allowed for the building = 0.75% per annum
11
Differentiate clearly between the following: (i) Capitalized value and year’s purchase (ii) Freehold property and leasehold property (iii) Depreciation and obsolescence.
BT-4
12 Examine in detail about various methaods of calculations Depreciation BT-4
13 Explain in detail about various methaods of Valuation BT-5
14
Write Short note on the following terms: (i)Sinking fund (ii)Outgoings (iii)Capitalized value (iv)Price and Cost
BT-6
UNIT V REPORT PREPARATION Principles for report preparation – report on estimate of residential building – Culvert – Roads – Water supply and sanitary installations – Tube wells – Open wells
Prat A
1 Define Engineer: BT-1 2 State Owner: BT-1 3 Define Analysis of work: BT-1 4 What are the principle of report preparation BT-1
5 List the factors involved in locating a site? BT-1
6 What are the set of drawings required for preparing a report? BT-1
7 Discuss Measurement Book BT-2 8 Identify the size of septic tank for 50 & 25 users? BT-2 9 Explain Work sheet BT-2
10 Discuss Drawings. BT-2 11 Illustrate the estimate for the Open well? BT-3 12 Write the estimate for the Tube well? BT-3 13 Why valuation is necessary? BT-3 14 Examine work BT-4 15 Define Site BT-4 16 Differentiate Open well and Tube well BT-4 17 Select any two principles for the preparation of Water supply schemes? BT-5
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18 Select any two principles for the preparation of residential building? BT-5 19 Write the estimate for the sanitary installation? BT-6 20 Write the estimate for the Water supply? BT-6
PART –B
1 (i) Define the Procedure for preparation for preparation of reports? (ii)State how will you prepare a report on estimate of Box culvert?
BT-1
2
Describe the following in Report Preparation? (i) Definition of reports (ii) Types of reports (iii) Necessity of report
BT-1
3
Prepare a report on estimate for the following items in Single storey Residential building ((i) Sub structure (ii) Super Structure
BT-1
4 Explain the report on estimation for construction of bituminous roads BT-1
5 Write a report on estimate for construction of cement concrete roads BT-2
6 Explain the report on estimation for construction of Water bound macadam roads BT-2
7 (i) List out the points to be considered while writing technical report? (ii) Differentiate Administrative report and technical report?
BT-2
8
Discuss the report on estimation for the construction of a multi storey building for the following (i) Sub structure (ii) Super Structure
BT-3
9 Write the report on estimate for construction structures : (i) Tube well (ii) Open well
BT-3
10 Examine the report on estimation for construction of Small bridge BT-4
11 Examine the report on estimation for construction of water supply & sanitary work BT-4
12 Examine the report on estimation for construction of following culverts (i) Bridge culvert (ii) Arch culvert
BT-4
13 Prepare the report on estimation for construction pipe line culvert BT-5
14 (i) Summarize the documents to be attached with a report (ii) Explain Status report
BT-6
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https://questions.examside.com/past-years/jee/question/a-boy-can-throw-a-stone-up-to-a-maximum-height-of-10-m-the-2012-marks-4-hapduvkfscdshejt.htm | 1,713,701,195,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00865.warc.gz | 427,954,190 | 47,978 | 1
AIEEE 2012
+4
-1
A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be
A
$$20\sqrt 2$$ m
B
10 m
C
$$10\sqrt 2$$ m
D
20 m
2
AIEEE 2011
+4
-1
A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :
A
$$\pi {{{v^4}} \over {{g^2}}}$$
B
$${\pi \over 2}{{{v^4}} \over {{g^2}}}$$
C
$$\pi {{{v^2}} \over {{g^2}}}$$
D
$$\pi {{{v^2}} \over g}$$
3
AIEEE 2011
+4
-1
An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :
$${{dv} \over {dt}} = - 2.5\sqrt v$$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :
A
2 s
B
4 s
C
8 s
D
1 s
4
AIEEE 2010
+4
-1
A particle is moving with velocity $$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$$, where K is a constant. The general equation for its path is
A
y = x2 + constant
B
y2 = x + constant
C
xy = constant
D
y2 = x2 + constant
EXAM MAP
Medical
NEET | 415 | 1,073 | {"found_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} | 3.4375 | 3 | CC-MAIN-2024-18 | latest | en | 0.838249 |
https://math.stackexchange.com/questions/2541471/minimal-surface-and-mean-curvature | 1,563,884,031,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529276.65/warc/CC-MAIN-20190723105707-20190723131707-00147.warc.gz | 452,597,756 | 36,873 | # Minimal Surface and Mean Curvature
In the frame of a course on instabilities, I am trying to prove that the soap film between two rings has the form of a catenoid. Since pressure is equal on either side of the film, we expect to have a surface that has zero mean curvature everywhere, i.e.:
$$\frac 1 {r_1}+\frac 1 {r_2}=0 \tag1$$
where $r_1$ and $r_2$ are the two principal radii. (Assume we have two perpendicular planes, the $(x,r)$ plane and the $(r,\theta)$ plane). In the $(x,r)$ plane, the radius is:
$$r_1=\frac{r''(x)}{\left(1+r'(x)^2\right)^{3/2}} \tag2$$
while the radius is "constant" at each section in $(r,\theta)$ with $r_2=-r(x) \tag3$
so that the first relation becomes: $$r(x)r''(x)=\left(1+r'(x)^2\right)^{3/2} \tag4$$
which is not the relation given by finding the minimal surface
$$rr''=1+r'^2 \tag 5$$ that leads to the correct catenoid form.
Where did I go wrong? I feel I'm using the wrong formula for $r_1$ but can't seem to find an alternative. Can someone throw a hint my way?
• The two principal radii are to be taken along perpendicular directions in the tangent plane at any point. The $(r,\theta)$ plane is not perpendicular to this tangent plane. So $-r(x)$ is not equal to $r_2$. However, by the fact that the surface is one of revolution, your $r_1$ is correct. – Chrystomath Nov 28 '17 at 18:04
• @MMS Agreed, the way I wrote it only applies in the middle. I should've taken an inclined plane making an angle $\alpha$ with respect to the $r$ axis. And since the curvature $k=||dT/ds||$ we should replace $dT$ with $$dT_2=dT cos\alpha=\frac{dT}{\sqrt{1+\tan^2\alpha}}=\frac{dT}{\sqrt{1+r'^2}}$$ so that $$r_{corrected}=r(x)\sqrt{1+r'^2}$$ – György von Fesz Nov 28 '17 at 18:58
For any curve $r(\theta)$, the curvature is given from the formula $$r(\theta)=r(0)+\dot{r}(0)\theta+\frac{1}{2}\ddot{r}(0)\theta^2+o(3)=r(0)+\dot{s}t+\frac{1}{2}\dot{s}^2\kappa n+o(3)$$ that is, $\kappa=\frac{\ddot{r}(0)\cdot n}{\dot{s}^2}$.
In the problem of the soap film, the surface is given by $$\mathbf{r}(x,\theta)=\begin{pmatrix}r(x)\cos\theta\\r(x)\sin\theta\\x\end{pmatrix}$$ There are two relevant perpendicular curves, those at constant $x$ and at constant $\theta$. The tangents in the different directions are $$t_x=\frac{d}{dx}\mathbf{r}=\begin{pmatrix}r'(x)\cos\theta\\r'(x)\sin\theta\\1\end{pmatrix}$$ $$t_\theta=\frac{d}{d\theta}\mathbf{r}=\begin{pmatrix}-r(x)\sin\theta\\r(x)\cos\theta\\0\end{pmatrix}$$ The normal vector is then $$n=\frac{t_x\times t_\theta}{|t_x\times t_\theta|}=\frac{1}{\sqrt{1+r'(x)^2}}\begin{pmatrix}-\cos\theta\\-\sin\theta\\r'(x)\end{pmatrix}$$
$\dot{s}=|t|$ in each case; so for the constant $\theta$ curve, $\dot{s}=|t_x|=\sqrt{1+r'(x)^2}$, while for the constant $x$ curve, $\dot{s}=|t_\theta|=r$.
Thus the two curvatures are given by $$\kappa_x=\frac{t_x'\cdot n}{\dot{s}^2}=\frac{-r''(x)}{\sqrt{1+r'(x)^2}^{3/2}}$$ $$\kappa_\theta=\frac{t_\theta'\cdot n}{\dot{s}^2}=\frac{r(x)}{r(x)^2\sqrt{1+r'(x)^2}}$$
For a minimal surface $\kappa_x+\kappa_\theta=0$, so $r''r=1+(r')^2$.
As you can see, the principal curvature in the $\theta$ direction is not $1/r$ as you might think.
• Agreed. I tried to fix it in the comment, but your way is much more rigorous. Thank you – György von Fesz Nov 28 '17 at 22:11
$$\frac 1 {r_1}+\frac 1 {r_2}=0 \rightarrow \frac {r_2}{r1}=-1 \tag{11}$$
where $r_1$ and $r_2$ are the same two principal radii of principal curvatures. $r_2$ should be taken correctly normal to meridian as:
$$r_2= {r(x)}{\left(1+r'(x)^2\right)^{1/2}} \tag{22}$$
$$r_1=\frac{-\left(1+r'(x)^2\right)^{3/2}}{r''(x)} \tag{33}$$
Divide above two equations
$$\frac{r(x) \, r''(x)}{1+r'(x)^2}=1 \tag{44/5}$$
Also you have written curvature incorrecty for radius of curvature at equation 2). Also your equation 3) incorrect. What you wrote is true only for a cylinder. | 1,382 | 3,850 | {"found_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} | 3.984375 | 4 | CC-MAIN-2019-30 | latest | en | 0.854653 |
http://mathhelpforum.com/advanced-algebra/111240-proof-part-2-fundamental-theorm-print.html | 1,524,141,057,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936914.5/warc/CC-MAIN-20180419110948-20180419130948-00575.warc.gz | 180,442,839 | 2,847 | # Proof for Part 2 of the fundamental theorm
• Oct 29th 2009, 03:54 PM
Avi06
Proof for Part 2 of the fundamental theorm
Part 2 of the Fundamental Theorm of Algebra says that null space of matrix A is orthogonal to row space of A and left null space is orthogonal to column space.
How can one prove this?
• Oct 29th 2009, 08:46 PM
tonio
Quote:
Originally Posted by Avi06
Part 2 of the Fundamental Theorm of Algebra says that null space of matrix A is orthogonal to row space of A and left null space is orthogonal to column space.
Uuh? The fundamental Theorem of Algebra, as far as I know, is the one that states that the field of complex numbers is an algebraically closed field, and has nothing to do directly with matrices.
And what you want about matrices is here in "relation to the null space":
Row space - Wikipedia, the free encyclopedia
Tonio
How can one prove this?
.
• Oct 30th 2009, 05:18 AM
HallsofIvy
Perhaps this is the "fundamental theorem of linear algebra"! | 256 | 982 | {"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} | 2.734375 | 3 | CC-MAIN-2018-17 | latest | en | 0.912137 |
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
1
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Sensor: The term sensor is used for an element which produces a signal relating to the quantity being measured
SENSORS AND TRANSDUCERS
Transducer: The term transducer is often used in place of the term sensor. Transducers are defined as elements that when subject to some physical change experience a related change.
SENSOR=TRANSDUCER
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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PERFORMANCE TERMINOLOGY
Range and Span: Range of a transducer defines the limits between which the input can vary. The span is the maximum value of the input minus the minimum value.
Example: a load cell is used to measure forces might have a range of 0 to 50 kN and ists span is 50 kN
Error: Difference between the result of the
measurement and the true value of the quantity being measured.
Error = measured value – true value
Example: If a measurement gives a temperature reading of 25 °C when the actual temperature is 24 °C, the error is +1 °C.
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Accuracy: Accuracy is extent to which the value
indicated by a measurement system might be wrong. It is thus summation of all possible errors that are likely to occur, as well as the transducer has been calibrated.
Example: A temperature-sensing instrument might be specified as
having an accuracy of ± 2 °C. This means that the measurements can be expected to lie between + or - 2 °C of the true value.
Accuracy is is often expressed as a percentage of the full range output or full scale deflection.
Example: accuracy of ± 5 % of full range output, full range is 0 to 200 °C
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Sensitivity: The relationship indicating how much output you get per unit input.
sensitivity = output/input
Example: A resistance termometer’s sensitivity is 0.5 Ω/ °C.
This term is frequently used to indicate the sensitivity to inputs other than that being measured. There can be sensitivy of a transducer to environmental
conditions, such as temperature, fluctuations of the supply voltage, etc.
Example: A pressure sensor might have a temperature sensitivity of ±1% of the reading per °C in temperature.
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Hysteresis Error: Maximum difference in output for increasinfg and decreasing values.
Non-Linearity Error: Maximum difference from the straight line (an linear characteristic).
Example: an nonlinearity error of a pressure sensor may be quoted as ±0.5% of the full range.
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Repeatability/Reproduceability: Ability to give the same output for repeated applications of the same input value. 100 range full n valuesgive . min . max ity repeatabil = − ×
Example: a transducer for the measurement of angular velocity typically might be quoted as having a repeatability of ±0.01% of the full range at particular angular velocity.
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Stability: Ability to give the same output when used to measure a constant input over a period of time. A term drift is often used to describe the change in
output that occurs over time. Drift may be expressed as a percentage of the full range output.
The term zero drift is used for the changes that occur in output when there is a zero input.
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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transducer is the range of inut values for which there is no output.
Example: in flow meters, due to bearing friction, there is no output until the flow reaches to a particular value.
The dead time is the lenght of time from the
application of an input until the output begins to respond and change.
Resolution: A smallest change in the input until an observable change in the output occurs. For a
sensor giving digital output the smallest change in the output is one bit.
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DYNAMIC CHARACTERISTICS
Resonse Time: The time which elapses after a constant input, a step input, ia applied to the
transducer up to a point at which the transducer gives an output corresponding to some specified percentage, e.g. 95%.
Time Constant: This is the 63.2% of the respnse time. Rise Time: This is the time taken for the output to
rise to some specified percentage of the steady state output.
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Displacement, Position and Proximity Sensors - Potentiometers
- Strain Gauged Element - Capacitive Element
- Differential Transformers
- Eddy Current Proximity Sensors - Inductive Proximity Switch
- Optical Encoders - Pneumatic Sensors - Proximity Switches - Hall Effect Sensor
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Linear Potentiometer
Moving direction
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Various Applications of Strain Gauged Elements
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Push-pull sensor Capacitive proximity sensor Capacitive Element Applications
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Hall Effect Proximity Sensor b Bi KH = Hall U + Output (V) - 0
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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Mechanical Proximity Switches
Lever Operated Roller Operated Cam Operated
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS 31
### Pyroelectric Sensors
t ∆ = ∆q kp Pyroelectric Effect
Lithium Tantalate Curie Temp. 610˚C
Equivalent Circuit
Dual Pyroelectric Sensor Application
Unpolarized
Crystal Polarized Crystal
External Electric Field Applied during the heating Process
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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### Pressure Sensors
Bellows
LVDT Bellows
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### Pressure Measurement
Orifice Plates
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### Temperature Sensors
Bimetalic Strip
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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### Temperature Sensors
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The magnetic flux created by the primary current IP is concentrated in a magnetic circuit and
measured in the air gap using a Hall device. The output from the Hall device is then signal
conditioned to provide an exact representation of the primary current at the output.
The magnetic flux created by the primary current lpis balanced by a complementary flux produced by driving a current through the secondary
windings. A hall device and associated electronic circuit are used to generate the secondary
(compensating) current that is an exact
### Current Sensors
Open Loop Technology
Closed Loop Technology
Hall Sensor
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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### Torque Transducers
Principles:
• Strain Gauge Torque Sensor • Deflection Torque Sensor • Reaction torque Sensors
Drive Element (Motor) Driven Element (Load) Torque Sensing Element
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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### Deflection Torque Transducers
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FALL 2004 MKM 503E MECHATRONIC SYSTEM COMPONENTS
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https://www.daniweb.com/programming/software-development/threads/307392/problems-with-my-3d-grid-creating-algorithm | 1,537,944,425,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163704.93/warc/CC-MAIN-20180926061824-20180926082224-00399.warc.gz | 708,232,871 | 13,557 | I've created an algorithm that draws a 2D grid in the most efficient way I could think of. Suggestions to improve it are encouraged though :)
public void DrawWireGrid(int xDivisions, int yDivisions, int cellSize, Color color)
{
Vector3 a = new Vector3();
Vector3 b = new Vector3();
for (int x = 0; x <= xDivisions; ++x)
{
a.X = x * cellSize;
b.X = a.X;
b.Y = yDivisions * cellSize;
DrawLine(a, b, color);
}
a.X = b.X = a.Y = b.Y = 0.0f;
for (int y = 0; y <= yDivisions; ++y)
{
a.Y = y * cellSize;
b.X = xDivisions * cellSize;
b.Y = a.Y;
DrawLine(a, b, color);
}
}
Using my original algorithm I've altered it to draw a 3 dimensional grid.
public void DrawWireGrid(int xDivisions, int yDivisions, int zDivisions, int cellSize, Color color)
{
Vector3 a = new Vector3();
Vector3 b = new Vector3();
for (int z = 0; z <= zDivisions; ++z)
{
for (int x = 0; x <= xDivisions; ++x)
{
a.X = x * cellSize;
a.Z = z * cellSize;
b.X = a.X;
b.Y = yDivisions * cellSize;
b.Z = z * cellSize;
DrawLine(a, b, color);
}
a.X = b.X = a.Y = b.Y = a.Z = b.Z = 0.0f;
for (int y = 0; y <= yDivisions; ++y)
{
a.Y = y * cellSize;
a.Z = z * cellSize;
b.X = xDivisions * cellSize;
b.Y = a.Y;
b.Z = z * cellSize;
DrawLine(a, b, color);
}
}
}
If I comment out either the yDivisions for loop, or the xDivisions for loop, the grid lines will draw correctly for that dimension.
When both loops inside the zDivisions loop are run, the grid refuses to draw the xDivisions correctly and no longer moves them out by the z value.
Can anyone spot where this algorithm is going wrong?
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Discussion Span
Last Post by DaveTran
Some optimizations for your loops could be done, take your xDivisions loop:
for (int x = 0; x <= xDivisions; ++x) {
a.X = x * cellSize;
a.Z = z * cellSize;
b.X = a.X;
b.Y = yDivisions * cellSize;
b.Z = z * cellSize;
DrawLine(a, b, color);
}
Lines 3, 6 and 7 never change values inside the loop, so move them outside the loops. Also 3 and 7 are the same equations, don't compute it twice:
a.Z = z * cellSize;
b.Y = yDivisions * cellSize;
b.Z = a.Z;
for (int x = 0; x <= xDivisions; x++) {
a.X = x * cellSize
b.X = a.X;
DrawLine(a, b, color);
}
Also (most of the time) multiplication is more 'expensive' in terms of CPU time than addition is, and you currently perform x multiplications in the loop. Change this to 1 multiplication and x additions:
float limit = xDivisions * cellSize;
a.Z = z * cellSize;
b.Y = yDivisions * cellSize;
b.Z = a.Z;
for (float x = 0.0f; x <= limit; x += cellSize) {
a.X = x;
b.X = x;
DrawLine(a, b, color);
}
This also saves you a call to the a.X property.
Good improvements, thank you
Nice!
kudos. Some very neat improvements :)
Thank you Momerath. A great improvement on what I had come up with originally.
I'm having a hard time with the 3D variation and still scratching my noodle over how to do it properly.
Vector3 a = new Vector3();
Vector3 b = new Vector3();
int endZ = zDivisions * cellSize;
for (int z = 0; z <= endZ; z += cellSize)
{
int end = xDivisions * cellSize;
b.Y = yDivisions * cellSize;
for (int x = 0; x <= end; x += cellSize)
{
a.X = x;
a.Z = z;
b.X = x;
b.Z = z;
DrawLine(a, b, color);
}
end = yDivisions * cellSize;
a.X = 0.0f;
b.X = xDivisions * cellSize;
for (int y = 0; y <= end; y += cellSize)
{
a.Z = z;
a.Y = y;
b.Y = y;
b.Z = z;
DrawLine(a, b, color);
}
}
The z values will only draw 2D grids along a third axis with the above equation and will not create a cube like grid.
Any suggestions on where to start?
After some hard thinking I've got an algorithm to draw a 3D grid in fastest way I can think of.
Vector3 a = new Vector3();
Vector3 b = new Vector3();
int eX = xDivisions * cellSize;
int eY = yDivisions * cellSize;
int eZ = zDivisions * cellSize;
for (int z = 0; z <= eZ; z += cellSize)
{
// X Axis
a.X = 0.0f;
b.X = eX;
for (int y = 0; y <= eY; y += cellSize)
{
a.Y = y;
a.Z = z;
b.Y = y;
b.Z = z;
DrawLine(a, b, color);
}
// Y Axis
a.Y = 0.0f;
b.Y = eY;
for (int x = 0; x <= eX; x += cellSize)
{
a.X = x;
a.Z = z;
b.X = x;
b.Z = z;
DrawLine(a, b, color);
}
}
// Z axis
for (int y = 0; y <= eY; y += cellSize)
{
for (int x = 0; x <= eX; x += cellSize)
{
a.X = x;
a.Y = y;
a.Z = 0.0f;
b.X = x;
b.Y = y;
b.Z = eZ;
DrawLine(a, b, color);
}
}
If anyone reading has any advice to improve my method, or perhaps has an alternate approach, please let it be known. :)
Edited by DaveTran: n/a
a.Z = 0.0f can be set once before the Z axis loop. I've just changed that.
EDIT:
Vector3 a = new Vector3();
Vector3 b = new Vector3();
int eX = xDivisions * cellSize;
int eY = yDivisions * cellSize;
int eZ = zDivisions * cellSize;
for (int z = 0; z <= eZ; z += cellSize)
{
for (int y = 0; y <= eY; y += cellSize)
{
// X axis
a.X = 0.0f;
a.Y = y;
a.Z = z;
b.X = eX;
b.Y = y;
b.Z = z;
DrawLine(a, b, color);
for (int x = 0; x <= eX; x += cellSize)
{
// Y axis
a.X = x;
a.Y = 0.0f;
a.Z = z;
b.X = x;
b.Y = eY;
b.Z = z;
DrawLine(a, b, color);
// Z axis
a.X = x;
a.Y = y;
a.Z = 0.0f;
b.X = x;
b.Y = y;
b.Z = eZ;
DrawLine(a, b, color);
}
}
}
Edited by DaveTran: n/a | 1,760 | 5,106 | {"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} | 2.890625 | 3 | CC-MAIN-2018-39 | longest | en | 0.657313 |
https://web2.0calc.com/questions/vector-question_3 | 1,653,307,947,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00523.warc.gz | 681,759,389 | 5,370 | +0
# Vector question
0
67
1
Let a and b be two vectors such that neither of a-b and a+b is the zero vector.
(a) Prove that if the vectors a and b are of equal length, then a-b and a+b are perpendicular.
(b) Prove that if a-b and a+b are perpendicular, then a and b are of equal magnitude.
Jan 26, 2022 | 90 | 307 | {"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} | 2.625 | 3 | CC-MAIN-2022-21 | longest | en | 0.883332 |
https://www.youphysics.education/archimedes-principle-anchor/ | 1,623,586,608,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00009.warc.gz | 987,226,125 | 6,204 | Archimedes' principle - Acceleration of an anchor
Problem statement:
If an anchor is made of iron, find its acceleration when it sinks after being dropped in the sea.
Givens: ρfl = 1.03 103 kg/m3; ρFe = 7.85 103 kg/m3; g = 10 m/s2
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Solution:
To solve this problem we are going to apply Archimedes’s principle as well as Newton’s second law.
In the figure below the forces acting on the anchor, its weigh P and the buoyant force F are shown:
Using Newton’s second law:
Since the density of iron is greater than the density of sea water, the magnitude of the anchor’s weight is greater than the magnitude of the buoyant force, so the acceleration of the anchor points downwards. Projecting onto the vertical axis: | 202 | 841 | {"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} | 3.28125 | 3 | CC-MAIN-2021-25 | latest | en | 0.93402 |
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Depth-first_search | 1,368,892,662,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382503/warc/CC-MAIN-20130516092622-00001-ip-10-60-113-184.ec2.internal.warc.gz | 318,900,873 | 5,719 | # All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary
# Science Fair Project Encyclopedia
For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.
# Depth-first search
Depth-first search (DFS) is an algorithm for traversing or searching a tree, tree structure, or graph. Intuitively, you start at the root (selecting some node as the root in the graph case) and explore as far as possible along each branch before backtracking.
Formally, DFS is an uninformed search that progresses by expanding the first child node of the search tree that appears and thus going deeper and deeper until a goal state is found, or it hits a node that has no children. Then the search backtracks and starts off on the next node.
All freshly expanded nodes are added to a LIFO queue (stack) for expansion.
• DFS is complete: if the tree is finite, then a solution would be found if one exists.
• DFS is not optimal , even with a finite tree and all non-negative path costs.
Space complexity of DFS is much lower compared to BFS (breadth-first search). It also lends itself much better to heuristic methods of choosing a likely-looking branch.
When searching large graphs that can not be fully contained in memory, DFS suffers from non-termination when the length of a path in the search tree is infinite. The simple solution of "remember which nodes I have already seen" doesn't work because there can be insufficient memory. This can be solved by maintaining an increasing limit on the depth of the tree, which is called iterative deepening depth-first search.
For the following graph:
a depth-first search starting at A, assuming that the left edges in the shown graph are chosen before right edges, and assuming the search remembers previously-visited nodes and will not repeat them (since this is a small graph), will result in the search visiting the nodes in the following order: A, B, D, F, E, C, G.
If the search did not remember previously visited nodes, but given the other conditions in the previous paragraph, the search would proceed as A, B, D, F, E, A, B, D, F, E, etc. forever, caught in the A, B, D, F, E cycle and never reaching C or G.
Iterative deepening prevents this loop and will reach the following nodes on the following depths, assuming it proceeds left-to-right as above:
• 0: A
• 1: A (repeated), B, C, E
(Note that iterative deepening has now seen C, when a conventional depth-first search did not.)
• 2: A, B, D, F, C, G, E, F
(Note that it still sees C, but that it came later. Also note that it sees E via a different path, and loops back to F twice.)
• 3: A, B, D, F, E, C, G, E, F, B
For this graph, as more depth is added, the two cycles "ABFE" and "AEFB" will simply get longer before the algorithm gives up and tries another branch.
## Pseudocode
```function DFS(Start, Goal)
push(Stack,Start)
while Stack is not empty
var Node := Pop(Stack)
if Node = Goal
return Node
for Child in Expand(Node)
push(Stack, Child)
``` | 752 | 3,232 | {"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} | 2.890625 | 3 | CC-MAIN-2013-20 | latest | en | 0.899951 |
https://poletoparis.com/how-do-you-write-a-negative-number-in-sas/ | 1,723,322,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00881.warc.gz | 372,086,590 | 9,442 | ## How do you write a negative number in SAS?
Use the ABS() function to get the absolute value for positive, and multiply it with -1 to get the negative.
### What is numeric format in SAS?
There are two components in a SAS numeric format. The number of columns (width) v of the output and the number of decimal places. The SAS system uses floating-point representation referred to us as W.D, where W is the width and D is the number of digits to the right of the decimal place.
#### How do you use absolute value in SAS?
Absolute value of the column in SAS is calculated using ABS() Function in SAS. ABS() Function in SAS takes column as argument and calculates absolute value of the column.
Which is an example of standard numeric data in SAS?
Examples of standard numeric values include: 26, 3.9, -13, +3.14, 314E-2, and 2.193E3.
What are SAS formats?
A format is a layout specification for how a variable should be printed or displayed. SAS contains many internal formats and informats, or user defined formats and informats can be constructed using PROC FORMAT.
## What is best format in SAS?
When a format is not specified for writing a numeric value, SAS uses the BEST w. format as the default format. The BEST w. format attempts to write numbers that balance the conflicting requirements of readability, precision, and brevity.
### What does ABS mean in SAS?
Returns the absolute value of a number.
#### How do I format a numeric variable in SAS?
SAS can handle a wide variety of numeric data formats. It uses these formats at the end of the variable names to apply a specific numeric format to the data….Output Numeric Formats.
Format Use
n. Write maximum “n” number of digits with no decimal point.
n.p Write maximum “n.p” number of columns with “p” decimal points.
What are formats in SAS?
What is BEST12 format in SAS?
Format. refers to the instructions that SAS uses when printing variable values. If no format is specified, the default format is BEST12. for a numeric variable, and \$w. for a character variable.
## How do you use ABS in SAS?
Returns the absolute value of a number. Note: The ABS function returns a nonnegative number that is equal in magnitude to the magnitude of the argument….Example 1.
Statements Results
x = abs(3.5) // outputs 3.5
x = abs(-7) // outputs 7
x = abs(-3*1.5) // outputs 4.5
### What is format and informat in SAS?
Informats is used to tell SAS how to read a variable whereas Formats is used to tell SAS how to display or write values of a variable. Informats is basically used when you read in sample data which is being created using CARDS/DATALINES statement or read or import data from either an external file (Text/Excel/CSV). | 618 | 2,703 | {"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} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.785199 |
http://slideplayer.com/slide/1404815/ | 1,526,959,564,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00371.warc.gz | 261,483,575 | 23,962 | # Measuring Volume Temperature Mass.
## Presentation on theme: "Measuring Volume Temperature Mass."— Presentation transcript:
Measuring Volume Temperature Mass
Reading the Meniscus Always read volume from the bottom of the meniscus. The meniscus is the curved surface of a liquid in a narrow cylindrical container.
Try to avoid parallax errors.
Parallax errors arise when a meniscus or needle is viewed from an angle rather than from straight-on at eye level. Correct: Viewing the meniscus at eye level Incorrect: viewing the meniscus from an angle
Graduated Cylinders The glass cylinder has etched marks to indicate volumes, a pouring lip, and quite often, a plastic bumper to prevent breakage.
Measuring Volume Determine the volume contained in a graduated cylinder by reading the bottom of the meniscus at eye level. Read the volume using all certain digits and one uncertain digit. Certain digits are determined from the calibration marks on the cylinder. The uncertain digit (the last digit of the reading) is estimated.
Use the graduations to find all certain digits
There are two unlabeled graduations below the meniscus, and each graduation represents 1 mL, so the certain digits of the reading are… 52 mL.
Estimate the uncertain digit and take a reading
The meniscus is about eight tenths of the way to the next graduation, so the final digit in the reading is 0.8 mL The volume in the graduated cylinder is 52.8 mL.
10 mL Graduate What is the volume of liquid in the graduate? 6 _ . _ _ mL 6 2
25mL graduated cylinder 1 _ _ . _ mL 1 5
What is the volume of liquid in the graduate? 1 _ _ . _ mL 1 5
100mL graduated cylinder 5 _ _ . _ mL 2 7
What is the volume of liquid in the graduate? 5 _ _ . _ mL 2 7
The cylinder contains:
Self Test Examine the meniscus below and determine the volume of liquid contained in the graduated cylinder. The cylinder contains: 7 _ _ . _ mL 6
The Thermometer Determine the temperature by reading the scale on the thermometer at eye level. Read the temperature by using all certain digits and one uncertain digit. Certain digits are determined from the calibration marks on the thermometer. The uncertain digit (the last digit of the reading) is estimated. On most thermometers encountered in a general chemistry lab, the tenths place is the uncertain digit.
Do not allow the tip to touch the walls or the bottom of the flask.
If the thermometer bulb touches the flask, the temperature of the glass will be measured instead of the temperature of the solution. Readings may be incorrect, particularly if the flask is on a hotplate or in an ice bath.
Determine the readings as shown below on Celsius thermometers: _ _ . _ C 8 7 4 _ _ . _ C 3 5
Measuring Mass - The Beam Balance
Our balances have 4 beams – the uncertain digit is the thousandths place ( _ _ _ . _ _ X)
Balance Rules In order to protect the balances and ensure accurate results, a number of rules should be followed: Always check that the balance is level and zeroed before using it. Never weigh directly on the balance pan. Always use a piece of weighing paper to protect it. Do not weigh hot or cold objects. Clean up any spills around the balance immediately.
Mass and Significant Figures
Determine the mass by reading the riders on the beams at eye level. Read the mass by using all certain digits and one uncertain digit. The uncertain digit (the last digit of the reading) is estimated. On our balances, the thousandths place is uncertain.
Determining Mass 1. Place object on pan
2. Move riders along beam, starting with the largest, until the pointer is at the zero mark
Check to see that the balance scale is at zero
1 1 4 ? ? ? _ _ _ . _ _ _ Read Mass
1 1 4 4 9 7 _ _ _ . _ _ _ Read Mass More Closely | 870 | 3,735 | {"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} | 3.46875 | 3 | CC-MAIN-2018-22 | latest | en | 0.815005 |
http://serc.carleton.edu/sp/ssac/national_parks/examples/Indiana_Dunes_Shifting_Sands.html | 1,503,012,200,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104172.67/warc/CC-MAIN-20170817225858-20170818005858-00635.warc.gz | 366,708,953 | 9,745 | # Shifting Sands: Quantifying Shoreline and Dune Migration at Indiana Dunes National Lakeshore
This material is replicated on a number of sites as part of the SERC Pedagogic Service Project
#### Summary
This Spreadsheets Across the Curriculum activity introduces Geology of National Parks students to the coastal zone of Indiana Dunes National Lakeshore at the southern end of Lake Michigan. The setting is one of beaches and dunes in a generally urban environment. The theme of the module is the interplay of engineering modifications and beach and dune dynamics. Students use data from air photos from 1939 and 2005, and three times in between, to study the retreat of the shoreline and the advance of the front of the dunes. The spreadsheet work involves calculating the average rate of change of the shoreline and dune front. The problem is solved using weighted averages, which are introduced with a problem involving the average driving speed in a car trip between Chicago and the park. The intent of the module is to have Geology of National Parks students make calculations that underscore the fact that the coastal zone is a dynamic environment in which engineering works have consequences.
This material is based upon work supported by the National Science Foundation under Grant Number NSF DUE-0836566. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
## Learning Goals
Slide 4 of the module.
### Students will:
• Read explanatory slides on the geologic setting of Indiana Dunes National Lakeshore, engineering modifications to the coastline, and basic background on beach erosion, beach nourishment, and dune growth and migration.
• View air photos of the shoreline and dune front from 1939, 1972, 1998, 2003, and 2005 and use them to calculate the rate that the shoreline has retreated and the rate that the dune front has advanced.
• Warm up for the calculation by working a weighted average problem to calculate the average speed traveled on a car trip from Chicago to the park given the speeds and travel times along the three legs of the trip (city traffic, interstate, country road). .
• Use a spreadsheet to determine the average movement of the shoreline and dune front (linear distance) from one date to the next from 14 distance measurements spaced equally along the length of the shoreline.
• Use the spreadsheet to calculate the rate (distance per time) of shoreline and dune-front migration for each time interval.
• Use the spreadsheet to calculate the average rate of shoreline and dune-front migration over the entire period from 1939 to 2005 (average of individual rates weighted by the length of time intervals).
### In the process the students will:
• See effects of a longshore current and anthropogenic modifications to the shoreline.
• Learn about beach erosion and beach nourishment.
• Work with real data concerning the changing lakeshore.
• Get practice calculating averages from data with unequal weights.
• Be convinced that the coastal zone is a dynamic environment and that cause-and-effect relations must be considered carefully before making modifications to a stretch of shoreline.
## Context for Use
Slide 10 of the module.
This module is designed for potential use in the Geology of National Parks service course at USF. The course is offered as an online course every semester. It includes readings from Parks and Plates, weekly quizzes based on that textbook, and weekly student activities designed to align the course with the University's general education requirements. This module is intended to be one of those activities, with the specific goal of meeting the gen-ed quantitative literacy dimension.
## Description and Teaching Materials
Slide 12 of the module.
• ### PowerPointSSACgnp.TC225.MHH1.1(PowerPoint 6.6MB Jun1 13)
Optimal results are achieved with Microsoft Office 2007 or later; the module will function in earlier versions with slight cosmetic compromises. If the embedded spreadsheets are not visible, save the PowerPoint file to disk and open it from there.
The above PowerPoint presentation file is the student version of the module. It includes a template for students to use to complete the spreadsheet(s) and answer the end-of-module questions, and then turn in for grading.
An instructor version is available by request. The instructor version includes the completed spreadsheet. Send your request to Len Vacher (vacher@usf.edu) by filling out and submitting the Instructor Module Request Form.
## Teaching Notes and Tips
The module is constructed to be a stand-alone resource. It can be used as a homework assignment, lab activity, or as the basis of an interactive classroom activity. It was used as an out-of-class activity in a senior-elective course, Environmental Geology of the National Parks (for geology majors and nonmajors), during development of the module in Spring 2010, and as an out-of-class activity in Computational Geology (a QL course for geology majors) in Fall 2010 and Fall 2011. In general, the students considered this module to be one of the more elementary modules in the collection. It is now one of the modules that is rotated into the online introductory-level Geology of National Parks course.
## Assessment
There is a slide at the end of the presentation that contains end-of-module questions. The end-of-module questions can be used to examine student understanding and learning gains from the module. Pre/post test, pre/post test answer key, and answer key for end-of-module questions are at the end of the instructor version of the module.
## References and Resources
US National Park Service (NPS)
Indiana Dunes National Lakeshore
Michigan City Harbor, Indiana
• http://www.lrc.usace.army.mil/co-o/Mich_City.htm
Indiana Dunes National Lakeshore: Future
Longshore currents
Beach Erosion
Jetties
Beach Renourishment
Dunes
Orthorectified
Georeferenced
Home | About the Center | Campus Visits | Events | Projects | Publications
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(360) 867-5611 | 1,266 | 6,150 | {"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} | 2.859375 | 3 | CC-MAIN-2017-34 | latest | en | 0.911798 |
https://aplusphysics.com/community/index.php?/blogs/blog/479-ivir-greats-physics/ | 1,656,450,158,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00536.warc.gz | 147,206,478 | 25,910 | # IVIR GREAT's Physics
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## Jenga
This past weekend, I saw a giant game of Jenga at MIT. Literally. The blocks were nearly 2x4s, and the structure was taller than I am. While I did not stay to watch, it is interesting to think about a few of the different strategies that I remember from my childhood days. First of all, I used to believe that the faster you pulled the object out, the less chance a collapse would occur. While I'm not sure of my logic behind this reasoning, I most likely imagined that hopefully the structure just w
## Horsepower
Cars, especially sports cars, often have their horsepower compared as a somewhat ignorant method to determining the faster car. However, this is not the most accurate way of making judgement because there are other variables, some much more important, in determining a cars acceleration or top speed. While an engine's horsepower provides the force, accelerating the car, it is important to remember the basic fact that F=ma. Therefore, a 4000 lb car with twice as much horsepower as a 2000 lb car wi
## Second Bounce
Have you ever dropped a ball and had it bounce once normally before taking a crazy second bounce? I'm always watching out for this phenomenon when I have a lacrosse ball on a hard surface, but I've never really understood what was going in. The main factor causing the crazy second bounce is actually the spin on the ball acquired during the first bounce. As the ball falls, it usually has a small amount of spin or is traveling at an angle that isn't 90 degrees. As the ball hits the ground, the ide
In the past few years, Madden NFL video games have adopted real time physics, an extremely complicated technology that allows the players in the game to react according the actual physics. In previous Maddens, a spin move would result in one of four tackles, and there were only about 10 different types of tackles, making the end of plays appear too staged. However, real time physics enables the characters in the video game to react to a force from an opposing player as they would in real life, c
## Frisbee
The simple art of Kan Jam or any other frisbee related game seems skills based, but it may help to have knowledge on the physics of the flight as well. The first aspect of a frisbee is the design. Most frisbees are relatively thin with curved edges, which make it more aerodynamic, creating lift while in flight. Similarly to cars, the curved shape at the front of the frisbee allows for air to pass over the frisbee faster than air passing under it, making the air above the frisbee at a lower press
## Shape and Dolphins
It is not surprising that shape has a lot to do with the movement and speed of objects, but it may be surprising that dolphins pose such a question to researchers when it comes to their advanced swimming techniques. The shape of an object can determine numerous factors of an object that are integral to its performance. For example, race-cars are built with a specific shape in order to produce high amounts of downforce while airplaines are built with their wings in order to create lift when pierc
## BMX
As usual, I found myself watching quite a bit of YouTube over break, especially different BMX videos. While I could only dream of pulling off the tricks the professionals do, watching it makes me feel nervous as I realize the potential consequences of a nasty fall. I have kind of a lot of blogs on landings/falling, so I figure I will use this blog to tackle one of the most crazy things a bmx rider will do: grind a downward sloping railing on one peg. For those not extremely comfortable with BMX
## Turf: the good, the bad, the ugly
Playing sports in high school, you start to look forward to any opportunities to leave the swampy grass for the fresh turf field. Since Pinegrove fields are uneven, rocky and often have large spots of grass missing, the turf is always a better option. However, playing on turf does not always end well for a variety of reasons. The good: traction. Turf provides excellent traction as it increases the coefficient of friction between the ground and cleats, enabling for sharp movements. On grass,
## Boxing vs UFC
Boxing and UFC are both entertaining (when there are available without the astronomical pay-per-view prices), but the physics of each one is pretty different. If anything, UFC is a more dangerous sport, making boxing look like a children's show. The major difference is due to the different types of gloves used in each one. Boxing gloves are pretty well known, but I've included pictures of both types below to make the comparison easier. While boxing gloves have quite a bit of padding, UFC gloves
## Tire Swing
Growing up, my best friend had a tire swing in his backyard. While it provided trivial entertainment, looking back, a tire swing involves quite a bit of physics. First of all, there is the connection to the tree. The swing needs to be far enough away from the base of the tree in order to prevent accidental collisions, but it also needs to be sturdy enough to withstand human weight. The further out from the base of the tree you put the swing, the more torque that is applied to the branch as the l
## Snow and Rain
Since we learn that objects consistently gain velocity in free fall motion due to the acceleration of gravity at 10 m/s^2, why doesn't rain and snow wreck havoc since it is falling from an insanely high distance? One reason for this lack of speed, especially with snowfall, is because of air resistance and drag forces. The net force on a snowflake would be the weight (mg) minus the drag forces acting on that object. Since snowflakes are relatively porous and non-aerodynamically shaped, the drag f
## Helmet to Helmet collisions
In the great game the Denver Broncos happened to win, there was a critical personal foul called in which a Denver player blasted the New England receiver shortly after the receiver caught the ball, incidentally hitting the receiver helmet to helmet. Considering that the average NFL player can run 15-20 mph and weighs around 200 pounds (90 kg), that is a lot of linear momentum that the receiver is being hit with. When the player is hit, a lot of the energy is dissipated as the players come to res
## Lets Go Broncos!
Since I'm sure everybody is watching the Denver Broncos vs. New England Patriots game right now (its currently halftime), I'm sure many of you are thinking how could the Patriots miss an extra point. Well, if you are curious, visit my previous blog on field goal kicking. More importantly, I'm sure some of you are like me, nervous due to Peyton Manning's lack of throwing ability in his old age. After his neck surgery, and since he is approaching 40, his arm strength is decreasing, requiring him t
## Trampoline Limits
A trampoline is a great example of spring potential energy and a restoring force, but it also brings up another question: why is there a 'maximum' height and why can you "double bounce" someone to make them fly higher than that maximum height. As a person begins jumping on a trampoline, their kinetic energy is converted to spring potential energy when they contact the trampoline, and then converted back to kinetic and gravitational potential energy as the person leaves the trampoline towards the
## Bulletproof Glass
As it turns out, the physics behind the ability of bulletproof glass to stop the momentum of a bullet without shattering is in fact a lot of chemistry. However, this makes a lot of sense considering that a large portion of chemistry is simple physics applied on the microscopic scale. A bullet typical travels 400 meters per second, which creates a large momentum despite the fact that the bullet is only a few grams. Normal glass is relatively fragile, shattering upon impact of most bullets while s
## Stealth Bomber
The B-2 bomber, commonly known as the Stealth Bomber, is an extremely expensive military aircraft capable over flying undetected across the globe. First of all, the bomber can travel over 6,000 miles without refueling due to its aerodynamic shape. The entire aircraft acts as a singular wing, due to its "flat" shape, which allows the resistance from air to hit the slightly angled bottom and push the aircraft up against the force of gravity. Since the aircraft can almost travel at the speed of sou
## Drifting-Without snow
People have started to worry about the roads in Rochester as a thin coating of snow, or ice, can cause cars to start sliding due to the lower coefficient of friction between rubber and ice rather than rubber and asphalt. This sensation, although scary for most drivers, is often sought after by drifting. Drifting is the process of purposefully kicking out the back-end of a car around a turn, for the thrill and awesomeness, an then corralling the car as the driver comes out of the turn. Although o
## The Slapshot
In honor of my Ovechkin's 500th goal (my favorite player on my favorite team), I decided to look into the physics behind the infamous slapshot in hockey. The basic physics of the slapshot include the windup that produces torque applied to the puck and the transfer of energy to the puck, but there is a lot more physics involved that launches the puck with such a high velocity. First of all, the collision with the puck is mostly elastic, but considering the huge noise produced during a shot,
## Snowboarding Rotation
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## Physics of Field Goal
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## Why we don't (or do?) fall
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## Barrett .50 cal
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## Phone crack part 2
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## #LaxBro
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• Create New... | 2,808 | 13,072 | {"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} | 3.140625 | 3 | CC-MAIN-2022-27 | longest | en | 0.972098 |
sozoenterprise.com | 1,656,122,556,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00155.warc.gz | 49,481,569 | 10,608 | # 7 Points You Probably Really Did Not Understand About Modelling
Modeling is the procedure of creating forecasts based on complex information. Versions might be in the kind of mathematical equations, computer system simulations, or even statistical designs. In all situations, designs require in-depth understanding about the materials as well as settings used to construct them. They are often the outcome of a variety of concepts coordinating in a particular configuration. For instance, modern climate modeling utilizes components of a range of various concepts. But, just how do versions function? ivy getty height
Modelling
Modelling is a technique of analytic and creative thinking. This innovative procedure requires imagination and ingenuity. Designing combines mathematical knowledge with real-world situations to develop a realistic representation of an intricate circumstance. The vital questions to ask before beginning any type of modelling process consist of: what problem are you attempting to fix? How much detail should you consist of? And also what objective are you trying to complete with this version? Listed here are some examples of modelling tasks and some info concerning its usages. christopher r getty
The importance of modelling is commonly accepted in the scientific neighborhood. Versions can be utilized to assess patterns in emerging infectious illness and also measure the prospective benefits and threats of public health treatments. On top of that, simulations help inform danger evaluations for emerging conditions as well as can offer a thorough understanding of specific situations. These simulations assist individuals create readiness plans, react faster to epidemics, and also keep an eye on the spread of diseases. Designing is an important component of neurolinguistic programs. william p getty
Modeling in the all-natural and also social scientific researches
In lots of methods, modeling in the natural and also social sciences resembles mathematics. They are both types of abstract calculus. One instance of a design is Newton’s formula of motion, which defines the interaction between a pendulum as well as its basic legislation. It uses the mathematics of differential formulas, however does not count on theory or population growth. But what are the distinctions in between these 2 types of modeling? This write-up will check out several of the distinctions.
One common sight of designs is the “fictional things” view, in which models amount imaginary things in literature, such as Sherlock Holmes or Center Earth. Both scientists and fiction authors presented designs, for example, when Bohr developed the atom and Conan Doyle produced Sherlock Holmes. While this sight does not line up with clinical practice, it still squares well with our daily methods. Unlike fiction writers, scientists commonly talk about versions as items. They typically take them to stand for imaginary atoms, populations, or economies.
Modeling in psychology
In psychology, modeling is an effective concept. When used properly, it allows us to discover just how effective individuals come close to situations and attain objectives. When used effectively, modeling can cause success in virtually any area of our lives. In company, for example, it can aid us build a winning mindset and also be a lot more efficient. Ultimately, this concept is applicable to a lot of elements of life, consisting of advertising and marketing as well as sales. But just how can we utilize modeling to make our very own choices? Allow’s check out the steps included.
Modeling involves replicating another person’s habits. Youngsters normally replicate the actions of people around them, including their parents. In psychology, modeling can aid us learn brand-new abilities, create a mindset, as well as accomplish our goals. The principle of modeling has been around for numerous years, and was created by Albert Bandura. For instance, it prevails expertise that children mimic their moms and dads’ bad habits. To maximize this all-natural tendency, parents can design etiquettes.
Modeling in economics
Modeling in economics is an integral part of the field of business economics. Unlike traditional business economics, this branch of scientific research has been around for centuries. The study of economics is important for recognizing the performance of markets and determining their possibility. Economists likewise utilize versions in policymaking, to assess the impact of new plans. However these designs are not best as well as they ought to be tempered by the realistic look of observed data and the credibility of the underlying concepts.
The mathematical modeling of business economics as well as financing is a fast-growing field with wide applications in lots of self-controls. For students of economics, mathematical modeling can work as a different textbook. This field includes numerous branches of mathematics with robust measurable reasoning. It focuses on business economics and also financing, while concurrently incorporating various contexts of possibility as well as stochastic processes. The end outcome is a set of devices for checking out modern monetary tools and markets. | 942 | 5,215 | {"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} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.93098 |
http://maths.crusoecollege.vic.edu.au/number-3-0/ | 1,701,904,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00522.warc.gz | 31,503,162 | 7,805 | # Number 3.0
Place Value
Choose six decimals (three with 2 decimal places, and three with 1 decimal place). Order them, label them with their place value and represent them using decimal squares. Explain how you ordered the decimals in a poster, a glog, or as I have done here, a Photostory.
Rounding
● They round numbers up and down to the nearest unit, ten, hundred, or thousand.
Counting
● They skip count forwards and backwards, from various starting points using multiples of 2, 3, 4, 5, 10 and 100.
Estimation
● They estimate the results of computations and recognise whether these are likely to be over-estimates or under-estimates.
You need to show you can add, subtract, multiply and divide, as shown in the video beneath. There are four worksheets (click to enlarge) – complete all four, and pick three sums from each sheet to do a screencast, and explain how you do each sum.
Times Tables
Take a photo of your tangled tables sheet where you have completed the sheet in ten minutes. | 239 | 1,018 | {"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} | 2.859375 | 3 | CC-MAIN-2023-50 | latest | en | 0.896752 |
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#### metamorphium
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##### Dijkstra puzzle
« on: November 06, 2004, 10:52:58 AM »
Hi muters.
I have a puzzle in which I have a grid 8x8 in which there are a few tiles. They can be moved according to the puzzle rules and the point is that if I choose tile start and tile destination, it has to slide around obstacles via the shortest path there (if it's of course possible).
Several of you told me, why not use built in path finding algorithm? Well I didn't found a way so maybe I spent some extra time reinventing the wheel. On the other hand I managed to hack Dijkstra's algorithm (breadth-first search) to work with wintermute and its single dimensional arrays. From my tests in 8x8 grid the worst case was 16 different paths with 10 elements, but usuallly it's much less.
For sake of speed I am using the array pathes which is devided to segments (in my case I use 64 nodes for each path), but all this is easily tweakable with the constants.
I cleaned and commented the code (from the mess Mnemonic saw ) and here it is for anyone who would need something like this.
Function for calling are initsearch(); and findroute(TileX,TileY); Source is ActiveTilie.x and ActiveTilie.y;
Enjoy
Code: [Select]
`global PuzzleArray; // The map with obstacles (0 is free, !=0 is obstacle)var open = new Array(64); // Open set for grid (max lenght of longest path)var pathes = new Array (1300); // All Pathes are stored herevar pathcnt = 0; // Number of pathes var correct_path = 0; // When path found, this one is the rightfulvar ActiveTilie; // Active TileActiveTilie.number = 0; // Info about selected TileActiveTilie.x = 0; ActiveTilie.y = 0;global tilies; // Definition of tilies as appears in scene_initconst GRIDX = 8; // Grid Size Xconst GRIDY = 8; // Grid Size Yconst OPENSIZE = 64; // Size of OPEN array// This is called whenever we need to use searchfunction initSearch(){ correct_path = 0; for (var a=0;a<OPENSIZE;a=a+1) open[a] = 0; for (a=0;a<1299;a=a+1) pathes[a] = -1; pathcnt = 0; return;}// Function for deleting path, which is a dead endfunction deletepath(pathc){ var startpath = pathc*OPENSIZE; var lastpath = (pathcnt-1)*OPENSIZE; if (pathcnt == 1) { // We have no more paths. There is noway to run... pathcnt = 0; return; } if (pathc == (pathcnt-1)) { // Deleted path is the last one pathes[startpath] = -1; pathcnt = pathcnt - 1; return; } // Move the last part in place of deleted. while (pathes[lastpath] != -1) { pathes[startpath] = pathes[lastpath]; lastpath = lastpath + 1; startpath = startpath + 1; } pathes[startpath] = -1; pathcnt = pathcnt - 1; return;}// Discard last node for path branchingfunction popnode(pth){ var spos = pth*OPENSIZE; while (pathes[spos] != -1) { spos = spos + 1; } pathes[spos-1] = -1; return;}// Add note on top of path stackfunction addnode(pth, x, y){ var spos = pth*OPENSIZE; while (pathes[spos] != -1) { spos = spos + 1; } if (spos>((pth*OPENSIZE) + OPENSIZE-1)) { // Too long path, it would overlap to different path = dead end deletepath(pth); return; } pathes[spos] = y * GRIDY + x; pathes[spos+1] = -1; return;}// Add path to path repositoryfunction addpath(oldpath,px,py){ pathcnt = pathcnt + 1; var startpath = (oldpath*OPENSIZE); var newpath = (pathcnt-1)*OPENSIZE; while (pathes[startpath] != -1) { pathes[newpath] = pathes[startpath]; newpath = newpath + 1; startpath = startpath + 1; } pathes[newpath-1] = -1; addnode(pathcnt-1,px,py); return;}// Returns last node from pathfunction getLastNode(pths){ var spos = pths*OPENSIZE; while (pathes[spos] != -1) { spos = spos + 1; } var news = pathes[spos-1]; return news;}// Gets tile number from Puzzle Gridfunction getTile(x,y){ return PuzzleArray[y*GRIDY+x];}// We have to check if two tiles (sx,sy) (dx,dy) are adjacentfunction isAdjacent(sx,sy,dx,dy){ if((dx < 0) || (dx >= GRIDY)) { return false; } if((dy < 0) || (dy >= GRIDY)) { return false; } if(sx == dx) { // The squares are adjacent if the source square is above the destination square if((sy - 1) == dy) return true; // Or if the source square is below the destination square else if((sy + 1) == dy) return true; } else if(sy == dy) { // The squares are adjacent if the source square is left of the destination square if((sx - 1) == dx) return true; // Or if the source square is right of the destination square else if((sx + 1) == dx) return true; } return false; // CELL (dest_x, dest_y) IS NOT adjacent to CELL (src_x, src_y) }// We have to check, if tile x,y is Open and not solidfunction isOpen(x,y){ if( PuzzleArray[y*GRIDY+x] != 0 ) return false; if(open[y*GRIDY+x] != 0) return false; return true; }// This is the Dijkstra's guts.function findroute(TileX, TileY){ initSearch(); var actpath = 0; // Actual processed path. var opener = 0; // Indicates how many branches are from current node. If 0, delete path. var tmp_node; // value (global value), x,y // Create the first path open[ActiveTilie.y*GRIDY + ActiveTilie.x] = 1; pathcnt = 1; addnode(actpath,ActiveTilie.x,ActiveTilie.y); while (pathcnt > 0) { // Get last node from actual path tmp_node.value = getLastNode(actpath); tmp_node.x = tmp_node.value % GRIDY; tmp_node.y = ToInt(tmp_node.value / GRIDY); opener = 0; // We have no open ends yet. for (var ox = tmp_node.x - 1; ox < tmp_node.x + 2; ox = ox + 1) for (var oy = tmp_node.y - 1; oy < tmp_node.y + 2; oy = oy + 1) { if (isOpen(ox,oy) && isAdjacent(tmp_node.x,tmp_node.y,ox,oy)) { opener = opener + 1; if (ox == TileX && oy == TileY) { //Path Found if (opener > 1) { popnode(actpath); } addnode(actpath,ox,oy); correct_path = actpath*OPENSIZE; return true; // Returning correct path } else { // Let's look some more // Set cell as visited open[oy*GRIDY+ox] = 1; if (opener == 1) { //Go down the road. addnode(actpath,ox,oy); } else { // Add new path addpath(actpath,ox,oy); } // else } // else } // If } // For if (opener == 0) deletepath(actpath); // Path is a dead end actpath = actpath + 1; // Let's walk down the next path if (actpath>=pathcnt) actpath=0; // We have to go to first path } return false;}`
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##### Re: Dijkstra puzzle
« Reply #1 on: November 06, 2004, 02:00:00 PM »
I think that when you said "pathfinding", people thought you were talking about pathfinding for the characters.
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#### metamorphium
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##### Re: Dijkstra puzzle
« Reply #2 on: November 06, 2004, 02:15:47 PM »
Sure could be the cause although I tried to say that it's puzzle with tiles
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#### Mnemonic
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##### Re: Dijkstra puzzle
« Reply #3 on: November 07, 2004, 09:40:07 AM »
Impressive scripting, I wouldn't believe it's possible Thanks for sharing.
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#### Jerrot
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##### Re: Dijkstra puzzle
« Reply #4 on: November 08, 2004, 11:12:37 PM »
I think that when you said "pathfinding", people thought you were talking about pathfinding for the characters.
Oops, so did I, that's why I answered exactly with these words. And to be honest - I still don't understand it completely, but the code looks impressive. Checking it some other day with some more time. (Did I mention, that I found out how the word "deadline" was created ? For sure a poor programmer died of a heart attack... at least I'm afraid of this currently. )
Anyway - thanks for sharing, metamorphium!
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Mooh!
#### metamorphium
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##### Re: Dijkstra puzzle
« Reply #5 on: November 08, 2004, 11:28:34 PM »
Ah,
http://www.volny.cz/kavani/puzzle.zip
I stripped the puzzle from it so it's just moving tiles, but you can see the implementation.
Just click on tile and then anywhere on grid.
Enjoy...
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#### SBOVIS
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##### Re: Dijkstra puzzle
« Reply #6 on: March 26, 2007, 02:35:53 PM »
Metamorphium,
Is it easy to put my own background picture made up of the pieces and to set the puzzle/picture as mixed up initially before the player works it out?
Also is there a restiction to size of picture/blocks as long as they are within the grid?
Great design!!
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#### Chris
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##### Re: Dijkstra puzzle
« Reply #7 on: June 22, 2010, 03:20:49 PM »
plese reupload file . thank you
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#### metamorphium
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##### Re: Dijkstra puzzle
« Reply #8 on: June 22, 2010, 03:28:49 PM »
I am afraid I don't have it anymore. This thread is 6 years old.
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#### Chris
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##### Re: Dijkstra puzzle
« Reply #9 on: June 22, 2010, 07:55:58 PM »
Please tell me. How do I use this. Help me more accurately. Thank you
« Last Edit: June 22, 2010, 07:58:42 PM by Chris »
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#### metamorphium
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##### Re: Dijkstra puzzle
« Reply #10 on: June 23, 2010, 01:25:09 PM »
Chris, I don't think you'd ever need this. It's an implementation of Dijkstra's algorithm. Unless you exactly know what you need to achieve using Dijkstra, you'd be better off without.
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#### piere
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##### Re: Dijkstra puzzle
« Reply #11 on: February 21, 2011, 11:16:19 PM »
Hello I put this script in and called the function but it says "Function 'initsearch' is referenced but not defined". Where should I put this script and how would I go about calling it so it would work? Thanks
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#### Mnemonic
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##### Re: Dijkstra puzzle
« Reply #12 on: February 22, 2011, 08:37:56 AM »
Hi, I don't know much about the script, but looking at it, the function is called "initSearch" (with capital S).
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#### piere
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##### Re: Dijkstra puzzle
« Reply #13 on: February 23, 2011, 03:01:32 AM »
Thanks Mnemonic, now I don't get the error but nothing happens. I put the script in an entity click, so when it says "On Look " or whatever, it calls the function.
here is my code:
Code: [Select]
`#include "scripts\base.inc"////////////////////////////////////////////////////////////////////////////////on "LookAt"{ actor.GoToObject(this); initSearch(); }////////////////////////////////////////////////////////////////////////////////on "Take"{ //Game.OpenDocument("http://www.dead-code.org");}////////////////////////////////////////////////////////////////////////////////on "Talk"{ actor.GoToObject(this); actor.Talk("Blah");}////////////////////////////////////////////////////////////////////////////////on "LeftClick"{ actor.GoToObject(this);} method Talk(srcString,srcActor,xPosition){ var twin; if (srcActor == null) twin = "talk"; // basic window with ghost else twin = srcActor; // NPC window var tmpState = Game.Interactive; Game.Interactive = false; // we save the interactivity state for later and turn it off var dlgWin = Game.LoadWindow("interface\system\caption.window"); // load the dialogue window var talkRobotEnt = Scene.CreateEntity(); // create the entity used for talking var tString = Game.ExpandString(srcString); // prepare the localized string to handle formatting var tLength = tString.Length; var lines = ToInt(tLength / 300) + 1; // find out how many lines will we need dlgWin.SetImage("interface/system/wme_logo.png"); // set the image dlgWin.Y = 425; // set the caption parameters talkRobotEnt.SubtitlePosRelative = false; talkRobotEnt.SubtitlesPosXCenter = false; talkRobotEnt.SubtitlesWidth = 680; talkRobotEnt.SubtitlesPosX = 90; if (xPosition != null) talkRobotEnt.SubtitlesPosX = xPosition; talkRobotEnt.SubtitlesPosY = 630 + 15* lines; // position the caption in the window based on number of lines talkRobotEnt.SetFont("fonts\verdana.font"); // set the speech font talkRobotEnt.SoundPanning = false; // make the sound centered talkRobotEnt.Talk(srcString, null, "", "", 0); // say the line Game.UnloadObject(dlgWin); // dispose of the window Scene.DeleteEntity(talkRobotEnt); // kill the talk entity Game.Interactive = tmpState; }global PuzzleArray; // The map with obstacles (0 is free, !=0 is obstacle)var open = new Array(64); // Open set for grid (max lenght of longest path)var pathes = new Array (1300); // All Pathes are stored herevar pathcnt = 0; // Number of pathes var correct_path = 0; // When path found, this one is the rightfulvar ActiveTilie; // Active TileActiveTilie.number = 0; // Info about selected TileActiveTilie.x = 0; ActiveTilie.y = 0;global tilies; // Definition of tilies as appears in scene_initconst GRIDX = 8; // Grid Size Xconst GRIDY = 8; // Grid Size Yconst OPENSIZE = 64; // Size of OPEN array// This is called whenever we need to use searchfunction initSearch(){ correct_path = 0; for (var a=0;a<OPENSIZE;a=a+1) open[a] = 0; for (a=0;a<1299;a=a+1) pathes[a] = -1; pathcnt = 0; return;}// Function for deleting path, which is a dead endfunction deletepath(pathc){ var startpath = pathc*OPENSIZE; var lastpath = (pathcnt-1)*OPENSIZE; if (pathcnt == 1) { // We have no more paths. There is noway to run... pathcnt = 0; return; } if (pathc == (pathcnt-1)) { // Deleted path is the last one pathes[startpath] = -1; pathcnt = pathcnt - 1; return; } // Move the last part in place of deleted. while (pathes[lastpath] != -1) { pathes[startpath] = pathes[lastpath]; lastpath = lastpath + 1; startpath = startpath + 1; } pathes[startpath] = -1; pathcnt = pathcnt - 1; return;}// Discard last node for path branchingfunction popnode(pth){ var spos = pth*OPENSIZE; while (pathes[spos] != -1) { spos = spos + 1; } pathes[spos-1] = -1; return;}// Add note on top of path stackfunction addnode(pth, x, y){ var spos = pth*OPENSIZE; while (pathes[spos] != -1) { spos = spos + 1; } if (spos>((pth*OPENSIZE) + OPENSIZE-1)) { // Too long path, it would overlap to different path = dead end deletepath(pth); return; } pathes[spos] = y * GRIDY + x; pathes[spos+1] = -1; return;}// Add path to path repositoryfunction addpath(oldpath,px,py){ pathcnt = pathcnt + 1; var startpath = (oldpath*OPENSIZE); var newpath = (pathcnt-1)*OPENSIZE; while (pathes[startpath] != -1) { pathes[newpath] = pathes[startpath]; newpath = newpath + 1; startpath = startpath + 1; } pathes[newpath-1] = -1; addnode(pathcnt-1,px,py); return;}// Returns last node from pathfunction getLastNode(pths){ var spos = pths*OPENSIZE; while (pathes[spos] != -1) { spos = spos + 1; } var news = pathes[spos-1]; return news;}// Gets tile number from Puzzle Gridfunction getTile(x,y){ return PuzzleArray[y*GRIDY+x];}// We have to check if two tiles (sx,sy) (dx,dy) are adjacentfunction isAdjacent(sx,sy,dx,dy){ if((dx < 0) || (dx >= GRIDY)) { return false; } if((dy < 0) || (dy >= GRIDY)) { return false; } if(sx == dx) { // The squares are adjacent if the source square is above the destination square if((sy - 1) == dy) return true; // Or if the source square is below the destination square else if((sy + 1) == dy) return true; } else if(sy == dy) { // The squares are adjacent if the source square is left of the destination square if((sx - 1) == dx) return true; // Or if the source square is right of the destination square else if((sx + 1) == dx) return true; } return false; // CELL (dest_x, dest_y) IS NOT adjacent to CELL (src_x, src_y) }// We have to check, if tile x,y is Open and not solidfunction isOpen(x,y){ if( PuzzleArray[y*GRIDY+x] != 0 ) return false; if(open[y*GRIDY+x] != 0) return false; return true; }// This is the Dijkstra's guts.function findroute(TileX, TileY){ initSearch(); var actpath = 0; // Actual processed path. var opener = 0; // Indicates how many branches are from current node. If 0, delete path. var tmp_node; // value (global value), x,y // Create the first path open[ActiveTilie.y*GRIDY + ActiveTilie.x] = 1; pathcnt = 1; addnode(actpath,ActiveTilie.x,ActiveTilie.y); while (pathcnt > 0) { // Get last node from actual path tmp_node.value = getLastNode(actpath); tmp_node.x = tmp_node.value % GRIDY; tmp_node.y = ToInt(tmp_node.value / GRIDY); opener = 0; // We have no open ends yet. for (var ox = tmp_node.x - 1; ox < tmp_node.x + 2; ox = ox + 1) for (var oy = tmp_node.y - 1; oy < tmp_node.y + 2; oy = oy + 1) { if (isOpen(ox,oy) && isAdjacent(tmp_node.x,tmp_node.y,ox,oy)) { opener = opener + 1; if (ox == TileX && oy == TileY) { //Path Found if (opener > 1) { popnode(actpath); } addnode(actpath,ox,oy); correct_path = actpath*OPENSIZE; return true; // Returning correct path } else { // Let's look some more // Set cell as visited open[oy*GRIDY+ox] = 1; if (opener == 1) { //Go down the road. addnode(actpath,ox,oy); } else { // Add new path addpath(actpath,ox,oy); } // else } // else } // If } // For if (opener == 0) deletepath(actpath); // Path is a dead end actpath = actpath + 1; // Let's walk down the next path if (actpath>=pathcnt) actpath=0; // We have to go to first path } return false;}`
« Last Edit: February 23, 2011, 04:08:17 AM by piagent »
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#### Mnemonic
• WME developer
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##### Re: Dijkstra puzzle
« Reply #14 on: February 23, 2011, 07:56:38 AM »
The question is, what do you expect it to do. It's quite a specific script and meta used it to implement a rather elaborate puzzle. I don't see much practical use for it, other than as a demonstration of advanced scripting.
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Please don't send me technical questions in private messages, use the forum.
Pages: [1]
Page created in 0.281 seconds with 22 queries. | 5,819 | 19,295 | {"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} | 2.78125 | 3 | CC-MAIN-2018-39 | longest | en | 0.892326 |
https://socratic.org/questions/the-sum-of-two-consecutive-odd-integers-is-16-what-are-the-two-integers | 1,638,178,110,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00286.warc.gz | 598,386,853 | 6,235 | # The sum of two consecutive odd integers is -16. What are the two integers?
Jan 8, 2017
The two integers are $- 9$ and $- 7$
#### Explanation:
We will let the first integer be $x$. The because these are consecutive ODD integers we need to add two to the first integer or $x + 2$.
We can now write and solve for $x$:
$x + \left(x + 2\right) = - 16$
$x + x + 2 = - 16$
$2 x + 2 = - 16$
$2 x + 2 - \textcolor{red}{2} = - 16 - \textcolor{red}{2}$
$2 x + 0 = - 18$
$2 x = - 18$
$\frac{2 x}{\textcolor{red}{2}} = - \frac{18}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = - 9$
$x = - 9$
So the first integer is $- 9$ and we know the second integer is $x + 2$ or $- 9 + 2 = - 7$ | 288 | 753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "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} | 4.625 | 5 | CC-MAIN-2021-49 | latest | en | 0.679003 |
https://blender.stackexchange.com/questions/271709/measuring-and-setting-exact-distance-between-two-objects | 1,713,346,501,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00837.warc.gz | 109,754,412 | 37,360 | # Measuring and setting exact distance between two objects
I need an exact distance of $$0.045$$ meters between the edges of two objects as seen here:
.
I can't figure out for the life of me how to:
1. Set the measurement tool's ruler to an exact length of $$0.045$$ meters, as it insists on literally any value above or below.
2. Snap the selected rectangle on the left to an exact distance of $$0.045$$ meters from the edge of the other object.
EDIT: While I've now figured out how to snap edges (make sure Affect>Move is enabled under the Snapping menu), entering the move distance as 0.045 meters from the edge manually winds up with the distance being 0.049244 meters instead:
The ruler on the y axis is also supposed to be 0.03 meters (I'd since tried to move both rectangles again, hence the odd placement).
Turns out, however, I was grabbing the wrong edge! I was selecting top edge of the prism, and not the bottom, which lead to the discrepancy. Since I was in top-down view I didn't notice.
In order to select the bottom edge for measurement, you have to click slightly away from the edge, like so:
Thanks to Harry for helping with this!
FULL SOLUTION: Okay, so Harry's answer allowed me to align the edges of the two objects.
In an attempt to make it more clear for other complete beginners like myself:
1. Enable snapping by clicking the snapping magnet button
1. Click the snapping mode button, set it to "Edge", and make sure "Affect>Move" is enabled. (I'm not sure if the "Snap With" setting is important in this case).
1. Select your object by clicking it using the select box tool.
2. Press G to move your object, and get your cursor to the edge of the other object you wish to align your select object to. Left-click to snap it to the edge.
1. With the right object selected, press G, then press the letter corresponding to the axis you need to move the object (X, Y, or Z). Then, type the exact distance you want. In my case, it's -0.045, for 0.045 meters of distance between the two edges. The - key is used for negative values.
2. When measuring to verify the correct distance was entered and accepted, make sure you select the right edge. See this image to understand why:
Very confusing if you're in 2D mode and don't understand why everything is slightly off.
To select and measure the bottom edge, click slightly away from the edge, like so:
• Just snap one edge to the other, then move it again with G, press X, Y or Z depending on in which direction you want to go, then enter .045 and press Return or LMB (or -.045 if you want to go in the opposite direction). Aug 11, 2022 at 6:41
With the Measure Tool active, hover your mouse over the first edge and then press and hold down Ctrl (which will display a small white circle snapping to the edge) and move along that edge then click LMB to confirm. This creates a 0m measurement annotation. Click LMB this annotation and drag it to the other edge while again holding down Ctrl. Notice how it snaps when it is horizontally aligned. | 705 | 3,029 | {"found_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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.944111 |
https://axiumlearn.co.za/question-6-21/ | 1,723,712,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00580.warc.gz | 92,986,847 | 21,754 | # Question 6
A light ray is incident on a glass prism. The angle of incidence is 38°, as shown below.
The refractive index of glass is 1,5 and that of air is 1.
6.1 Define the term angle of refraction. (2)
6.2 Calculate the angle of refraction inside the glass prism. (3)
6.3 Redraw the glass prism in the ANSWER BOOK. Complete the path of the light ray inside the prism and label the angle of refraction. (2)
A second prism, Q, of unknown material, is now placed next to the glass prism, as shown in the diagram below.
The light ray travels from the glass prism and enters prism Q at an angle of incidence of 36°. The angle of refraction inside prism Q is 41°.
6.4 Calculate the refractive index of prism Q. (2)
6.5 How does the speed of light in the glass prism compare to the speed of light in prism Q? Write only GREATER THAN, SMALLER THAN or REMAINS THE SAME. (1)
6.6 Explain the answer to QUESTION 6.5 by referring to the refractive indices of the materials. (2)
The critical angle for the glass prism Q boundary is 63,3°. The angle of incidence when the light ray travels from the glass prism to prism Q is increased to 65°.
6.7 Define the term critical angle. (2) | 314 | 1,189 | {"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} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.850991 |
https://www.thestudentroom.co.uk/showthread.php?page=12&t=4097287 | 1,513,514,554,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00706.warc.gz | 823,784,370 | 41,022 | You are Here: Home
# National 5 Modern Studies Watch
1. (Original post by Ethan100)
lol
"Well this person has done more work than they should have, I'm going to remove some mark"
Don't worry, you won't lose marks automatically for doing more paras. All that happens is if for a 6 marker u do 3 paras instead of 2 then the marker would read the paras and then award the marks to the best two. So u Might have a para worth 3 marks, a para worth 2 marks and a para worth 1 mark. You would get 5 marks as they award the marks to the best two. It's different from last year and 2014 where a para was expected to be worth 2 marks whereas now a para is expected to be worth 3 marks. If you click on the links below u can see an answer for a 2014 paper and how they did 3 very basic points and got the marks. (Note this person wrote it as one para but still has 3 points. Its not really how many paras you do but how many different points you raise). (It's sqa and a real candidates answer so legit btw).
ANSWER (in order to see the marks they got u have to click your mark and say what u thing they got and then click reveal examiner feedback to see what they really got.
http://www.understandingstandards.or.../QuestionPaper
So ye all that's changed is ur answers need to be more detailed than they used to be but u need less answers.
2. (Original post by Puppo)
Don't worry, you won't lose marks automatically for doing more paras. All that happens is if for a 6 marker u do 3 paras instead of 2 then the marker would read the paras and then award the marks to the best two. So u Might have a para worth 3 marks, a para worth 2 marks and a para worth 1 mark. You would get 5 marks as they award the marks to the best two. It's different from last year and 2014 where a para was expected to be worth 2 marks whereas now a para is expected to be worth 3 marks. If you click on the links below u can see an answer for a 2014 paper and how they did 3 very basic points and got the marks. (Note this person wrote it as one para but still has 3 points. Its not really how many paras you do but how many different points you raise). (It's sqa and a real candidates answer so legit btw).
ANSWER (in order to see the marks they got u have to click your mark and say what u thing they got and then click reveal examiner feedback to see what they really got.
http://www.understandingstandards.or.../QuestionPaper
So ye all that's changed is ur answers need to be more detailed than they used to be but u need less answers.
It's **** compared to the last years I must say
3. Here's what I put and my modern teacher said she was happy with what I'd told her:
Democracy in the UK
Q1 - 1st point: Parties use local media to talk about their aims and ambitions to a wide variety of potential voters. Listeners can also interact with the candidate through a phone call or text conversation
2nd point: Parties hand out leaflets to influence potential voters. Leaflets are cost effective and very simple on the eye, making voters liable to change their views in harmony with the party's views.
Q2 - 1st point: FPTP can produce a coalition government. Coalition governments can create disillusion and confusion in the British public, meaning we are likely to not develop and prosper to what has been originally promised. Example was the 2010 coalition formed between the Tories and Lib Dems
2nd point: Smaller parties find it hard to gain seats in the Commons if they are out with the big three parties. This is considered undemocratic because it means that the Commons is not representing a significant number of people's favoured parties and views. An example of this is how the Green Party struggle to represent in the Commons in General Elections.
3rd point: Many votes are wasted in the FPTP system. This is due to how there is only one vote allowed per person and how many parties are allowed to take part. An example of this is the results from the 2010 election, in which the majority of voters did not see their favoured party get voted in
Q3 - Selective in the use of Facts. It was okay but the source was a bit of a shambles in my opinion.
Crime and Law
Q1 - 1st point: Police can get involved in the community in an attempt to tackle crime. Police can do this by teaching the youth about crime and the consequences of crime as well as developing relationships neighbourhoods.
2nd point: Police can roll out in high numbers in an attempt to tackle crime. Police can do this to scare potential criminals away from committing crimes at big events such as football games and concerts.
Q2 - 1st point: Geographical location is one factor. This is because of how crime rate is much higher in city areas than rural areas, and that people are more likely to be directly affected in cities than rural areas. An example of this is the London Riots in (2011?).
2nd point: Age is another factor. This is because many criminals target the elderly due to their fragility and vulnerability and old age. An example of this is the mental trauma caused by the crime inflicted on them which can cause low morale and confidence in the elderly citizen.
Q3 - 10 marker was okay. Can't really remember it haha
USA
Q1 - 1st point: Americans have the right to run as a candidate for their political party. Americans can run for a political role at state to federal level. An example of this is Bernie Sanders running for president for the Democratic party.
2nd point: Americans can freely campaign for their favoured candidate/party. Americans can freely fund and campaign for their party in an attempt to boost the party's chances in influencing potential voters.
3rd point: Americans can attend political movements for their candidate/party. Americans can attend political rallies and walks to support their candidate and gain media attention.
Q2 - 1st point - America's economy and infrastructure have a big impact on foreign countries. This is because of their global recognition and their strong trade links with other Western countries, which take are able to prosper from the links. An example of this is Britain's trade link with America and how American companies like Apple and Google are able to thrive in a British economy as well as contribute to it.
2nd point - America's strong military has a big impact on foreign countries. This is because many countries cannot match the US's military force and succumb under invasion and threat of a war. An example of this is when America took over Libya when Colonel Gadaffi was murdered by the civilians against them, meaning Libya was under American leadership.
Q3- 1st conclusion: Japan was had less of a problem with crime than other countries
2nd conclusion: Can't rememeber it
3rd conclusion: South Korea
4. Thank god for that
5. For democracy in Scotland Q1 I said televised debates, would that be right? I'm quite worried because I haven't seen anybody else mention it
And for the US influence I said culture and financially, I'm quite worried I won't get the marks for culture
6. (Original post by jj1814)
For democracy in Scotland Q1 I said televised debates, would that be right? I'm quite worried because I haven't seen anybody else mention it
And for the US influence I said culture and financially, I'm quite worried I won't get the marks for culture
Me and my friend both did debates and flyers for question 1!
7. (Original post by Puppo)
Same lmao in all the writing exams I keep asking for more paper and finsish at the very end. God help me for higher
The woman came up to me when I asked for my third extra sheet and she was really mad 😂 and yeah I think its going to be so hard 😂 hopefully it'll be fine next year but there are way more skills questions I think 😁
8. (Original post by jj1814)
For democracy in Scotland Q1 I said televised debates, would that be right? I'm quite worried because I haven't seen anybody else mention it
And for the US influence I said culture and financially, I'm quite worried I won't get the marks for culture
I said televised debates and posters so hopefully we'll be fine!!
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 2,117 | 9,411 | {"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} | 2.546875 | 3 | CC-MAIN-2017-51 | latest | en | 0.94837 |
http://www.thefullwiki.org/B%C3%A9zier_curve | 1,553,066,718,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202303.66/warc/CC-MAIN-20190320064940-20190320090940-00085.warc.gz | 383,239,348 | 18,111 | # Bézier curve: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
Cubic Bézier curve
In the mathematical field of numerical analysis, a Bézier curve is a parametric curve important in computer graphics and related fields. Generalizations of Bézier curves to higher dimensions are called Bézier surfaces, of which the Bézier triangle is a special case.
Bézier curves were widely publicized in 1962 by the French engineer Pierre Bézier, who used them to design automobile bodies. The curves were first developed in 1959 by Paul de Casteljau using de Casteljau's algorithm, a numerically stable method to evaluate Bézier curves.
In vector graphics, Bézier curves are an important tool used to model smooth curves that can be scaled indefinitely. "Paths," as they are commonly referred to in image manipulation programs[note 1] are combinations of linked Bézier curves. Paths are not bound by the limits of rasterized images and are intuitive to modify. Bézier curves are also used in animation as a tool to control motion.[note 2]
Bézier curves are also commonly used over the time domain, particularly in animation and interface design. Thus, a Bézier curve is often used to describe or control the velocity over time of an object moving from A to B. For example, an icon might "ease-in-out" or follow a "cubic Bézier" in moving from A to B, rather than simply moving at a fixed number of pixels per step. Indeed, when animators or interface designers discuss the "physics" or "feel" of an operation, they often are referring to the particular Bézier curve used to control the velocity over time of the move in question.
## Applications
### Computer graphics
Bézier curves are widely used in computer graphics to model smooth curves. As the curve is completely contained in the convex hull of its control points, the points can be graphically displayed and used to manipulate the curve intuitively. Affine transformations such as translation, scaling and rotation can be applied on the curve by applying the respective transform on the control points of the curve.
Quadratic and cubic Bézier curves are most common; higher degree curves are more expensive to evaluate. When more complex shapes are needed, low order Bézier curves are patched together. This is commonly referred to as a "path" in programs like Adobe Illustrator or Inkscape. These poly-Bézier curves can also be seen in the SVG file format. To guarantee smoothness, the control point at which two curves meet must be on the line between the two control points on either side.
The simplest method for scan converting (rasterizing) a Bézier curve is to evaluate it at many closely spaced points and scan convert the approximating sequence of line segments. However, this does not guarantee that the rasterized output looks sufficiently smooth, because the points may be spaced too far apart. Conversely it may generate too many points in areas where the curve is close to linear. A common adaptive method is recursive subdivision, in which a curve's control points are checked to see if the curve approximates a line segment to within a small tolerance. If not, the curve is subdivided parametrically into two segments, 0 ≤ t ≤ 0.5 and 0.5 ≤ t ≤ 1, and the same procedure is applied recursively to each half. There are also forward differencing methods, but great care must be taken to analyse error propagation. Analytical methods where a spline is intersected with each scan line involve finding roots of cubic polynomials (for cubic splines) and dealing with multiple roots, so they are not often used in practice.
### Animation
In animation applications, such as Adobe Flash and Synfig, or in applications like Game Maker, Bézier curves are used to outline, for example, movement. Users outline the wanted path in Bézier curves, and the application creates the needed frames for the object to move along the path. For 3D animation Bézier curves are often used to define 3D paths as well as 2D curves for keyframe interpolation.
## Examination of cases
### Linear Bézier curves
Given points P0 and P1, a linear Bézier curve is simply a straight line between those two points. The curve is given by
$\mathbf{B}(t)=\mathbf{P}_0 + t(\mathbf{P}_1-\mathbf{P}_0)=(1-t)\mathbf{P}_0 + t\mathbf{P}_1 \mbox{ , } t \in [0,1]$
and is equivalent to linear interpolation.
A quadratic Bézier curve is the path traced by the function B(t), given points P0, P1, and P2,
$\mathbf{B}(t) = (1 - t)^{2}\mathbf{P}_0 + 2(1 - t)t\mathbf{P}_1 + t^{2}\mathbf{P}_2 \mbox{ , } t \in [0,1].$
A quadratic Bézier curve is also a parabolic segment.
TrueType fonts use Bézier splines composed of quadratic Bézier curves.
### Cubic Bézier curves
Four points P0, P1, P2 and P3 in the plane or in three-dimensional space define a cubic Bézier curve. The curve starts at P0 going toward P1 and arrives at P3 coming from the direction of P2. Usually, it will not pass through P1 or P2; these points are only there to provide directional information. The distance between P0 and P1 determines "how long" the curve moves into direction P2 before turning towards P3.
The parametric form of the curve is:
$\mathbf{B}(t)=(1-t)^3\mathbf{P}_0+3(1-t)^2t\mathbf{P}_1+3(1-t)t^2\mathbf{P}_2+t^3\mathbf{P}_3 \mbox{ , } t \in [0,1].$
Since the lines $\mathbf{P}_0\mathbf{P}_1$ and $\mathbf{P}_2\mathbf{P}_3$ are the tangents of the Bézier curve at $\mathbf{P}_0$ and $\mathbf{P}_3$, respectively, cubic Bézier interpolation is essentially the same as cubic Hermite interpolation.
Modern imaging systems like PostScript, Asymptote and Metafont use Bézier splines composed of cubic Bézier curves for drawing curved shapes.
## Generalization
The Bézier curve of degree n can be generalized as follows. Given points P0, P1,..., Pn, the Bézier curve is
${\mathbf{B}(t)=\sum_{i=0}^n {n\choose i}(1-t)^{n-i}t^i\mathbf{P}_i =(1-t)^n\mathbf{P}_0+{n\choose 1}(1-t)^{n-1}t\mathbf{P}_1+\cdots +{n\choose n-1}(1-t)t^{n-1}\mathbf{P}_{n-1}+t^n\mathbf{P}_n,\quad t \in [0,1],}$
where $\scriptstyle {n \choose i}$ is the binomial coefficient.
For example, for n = 5:
${\mathbf{B}(t)=(1-t)^5\mathbf{P}_0+5t(1-t)^4\mathbf{P}_1+10t^2(1-t)^3\mathbf{P}_2+10t^3(1-t)^2\mathbf{P}_3+5t^4(1-t)\mathbf{P}_4+t^5\mathbf{P}_5,\quad t \in [0,1].}$
This formula can be expressed recursively as follows: Let $\mathbf{B}_{\mathbf{P}_0\mathbf{P}_1\ldots\mathbf{P}_n}$ denote the Bézier curve determined by the points P0, P1,..., Pn. Then
$\mathbf{B}(t) = \mathbf{B}_{\mathbf{P}_0\mathbf{P}_1\ldots\mathbf{P}_n}(t) = (1-t)\mathbf{B}_{\mathbf{P}_0\mathbf{P}_1\ldots\mathbf{P}_{n-1}}(t) + t\mathbf{B}_{\mathbf{P}_1\mathbf{P}_2\ldots\mathbf{P}_n}(t)$
In other words, the degree n Bézier curve is a linear interpolation between two degree n − 1 Bézier curves.
### Terminology
Some terminology is associated with these parametric curves. We have
$\mathbf{B}(t) = \sum_{i=0}^n \mathbf{b}_{i,n}(t)\mathbf{P}_i,\quad t\in[0,1]$
where the polynomials
$\mathbf{b}_{i,n}(t) = {n\choose i} t^i (1-t)^{n-i},\quad i=0,\ldots n$
are known as Bernstein basis polynomials of degree n, defining t0 = 1 and (1 - t)0 = 1. The binomial coefficient, $\scriptstyle {n \choose i}$, has the alternative notation,
$^n\mathbf{C}_i = {n \choose i} = \frac{n!}{i!(n-i)!}.$
The points Pi are called control points for the Bézier curve. The polygon formed by connecting the Bézier points with lines, starting with P0 and finishing with Pn, is called the Bézier polygon (or control polygon). The convex hull of the Bézier polygon contains the Bézier curve.
• The curve begins at P0 and ends at Pn; this is the so-called endpoint interpolation property.
• The curve is a straight line if and only if all the control points are collinear.
• The start (end) of the curve is tangent to the first (last) section of the Bézier polygon.
• A curve can be split at any point into 2 subcurves, or into arbitrarily many subcurves, each of which is also a Bézier curve.
• Some curves that seem simple, such as the circle, cannot be described exactly by a Bézier or piecewise Bézier curve; though a four-piece cubic Bézier curve can approximate a circle, with a maximum radial error of less than one part in a thousand, when each inner control point (of offline point) is the distance $\textstyle\frac{4\left(\sqrt {2}-1\right)}{3}$ horizontally or vertically from an outer control point on a unit circle. More generally, an n-piece cubic Bézier curve can approximate a circle, when each inner control point is the distance $\textstyle\frac{4}{3}\tan(t/4)$ from an outer control point on a unit circle, where t is 360/n degrees, and n > 2.
• The curve at a fixed offset from a given Bézier curve, often called an offset curve (lying "parallel" to the original curve, like the offset between rails in a railroad track), cannot be exactly formed by a Bézier curve (except in some trivial cases). However, there are heuristic methods that usually give an adequate approximation for practical purposes.
• Every quadratic Bézier curve is also a cubic Bézier curve, and more generally, every degree n Bézier curve is also a degree m curve for any m > n. In detail, a degree n curve with control points P0, …, Pn is equivalent (including the parametrization) to the degree n + 1 curve with control points P'0, …, P'n + 1, where $\mathbf P'_k=\tfrac{k}{n+1}\mathbf P_{k-1}+\left(1-\tfrac{k}{n+1}\right)\mathbf P_k$.
## Constructing Bézier curves
### Linear curves
Animation of a linear Bézier curve, t in [0,1]
The t in the function for a linear Bézier curve can be thought of as describing how far B(t) is from P0 to P1. For example when t=0.25, B(t) is one quarter of the way from point P0 to P1. As t varies from 0 to 1, B(t) describes a curved line from P0 to P1.
For quadratic Bézier curves one can construct intermediate points Q0 and Q1 such that as t varies from 0 to 1:
• Point Q0 varies from P0 to P1 and describes a linear Bézier curve.
• Point Q1 varies from P1 to P2 and describes a linear Bézier curve.
• Point B(t) varies from Q0 to Q1 and describes a quadratic Bézier curve.
Construction of a quadratic Bézier curve Animation of a quadratic Bézier curve, t in [0,1]
### Higher-order curves
For higher-order curves one needs correspondingly more intermediate points. For cubic curves one can construct intermediate points Q0, Q1 & Q2 that describe linear Bézier curves, and points R0 & R1 that describe quadratic Bézier curves:
Construction of a cubic Bézier curve Animation of a cubic Bézier curve, t in [0,1]
For fourth-order curves one can construct intermediate points Q0, Q1, Q2 & Q3 that describe linear Bézier curves, points R0, R1 & R2 that describe quadratic Bézier curves, and points S0 & S1 that describe cubic Bézier curves:
Construction of a quartic Bézier curve Animation of a quartic Bézier curve, t in [0,1]
## Polynomial form
Sometimes it is desirable to express the Bézier curve as a polynomial instead of a sum of less straightforward Bernstein polynomials. Application of the binomial theorem to the definition of the curve followed by some rearrangement will yield:
$\mathbf{B}(t) = \sum_{j = 0}^n {t^j \mathbf{C}_j}$
where
$\mathbf{C}_j = \frac{n!}{(n - j)!} \sum_{i = 0}^j \frac{(-1)^{i + j} \mathbf{P}_i}{i! (j - i)!} = \prod_{m = 0}^{j - 1} (n - m) \sum_{i = 0}^j \frac{(-1)^{i + j} \mathbf{P}_i}{i! (j - i)!} .$
This could be practical if $\mathbf{C}_j$ can be computed prior to many evaluations of $\mathbf{B}(t)$; however one should use caution as high order curves may lack numeric stability (de Casteljau's algorithm should be used if this occurs). Note that the product of no numbers is 1.
## Rational Bézier curves
Sections of conic sections represented exactly by rational Bézier curves
The rational Bézier curve adds adjustable weights to provide closer approximations to arbitrary shapes. The numerator is a weighted Bernstein-form Bézier curve and the denominator is a weighted sum of Bernstein polynomials. Rational Bézier curves can, among other uses, be used to represent segments of conic sections exactly.[1]
Given n + 1 control points Pi, the rational Bézier curve can be described by:
$\mathbf{B}(t) = \frac{ \sum_{i=0}^n b_{i,n}(t) \mathbf{P}_{i}w_i } { \sum_{i=0}^n b_{i,n}(t) w_i }$
or simply
$\mathbf{B}(t) = \frac{ \sum_{i=0}^n {n \choose i} t^i (1-t)^{n-i}\mathbf{P}_{i}w_i } { \sum_{i=0}^n {n \choose i} t^i (1-t)^{n-i}w_i }.$
## Notes
1. ^ Image manipulation programs such as Inkscape, Adobe Illustrator, Adobe Photoshop, and GIMP.
2. ^ In animation applications such as Adobe Flash, Adobe After Effects, Microsoft Expression Blend, Blender, Autodesk Maya and Autodesk 3ds max.
## References
1. ^ Neil Dodgson (2000-09-25). "Some Mathematical Elements of Graphics: Rational B-splines". Retrieved 2009-02-23.
# Wiktionary
Up to date as of January 15, 2010
## English
Wikipedia has an article on:
Wikipedia
### Etymology
Popularized by Pierre Bézier.
### Noun
Singular Bézier curve Plural Bézier curves
Bézier curve (plural Bézier curves)
1. A cubic curve used as a spline, defined by four control points, of which two are the ends, and the other two determine the velocity with respect to the parameter at the ends. | 3,774 | 13,368 | {"found_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": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-13 | latest | en | 0.93954 |
https://www.cut-the-knot.org/m/Geometry/ThreeLinesAt60Degrees2.shtml | 1,516,725,914,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891980.75/warc/CC-MAIN-20180123151545-20180123171545-00481.warc.gz | 906,323,149 | 21,580 | # Equilateral Triangles Formed by Circumcenters
### What Might This Be About?
21 December 2013, Created with GeoGebra
### Problem
Segments $AA',$ $BB',$ $CC'$ are equal in length and at $60^\circ$ to each other. The lines they are on intersect at points $A_1,$ $B_1,$ and $C_1,$ as shown below:
Prove that the circumcenters of triangles $AB_1C',$ $A'BC_1,$ $A_1B'C$ form an equilateral triangle, and so are the circumcenters of triangles $A'B_1C,$ $AB'C_1,$ $A_1BC'.$
### Hint
The configuration in the problem is practically the same as that of Miquel's circumcenters. Here, too, we can detect three pairs of parallel lines with the same distance between the lines in a pair.
### Solution
$\Delta A_1B_1C_1$ is equilateral.
The distance between the perpendicular bisectors of $A'C_1$ and $AB_1$ equals $(B_1C_1 - AA')/2$ which is the same for the other two pairs of bisectors.
You are now asked to finish the proof.
### Acknowledgment
The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page. | 295 | 1,029 | {"found_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} | 3.578125 | 4 | CC-MAIN-2018-05 | longest | en | 0.892319 |
http://math.stackexchange.com/questions/146039/finding-fx-when-fx-int1-0-exyy2dy | 1,469,788,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830064.24/warc/CC-MAIN-20160723071030-00035-ip-10-185-27-174.ec2.internal.warc.gz | 166,963,006 | 20,108 | # Finding $f'(x)$ when $f(x)=\int^1_0 e^{xy+y^2}dy$
If $f(x) = \int^1_0 e^{xy+y^2}dy$, find $f'(0)$.
I understand that this is function defined by an integral, and $e^{y^{2}}$ does not integrate into an elementary function. So, I will need to take $f'(x)$ which yields:
$$\int^1_0 ye^{xy+y^2}dy$$
I am trying to integrate this, but I am failing. I take it I should use integration by parts, but I can't because I still have $e^{y^2}$ term. Any help?
-
You want to calculate $f^{\prime}(0)$, so plug $x = 0$ and it will be an easy integral. – Ayman Hourieh May 16 '12 at 23:22
Derive under the integration sign, this is smooth and it will be allowed. – checkmath May 16 '12 at 23:27
Thank you Ayman – Donnie May 16 '12 at 23:33
You're asked to calculate $f^{\prime}(0)$. You don't have to calculate $f^{\prime}(x)$ first.
\begin{align} f^{\prime}(0) = \int_0^1 y e^{y^2}dy \end{align}
Let $u = y^2$, $du = 2ydy$:
\begin{align} f^{\prime}(0) &= \frac{1}{2} \int_0^1 e^u du = \left. \frac{1}{2}e^u \right|_0^1 \\ &= \frac{1}{2}(e - 1) \end{align}
As for the general case of $f^\prime(x)$, the integral cannot be solved using elementary functions only. One will have to use the error function.
-
You are given
$$f(x) =\int_0^1 \exp(xy+y^2)dy$$
Differentiating gives
$$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy$$
Since we need $f'(0)$ we might as well plug in $x=0$. This gives
$$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy$$
But this integral is quite striaghtforward,
$$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy =\left. \frac 1 2 e^{y^2} \right|_0^1 = \frac 1 2 (e-1)$$
By the OP's request:
$$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy$$
We complete the square
$$f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_0^1 {y\exp \left[ {{{\left( {y + \frac{x}{2}} \right)}^2}} \right]dy}$$
We change variables
\eqalign{ & y + \frac{x}{2} = u \cr & dy = du \cr}
\eqalign{ & f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\left( {u - \frac{x}{2}} \right)} \exp \left( {{u^2}} \right)du \cr & f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du} \cr}
Let's focus on the first integral:
$${I_1} = \int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du = \left. {\frac{1}{2}\exp {u^2}} \right|_{x/2}^{1 + x/2} = \frac{1}{2}\exp \frac{{{x^2}}}{4}\left[ {\exp \left( {x + 1} \right) - 1} \right]$$
So we have
$$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du}$$
The other integral evaluates in terms of the error function
$$\operatorname{erf} (x) = \frac{2}{{\sqrt \pi }}\int_0^x {{e^{ - {t^2}}}} dt.$$
with a change or variables $u=-v$,
$$\int_{ - \left( {1 + \frac{x}{2}} \right)}^{ - \frac{x}{2}} {\exp \left( { - {v^2}} \right)dv} = \frac{{\sqrt \pi }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$$
So the function is
$$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\frac{{\sqrt \pi }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$$
For large values of $x$, you can neglect the last result (since the value will be smaller and smaller), and you can estimate $f'$ with
$$f'(x) \approx \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right]$$
Cheers.
-
If you put $x = 0$ in immediately after differentiating, you'll be able to skip a few steps... (edit: looks like you've done this now) – Ben Millwood May 16 '12 at 23:31
@benmachine I was thinking about finding the general $f'$ but then I changed my mind and that's why that mid step was kind of stranded there. – Pedro Tamaroff May 16 '12 at 23:31
@Peter, how would one go about finding the general f'? Is it possible? Thanks! – Donnie May 16 '12 at 23:33
@Donnie There is no elementary closed form of $f'$, but you can get a good approximation for large values of $x$. – Pedro Tamaroff May 16 '12 at 23:49
@PeterTamaroff Thank you very much. This is quite interesting. – Donnie May 17 '12 at 0:08
You have got f'(x), so let x be 0, and integrate by parts. Then you get the answer (e-1)/2.
-
Pan Yan, if you look at the other two anwers, you can see you are simply repeating in a summarized way what has been already been explained. It isn't useful to repeat what other questions say, but rather to add extra remarks, ideas or information. This question, as it is, is not useful for the OP. – Pedro Tamaroff May 17 '12 at 17:03 | 1,870 | 4,843 | {"found_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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-30 | latest | en | 0.754224 |
http://sepwww.stanford.edu/public/docs/sep97/morgan1/paper_html/node5.html | 1,508,759,291,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00018.warc.gz | 309,257,422 | 3,921 | Next: 2: Warping to match Up: Analysis Previous: Analysis
1: In-Fill and extrapolation of irregular picked seismic horizon
The irregular picked seismic horizons of Figures 1 and 2 are not defined at many x and y locations. So to create , we must fill all the interior holes'' in the irregular picked seismic horizons and also extrapolate them to the edges of the grid.
A human attempting to fill and extrapolate the data by hand would first discern, then manually extend, its dominant trends into the empty regions. Unfortunately, the human approach is a non-linear'' one; tough to reproduce and even tougher to encode into a computer algorithm. However, by computing a 2-D Prediction Error Filter (PEF) from the irregular picked seismic horizon, we encapsulate the spatial spectrum of the known data, and thus systematically extrapolate by imposing this spectrum on the output model. Crawley (1995) discusses a closely related example using sparse side-scan sonar bathymetry data.
First we must contend with a detail: the data used to estimate a PEF must obey the stationarity assumption. In other words, the spectrum of the data must be spatially invariant in order to encapsulate the inverse spectrum with a single PEF. Though the spatial spectra of the irregular picked seismic horizons in this example are roughly constant, the stationarity assumption is commonly violated for real-world problems. I make use of data patching'' to subdivide the data into smaller regions where the assumption is assumed to hold, and then estimate a PEF from the data contained in each patch Claerbout (1992). The dashed lines on Figures 1 and 2 delimit the four equal-sized patches I use in this example.
The problem of finding and is underdetermined, since we have only 578 known seismic data values, but 1600 model points. However, the classical least squares solution to the problem is valid only for overdetermined Strang (1986) systems. To convert this underdetermined problem to an overdetermined one, we must constrain the output model with additional regression equations, a process known as regularization. Normally the regularization operator imposes a minimum-energy,'' or other similarly safe'' constraint on the free model variables, but adds little or no meaningful statistical information to the problem. However, by using a PEF as the regularization operator, we impose a fundamental statistical property of the known data on the model. In symbols, the problem can be stated through the following least squares fitting goals.''
(1) (2)
In Equations (1) and (2), the output is . The '' means that we minimize the squared L2 norm of the residual. Equation (1) forces to match the irregular picked seismic horizon , where it is known. is known-data selector'' operator, which effectively ignores the difference wherever does not exist. is a so-called damping factor,'' which weights the effective strength of the regularization equations (2) relative to the data-matching'' equations (1). The operator is convolution with the patch-variant PEF. The problem is solved iteratively, using a conjugate direction-type (CD) algorithm.
The result is shown in Figure 3. Now that we have the surfaces and .
pef-seisfill-both
Figure 3
Top: and . Bottom: contour plots of surfaces shown above. Note that the predominant trend in both figures is roughly east-north-east'', consistent with the trends predicted by the patch-variant PEF.
Next: 2: Warping to match Up: Analysis Previous: Analysis
Stanford Exploration Project
7/5/1998 | 740 | 3,531 | {"found_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} | 2.578125 | 3 | CC-MAIN-2017-43 | latest | en | 0.893194 |
https://number.academy/3282450 | 1,718,936,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00142.warc.gz | 378,481,916 | 11,850 | # Number 3282450 facts
The even number 3,282,450 is spelled 🔊, and written in words: three million, two hundred and eighty-two thousand, four hundred and fifty, 3.3 million. The ordinal number 3282450th is said 🔊 and written as: three million, two hundred and eighty-two thousand, four hundred and fiftieth. The meaning of the number 3282450 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 3282450. What is 3282450 in computer science, numerology, codes and images, writing and naming in other languages
## What is 3,282,450 in other units
The decimal (Arabic) number 3282450 converted to a Roman number is (M)(M)(M)(C)(C)(L)(X)(X)(X)MMCDL. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
3282450 seconds equals to 1 month, 1 week, 2 days, 23 hours, 47 minutes, 30 seconds
3282450 minutes equals to 6 years, 9 months, 1 week, 4 days, 11 hours, 30 minutes
### Codes and images of the number 3282450
Number 3282450 morse code: ...-- ..--- ---.. ..--- ....- ..... -----
Sign language for number 3282450:
Number 3282450 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 3282450 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 3282450 are 2 * 3 * 5 * 5 * 79 * 277
The factors of 3282450 are
1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 79, 150, 158, 237, 277, 395, 474, 554, 790, 831, 1185, 1385, 1662, 1975, 2370, 2770, 3950, 4155, 5925, 6925, 8310, 3282450. show more factors ...
Total factors 48.
Sum of factors 8273280 (4990830).
#### Powers
The second power of 32824502 is 10.774.478.002.500.
The third power of 32824503 is 35.366.685.319.306.125.312.
#### Roots
The square root √3282450 is 1811,753294.
The cube root of 33282450 is 148,616161.
#### Logarithms
The natural logarithm of No. ln 3282450 = loge 3282450 = 15,004101.
The logarithm to base 10 of No. log10 3282450 = 6,516198.
The Napierian logarithm of No. log1/e 3282450 = -15,004101.
### Trigonometric functions
The cosine of 3282450 is 0,623024.
The sine of 3282450 is 0,782203.
The tangent of 3282450 is 1,255495.
## Number 3282450 in Computer Science
Code typeCode value
3282450 Number of bytes3.1MB
Unix timeUnix time 3282450 is equal to Saturday Feb. 7, 1970, 11:47:30 p.m. GMT
IPv4, IPv6Number 3282450 internet address in dotted format v4 0.50.22.18, v6 ::32:1612
3282450 Decimal = 1100100001011000010010 Binary
3282450 Decimal = 20011202200020 Ternary
3282450 Decimal = 14413022 Octal
3282450 Decimal = 321612 Hexadecimal (0x321612 hex)
3282450 BASE64MzI4MjQ1MA==
3282450 MD570c441028e01d37030ab12116e2b6e9e
3282450 SHA1d1a0071d04c63dac9d830128ab790c6fa9f9b5dc
3282450 SHA25628dab026aa5994cfb3cb0933b7f95a40fa0296e29a7f9cc31b9933c1d67106fe
3282450 SHA384c59b2190601c0672297dd8a3f4b92db7be65f2cb794392bd493edc4fd429eae56f538c0e0a08997040443d85822a223c
More SHA codes related to the number 3282450 ...
If you know something interesting about the 3282450 number that you did not find on this page, do not hesitate to write us here.
## Numerology 3282450
### Character frequency in the number 3282450
Character (importance) frequency for numerology.
Character: Frequency: 3 1 2 2 8 1 4 1 5 1 0 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 3282450, the numbers 3+2+8+2+4+5+0 = 2+4 = 6 are added and the meaning of the number 6 is sought.
## № 3,282,450 in other languages
How to say or write the number three million, two hundred and eighty-two thousand, four hundred and fifty in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 3.282.450) tres millones doscientos ochenta y dos mil cuatrocientos cincuenta German: 🔊 (Nummer 3.282.450) drei Millionen zweihundertzweiundachtzigtausendvierhundertfünfzig French: 🔊 (nombre 3 282 450) trois millions deux cent quatre-vingt-deux mille quatre cent cinquante Portuguese: 🔊 (número 3 282 450) três milhões, duzentos e oitenta e dois mil, quatrocentos e cinquenta Hindi: 🔊 (संख्या 3 282 450) बत्तीस लाख, बयासी हज़ार, चार सौ, पचास Chinese: 🔊 (数 3 282 450) 三百二十八万二千四百五十 Arabian: 🔊 (عدد 3,282,450) ثلاثة ملايين و مئتان و اثنان و ثمانون ألفاً و أربعمائة و خمسون Czech: 🔊 (číslo 3 282 450) tři miliony dvěstě osmdesát dva tisíce čtyřista padesát Korean: 🔊 (번호 3,282,450) 삼백이십팔만 이천사백오십 Danish: 🔊 (nummer 3 282 450) tre millioner tohundrede og toogfirstusindfirehundrede og halvtreds Hebrew: (מספר 3,282,450) שלושה מיליון מאתיים שמונים ושניים אלף ארבע מאות וחמישים Dutch: 🔊 (nummer 3 282 450) drie miljoen tweehonderdtweeëntachtigduizendvierhonderdvijftig Japanese: 🔊 (数 3,282,450) 三百二十八万二千四百五十 Indonesian: 🔊 (jumlah 3.282.450) tiga juta dua ratus delapan puluh dua ribu empat ratus lima puluh Italian: 🔊 (numero 3 282 450) tre milioni e duecentottantaduemilaquattrocentocinquanta Norwegian: 🔊 (nummer 3 282 450) tre million to hundre og åttito tusen fire hundre og femti Polish: 🔊 (liczba 3 282 450) trzy miliony dwieście osiemdziesiąt dwa tysiące czterysta pięćdziesiąt Russian: 🔊 (номер 3 282 450) три миллиона двести восемьдесят две тысячи четыреста пятьдесят Turkish: 🔊 (numara 3,282,450) üçmilyonikiyüzseksenikibindörtyüzelli Thai: 🔊 (จำนวน 3 282 450) สามล้านสองแสนแปดหมื่นสองพันสี่ร้อยห้าสิบ Ukrainian: 🔊 (номер 3 282 450) три мільйони двісті вісімдесят дві тисячі чотириста п'ятдесят Vietnamese: 🔊 (con số 3.282.450) ba triệu hai trăm tám mươi hai nghìn bốn trăm năm mươi Other languages ...
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https://www.jiskha.com/display.cgi?id=1298787056 | 1,516,475,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00242.warc.gz | 943,237,994 | 3,767 | # Math
posted by .
Find the perimeter of the cord r=a(1+cos theta)
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http://javaexplorer03.blogspot.com/2015/09/space-complexity.html | 1,526,893,395,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00572.warc.gz | 145,528,727 | 17,046 | ## Monday, 28 September 2015
### Space Complexity
Space Complexity:
Auxiliary Space is the extra space or temporary space used by an algorithm. Space Complexity of an algorithm is total space taken by the algorithm with respect to the input size.
Space complexity includes both Auxiliary space and space used by input.
Space complexity = Auxiliary space + space used by input
For example, if we want to compare standard sorting algorithms on the basis of space, then Auxiliary Space would be better criteria than Space Complexity.
Merge Sort uses O(n) auxiliary space (extra space), Insertion sort and Heap Sort use O(1) auxiliary space. Space complexity of all these sorting algorithms is O(n) though. | 143 | 708 | {"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} | 2.546875 | 3 | CC-MAIN-2018-22 | latest | en | 0.861084 |
https://quant.stackexchange.com/questions/55403/compute-the-price-of-a-derivative-which-pays-logs-ts-t-in-the-black-scholes/55601#55601 | 1,632,648,443,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00424.warc.gz | 517,764,464 | 40,743 | # Compute the price of a derivative which pays $\log(S_T)S_T$ in the Black Scholes world
Compute the price of a derivative which has pays $$\log(S_T)S_T$$, you can assume that the Black Scholes model is valid.
Using the stock measure we can write the expectation as
$$D(0) = S_0 \mathbb{E}_S(\log S_T)$$
with the expectation in the stock measure. In this measure,
$$dS_t = (r + \sigma^2)S_t dt + \sigma S_t dW_t$$
How has this been derived?
and it follows from Ito's lemma that
$$d \log S_t = (r+0.5\sigma^2)dt + \sigma dW_t$$
Why are we using Ito's lemma here?
Following this answer, let $$\mathbb Q$$ be the probability measure associated to the risk-free bank account as numeraire and $$\mathbb Q^1$$ the probability measure associated to the stock as numeraire.
You know that the standard equation $$\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^\mathbb{Q}$$ can be written as $$\mathrm{d}S_t=(r+\sigma^2)S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb{Q}^1}$$ under the stock measure by applying Girsanov's theorem (this is example 1 of section 3 of this answer). We simply use $$\mathrm{d}W_t^\mathbb{Q}=(\sigma\mathrm{d}t+\mathrm{d}W_t^{\mathbb{Q}^1})$$.
Similarly, applying Ito's Lemma to $$f(t,x)=\ln(x)$$, we have $$\mathrm{d}\ln(S_t)=\left(r-\frac{1}{2}\sigma^2\right)\mathrm{d}t+\sigma \mathrm{d}W_t^{\mathbb{Q}}$$ which translates to $$\mathrm{d}\ln(S_t)=\left(r+\frac{1}{2}\sigma^2\right)\mathrm{d}t+\sigma \mathrm{d}W_t^{\mathbb{Q}^1}$$ under the new measure. The latter equation is equivalent to $$\ln(S_t)= \ln(S_0)+\left(r+\frac{1}{2}\sigma^2\right)t+\sigma W_t^{\mathbb{Q}^1}.$$ Because $$W_t^{\mathbb{Q}^1}$$ is a standard Brownian motion under the stock measure $$\mathbb{Q}^1$$ (by construction) and thus has zero expectation, we have $$\mathbb{E}^{\mathbb{Q}^1}[\ln(S_t)]=\ln(S_0)+\left(r+\frac{1}{2}\sigma^2\right)t.$$
Turning now to the claim paying $$S_T\ln(S_T)$$, we can derive its price as follows \begin{align*} e^{-rT}\mathbb{E}^\mathbb{Q}[S_T\ln(S_T)] &= e^{-rT}\mathbb{E}^{\mathbb{Q}^1}\left[S_T\ln(S_T)\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{Q}^1}\right] \\ &= S_0 \mathbb{E}^{\mathbb{Q}^1}\left[\ln(S_T)\right] \\ &= S_0 \left(\ln(S_0)+\left(r+\frac{1}{2}\sigma^2\right)T\right). \end{align*} Here, I used $$\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{Q}^1}=\frac{S_0e^{rT}}{S_T}$$.
Of course, this value can be negative (just like the payoff this claim can be negative).
• lovely method!! Jul 12 '20 at 14:57
Part 1: deriving the drift of the stock price process under the stock Numeraire.
Under the risk-neutral measure, the process for $$S_t$$ is as follows:
$$S_t = S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$$
In the above model, the Numeraire is $$N(t)=e^{rt}$$ with $$N(t_0):=1$$. Specifically, $$W(t)$$ is a standard Brownian motion under the risk-neutral measure associated with the Numeraire $$N(t)$$.
The change of Numeraire formula is (I wanna change from $$N(t)$$ to some $$N_1(t)$$):
$$\frac{dN_1(t)}{dN(t)}= \frac{N(t_0)N_1(t)}{N(t)N_1(t_0)}$$
Using the stock as numeraire gives:
$$\frac{dN_{S}}{dN}(t) = \frac{1*S_t}{e^{rt}S_0}=\frac{S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]}{e^{rt}S_0}=e^{-0.5\sigma^2t+\sigma W_t}$$
The radon-nikodym derivative above is directly applicable to $$W(t)$$ using the Cameron-Martin-Girsanov Theorem.
Diving into the detail of how changing probability measure actually works, let's consider the probability distribution of $$W(t)$$ under the risk-neutral measure:
$$\mathbb{P}^Q(W_t \leq k)=\int_{h=-\infty}^{h=k}\frac{1}{\sqrt{2\pi}}e^{\frac{-h^2}{2t}}dh$$
We can define some new probability measure $$\mathbb{P}^2$$ using the Radon-Nikodym derivative $$y(W_t,t):=e^{-0.5\sigma^2t+\sigma W_t}$$ as follows:
$$\mathbb{P}^2(W_t\leq k):=\mathbb{E}^Q[y(W_t,t)I_{W(t) \leq k}]$$
Evaluating the expectation gives:
$$\mathbb{E}^Q[y(W_t,t)I_{W(t) \leq k}] = \int_{h=-\infty}^{h=k}y(W_t,t) f_{W_t}(h)dh = \\ = \int_{h=-\infty}^{h=k}e^{-0.5\sigma^2t+\sigma h} \frac{1}{\sqrt{2\pi}}e^{\frac{-h^2}{2t}}dh= \\ =\int_{h=-\infty}^{h=k}\frac{1}{\sqrt{2\pi}}e^{\frac{-(h^2-\sigma t)}{2t}}dh$$
Therefore we can see that applying the Radon-Nikdym derivative adds the drift $$\sigma t$$ to $$W_t$$ under the probability meaure $$\mathbb{P}^2$$ (we can see that via the probability distribution of $$W_t$$ under $$\mathbb{P}^2$$).
So in our case, $$\mathbb{P}^2$$ is the probability measure defined by using $$S_t$$ as numeraire, we can call it $$\mathbb{P}^{S_t}$$. The final step is to figure out the process of $$S_t$$ under $$\mathbb{P}^{S_t}$$:
Let's use the following algebric "trick": I am going to define a new process under the original risk-neutral measure $$Q$$, called $$\tilde{W_t}$$ as follows: $$\tilde{W_t}:=W_t-\sigma t$$.
Therefore, under the original measure $$Q$$, the process $$\tilde{W_t}$$ has a "negative" drift equal to $$-\sigma t$$.
Let's now insert $$\tilde{W_t}$$ into the original process equation for $$S_t$$ using $$W_t = \tilde{W_t} + \sigma t$$:
$$S_t=S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]= \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma (\tilde{W(t)}+\sigma t) \right] = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma^2 t + \tilde{W(t)} \right] = \\ = S_0exp\left[ (r+0.5 \sigma^2)t+ \tilde{W(t)} \right]$$
We know that applying the radon-nikodym derivative from before (i.e $$e^{-0.5\sigma^2t+\sigma W_t}$$ ) adds drift $$\sigma t$$, and we defined $$\tilde{W_t}$$ to have drift $$-\sigma t$$. Therefore applying the radon-nikodym to $$\tilde{W_t}$$ will remove the drift from $$\tilde{W_t}$$ and the process $$\tilde{W_t}$$ will become a driftless Standard Brownian motion under $$\mathbb{P}^{S_t}$$.
So we have the process for $$S_t$$ under $$\mathbb{P}^{S_t}$$ as:
$$S_0exp\left[ (r+0.5 \sigma^2)t+ \tilde{W(t)} \right]$$
Wehere $$\tilde{W(t)}$$ is a Standard Brownian motion without a drift.
Part 2: Ito's lemma to derive the process for $$log(S_t)$$.
I assume you know how to apply Ito's lemma to solve the standard GBM model for a stock price, i.e. our starting eqution above. Then by inspection, one can see that applying Ito's lemma to $$ln(S_t)$$ under measure $$\mathbb{P}^{S_t}$$ will produce the same result, but with a different drift. Indeed under $$\mathbb{P}^{S_t}$$:
$$S_t=S_0exp\left[ (r+0.5 \sigma^2)t+\sigma \tilde{W(t)} \right]$$
Therefore:
$$ln \left( \frac{S_t}{S_0} \right)= (r+0.5 \sigma^2)t+\sigma \tilde{W(t)}$$
I.e. the probability measure does not affect the way that Ito's lemma can be applied.
• I dont understand the applying radon nikodym bit Jul 3 '20 at 17:16
• $S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$ how was this solved? How was the third term in the LHS evaluated? Jul 3 '20 at 18:57
• $S_t$ is an Ito Process (by definition of Ito Process). By Ito's Lemma for any twice-differentiable function $F(S_t,t)$, we have: $F(S_t,t)=F(S_0,t_0)+\int_{h=t_0}^{h=t}\left( \frac{\partial F}{\partial t}+ \frac{\partial F}{\partial S}\mu S_h + 0.5\frac{\partial^2 F }{\partial S^2} \sigma^2S_h^2 \right) dh + \int_{h=t_0}^{h=t}\frac{\partial F}{\partial S}\sigma S_h dW_h$. Take $F(S_t,t)=ln(S_t)$, compute the derivatives, and you get: $ln(S_t)=ln(S_0)+\int_{h=t_0}^{h=t}\left(\mu + 0.5\sigma^2 \right) dh + \int_{h=t_0}^{h=t}\sigma dW_h$. The two integrals then evaluate directly to the result. Jul 5 '20 at 12:50
• @Permian: Radon-Nikodym: look at Cameron-Martin-Girsanov theorem. Radon-Nikdodym "derivative" is a random variable that allows you to change probability measure from some $\mathbb{P^1}$ to some $\mathbb{P^2}$. Jul 5 '20 at 12:52
• i know these things individually, im just struggling to see how to fit them together Jul 5 '20 at 16:16 | 2,930 | 7,824 | {"found_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": 74, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-39 | latest | en | 0.727592 |
http://mathhelpforum.com/discrete-math/76644-don-t-understand-nonnegative-integers-n-n-n-prove-your-answer.html | 1,529,873,701,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867055.95/warc/CC-MAIN-20180624195735-20180624215735-00238.warc.gz | 204,445,687 | 9,801 | # Thread: Don't understand this For which nonnegative integers n is nē < n!? prove your answer?
1. ## Don't understand this For which nonnegative integers n is nē < n!? prove your answer?
This problem been giving me fits;
Q: For which nonnegative integers n is nē < n!? prove your answer?
__________________________________________________ ______________
I know I have to use Basis Step, Inductive Hypothesis and Inductive Step. But when doing the basis step, if you let n > 1, then n^(2) is larger than n!, so how can the Basis hold up? I though Basis is when the LHS is equal to the RHS or it has to be true like 4 > 2.
This is what I am stuck at?
2. Originally Posted by Grillakis
This problem been giving me fits;
Q: For which nonnegative integers n is nē < n!? prove your answer?
__________________________________________________ ______________
I know I have to use Basis Step, Inductive Hypothesis and Inductive Step. But when doing the basis step, if you let n > 1, then n^(2) is larger than n!, so how can the Basis hold up? I though Basis is when the LHS is equal to the RHS or it has to be true like 4 > 2.
This is what I am stuck at?
Consider the function:
$\displaystyle f(k)=k^2(k+1)-(k+1)^2=k^3-2k+1$
$\displaystyle k \in \mathbb{R}$ , then:
$\displaystyle f'(k)=3k^2-2$
Now if $\displaystyle k\ge 1,\ f'(k)>0$ and so $\displaystyle f(k)$ is increasing, and as $\displaystyle f(1)=0,\ f(k)\ge0$ for all $\displaystyle k\ge 1$
Hence for all natural numbers $\displaystyle k\ge 1$ we have:
$\displaystyle f(k)=k^2(k+1)-(k+1)^2 \ge 0$
or:
$\displaystyle k^2(k+1)\ge (k+1)^2 \ \ \ \ \ \ \ \ \ \ \ ...(1)$
Now for the induction step of the proof assume that for some integer $\displaystyle k>1$ that $\displaystyle k^2\le k!$, then:
$\displaystyle k^2(k+1)\le (k+1)!$
but by $\displaystyle (1)$ above:
$\displaystyle (k+1)^2 \le k^2(k+1)\le (k+1)!$
etc...
CB
3. I see...thanks CB | 577 | 1,913 | {"found_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} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.789144 |
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# My debrief - 750 (Q51, V40) - IR 8
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GMAT 1: 750 Q51 V40
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My debrief - 750 (Q51, V40) - IR 8 [#permalink]
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19 Jan 2014, 08:06
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A day ago, I finished the GMAT around this time. I don't remember a lot of details from the test experience anymore, except for the painful few minutes entering your data right before you get the test score. Admittedly, I thought I screwed up big time to the extent that I considered cancelling my score. Then, a 750 flashed in front of my eyes. The euphoric feeling is a little absent, as I secretly hoped for 770+ (in line with my mock exams from GMATPrep), but overall I am very happy with the result.
To help future test takers, here is some advice.
Materials
I used the MGMAT Guides and MGMAT Advanced Quant for my theoretic foundation of Quant and e-gmat (in addition) for Verbal. For CATS, I used GMATprep, MGMAT and GMATclub tests.
MGMAT
The MGMAT guides were great as a solid basic foundation of math, yet were sometimes a little bit out of scope. The MGMAT Advanced Quant were great to take your theoretic foundation to the next level, teaching you dirty tricks that you haven't learned anywhere in high school. The math questions were quite difficult, but not very representative for the questions you will face at the GMAT.
GMATClub
Afterwards, I switched to the notorious GMATClub tests. As I was so used to MGMAT style, I screwed up the first two tests. GMATClub tests rely on a complete different concept, namely very advanced trickery, whereas MGMAT relies (often) on difficult calculation. Both of them were great, but if you want a true reflection of (difficult) GMAT-like questions, GMATClub beats any other company by far.
e-gmat
Furthermore, e-gmat helped me a lot for Sentence Correction. I purchased the entire Verbal Online pack, yet only ended up using the SC and CR part. SC was great and definitely made me feel secure in answering questions during CATs and the actual GMAT more than any other prep company could do. I was not so enthousiastic about their CR section - quite a few questions seemed a little ambigious (as is the case with almost every prep company). Thus, highly recommended for SC, neutral about CR.
The actual GMAT
Essay
I started off with the essay. Easy topic, flaws were obvious and could finish in time with 5 minutes left. I watched around the room and relaxed a little, before moving on to IR.
Integrated Reasoning
IR went surprisingly well - I hadn't practiced a single IR question in my life (apart from my last CAT), so I was not expecting a high score here. I can't really give people advice on this topic, as I haven't practiced this section.
Quant
The first 4 questions were very easy, probably below level 600. Then, all of a sudden the level went up in a rapid pace. The questions were hard, very hard - on par with GMATClub questions (I wonder if anybody else experienced something similar lately). Time flew by: by the time I hit question 28, I only had 10 minutes left. I completely rushed the last 10 questions (skipped two lame geometry questions I expected not to be able to answer within < 2 minutes). Quite frankly, I thought I screwed up. During the break, the MGMAT algorythm that makes your score drop tremendously if you miss a few of the last 10 questions kept spooking through my head.
Verbal
Started off with a few SC questions that were quite easy. Same goes for the CR questions. Around question 7, I got a huge RC passage with four very difficult questions. I didn't answer any of them with full confidence. It was hard to say what way I was going - SC seemed easy, CR and RC seemed moderately difficult.
At the end, I was quite sure my score would be < 700. I had no idea how I performed on Verbal and I was quite confident I scored low on Quant. Then a 750 (Q51, V40) appeared. I was especially shocked at my quant score. I am convinced that I missed quite a few questions, yet I maxed my score. I still think I could have scored a bit higher on verbal - honestly think the first 10 questions screwed me up. In general, I am happy that I am done with the GMAT. I will only apply to b-school in 2 years from now, which gives me plenty of time to focus on other stuff for now
Good luck to everybody.
My debrief - 750 (Q51, V40) - IR 8 [#permalink] 19 Jan 2014, 08:06
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My GMAT experience - 750 (Q51, V40) 7 22 Aug 2007, 16:50
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### 003 Christmas Mindbender 3
Festive PowerPoint Presentation
Please note that the Mindbender Presentations are 10-12 MB in size so they may take a couple of minutes to download.
MINDBENDER 1 (Level: Fiendish)
NUMBERS, NUMBERS, EVERYWHERE!!!!!
Prepare to stretch your student’s minds (and probably your own as well) with this seasonal piece of festive fun. Each Mindbender has 24, out -of - the ordinary mathematical questions. Each question contains the initial letters of one or more words that will make the GIVEN MATHEMATICAL STATEMENT TRUE. Some questions are relatively straightforward whilst others are fiendishly difficult. If things get tricky (and they will), you can click on the question number button which will take you to one of 24 different CHRISTMAS SCENES, each of which contains a CLUE to the answer. The clue may be written or graphical or in some cases it may be the scene itself. If the clue doesn’t do it then you can click to view the answer. It is the clues that make the Mindbenders (particularly 1 and 2) suitable for all abilities. It is not important for a student to answer all the questions or to do them in order. The idea is to have some number fun.
Examples Include:
12 = S of Z to give Signs of the Zodiac
21 = S on a D to give spots on a Dice
6 = F of a C to give faces of a cube.
There are lots of ways to present this:
1. Print worksheet and distribute. Students work individually/pairs/groups.
2. Brainstorm with the class for maximum excitement.
3. Perhaps do 2 with Mindbender 1 then 1 with other Mind Benders.
4. Think about a group/class competition/time limit?
Whichever approach you adopt it is sure to generate a lot of fun and interest.
Slide 2 gives some teacher guidance and slide 3 gives an introduction to students.
Note:
The Enrichment Area contains a “12 Days of Christmas” presentation which is ideal for more able students. This is an investigation that leads to the numbers and formulae for calculating how many presents were given over the Christmas period. The presentation shows animated balls for depicting triangular and tetrahedral numbers. Also includes a worksheet.
SPECIAL OFFER:
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Slides Animation Sound Worksheets Rating 30 n n y
£3.00 | 528 | 2,338 | {"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} | 3.703125 | 4 | CC-MAIN-2022-49 | latest | en | 0.910973 |
https://goprep.co/suppose-that-the-electric-field-part-of-an-electromagnetic-i-1njza2 | 1,620,952,454,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989616.38/warc/CC-MAIN-20210513234920-20210514024920-00058.warc.gz | 309,727,871 | 28,209 | Q. 11
# Suppose tha
(a) Given: E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}
As it is clear from the above electric field vector, that electric field is in the negative x direction. Therefore, the direction of propagation of the vector will be in negative y-direction i.e. –j.
(b) As we know that the general equation for electric field vector in the positive x-direction is:
..........(i)
The electric field part of an electromagnetic wave in vacuum is
E = {(3.1 N/C) cos [(1.8 rad /m) y + (5.4 × 106 rad/s)t..............(ii)
Comparing equation (i) and (ii), we get
E0 = 3.1 N
(c) Angular frequency ω = 5.4 × 106 rad/s
Wave number i.e. k = 1.8 rad/m
Wavelength of the wave can be calculated as follows:
λ = 2π/k
(d) Angular frequency of the wave can be expressed as follows:
ω = 2π v
5.4 × 106 = 2 × 3.14 × v
V = 0.86 × 106 Hz
(e) Magnetic field strength can be calculated as follows:
B0 = E0/c
B0 = 1.03 × 10-7 T
(f) Since magnetic field vector is in the negative z-direction, the general equation for magnetic field vector can be written as follows:
The expression for the magnetic field part of the wave can be written as follows:
B = B0 cos (ky + wt) k
B = {1.03 × 10-7 cos [(1.8 rad/m)y + (5.4 × 106 rad/s)t]}k
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You are given a 2Physics - Exemplar | 585 | 2,001 | {"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} | 3.453125 | 3 | CC-MAIN-2021-21 | latest | en | 0.833523 |
http://oeis.org/A056914 | 1,511,027,163,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805008.39/warc/CC-MAIN-20171118171235-20171118191235-00100.warc.gz | 229,554,504 | 4,359 | This site is supported by donations to The OEIS Foundation.
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A056914 a(n) = L(4n+1) where L() are the Lucas numbers. 7
1, 11, 76, 521, 3571, 24476, 167761, 1149851, 7881196, 54018521, 370248451, 2537720636, 17393796001, 119218851371, 817138163596, 5600748293801, 38388099893011, 263115950957276, 1803423556807921, 12360848946698171 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 REFERENCES V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers, A Publication of the Fibonacci Association, Houghton Mifflin Co., 1969, pps. 27-29. LINKS Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (7,-1) FORMULA a(n) = 7a(n-1) - a(n-2); a(0)=1, a(1)=11. G.f.: (1-4*x)/(1-7*x+x^2). - Philippe Deléham, Nov 02 2008 EXAMPLE a(n)={11*[((7+3*sqrt(5))/2)^n - ((7-3*sqrt(5))/2)^n]-[((7+3*sqrt(5))/2)^(n-1) - ((7-3*sqrt(5))/2)^(n-1)]}/3*sqrt(5). CROSSREFS Cf. (A056914)=sqrt{5*(A033889)^2-4}. Cf. quadrisection of A000032: A056854 (first), this sequence (second), A246453 (third, without 11), A288913 (fourth). Sequence in context: A034269 A256597 A245561 * A232032 A272395 A039674 Adjacent sequences: A056911 A056912 A056913 * A056915 A056916 A056917 KEYWORD easy,nonn AUTHOR Barry E. Williams, Jul 11 2000 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 617 | 1,809 | {"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} | 3.71875 | 4 | CC-MAIN-2017-47 | latest | en | 0.580731 |
https://toolshed.g2.bx.psu.edu/repos/xuebing/seq_to_meme/diff/tip/seq_to_meme.py | 1,656,941,001,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00501.warc.gz | 620,635,339 | 3,842 | ### diff seq_to_meme.py @ 1:54f508fcdc40defaulttip
author xuebing Sat, 31 Mar 2012 21:50:15 -0400
line wrap: on
line diff
```--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/seq_to_meme.py Sat Mar 31 21:50:15 2012 -0400
@@ -0,0 +1,102 @@
+# -*- coding: iso-8859-1 -*-
+import random,sys,math
+
+#import pylab
+
+
+ try:
+ f=open(filename)
+ except IOError:
+ print "could not open",filename,"Are you sure this file exists?"
+ sys.exit(1)
+
+ seqs=[]
+ maxL = 0
+ for line in f:
+ if '>' in line or 'N' in line:
+ next
+ else:
+ seq = line.strip().upper()
+ if maxL < len(seq):
+ maxL = len(seq)
+ seqs.append(seq)
+ f.close()
+
+ print 'max seq length:',maxL
+ for i in range(len(seqs)):
+ if len(seqs[i]) < maxL:
+ del seqs[i]
+ print len(seqs),'sequences with length = ',maxL
+ return seqs
+
+
+def createWeightMatrix(seqs,psuedocont):
+
+ motifWidth = len(seqs[0])
+ weightMatrix = []
+ for i in range(motifWidth):
+ weightMatrix.append({'A':psuedocont,'C':psuedocont,'G':psuedocont,'T':psuedocont})
+
+ #Use a for loop to iterate through all the sequences. For each sequence, begin at the start site in starts, and look at motifWidth bases. Count how many times each base appears in each position of the motif
+ for seq in seqs:
+ for pos in range(motifWidth):
+ weightMatrix[pos][seq[pos]] = weightMatrix[pos][seq[pos]] + 1.0
+
+ #Normalize your weight matrix (so that it contains probabilities rather than counts)
+ #Remember the added psuedocounts when you normalize!
+ for pos in range(motifWidth):
+ totalCount = sum(weightMatrix[pos].values())
+ for letter in weightMatrix[pos].keys():
+ weightMatrix[pos][letter] = weightMatrix[pos][letter]/totalCount
+
+ return weightMatrix
+
+def printMemeFormat(weightMatrix,motifName,filename,nSite,background):
+ f = open(filename,'w')
+ f.write('MEME version 4.4\n\n')
+
+ f.write('ALPHABET= ACGT\n\n')
+
+ f.write('strands: + -\n\n')
+
+ f.write('Background letter frequencies (from:\n')
+ f.write(background+'\n\n')
+
+ f.write('MOTIF '+motifName+'\n\n')
+
+ f.write('BL MOTIF '+motifName+' width='+str(len(weightMatrix))+' seqs='+str(nSite)+'\n')
+ f.write('letter-probability matrix: alength= 4 '+'w= '+str(len(weightMatrix))+' nsites= '+str(nSite)+' E= 0\n')
+ for position in range(len(weightMatrix)):
+ probsThisPosition=weightMatrix[position]
+ f.write(' '+"%.6f" %(probsThisPosition['A'])+'\t '+"%.6f" %(probsThisPosition['C'])+'\t '+"%.6f" %(probsThisPosition['G'])+'\t '+"%.6f" %(probsThisPosition['T'])+'\t\n')
+ f.write('\n\n')
+ f.close()
+
+#get a two decimal-place string representation of a float f
+def twoDecimal(f):
+ return "%.6f" %(f)
+
+def run():
+
+ #Get file name from command line
+ if len(sys.argv) < 3:
+ print "python seq2meme.py motif_fasta outputfile motifName psuedocont background"
+ sys.exit(1)
+ else:
+ motifFile=sys.argv[1] #
+ outFile=sys.argv[2]
+ motifName=sys.argv[3]
+ psuedocont = float(sys.argv[4])
+ background=' '.join(sys.argv[5].strip().split(','))
+ | 1,019 | 3,289 | {"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} | 2.640625 | 3 | CC-MAIN-2022-27 | latest | en | 0.58885 |
https://gmatclub.com/forum/store-s-sold-a-total-of-90-copies-of-a-certain-book-during-68590.html | 1,521,651,337,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647671.73/warc/CC-MAIN-20180321160816-20180321180816-00071.warc.gz | 605,542,526 | 40,679 | It is currently 21 Mar 2018, 09:55
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Store S sold a total of 90 copies of a certain book during
Author Message
Current Student
Joined: 11 May 2008
Posts: 552
Store S sold a total of 90 copies of a certain book during [#permalink]
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08 Aug 2008, 03:24
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Store S sold a total of 90 copies of a certain book during the seven days of last week, and it sold different numbers of copies on any two of the days. If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did Store S sell more than 11 copies on Friday?
1) Last Week Store S sold 8 copies of the book on Thursday
2) Last week Store S sold 38 copies of the book on Saturday
SVP
Joined: 17 Jun 2008
Posts: 1504
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08 Aug 2008, 03:37
Stmt 1 is sufficient. If the store sold 8 copies on Thursday and second highest number of books on Friday, then on Friday he must have sold more than 11 so that on other days he could have sold at least 11 and on Saturday the remaining.
Stmt 2 is sufficient. Similar logic as for stmt 1.
Current Student
Joined: 11 May 2008
Posts: 552
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08 Aug 2008, 04:02
hmm, if D is correct, then
1,2,3,4,8,9,63 is one option
1,2,3,4,8,12,60 is another option... out of whihc on friday 9 and 12 copies can be sold.
i think this is wrong...
Manager
Joined: 15 Jul 2008
Posts: 205
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08 Aug 2008, 06:11
B should be the answer. 38 on Saturday means it has to be greater than 11. With 11 on Friday, and 11 being second highest, we cannot get to 90. Only till 89
no fixed approach.. trial and error.
VP
Joined: 17 Jun 2008
Posts: 1325
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08 Aug 2008, 21:10
arjtryarjtry wrote:
Store S sold a total of 90 copies of a certain book during the seven days of last week, and it sold different numbers of copies on any two of the days. If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did Store S sell more than 11 copies on Friday?
1) Last Week Store S sold 8 copies of the book on Thursday
2) Last week Store S sold 38 copies of the book on Saturday
good question
total no. of copies =90
(1) 8 on thursday hence 90-8(5)=50 => minimum to be sold on sat to make gretest set sold on sat is 26 hence 24 is maximum that could be sold on friday
now consider min that could be sold on friday to make it second largest ,then 9 can be possible,10,11 15 etc rest sold on sat.
INSUFFI
(2)38 on sat means 52 sold on rest 6 days => maximum that could be sold ion 5 days is 40 so that 12 on friday is second greatest.
hence minimum sold on friday is 12 which > than 11hence SUFFI
IMO B
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Its Now Or Never
Manager
Joined: 27 May 2008
Posts: 139
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15 Aug 2008, 23:11
arjtryarjtry wrote:
Store S sold a total of 90 copies of a certain book during the seven days of last week, and it sold different numbers of copies on any two of the days. If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did Store S sell more than 11 copies on Friday?
1) Last Week Store S sold 8 copies of the book on Thursday
2) Last week Store S sold 38 copies of the book on Saturday
Frds,
I think the key over here is "and it sold different numbers of copies on any two of the days"
From 1)
If 8 copies were sold on Thursday, then:
Case 1
5,6,7,8,12,50,2
Case 2
5,6,7,8, 10, 50, 4
Thus insufficient
From 2)
If 38 copies were sold on Saturday, thn:
Case 1
5,6,7,8,16,38,10
Case 2
6,7,8,9,11,38,10
sum = 89
Thus not possible
Re: books and days [#permalink] 15 Aug 2008, 23:11
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Store S sold a total of 90 copies of a certain book during
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,345 | 4,802 | {"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} | 4.0625 | 4 | CC-MAIN-2018-13 | latest | en | 0.917179 |
https://worldbuilding.stackexchange.com/questions/81352/method-for-determining-manpower-needed-to-crew-a-starship-based-on-size | 1,653,553,985,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00668.warc.gz | 694,138,125 | 65,737 | # Method for determining manpower needed to crew a starship based on size
In the setting I'm building up, there are starships ranging from several meters in length to 1.6 kilometers. I'm trying to devise a means to determine what sort of manpower each ship would likely need.
My current method is looking at naval vessels of similar roles to their fictional counterparts (anywhere from World War II to cutting edge modern ships) and scaling their crew by the difference in size of the two ships.
I was curious to hear of other methods for determining crew requirements and to see if the numbers from my method sounded plausible.
.
Example of my method:
• Fictional ship: Heavily armed battleship, with a focus on firepower, length of 600 meters.
• Comparable ship: Yamato-class battleship, length of 256 meters (crew 2400-2700)
• 256/600 = 2.34
• Fictional ship crew: 2.34 * 2500 = 5859 maximum crew
• I don't think you need linear scaling for the personal to run ships solely based on their length. the total volume that needs to be taken care of doesn't increase by the same factor. I don't know the exact math, hence the commet but my estimation is to increase it with a factor 1.5 or something (2400|2700 * 1.5 = 3600|4050. May 20, 2017 at 18:48
• Except not everything scales with meters. You only need 1 captain, 1 navigator, etc. In a small ship those could be combined. Propulsion engineers probably scale with the cube root of the size; environmental engineers with the square root. So more like (sqrt(600 / 256 ) = 1.53, * 2500=3827. But the 1.6 kilometer ship scaling factor is sqrt(1600/256)=2.5, *2500=6250. One reason we make bigger ships is for economies of scale, including in personnel. Enough for it to be worth the risk of it sinking all at once, vs say 10 smaller ships. May 20, 2017 at 20:12
• @Amadeus bigger ships are more robust, if you need one shot to sink a smaller ship and have time for 3 shots you will destroy 30% of fleet, but the same 3 shots for bigger ship they will do the damage but not necessary 30% less capabilities and most likely it will not affect the bigger ship too much. May 20, 2017 at 20:46
Expanding on Sasha's answer, warship crews are not so much about the length of the ship, but about their displacement and whether they're specialized or not. But really it's all about technology level.
For example, HMS Victory, a 1st Rate Ship Of The Line launched in 1765 with a length of 70 m, it had a crew of about 850 men to sail the ship and work the over 104 guns. Most of the crew were necessary to work the sails and guns, but nearly everyone pulled double or triple duty.
Admiralty regulations stipulated the crew required by each type of gun, based on a rule-of-thumb of one man per 5 cwt [about 100 lbs] of gun, which worked out as 14 men for a 32-pounder, 11 men for an 18-pounder, and so on. These were the absolute minimum required and if, as generally happened, only one side of the ship was engaged with the enemy, then the crews on the unengaged side crossed the deck to assist... The crew of each gun was lead by a gun captain but most of his crew had another designated task and when called for by the pipe had to leave the gun. Such tasks included trimming the sails, manning a boarding party, fetching powder, acting as a boat's crew, and so on.
Patrick O'Brian's Navy by Richard O'Neill, Chris Chant, David Miller, and Dr Clive Wilkinson, p 76
Fast forward to a South Dakota class battleship [I'm using them as an example of a WWII battleship because they were never upgraded]. At 210 m long if we just scaled up the crew from HMS Victory based on length we'd need 2550 men... which is surprisingly close to the actual 2364 number, but this is a ruse.
Length is not a good choice for scaling a ship, displacement is. It's a measure of how much water the ship displaces when floating, and it gives you a rough idea of the size of the ship. HMS Victory is 3500 tons, but USS South Dakota was 45,000 tons! Over 10 times the size of Victory, but only 3 times the crew. Naval warfare changed a lot between Napoleon and WWII.
What about a modern ship? We don't use battleships anymore, our modern "do it all" surface vessel is an Arleigh Burke class destroyer. 155 m long and displacing 9,000 tons, highly automated, it has a crew of just 300 men.
But an Arleigh Burke is a design from the 90s. The new Zumwalt class destroyers just being commissioned are state of the art. They're 180 m long displacing 14,000 tons with a compliment of just 140.
A closer comparison to a starship might be a submarine. A nuclear submarine is required to be self-contained for long periods of time, must be sealed off against the environment, is very tight on space, and has a highly technical engine and weapons. A Virginia class, state of the art, has a length of 115 m, displaces 8000 tons, and a crew of just 135. The trend line of submarines is not so clearly downward as their size and capability have increased dramatically over the decades, but as you can see it's a pretty small crew.
You could make a plot of year, displacement, and crew and extrapolate, but as you can see the trend line for warship crew is inexorably downward.
This is of great benefit for warship designers. Less crew means less weight and space for that crew and all their amenities: less food, water, bunks, storage, kitchens, mess halls, entertainment, air conditioning, laundry, chaplains, doctors, medical facilities, medicines, etc...
On the other hand, less crew means more electrical power to run all the automated devices on the ship; overloaded electrical systems are a big hassle for upgrading navy ships. Less crew also means less people available for damage control and maintenance, until you have reliable repair robots.
So it depends on your tech level. If you're in a starship you can reliably assume a tech level at least equal to our own, but maybe you're in a retro-future Apollo-era space-race-run-wild setting with 60s level automation.
If you want to be realistic, you'd use as small a crew as possible. Starships are all about doing as much as you can with as little mass as possible, because $Force = mass \times acceleration$. Less mass means less force for the same delta-V means less fuel means less mass means less force.... More crew means more mass; not just for the crew itself, but for all the support structure that goes along with them (see above).
• You make a good point. I didn't have the means to calculate displacement for my starships, so I was going by length since it was something I could measure (and adjusting based on other dimensions as needed.) Still, if increasing automation can bring modern warships to use such a low crew count, I could easily bring these ships down as well. I didn't realize it had THAT great an effect. May 21, 2017 at 3:54
• @Arvex I thought about it and submarines might be a better choice to work from; I added a paragraph about them. I'm gonna say a crew of 100 is a good starting point, go down from there. May 21, 2017 at 4:14
I guess the biggest question you need to answer is the level of automation available and how advanced is the drone and robot technology of the setting?
With enough tech you could easily have a crew of say 12 people on the command center, with every system automated and using drones when they need to go anywhere.
• I was planning for crew counts to be reduced a bit based on the tech levels of the various fleets, though nowhere near that heavy. I'm just looking to have base numbers that I can adjust based on technological level. May 20, 2017 at 19:32
Rather than focus on the size of the ship first I would suggest you focus on what you want your crew to do. Break down your crew into it's component departments and work from there.
Presumably you're going to have a set of officers who run things from the "Bridge", some sort of engineering crew to run the engine rooms, other staff for things like medical and food prep and stuff like that. Do you need to carry troops? Add a "marine" squad. Fighters? You need pilots and people to maintain them. If your weapons aren't controlled from the bridge then you're going to need on mount crews for each gun.
Work out how many people you want for each. Presumably the story is going to be about these people anyway, right? Once you have that sort of detail nutted out you start working out how many people you need for each. Bigger ships have multiple engine rooms? More engineers. More guns? More on mount crews and so on.
• Was going to write something similiar. I agree that if you are telling a story involving these crews, the "Why" you need a certian size of crew should drive your reasoning. If you want a large crew because it makes for a better story, then that should be the reason you go with in the end, no matter what kind of hand wavium you need to justify it. May 22, 2017 at 20:11 | 2,104 | 8,896 | {"found_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} | 3.421875 | 3 | CC-MAIN-2022-21 | longest | en | 0.951276 |
https://books.google.com/books/about/The_Descent_Map_from_Automorphic_Represe.html?id=gAaj4zEeDyUC&hl=en | 1,568,954,489,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00199.warc.gz | 376,858,307 | 10,438 | # The Descent Map from Automorphic Representations of GL(n) to Classical Groups
World Scientific, 2011 - Mathematics - 352 pages
8. Non-vanishing of the descent I. 8.1. The Fourier coefficient corresponding to the partition (m, m, m' - 2m). 8.2. Conjugation of S[symbol] by the element [symbol]. 8.3. Exchanging the roots y[symbol] and x[symbol]. 8.4. First induction step : exchanging the roots y[symbol] and x[symbol], for 1[symbol]; dim[symbol]V = 2m. 8.5. First induction step : odd orthogonal groups. 8.6. Second induction step : exchanging the roots y[symbol] and x[symbol], for i + j[symbol]. 8.7. Completion of the proof of Theorems 8.1., 8.2.; dim[symbol]V = 2m. 8.8. Completion of the proof of theorem 8.3. 8.9. Second induction step : odd orthogonal groups. 8.10. Completion of the proof of Theorems 8.1, 8.2; h(V) odd orthogonal -- 9. Non-vanishing of the descent II. 9.1. The case H[symbol]. 9.2. The case H = SO[symbol]. 9.3. Whittaker coefficients of the descent corresponding to Gelfand-Graev coefficients : the unipotent group and its character; h(V) [symbol]. 9.4. Conjugation by element [symbol]. 9.5. Exchanging roots : h(V) = SO[symbol]. 9.6. Nonvanishing of the Whittaker coefficient of the descent corresponding to Gelfand-Graev coefficients : h(V) = SO[symbol], U[symbol]. 9.7. Nonvanishing of the Whittaker coefficient of the descent corresponding to Gelfand-Graev coefficients : h(V) = U[symbol], SO[symbol]. 9.8. The Whittaker coefficient of the descent corresponding to Fourier-Jacobi coefficients : H[symbol]. 9.9. The nonvanishing of the Whittaker coefficient of the descent corresponding to Fourier-Jacobi coefficients : H[symbol] = S[symbol]. 9.10. Nonvanishing of the Whittaker coefficient of the descent corresponding to Fourier-Jacobi coefficients : h(V) = U[symbol] -- 10. Global genericity of the descent and global integrals. 10.1. Statement of the theorems. 10.2 Proof of Theorem 10.3. 10.3. Proof of Theorem 10.4. 10.4. A family of dual global integrals I. 10.5. A family of dual global integrals II. 10.6. L-functions -- 11. Langlands (weak) functorial lift and descent. 11.1. The cuspidal part of the weak lift. 11.2. The image of the weak lift. 11.3. On generalized endoscopy. 11.4. Base change. 11.5. Automorphic induction
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### Contents
1 Introduction 1 2 On Certain Residual Representations 17 3 Coefficients of GelfandGraev Type of FourierJacobi Type and Descent 41 4 Some double coset decompositions 65 GelfandGraev characters 81 FourierJacobi characters 121 7 The tower property 151
8 Nonvanishing of the descent I 187 9 Nonvanishing of the descent II 235 10 Global genericity of the descent and global integrals 281 11 Langlands weak functorial lift and descent 313 Bibliography 335 Index 339 Copyright | 858 | 2,850 | {"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} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.619603 |
https://mathoverflow.net/questions/348438/when-do-the-spectra-of-overrings-glue-to-a-proper-morphism | 1,603,350,734,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00035.warc.gz | 414,207,771 | 26,870 | # When do the spectra of overrings glue to a proper morphism?
This question is motivated by the construction of blowups.
Let $$A \subset K$$ be a commutative domain and its fraction field, and let $$\{A_i\}$$ be some finite collection of overrings in between.
Let $$X = \mathrm{Spec}A$$ and let $$\tilde{X} = \bigcup U_i$$ be the $$X$$-scheme formed by gluing the overing spectra via inclusions of (generated) subrings.
So $$\mathrm{Spec}(A_i) \cap \mathrm{Spec}(A_j) = \mathrm{Spec}(A_i+A_j)$$, and so on.
If an overring is a localization, then its induced morphism to $$X$$ is an open immersion, but this need not be the case. However, I believe it is the case that the ideal class group of $$A$$ is trivial if and only if each such overring is a localization.
For example, if $$A = k[x,y], A_0 = k[x,y,\frac{x}{y}], A_1 = k[x.y.\frac{y}{x}], A_{01} = k[x^{\pm 1}, y^{\pm 1}]$$, then the the overrings glue to the blowup of the plane at the origin.
Question: When do the overrings glue to a proper morphism $$\tilde{X} \to X$$? What are some good criteria for properness, and how to check it?
Bonus question: When is the morphism finite?
In the blowup example, $$\tilde{X}$$ is proper but not finite, and upon removal of either overring from the collection, the remaining subscheme is not proper.
• @RizaHawkeye Yes, that's true. The reason I mentioned removing some of the $A_i$ from the collection is to give an example of subsets of the $A_i$ that provide different answers. Instead I could add in $A_2 = k[x,y,\frac{x}{y+1}], A_{02} = k[x,y,\frac{1}{x}, \frac{1}{y+1}]$ and obtain a scheme with more affine patches which is not proper. So I'm looking for nice criteria to identify which of $\{A_0, A_1, A_{01}\}, \{A_0, A_{01}\}, \{A_1, A_{01}\}, \{A_0, A_1, A_{01}, A_2, A_{02}\}$ define a proper morphism to $A$. – PrimeRibeyeDeal Jul 20 at 19:49 | 586 | 1,864 | {"found_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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-45 | longest | en | 0.884717 |
https://gusatekugibod.lausannecongress2018.com/introduction-to-the-metric-system-the-most-used-system-of-measurement-in-the-world-used-by-everyone-23097lu.html | 1,603,894,816,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00687.warc.gz | 346,574,177 | 4,455 | # Introduction to the metric system the most used system of measurement in the world used by everyone
To help students explore the metric system, they participate in a series of activities designed to introduce and build knowledge: How does basing a measurement system based the on number 10 make using the system more efficient? Have students view slide 5 of the Metric System Presentation to illustrate the power of 10 and preview metric unit prefixes.
Since the s, the International System of Units has been internationally agreed upon as the standard metric system.
The SI base units are based on physical standards. The definitions of the SI base units have been and continue to be modified and new base units added as advancements in science are made. Each SI base unit except the kilogram is described by stable properties of the universe. Although we often use mass and weight interchangeably, each one has a specific definition and usage.
The mass amount of matter of an object remains the same regardless of where the object is placed. For example, moving a brick to the moon does not cause any matter in it to disappear or be removed.
We call this force of attraction, the force of gravity. The gravitational pull on the object varies depending on where the object is with respect to the Earth or other gravity-producing object.
• Making Measurement Simple: The Metric System
• The Metric System and Measurement - Biology LibreTexts
• SI Units and Prefixes - Freshman Science Textbook
For example, a man who weighs pounds on Earth would weigh only pounds if he were in a stationary position, miles above the Earth's surface. This same man would weigh only pounds on the moon because the moon's gravity is only one-sixth that of Earth. The mass of this man, however, would be the same in each situation because the amount of matter in him is constant.
We measure weight with a scale, which is a spring that compresses when an object is placed on it.
## The Metric System | General Science | Visionlearning
If the gravitational pull is less, the spring compresses less and the scale shows less weight. We measure mass with a balance.
A balance compares the unknown mass to known masses by balancing them on a lever. If we take our balance and known masses to the moon, an object will have the same measured mass that it had on the Earth. The weight, of course, would be different on the moon.
Consistency requires that scientists use mass and not weight in its measurements of the amount of matter. The basic unit of mass in the International System of Units is the kilogram. A gram is a relatively small amount of mass and so larger masses are often expressed in kilograms.
When very tiny amounts of matter are measured, we often use milligrams which are equal to gram. There are numerous larger, smaller, and intermediate mass units that may also be appropriate.
At the end of the century, a kilogram was the mass of a cubic decimeter of water. Ina new international prototype of the kilogram was made of a platinum-iridium alloy.country in the world that does not use the metric system as its predominant system of measurement.
Most Americans think that our involvement with metric measurement is relatively new.
In fact, the United States has been increasing its use of metric rest of the world in the use of the metric system of measurement. The study found that. In this lesson, students are introduced to the metric system and practice using the metric system to measure many different quantities.
Because the metric system is the language of measurement for scientific practice world wide (almost!), learning this system is important for understanding and participating in global science practice. metric units.
STUDY. PLAY. the primary system of measurement used by most countries in the world; the modern metric system of measurement; abbreviated SI. International System of Units. a metric unit of length. meter. a system of measurement based on the number ten.
metric system. 2. METRIC SYSTEM 2. VOLUME The basic unit of volume in the metric system is the liter (1). The most common derived unit is the milliliter (ml) (10~3 or 1/ of a liter).
The metric system is the primary system of measurement used through much of the world and in science.
## Commonly used metric system units, symbols, and prefixes
Each type of measurement has a base unit to which . The system of weights and measures is the collection of units and how they relate to each other. Example of The majority of the world uses the metric system: CENTIMETER The measurement most often used for length was the vara. It was a wood stick that measured between 32 and 35 inches.
Metric system - Wikipedia | 946 | 4,713 | {"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} | 4.40625 | 4 | CC-MAIN-2020-45 | latest | en | 0.928371 |
http://2dengine.com/doc/gs_physics3.html | 1,503,460,989,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.92/warc/CC-MAIN-20170823035753-20170823055753-00360.warc.gz | 1,883,549 | 3,765 | # Physics part 3
In the last two physics tutorials, we looked at how to create bodies, shapes and how to move them around. Generally, we told Box2D what to do. Conversely, this tutorial focuses on getting information back from Box2D using queries and contact listeners.
## Querying the world
Searching through a world is done using the "Query" function. The QueryAABB function accepts a rectangular area in the form of a b2.AABB. Box2D then finds all shapes that "potentially" intersect with the supplied rectangular area. It is important here to emphasize the word "potentially". The query may contain some shapes that do not overlap the supplied AABB. The reason why the QueryAABB function is so inconsistent is because it was designed as a quick filter used during Box2D's broad collision phase.
In the next bit of code, we are going to implement a "QueryPoint" function. This sort of function is useful in games where you want to select shapes or bodies using the mouse.
```local slop = b2.linearSlop/2
local aabb = b2.AABB()
function b2.World:QueryPoint ( pt )
-- query potential shapes
local q = {}
local qcb = b2.QueryCallback()
qcb.ReportFixture = function(qcb, f)
-- ignore shapes if the point is not inside them
if f:TestPoint(pt) then
table.insert(q, f)
end
return true
end
aabb.lowerBound:Set(pt.x - slop, pt.y - slop)
aabb.upperBound:Set(pt.x + slop, pt.y + slop)
self:QueryAABB(qcb, aabb)
return q
end```
Next, we are going to populate the Box2D world with some arbitrary shapes. Notice that we are using some Lua functions (such as "NewBody", "NewBox" and "OnCreate") that were written and explained in the previous physics tutorial.
```-- create a new world
gravity = b2.Vec2 ( 0, 0 )
world = b2.World ( gravity )
-- create a bunch of bodies and shapes
for x = -100, 100, 10 do
for y = -100, 100, 10 do
local body = world:NewBody ( "staticBody", x, y, 0, 2, 2, false, false, false, false )
local shape = body:NewBox ( 2, 2, 0.1, 0.5, 0.5, false )
body:OnCreate ( )
end
end```
Using the "QueryPoint" function, we will allow the user select one body at a time by pressing the mouse. When selecting a body, its sprite changes color to RED. The tricky part here is transforming the mouse cursor position to world coordinates. This is necessary because the mouse cursor is in a different coordinate system than the physics simulation. In other words, the bodies and shapes rendered on screen are usually transformed and often scaled.
```-- currently selected body
selected = nil
mouse.on_press = function ( mouse, button )
-- deselect the previously selected body
if selected then
selected.sprite.color = WHITE
selected = nil
end
-- make a new query at the cursor position
local pt = b2.Vec2 ( mouse.xaxis, mouse.yaxis )
pt.x, pt.y = camera:get_world_point ( pt.x, pt.y )
local q = world:QueryPoint ( pt )
-- color the selected body red
if #q > 0 then
selected = q[1]:GetBody ( )
selected.sprite.color = RED
end
end```
## Collision callbacks
In most games, you need to know when two particular objects have collided. Box2D reports these events to Lua using "contact listeners". b2.ContactListener is simply a class containing some callback functions that are evoked by Box2D. Let's look at some Lua code that creates and activates a new contact listener:
```local listener = b2.ContactListener ( )
listener.BeginContact = function ( listener, ct )
-- called when two fixtures begin to touch
end
listener.EndContact = function ( listener, ct )
-- called when two fixtures cease to touch
end
listener.PreSolve = function ( listener, ct, om )
-- called before a collision is resolved
end
listener.PostSolve = function ( listener, ct, ci )
-- called after a collision is resolved
end
world:SetContactListener ( listener )```
As you can see, b2.ContactListener has four callback functions. "PostSolve" would probably be the most useful callback in making an action game since it's called after the collision has been resolved. Notice that sensor shapes do not raise the "PreSolve" or "PostSolve" callbacks since they do not respond to collisions. I suggest looking at the Box2D documentation for the full details on the b2.Contact object.
There is a catch to using contact listeners. Box2D does not allow you to make changes to the world during a collision callback. This means that you cannot destroy or create a body or shape until the collision callback function has returned. What you usually have to do is mark the particular body or shape for deletion and destroy it later. Next, we will look at some example code which shows how this is done:
```-- set up a contact listener
listener = b2.ContactListener ( )
listener.PostSolve = function ( listener, ct, ci )
local fix1 = ct:GetFixtureA ( )
local fix2 = ct:GetFixtureB ( )
local body1 = fix1:GetBody ( )
local body2 = fix2:GetBody ( )
-- mark bodies for deletion
if (body1.type == 'player' and body2.type == 'enemy') then
elseif (body1.type == 'enemy' and body2.type == 'player') then
end
end
world:SetContactListener ( listener )```
There are several things to note at this point. First off, collision callbacks may be evoked at the moment a new shape is created (assuming that the newly created shape is initially in contact with another shape). This suggests that contact listeners should be set before any shapes have been generated. Secondly, the collision callbacks always work with shapes. b2.Contact contains the two colliding shapes, but you need to get the associated bodies yourself.
Going back to our example, we will now add some bodies to the world. After creating each body, we assign it its 'type' property (in this case a string value, either 'player' or 'enemy'). In this example, the collision callback looks up the 'type' property so it must be assigned before any of the callbacks could be evoked.
```-- create a bunch of bodies and shapes
for x = -100, 100, 10 do
for y = -100, 100, 10 do
local body = world:NewBody ( "staticBody", x, y, 0, 2, 2, false, false, false, false )
body.type = 'enemy'
local shape = body:NewBox ( 2, 2, 0.1, 0.5, 0.5, false )
body:OnCreate ( )
end
end
-- create a user-controlled body
player = world:NewBody ( "dynamicBody", 0, 0, 0, 2, 0, false, false, false, false )
player.type = 'player'
local shape = player:NewCircle ( 1, 0, 0, 0.1, 0.5, 0.5, false )
player:OnCreate ( )```
The last part of the code shouldn't come as a surprise, especially if you have read the last Box2D tutorial. A timer object is set up to continuously destroy the marked bodies, move the player and to update the world simulation. By running the script, you can see how any bodies that come in contact with our little 'player' object get destroyed.
```timer = Timer ( )
timer:start ( 16, true )
timer.on_tick = function ( timer )
-- destroy marked bodies
local body = world:GetBodyList ( )
while body do
local next = body:GetNext ( )
if body.state == 'dead' then
body:OnDestroy ( )
world:DestroyBody ( body )
end
body = next
end
-- check for player input
local position = b2.Vec2 ( )
player:GetPosition ( position )
if keyboard:is_down ( KEY_UP ) == true then
player:ApplyForce ( b2.Vec2 ( 0, 10 ), position, true )
end
if keyboard:is_down ( KEY_DOWN ) == true then
player:ApplyForce ( b2.Vec2 ( 0, -10 ), position, true )
end
if keyboard:is_down ( KEY_LEFT ) == true then
player:ApplyForce ( b2.Vec2 ( -10, 0 ), position, true )
end
if keyboard:is_down ( KEY_RIGHT ) == true then
player:ApplyForce ( b2.Vec2 ( 10, 0 ), position, true )
end
-- update the physics simulation
local seconds = timer:get_delta_ms ( ) / 1000
world:Step ( seconds, 10, 8 )
-- iterate all bodies and update their sprites
local body = world:GetBodyList ( )
while body do
body:OnUpdate ( )
body = body:GetNext ( )
end
end``` | 1,995 | 7,720 | {"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} | 3.125 | 3 | CC-MAIN-2017-34 | latest | en | 0.806893 |
https://socratic.org/questions/how-do-you-find-the-range-of-f-x-5cos-x-pi-3#138169 | 1,726,553,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00232.warc.gz | 495,825,842 | 5,757 | # How do you find the range of f(x)=5cos(x + pi) + 3?
##### 1 Answer
Apr 15, 2015
$\cos \left(x + \pi\right)$ has a range of $\left[- 1 , + 1\right]$
(because cos(anything) has a range of $\left[- 1 , + 1\right]$)
$5 \cos \left(x + \pi\right)$ has a range of $\left[- 5 , + 5\right]$
$f \left(x\right) = 5 \cos \left(x + \pi\right) + 3 h a s a r a n \ge o f$[-2,+8]# | 158 | 370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.588792 |
https://cracku.in/162-in-a-certain-code-detail-is-written-as-bjmufe-how--x-sbi-clerk-2008-1 | 1,725,921,920,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00422.warc.gz | 166,912,998 | 26,202 | Question 162
# In a certain code DETAIL is written as BJMUFE. How is SUBMIT written in that code?
Solution
DETAIL is written as BJMUFE
The pattern followed is that the word is split into two halves = (DET) (AIL) and then the second half is written first and then the first half in reverse order = (AIL) (TED)
Similarly, for SUBMIT :
=> Ans - (B) | 92 | 351 | {"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} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.969149 |
http://forums.devshed.com/python-programming-11/ensure-input-range-143327.html | 1,524,611,592,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947421.74/warc/CC-MAIN-20180424221730-20180425001730-00023.warc.gz | 125,265,309 | 10,767 | ### Thread: How to ensure input number range?
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James
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#### How to ensure input number range?
I am writing my first Python program, which is a simple little lottery thing.
It runs well, but the user can input any numbers he wishes, and I want to limit him to a specific bit range.
I have tried various ways but can't seem to understand how to do this.
The problem is:
I have a seven-digit number. The first six digits can be between 1 and 35, while the seventh digit can only be between 1 and 10.
A gentle shove in the right direction would be much appreciated. Maybe even a reference for me to read? I do have the Reference Manual.
Thanks
James
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Contributing User
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I presume that this is a text-mode program, in which case the best you can do is ask for a number, and if it is not in the right range then print an error message and try again.
e.g.
Code:
```def getNumberInRange(min, max):
while 1:
val = raw_input('Please enter a number between %s and %s' % (min, max) )
try:
val = int(val)
if min <= val <= max:
return val
except ValueError:
pass
print 'Incorrect - please try again'
print getNumberInRange(1, 35)```
If it is a GUI program such as Tkinter or wxWindows then you have a number of other options, such as using a spin control or a drop-down list of possible numbers.
Dave - The Developers' Coach
Last edited by DevCoach; April 28th, 2004 at 06:56 AM.
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James
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Excellent, thank you very much.
Solved my immediate problem and led me into a new area (GUI). So my studies continue.
Very grateful, for sure!
James
4. another common solusion would be to do an 'if number in range' check i.e.
Code:
```imaginary python shell
>>> min = 1
>>> max = 50
>>> num = 20
>>> if num in range(max, min): print True
True
>>>```
I also believe you can use xrange() aswell as range() for this.
Code:
```#!/usr/bin/env python
def inrange(num, min, max):
try:
return int(num) in range(max, min)
except ValueError, message:
print message
if __name__ == '__main__':
num = raw_input('Please enter a number')
min = 1
max = 50
while not inrange(num, min, max): num = raw_input('Please enter a number')```
Note: none of this is tested in particular the last section in the second example may not work however the rest is probably fine . Let me know if it doesnt work for you.
Hope this helps,
Mark. | 719 | 2,679 | {"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} | 2.515625 | 3 | CC-MAIN-2018-17 | latest | en | 0.875512 |
http://perplexus.info/show.php?pid=5067 | 1,544,713,261,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824912.16/warc/CC-MAIN-20181213145807-20181213171307-00026.warc.gz | 225,324,595 | 4,158 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Puzzle of Sphinx (Posted on 2006-10-11)
Here is a little sphinx, small version of the big one found in the Egyptian desert.( Its dimensions are written next to each side )
``` /\
1/ \
/-1--/ \2
1/ \
/------3------\
```
What is the minimum pieces of little sphinx required to make a larger sphinx?
How many such irregular shapes can you think of which follow this logic?
No Solution Yet Submitted by Salil Rating: 3.0000 (1 votes)
Subject Author Date re(2): Bless the Pharoah Dej Mar 2006-10-12 12:04:13 re: Bless the Pharoah brianjn 2006-10-11 21:04:17 Solution (Part I) Dej Mar 2006-10-11 20:09:23 Bless the Pharoah brianjn 2006-10-11 08:12:58
Search: Search body:
Forums (0) | 249 | 814 | {"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} | 2.609375 | 3 | CC-MAIN-2018-51 | latest | en | 0.769208 |
https://www.pokertracker.com/forums/viewtopic.php?f=58&t=99548&view=print | 1,607,216,142,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00165.warc.gz | 805,827,201 | 6,075 | Page 1 of 1
### Net won / C Net won calculations, rebuy effects
Posted: Wed Oct 14, 2020 11:23 pm
Background: I have to manually enter buy-in amount for ACR tourneys but rebuys field is automatically populated. Noticing some issues.
Does net won = (buy-in + fee) * rebuys + prize won, excluding bounties for simplicity?
Or does net won = (buy-in + fee) + prize won?
I think it is the latter of the above based on looking at my data. THis would mean that I need to enter the total buy ins + fees instead of cost for single buy-in and letting the rebuy multiply it which would lead me to ask what the point of the rebuy field is.
Also, why is there net won vs. C net won? Do you have to enter the "\$" symbol for this to show in both fields? Graph of results just uses C net won so I would have to go back and enter \$ where I forgot. Is ROI dependent on C Net won vs. net won?
### Re: Net won / C Net won calculations, rebuy effects
Posted: Thu Oct 15, 2020 6:15 am
carter665 wrote:Background: I have to manually enter buy-in amount for ACR tourneys but rebuys field is automatically populated. Noticing some issues.
Does net won = (buy-in + fee) * rebuys + prize won, excluding bounties for simplicity?
Or does net won = (buy-in + fee) + prize won?
I think it is the latter of the above based on looking at my data.
Net won is:
carter665 wrote:THis would mean that I need to enter the total buy ins + fees instead of cost for single buy-in and letting the rebuy multiply it which would lead me to ask what the point of the rebuy field is.
carter665 wrote:Also, why is there net won vs. C net won? Do you have to enter the "\$" symbol for this to show in both fields?
When editing a tournament you should enter the results in the game currency (the currency used on the poker site for the buyin/prizes). Any tournaments that are played in a currency that differs from your selected 'My Currency' in PokerTracker 4 will automatically be converted:
Currency Guide
carter665 wrote:Is ROI dependent on C Net won vs. net won?
The ROI in PokerTracker 4 uses 'My Currency' however ROI doesn't change based on currency. For example if you play a tournament in €EUR but have 'My Currency' in \$USD:
Prize Won €35
(€15 Net Won / €20 Buy-in + Fee + Rebuys + Add-ons) * 100 = 75% ROI
assuming an exchange rate of 1.5 (could be anything):
Prize Won \$52.5
(\$22.50 Net Won / \$30 Buy-in + Fee + Rebuys + Add-ons) * 100 = 75% ROI
### Re: Net won / C Net won calculations, rebuy effects
Posted: Sat Oct 17, 2020 11:58 pm
Thanks.
The below formula doesn't make sense. Rebuys and Add-ons are just integers for # of times it happens. So if there is a buy-in of \$10 and fee of \$1 and I rebuy once, this would equal 10 + 1 + 1 = 12 instead of (10 +1)*2 = \$22
Is there a way to insert \$ before buy-in and fee for a large amount of tourneys instead of doing it manually for every single one where this is missing?
### Re: Net won / C Net won calculations, rebuy effects
Posted: Sun Oct 18, 2020 4:17 am
In the expression:
carter665 wrote:Is there a way to insert \$ before buy-in and fee for a large amount of tourneys instead of doing it manually for every single one where this is missing?
Do you mean that some of your tournaments were entered as play money because you didn't set the currency?
If you edit all of those tournaments at once and select them all in the upper section you should be able to change the currency selector for them all at once.
### Re: Net won / C Net won calculations, rebuy effects
Posted: Thu Oct 22, 2020 11:31 pm
See attached screen shots.
The re-buy = 1 while buy-in = 8 and fee = 0.8. prize = 0. This is not what was explained in the formula.
Net won = -8.80 so re-buy is not being incorporated. C net won = 0, ROI is blank. If I add a "\$" to buy-in and fee, Cnet won & ROI will be correct but my question was if there is another way to do this besides manually do for hundreds of tourneys?
### Re: Net won / C Net won calculations, rebuy effects
Posted: Fri Oct 23, 2020 6:26 am
carter665 wrote:See attached screen shots.
There are no screenshots attached.
Two entries are required for rebuys - are you using both? One is for the number of rebuys made and the other is the cost of each rebuy (under the R&A / Bounty tab) - see this screenshot:
rebuy.PNG (20.5 KiB) Viewed 177 times
carter665 wrote:but my question was if there is another way to do this besides manually do for hundreds of tourneys?
If that data isn't in the hand history/summary files provided by your poker site then the data will have to be added manually. Initial support was added for WPN tournament summaries in PokerTracker v4.15.23 but these files are still incomplete and you already have a ticket in our system so you will be notified of any changes in this regard.
### Re: Net won / C Net won calculations, rebuy effects
Posted: Sat Oct 24, 2020 12:38 am
Thanks for the help but I just have to say god fucking damnit.
Some of these features are so fucking stupid. Wouldn't be an issue if the summaries imported correctly in the first place. Just an unbelievable monstrosity of annoyances. Unnecessarily intricate with separate field for rebuy cost that also needs to be filled out manually which again, wouldn't be an issue if the tournament summary functionality worked.
PS: Screen shot file transfer isn't completing for some reason but it's OK - I understand now.
highfalutin | 1,391 | 5,420 | {"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} | 2.8125 | 3 | CC-MAIN-2020-50 | latest | en | 0.971593 |
http://www.candlepowerforums.com/vb/showthread.php?178086-Converting-lux-to-lumen | 1,527,315,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867311.83/warc/CC-MAIN-20180526053929-20180526073929-00331.warc.gz | 352,099,072 | 18,250 | # Thread: Converting lux to lumen
1. ## Converting lux to lumen
Being a photographer, I have a professional light meter which also measures light in luxes. How should I multiply or divide the quantity of lux units in order to get lumens.
1 lx = 1 lm/m2
But no datasheet of LEDs shows lumens per square meter, there are just lumens.
I have found this conversion but I also get weird results, no matter what I convert to.
http://www.unitconversion.org/unit_c...umination.html
For example, my room is lit by a 23W Philips bulb (tungsten). The meter shows 12lx from a 1m distance. When I try to convert it, I get:
12 lumen/m2
0.0012 lumen/cm2
etc...
My room is quite big and I don't believe that a 12lm/m2 LED will replace my 23W bulb (equivalent to a 100W filament bulb).
So what's the deal with this conversion? I would like to perform some experiments but I simply don't know what are these lumens without any suffix. All the lumens I get are 'per square something'
2. ## Re: Converting lux to lumen
Speaking of...I noticed the other day at the store that they are starting to put lumens ratings on light bulbs. They had one...I think it was a 100W bulb that said 3000 lumens, or something ridiculous like that.
I figured I'd just send a batch of thse to the Army, so they could save money on those Surefire Hellfighter lights.
Maybe they use a similar conversion?
3. ## Re: Converting lux to lumen
Lumens is a light quantity, and lux is a light quantity over an area (luminous flux.) They are two different measurements and there is no direct conversion.
4. ## Re: Converting lux to lumen
You can find good info here.
-N
-----
6. ## Re: Converting lux to lumen
Originally Posted by Delta
Speaking of...I noticed the other day at the store that they are starting to put lumens ratings on light bulbs. They had one...I think it was a 100W bulb that said 3000 lumens, or something ridiculous like that.
I figured I'd just send a batch of thse to the Army, so they could save money on those Surefire Hellfighter lights.
Maybe they use a similar conversion?
I'd guess that a 100W lightbulb could be up to 3000 lumen. (30 lumens per watt)
Notice how the whole room lights up when you turn on a 100w lamp - an effect that can only be replicated with a bounce test from a very powerful light.
7. ## Re: Converting lux to lumen
For decades they have been putting lumen ratings on light bulb packaging. Takes a pretty powerful incan flashlight light to equal the output of a 100 watt light bulb, and I believe that those are accurate lab figures for light bulb lumens.
Bill
8. ## Re: Converting lux to lumen
Likely that 100W bulb is more like 1,500 lumens.
Semiman
9. ## Re: Converting lux to lumen
Originally Posted by SemiMan
Likely that 100W bulb is more like 1,500 lumens.
Semiman
Yep! Phillips rates theirs at 1600 initial lumens.
10. ## Re: Converting lux to lumen
Go find a package of 100 watt light bulbs. The lumen rating will be on the packiage. It may be 1500 lumens, I'm out of spare bulbs, but that 1500 lumens, or so will sure "light up a room". Cheap too. Not too portable.
Bill
11. ## Re: Converting lux to lumen
There has been lots of talks about LEDbulbs in the market now. In my opinion its not possible to make good LEDbulb yet.If its has to give same kind of light as incandescent.
Some models i have seen ,example:
8w e27 up tp +400lm.Manufacturer is saying that is the same lightoutput as 40w incandescent.Well..it might almost be,BUT..Its about 140deg angle.So you cant really compare that to incandescent, because of that angle.And...that lumen is from the LEDs not from the bulb,so there is difference.
This 8w e27-model has 3pcs of CREE XR-E.Ithink this UP TO.. is meaning if you drive every LED with 700mA. But they are not driving it with 700mA.In the datasheet they tell that LEDs are 3x2w.So 2w is wasted in the other parts in the bulb.If they drive every LED with 700mA, it should be more than 10w..
I think there is lots of different kind of E-base LEDbulbs coming to the markets soon. So, im affraid the marketing its going to be once again, with wrong informations. This wrong kind of marketing is problem in the MR16 now.If this continues,there is a hard way for the products ,in the general lighting
12. ## Re: Converting lux to lumen
Originally Posted by SemiMan
Likely that 100W bulb is more like 1,500 lumens.
Semiman
Unless it is a Mercury Vapor bulb (100W rated at 4300 lumens):
http://www.prolighting.com/h310mevabume.html
but we digress....
-----
14. ## Re: Converting lux to lumen
We have diverted a little bit from the main subject, but I will tell you what I am planning to do.
I needed a simple conversion to lumens in order to test the real output of different LEDs. What people claim is one thing and the truth is another. Having the possibility to measure light would let me do some sort of comparison.
So far I only see lux meters all around but no lumen meters. So how do they measure the output at the factory?
15. ## Re: Converting lux to lumen
They most likely use an Integrated Sphere calibrated for accuracy by a reputable lab. SureFire does this. Arc does this. Not sure about other companies. If you can capture all of the lux of a light in a lightbox such as made by Quickbeam at flashlightreviews.com you can get an approximate value for lumens if you work out a formula based on the lumen output of a known light. People who were involved in the Lightmeter Benchmark Testing thread (see threads of interest in battery forum) were able to receive lumen info on the passaround light from a certified lab using an IS, and those using lightbox or other contraptions to test approximate lumen output were able to use this info. Using that info people could work out a formula to get approximate lumen output info for various lights that they test.
Of course if we all had IS's we would be more accurate, but even IS's have to be calibrated every so often for accuracy.
Bill
16. ## Re: Converting lux to lumen
Originally Posted by MikePL
We have diverted a little bit from the main subject, but I will tell you what I am planning to do.
I needed a simple conversion to lumens in order to test the real output of different LEDs. What people claim is one thing and the truth is another. Having the possibility to measure light would let me do some sort of comparison.
So far I only see lux meters all around but no lumen meters. So how do they measure the output at the factory?
As noted already, there is no accurate way to convert Lux to Lumens because these are two very different dimensions and quantities.
Maybe a photographic analogy....
Take a picture with studio lights in your room. Set your digital camera on ISO 100, F11, with shutter speed 1/125 sec. Now adjust your studio lights so that you get a good exposure, either by adjusting the output of the lights and/or moving the lights closer to the subject. Now go outside in the sunlight with that same subject and take a picture with the same settings...ISO 100, F11, 1/125 sec. and you get a proper exposure.
Both pictures look good because they both have the same Lux (light per area). But I'm sure you'll agree that there's no way you're studio lights has the same Lumens (total output) as the Sun.
So is there a simple conversion....the answer is no.
Is there a way to measure Lumens...the answer is yes...BUT...it'll is very difficult to get accurate measurements and it would take a lot of very precise and expensive equipment to do it...and then it would have to be adapted to the different characteristics of various LEDs
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• | 1,893 | 7,748 | {"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} | 2.8125 | 3 | CC-MAIN-2018-22 | latest | en | 0.942964 |
https://math.stackexchange.com/questions/4527037/composition-and-infinitely-differentiable-in-normed-vector-space | 1,722,930,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00064.warc.gz | 311,719,939 | 35,319 | Composition and Infinitely differentiable in Normed vector space
Say, $$U, V, W$$ are normed vector spaces that has differentiation. We know that if $$f : U \rightarrow V$$, $$g : V \rightarrow W$$ are differentiable, then so is $$g \circ f$$ and $$D (g \circ f) (x) = Dg(f(x)) \circ Df(x)$$
Question is, if $$f, g$$ are infinitely differentiable, is $$g \circ f$$ infinitely differentiable as well? Specifically, how do I go with induction step for $$C^n$$?
• from this formula and induction, you can show that if $f$ and $g$ are $C^r$, then so is $g\circ f$. Commented Sep 8, 2022 at 1:24
• I am lost on the induction step. How do I compute differential of $x \mapsto Dg(f(x)) \circ Df(x)$ ? Commented Sep 8, 2022 at 1:35 | 228 | 726 | {"found_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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.878093 |
https://web2.0calc.com/questions/right-triangle-problem-hurry-please | 1,716,407,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00616.warc.gz | 543,316,375 | 6,309 | +0
# Right Triangle Problem (Hurry please)
+1
7
3
+45
Find the Pythagorean triple that has 97 as the length of the hypotenuse without using a computer or calculator.
Nov 4, 2023
#1
+1199
0
Let (a,b,97) be the Pythagorean triple we are looking for. We know that a2+b2=972=9409. We also know that a and b must be relatively prime, since the greatest common divisor of the legs of a Pythagorean triple is always 1.
One way to find a primitive Pythagorean triple is to use the Pythagorean formula and experiment with different values for a and b. In this case, we can start by trying to find a value for a that is relatively close to the square root of 9409. We know that 8100<9409<10000, so we can try a=90. Substituting into the Pythagorean formula, we get 902+b2=9409, so b2=9409−902=329. Since 329 is a prime number, we know that b=17. Therefore, the Pythagorean triple with 97 as the hypotenuse is (90,17,97).
Nov 5, 2023
#2
+806
+1
972 = 9409 (c)
902 = 8100 (a or b)
172 = 289 (b or a)
a2 + b2 must = c2 but, as you can see from the above, it does not.
The Pythagorean triple with 97 as the hypotenuse is not (90,17,97).
I recommend always checking an answer, if possible, before posting.
.
Nov 5, 2023
#3
+397
+1
Let,
\(\displaystyle a = m^{2}-n^{2},\\\displaystyle b=2mn,\\ \displaystyle c=m^{2}+n^{2},\)
then a, b and c form a Pythagorean triple.
\(\displaystyle a^{2}+b^{2}=(m^{2}-n^{2})^{2}+(2mn)^{2} \\ \displaystyle =m^{4}+n^{4}-2m^{2}n^{2}+4m^{2}n^{2} \\ \displaystyle =m^{4}+n^{4}+2m^{2}n^{2}=(m^{2}+n^{2})^{2} \\ \displaystyle = c^{2}.\)
What we are looking for then are values of m and n such that m^2 + n^2 = 97.
m^2 = 97 - n^2, so try
n =1 , m^2 = 97 - 1 = 96, m not an integer,
n = 2, m^2 = 97 - 4 = 93 mnai,
n = 3, m^2 = 97 - 9 = 88 mnai,
n = 4, m^2 = 97 - 16 = 81, m = 9, got it.
So, a = 9^2 - 4^2 = 81 - 16 = 65,
b = 2*9*4 = 72.
Check, (for silly mistakes), a^2 + b^2 = 65^2 + 72 ^2 = 4225 + 5184 = 9409 = 97^2.
Nov 6, 2023 | 782 | 1,988 | {"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} | 4.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.821288 |
http://www.roman10.net/2011/03/ | 1,519,351,452,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00167.warc.gz | 531,533,534 | 17,257 | ## Real-Time Video Bilayer Segmentation–Theory–Part 2 Producing the Trimap
Side Note: First draft on Mar 31 2011.
2. Producing the Trimap
Below is an example of a trimap,
Figure 1. Trimap based on Iprob
How do we get this trimap?
Part 1 has covered how to compute the Iprob, once we have the Iprob for every pixel of a video frame, we compare the Iprob with a small real value epsilon.
here B stands for background, F stands for foreground and U refers to unknown. In figure 1, the white area are F, black area are B and gray area are U.
For the white and black areas, it’s confident to decide whether they’re foreground or background based on Iprob alone. The gray regions are further processed through MRF to decide it’s foreground or background.
As a matter of fact, the benefit of this trimap is to reduce the MRF processing. With the trimap, MRF processing is only done for pixels in unknown region.
3. GraphCut for Unkown Regions
There are many problems can be described using MRF in the form of
E = Ed+λEs
where the Ed is the data cost and Es is the smoothness cost. In essence, Ed describes the energy of a pixel being classified as certain category, Es describes the dependence of this classification’s dependency on its neighbors. The solution for these problems are usually finding a classification that results in minimum cost for all pixels.
There’re many ways to solve the cost (some refer as energy) minimization problem, one of them is GraphCut.
For this real-time bilayer segmentation, the Ed and Es function can be expressed as,
here Data(Seg) is the Ed for all pixels under unknown region. And,
here Sth(Seg) is the Es for all neighboring pixels under unknown region. dist(p,q) refers Euclidean distance between pixels p and q.
Note that Seg can be either F or B in the two equations above. For smoothness energy, if the two neighboring pixels p and q are classified as the same category (both F or both B), the energy is 0.
P(Cp|F), P(Cp|B) and Iprob are already calculated in part 1, smoonthness energy is also straightforward to calculated for two given pixels.
The method of calculating L(pi|F) and L(pi|B) will be covered in Part 3.
## Real-Time Video Bilayer Segmentation–Theory–Part 1 Bayesian Estimation
### Side Note: First draft on Mar 30 2011.
Video bilayer segmentation refers to the process of dividing the video frames into foreground and background. Here we introduce a video bilayer segmentation process which is close to real-time.
The entire process can be illustrated as the figure below,
Figure 1. Process Overview of the real-time Bilayer Segmentation
The input includes the video and the segementation mask for the first frame. The segmentation of the first frame can be done using background subtraction, interactive graph cut, image snapping, lazy snapping and so on.
The segmentation for the rest of the video frames are done one by one automatically by the process illustrated above.
1. Bayesian Estimation
For each pixel p in a video frame, a probability Iprob (p) of a pixel belongs to foreground can be expressed as,
where Cp is the color vector of pixel p, F and B are foreground and background respectively. The likelihood P(Cp|F) and P(Cp|B) are calculated by accumulating background and foreground color historgrams. The prior P (F) and P (B) are computed from the previous segmentation mask.
1.1 Calculation of likelihood P(Cp|F) and P(Cp|B)
The likelihood basically indicates the probability of a color being foreground (as in P(Cp|F)) or background (as in P(Cp|B)) based on color distribution of all previous segmentation results.
To build a color histogram (here we use 2 dimensional histogram as example), we set a two dimentional grid, each bin in the grid with certain value ranges. For example,
[0…10, 0..10][0..10, 11…20]….[0..10, 251..260]
[11..20, 0..10][11..20, 11..20]…[11..20, 251..260]
[251..260, 0..10][251..260, 11..20]…[251..260, 251..260]
And we count the number of pixels that falls into this 2-dimentional grids. As the video frame pixel normally contains 3 components, therefore, a 3-dimensional can be built for it.
There’re two ways of creating the color histograms for likelihood calculation. The first one is the accumlative histograms. As the foreground and background changes, the accumlative histogram can incorporate these changes into the calculation. As segmentation always contain some errors, the accumlation process can be improved by only updating the histogram for the bins which have zero values. In this case, the error pixels doesn’t accumulate and only have very small values in the histogram with limited influence.
The other method is to use the first segmentation result to build the color histogram and use it for subseqent processing. This is useful if we know the foreground and background color distribution is not going to change much.
Once the color histogram is built, We can normalize them and read the values for the likelihoods P(Cp|F) and P(Cp|B).
1.2 Compute Priors P (F) and P(B)
The priors are computed based on the previous frame, in consideration of temporal correlations. The computation can be expressed as the formula below,
where a(t-1) is the previous segmentation mask, with 255 indicates the foreground and 0 for background. G3x3 and G7x7 are Gaussian filters. Resize are scaling transformation operations. The result Mt is a smoothed mask.
With Mt, the P (F) and P(B) are be calculated by,
With the priors and likehoods, Iprob can be calculated.
Reference:
## Use OpenCV2.1.0 with Matlab 2010a
Side Note: First draft on Mar 27 2011.
OpenCV and Matlab are often “competitors”, as a lot of image/video processing techniques are implemented in both of them. However, there’re times you want to combine the strength/functions of the two. And luckily Matlab provides C/C++ interface.
Below are the step-by-step configurations for setting up Matlab environment for calling OpenCV functions.
2. Set up C/C++ compiler for Matlab.
• Start up matlab 2010a, enter “mex –setup”. The screen capture below shows this step in detail,
3. Edit the mexopts.bat file.
• The file location can be found at the second last line of the screen capture.
• Add “set OPENCVDIR=C:OpenCV2.1” at the beginning of the file just below the line “set VCINSTALLDIR=%VSINSTALLDIR%VC”.
• Add “%OPENCVDIR%includeopencv;” at the begnning of the the “set INCLUDE=” line
• Add “%OPENCVDIR%lib;%OPENCVDIR%bin;” at the beginning of the “set LIB=” line
• Add “cv210.lib highgui210.lib cvaux210.lib cxcore210.lib” at the line “set LINKFLAGS=” at the position before “/nologo”.
4. Edit PATH environment variable
Append “C:OpenCV2.1lib” at the end of the PATH.
5. Restart Matlab and you can start using OpenCV from Matlab by using the Matlab C++ interface.
## Image Histogram Equalization
Side Note: First draft on Mar 26 2011.
1. Concepts
Image Histogram refers to a graphical representation of tonal distribution of a digital image. It plots the number of pixels at each tonal level. (according to wikipedia)
Image histogram equalization is an image processing method for image contrast adjustment. The nice property of image histogram equalization is it makes use of the input image property. It can be very effective for image that has low contrast. Below is an example,
2. Matlab Implementation
Matlab has histogram equalization function built-in, histeq. But one can easily implement it himself/herself.
A simple histogram equalization makes use of the Cumulative Distribution Function (CDF),
where Sk is the transformed tonal level for an input tonal level rk. k = 0, 1, …2^numOfBitsPerChannelOfAPixel-1 is the tonal level range. nj is the total number of pixels with jth tonal level. The equation says the total number of pixels have gray level no more than kth level divide by the total number of pixels will be the CDF of input tonal level k.
The CDF is able to map the input tonal level for every pixel to a range of [0, 1], once we scale this range to [0, 255], we can produce the output tonal level for every pixel.
The matlab implementation for Gray image and explanation is as below,
```function[O]=feipeng_histeq(fpath)
%1. read the image file into a matrix
%2. get the total number of pixel
N=numel(I);
%3. get a one-dimensional matrix indicating number of
%pixels at each
%grey-level
A=zeros(1,256);
for i=1:1:256
A(1,i)=numel(I(I==i-1));
end
figure, imhist(I); %for test
%4. get the cumulative number of pixels at each grey-level
%[note]can be improved by using the previous calculated
% B for calculation of
%current B
B=zeros(1,256);
for i=1:1:256
for j=1:1:i
B(1,i)=B(1,i)+A(1,j);
end
end
%5. get the normalized cumulative distribution at each
% grey-level
for i=1:1:256
B(1,i)=B(1,i)/N;
end
figure, plot(0:255, B);
%6. for each pixel, compute it's output intensity
[nrows,ncols]=size(I);
O=zeros(nrows,ncols, 'uint8');
for i=1:1:nrows
for j=1:1:ncols
O(i,j)=B(1,I(i,j)+1)*255;
end
end
AO = zeros(1, 256);
for i=1:1:256
AO(1, i) = numel(O(O==i-1));
end
figure, imhist(O);
figure,imshow(O);
%for comprison with built-in histogram equalization
O1 = histeq(I);
figure, imhist(O1);
figure, imshow(O1);
end```
The plots for the histogram are shown below,
The 3 plots are input image, the output image histogram plot for the implementation above and the matlab built-in implementation histogram.
3. Photoshop
Photoshop (not sure starts with which version, but CS5 definitely) has histogram equalization built-in at Image=>Adjustments=>Curves.
## Pre-increment and Post-increment
Side Note: First draft on Mar 26 2011. Back in undergraduate, I was in a programming competition team. We were writing small programs that have to be extremely efficient as there’re time constraint for running these programs. One night, a teammate of mine told me that “you know, pre-increment is more efficient than post-increment”.
Well. The different of pre-increment and post-increment is not difficult to tell. People has basic programming background usually know pre-increment returns the value after increment (think it as increment first, then return), while post-increment returns the value before increment (think it as return the value first, then increment). For example (in C programming language),
```#include <stdio.h>
int main(int argc, char **argv)
{
int x = 5;
printf("x=%dn", ++x); //pre-increment
printf("x=%dn", x++); //post-increment
printf("x=%dn", x);
return 0;
}```
The program will give you the following result,
```x = 6
x = 6
x = 7```
That’s the difference most people know about. If we don’t care about the value returned by the increments, instead using it for increasing the value only, is any difference between them?
The answer is YES. And this goes down to the implementation of the two operations. Here we just use C syntax to illustrate the difference, the actual implementation for different languages may vary, but you should get the idea that pre-increment is more efficient.
```int pre_increment(int* _input) {
*_input = *_input + 1;
return *_input;
}
int post_increment(int* _input) {
int temp = *_input;
*_input = *_input + 1;
return temp;
}```
As you can tell from the code, post_increment needs to introduce a temporary variable to hold the value before the increment operation. That’s why it is less efficient.
So next time, you are writing a loop increment operation, you know pre-increment is a better choice.
```for (int i = 0; i < N; ++i) {
//do some processing
}```
## MPEG4–Overview of Video Decoding
Side note: First draft on Mar 23 2011.
MPEG4 Part2 (Visual) standard defines the bitstream syntax for MPGE4 Part2 compatible video, but let the manufactuers to customize and figure out their encoding process and implementation. On the other hand, the standard defines the decoding process in detail. This article gives an overview of the decoding process.
The entire process can be illustrated as the diagram below,
The decoding consists of 3 main procedures, shape decoding, texture decoding and motion decoding. For MPEG-4 SP video, arbitrary shape is not supported and therefore the shape decoding is not applicable.
The decoding starts with Entropy decoding, which is not shown in the above diagram. The decoded bits will go through different decoding procedures according to bitstream syntax.
The texture decoding will decode the run length symbols, then the inverse scan is carried out to recover the quantized DCT coefficients. Then inverse quantization and IDCT is done to recover the pixel values for DCT blocks. The texture decoding operates on the spatial domain.
Motion decoding is then carried out to get the motion vectors,. These motion vectors, with the decoded texture results and previous reconstructed VOP as reference, motion compensation is carried out to reconstruct the VOP. The motion decoding operates on the time domain.
## The Evil of Trial-and-Error
Side note: First draft on Mar 23 2011. If something good becomes cheap, it may not be good at all.
Well, maybe the title is a bit overwhelming. Trial is a good thing in general. You have the chances to test out your ideas, collect feedbacks, and correct the errors. In traditional wisdom, it’s always good to try it out, especially when it takes a lot of courage.
But things turn upside down when trail is cheap. It costs almost nothing for you to try. Ya, that’s often true in programming. Actually, that’s an attractive thing about it. You have an idea, you always implement it and try it out. Computers give you feedback instantly in most cases. Isn’t that great?
Ya, I love it too. But it turns out we love trial-and-error so much that we become additive to it and simply give up thinking. Well, I believe many programmers have experience debugging a program without much thinking. Just modify something and try it out, see if it works. It takes no time anyway. Error? Modify something else, try it out. Thinking is too troublesome and sometimes we’re just lazy (busy typing is not really working hard by the way). In the end, hours have gone and we’re still stuck.
Actually, those bugs may just require a little bit of thinking, sometimes a little bit of reading. But we are too busy with trial-and-error. Ya, we work extra hours to do it.
That’s the evil of trial-and-error.
## MPEG4–DCT and IDCT
Side note: First Draft on Mar 22 2011.
This article covers discrete cosine transform and inverse discrete cosine transform used in MPEG4 Simple Profile.
MPEG-4 part 2(also called MPEG4 Visual) defines 3 different types of inverse DCT operations, standard 8×8 IDCT, SA-IDCT (Shape Adaptive DCT), and ΔDC-SA-DCT (SA-DCT with DC separation and ΔDC Correction). As MPEG-4 Simple Profile doesn’t support arbitary shape encoding, only standard DCT is applicable for it.
1. DCT
According to MPEG4 standard, the NxN 2-Dimentional DCT is defined as below,
where x, y = 0, 1, …N-1, they’re coordinates in the sample domain (spatial domain).
u, v = 0,1,…N-1, they’re the coordinates in the transform domain, and
The above equations can be expressed in matrix equations. For DCT,
where f is the matrix of samples, A is the transform matrix, and F is the transformed DCT coefficients.
The values of A can be derived from the previous equation,
where
The first matrix multiplication Af can be seen as 1-D DCT of each column of f, and the second matrix multiplication (Af)At can be seen as a 1-D DCT on each row of (Af). Note that the order of the two multiplications can be exchanged.
2. IDCT
The inverse DCT is defined as,
Supposed a pixel is represented in n bits, then the input to the DCT and output from the IDCT are represented with (n+1) bits. The DCT coefficients are represented in (n+4) bits. The range for the coefficients is [-2^(n+3), +2^(n+3)-1].
Again, the equation can be expressed in matrix form,
The meanings and values follow DCT operation above.
## A Simple Note on Engery Minimization for Markov Random Field
Side note: First Draft on Mar 22 2011. I don’t like Math, probably because I never find it is useful until my college ends. Then I realized it’s in fact very useful in lots of area, well, at least in multimedia area. To get rid of the afraid I had about math, everything math related is put under Machinery. It’s a cool name.
Markov Random Field is a new branch of probability theory that seems to be important in theory and applications of probability. It’s a branch, so definitely I am not going to/cannot cover it here. Please refer to Markov Random Fields and Their Applications for detailed introduction.
In image and video processing, a lot of problems can be expressed as a energy minization problem with the equation below,
E = Ed+λEs
where Ed is the data cost for every pixel and Es is the inter-pixel cost also called smoothness term. The solution to these problems can often be transformed to minimization of the energy E.
There is very good implementation of energy minization library online at Middlebury Computer Vision Lab. The implementation provides a simple to use API to minimize the energy function with various algorithms, including Iterated Conditional Modes (ICM), Graph Cuts (Swap-move and Expansion-move), Max-Product Loopy Belief Propagation (LBP) and Tree-Reweighted Message Passing (TRM).
With this library, all you have to do is to define the data cost function, smoothness function, pass in the graph elements, and call “optimze()” method. Details can be found at the README.txt of the library.
According to the paper from the lab, the expansion-move graph cuts seem to be a good algorithm to use in many cases, but other algorithms may also perform better under some cases. | 4,242 | 17,653 | {"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} | 2.96875 | 3 | CC-MAIN-2018-09 | latest | en | 0.907234 |
https://www.jiskha.com/display.cgi?id=1190339352 | 1,503,135,577,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00245.warc.gz | 929,061,971 | 3,853 | # Calculus...
posted by .
How can I use the "angle addition formula" and the formula "limit of x approaching 0 of sin(x)" to show that the limit of x approaching 0 sin(a+x) = sin(a) for all a?
• Calculus... -
sin (a+x) = sin a cos x + sin x cos a
As x-> 0, cos x -> 1 and sin a -> 0
Therefore sin (a+x) approaches sin a
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More Similar Questions | 740 | 2,215 | {"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} | 4.0625 | 4 | CC-MAIN-2017-34 | latest | en | 0.673851 |
http://math.stackexchange.com/questions/98804/find-the-least-natural-number-for-which-the-statement-is-true?answertab=active | 1,462,266,141,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121090.75/warc/CC-MAIN-20160428161521-00092-ip-10-239-7-51.ec2.internal.warc.gz | 186,899,148 | 19,874 | # Find the least natural number, for which the statement is true
could anyone help me with a small problem?
I need to find the least natural number $n$, $n>1$ for which the statement is true.
Statement: For any $n$ natural numbers, we can find two, $a$ and $b$, such that $a^2-b^2$ can be divided by 595
-
Do you have any ideas? Can you see that you need to know the number of squares modulo 595 (that is, the number of squares in $\mathbb{Z}/595\mathbb{Z}$)? What would the answer to this question be if instead of 595 you had an odd prime number $p$? – Jonas Kibelbek Jan 13 '12 at 17:01
This is mainly a Pigeonhole Principle problem with a bit of number theory thrown in.
Note that $595=5\times 7\times 17$. Let's consider first the problem if we replace $595$ with $5$:
In order for $a^2-b^2$ to be divisible by $5$, you need $a^2$ and $b^2$ to have the same remainders modulo $5$. There are three possible remainders a square can have when divided by $5$: $0$, $1$, and $4$. That means that if you have at least four numbers in your set, you are guaranteed that two of them will have squares whose remainders are the same.
Similarly, for $7$, you have 4 possible remainders for squares: $0$, $1$, $2$, and $4$. So you would need at least $5$ numbers to ensure the difference of the squares of two of them is divisible by $7$.
What about divisible by $35$? You need both their remainders modulo $5$ and modulo $7$ to match. There are 3 possible remainders modulo $5$, and $4$ possible remainders modulo $7$, so there are 12 possible combinations; so you need at least $13$ numbers to ensure you can get $a^2-b^2$ divisible by $35$.
With $17$, there are $9$ possible remainders for squares.
So: a number's square can have any of $3$ remainders modulo $5$, any of $4$ remainders modulo $7$, and any of $9$ remainders modulo $17$. That means that it can have any of $3\times 4\times 9 = 108$ remainders modulo $595$.
So $109$ numbers will definitely be enough: if you have $109$ natural numbers, then there must be two of them, $a$ and $b$, with $a\neq b$ such that $a^2$ and $b^2$ have the same remainders modulo $5$, $7$, and $17$ (and hence modulo $595$).
Now the question is: can you get away with fewer than $109$?
No. For each possible combination of remainders modulo $5$, $7$, and $17$, the Chinese Remainder Theorem ensures the existence of an integer $m$ that has exactly those remainders; so you can find a set with exactly $108$ natural numbers, and where the remainders when dividing their square modulo $595$ are all pairwise distinct.
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Wow, so simple, firstly I thought I'd never understand a solution. Thanks a lot! – user1131662 Jan 13 '12 at 17:10
@user1131662 Since many students often have difficulty eliciting the essential points in arguments of this type, and since such points are not made explicit above, I have added an answer which explicitly addresses such. – Bill Dubuque Jan 13 '12 at 20:21
HINT $\$ The problem reduces to counting the number of squares in $\rm\:\mathbb Z_{595}\ :=\ \mathbb Z\:\ (mod\ 595)\:.\:$
By CRT (Chinese Remainder) we have a ring isomorphism $\rm\:h\!:\ \mathbb Z_{595} \cong \mathbb Z_5 \times \mathbb Z_7 \times \mathbb Z_{17}$
This isomorphism $\rm\:h\:$ restricts to a bijection on squares, namely $\rm\ \mathbb Z_{595}^2 \cong \mathbb Z_5^2 \times \mathbb Z_7^2 \times \mathbb Z_{17}^2$
since, denoting $\rm\: h(n) = (a,b,c),\:$ we have $\rm\ h(n^2) = (h(n))^2 = (a,b,c)^2 = (a^2,b^2,c^2)\:.$
So an element of $\rm\:\mathbb Z_{595}\:$ is a square iff this is true of its images in each component factor $\rm\:\mathbb Z_p\:.\:$
Thus the problem reduces to counting the squares in each component $\rm\:\mathbb Z_p\:,\:$ which is easy.
NOTE The same product decomposition works for counting the size of the image of any polynomial map $\rm\:f(x)\in \mathbb Z[x]\:$ since such polynomial maps commute with ring homomorphisms $\rm\ h(f(n)) = f(h(n)) =\: f(a,b,c) = (f(a),f(b),f(c))\:.\$ Hence an element of $\rm\:\mathbb Z_{595}\:$ has form $\rm\:f(n)\:$ iff it has that form in each factor in the product decomposition. The decomposition reduction works precisely because ring homomorphisms preserve the form of polynomials, e.g. squares.
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The condition that 595 divides $a^2-b^2$ can be translated to $a^2\equiv b^2$ mod 595. If $m$ is the number of squares in $\mathbb{Z}/595\mathbb{Z}$ then if we take m+1 natural numbers by the pigeon hole principle two of them must have the same square mod 595. We can also give a set of integers of size $m$ such that the squares fall into all the different quadratic residue classes mod 595, so m+1 will be minimal.
With the chinese remainder theorem you can calculate the number of squares. Unfortunately I have to go, if necessary I'll expand on my answer tomorrow!
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I'm sorry, but I'm too stupid to understand that one. But as I hope, the solution by "Arturo Magidin" is the right one because it's wway simplier. – user1131662 Jan 13 '12 at 17:16
@user1131662: Well, I explained why it suffices to count the number of squares mod 595 and mentioned that you can do it with the chinese remainder theorem. Arturo explicitly told you how to do it and Bill's answer explains why this counting process works. Maybe my answer would have been better suited as a comment (I wanted to be more detailed and unexpectedly had to go) – Michalis Jan 14 '12 at 12:32
Note that all answers suggest the same solution to the problem though. – Michalis Jan 14 '12 at 12:33 | 1,563 | 5,484 | {"found_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} | 4.46875 | 4 | CC-MAIN-2016-18 | latest | en | 0.916378 |
https://www.analyticsvidhya.com/blog/2021/05/20-questions-to-test-your-skills-on-logistic-regression/?utm_source=reading_list&utm_medium=https://www.analyticsvidhya.com/blog/2022/06/hypothesis-testing-in-inferential-statistics/ | 1,701,245,754,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00749.warc.gz | 743,994,416 | 53,148 | CHIRAG GOYAL — Updated On June 24th, 2022
This article was published as a part of the Data Science Blogathon
## Introduction
Logistic Regression, a statistical model is a very popular and easy-to-understand algorithm that is mainly used to find out the probability of an outcome.
Therefore it becomes necessary for every aspiring Data Scientist and Machine Learning Engineer to have a good knowledge of Logistic Regression.
In this article, we will discuss the most important questions on Logistic Regression which is helpful to get you a clear understanding of the techniques, and also for Data Science Interviews, which covers its very fundamental level to complex concepts.
## 1. What do you mean by the Logistic Regression?
It’s a classification algorithm that is used where the target variable is of categorical nature. The main objective behind Logistic Regression is to determine the relationship between features and the probability of a particular outcome.
For Example, when we need to predict whether a student passes or fails in an exam given the number of hours spent studying as a feature, the target variable comprises two values i.e. pass and fail.
Therefore, we can solve classification problem statements which is a supervised machine learning technique using Logistic Regression.
## 2. What are the different types of Logistic Regression?
Three different types of Logistic Regression are as follows:
1. Binary Logistic Regression: In this, the target variable has only two 2 possible outcomes.
For Example, 0 and 1, or pass and fail or true and false.
2. Multinomial Logistic Regression: In this, the target variable can have three or more possible values without any order.
For Example, Predicting preference of food i.e. Veg, Non-Veg, Vegan.
3. Ordinal Logistic Regression: In this, the target variable can have three or more values with ordering.
For Example, Movie rating from 1 to 5.
## 3. Explain the intuition behind Logistic Regression in detail.
Given:
By using the training dataset, we can find the dependent(x) and independent variables(y), so if we can determine the parameters w (Normal) and b (y-intercept), then we can easily find a decision boundary that can almost separate both the classes in a linear fashion.
Objective:
In order to train a Logistic Regression model, we just need w and b to find a line(in 2D), plane(3D), or hyperplane(in more than 3-D dimension) that can separate both the classes point as perfect as possible so that when it encounters with any new unseen data point, it can easily classify, from which class the unseen data point belongs to.
For Example, Let us consider we have only two features as x1 and x2.
Let’s take any of the +ve class points (figure below) and find the shortest distance from that point to the plane. Here, the shortest distance is computed using:
di = wT*xi / ||w||
If weight vector is a unit vector i.e, ||w||=1. Then,
di = wT*xi
Since w and xi are on the same side of the decision boundary therefore distance will be +ve. Now for a negative point, we have to compute dj = wT*xj. For point xj, distance will be -ve since this point is the opposite side of w.
Thus we can conclude, points that are in the same direction of w are considered as +ve points and the points which are in the opposite direction of w are considered as -ve points.
Now, we can easily classify the unseen data points as -ve and +ve points. If the value of wT*xi>0, then y =+1 and if value of wT*xi < 0 then y = -1.
• If yi = +1 and wT*xi > 0, then the classifier classifies it as+ve points. This implies if yi*wT*xi > 0, then it is a correctly classified point because multiplying two +ve numbers will always be greater than 0.
• If yi = -1 and wT*xi < 0, then the classifier classifies it as -ve point. This implies if yi * wT*xi > 0 then it is a correctly classified point because multiplying two -ve numbers will always be greater than zero. So, for both +ve and -ve points the value of yi* wT*xi is greater than 0. Therefore, the model classifies the points xi correctly.
• If yi = +1 and wT*xi < 0, i.e, yi is +ve point but the classifier says that it is -ve then we will get -ve value. This means that point is classified as -ve but the actual class label is +ve, then it is a miss-classified point.
• If yi = -1 and wT*xi > 0, this means actual class label is -ve but classified as +ve, then it is miss-classified point( yi*wT*xi < 0).
Now, by observing all the cases above now our objective is that our classifier minimizes the miss-classification error, i.e, we want the values of yi*wT*xi to be greater than 0.
In our problem, xi and yi are fixed because these are coming from the dataset.
As we change the values of the parameters w, and b the sum will change and we want to find that w and b that maximize the sum given below. To calculate the parameters w and b, we can use the Gradient Descent optimizer. Therefore, the optimization function for logistic regression is:
## 4. What are the odds?
Odds are defined as the ratio of the probability of an event occurring to the probability of the event not occurring.
For Example, let’s assume that the probability of winning a game is 0.02. Then, the probability of not winning is 1- 0.02 = 0.98.
• The odds of winning the game= (Probability of winning)/(probability of not winning)
• The odds of winning the game= 0.02/0.98
• The odds of winning the game are 1 to 49, and the odds of not winning the game are 49 to 1.
## 5. What factors can attribute to the popularity of Logistic Regression?
Logistic Regression is a popular algorithm as it converts the values of the log of odds which can range from -inf to +inf to a range between 0 and 1.
Since logistic functions output the probability of occurrence of an event, they can be applied to many real-life scenarios therefore these models are very popular.
## 6. Is the decision boundary Linear or Non-linear in the case of a Logistic Regression model?
The decision boundary is a line or a plane that separates the target variables into different classes that can be either linear or nonlinear. In the case of a Logistic Regression model, the decision boundary is a straight line.
Logistic Regression model formula = α+1X1+2X2+….+kXk. This clearly represents a straight line.
It is suitable in cases where a straight line is able to separate the different classes. However, in cases where a straight line does not suffice then nonlinear algorithms are used to achieve better results.
## 7. What is the Impact of Outliers on Logistic Regression?
The estimates of the Logistic Regression are sensitive to unusual observations such as outliers, high leverage, and influential observations. Therefore, to solve the problem of outliers, a sigmoid function is used in Logistic Regression.
## 8. What is the difference between the outputs of the Logistic model and the Logistic function?
The Logistic model outputs the logits, i.e. log-odds; whereas the Logistic function outputs the probabilities.
Logistic model = α+1X1+2X2+….+kXk. Therefore, the output of the Logistic model will be logits.
Logistic function = f(z) = 1/(1+e-(α+1X1+2X2+….+kXk)). Therefore, the output of the Logistic function will be the probabilities.
## 9. How do we handle categorical variables in Logistic Regression?
The inputs given to a Logistic Regression model need to be numeric. The algorithm cannot handle categorical variables directly. So, we need to convert the categorical data into a numerical format that is suitable for the algorithm to process.
Each level of the categorical variable will be assigned a unique numeric value also known as a dummy variable. These dummy variables are handled by the Logistic Regression model in the same manner as any other numeric value.
## 10. Which algorithm is better in the case of outliers present in the dataset i.e., Logistic Regression or SVM?
SVM (Support Vector Machines) handles the outliers in a better manner than the Logistic Regression.
Logistic Regression: Logistic Regression will identify a linear boundary if it exists to accommodate the outliers. To accommodate the outliers, it will shift the linear boundary.
SVM: SVM is insensitive to individual samples. So, to accommodate an outlier there will not be a major shift in the linear boundary. SVM comes with inbuilt complexity controls, which take care of overfitting, which is not true in the case of Logistic Regression.
## 11. What are the assumptions made in Logistic Regression?
Some of the assumptions of Logistic Regression are as follows:
1. It assumes that there is minimal or no multicollinearity among the independent variables i.e, predictors are not correlated.
2. There should be a linear relationship between the logit of the outcome and each predictor variable. The logit function is described as logit(p) = log(p/(1-p)), where p is the probability of the target outcome.
3. Sometimes to predict properly, it usually requires a large sample size.
4. The Logistic Regression which has binary classification i.e, two classes assume that the target variable is binary, and ordered Logistic Regression requires the target variable to be ordered.
For example, Too Little, About Right, Too Much.
5. It assumes there is no dependency between the observations.
## 12. Can we solve the multiclass classification problems using Logistic Regression? If Yes then How?
Yes, in order to deal with multiclass classification using Logistic Regression, the most famous method is known as the one-vs-all approach. In this approach, a number of models are trained, which is equal to the number of classes. These models work in a specific way.
For Example, the first model classifies the datapoint depending on whether it belongs to class 1 or some other class(not class 1); the second model classifies the datapoint into class 2 or some other class(not class 2) and so-on for all other classes.
So, in this manner, each data point can be checked over all the classes.
## 13. How can we express the probability of a Logistic Regression model as conditional probability?
We define probability P(Discrete value of Target variable | X1, X2, X3…., Xk) as the probability of the target variable that takes up a discrete value (either 0 or 1 in the case of binary classification problems) when the values of independent variables are given.
For Example, the probability an employee will attain (target variable) given his attributes such as his age, salary, etc.
## 14. Discuss the space complexity of Logistic Regression.
During training: We need to store four things in memory: x, y, w, and b during training a Logistic Regression model.
• Storing b is just 1 step, i.e, O(1) operation since b is a constant.
• x and y are two matrices of dimension (n x d) and (n x 1) respectively. So, storing these two matrices takes O(nd + n) steps.
• Lastly, w is a vector of size-d. Storing it in memory takes O(d) steps.
Therefore, the space complexity of Logistic Regression while training is O(nd + n +d).
During Runtime or Testing: After training the model what we just need to keep in memory is w. We just need to perform wT*xi to classify the points.
Hence, the space complexity during runtime is in the order of d, i.e, O(d).
## 15. Discuss the Test or Runtime complexity of Logistic Regression.
At the end of the training, we test our model on unseen data and calculate the accuracy of our model. At that time knowing about runtime complexity is very important. After the training of Logistic Regression, we get the parameters w and b.
To classify any new point, we have to just perform the operation wT * xi. If wT*xi>0, the point is +ve, and if wT*xi < 0, the point is negative. As w is a vector of size d, performing the operation wT*xi takes O(d) steps as discussed earlier.
Therefore, the testing complexity of the Logistic Regression is O(d).
Hence, Logistic Regression is very good for low latency applications, i.e, for applications where the dimension of the data is small.
## 16. Why is Logistic Regression termed as Regression and not classification?
The major difference between Regression and classification problem statements is that the target variable in the Regression is numerical (or continuous) whereas in classification it is categorical (or discrete).
Logistic Regression is basically a supervised classification algorithm. However, the Logistic Regression builds a model just like linear regression in order to predict the probability that a given data point belongs to the category numbered as “1”.
For Example, Let’s have a binary classification problem, and ‘x’ be some feature and ‘y’ be the target outcome which can be either 0 or 1.
The probability that the target outcome is 1 given its input can be represented as:
If we predict the probability by using linear Regression, we can describe it as:
where, p(x) = p(y=1|x)
Logistic regression models generate predicted probabilities as any number ranging from neg to pos infinity while the probability of an outcome can only lie between 0< P(x)<1.
However, to solve the problem of outliers, a sigmoid function is used in Logistic Regression. The Linear equation is put in the sigmoid function.
## 17. Discuss the Train complexity of Logistic Regression.
In order to train a Logistic Regression model, we just need w and b to find a line(in 2-D), plane(in 3-D), or hyperplane(in more than 3-D dimension) that can separate both the classes point as perfect as possible so that when it encounters with any new point, it can easily classify, from which class the unseen data point belongs to.
The value of w and b should be such that it maximizes the sum yi*wT*xi > 0.
Now, let’s calculate its time complexity in terms of Big O notation:
• Performing the operation yi*wT*xi takes O(d) steps since w is a vector of size-d.
• Iterating the above step over n data points and finding the maximum sum takes n steps.
Therefore, the overall time complexity of the Logistic Regression during training is n(O(d))=O(nd).
## 18. Why can’t we use Mean Square Error (MSE) as a cost function for Logistic Regression?
In Logistic Regression, we use the sigmoid function to perform a non-linear transformation to obtain the probabilities. If we square this nonlinear transformation, then it will lead to the problem of non-convexity with local minimums and by using gradient descent in such cases, it is not possible to find the global minimum. As a result, MSE is not suitable for Logistic Regression.
So, in the Logistic Regression algorithm, we used Cross-entropy or log loss as a cost function. The property of the cost function for Logistic Regression is that:
• The confident wrong predictions are penalized heavily
• The confident right predictions are rewarded less
By optimizing this cost function, convergence is achieved.
## 19. Why can’t we use Linear Regression in place of Logistic Regression for Binary classification?
Linear Regressions cannot be used in the case of binary classification due to the following reasons:
1. Distribution of error terms: The distribution of data in the case of Linear and Logistic Regression is different. It assumes that error terms are normally distributed. But this assumption does not hold true in the case of binary classification.
2. Model output: In Linear Regression, the output is continuous(or numeric) while in the case of binary classification, an output of a continuous value does not make sense. For binary classification problems, Linear Regression may predict values that can go beyond the range between 0 and 1. In order to get the output in the form of probabilities, we can map these values to two different classes, then its range should be restricted to 0 and 1. As the Logistic Regression model can output probabilities with Logistic or sigmoid function, it is preferred over linear Regression.
3. The variance of Residual errors: Linear Regression assumes that the variance of random errors is constant. This assumption is also not held in the case of Logistic Regression.
## 20. What are the advantages of Logistic Regression?
The advantages of the logistic regression are as follows:
1. Logistic Regression is very easy to understand.
2. It requires less training.
3. It performs well for simple datasets as well as when the data set is linearly separable.
4. It doesn’t make any assumptions about the distributions of classes in feature space.
5. A Logistic Regression model is less likely to be over-fitted but it can overfit in high dimensional datasets. To avoid over-fitting these scenarios, One may consider regularization.
6. They are easier to implement, interpret, and very efficient to train.
## 21. What are the disadvantages of Logistic Regression?
The disadvantages of the logistic regression are as follows:
1. Sometimes a lot of Feature Engineering is required.
2. If the independent features are correlated with each other it may affect the performance of the classifier.
3. It is quite sensitive to noise and overfitting.
4. Logistic Regression should not be used if the number of observations is lesser than the number of features, otherwise, it may lead to overfitting.
5. By using Logistic Regression, non-linear problems can’t be solved because it has a linear decision surface. But in real-world scenarios, the linearly separable data is rarely found.
6. By using Logistic Regression, it is tough to obtain complex relationships. Some algorithms such as neural networks, which are more powerful, and compact can easily outperform Logistic Regression algorithms.
7. In Linear Regression, there is a linear relationship between independent and dependent variables but in Logistic Regression, independent variables are linearly related to the log odds (log(p/(1-p)).
## End Notes
I hope you enjoyed the questions and were able to test your knowledge about Logistic Regression.
If you liked this and want to know more, go visit my other articles on Data Science and Machine Learning by clicking on the Link
Something not mentioned or want to share your thoughts? Feel free to comment below And I’ll get back to you.
## Chirag Goyal
Currently, I am pursuing my Bachelor of Technology (B.Tech) in Computer Science and Engineering from the Indian Institute of Technology Jodhpur(IITJ). I am very enthusiastic about Machine learning, Deep Learning, and Artificial Intelligence.
The media shown in this article are not owned by Analytics Vidhya and is used at the Author’s discretion.
###### CHIRAG GOYAL
I am currently pursuing my Bachelor of Technology (B.Tech) in Computer Science and Engineering from the Indian Institute of Technology Jodhpur(IITJ). I am very enthusiastic about Machine learning, Deep Learning, and Artificial Intelligence. Feel free to connect with me on Linkedin. | 4,070 | 18,807 | {"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} | 3.6875 | 4 | CC-MAIN-2023-50 | latest | en | 0.907095 |
https://chouprojects.com/working-with-elapsed-time/ | 1,685,804,793,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649293.44/warc/CC-MAIN-20230603133129-20230603163129-00493.warc.gz | 199,623,104 | 30,494 | # Working With Elapsed Time In Excel
by Jacky Chou
Updated on
## Key Takeaways:
• Elapsed time is an important concept in Excel that involves calculating the duration between two timestamps or time intervals, which can be represented in different time formats.
• To convert time formats in Excel, users can utilize built-in functions like TEXT, VALUE, and TIME, and ensure that the cells are correctly formatted as date/time values.
• To calculate time differences in Excel, users can choose from different methods like the DATEDIF function, which can calculate the number of days, months, or years between two dates, or the HOUR and MINUTE functions, which can extract the respective time units from a time value.
• Performing time arithmetic in Excel involves adding or subtracting time intervals from one another, and utilizing functions like SUM, SUMIF, and SUMIFS, which can perform time-based calculations based on criteria or conditions, and rounding and formatting time values to optimize their readability and accuracy.
• The effective management of elapsed time in Excel is crucial for many industries and applications, including project management, scheduling, and financial analysis.
Do you struggle with calculating elapsed time in Excel? This blog will show you how to easily and accurately work with elapsed time without having to resort to complex formulas. Give yourself a break and learn the quick and easy way to calculate elapsed time using Excel.
## Converting Time Formats in Excel
Converting time formats in Excel can be easy! To do this, you have to know how to convert from text and decimal formats to time format. Your worksheet might be full of unnecessary data if you don’t work with a time format. In this section, we’ll discuss how to do this. We’ll look at converting time from text format to time format and from decimal format to time format.
Image credits: chouprojects.com by David Jones
### Converting Time from Text Format to Time Format
When working with time in Excel, sometimes you may find yourself needing to convert a text format into a time format. This can be useful for sorting and filtering data or performing calculations.
To convert time from a text format to a time format in Excel, follow these 5 simple steps:
1. Select the cells that contain the text-formatted time values.
2. Right-click on the selected cells and choose ‘Format Cells’.
3. In the Format Cells dialog box, under the ‘Number’ tab, select ‘Time’ from the Category list.
4. Select the appropriate Type option for the way your time values are formatted (i.e. 12-hour clock or 24-hour clock).
5. Click OK to apply the changes and convert your text-formatted time values into proper time values in Excel.
It’s important to note that if your text-formatted time values include additional characters such as spaces or letters, you may need to clean up your data before attempting to convert it using these steps.
Remember that being able to efficiently work with elapsed time in Excel is an essential skill for many business professionals. By mastering this conversion process, you can save valuable time and improve accuracy in your work.
Don’t miss out on developing this important skill – try converting some of your own data today using these easy-to-follow steps!
Time may be money, but converting decimals to hours and minutes in Excel is a priceless skill.
### Converting Time from Decimal Format to Time Format
The process of converting the decimal format to time format in Excel is a crucial aspect of efficient data management. Here’s how you can achieve it seamlessly:
1. Identify the cell(s) that hold the decimal format data.
2. Highlight the selected cell(s) and navigate to the ‘Format Cells’ option.
3. Under the ‘Category’ section, choose ‘Time’, followed by your preferred 24/12-hour time format.
4. Click ‘OK’, and voila! Your decimals are now converted to a recognisable time format.
It’s worth noting that this method only works if your decimal values represent elapsed time (not regular numbers), usually in terms of days, hours, minutes, or seconds.
In addition, using Excel’s built-in date and time functions can increase efficiency when working with elapsed or regular time formats.
Did you know? A study conducted by Microsoft found that 41% of respondents believed lack of proficiency in Excel hindered their career opportunities.
Why waste time trying to calculate time differences in your head when Excel can do it for you? #MathIsHard
## Calculating Time Differences in Excel
Calculate time differences in Excel with no trouble! Use these sub-sections for effective solutions:
1. DATEDIF Function
2. MINUTE and HOUR Functions
These functions are simple and strong. They help you work out time intervals between two different points accurately.
Image credits: chouprojects.com by David Arnold
### Using the DATEDIF Function
Calculating the Time Differences in Excel can be achieved using various tools available at your disposal. Amongst those tools is the DATEDIF function, a remarkable function that calculates date differences between two dates. It has three parameters: start date, end date, and unit. By using Unit as `'d'` for days or `'m'` for months or `'y'` for years, the function automatically calculates the elapsed time between two dates.
Using the DATEDIF Function is a convenient way to calculate time differences. The elapsed time from one date to another is often required in several work environments. When calculating data trends, budgeting cycles and quoting project timelines effectively use of this function assists with accurate results.
The DATEDIF Function has its limitations; it does not differentiate between weekdays or weekends while calculating differences, yet some projects require more attention on working hours as other hours might delay project deliveries. Despite these limitations, its convenience cannot be overlooked since it provides an easy calculator without having to remember complex equations.
Invented in 1985 by Bertrand Fischer as part of Lotus 1-2-3 spreadsheet software, which introduced several improvements aimed at enhancing the functionality of existing spreadsheets while also introducing new ones like DATEDIF Function.
“Time flies when using Excel’s MINUTE and HOUR functions, but at least you’ll have a second to spare for a dark joke.”
### Using the MINUTE and HOUR Functions
When dealing with time differences on Excel, using the MINUTE and HOUR Functions can come in handy. These functions allow users to extract the minutes or hours from a given time value, making it easier to calculate elapsed time.
Here’s a simple 4-step guide to using the functions:
1. Select the cell where you want to display the result.
2. Enter the formula `=MINUTE(time)` or `=HOUR(time)`, depending on what you need.
3. Replace “time” with the cell reference for the starting time value.
4. Press Enter and voila! The extracted minute/hour value will appear in your selected cell.
It’s worth noting that both of these functions work based on a 24-hour clock, meaning that any values above 23:59 will not be calculated accurately.
However, what happens when calculating elapsed time that goes over multiple days? In such cases, it’s better to use more advanced formulas like DATEDIF or even consider external plugins.
Interestingly, while Excel was first released back in 1985, it wasn’t until ten years later (1995) that Microsoft added support for calculating elapsed time in hours and minutes. Before then, users would have had to do all their calculations manually – a cumbersome process indeed!
Overall, using the MINUTE and HOUR Functions is a simple yet effective way of working with elapsed time in Excel. With just a few clicks, you can quickly extract vital information from your data and make informed decisions based on accurate calculations.
Who knew Excel could be the ultimate time traveler? Performing time arithmetic has never been so easy (or potentially dangerous).
## Performing Time Arithmetic in Excel
Want to do math with time in Excel? To make adding and subtracting time intervals effortless, you need to know how to do time rounding and formatting. This ensures accuracy and the output looks just how you want it.
Image credits: chouprojects.com by Harry Woodhock
### Adding and Subtracting Time Intervals
When it comes to manipulating time in Excel, arithmetic operations such as adding and subtracting intervals are crucial for any user trying to summarize or analyze data. Here’s how you can perform time arithmetic in Excel.
1. Start by inputting the start and end times of each interval into separate cells.
2. Subtract the start time from the end time of both intervals to get the elapsed times using the formula `=end-start`.
3. Convert these elapsed times into total minutes by multiplying them with 1440 (the number of minutes per day).
4. Add or subtract the total minutes together or from another given value.
5. Convert back to hours and minutes format by dividing by 60 (the number of minutes per hour) and formatting accordingly.
Manipulating elapsed time in Excel requires a little bit of mathematical understanding. Once a sound grounding has been formed, users will be able to carry out calculations at great speeds without compromising on precision.
Pro tip: Make sure that all cells are formatted correctly when working with time functions, as incorrect formatting might affect formula outputs.
Time may be money, but rounding it in Excel is priceless.
### Time Rounding and Formatting
Precision Time Calculation and Representation on Excel
The accurate representation of time is essential in Excel, and rounding it to the nearest minute or second can be useful for efficient calculation. Formatting the cells with the right format codes is key to displaying a specific time format.
Below is a table showing how to round up to the nearest minute and second, as well as formatting cells for AM/PM displays or military time.
OperationFormulaFormat Code
Round Up to Nearest Minute=CEILING(A2,”0:01″)[h]:mm:ss
Round Up to Nearest Second=CEILING(A2,”0:00:01)mm:ss.0
AM/PM Display=TEXT(A2,”hh:mm:ss AM/PM”)hh:mm:ss;@”AM/PM”
Military Time Display=TEXT(A2,”hh:mm:ss”)hh:mm:ss
An additional way of formatting would be using the TEXT Function that supplies an optional formatting string.
Excel has several ways of working with time, including performing operations such as addition and subtraction of different time frames. Using specific codes when formatting numeric values can help ensure that users get precise presentations.
Don’t miss out on optimizing your Excel workflow by learning about precision time calculation today!
## Five Facts About Working with Elapsed Time in Excel:
• ✅ Excel stores time as a decimal value representing the fraction of a 24-hour day. (Source: Microsoft Support)
• ✅ To calculate elapsed time in Excel, subtract the end time from the start time and format the result as a time value. (Source: Excel Easy)
• ✅ The SUM function can be used to add up elapsed time values in Excel. (Source: Ablebits)
• ✅ Excel provides several built-in formats for displaying elapsed time, such as [h]:mm:ss or [m]:ss. (Source: Spreadsheeto)
• ✅ In Excel, the NOW() function can be used to calculate the current date and time, which can be useful for tracking elapsed time from a specific point. (Source: Excel Campus)
## FAQs about Working With Elapsed Time In Excel
### What is “Elapsed Time” in Excel?
“Elapsed time” refers to the duration between two specific points in time. In Excel, elapsed time can be calculated using built-in functions or by subtracting one time value from another. This is especially useful for tracking the duration of a process or activity, or for calculating how much time has passed between two events.”
### How do I calculate elapsed time in Excel?
To calculate elapsed time in Excel, use the formula: End Time – Start Time = Elapsed Time. This formula subtracts the start time from the end time, resulting in the elapsed time between the two.
### Can I format elapsed time in Excel?
Yes, you can format elapsed time in Excel to appear in a number of different ways. For example, you can display elapsed time in hours, minutes, and seconds, or in decimal format. To format elapsed time, right-click on the cell containing the formula, select “Format Cells,” and choose the desired time format from the list.
### What are some other useful functions for working with elapsed time in Excel?
Aside from basic subtraction, there are several functions in Excel that can help you work with elapsed time. These include:
• NETWORKDAYS – calculates the number of workdays between two dates, excluding weekends and holidays
• ROUNDDOWN – rounds a time value down to a specific interval (e.g. to the nearest 15 minutes)
• TODAY – returns the current date and time in Excel format
### What if I need to calculate elapsed time across multiple days?
If your elapsed time calculation spans multiple days, you may need to use a more complex formula that takes into account both the date and time values. One way to do this is to convert each date/time value to a number using the VALUE function, subtract the start value from the end value, and then convert the result back to a date/time format using the TEXT function. For example:
=(TEXT(VALUE(End Date & ” ” & End Time),”mm/dd/yyyy hh:mm:ss AM/PM”)-TEXT(VALUE(Start Date & ” ” & Start Time),”mm/dd/yyyy hh:mm:ss AM/PM”))
### How can I use elapsed time data for analysis and reporting?
Elapsed time data can provide valuable insights into the efficiency and effectiveness of business processes. To use elapsed time data for analysis and reporting, you can use Excel’s built-in chart and graph features to create visualizations that highlight trends or outliers in the data. You can also use pivot tables to group and summarize data by specific criteria, such as department or project.
Auther name
Jacky Chou is an electrical engineer turned marketer. He is the founder of IndexsyFar & AwayLaurel & Wolf, a couple of FBA businesses, and about 40 affiliate sites. He is a proud native of Vancouver, BC, who has been featured on Entrepreneur.comForbesOberlo, and GoDaddy. | 2,927 | 14,249 | {"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} | 3.109375 | 3 | CC-MAIN-2023-23 | longest | en | 0.887465 |
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# CS 354 Object Viewing and Representation
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CS 354 Computer Graphics
CS 354 Computer Graphics
University of Texas, Austin
February 9, 2012
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## CS 354 Object Viewing and RepresentationPresentation Transcript
• CS 354 Object Viewing and Representation Mark Kilgard University of Texas February 9, 2012
• Today’s material
• In-class quiz
• Lecture topic: viewing + object representation
• Looking at an object
• Representing an object
• Assignment
• Chapter 4, page 238-241; Chapter 6, 331-341 (Tuesday)
• Chapter 4, pages 241-249
• You should be working on Project #1
• I was sick Tuesday
• Thanks to Dr. Fussell for lecturing in my absence
• Final exam is during the exam period
• (Instead of last day of class)
• Making last day of class a review session instead
• Date & time: Thursday, May 10, 2:00-5:00 pm
• Location: TBD
• Quiz feedback started
• Look on piazza
• My Office Hours
• Tuesday, before class
• Painter (PAI) 5.35
• 8:45 a.m. to 9:15
• Not last Tuesday
• Thursday, after class
• ACE 6.302
• 11:00 a.m. to 12:00
• Last time, this time
• Last two lectures, we discussed
• Consequences of homogeneous coordinates
• Thursday:
• Projection transforms: ortho and frustum
• Tuesday:
• Dr. Fussell discussed rotations with you
• This lecture
• How do we look at objects
• How do we represent objects for interactive rendering
• Daily Quiz
• Multiple choice: Renaissance artists incorporated perspective into their artwork. Which OpenGL command best corresponds to this type of transform: a) glFrustum b) glOrtho c) glTranslatef d) glScalef e) glRotatef
• Answer “Clipped” or “Unclipped”: Would the following position in homogenous clip space be clipped or not? (40, -10, 20, 100)
• On a sheet of paper
• Write your EID, name, and date
• Write #1, #2, #3, followed by its answer
• Conceptual Vertex Transformation glVertex* API commands Modelview matrix User-defined clip planes View-frustum clip planes to primitive rasterization object-space coordinates (x o ,y o ,z o ,w o ) eye-space coordinates (x e ,y e ,z e ,w e ) clipped eye-space coordinates clipped clip-space coordinates Perspective division Projection matrix Viewport + Depth Range transformation (x c ,y c ,z c ,w c ) window-space coordinates (x w ,y w ,z w ,1/w c ) normalized device coordinates (NDC) (x n ,y n ,z n ,1/w c ) clip-space coordinates (x c ,y c ,z c ,w c ) (x e ,y e ,z e ,w e ) (x e ,y e ,z e ,w e )
• Clip Space Clip Cube (x min /w,y min /w,z min /w) Pre-perspective divide puts the region surviving clipping within -w ≤ x ≤ w, -w ≤ y ≤ w, -w ≤ z ≤ w (x max /w,y min /w,z min /w) (x max /w,y min /w,z max /w) (x min /w,y min /w,z max /w) (x max /w,y max /w,z max /w) (x max /w,y max /w,z min /w) (x min /w,y max /w,z min /w) (x min /w,y max /w,z max /w) Constraints x min = -w x max = w y min = -w y max = w z min = -w z max = w w>0
• Vertex Transformation
• Object-space vertex position transformed by a general linear projective transformation
• Expressed as a 4x4 matrix
• Two Transforms in Sequence
• OpenGL thinks of the projective transform as really two 4x4 matrix transforms
FIRST object-space to eye-space SECOND eye-space to clip-space 16 Multiply-Add operations Another 16 Multiply-Add operations
• Modelview-Projection Transform
• Matrixes can associate (combine)
• Combination of the modelview and projection matrix = modelview-projection matrix
• or often simply the “MVP” matrix
concatenation is 64 Multiply-Add operations, done by OpenGL driver Matrix multiplication is associative (but not commutative) A(BC) = (AB)C, but ABC≠CBA
• Specifying the Projection and Modelview Transforms
• Specified in two parts
• First the projection
• glMatrixMode ( GL_PROJECTION );
• glFrustum (-4, +4, // left & right -3, +3, // top & bottom 5, 80); // near & far
• Second the model-view
• glMatrixMode ( GL_MODELVIEW );
• glTranslatef (0, 0, -14);
• So objects centered at (0,0,0) would be at (0,0,-14) in eye-space
Resulting projection matrix Resulting modelview matrix
• Frustum Transform
• Prototype
• glFrustum ( GLdouble left, GLdouble right, GLdouble bottom, GLdouble top, GLdouble near, GLdouble far)
• Post-concatenates a frustum matrix
• glFrustum Matrix
• Projection specification
• glLoadIdentity (); glFrustum (-4, +4, -3, +3, 5, 80)
• left=-4, right=4, bottom=-3, top=3, near=5, far=80
• Matrix
= -Z axis symmetric left/right & top/bottom so zero 80 5
• Translate Transform
• Prototype
• glTranslatef ( GLfloat x, GLfloat y, GLfloat z)
• Post-concatenates this matrix
• glTranslatef Matrix
• Modelview specification
• glLoadIdentity (); glTranslatef (0,0,-14)
• x translate=0, y translate=0, z translate=-14
• Point at (0,0,0) would move to (0,0,-14)
• Down the negative Z axis
• Matrix
= the translation vector
• Resulting Modelview-Projection Transform Matrix
• Transform composition via matrix multiplication
Resulting modelview-projection matrix
• Now Draw Some Objects
• Draw a wireframe cube
• glColor3f (1,0,0); // red
• glutWireCube (6);
• 6x6x6 unit cube centered at origin (0,0,0)
• Draw a teapot in the cube
• glColor3f (0,0,1); // blue
• glutSolidTeapot (2.0);
• centered at the origin (0,0,0)
• handle and spout point down the X axis
• top and bottom in the Y axis
• As we’d expect given a frustum transform, the cube is in perspective
• The teapot is too but more obvious to observe with a wireframe cube
• What We’ve Accomplished
• Simple perspective
• With glFrustum
• Establishes how eye-space maps to clip-space
• Simple viewing
• With glTranslatef
• Establishes how world-space maps to eye-space
• All we really did was “wheel” the camera 14 units up the Z axis
• No actual “modeling transforms”, just viewing
• Modeling would be rotating, scaling, or otherwise transform the objects with the view
• Arguably the modelview matrix is really just a “view” matrix in this example
(0,0,14) (0,0,0)
• Let’s Add Some Simple Modeling
• Try some modeling transforms to move teapot
• But leave the cube alone for reference
glPushMatrix (); { glTranslatef (1.5, -0.5, 0); glutSolidTeapot (2.0); } glPopMatrix (); glPushMatrix (); { gl Scalef (1.5, 1.0, 1.5); glutSolidTeapot (2.0); } glPopMatrix (); glPushMatrix (); { gl Rotatef ( 30, 1,1,1 ); glutSolidTeapot (2.0); } glPopMatrix (); Notice: We “bracket” the modeling transform with glPushMatrix / glPopMatrix commands so the modeling transforms are “localized” to the particular object
• Add Some Lighting
• Some lighting makes the modeling more intuitive
glPushMatrix (); { glTranslatef (1.5, -0.5, 0); glutSolidTeapot (2.0); } glPopMatrix (); glPushMatrix (); { gl Scalef (1.5, 1.0, 1.5); glutSolidTeapot (2.0); } glPopMatrix (); glPushMatrix (); { gl Rotatef ( 30, 1,1,1 ); glutSolidTeapot (2.0); } glPopMatrix (); We’ve not discussed lighting yet but per-vertex lighting allows a virtual light source to “interact” with the object’s surface orientation and material properties
• Resulting Modelview-Projection Matrix
• Let’s consider the “combined” modelview matrix with the rotation
• glRotate(30, 1,1,1) defines a rotation matrix
• Rotating 30 degrees…
• … around an axis in the (1,1,1) direction
projection view model
• Combining All Three projection view model modelview modelview-projection Matrix-by-matrix multiplication is associative so PVM = P (V M) = (P V) M OpenGL keeps V and M “together” because eye-space is a convenient space for lighting
• Equivalent Math Paths from Object- to Clip-space object-to-world-to-eye-to-clip object-to-eye-to-clip object-to-clip modelview projection projection modelview-projection model view
• A Better Viewing Matrix
• “ Look at” Transform
• Concept
• Given the following
• a 3D world-space “eye” position
• a 3D world-space center of view position (looking “at”), and
• an 3D world-space “up” vector
• Then an affine (non-projective) 4x4 matrix can be constructed
• For a view transform mapping world-space to eye-space
• A ready implementation
• The Open GL U tility library ( GLU ) provides it
• gluLookAt ( GLdouble eyex, GLdouble eyey, GLdouble eyez, GLdouble atx, GLdouble atz, GLdouble atz, GLdouble upx, GLdouble upy, GLdouble upz);
Primary OpenGL libraries Link with –lglut for GLUT Link with –lGLU for GLU Link with –lGL for OpenGL
• “Look At” Concept
• High-level goal
• Puts (eyex,eyey,eyez) at the origin of clip space
• Essentially a translate
• Then rotates so
• “ eye – at” direction vector looks down the negative Z (-Z) axis, AND
• Orthogonalize “up” vector to correspond to positive Y (+Y) axis
• Application
• Useful when you want the camera to maintain observation of a point of interest in the scene
• Whether the camera or the point of interest is moving, or both!
• A lot like your object viewer Project #1
• Study this topic
• Section 4.3 of your text (pages 204-214)
• “Look At” Diagram E. Angel and D. Shreiner: Interactive Computer Graphics 6E © Addison-Wesley 2012
• “Look At” in Practice
• Consider our prior view situation
• Instead of an arbitrary view…
• … we just translated by 14 in negative Z direction
• glTranslatef (0,0,14)
• What this means in “Look At” parameters
• (eyex,eyey,eyez) = (0,0,14)
• (atx,aty,atz) = (0,0,0)
• (upx,upy,upz) = (0,1,0)
glTranslatef(0,0,-14) gluLookAt(0,0,14, 0,0,0, 0,1,0) Same matrix; same transform Not surprising both are “just translates in Z” since the “Look At” parameters already have use looking down the negative Z axis
• The “Look At” Algorithm
• Vector math
• Z = eye – at
• Z = normalize(Z) /* normalize means Z / length(Z) */
• Y = up
• X = Y × Z /* × means vector cross product! */
• Y = Z × X /* orthgonalize */
• X = normalize(X)
• Y = normalize(Y)
• Then build the following affine 4x4 matrix
• Warning: Algorithm is prone
• to failure if normalize divides
• by zero (or very nearly does)
• So
• Don’t let Z or up be zero length vectors
• Don’t let Z and up be coincident vectors
• “Look At” Examples gluLookAt (0,0,14, 0,0,0, 0,1,0); // eye (x,y,z) // at (x,y,z) // up (x,y,z) gluLookAt (1,2.5,11, 0,0,0, 0,1,0); // eye (x,y,z) // at (x,y,z) // up (x,y,z) Same as the glTranslatef(0,0,-14) as expected Similar to original, but just a little off angle due to slightly perturbed eye vector
• “Look At” Major Eye Changes gluLookAt (-2.5, 11 ,1, 0,0,0, 0,1,0); // eye (x,y,z) // at (x,y,z) // up (x,y,z) gluLookAt (-2.5, - 11 ,1, 0,0,0, 0,1,0); // eye (x,y,z) // at (x,y,z) // up (x,y,z) Eye is “above” the scene Eye is “below” the scene
• “ Look At” Changes to AT and UP gluLookAt (0,0,14, 2,-3,0, 0,1,0); // eye (x,y,z) // at (x,y,z) // up (x,y,z) gluLookAt (0,0,14, 0,0,0, 1,1,0); // eye (x,y,z) // at (x,y,z) // up (x,y,z) Original eye position, but “at” position shifted Eye is “below” the scene
• Complex Scene Example Each character, wall, ceiling, floor, and light have their own modeling transformation
• Representing Objects
• Interested in object’s boundary (or manifold)
• Various approaches
• Procedural representations
• Often fractal
• Explicit polygon (triangle) meshes
• By far, the most popular method
• Curved surface patches
• Often displacement mapped
• Implicit representation
• Blobby, volumetric
Sierpinski gasket Utah Teapot Blobby modeling in RenderMan Quake 2 key frame triangle meshes Fractal tree [Philip Winston]
• Focus on Triangle Meshes
• Easiest approach to representing object boundaries
• So what is a mesh and how should it be stored?
• Simplest view
• A set of triangles, each with its “own” 3 vertices
• Essentially “triangle soup”
• Yet triangles in meshes share edges by design
• Sharing edges implies sharing vertices
• More sophisticated view
• Store single set of unique vertexes in array
• Then each primitive (triangle) specifies 3 indices into array of vertexes
• More compact
• Vertex data size >> index size
• Avoids redundant vertex data
• Separates “topology” (how the mesh is connected) from its “geometry” (vertex positions and attributes)
• Connectivity can be deduced more easily
• Makes mesh processing algorithms easier
• Geometry data can change without altering the topology
• Consider a Tetrahedron
• Simplest closed volume
• Consists of 4 triangles and 4 vertices
• (and 4 edges)
v0 v1 v3 v2 triangle list 0: v0,v1,v2 1: v1,v3,v2 2: v3,v0,v2 3: v1,v0,v3 (x0,y0,z1) (x1,y1,z1) (x2,y2,z2) (x3,y3,z3) vertex list 0: (x0,y0,z0) 1: (x1,y1,z1) 2: (x2,y2,z2) 3: (x3,y3,z3) topology geometry potentially on-GPU!
• Benefits of Vertex Array Approach
• Unique vertices are stored once
• Saves memory
• On CPU, on disk, and on GPU
• Matches OpenGL vertex array model of operation
• And this matches the efficient GPU mode of operation
• The GPU can “cache” post-transformed vertex results by vertex index
• Saves retransformation and redundant vertex fetching
• Direct3D has the same model
• Allows vertex data to be stored on-GPU for even faster vertex processing
• OpenGL supported vertex buffer objects for this
• See “Modern OpenGL Usage: Using Vertex Buffer Objects Well”
• http://www.slideshare.net/Mark_Kilgard/using-vertex-bufferobjectswell
• Next Lecture
• More about triangle mesh representation
• Blending, Compositing, Anti-aliasing
• More than simply writing pixels into the framebuffer
• Also information on GLUT input callbacks
• As usual, expect a short quiz on today’s lecture
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### Why does splitting $n$ bit integers into chunks of size $\log(n)$ specifically, help in multiplying them
In integer multiplication algorithms such as the Schonhage-Strassen algorithm (and the recently described Harvey and van der Hoeven algorithm), integers of size $n$ are reduced to polynomials with ...
200 views
### Subquadratic 3SUM when one set is in [n^1.99]
Chan and Lewenstein (STOC 2015) said: 3SUM for three integer sets where only one set is assumed to be in $[n^{2−\delta}]$ can still be solved in subquadratic time (by doing several FFTs, without ...
1 vote
33 views
### Are there uses for a Fourier transform of length $n^m$ with elements of maximum size $n$?
In essence, I'm trying to get a better feel for when there is a use for FFT with small coefficients, compared to the length, assuming that we get a better runtime. I've been toying with an idea for a ...
• 2,080
131 views
### What would faster Fourier Transform(FFT?) and/or multiplication algorithms imply?
There are many problems which have implications on P vs. NP and other complexity classes. Supposing that we're interested in Fourier transforms and/or multiplication algorithms, do faster Fourier ...
• 2,080
92 views
### Is gaussian smoothing possible in less operations than O(N log N)
Gaussian filtering is popular in applications, for my question it can be written as (I've fixed the size of window): $$y_i = \sum_{j = 1}^{n} x_j e^{(i - j)^2}, \qquad i = 1, 2, ..., n$$ One can ...
• 121
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### Finding witness in minkowski sum of integers
Let $A$ and $B$ be subsets of $\{0,\ldots,n\}$. We are interested in finding the Minkowski sum $A+B=\{a+b~|~a\in A,b\in B\}$. $\chi_X:\{0,\ldots,2n\}\to \{0,1\}$ is a characteristic function of $X$ ...
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### Complexity of convolution in the max/plus ring
We can do convolution in $O(n\log n)$ for plus/multiply polynomials with FFT. However, the approach doesn't seem very generalisable to rings in general. Has there been any progress over the naive \$O(...
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## Problem 2
The polynomial $x^3+ax^2+bx+c$ has real roots $\alpha$, $-\alpha$ and $\frac{1}{\alpha}$. Thus the value of the sum of $b+c^2+ac+\frac{b}{c^2}$ is:
(a) $-2$
(b) $-1$
(c) $0$
(d) $1$
(e) $2$
## Solution
By Girard's relations (also called Vieta's formulas or Newton's identities),
$\begin{cases}a=-\frac{1}{\alpha}\\b=-\alpha^2\\c=\alpha\end{cases}$
$b+c^2+ac+\frac{b}{c^2}$
$-\alpha^2+\alpha^2-\frac{1}{\alpha}\cdot\alpha+\frac{-\alpha^2}{\alpha^2}=\boxed{-2}~~~~\boxed{\text{A}}$ | 226 | 565 | {"found_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": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-51 | longest | en | 0.596234 |
http://mathhelpforum.com/math-software/83598-matlab-differene-equation-question.html | 1,500,867,114,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424683.39/warc/CC-MAIN-20170724022304-20170724042304-00537.warc.gz | 200,956,173 | 11,285 | # Thread: Matlab differene equation question
1. ## Matlab differene equation question
I hav a question im not quiet sure how to do on matlab. Here it is:
Consider the following difference equation:
q(n) = 2*q(n-1) + q(n-2) - 2*q(n-3)
The analytical solution of the difference equation will be in the form
q(n) = a(1) + a(2)*(-1)^n + a(3)*2^n
Write down the system of linear equations in a(1), a(2), a(3) so that q(0) = 1
q(1) = 2, q(2) = 2.5. Use Matlab to find the values for a(1), a(2), a(3). Hand in your Matlab commands.
Usually I would do this by coming up with three simultaneous equations and solving them manually (or with a calculator) but this asks specifically for Matlab, and we havnt done simultanious equations, so i'm assuming theres another way, and if not mabe someone could run me through simultanious equations on Matlab.
Any help is greatly appreciated.
2. Originally Posted by mrtwigx
I hav a question im not quiet sure how to do on matlab. Here it is:
Consider the following difference equation:
q(n) = 2*q(n-1) + q(n-2) - 2*q(n-3)
The analytical solution of the difference equation will be in the form
q(n) = a(1) + a(2)*(-1)^n + a(3)*2^n
Write down the system of linear equations in a(1), a(2), a(3) so that q(0) = 1
q(1) = 2, q(2) = 2.5. Use Matlab to find the values for a(1), a(2), a(3). Hand in your Matlab commands.
Usually I would do this by coming up with three simultaneous equations and solving them manually (or with a calculator) but this asks specifically for Matlab, and we havnt done simultanious equations, so i'm assuming theres another way, and if not mabe someone could run me through simultanious equations on Matlab.
Any help is greatly appreciated.
What have you done in matlab?
CB
3. Originally Posted by CaptainBlack
What have you done in matlab?
CB
Im only halfway through my first applied maths course (stage one) so probably not much. We'v done basic stuff like solving simple equations and drawing graphs. we'v done difference equations which involves analytical technique, for loops to do things like growh models and predator/prey models, and a few other things. Wev done Nonlinear equations which involves Newton-Raphson method, and have just started the bisection method, which involves using booleons. And I think thats about it.
4. Originally Posted by mrtwigx
Im only halfway through my first applied maths course (stage one) so probably not much. We'v done basic stuff like solving simple equations and drawing graphs. we'v done difference equations which involves analytical technique, for loops to do things like growh models and predator/prey models, and a few other things. Wev done Nonlinear equations which involves Newton-Raphson method, and have just started the bisection method, which involves using booleons. And I think thats about it.
Have you done any matrix calculations? | 734 | 2,860 | {"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} | 3.890625 | 4 | CC-MAIN-2017-30 | longest | en | 0.897355 |
systemvoidgames.com | 1,631,858,369,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00166.warc.gz | 56,816,201 | 9,446 | ## A Probabilistic Shot Accuracy Model for FPS Enemy AI
I’m currently working on an FPS called DEBASER. In this game, until very recently in development, the enemies had 100% accuracy when shooting, so long as you were visible by the time they pulled the trigger. Obviously, this is both unrealistic and makes the game unreasonably hard. After some playtesters complained, I began searching for a solution. For simplicity, I decided that I would use a probabilistic model to determine if the enemy misses or hits to player according to RNG.
## The Variables
As a starting point, I figured that a good model for shot accuracy should take two variables into account: distance and speed of the target. If you were trying to shoot something, you would much rather it be close to you and not moving, than far away and zig-zagging.
An immediate problem with this model is that speed can be misleading. It’s difficult to hit a target running around you in a circle, but it’s easy if they’re running directly towards you. This is because you only have to account for movement that changes the angle you must shoot at. It follows that we would be better suited to consider the angular velocity of the target relative to the shooter.
Above is a basic diagram showing how we get the angular velocity of the player relative to the enemy. Note that v is the tangential velocity, which is found by projecting the velocity vector of the player onto the plane whose normal is the facing vector of the enemy:
## The Difficulty Model
As an intermediate to the Accuracy model, let’s first model “difficulty.” Difficulty (D) will be a number from 0 to infinity, where 0 means it is 100% likely that the enemy will hit the player and infinity means it is 0% likely that the enemy will succeed. Difficulty should increase as range and angular velocity increase, like so:
To get an intuition for how this function behaves, let’s graph it with D on the y axis and r on the x axis. v will raise and lower.
Notice that the difficulty starts at 0 when the distance is zero. Intuitively, this makes sense. However, the function is asymptotic: it converges to a number, not infinity. This makes no sense, as difficulty should always be rising as distance increases. The reason for this convergence is that, as you get farther away, angular velocity shrinks so fast that it cancels out the growth of r. The solution to this is to add a “helper” constant (c) to the angular velocity at all times. This will ensure that it always approaches infinity.
Above, you can see the effect that increasing the helper constant has on the limits of the Difficulty model.
With that, our Difficulty model is complete:
## The Accuracy Model
Our final accuracy model should represent the probability from 0 to 1 that the shot will be accurate. Because Difficulty can be anywhere from 0 to infinity, we should start by establishing two more constants to confine our results: L = the difficulty below which accuracy is 100%. U is the difficulty above which accuracy is 0%.
Above we see L and U in the context of the current difficulty model. To create an accuracy model, we need to form a piecewise function. The behavior when D is above and below L and U has already been specified above, but what about in-between? We don’t want this function to have any gaps in it, so what we can do is interpolate from 1 to 0 based on how far D has traveled throughout the region between U and L. The math behind this interpolation gets a bit messy but the overall concept is simple. The resulting Accuracy model is as follows:
## Accounting for Human Stupidity
Humans are bad at probability (well, maybe it’s just me…); as I play-tested this model, I noticed that the AI was still very accurate. At 50% accuracy they were still very often taking me out on the first try. In a deliberate departure from realism, I added another constant, h, which I’ve affectionately named the Constant of Human Stupidity. The final model simply applies h as an exponent to A. Along with making it easier to avoid getting shot, this constant has the effect of making the model look more exponential for higher values of h. Below is the result of raising and lowering h.
## The Ping Pong Modulo Function
Earlier today I had to spend a whole 20 minutes coming up with this function. It’s like a modulo function, but instead of looping back to zero every cycle, it transitions back down to zero. It cycles every 2(n-1) elements. There’s probably a proper name for it, but I can’t figure it out.
In case anyone else would rather google it than spend time figuring it out, here it is:
```def ping_pong_mod(i, n):
cycle = 2 * (n - 1)
i = i % cycle
if i >= n:
return cycle - i
return i```
I hope this is helpful to a fellow lazy person in the future. | 1,049 | 4,786 | {"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} | 3.8125 | 4 | CC-MAIN-2021-39 | latest | en | 0.963257 |
https://www.experts-exchange.com/questions/21829061/matlab-syntax-question.html | 1,513,295,204,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948551162.54/warc/CC-MAIN-20171214222204-20171215002204-00213.warc.gz | 699,516,345 | 25,236 | Solved
# matlab- syntax question
Posted on 2006-04-26
Medium Priority
126 Views
Hi,
I saw a following matlab code. there is a "." (dot) after "w1". what is the meaning of that dot? I try to remove it, and I don't see any different in the result.
w1=0.5*pi;
td=0:ts:4;
x_sin=sin(w1.*td);
plot (x_sin)
0
Question by:hsuyf
LVL 11
Accepted Solution
dbkruger earned 1000 total points
ID: 16549545
In matlab,* is matrix multiplication.
so:
a=[1 2 3]; % row vector
b = [1;2;3]; % column vector
b*a =
1 * *
* 4 *
* * 9
I'm not going to write it out, but you get the picture.
If you just want to multiply the first element of a times b, the second, etc. then use .*
b.*a = [1 4 9]
0
Author Comment
ID: 16549607
thank you very much!!!
0
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https://mathoverflow.net/questions/227238/do-the-analogies-between-metamathematics-of-set-theory-and-arithmetic-have-some | 1,560,873,724,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998755.95/warc/CC-MAIN-20190618143417-20190618165417-00337.warc.gz | 518,524,077 | 33,956 | Do the analogies between metamathematics of set theory and arithmetic have some deeper meaning?
By "formal analogies" between the metamathematics of $\mathsf{ZFC}$/set theory and $\mathsf{PA}$(=Peano Arithmetic)/first order arithmetic, I mean facts such as the following:
• We are considering a first-order theory ($\mathsf{ZFC}$ or $\mathsf{PA}$) motivated as a first-order approximation to a second-order theory (second-order $\mathsf{ZFC}^2$ or $\mathsf{PA}^2$) which is "more or less" categorical; because of this, some axioms (the separation and replacement axioms on the one hand, the induction axiom on the other) have to be stated as axiom schemes in the first-order theory.
• There is an interesting hierarchy of formulæ, $\Sigma_n$ or $\Pi_n$, based on alternations of quantifiers (viz.: the arithmetic hierarchy vs. the Lévy hierarchy); at the lowest ($\Delta_0$) level of this hierarchy are formulæ with only "bounded" quantifiers.
• There is a uniform truth predicate for any (concrete) given level of the hierarchy, which is built upon some kind of absoluteness of $\Delta_0$ formulæ. As a related fact, the infinite axiom schemes are naturally stratified along the hierarchy (they can be cut off at the $\Sigma_n$ level and stated as a single formula for each concrete $n$).
• There is a reflection theorem which ensures that any finite set of true statements (or one bounded in the hierarchy of formulæ) is consistent. In particular, the full theory proves the consistency of the subtheory with the axiom schemes cut off at the $\Sigma_n$ level: that is, the theory ($\mathsf{ZFC}$ or $\mathsf{PA}$) is reflexive. In fact, it is even essentially reflexive (every consistent extension is reflexive).
• There is a conservative two-kinded extension (Gödel-Bernays on the set theoretical side, $\mathsf{ACA}_0$ on the arithmetical side) which is obtained by allowing formation of classes but only with a comprehension scheme for such classes that does not involve quantifying over classes; remarkably, this conservative two-kinded extension is finitely axiomatizable. There is also a standard strictly stronger two-kinded extension (Morse-Kelley on the set-theoretical side, second-order arithmetic $\mathsf{Z}_2$ seen as a first-order theory on the arithmetical side).
(I hope I didn't mess things up too much, but all of these facts are standard and can be found in standard textbooks such as Jech's Set Theory for the set-theoretical side and Hájek and Pudlák's Metamathematics of First-Order Arithmetic plus Simpson's Subsystems of Second-Order Arithmetic for the arithmetical side.)
I'm sure many more examples can be found. Maybe I should also nod to the similarity between proof theory of extensions of $\mathsf{PA}$ by analysing ordinal notations made by collapsing recursively large ordinals, and large cardinal extensions of $\mathsf{ZFC}$ — or maybe not.
Yet as striking as this analogy seems, nobody seems to comment upon it as far as I know. (At the very least, this seems pedagogically regrettable: I'm sure all of the above statements would be more memorable to students if the analogous statements were made explicit.)
So: is there some deeper truth to be found behind this parallelism? (Or are all my sample facts just aspects of a single phenomenon? Or is this just a red herring?) Might it make sense to bring $\mathsf{ZFC}$ and $\mathsf{PA}$ under an umbrella metatheory so that the above facts can be proved in a common formalism? At the very least, is there a textbook I missed where the analogy is played out in some detail?
And perhaps more thought provokingly: can one give an example of a completely different kind of theory that is just as similar to $\mathsf{ZFC}$ and $\mathsf{PA}$ as they are to each other?
(I'm of course aware that are also huge differences between $\mathsf{ZFC}$ and $\mathsf{PA}$; but I would tend to say that they make the similarities all the more striking.)
• The part about axiom schemata is a red herring, IMO: there is always a way to build a conservative extension to the theory with only finitely many axioms. One fact that might help understanding the analogy is the following fact: ZF+negation of infinity is equivalent to PA, in the sense that each is interpretable in the other and the interpretations are inverse to each other. – cody Dec 29 '15 at 21:55
• Perhaps this is too glib, but several of these similarities seem to stem from the fact that both systems were, in part, set up to provide a formalism for the foundations of (a lot of) mathematics, in a hierarchical (as opposed to a structuralist) way. Truth, reflection, and hierarchy seem to emerge rather naturally from such a project. – Timothy Chow Dec 30 '15 at 2:57
• Gro-Tsen, does Ed Nelson's IST qualify? – Mikhail Katz Dec 30 '15 at 17:35
A significant amount of the parallelism can be explained by the bi-interpretabiity of $PA$ with $ZF^{-\infty}$ ("finite set theory"), which is the theory obtained from $ZF$ by replacing the axiom of infinity by its negation, and adding the sentence asserting that every set has a transitive closure. For more detail and references, see my joint paper with Schmerl and Visser entitled $\omega$-Models of Finite Set Theory here.
But that is only part of the story since the parallelism between arithmetic and set theory is deep and mysterious. I can even say that my career as a logician has been greatly shaped by comparing and contrasting the metamathematics of arithmetic and set theory.
My 1998-paper Analogues of MacDowell-Specker Theorem for Set Theory, available here, gives a synopsis of the similarities and differences between $PA$ (equivalently: $ZF^{-\infty}$) and $ZF$ through the lens of model theory.
Added in the third edit: Here is a relevant quote from the above paper:
The axiom of infinity is of course only the first step in the progression of ever bolder large cardinal axioms. As we shall see below, the negation of the axiom of infinity endows $ZF^{-\infty}$ with a model theoretic behavior that $ZF$ can only imitate with the help of additional axioms asserting the existence of large cardinals. This is partially explainable by noting that the negation of the axiom of infinity in finite set theory itself can be viewed as a large cardinal axiom, not positing the existence of a large set - indeed denying it - but attributing a large cardinal character to the universe itself. Of course the axioms of power set and replacement give a ”strong inaccessibility” character to the class of ordinals which allows models of $ZF$ to share some of the model theoretic properties of $ZF^{-\infty}$.
Added in the second edit: In this 2009-presentation I describe a scheme $\Lambda$ (named in honor of Azriel Levy) consisting of set-theoretic sentences of the form "there is an $n$-reflective, $n$-Mahlo cardinal" which has the surprising property that:
$ZFC + \Lambda$ is the weakest extension of $ZFC$ whose model-theoretic behavior matches that of $PA$, in several surprising respects. So $PA$ is used here as a guide to find an improvement of $ZFC$.
Most, but not all of the results in the presentation have appeared in print.
• @Prof. Enayat: Since Robinson arithmetic $Q$ can also be formulated in finite $ZF$, does $Q$+"Second-order Mathematical Induction " imply some form of the Axiom of Infinity? – Thomas Benjamin Dec 30 '15 at 16:19
• @Thomas Benjamin: when mathematical induction is added to $Q$ (within second order logic), we obtain full second order arithmetic, which is a categorical theory in second order logic, whose only model up to isomorphism is bi-interpretable with the structure $(H(\omega_1), \in)$, which of course is a model of axiom of infinity, more specifically, it is a model of all the axioms of ZFC except the power set axiom. However, most people think of second order logic as a reformulation of set theory, so the fact that we get a model of the Axiom of Infinity is not surprising. – Ali Enayat Dec 30 '15 at 16:42
• @Prof. Enayat: Is there a difference then between second-order $PA$ and full second-order arithmetic? (Sorry for seeming obtuse....) – Thomas Benjamin Dec 30 '15 at 22:24
• @Thomas Benjamin: in my usage, full second order arithmetic is formulated in second order logic, whereas what is commonly referred to as second arithmetic by many logicians, is a first order theory $Z_2$ (e.g., in Simpson' s comprehensive textbook on the subject). It is well-known that $Z_2$ can interpret the theory obtained by deleting the power-set axiom from $ZF$ and adding the sentence "every set is finite or countable". The details are in Simpson' s book. – Ali Enayat Dec 31 '15 at 8:35 | 2,055 | 8,642 | {"found_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} | 2.65625 | 3 | CC-MAIN-2019-26 | longest | en | 0.894245 |
https://spreadcheaters.com/how-to-square-a-column-in-excel/ | 1,717,075,565,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00326.warc.gz | 469,698,487 | 18,830 | # How to square a column in Excel
In Excel, squaring a column means taking each value in the column and raising it to the power of 2 (i.e., multiplying it by itself). Squaring a column in Excel can help you to better understand and analyze your data, perform calculations, and create visualizations that can aid in decision-making.
In this tutorial, we will learn how to square a column in excel. In our data set numbers are shown in the first column and their square is required in the next column. For this we have two methods, using Function and using formula. The following steps will guide you to use the Function and Formula.
## Step 1 – click on the first cell
• Click on the first cell of the column where the squared values are to be shown
## Step 2 – Type the formula
• After clicking on the cell, type “=” and then type the formula in it
• Where the formula consists of the address of the cell “having the value to be squared” multiplied by itself i.e. =A2*A2
## Step 3 – press Enter key
• After typing the formula, press Enter key to get the squared value in the cell
## Step 4 – Apply the Formula to the Entire Column
• After getting the square of the first cell, drag the first cell to get the required result
## Method 2: Square a column using Power Function
The POWER function in Excel is a mathematical function that raises a number to a specified power. The syntax of the POWER function is as follows:
• =POWER(number, power)
• The “number” argument is the base number that you want to raise to a power, and the “power” argument is the exponent to which you want to raise the base number
## Step 1 – Click on the cell
• Click on the first cell of the column where the squared values are to be shown
## Step 2 – Use Function to create the formula
• After clicking on the cell, use power function by typing “=Power(
## Step 3 – Type the Arguments of the function
• After typing in the =Power, type the arguments of the function.
• Arguments are number and power. Here the number is the address of the cell to be squared and the power is 2.
• After typing arguments, type a closing bracket “)
## Step 4 – press Enter key
• After typing the argument, press Enter key to get the squared value
## Step 5 – Apply the Function on the Entire Column
• After getting the square of the first cell, drag the first cell to get the required result | 543 | 2,373 | {"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} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.846369 |
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# Thermal Physics Urgent doubt
Asked by scholar1 23rd March 2010, 1:04 PM
Dear student,
The expansion in length is called linear expansion. The expansion in area is called area expansion. The expansion in volume is called volume expansion.
Linear expansion - l/l a1 T
Area expansion - A/A 2 a1 T
Volume expansion - V/V 3 a1 T
The coefficient of linear expansion is denoted by the letter 'a' .
Where l = original length
The coefficient of linear expansion is a characteristic property of the material of the rod.
Expansion in area of a Laminar surface due to heating is known as superficial expansion.
Let a1 = area of a surface
t1 = its initial temperature
a2 = area of the surface after heating
t2 = final temperature
The coefficient of superficial expansion is defined as the ratio of increase in area to its original area for every degree increase in temperature.
### Increase in area = a2-a1
Rise in temperature = (t2-t1)
Coefficient of superficial expansion (b)
.
Increase in volume of a solid on heating is cubical expansion.
Let V1 = Initial volume at t10C
V2 = its volume at t20C (t2>t1)
Then increase in volume 'v' = v2-v1
Increase in temperature 't' = t2-t1
The coefficient of cubical expansion is defined as the ratio of the increase in volume to its original volume for every degree increase in temperature. It is denoted by 'g'.
We can show that g = 3a
.
Regards
Topperlearning.com
Answered by Expert 27th March 2010, 12:46 PM
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https://math.stackexchange.com/questions/975294/why-wouldnt-someone-accept-gentzens-consistency-proof | 1,701,889,404,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00553.warc.gz | 448,827,808 | 37,047 | # Why wouldn't someone accept Gentzen's consistency proof?
The vast majority of contemporary mathematicians believe that Peano's axioms are consistent, relying either on intuition or the acceptance of a consistency proof such as Gentzen's proof.
If Gentzen's proof shows the consistency of PA, then what is there to "accept"? Is the point that some mathematicians might not accept that Gentzen's proof shows the consistency of PA because we don't know whether the system Gentzen uses in his proof is consistent?
• You can see also Gentzen's consistency proof with Kleene's comment, and also Annika Siders, Gentzen’s Consistency Proofs for Arithmetic : "It should be noted that the theory, in which the proof is formalizable, is incomparable to Peano arithmetic. The theory is not stronger than Peano arithmeic, since complete induction cannot be proved for all formulas. But on the other hand, neither is the theory weaker, since it proves the consistency of Peano arithmetic." Oct 16, 2014 at 7:37
• If PA are inconsistent, then all sets must be Dedekind-finite. Oct 17, 2014 at 2:48
Essentially due to Gödel's Incompleteness Theorems, proofs of the consistency of $\mathsf{PA}$ must involve methods that transcend $\mathsf{PA}$ itself. If one has any doubts about the consistency of $\mathsf{PA}$, those doubts are likely only to be amplified concerning the methods used to prove the consistency of $\mathsf{PA}$. (For example, from $\mathsf{ZF}$, then you can easily construct a model of $\mathsf{PA}$, but the consistency of $\mathsf{ZF}$ is "more debatable" than that of $\mathsf{PA}$, so you haven't really gained anything.)
Gentzen's proof relies on infinitary processes (in particular, induction up to $\varepsilon_0$; see Wikipedia), and may not have been accepted by the Hilbert school (who sought purely finitary proofs of consistency). The ordinal $\varepsilon_0$ is important here because (assuming its consistency) $\mathsf{PA}$ cannot prove that it is well-founded, and it is basically this move that transcends $\mathsf{PA}$.
• @Dennis: It might be worth adding that Gentzen used the transfinite induction as the main tool here, but as far as $\sf PA$ goes, he used only a small fragment of it. Oct 16, 2014 at 12:40 | 542 | 2,236 | {"found_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} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.936821 |
https://ru.scribd.com/document/318906510/PAIT1517-A01-PT-3-P-2 | 1,569,094,835,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574662.80/warc/CC-MAIN-20190921190812-20190921212812-00296.warc.gz | 643,225,176 | 77,253 | Вы находитесь на странице: 1из 7
# PGSFinal Draft of Azam 11th Phase - 3
SECTION - 1: PHYSICS
Part - A (One or more options correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and
(D) out of which ONE or MORE are correct.
1.
The potential energy U in joule of a particle of mass 1 kg moving in xy plane, obeys the law
U 3 x 4 y , when x, y are the co-ordinates of the particle in meter. If the particle is at rest at
## (6, 4) at time t = 0, then
(A) the particle has constant acceleration
(B) the work done by the external force from the position of rest of the particle and the instant at
which the particle crossing x axis is 25 joule
(C) the speed of the particle when it crosses the y axis is 10 m/s
(D) none of these
2.
## (A) The particle has stable equilibrium at points 3 & b
(B) The particle is in neutral equilibrium at point b& 2
(C) No power is delivered by the force on the particle at the points 1, 3 & b
(D) The particle has unstable equilibrium at point 1.
3.
An object of mass m is sliding down a hill of arbitrary shape and, after traveling down on the path,
the mass stops because of friction. The friction coefficient may be different for different segments
of the entire path but it is independent of the velocity and direction of motion.
(A) The work that a force must perform just to return the object to its initial position along the
same path is 2mgl.
(B) The work done against force of friction while returning to its initial position is mgl.
(C) The work that a force must perform just to return the object to its initial position along the
same path is mgh.
(D) The work done against force of friction while returning to its initial position is 2mgh.
4.
The system shown in figure is in equilibrium and at rest. The spring and string are massless. Now
the string is cut. The acceleration of mass 2m and m just after the string is cut will be:
(A) Acceleration of 2m is g/2 upwards
(B) Acceleration of 2m is g upwards
(C) Acceleration of m is g downwards
(D) Acceleration of m
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5.
6.
A block of mass2 kg moves over a fixed track as shown in the figure. AB = 2 m rough (=0.4)
track AC and BD are frictionless. Then which of the following is /are correct (g = 10 m/s2)
## (A) the block stops at B
(B) Kinetic energy left over by the time block moves to
and from once between AB is 48 J
(C) Block slides along AB 4 times before it comes to rest
(D) None of these
Two bodies undergo perfectly elastic collision then
(A) Kinetic energy of the system of two bodies is the same before and after the collision
(B) Kinetic energy of the system of two bodies is constant during the collision.
(C) Kinetic energy of the system of two bodies during collision is more than their kinetic energy
before collision.
(D) Kinetic energy of the system of two bodies during collision is less than their kinetic energy
before collision.
7.
A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper
end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched)
length, comes to rest again after descending through a distance x.
(A) x = mg/k
(B) x = 2 mg/k
(C) The ball will have no acceleration at the position where it has descended through x/2
(D) The ball will have an upward acceleration equal to g at its lowermost position
8.
## Kinetic energy of a particle is continuously increasing with time. It means :
(A) resultant force always acts along the direction of motion
(B) resultant force always acts normal to the direction of motion
(C) resultant force is at an angle less than 90 to the direction of motion
(D) its height above the ground level may be decreasing
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## Part - A (Paragraph Type)
This section contains 4 paragraphs each describing theory, experiment, date etc. Eight questions
relate to four paragraphs with two questions on each paragraph. Each question of paragraph has only
one correct answer along the four choice (A), (B), (C) and (D).
Paragraph for Questions 9 to 10
u 4 gR
A particle is projected with a velocity
along a smooth semicircular track of radius R.
(See figure). At some angular position particle will leave contact with the track. After that particle
will become a projectile.
9.
The angle made by position vector of particle with vertical when particle leaves contact with track
is
(A) sin12/3
(B) tan12/3
(C) cos12/3 (D) cot12/3
10.
(A) 2gR (1 cos )
(B) 2gR (1 sin )
(C)
2gR (1 tan )
(D)
2gR (1 cot )
## Paragraph for Questions 11 to 12
At the moment t = 0 the force F kt is applied to a small body of mass m resting on a smooth
horizontal plane (k is a constant). The permanent direction of this force forms an angle with the
horizontal as shown.
11.
12.
Find the time t at which moment, the body breaks off the plane.
mg
mg
mg
2
(A) 2k sin
(B) k sin
(C) k sin
2mg
(D) k sin
The velocity of the body at the moment of its breaking off the plane is:
mg 2 cos
2mg 2 cos
mg cos
2mg cos
2
2
2
2
(A) 2k sin
(B) 2k sin
(C) k sin
(D) k sin
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## Paragraph for Questions 13 to 14
The figure shows the variation of potential energy of a particle as a function of x, the x-coordinate
of the region. It has been assumed that potential energy depends on x. for all other values of x, U
is zero i.e. for x < - 10 and x > 15 U = 0
11.
## If total mechanical energy of the particle is 25 J, then it can be found in region.
(A) 10 < x < - 5 and 6 < x < 15
(B) 10 < x < 0 and 6 < x < 10
(C) 5 < x < 6
(D) 10 < x < 10
12
If total mechanical energy of the particle is 40J, then it can be found in region.
(A) x< - 10 and x > 15
(B) 10 < x < - 5 and 6 < x < 15
(C) 10 < x < 15
(D) is not possible
Paragraph for Questions 15 to 16
0
A rough inclined plane of inclined angle 53 has Coefficient of friction ( 0.2) with a block on
2
it of 2kg .Block is released from rest from the top of inclined plane ( g 10m / s )
15.
What is the work done by the friction to slide along the incline by 1m
(A) 2.4 J
(B) 8 J
(C) 2 J
(D) 6 J
16.
(A) 185 J/S
(B) 190 J/S
## (D) 120 J/S
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## Part - A (Matching List Type)
This section contains 4 multiple choice questions. Each question has matching lists. The codes for the
lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
17. ACB is a smooth quarter circular path of radius R. Four forces are acting at a particle placed at A.
F1 is always horizontal, F2 is always vertical, F3 is always tangential to the path, F4 is always
directed from particles position to point B. Magnitude of all forces are equal to F.
B
F2
F4
F3
F1
Column I
work done by F1 is
work done by F2 is
work done by F3 is
(Q)
(R)
work done by F4 is
(A) P2
(C) P 1
18.
Column II
F.R
F.R 2
FR
2
F.R
2
Q 1 R3 S 4
Q 1 R3 S 2
(B) P1 Q2 R3 S4
(D) P 4 Q3 R1 S2
Consider adjacent figure. All surfaces are smooth, thread & pulleys are ideal.
Then match the following:
Column I
(Q)
(R)
## Acceleration of any of the blocks
g
2 3 2
Acceleration of wedge
g 2 2
2 3 2
Normal reaction on any of the blocks
(A) P2
(C) P 2
Column II
Q 1 R3 S 4
Q 3 R1 S 4
g 2 2 1
2 3 2
3g
2 1
2 3 2
(B) P1 Q4 R2 S3
(D) P 4 Q1 R3 S1
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19.
Force acting on a block versus time and acceleration versus time graph are as shown in figure.
Talking value of g = 10 ms 2, match the following two columns.
Column I
Column II
Coefficient of static friction
0.2
(Q)
Coefficient of kinetic friction
0.3
(R)
Force of friction at t=0.1 s
0.4
a
0.5
Value of
, where a is acceleration of block at
10
t=8 s
(A) P3 Q 4 R2 S 1
(B) P1 Q4 R2 S3
(C) P 2 Q 3 R1 S 4
(D) P 3 Q2 R1 S4
20.
## In the diagram shown, match the following (g = 10m/s2)
Blocks are on smooth incline. F1 and F2 are parallel to the inclined plane. The motion of the
blocks is along the incline the surface.
Column I
Column II
Acceleration of 2kg block
39 SI unit
(Q)
Net force on 3kg block
25 SI unit
(R)
Normal reaction between 2kg and 1kg
2 SI unit
Normal reaction between 3kg and 2kg
6 SI unit
(A) P2 Q 1 R3 S 4
(B) P3 Q1 R2 S4
(C) P 2 Q 3 R1 S 4
(D) P 4 Q3 R1 S2
Multiple
1. ABC
2. ACD
7. BCD
Paragraph
9. C
15. A
10. A
16. A
Matrix Match
17. C
18. A
8. CD
4. AC
11. B
12.A
19.D
20. B
5. AB
13.A
14.D
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Pune-411037, ph: (020) 41418686 email:info.pune@fiitjee.com, www.fiitjee.com | 2,925 | 9,489 | {"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} | 3.21875 | 3 | CC-MAIN-2019-39 | latest | en | 0.912365 |
https://en.wikipedia.org/wiki/Multi-commodity_flow_problem | 1,709,510,195,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00624.warc.gz | 227,760,349 | 19,766 | # Multi-commodity flow problem
The multi-commodity flow problem is a network flow problem with multiple commodities (flow demands) between different source and sink nodes.
## Definition
Given a flow network ${\displaystyle \,G(V,E)}$, where edge ${\displaystyle (u,v)\in E}$ has capacity ${\displaystyle \,c(u,v)}$. There are ${\displaystyle \,k}$ commodities ${\displaystyle K_{1},K_{2},\dots ,K_{k}}$, defined by ${\displaystyle \,K_{i}=(s_{i},t_{i},d_{i})}$, where ${\displaystyle \,s_{i}}$ and ${\displaystyle \,t_{i}}$ is the source and sink of commodity ${\displaystyle \,i}$, and ${\displaystyle \,d_{i}}$ is its demand. The variable ${\displaystyle \,f_{i}(u,v)}$ defines the fraction of flow ${\displaystyle \,i}$ along edge ${\displaystyle \,(u,v)}$, where ${\displaystyle \,f_{i}(u,v)\in [0,1]}$ in case the flow can be split among multiple paths, and ${\displaystyle \,f_{i}(u,v)\in \{0,1\}}$ otherwise (i.e. "single path routing"). Find an assignment of all flow variables which satisfies the following four constraints:
(1) Link capacity: The sum of all flows routed over a link does not exceed its capacity.
${\displaystyle \forall (u,v)\in E:\,\sum _{i=1}^{k}f_{i}(u,v)\cdot d_{i}\leq c(u,v)}$
(2) Flow conservation on transit nodes: The amount of a flow entering an intermediate node ${\displaystyle u}$ is the same that exits the node.
${\displaystyle \forall i\in \{1,\ldots ,k\}:\,\sum _{w\in V}f_{i}(u,w)-\sum _{w\in V}f_{i}(w,u)=0\quad \mathrm {when} \quad u\neq s_{i},t_{i}}$
(3) Flow conservation at the source: A flow must exit its source node completely.
${\displaystyle \forall i\in \{1,\ldots ,k\}:\,\sum _{w\in V}f_{i}(s_{i},w)-\sum _{w\in V}f_{i}(w,s_{i})=1}$
(4) Flow conservation at the destination: A flow must enter its sink node completely.
${\displaystyle \forall i\in \{1,\ldots ,k\}:\,\sum _{w\in V}f_{i}(w,t_{i})-\sum _{w\in V}f_{i}(t_{i},w)=1}$
## Corresponding optimization problems
Load balancing is the attempt to route flows such that the utilization ${\displaystyle U(u,v)}$ of all links ${\displaystyle (u,v)\in E}$ is even, where
${\displaystyle U(u,v)={\frac {\sum _{i=1}^{k}f_{i}(u,v)\cdot d_{i}}{c(u,v)}}}$
The problem can be solved e.g. by minimizing ${\displaystyle \sum _{u,v\in V}(U(u,v))^{2}}$. A common linearization of this problem is the minimization of the maximum utilization ${\displaystyle U_{max}}$, where
${\displaystyle \forall (u,v)\in E:\,U_{max}\geq U(u,v)}$
In the minimum cost multi-commodity flow problem, there is a cost ${\displaystyle a(u,v)\cdot f(u,v)}$ for sending a flow on ${\displaystyle \,(u,v)}$. You then need to minimize
${\displaystyle \sum _{(u,v)\in E}\left(a(u,v)\sum _{i=1}^{k}f_{i}(u,v)\cdot d_{i}\right)}$
In the maximum multi-commodity flow problem, the demand of each commodity is not fixed, and the total throughput is maximized by maximizing the sum of all demands ${\displaystyle \sum _{i=1}^{k}d_{i}}$
## Relation to other problems
The minimum cost variant of the multi-commodity flow problem is a generalization of the minimum cost flow problem (in which there is merely one source ${\displaystyle s}$ and one sink ${\displaystyle t}$). Variants of the circulation problem are generalizations of all flow problems. That is, any flow problem can be viewed as a particular circulation problem.[1]
## Usage
Routing and wavelength assignment (RWA) in optical burst switching of Optical Network would be approached via multi-commodity flow formulas.
Register allocation can be modeled as an integer minimum cost multi-commodity flow problem: Values produced by instructions are source nodes, values consumed by instructions are sink nodes and registers as well as stack slots are edges.[2]
## Solutions
In the decision version of problems, the problem of producing an integer flow satisfying all demands is NP-complete,[3] even for only two commodities and unit capacities (making the problem strongly NP-complete in this case).
If fractional flows are allowed, the problem can be solved in polynomial time through linear programming,[4] or through (typically much faster) fully polynomial time approximation schemes.[5]
## Applications
Multicommodify flow is applied in the overlay routing in content delivery.[6]
## References
1. ^ Ahuja, Ravindra K.; Magnanti, Thomas L.; Orlin, James B. (1993). Network Flows. Theory, Algorithms, and Applications. Prentice Hall.
2. ^ Koes, David Ryan (2009). "Towards a more principled compiler: Register allocation and instruction selection revisited" (PhD). Carnegie Mellon University. S2CID 26416771.
3. ^ S. Even and A. Itai and A. Shamir (1976). "On the Complexity of Timetable and Multicommodity Flow Problems". SIAM Journal on Computing. SIAM. 5 (4): 691–703. doi:10.1137/0205048.Even, S.; Itai, A.; Shamir, A. (1975). "On the complexity of time table and multi-commodity flow problems". 16th Annual Symposium on Foundations of Computer Science (SFCS 1975). pp. 184–193. doi:10.1109/SFCS.1975.21. S2CID 18449466.
4. ^ Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein (2009). "29". Introduction to Algorithms (3rd ed.). MIT Press and McGraw–Hill. p. 862. ISBN 978-0-262-03384-8.{{cite book}}: CS1 maint: multiple names: authors list (link)
5. ^ George Karakostas (2002). "Faster approximation schemes for fractional multicommodity flow problems". Proceedings of the thirteenth annual ACM-SIAM symposium on Discrete algorithms. pp. 166–173. ISBN 0-89871-513-X.
6. ^ Algorithmic Nuggets in Content Delivery sigcomm.org
Add: Jean-Patrice Netter, Flow Augmenting Meshings: a primal type of approach to the maximum integer flow in a muti-commodity network, Ph.D dissertation Johns Hopkins University, 1971 | 1,629 | 5,697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 32, "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} | 3.328125 | 3 | CC-MAIN-2024-10 | latest | en | 0.756314 |
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Question
# To make the perfect shot and win the game, a pool player will need
to direct the eight ball at an angle of 45 degrees below the horizontal. In an attempt to make the perfect shot , the player lines up the cue ball and takes the shot. The cue ball with a mass of 0.17 kg teachers a velocity of 1.1 m/s at 0 degrees, just before it collided with the stationary eight ball with a mass of 0.16kg. Immediately after the collision, the cue ball then moves with velocity of 0.31m/s at 60 degrees. Determine whether or not the player will make the perfect shot and win the game. Include a vector addition diagram and use physics principles and formulas included.
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Subject: Physics, Science
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Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed! | 1,240 | 3,995 | {"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} | 3.03125 | 3 | CC-MAIN-2020-34 | latest | en | 0.512309 |
http://programmersheaven.com/discussion/262808/can-u-help-me-plz | 1,521,768,733,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00341.warc.gz | 233,988,750 | 11,706 | # Can u help me plz.
Hi there, 1st , sorry about my bad english but i'm from mexico ^^
here's the question ...
i'm doing a program to analize an expretion, like ((44/12)*15)
when the program starts , it asks the user to 'press' the correct order of the equation(without the parentesis) 1 by 1, so everytime a caracter is readed , it's temp. saved on a variable (type string).
i can use the integer values saved there, but i can not do operations like plus, div, etc.., so dats my question ,
how can i assing integer values saved on a string variable into a string variable
example:
var1:string;
var2:integer;
write('character? ');
readln(var1); {i need to read it on a string variable because of the simbols , + , ^ , * , ... etc }
var2:=var1; {obviously,here's when error 25 appears}
i hope some1 can help me with this maybe 'dumb' question,
^^ thanx anyways , because of this excellent website
• : Hi there, 1st , sorry about my bad english but i'm from mexico ^^
:
: here's the question ...
:
: i'm doing a program to analize an expretion, like ((44/12)*15)
: when the program starts , it asks the user to 'press' the correct order of the equation(without the parentesis) 1 by 1, so everytime a caracter is readed , it's temp. saved on a variable (type string).
: i can use the integer values saved there, but i can not do operations like plus, div, etc.., so dats my question ,
:
: how can i assing integer values saved on a string variable into a string variable
:
: example:
:
: var1:string;
: var2:integer;
:
: write('character? ');
: readln(var1); {i need to read it on a string variable because of the simbols , + , ^ , * , ... etc }
:
:
: var2:=var1; {obviously,here's when error 25 appears}
:
:
: i hope some1 can help me with this maybe 'dumb' question,
:
: ^^ thanx anyways , because of this excellent website
:
:
:
:
hi!
I find your question interesting,although I can`t find a solution for your problem.
The error is in this line:var2:=var1;
It is simply not possible that you assign an integer variable with a string value.It a type mischmash.
I suggest that you store the nimbers in integer variables and the sybmols'+','-',etc in strings.then you can use if-then or case to comply the operation requested.
Looking forward to hearing from you.
li-zi
• : Hi there, 1st , sorry about my bad english but i'm from mexico ^^
:
: here's the question ...
:
: i'm doing a program to analize an expretion, like ((44/12)*15)
: when the program starts , it asks the user to 'press' the correct order of the equation(without the parentesis) 1 by 1, so everytime a caracter is readed , it's temp. saved on a variable (type string).
: i can use the integer values saved there, but i can not do operations like plus, div, etc.., so dats my question ,
:
: how can i assing integer values saved on a string variable into a string variable
:
: example:
:
: var1:string;
: var2:integer;
:
: write('character? ');
: readln(var1); {i need to read it on a string variable because of the simbols , + , ^ , * , ... etc }
:
:
: var2:=var1; {obviously,here's when error 25 appears}
:
:
: i hope some1 can help me with this maybe 'dumb' question,
:
: ^^ thanx anyways , because of this excellent website
If var1 is going to have non-numerical characters (like +, ^, *, etc), then you'll have to write a function to split that string so you can separate the numbers from the operators.
Then to convert a string containing a number to an integer, you can use the Val() function. | 933 | 3,480 | {"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} | 2.625 | 3 | CC-MAIN-2018-13 | latest | en | 0.794503 |
http://hashtaginfosolution.in/whats-the-device-of-cost-in-physics-lessons-p/ | 1,713,987,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00734.warc.gz | 17,317,244 | 31,101 | Posted on : by :
# What’s the Unit Of Cost In Psychology Lessons?
So that you wish to take a component of control? That’s fine. You may grow to be a physicist, if you analyze the machine of measure at Berkeley Physics courses.
The system of measure is popularly known as the unit of energy or induce. The machine of energy is described as being”positive” summarizing and paraphrasing helps the writer in a few components and”negative” in others. Certain means to function as energy or to be an induce. “Positive” implies that it has been added into another person.
The present energy of one can be converted by An individual . The energy that was present must be altered to do the job. Carrying out work does this conversion.
Do the job is a thing is inserted to something else. That’s all there is to it. This is the system of meter in statistics. paraphrasingservice com This will be the unit of force in physics.
The meter is utilised to express exactly the very same amount as Newton’s m.a. O. The meter is the system of power in physics. The system of meter is not a great component of power.
What’s rust in math? Should I have some thing on my own hand, does it look like an ongoing? It’s the flow of the drive. Because it resembles a current, it truly is called the flow of the pressure. It is energy in physics.
Even the person that holds upward the thing in the air, looks like it is a gift? Certainly. The thing is energy in physics. Has to be altered to induce.
One has to make utilize of the drive of the thing to take it, to get this done. This is to convert electricity into work. It’s mandatory that you interpret energy and use the work to translate .
Gravity holds the object. Gravity is a force which holds this up from the air. It’s this particular force that you uses to change energy.
Gravity is just one of those basics in math. http://www4.ncsu.edu/~dsbeckma/222scoringrubric.html It’s the reason behind repulsion and fascination. It causes also the push of repulsion and the flow of attraction. It also is the reason behind movement. It is also the basis for heat.
This one force is the one that keeps the item and allows it to hang from the atmosphere at 1 place while allowing it to hang from the atmosphere at another place. One has to find an easy method to interpret this pressure into power and use the electricity in 1 location to create the identical force and heat in another place. This can be the fashion.
The gap between operate and one’s present energy could be that the transformation of their present energy . The conversion of electricity to function is how one gets a physicist. That is what the machine of vitality Berkeley Physics paths is. This is actually the conversion from vitality to do the job . | 597 | 2,758 | {"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} | 3.65625 | 4 | CC-MAIN-2024-18 | latest | en | 0.953564 |
https://quizizz.com/en-in/division-with-unit-fractions-worksheets-class-6?page=1 | 1,726,347,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00645.warc.gz | 441,029,230 | 28,467 | Mixed Number to Improper Fraction & vice-versa
15 Q
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Unit 2 Review (Fractions and Decimals)
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11/17 AM: Module 4 Review
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5.NF.7 Dividing Unit Fractions and Whole Numbers
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Rates, Ratios, Unit Rates, and more! (2023)
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Ratios Vocabulary Quiz
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1st Nine Weeks Vocab
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Jeopardy form 1.3
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Unit 3- Fraction Operation Vocabulary
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Game Master - Math Vocabulary 29
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Division with Fractions and Mixed numbers
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Fraction Division
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Fraction Division
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Fraction Division
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Multiplying and Dividing Fractions Unit Test
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Unit 3- Division of Fractions
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Unit 1 Vocabulary Part 1
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Skills and Strategies Match Up
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Ratio and rate pre-assessment
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Practice quizz (10.12.2020)
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10/13 warm up- vocab
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PA 2 Assessment- Halves and Quarters
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Unit 8 Fraction Multiplication & Division
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## Explore printable Division with Unit Fractions worksheets for 6th Class
Division with Unit Fractions worksheets for Class 6 are an essential resource for teachers looking to help their students master the concept of fractions in math. These worksheets provide a variety of problems that focus on the skills required for multiplying and dividing fractions, ensuring that students have a solid foundation in this crucial area of mathematics. By incorporating these worksheets into their lesson plans, teachers can provide ample opportunities for students to practice and apply their knowledge of fractions, ultimately leading to a deeper understanding of the subject matter. With a range of difficulty levels and problem types, Division with Unit Fractions worksheets for Class 6 cater to the diverse needs of students, making them an invaluable tool for any math teacher.
In addition to worksheets, Quizizz offers a comprehensive platform for teachers to create engaging and interactive assessments that can be used alongside Division with Unit Fractions worksheets for Class 6. With Quizizz, teachers can easily design quizzes and games that test students' understanding of math concepts such as fractions, multiplication, and division. The platform also provides valuable data and insights on student performance, allowing teachers to identify areas where students may need additional support or practice. By incorporating Quizizz into their teaching strategies, educators can create a dynamic and effective learning environment that not only reinforces the skills taught through worksheets but also encourages students to take an active role in their education. With its diverse offerings and user-friendly interface, Quizizz is an excellent resource for teachers looking to enhance their instruction and assessment of math concepts like fractions. | 696 | 2,929 | {"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} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.923975 |
https://discuss.leetcode.com/topic/67900/dp-solution-in-python | 1,516,568,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890874.84/warc/CC-MAIN-20180121195145-20180121215145-00465.warc.gz | 650,101,577 | 8,181 | # DP solution in Python
• ``````class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
if m > 1 and n > 1:
res = [[0 for p in range(n)] for q in range(m)]
for i in range(1, m):
res[i][0] = 1
for j in range(1, n):
res[0][j] = 1
for i in range(1, m):
for j in range(1, n):
res[i][j] = res[i][j - 1] + res[i - 1][j]
return res[m - 1][n - 1]
elif m == 0 or n == 0:
return 0
else:
return 1
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 202 | 533 | {"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} | 2.59375 | 3 | CC-MAIN-2018-05 | latest | en | 0.768102 |
https://physics.stackexchange.com/questions/674147/how-should-be-the-charges-get-distributed-in-this-question | 1,718,321,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00353.warc.gz | 420,432,568 | 39,517 | # How should be the charges get distributed in this question?
My teacher gave us a question today to just draw the charge distribution. The question goes like this
Suppose we have a neutral thick metallic shell of inner radius $$r$$ and outer radius $$2r$$. Now at distance of $$\frac{r}{2}$$ from its center , a positive charge of magnitude $$Q$$ is placed. How are charges induced (in terms of uniform and non - uniform distribution) on the inner and outer surface of the shell ?
He then explained that the charge on the inner surface of the shell is non uniform since the external charge is not at the center and I think it was quite intuitive but for the outer shell he suggested that the charge distribution is uniform but this time he didn't explain why he said this.
So please explain why the charge distribution should be uniform on the outer surface of the thick shell ?
• physics.stackexchange.com/q/505977/247580 hope this helps Commented Oct 28, 2021 at 18:19
– hft
Commented Oct 28, 2021 at 19:25
So please explain why the charge distribution should be uniform on the outer surface of the thick shell ?
Because the sphere is a metal, the outer surface will be an equipotential surface. The electric potential is the same constant value over the entire outer surface (and at the inner surface and within the metal too).
There is no additional charge outside the outer surface of the sphere, therefore, in that region, the electric potential $$\phi$$ must satisfy the Laplace equation: $$-\nabla^2 \phi = 0 \quad (\text{outside the sphere})$$
Now we see that, outside the sphere, we have a spherically symmetric equation with spherically symmetric boundary conditions.
We already know one spherically symmetric solution to the Laplace equation: $$\phi(r) \propto \frac{1}{r}\quad (\text{outside the sphere})$$
This solution can be made to satisfy the boundary conditions by choosing $$\phi(r) = \frac{Q}{r}\quad (\text{outside the sphere, gaussian units})\;,$$ or in SI units, $$\phi(r) = \frac{Q}{4\pi \epsilon_0 r}\quad (\text{outside the sphere, SI units})\;.$$
Because the solution to the Laplace equation is unique, this must be the correct solution.
The associated electric field outside the sphere is $$\vec E(r) = \hat r \frac{Q}{4\pi \epsilon_0 r^2}\quad (\text{outside the sphere, SI units})\;.$$
The charge distribution on the outer side of shell can be found by using the pillbox method. I'll leave that as an exercise, but it should be pretty clear that this charge distribution is also uniform on the outer surface since the field is spherically symmetric outside and zero inside. | 621 | 2,619 | {"found_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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-26 | latest | en | 0.920255 |
https://www.sexyloops.co.uk/theboard/viewtopic.php?p=15017 | 1,620,406,735,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00528.warc.gz | 1,033,841,538 | 14,617 | PLEASE NOTE: In order to post on the Board you need to have registered. To register please email paul@sexyloops.com including your real name and username. Registration takes less than 24hrs, unless Paul is fishing deep in the jungle!
## Leader Butt Diameter
Moderator: Torsten
gordonjudd
Posts: 1411
Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California
### Leader Butt Diameter
If I remember well, the leader diameter at the junction should be 85% of the line diameter. Look after post 39 by Gordon Judd in the thread "leader hinging in slow motion"
Merlin,
If the aim is to match the leader butt's impedance to the impedance of a floating fly line then the ratio will depend on the type of leader material as well. 85% is a rule of thumb for nylon leaders that have a specific gravity of 1.14. If you use fluorcarbon then its higher specific gravity of 1.78 would reduce that matching impedance diameter to around 2/3 of the line diameter as shown below.
As Jack Moray explained in his leader design approach (that was mentioned by Phillip in the hinging thread) to get the best energy transfer in the leader you want to minimize the difference in the diameter squared ratios from section to section. That generally requires smaller step sizes in the butt and larger step sizes near the tippet end. That approach is opposite of what you see in most leader formulas. The above "ideal" butt diameters are also much larger than the typical diameters used in commercial leaders.
Here is an example of a Moray design where you are trying to get equal step sizes in the linear mass density changes from section to section in a stepped leader.
Producing those linear mass density values in a Nylon 6 leader material would use these diameter values where the small changes in diameter are at the butt and the large steps are at the tippet end as shown below.
However as cautioned in the other thread:
Since we are trying to smooth out the layout of the fly getting a maximum transmission of the wave energy through the leader may not be ideal from a fishing standpoint so all of this may be academic.
Gordy
John Waters
Posts: 1039
Joined: Mon Jan 14, 2013 9:16 pm
### Leader Butt Diameter
I remember many lengthy and instructive discussions with Jack Morey, a great fly fisherman and innovative designer. His use of relative specific gravity and leaders tapered by weight, not diameter and modified for stiffness have underpinned my leader designs for both fishing and casting. For example, his 9 feet leader tapered to 4X, for a 5 weight floating line with 6 inches of level line tip, started with 44 inches of 0.024 inch diameter mono. For the same line with an 18 inch level tip on the flyline he used 0.022 inch mono. Some like to remove the level line section from their line, others leave it on. I remove it.
Turnover is important for tournament casting, so why not for fishing.
John
gordonjudd
Posts: 1411
Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California
### Leader Butt Diameter
His use of relative specific gravity and leaders tapered by weight, not diameter
John,
His taper depends on the linear mass density of each section, which depends on the volume mass density of the material and the square of the diameter of a given section. (i.e., Rho_l=rho*area) Thus his design relies on both the specific gravity of the material and the diameter squared of each section.
I think he based the shape of the normalized rho_l vs length curve on his experience with different tapers as Phillip and I modified his stated procedure a bit to come up with section lengths that matched the ones he actually used in some of his examples.
Nevertheless his recommendation that the taper diameters be set based on the the square of the diameter of the diameters rather than just the diameter was new to me. From an impedance standpoint that makes perfect sense in order to maximize the energy transmission down the leader if that is your goal.
I wish that I would have had the opportunity to interact with him, as I think he was before his time in regards to applying some physics to the design of leaders.
and modified for stiffness have underpinned my leader designs for both fishing and casting.
Do you know how the bending stiffness of each section was taken into account in order to modify the values given by equal steps in the linear mass density of each section? I would think that would be another important factor, but the above example for a 7wt line did not take bending stiffness into account.
Gordy
Paul Arden
Posts: 14688
Joined: Thu Jan 03, 2013 11:20 am
Location: Belum Rainforest
Contact:
### Leader Butt Diameter
Thanks guys I didn't know about stepping up leader diameter with sinking lines. I must have missed that!
Cheers, Paul
It's an exploration; bring a flyrod.
Flycasting Definitions
gordonjudd
Posts: 1411
Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California
### Leader Butt Diameter
Thanks guys I didn't know about stepping up leader diameter with sinking lines. I must have missed that!
Paul,
You are in good company. I don't think that most people think about the linear mass density matching of line to leader when it comes to sizing a leader butt diameter. The importance of how linear mass density impacts the dynamics of a flyline does not receive the attention it deserves.
The rho_l and diameter of the sinking line will depend on its sink rate (which is associated with its specific gravity) but here is what the matching curve would like for a fast sink line having a specific gravity of 2.25 assuming it would meet the AFTMA design standards for the mass of 30 feet of line.
I don't know how much trouble you might have in connecting a large diameter mono line to a smaller diameter sinking line, so there may be some problems in actually matching up the rho_l of the leader butt to the tip of the sinking line as discussed in this article at the CastFlys.com site.
Gordy
Merlin
Posts: 1575
Joined: Wed Jan 09, 2013 8:12 pm
Location: France
### Leader Butt Diameter
I repeat here the post I made on the other thread, for the consistency of this one.
There are two important parameters for line flight which are the diameter of the line and its « linear density » (in other words the weight by unit of length). The higher they are, the better for the roll over.
Since the loop can be considered as a wave, Gordy explained that the transmission of energy will be maintained if the linear density is conserved as the material change (string wave theory). The linear density is the product of the volumetric density (usually named the density) by the section area of the line. If the density of the material changes for a higher one, then the diameter must decrease. The section area of the line and the leader being proportional to the square of their diameter, the calculation is easy.
Let’s say my line is has a 2 mm diameter. I take 1.2 g/cm3 for the leader and 0.85 g/cm3 for the floating line. Then the diameter of my leader should be equal to the square root of (0.85*2^2)/(1.2) = 1.68 mm. In such condition the linear density of the line is the same as the linear density of the leader. I made the calculation for you; it is equal to 2.67 g/m.
The thing is although I maintain the linear density, I cannot keep the same diameter, so for line flight, I am going to lose something in terms of air drag.
If the situation is reversed, for example if you use a sinking line, then the leader diameter should be larger than the line diameter, because of the respective values of density of the materials.
Ben’s remark appears to be a result from experience matching the theoretical expectations for sinking lines, I wonder if other people have the same experience since it is quite unusual, as noted by Paul.
Gordy
I’m not sure to follow the logic of Jack Moray : where do the guideline curves come from? A best guess? Something theoretical? Since leader density is a constant (if you don't mix materials), then I cant' see impedance conservation here. OK, I'm not going to open an issue on "COI" (conservation of inpedance), it is just to understand the logic behind.
Merlin
Fly rods are like women, they won't play if they're maltreated
Charles Ritz, A Flyfisher's Life
Bryan
Posts: 6
Joined: Mon Mar 10, 2014 8:52 am
### Leader Butt Diameter
Enlightening, indeed.
So, then the challenge of how to attach a larger diameter leader to the end of a sinking line? Reading the article and watching the videos at the Castflys.com website, raises a couple of initial quesitons:
1. The twisted leader solution to the larger diameter leader conundrum relies on a loop to loop connection. Won't that itself cause hinging and offset (or worse) the benefit of using a larger diameter leader butt?
2. The No Name Knot used to attach the twisted leader to the tippet seems like a one-shot deal. Once you've tied the tippet to the loop end of the twisted leader using that knot, it seems you can't replicate it on that twisted leader---the loop end has been used. So, what next when you have to replace your tippet? Cut the tippet some distance away from the No Name Knot and attach new tippet? Or, use a new twisted leader?
I appreciate the input folks are putting in here. Improving my leader design and construction is something I'm keenly interested in learning more about.
Thank you.
Cheers,
Bryan
Paul Arden
Posts: 14688
Joined: Thu Jan 03, 2013 11:20 am
Location: Belum Rainforest
Contact:
### Leader Butt Diameter
I suppose with the higher density lines it makes more sense to use a polyleader. But then you have the cursed loop to loop connection again. Lots to think about!
Thanks, Paul
It's an exploration; bring a flyrod.
Flycasting Definitions
gordonjudd
Posts: 1411
Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California
### Leader Butt Diameter
I’m not sure to follow the logic of Jack Moray : where do the guideline curves come from? A best guess? Something theoretical?
Merlin,
I think his normalized curve shape came from experience. As I said in order to get section lengths that matched up with his examples I had to use the "experience-based" normalized curve rather than strictly following the iteration procedure he spelled out in his method. He was doing his calculations with a calculator and graph paper, so cutting corners on calculations might have been a practical necessity. Also he may have found that to get a better lay out of the fly he lengthened the section lengths of the tippet end of the leader to eliminate the splash down effect when fishing dry flies.
Good tournament casters will lengthen (or shorten) the tippet on their dry fly leaders to get the fly to "hover" over the ring at the end of the cast so that they can better judge distance. A 1-2" difference in tippet length will make a big difference in how the fly hovers for them, but I was never that sophisticated. Thanks to John Napoli at the Long Beach Casting Club for showing me that adjustment trick many years ago.
I would expect that based on practical experience Jack Moray used different normalized curve shapes for different situations as well. I have noticed that Rajeff's Bass Bug leader design has a convex slope from beginning to end to get a good turn over of the bug. Harvey used a S-shaped curve with longer flatter transitions at the tippet end. That tends to produce wiggles in the layout of a long leader to help reduce the drag on dry flies. I expect Moray did the same thing, i.e. developed different normalized rho_l vs distance curves for different types of fishing situations. John may have some examples of the different curves he used.
Since leader density is a constant (if you don't mix materials), then I cant' see impedance conservation here.
If you assume the velocity of the wave at the boundaries is the same and knowing that the Tension is proportional to rho_l*v.^2 then the impedance will change whenever you change the linear mass density as shown below.
http://img12.imageshack.us/img12/6982/impedanceequation.jpg
Thus the relative impedance for two adjacent sections will be proportional to their rho_l values which will be proportional to the square of their diameters for a leader made with the same material.
Moray's point was it is better to have a series of nominally equal impedance ratios in each of the transmission loss values rather than having big losses that would result from big impedance ratio differences at the butt end of the leader followed by smaller impedance ratios as you get to the tippet end of the leader.
That approach is in keeping with the guideline given in the Harvard transverse wave paper where the emphasis is on the gradual change in the impedance difference between each section.
The two basic ways to match two impedances are to (1) simply make one of them equal
to the other, or (2) keep them as they are, but insert a large number of things (whatever
type of things the two original ones are) between them with impedances that gradually
change from one to the other.
Since the linear mass density varies as the square of the diameter that means you have smaller diameter steps at the butt end of the leader than you do at the tippet end. That logic makes some sense to me, and is counter to the prescription given in most leader formulas where the big steps are at the butt end followed by smaller steps at the tippet end.
OK, I'm not going to open an issue on "COI" (conservation of inpedance), it is just to understand the logic behind.
You can breathe a sigh of relief. Just as COM does not apply to the predicting the fly velocity history in a normal tethered cast, there is no COI going on with Moray's leader design approach either.
Gordy
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Posts: 2950
Joined: Thu Jan 10, 2013 12:04 pm
### Leader Butt Diameter
Gordy
in your assumptions, are you assuming that the fly line/leader is a string wave?
Bright but shite
IANACI - There’s no such thing as absolutes
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# A, B C and D are four points in a plane with position vectors, veca, vecb vecc and vecd respectively, such that (veca-vecd).(vecb-vecc)= (vecb-vecd).(vecc-veca)=0 then point D is the ______ of triangle ABC.
Question from Class 11 Chapter Different Products Of Vectors And Their Geometrical Applications
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Answer :
orthocenter
Solution :
Given that `veca, vecb , vecc and vecd` are position vectors of points A,B,C and D, respectively, such that <br> `(veca - vecd) . (vecb - vecc) = (vecb.vecd) . (vecc- veca) =0` <br> `Rightarrow vec(DA).vec(CB) = vec(DB).vec(AC) =0` <br> ` Rightarrow vec(DA) bot vec(CB) and vec(DB) bot vec(AC)` <br> Clerly, D is the orthocentre of `triangleABC` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_V_3DG_C02_E01_286_S01.png" width="80%"> | 340 | 1,026 | {"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} | 3.203125 | 3 | CC-MAIN-2020-24 | latest | en | 0.484868 |
https://fosterjee.com/topic/tautology/ | 1,660,734,940,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00688.warc.gz | 261,646,764 | 30,009 | ### Mathematics Class XI
Unit-I: Sets and Functions
Chapter 1: Sets
Unit-II: Algebra
Chapter 5: Binomial Theorem
Chapter 6: Sequence and Series
Unit-III: Coordinate Geometry
Chapter 1: Straight Lines
Chapter 2: Conic Sections
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability
Chapter 1: Statistics
Chapter 2: Probability
# Tautology
If a composite proposition p is always true for all possible assignment of truth values to its prime components, then it is called tautology.
The following are some examples of tautology:
1. (Law of excluded middle)
2. (Law of contradiction)
3. (Law of double negation). (The above 3 tautologies are widely known laws of classical logic.)
4. (principle of syllogism)
5. (law of contrapositive)
If a condition is a tautology, then we say that p implies q i.e .
Truth table
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http://newbusinessmath.com/2017/06/ | 1,500,568,150,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423269.5/warc/CC-MAIN-20170720161644-20170720181644-00163.warc.gz | 237,879,529 | 10,964 | # Monthly Archive: June 2017
## Use Of Common Deviation In Business How To Calculate Normal Deviation Applying Microsoft Excel
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A nation with a trade deficit will practical experience a reduction in its foreign exchange reserves , which ultimately lowers (depreciates) the value of its currency. Reading and calculating exchange rates aren’t incredibly complicated, but small errors can lead to major errors in some circumstances.… | 1,898 | 9,690 | {"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} | 2.96875 | 3 | CC-MAIN-2017-30 | latest | en | 0.952611 |
http://ilcasarosf.com/reading-comprehension-kindergarten-worksheets-free/20-luxury-kindergarten-reading-comprehension-worksheets-worksheet/ | 1,531,680,563,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676588961.14/warc/CC-MAIN-20180715183800-20180715203800-00597.warc.gz | 173,458,438 | 9,906 | # Reading Comprehension Kindergartens Free New
By Jill J. Williams on July 12 2018 04:53:09
On this page you will find: a complete list of all of our math worksheets relating to algebra. Choose a specific picture topic below to view all of our worksheets in that content area. You will find addition lessons, worksheets, homework, and quizzes in each section.
If the answers numerator is greater than the denominator, then the answer is an improper fraction. This fraction should be turned into a proper fraction by taking wholes out of the numerator until the numerator is less than the denominator.
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If a child is really confused about decimals, converting decimal numnbers into money is a great way to make things clearer. For example: a child may be asked to say how much bigger 1.3 is than 0.9. If they convert these decimals into money (£1.30 and 90p) they may find that they can do this calculation very quickly in their head, getting the answer 40p which they convert back into the decimal, 0.4. Demonstrating that money maths depends on decimal understanding is also an easy way to prove that decimals are actually useful in real life! | 390 | 1,897 | {"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} | 2.734375 | 3 | CC-MAIN-2018-30 | latest | en | 0.94235 |
https://us.metamath.org/nfegif/aev.html | 1,716,041,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00403.warc.gz | 530,867,402 | 3,661 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > aev Unicode version
Theorem aev 1991
Description: A "distinctor elimination" lemma with no restrictions on variables in the consequent. (Contributed by NM, 8-Nov-2006.)
Assertion
Ref Expression
aev
Distinct variable group: ,
Proof of Theorem aev
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 hbae 1953 . 2
2 ax10lem5 1942 . . 3
3 ax-8 1675 . . . 4
43spimv 1990 . . 3
52, 4syl 15 . 2
61, 5alrimih 1565 1
Colors of variables: wff setvar class Syntax hints: wi 4 wal 1540 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 This theorem depends on definitions: df-bi 177 df-an 360 df-tru 1319 df-ex 1542 df-nf 1545 This theorem is referenced by: ax16ALT2 2048 a16gALT 2049
Copyright terms: Public domain W3C validator | 402 | 994 | {"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} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.348637 |
https://utzx.com/units/convert-6-75-inches-to-cm/ | 1,726,386,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00045.warc.gz | 545,892,199 | 22,595 | # Convert 6.75 Inches to CM
## How much cm is equal to 1 inch?
cminchesconverter is a useful converter tool for conversion that will help you achieve inch to cm conversion.
We all are aware that centimeters and inches are units used to measure length.
Before you calculate the ratio of 6.75 inches in cm, you must first know the amount of one inch to centimeters
Centimeters, also known as centimetres, is the measurement unit used to measure length in metric systems. English symbols are abbreviated as cm. The meter is internationally defined as an SI unit, while the centimeter is not. But one centimeter is one hundredth of meter. It is also approximately 39.37 inches.
## Inch Definition
An inch is an Anglo-American unit of length measurement. The symbol is in. In many European local languages, “inch” can be used interchangeably with or derived from “thumb”. Because the thumb of a human is about one-inch in width.
• Electronic components like the dimensions of the PC screen.
• Dimensions of car/truck tires.
## How Do I Change 6.75 inches into centimeters?
You can solve any problem in this inches to cm formula.
We can directly utilize the formula to calculate 6.75 in to cm.
1 inch = 2.54 cm
Below is an example to aid you understand this better.6.75 inches to cm= 2.54 × 6.75 = 17.145 cm.
inches cm 6.55 inches 16.637 cm 6.575 inches 16.7005 cm 6.6 inches 16.764 cm 6.625 inches 16.8275 cm 6.65 inches 16.891 cm 6.675 inches 16.9545 cm 6.7 inches 17.018 cm 6.725 inches 17.0815 cm 6.75 inches 17.145 cm 6.775 inches 17.2085 cm 6.8 inches 17.272 cm 6.825 inches 17.3355 cm 6.85 inches 17.399 cm 6.875 inches 17.4625 cm 6.9 inches 17.526 cm 6.925 inches 17.5895 cm | 481 | 1,691 | {"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} | 3.28125 | 3 | CC-MAIN-2024-38 | latest | en | 0.871944 |
https://conversioncalculator.org/tag/grams-to-cups-flour-south-africa/ | 1,685,359,443,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00593.warc.gz | 209,530,747 | 46,972 | ## Grams To Cups Flour
1 grams = 0.01 cups = ¼ tsp cups = grams ÷ 125 Since one cup of flour is equal … | 35 | 104 | {"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} | 2.890625 | 3 | CC-MAIN-2023-23 | longest | en | 0.786345 |
https://www.physicsforums.com/threads/partial-derivative-of-a-square-root.796244/ | 1,653,460,891,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662580803.75/warc/CC-MAIN-20220525054507-20220525084507-00061.warc.gz | 1,016,672,232 | 15,373 | # Partial derivative of a square root
Hi,
I'm using partial derivatives to calculate propagation of error. However, a bit rusty on my calculus.
I'm trying to figure out the partial derivative with respect to L of the equation:
2pi*sqrt(L/g)
(Yep, period of a pendulum). "g" is assumed to have no error. I know I can use the chain rule...
so, 2pi*(L/g)^(1/2) --> 2pi*1/2*(L/g)^(-1/2) , or pi*(L/g)^(-1/2).
I am doing this correctly? Or did I just take the derivative (and not the partial derivative)?
Thanks!
## Answers and Replies
SteamKing
Staff Emeritus
Homework Helper
Hi,
I'm using partial derivatives to calculate propagation of error. However, a bit rusty on my calculus.
I'm trying to figure out the partial derivative with respect to L of the equation:
2pi*sqrt(L/g)
(Yep, period of a pendulum). "g" is assumed to have no error. I know I can use the chain rule...
so, 2pi*(L/g)^(1/2) --> 2pi*1/2*(L/g)^(-1/2) , or pi*(L/g)^(-1/2).
I am doing this correctly? Or did I just take the derivative (and not the partial derivative)?
Thanks!
Partial derivative of what w.r.t. L? You haven't provided an equation, just an expression.
2pi*sqrt(L/g) = T, which is a function of L.
SteamKing
Staff Emeritus
Homework Helper
Rewrite as T = 2π*(L/g)1/2. Does that give you any ideas? Remember, g is a constant.
Using the chain rule, I can bring down the 1/2 and subtract 1 from the exponent, so
dL/dT = 1/2*2π*(L/g)-1/2 or dL/dT = π*(L/g)-1/2
Though, now it seems that I'm not treating "g" as a constant.
SteamKing
Staff Emeritus
Homework Helper
Using the chain rule, I can bring down the 1/2 and subtract 1 from the exponent, so
dL/dT = 1/2*2π*(L/g)-1/2 or dL/dT = π*(L/g)-1/2
Though, now it seems that I'm not treating "g" as a constant.
Well, factor g out of the square root before taking the derivative.
Technically, you are not using the chain rule. You are using the power rule.
Fredrik
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# Rotational Motion
## Physics of Rotational Motion
The laws and equations that govern nature and natural phenomena are described by physics. One prime focus of physics is the study of motion. We have dealt in detail about translational motion (objects that move along a straight or curved line) in the previous chapters, and now we will expand our view towards other types of motions as well.
We see rotational motion in almost everything around us. Every machine, all celestial bodies, most of the fun games in amusement parks and if you are a FIFA fan, and when you watch David Beckham’s familiar shot, the ball is actually executing rotational motion.
Objects turn about an axis. All the particles and the mass center do not undergo identical motions. All the particles of the body undergo identical motion. By definition, it becomes essential for us to explore how the different particles of a rigid body move when the body rotates.
## Rotational Kinematics
In rotational kinematics, we will investigate the relation between kinematical parameters of rotation. We shall now revisit angular equivalents of the linear quantities: position, displacement, velocity and acceleration, which we have already dealt in a circular motion.
Linear Kinematic Parameters Angular Kinematic Parameters Position s Angular position θ Displacement $$\begin{array}{l}\Delta s={{s}_{1}}-{{s}_{2}}\end{array}$$ Angular displacement $$\begin{array}{l}\Delta \theta ={{\theta }_{1}}-{{\theta }_{2}}\end{array}$$ Average velocity $$\begin{array}{l}{{v}_{avg}}=~\frac{\Delta s}{\Delta t}\end{array}$$ Average angular velocity $$\begin{array}{l}{{\omega }_{avg}}=~\frac{\Delta \theta }{\Delta t}\end{array}$$ Instantaneous velocity $$\begin{array}{l}{{v}_{ins}}\,\underset{\triangle t\to 0}{\mathop{\lim }}\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt}\end{array}$$ Instantaneous angular velocity $$\begin{array}{l}\underset{\Delta t\to 0}{\mathop{{{\omega }_{ins}}\lim }}\,\frac{\Delta \theta }{\Delta t}=\frac{d\theta }{dt}\end{array}$$ Average acceleration $$\begin{array}{l}{{a}_{avg}}=~\frac{\Delta v}{\Delta t}\end{array}$$ Average angular acceleration $$\begin{array}{l}{{\alpha }_{avg}}=~\frac{\Delta s}{\Delta t}\end{array}$$ Instantaneous acceleration $$\begin{array}{l}{{a}_{ins}}\,\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta v}{\Delta t}=\frac{dv}{dt}\end{array}$$ Instantaneous angular acceleration $$\begin{array}{l}{{\alpha }_{ins}}\,\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta \omega }{\Delta t}=\frac{d\omega }{dt}\end{array}$$
A case of constant angular acceleration is of great importance, and a parallel set of equations holds for this case just as in constant linear acceleration.
Linear Equations of Motion Angular Equations of Motion $$\begin{array}{l}v = v_{o} + at\end{array}$$ $$\begin{array}{l}\omega ={{\omega }_{0}}+\alpha t\end{array}$$ $$\begin{array}{l}x-{{x}_{0}}=~{{v}_{0}}+\frac{1}{2}a{{t}^{2}}\end{array}$$ $$\begin{array}{l}\theta -{{\theta }_{0}}=~{{\omega }_{0}}+\frac{1}{2}\alpha {{t}^{2}}\end{array}$$ $$\begin{array}{l}{{v}^{2}}=v_{0}^{2}+2a\left( x-{{x}_{0}} \right)\end{array}$$ $$\begin{array}{l}{{\omega }^{2}}=\omega _{0}^{2}+2\alpha \left( \theta -{{\theta }_{0}} \right)\end{array}$$
## Axis of Rotation
A rigid body of an arbitrary shape in rotation about a fixed axis (axis that does not move) called axis of rotation or rotation axis is shown in the below figure.
### Types of Motion involving Rotation
1. Rotation about a fixed axis (Pure rotation)
2. Rotation about an axis of rotation (Combined translational and rotational motion)
3. Rotation about an axis in the rotation (rotating axis – out of the scope of JEE)
#### Rotation About a Fixed Axis
Rotation of a ceiling fan, opening and closing of the door, rotation of our planet, rotation of hour and minute hands in analogue clocks are a few examples of this type.
#### Rotation about an axis of rotation
Rolling is an example of this category. Arguably, the most important application of rotational physics is in the rolling of wheels and wheel-like objects as our world now is filled with automobiles and other rolling vehicles.
The Rolling Motion of a body is a combination of both translational and rotational motion of a round-shaped body placed on a surface. When a body is set in a rolling motion, every particle of the body has two velocities – one due to its rotational motion, and the other due to its translational motion (of the centre of mass), and the resulting effect is the vector sum of both velocities at all particles.
There is a wheel that rotates with an angular acceleration which is given by α = 4at3 — 3bt2, where t is the time and a and b are constants. If the wheel has an initial angular speed ω0, what are the equations for (a) angular speed (b) angular displacement.
1.
$$\begin{array}{l}\alpha = \frac{d\omega}{dt} \Rightarrow d\omega = \alpha dt\end{array}$$
$$\begin{array}{l}\Rightarrow \int_{\omega_{o}}^{\omega} = \int_{o}^{t} \alpha dt = \int_{o}^{t} (4at^{3}- 3bt^{2})dt\end{array}$$
$$\begin{array}{l}\Rightarrow \omega = \omega _{o} + at^{4} – bt^{3}\end{array}$$
2. Further,
$$\begin{array}{l}\omega = \frac{d\Theta }{dt} \Rightarrow d\Theta = \omega dt\end{array}$$
$$\begin{array}{l}\Rightarrow \int_{o}^{\Theta } = \int_{o}^{t} \omega dt = \int_{o}^{t} (\omega _{o}+at^{4}- bt^{3})dt\end{array}$$
$$\begin{array}{l}\Rightarrow \Theta = \omega _{o} + t + \frac{at^{5}}{5} – \frac{bt^{4}}{4}\end{array}$$
### Kinetic Energy of Rotation
The rapidly rotating blades of a table saw machine and the blades of a fan certainly have kinetic energy due to rotation. If we apply the familiar equation to the saw machine as a whole, it would give us kinetic energy of its centre of mass only, which is zero.
The right approach:
We shall treat the saw machine or any rotating rigid body as a collection of particles at different speeds. We shall sum up all the kinetic energy of the particles to find the rotational kinetic energy of the whole body.
If m1, m2,…. mn are the masses of the constituent particles moving on circular paths with radii r1, r2,…. ro with velocities v1, v2,…. vo, then the kinetic energy of the body is given by;
$$\begin{array}{l}KE = \sum_{1}^{n}\frac{1}{2}m_{i}v_{i}^{2} = \sum_{1}^{n}\frac{1}{2}m_{i}r_{i}^{2}\omega ^{2} = \frac{1}{2}\omega ^{2} \sum_{1}^{n}m_{i}r_{i}^{2} = \frac{1}{2}I\omega ^{2}\end{array}$$
The term
$$\begin{array}{l}\sum_{1}^{n}m_{i}r_{i}^{2}\end{array}$$
is called rotational inertia or moment of inertia of the system of particles. For a continuous distribution of mass
$$\begin{array}{l}I = \int_{Body} r^{2}dm\end{array}$$
where dm is the mass of a particle at a distance r from the axis of rotation.
## What Is Torque
Torque is a rotational analogue of force and expresses the tendency of a force applied to an object that causes the object to rotate about a given point.
If you want to open a door, you will apply a force on the doorknob which is located as far as possible from the hinges of the door. If you try to apply the force nearer to the hinge line than the knob, or at any other angle other than 90ᴼ to the plane of the door, you must apply greater force than the former to rotate the door.
To determine how the applied force results in a rotation of the body about an axis, we resolve the Force (F) into two components. The tangential component (Fsinθ) is perpendicular to r and it does cause rotation whereas the radial component (Fcosθ) does not cause rotation because it acts along the line that intersects with the axis or pivot point.
The ability to rotate the body depends on the magnitude of the tangential component and also on how far from the axis the force (r – moment of an arm) is applied. Therefore, mathematically it can be represented as
$$\begin{array}{l}\vec{\tau }=\vec{r}\times \vec{F}\end{array}$$
SI unit of torque is Nm.
$$\begin{array}{l}\text{To find the direction of}\ \vec{\tau },\end{array}$$
we use right hand thumb rule sweeping the fingers from
$$\begin{array}{l}\vec{\tau }\ \text{(the first vector in the product) into}\ \vec{F}\ \text{(the second vector in the product),}\end{array}$$
$$\begin{array}{l}\text{the outstretched thumb will give the direction of}\ \vec{\tau }.\end{array}$$
### Stability of the Object
• Greater the rotational inertia of the object, an object is more stable because it is difficult to move.
• The stability of an object depends on the torques produced by its weight. Larger the torques produced if the mass is from COM, therefore more force is required to alter the stability.
Example: If you consider disk and bike wheel of the same mass, a bike wheel is more stable.
• When the object’s rotation is greater it is more stable.
### Newton’s Second Law of Rotation
$$\begin{array}{l}\text{If the net torque acting on a body about any inertial axis is}\ \vec{\tau }\end{array}$$
and the moment of inertia about that axis is I, then the angular acceleration of the body is given by the relation:
$$\begin{array}{l}\overrightarrow{~\tau }=I\overrightarrow{\alpha ~}\end{array}$$
### Rotational Equilibrium
The centre of mass of a body remains in equilibrium if the total external force acting on the body is zero. This follows from the equation F = Ma.
Similarly, a body remains in rotational equilibrium if the total external torque acting on the body is zero. This follows from the equation τ = Iα. Therefore, a body in rotational equilibrium must either be in rest or rotation with constant angular velocity.
Thus, if a body remains at rest in an inertial frame, the total external force acting on the body should be zero in any direction and the total external torque should be zero about any line.
Under the action of several coplanar forces, the net torque is zero for rotational equilibrium.
Note: If the net force on the body is zero, then the net torque may or may not be zero.
Ex. Determine the point of application of third force for which the body is in equilibrium when forces of 20N & 30N are acting on the rod as shown in the figure.
Solution:
Let the magnitude of the third force is F, is applied in upward direction then the body is in equilibrium when
$$\begin{array}{l}(i)\ \vec{F}_{net}=0\end{array}$$
(Translational Equilibrium)
⇒ 20 + F = 30 ⇒ F = 10N
So, the body is in translational equilibrium when 10N force acts on it in the upward direction.
(ii) Let us assume that this 10N force acts.
Then keep the body in rotational equilibrium.
So, torque about C = 0
i.e τC = 0
⇒ 30 × 20 = 10x
x = 60cm
So, 10N force is applied at 70cm from point A to keep the body in equilibrium.
Ex: A stationary uniform rod of mass ‘m’, length ‘l’ leans against a smooth vertical wall making an angle q with a rough horizontal floor. Find the normal force and frictional force that is exerted by the floor on the rod.
Solution:
As the rod is stationary, the linear acceleration and angular acceleration of the rod are 0.
i.e. acm = 0; ɑ = 0
N2 = f
N1 = mg [∵ acm = 0]
The torque about any point of the rod should also be zero.
∴ ɑ = 0
τA = 0 ⇒ mg cos θ (l / 2) + fl sin θ = N1 cos θ. l
N1 cos θ = sin θ f + (mg cos θ) / 2
f = [mg cos θ] / [2 sin θ] = mg cot θ / 2
Ex: The ladder shown in the figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder in the middle. The angle between the two legs is 60o. The fat person sitting on the ladder has a mass of 80 kg. Find the contact force exerted by the floor on each leg and the tension in the crossbar.
Solution:
The forces acting on different parts are shown in the figure. Consider the vertical equilibrium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg) g and the contact force N + N = 2N due to the floor. Thus, 2N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 N
Next, consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end,
N (2m) tan 30o = T (1m) or
T = N (2/√3) = [(392) N] * (2/√3) = 450N
Ex. The pulley shown in the figure has a moment of inertia l about the axis and radius R. Find the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley.
Solution:
Suppose the tension in the left string is T1 and that in the right string is T2. Suppose the block of mass M goes down with an acceleration a and the other block moves up with the same acceleration. This is also the tangential acceleration of the rim of the wheel as the string does not slip over the rim.
The angular acceleration of the wheel ɑ = a / R.
The equations of motion for the mass M, the mass m and the pulley are as follows;
Mg – T1 = Ma —- (i)
T2 – mg = ma —- (ii)
T1R – T2R = Iɑ = Ia / R —- (iii)
Substituting for T1 and T2 from equations (i) and (ii) in equation (iii)
[M (g – a)] – m (g + a)] R = Ia / R
Solving we get, a = [(M – m)gR2] / [I + (M + m)R2]
Angular momentum of a particle describing circular motion:
$$\begin{array}{l}\vec{L}=\vec{r}\times\vec{p}\end{array}$$
; as linear momentum
$$\begin{array}{l}\vec{p}\end{array}$$
is along with the tangent, hence
$$\begin{array}{l}\vec{r}\times\vec{p}=rpn\end{array}$$
, where n is the unit vector perpendicular to the plane of the circle.
$$\begin{array}{l}\Rightarrow \vec{|L|}=mvr=m\omega r^{2},\text{where}\ \vec{|p|}=mv=m\omega r\end{array}$$
Angular momentum of a rigid in a fixed axis rotation
In a fixed axis rotation, all the constituent particles describe circular motion. Hence, the angular momentum of the particles about corresponding centres are
L1 = m1𝛚r12, L2 = m2𝛚r22
and similarly, Ln = mn𝛚rn2 where 𝛚 is the angular speed of the body.
Since the angular momentum of all the particles have the same direction, therefore angular momentum of the whole body is given by
L = L1 + L2 +…… + Ln
= (m1r12 + m2r22 +….. + mn𝛚rn2) 𝛚
⇒ L = I𝛚
## Angular Momentum
The concept of linear momentum and conservation of linear momentum are extremely powerful tools to predict the collision of two objects without any other details of the collision. Thus, the angular counterpart, angular momentum plays a crucial role in orbital mechanics.
Angular momentum of a particle about a given point is given by,
$$\begin{array}{l}\vec{l}=\vec{r}\times \vec{p}=m\left( \vec{r}\times \vec{v} \right)\end{array}$$
The direction of angular momentum is also given by the right-hand rule. (refer torque)
Newton’s law in the angular form:
The vector sum of all the torques acting on a particle is equal to the time rate of change of the angular momentum of that particle.
$$\begin{array}{l}{{\overrightarrow{~\tau }}_{net}}=~\frac{d\vec{l}}{dt}\end{array}$$
### Conservation of Angular Momentum
By the definition of torque,
$$\begin{array}{l}{{\overrightarrow{~\tau }}_{net}}=~\frac{d\vec{l}}{dt},if{{\overrightarrow{~\tau }}_{net}}=0,~then~\frac{d\vec{l}}{dt}=0,\vec{l}=constant\end{array}$$
When the resultant torque acting on a system is zero, then the total vector angular momentum of the system remains constant. This is called the principle of conservation of angular momentum.
Examples of conservation of angular momentum
A planet revolves around the sun in an elliptical orbit when it goes near the sun, its speed increases and when it goes farther from the sun, its speed decreases. It happens because angular momentum is conserved, and hence time rate of the area swept by the planet is constant. A planet revolves around the sun in orbit in a fixed plane only. This is due to the conservation of angular momentum.
A uniform rod of mass m and length ‘l’ can rotate freely on a smooth hor pine about a vertical axis hinged at point H. A point mass having the same mass ‘m’ coming with an initial speed u perpendicular to the rod strikes the rod in-elastically at its free end. Find out the angular velocity of the rod just after collision.
Solution
Angular momentum is conserved about H because no external force is present in horizontal plane which is producing torque about H.
mul = (ml2/3 + ml2
⇒ ω = 3u/4l
Angular Impulse (Moment of Impulse)
Angular impulse of a torque in a given time is equal to the change in angular momentum. If the angular momentum of a body is changed by a torque τ in time dt then
### Combined Translational and Rotational Motion
Rolling motion is one such example.
Rolling motion is one such example.
To understand the concept of combined translational and rotational motion, we consider a uniform disc rolling on a horizontal surface. The velocity of its centre of mass is Vcom, and its angular speed is shown in the figure.
Let us take point A on the disc and concentrate on its motion.
Path of point A with respect to the ground will be a cycloid, as shown in the figure.
Motion of point A with respect to centre of mass is pure rotational while centre of mass itself is moving in a straight line. So, for the analysis of rolling motion, we deal with translational motion separately and rotational motion separately and then we combine the result to analyse the overall motion.
The velocity of any point A on the rigid body can be obtained as
in the direction perpendicular to the line OA.
Thus, the velocity of point A is the vector sum of Vcom, and Vp.com shown in figure
JEE FOCUS
1. Velocity of any point of the rigid body in combined R + T motion is the vector sum of v(velocity of the centre of mass) and rω.
2, In combined rotational and translational motion, angular velocity of any point of a rigid body with respect to other point in the rigid body is always the same.
3. Distance moved by the centre of mass of the rigid body in one full rotation is 2πR.
4. The speed of a point on the circumference of the body at any instant t is 2Rωsinθ.
Angular momentum of a rigid body in combined rotation and translation
Let O be a fixed point in an inertial frame of reference. Angular momentum of the body about O is.
The first term represents the angular momentum of the body as seen from the centre of mass frame.
The second term equals the angular momentum of centre of mass about point O.
Example. A circular disc of mass m and radius R is set into motion on a horizontal floor with a linear speed v in the forward direction and an angular speed ω = v/R in the clockwise direction, as shown in the figure. Find the magnitude of the total angular momentum of the disc about the bottommost point O of the disc.
Solution
Kinetic Energy of a Rolling Body
If a body of mass M is rolling on a plane such that the velocity of its centre of mass is V and its angular speed is w, its kinetic energy is given by
KE= (1/2)Mv2 + (1/2)Iω2
I is moment of inertia of body about axis passing through centre of mass.
In case of rolling without slipping,
KE= (1/2)Mω2R2 + (1/2)Iω2 [v = ωR]
= (1/2) [ MR2 + I]ω2 = (1/2)I0ω2
Ic is moment of inertia of the body about the axis passing through point of contact.
Uniform Pure Rolling
Pare rolling means no relative motion (or no slipping at point of contact between two bodies.)
For example, consider a disc of radius R moving with linear velocity v and angular velocity ω on a horizontal ground. The dis is said to be moving without slipping if velocities of points P and Q (shown in
figure b) are equal, i.e.,
vP= vQ
v – Rω = 0
v = Rω
If vp > vQ or v > Rω, the motion is said to be forward slipping and if vp < vQ < Rω, the motion is said to be backward slipping.
Now, suppose an external force is applied to the rigid body, the motion will no longer remain uniform.
The condition of pure rolling on a stationary ground is,
a=Rα
is the condition of pure rolling on the stationary ground. Sometimes it is simply said rolling.
Thus, v = Rω and a = Rα
Pure Rolling on an Inclined Plan
A rigid body of radius R and mass m is released at rest from height h on the incline whose inclination with horizontal is q and assume that friction is sufficient for pure rolling, then.
v = Rω and a = Rα
From figure
mgsinθ – f = ma ——(1) [ Fnet = ma]
f.R = cm R2.(a/R) —–(2) [Fnet = Iα]
from equation (1) and (2) we get
a = gsinθ/(1+k)
Binary Systems in Nature
In a binary system, two bodies revolve about their common centre of mass due to their mutual attractive forces, i.e., no external force acts on them.
Let there be two celestial bodies of mass m1 and m2 revolving about their common centre of mass.
By definition of centre of mass
Xcm = (m1x1 + m2x2)/(m1 +m2)
Taking centre of mass1 , as origin,
d1 = m2/(m1 +m2)d
d2=d—d1, where d1, d2 and d are as shown in the figure.
Gravitational force provides the required centripetal force, hence,
Gm1m2/d2 = m1v12/d1
Gm2/d2 = v12/d1 = (2πd1/T)(1/d1) [ since 2πd1/v1 = T]
where T is the time period of one revolution,
= 4π2d1/T2 = 4π2/T2 (m2/(m1 + m2))d
m1 + m2 = (4π2/T2G)d3
If one of the masses is known, the other mass can be calculated knowing T and d.
Typical examples of such systems are earth-moon system, binary stars.
Example. A solid sphere of radius r is gently placed on rough horizontal ground with an initial angular speed ω0 and no linear velocity. If the coefficient of friction is μ, find the linear velocity v and angular velocity w at the end of slipping.
Solution
m be the mass of the sphere.
Since it is a case of backward slipping, the force of friction is in forward direction. Limiting friction will act in this case.
Net torque on the sphere about the bottommost point is zero. Therefore, angular momentum of the sphere will remain conserved about the bottommost point.
Li = Lr
0 = Iω + mrv
(2/5)mr2ω0 = (2/5)mr2ω + mr (ωr)
ω = (2/7)ωand v = rω = (2/7)ω0
Example
A sphere of radius r kept on a rough inclined plane is in equilibrium by a string wrapped over it. If the angle of inclination is θ the tension in the string will be equal to
(A)mgsinθ
(B) 2mg/5
(C) mgsinθ/2
(D) None of these
Since f & N pass through the point P, their torque about P will be zero.
For the equilibrium of the sphere, the torques of tension T and weight mg about P will be zero.
= mgsinθ x r— T(2r)=0
=> 2Tr=mgr sinθ
T = mgsinθ/2
Hence (C) is correct
## Frequently Asked Questions on Rotational Motion
Q1
### What is rotational motion? Give an example.
Rotational motion is a type of motion in which the body follows a circular path. An example is the car wheel.
Q2
### What is the reason for rotational motion?
The torque or rotational analogue force is a reason for rotational motion. When torque is applied to the system of the particle about to its axis, it gives a twist, which is the reason for rotational motion.
Q3
### Is circular motion the same as rotational motion? Explain.
Circular motion: It is a motion of the body around a fixed point. In this case, a fixed point lies outside the body. Here the centripetal force is the reason for the circular motion.
Rotational motion: In the case of rotational motion, the fixed point lies inside the body. Rotational motion is due to the torque acting on the system of the particles.
Test Your Knowledge On Rotational Motion! | 6,311 | 23,196 | {"found_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} | 4.28125 | 4 | CC-MAIN-2023-50 | latest | en | 0.815762 |
https://ncatlab.org/nlab/show/MV+algebras | 1,529,831,056,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2018-26/segments/1529267866926.96/warc/CC-MAIN-20180624083011-20180624103011-00135.warc.gz | 681,959,752 | 7,480 | # MV algebras
## Idea
An MV-algebra is an algebraic structure which models Luzasiewicz multivalued logic, and the fragment of that calculus which deals with the basic logical connectives “and”, “or”, and “not”, but in a multivalued context.
## Definitions
An MV-algebra consists of
• a non-empty set, $A$;
• a binary operation, $\oplus$, on $A$;
• a unary operation, $\neg$, on $A$; and a constant, $0$, such that
1. $\langle A, \oplus, 0\rangle$ is a commutative monoid;
2. $\neg\neg x = x$ for all $x \in A$;
3. $x\oplus \neg 0 = \neg 0$ for all $x\in A$; and
4. $\neg(\neg x\oplus y)\oplus y = \neg(\neg y\oplus x)\oplus x$ for all $x, y$ in $A$.
The last axiom is more difficult to interpret but is clarified by some examples.
###### Example
Let $A = [0,1]$ be the unit interval. Define
$x\oplus y = min(1,x+y)$
$\neg x = 1-x.$
This gives a commutative monoid easily enough and the double negation and absorption axioms are easy to check. Finally the last axiom divides into two cases: $x \lt y$ and $y \lt x$ and only one of these needs checking!
###### Example
Any Boolean algebra defines a MV-algebra with $\oplus = \vee$, and $\neg$ being the complement operation. The expressions in the last axiom evaluate to $x\vee y$.
## Properties
Define $x \odot y \coloneqq \neg (\neg x \oplus \neg y)$, and define $x \Rightarrow y \coloneqq \neg x \oplus y$.
###### Proposition 1
Each MV algebra carries a lattice structure, where the join and meet operations are defined by
$x \vee y = (x \odot \neg y) \oplus y$
$x \wedge y = x \odot (\neg x \oplus y)$
and where $0$ and $1 = \neg 0$ are the bottom and top elements.
Let $\leq$ be the partial ordering for the lattice structure, where $x \leq y$ if $x = x \wedge y$.
###### Proposition 2
Each MV algebra is a residuated lattice?, i.e., a closed monoidal poset, where the monoidal product is $\odot$ and
$x \odot y \leq z \qquad iff \qquad x \leq y \Rightarrow z.$
In particular, for all elements $x$ and $y$ we have $1 = y \Rightarrow y$, $1 \Rightarrow y = y$, and $x \leq (x \Rightarrow y) \Rightarrow y$.
In fact, an MV algebra is a $\ast$-autonomous poset, where $\neg x = x \Rightarrow 0$.
###### Proposition 3
The equational variety of MV algebras is a Mal'cev variety, where the Mal’cev operation is defined by
$t(x, y, z) = ((x \Rightarrow y) \Rightarrow z) \wedge ((z \Rightarrow y) \Rightarrow x).$
In other words, $t(x, y, y) = x = t(y, y, x)$.
###### Completeness theorem
The variety of MV algebras is generated by $[0, 1]$ with its standard MV algebra structure. Consequently, an identity holds for all MV algebras if and only if it holds in $[0, 1]$.
## References
• D. Mundici, MV-algebras, a short tutorial, available here.
A recent preprint is
Last revised on July 4, 2015 at 16:47:55. See the history of this page for a list of all contributions to it. | 891 | 2,861 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 44, "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} | 3.875 | 4 | CC-MAIN-2018-26 | latest | en | 0.670065 |
https://www.physicsforums.com/threads/computing-determinants-allowed-shortcuts.686630/ | 1,527,033,574,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005116-00132.warc.gz | 817,467,851 | 15,020 | # Computing determinants: Allowed shortcuts?
1. Apr 19, 2013
### MrMultiMedia
I had a question about computing determinants and just was wondering what was allowed. So I know that for an n x n matrix, you can go across a row and choose the matrix element as your determinant coefficient for the (n-1) x (n-1) determinant and you go across the row and do this until you're finished with the row. I also know that this process is recursive until you get down to 2x2 from which point you work your way out of this nested maze of determinants you've created for yourself and found the determinant.
I also know that you can go down a column and use its matrix elements as coefficients instead of going down the row. This is useful if there are more zeros down the column than across the row and it simplifies the calculation.
I am aware that the sign of the coefficient depends on its location in the matrix. The top left element is positive and the rest of the signs are arranged like a checkerboard; no same signs have edges touching.
What I wanted to know is, if instead of using the rows or columns as the coefficients for your lower order determinants, could you use diagonals? If you got the signs correct, cancelled out the row and column that it's in and used the remaining matrix elements for your lower order matrix. Could you do this to find the determinant? I do realize that if you used the diagonal, all the coefficients would have the same sign. But does this matter? Is this against the rules? As long as every row or every column is represented, it shouldn't matter right? Or is that incorrect?
2. Apr 19, 2013
### poissonspot
Hey, you can compute determinants like so. What I think you are suggesting can be generalized by Laplace's formula.
3. Apr 23, 2013
### HallsofIvy
No, you cannot use diagonals. That should have been clear if you had looked at some simple examples.
For example, the determinant
$$\left|\begin{array}{cc}3 & 1 \\ 2 & 3\end{array}\right|= 3(3)- 1(2)= 7$$
If you "expand by the first column" you get 3(3)- 2(1)= 7. If you "expand by the second column" you get -1(2)+ 3(3)= 7. If you "expand by the first row" you get 3(3)- 1(2)= 7. If you "expand by the second row" you get -2(1)+ 3(3)= 7.
But if you "expand by the diagonal" you get 3(3)- 3(3)= 0.
One good method of finding a determinant is to "row reduce" to a triangular matrix. The determinant of a triangular matrix is, of course, the product of the numbers on the diagonal.
There are three row operations:
1) Swap two rows. That will multiply the determinant by -1.
2) Add a multiple of one row to another. That does not change the determinant.
3) Multiply a row by a number. That will multiply the determinant by by that number.
For example, I can easily reduce the matrix
$$\begin{bmatrix}3 & 1 \\ 2 & 3\end{bmatrix}$$
to a triangular matrix by adding -2/3 times the first row to the second:
$$\begin{bmatrix}3 & 1 \\ 2- (2/3)(3) & 3- (2/3)(1)\end{bmatrix}= \begin{bmatrix} 3 & 1 \\0 & 7/3 \end{bmatrix}$$
Which is very easily seen to have determinant 3(7/3)= 7.
A slightly more difficult example is
$$\begin{bmatrix} 2 & 1 & 3 \\ 1 & 1 & 2 \\ 0 & 1 & 1\end{bmatrix}$$
There is already a 0 in the first column, third row, so we can get the first column in the form we want by adding -1/2 to the second row:
$$\begin{bmatrix} 2 & 1 & 3 \\ 1- 1 & 1- 1/2 & 2- 3/2 \\ 0 & 1 & 1\end{bmatrix}= \begin{bmatrix}2 & 1 & 3 \\ 0 & 1/2 & 1/2 \\ 0 & 1 & 1\end{bmatrix}$$
and then add -2 times the second row to the third row to get
$$\begin{bmatrix}2 & 1 & 3 \\ 0 & 1/2 & 1/2 \\ 0 & 1- 2(1/2) & 1- 2(1/2)\end{bmatrix}= \begin{bmatrix}2 & 1 & 3 \\ 0 & 1/2 & 1/2 \\ 0 & 0 & 0\end{bmatrix}$$
Since the only row operations used were "add a multiple of one row to another", the determinant of the original matrix is exactly the same as the determinant of this "upper triangular matrix" which is obviously 2(1/2)(0)= 0.
If, in the row operations, you "swap rows" an odd number of times, you multiply the determinant of the final triangular matrix by -1. If you multiply a row by a number, nou have to divide the determinant of the final trianguar matrix by that number.
Last edited by a moderator: Apr 23, 2013 | 1,244 | 4,206 | {"found_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} | 3.65625 | 4 | CC-MAIN-2018-22 | latest | en | 0.952885 |
https://drsimonj.svbtle.com/a-tidy-model-pipeline-with-twidlr-and-broom | 1,725,874,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00687.warc.gz | 192,368,927 | 7,409 | # A tidy model pipeline with twidlr and broom
@drsimonj here to show you how to go from data in a data.frame to a tidy data.frame of model output by combining twidlr and broom in a single, tidy model pipeline.
## The problem #
Different model functions take different types of inputs (data.frames, matrices, etc) and produce different types of output! Thus, we’re often confronted with the very untidy challenge presented in this Figure:
Thus, different models may need very different code.
However, it’s possible to create a consistent, tidy pipeline by combining the twidlr and broom packages. Let’s see how this works.
## Two-step modelling #
To understand the solution, think of the problem as a two-step process, depicted in this Figure:
### Step 1: from data to fitted model #
Step 1 must take data in a data.frame as input and return a fitted model object. twidlr exposes model functions that do just this!
To demonstrate:
``````#devtools::install_github("drsimonj/twidlr") # To install
library(twidlr)
lm(mtcars, hp ~ .)
#>
#> Call:
#> stats::lm(formula = formula, data = data)
#>
#> Coefficients:
#> (Intercept) mpg cyl disp drat
#> 79.048 -2.063 8.204 0.439 -4.619
#> wt qsec vs am gear
#> -27.660 -1.784 25.813 9.486 7.216
#> carb
#> 18.749
``````
This means we can pipe data.frames into any model function exposed by twidlr. For example:
``````library(dplyr)
mtcars %>% lm(hp ~ .)
#>
#> Call:
#> stats::lm(formula = formula, data = data)
#>
#> Coefficients:
#> (Intercept) mpg cyl disp drat
#> 79.048 -2.063 8.204 0.439 -4.619
#> wt qsec vs am gear
#> -27.660 -1.784 25.813 9.486 7.216
#> carb
#> 18.749
``````
### Step2: fitted model to tidy results #
Step 2 must take a fitted model object as its input and return a tidy data frame of results. This is precisely what the broom package does via three functions: `glance`, `tidy`, and `augment`! To demonstrate:
``````#install.packages("broom") # To install
library(broom)
fit <- mtcars %>% lm(hp ~ .)
glance(fit)
#> r.squared adj.r.squared sigma statistic p.value df logLik
#> 1 0.9027993 0.8565132 25.97138 19.50477 1.89833e-08 11 -142.8905
#> AIC BIC deviance df.residual
#> 1 309.7809 327.3697 14164.76 21
tidy(fit)
#> term estimate std.error statistic p.value
#> 1 (Intercept) 79.0483879 184.5040756 0.4284371 0.672695339
#> 2 mpg -2.0630545 2.0905650 -0.9868407 0.334955314
#> 3 cyl 8.2037204 10.0861425 0.8133655 0.425134929
#> 4 disp 0.4390024 0.1492007 2.9423609 0.007779725
#> 5 drat -4.6185488 16.0829171 -0.2871711 0.776795845
#> 6 wt -27.6600472 19.2703681 -1.4353668 0.165910518
#> 7 qsec -1.7843654 7.3639133 -0.2423121 0.810889101
#> 8 vs 25.8128774 19.8512410 1.3003156 0.207583411
#> 9 am 9.4862914 20.7599371 0.4569518 0.652397317
#> 10 gear 7.2164047 14.6160152 0.4937327 0.626619355
#> 11 carb 18.7486691 7.0287674 2.6674192 0.014412403
#> .rownames hp mpg cyl disp drat wt qsec vs am gear carb
#> 1 Mazda RX4 110 21.0 6 160 3.90 2.620 16.46 0 1 4 4
#> 2 Mazda RX4 Wag 110 21.0 6 160 3.90 2.875 17.02 0 1 4 4
#> 3 Datsun 710 93 22.8 4 108 3.85 2.320 18.61 1 1 4 1
#> 4 Hornet 4 Drive 110 21.4 6 258 3.08 3.215 19.44 1 0 3 1
#> 5 Hornet Sportabout 175 18.7 8 360 3.15 3.440 17.02 0 0 3 2
#> 6 Valiant 105 18.1 6 225 2.76 3.460 20.22 1 0 3 1
#> .fitted .resid .hat .sigma .cooksd .std.resid
#> 1 148.68122 -38.681220 0.2142214 24.75946 0.069964902 -1.6801773
#> 2 140.62866 -30.628664 0.2323739 25.43881 0.049861042 -1.3460408
#> 3 79.99158 13.008418 0.3075987 26.38216 0.014633059 0.6019364
#> 4 125.75448 -15.754483 0.2103960 26.31579 0.011288712 -0.6826601
#> 5 183.21756 -8.217565 0.2016137 26.53317 0.002878707 -0.3541128
#> 6 111.38490 -6.384902 0.3147448 26.55680 0.003682813 -0.2969840
``````
## A single, tidy pipeline #
So twidlr and broom functions can be combined into a single, tidy pipeline to go from data.frame to tidy output:
``````library(twidlr)
library(broom)
mtcars %>%
lm(hp ~ .) %>%
glance()
#> r.squared adj.r.squared sigma statistic p.value df logLik
#> 1 0.9027993 0.8565132 25.97138 19.50477 1.89833e-08 11 -142.8905
#> AIC BIC deviance df.residual
#> 1 309.7809 327.3697 14164.76 21
``````
Any model included in twidlr and broom can be used in this same way. Here’s a `kmeans` example:
``````iris %>%
select(-Species) %>%
kmeans(centers = 3) %>%
tidy()
#> x1 x2 x3 x4 size withinss cluster
#> 1 5.901613 2.748387 4.393548 1.433871 62 39.82097 1
#> 2 5.006000 3.428000 1.462000 0.246000 50 15.15100 2
#> 3 6.850000 3.073684 5.742105 2.071053 38 23.87947 3
``````
And a ridge regression with cross-fold validation example:
``````mtcars %>%
cv.glmnet(am ~ ., alpha = 0) %>%
glance()
#> lambda.min lambda.1se
#> 1 0.2284167 0.8402035
``````
So next time you want to do some tidy modelling, keep this pipeline in mind:
## Limitations #
Currently, the major limitation for this approach is that a model must be covered by twidlr and broom. For example, you can’t use `randomForest` in this pipeline because, although twidlr exposes a data.frame friendly version of it, broom doesn’t provide tidying methods for it. So if you want to write tidy code for a model that isn’t covered by these packages, have a go at helping out by contributing to these open source projects! To get started creating and contributing to R packages, take a look at Hadley Wickham’s free book, “R Packages”.
## Sign off #
Thanks for reading and I hope this was useful for you.
For updates of recent blog posts, follow @drsimonj on Twitter, or email me at drsimonjackson@gmail.com to get in touch.
If you’d like the code that produced this blog, check out the blogR GitHub repository. | 2,329 | 6,317 | {"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} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.761302 |
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Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
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Question From class 12 Chapter CORDINATES AND STARIGHT LINES - FOR COMPETITION
# The straight lines form a triangle which is (A) isosceles (B) right angled (C) equilateral (D) scalene
The straight lines , and form (A) an isosceles triangle (B) an equilateral triangle (C) a scalene triangle (D) a right angled triangle
5:40
The straight lines and form a triangle which is :
4:22
The pair of straight lines together with the line form a triangle which is: (A) right angled but not isosscles (C) scalene (B) right isosceles (D) equilateral
9:41
The straight lines and form a triangle, which is
4:42
The straight line , , form a triangle which is
4:29
Determine whether the triangle formed by the lines and is (a) equilateral (c) isosceles 6. (b) right-angled (d) none
1:15
and , are three real distinct lines forming a triangle. Then the triangle is isosceles (b) scalene equilateral (d) right angled
2:07
The triangle formed by the pair of lines and the line is always 1) Equilateral 2) Isosceles 3) Right angled 4) Scalene
2:49
The line and forms a triangle which is (i) isosceles (ii) Equilateral (iii) Right Angled Triangle (iv) Right Angled isosceles
4:20
The points and form a (1) Right-angled triangle (3) Both (1) & (2) (2) Isosceles triangle (4) Equilateral triangle
2:30
Four lines form a figure which is.
4:06
Four lines form a figure which is.
2:30
26 Which of the following lines have the intercepts of equal lengths on the circle, (A) (B) (C) (D)
7:26
The equations of the sides respectively. The equation of the altitude through B is b. c. d. none of these
2:48
Latest Blog Post | 531 | 1,786 | {"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} | 3.640625 | 4 | CC-MAIN-2020-40 | latest | en | 0.805323 |
https://math.stackexchange.com/questions/1805731/the-extended-dot-product | 1,560,703,575,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998288.34/warc/CC-MAIN-20190616162745-20190616184745-00146.warc.gz | 512,389,739 | 33,435 | # The “extended” dot product
The standard dot product of two vectors $\vec{u},\vec{v} \in \mathbb{R}^n$ is given by:
$\vec{u}\cdot \vec{v} = \sum_{i=1}^n u_i v_i$
Assuming now that I have $m$ vectors in $\mathbb{R}^n$, does the following product have a name, and does it have any interesting properties:
$P = \sum_{i=1}^n \prod_{j=1}^m v_{j,i}$
In other words, it's the extension to the dot product, which is defined as the sum of element-wise products of all vectors.
• Isn't it just $v_1 \cdot v_2 \cdots v_m$? – WhatsUp May 30 '16 at 13:45
• @WhatsUp $v_1 \cdot v_2$ is a number, how do you take the dot product of a number and a vector? – Najib Idrissi May 30 '16 at 13:45
• I don't think so, for the reason mentioned by @Najib Idrissi – Michael Stachowsky May 30 '16 at 13:45
• Sorry, I was wrong. This is an $m$-linear map from $(\mathbb{R}^n)^m$ to $\mathbb{R}$. – WhatsUp May 30 '16 at 13:46
• This is related. You can define whatever operation on vectors you want, but if it is not "geometric" (in the sense that it does not depend on the choice of coordinates) then it will probably be of marginal use. – Giuseppe Negro May 30 '16 at 13:47 | 397 | 1,155 | {"found_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} | 2.84375 | 3 | CC-MAIN-2019-26 | longest | en | 0.892334 |
https://blogs.msdn.microsoft.com/craigfr/2007/09/21/aggregation-with-rollup/ | 1,516,539,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890582.77/warc/CC-MAIN-20180121120038-20180121140038-00093.warc.gz | 659,389,531 | 17,021 | # Aggregation WITH ROLLUP
In this post, I'm going to discuss how aggregation WITH ROLLUP works. The WITH ROLLUP clause permits us to execute multiple "levels" of aggregation in a single statement. For example, suppose we have the following fictitious sales data. (This is the same data that I used for my series of posts on the PIVOT operator.)
CREATE TABLE Sales (EmpId INT, Yr INT, Sales MONEY)
INSERT Sales VALUES(1, 2005, 12000)
INSERT Sales VALUES(1, 2006, 18000)
INSERT Sales VALUES(1, 2007, 25000)
INSERT Sales VALUES(2, 2005, 15000)
INSERT Sales VALUES(2, 2006, 6000)
INSERT Sales VALUES(3, 2006, 20000)
INSERT Sales VALUES(3, 2007, 24000)
We can write a simple aggregation query to compute the total sales by year:
SELECT Yr, SUM(Sales) AS Sales
FROM Sales
GROUP BY Yr
As expected, this query returns three rows - one for each year:
```Yr Sales
----------- ---------------------
2005 27000.00
2006 44000.00
2007 49000.00```
The query plan is a simple stream aggregate:
|--Compute Scalar(DEFINE:([Expr1004]=CASE WHEN [Expr1010]=(0) THEN NULL ELSE [Expr1011] END))
|--Stream Aggregate(GROUP BY:([Sales].[Yr]) DEFINE:([Expr1010]=COUNT_BIG([Sales].[Sales]), [Expr1011]=SUM([Sales].[Sales])))
|--Sort(ORDER BY:([Sales].[Yr] ASC))
|--Table Scan(OBJECT:([Sales]))
Now suppose that we want to compute not just the sale by year but the total sales as well. We could write a UNION ALL query:
SELECT Yr, SUM(Sales) AS Sales
FROM Sales
GROUP BY Yr
UNION ALL
SELECT NULL, SUM(Sales) AS Sales
FROM Sales
This query works and does give the correct result:
```Yr Sales
----------- ---------------------
2005 27000.00
2006 44000.00
2007 49000.00
NULL 120000.00```
However, the query plan performs two scans and two aggregations (one to compute the sales by year and one to compute the total sales):
|--Concatenation
|--Compute Scalar(DEFINE:([Expr1004]=CASE WHEN [Expr1023]=(0) THEN NULL ELSE [Expr1024] END))
| |--Stream Aggregate(GROUP BY:([Sales].[Yr]) DEFINE:([Expr1023]=COUNT_BIG([Sales].[Sales]), [Expr1024]=SUM([Sales].[Sales])))
| |--Sort(ORDER BY:([Sales].[Yr] ASC))
| |--Table Scan(OBJECT:([Sales]))
|--Compute Scalar(DEFINE:([Expr1010]=NULL))
|--Compute Scalar(DEFINE:([Expr1009]=CASE WHEN [Expr1025]=(0) THEN NULL ELSE [Expr1026] END))
|--Stream Aggregate(DEFINE:([Expr1025]=COUNT_BIG([Sales].[Sales]), [Expr1026]=SUM([Sales].[Sales])))
|--Table Scan(OBJECT:([Sales]))
We can do better by adding a WITH ROLLUP clause to the original query:
SELECT Yr, SUM(Sales) AS Sales
FROM Sales
GROUP BY Yr WITH ROLLUP
This query is simpler to write and uses a more efficient query plan with only a single scan:
|--Compute Scalar(DEFINE:([Expr1004]=CASE WHEN [Expr1005]=(0) THEN NULL ELSE [Expr1006] END))
|--Stream Aggregate(GROUP BY:([Sales].[Yr]) DEFINE:([Expr1005]=SUM([Expr1007]), [Expr1006]=SUM([Expr1008])))
|--Stream Aggregate(GROUP BY:([Sales].[Yr]) DEFINE:([Expr1007]=COUNT_BIG([Sales].[Sales]), [Expr1008]=SUM([Sales].[Sales])))
|--Sort(ORDER BY:([Sales].[Yr] ASC))
|--Table Scan(OBJECT:([Sales]))
The bottom stream aggregate in this query plan is the same as the stream aggregate in the original non-ROLLUP query. This aggregation is a normal aggregation and, as such, it can be implemented using a stream aggregate (as in this example) or a hash aggregate (try adding an OPTION (HASH GROUP) clause to the above query). It can also be parallelized.
The top stream aggregate is a special aggregate that computes the ROLLUP. (Unfortunately, in SQL Server 2005 there is no way to discern from the query plan that this aggregate implements a ROLLUP. This issue will be fixed in SQL Server 2008 graphical and XML plans.) A ROLLUP aggregate is always implemented using stream aggregate and cannot be parallelized. In this simple example, the ROLLUP stream aggregate merely returns each pre-aggregated input row while maintaining a running total of the Sales column. After outputting the final input row, the aggregate also returns one additional row with the final sum. Since SQL lacks a concept of an ALL value, the Yr column is set to NULL for this final row. If NULL is valid value for Yr, we can identify the ROLLUP row using the GROUPING(Yr) construct.
SELECT
CASE WHEN GROUPING(Yr) = 0
THEN CAST (Yr AS CHAR(5))
ELSE 'ALL'
END AS Yr,
SUM(Sales) AS Sales
FROM Sales
GROUP BY Yr WITH ROLLUP
```Yr Sales
----- ---------------------
2005 27000.00
2006 44000.00
2007 49000.00
ALL 120000.00```
We can also compute multiple ROLLUP levels in a single query. For example, suppose that we want to compute the sales first by employee and then for each employee by year:
SELECT EmpId, Yr, SUM(Sales) AS Sales
FROM Sales
GROUP BY EmpId, Yr WITH ROLLUP
```EmpId Yr Sales
----------- ----------- ---------------------
1 2005 12000.00
1 2006 18000.00
1 2007 25000.00
1 NULL 55000.00
2 2005 15000.00
2 2006 6000.00
2 NULL 21000.00
3 2006 20000.00
3 2007 24000.00
3 NULL 44000.00
NULL NULL 120000.00```
There are a couple of points worth noting about this query. First, since the combination of the EmpId and Yr columns is unique, in the absence of the WITH ROLLUP clause, this query would just return the original data. However, with the WITH ROLLUP clause the query produces a useful result. Second, the order of the columns in the GROUP BY clause is relevant with the WITH ROLLUP clause. To see why simply try the same query but reverse the EmpId and Yr columns. Instead of computing the sales first by employee it will compute the sales first by year.
The query plan for this query is identical to the query plan for the prior query except that it groups on both the EmpId and Yr columns instead of on just the EmpId column. Like the prior query plan, this query plan includes two stream aggregates: the bottom one which is a normal stream aggregate and the top one which computes the ROLLUP. This ROLLUP stream aggregate actually computes two running totals: one which computes the total sales for an employee for all years and one which compute the total sales for all employees and all years. This table shows how the ROLLUP computation proceeds:
EmpId Yr SUM(Sales) BY EmpId, Yr SUM(Sales) BY EmpId SUM(Sales) 1 2005 12000.00 12000.00 12000.00 1 2006 18000.00 30000.00 30000.00 1 2007 25000.00 55000.00 55000.00 1 NULL 55000.00 55000.00 2 2005 15000.00 15000.00 70000.00 2 2006 6000.00 21000.00 76000.00 2 NULL 21000.00 76000.00 3 2006 20000.00 20000.00 96000.00 3 2007 24000.00 44000.00 120000.00 3 NULL 44000.00 120000.00 NULL NULL 120000.00
In my next post, I'll take a look at the WITH CUBE clause. I'll discuss how it differs from WITH ROLLUP both in terms of function and in terms of its implementation.
1. In my last post, I wrote about how aggregation WITH ROLLUP works. In this post, I will discuss how aggregation
2. mknee says:
Great post – I’m sure many people would consder this "the basics" but it was new to me & VERY useful.
Thanks!
Mike Knee
3. In my last two posts, I gave examples of aggregation WITH ROLLUP and CUBE . SQL Server 2008 continues
4. Rakesh Dewangan says:
Very helpful, thanks for writing blog.
gr8 work!!!
Keep it up
5. Sonal Bhosale says:
Thanx a lot!!!!
6. Deepak Ugale says:
Post really helps to understand the basic use of ROLLUP
7. Cool article on WITH ROLLUP ..Thanks 🙂
8. Supriya says:
Very well explained. It cleared all my doubts. Thanks Craig 🙂
9. gopal says:
How can i display the final result as in the column "SUM(Sales) BY EmpId" of above table
10. Starting with SQL Server 2012, you can use a window aggregate function as follows:
SELECT EmpId, Yr, SUM(Sales) OVER (PARTITION BY EmpId ORDER BY Yr) AS Sales
FROM Sales
HTH
Craig
11. Waldemar Sarmiento says:
Actually the Aggregate Windows Function SUM(Sales) OVER (PARTITION BY EmpId)
exists since SQL 2005 but without the Order By
Waldemar | 2,286 | 8,117 | {"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} | 2.765625 | 3 | CC-MAIN-2018-05 | longest | en | 0.623116 |
http://cn.metamath.org/mpeuni/pmtridfv2.html | 1,660,064,419,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00084.warc.gz | 10,791,971 | 5,661 | Mathbox for Thierry Arnoux < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > pmtridfv2 Structured version Visualization version GIF version
Theorem pmtridfv2 30192
Description: Value at Y of the transposition of 𝑋 and 𝑌 (understood to be the identity when X = Y ). (Contributed by Thierry Arnoux, 3-Jan-2022.)
Hypotheses
Ref Expression
pmtridf1o.a (𝜑𝐴𝑉)
pmtridf1o.x (𝜑𝑋𝐴)
pmtridf1o.y (𝜑𝑌𝐴)
pmtridf1o.t 𝑇 = if(𝑋 = 𝑌, ( I ↾ 𝐴), ((pmTrsp‘𝐴)‘{𝑋, 𝑌}))
Assertion
Ref Expression
pmtridfv2 (𝜑 → (𝑇𝑌) = 𝑋)
Proof of Theorem pmtridfv2
StepHypRef Expression
1 pmtridf1o.y . . . . 5 (𝜑𝑌𝐴)
2 fvresi 6582 . . . . 5 (𝑌𝐴 → (( I ↾ 𝐴)‘𝑌) = 𝑌)
31, 2syl 17 . . . 4 (𝜑 → (( I ↾ 𝐴)‘𝑌) = 𝑌)
43adantr 466 . . 3 ((𝜑𝑋 = 𝑌) → (( I ↾ 𝐴)‘𝑌) = 𝑌)
5 pmtridf1o.t . . . . 5 𝑇 = if(𝑋 = 𝑌, ( I ↾ 𝐴), ((pmTrsp‘𝐴)‘{𝑋, 𝑌}))
6 simpr 471 . . . . . 6 ((𝜑𝑋 = 𝑌) → 𝑋 = 𝑌)
76iftrued 4231 . . . . 5 ((𝜑𝑋 = 𝑌) → if(𝑋 = 𝑌, ( I ↾ 𝐴), ((pmTrsp‘𝐴)‘{𝑋, 𝑌})) = ( I ↾ 𝐴))
85, 7syl5eq 2816 . . . 4 ((𝜑𝑋 = 𝑌) → 𝑇 = ( I ↾ 𝐴))
98fveq1d 6334 . . 3 ((𝜑𝑋 = 𝑌) → (𝑇𝑌) = (( I ↾ 𝐴)‘𝑌))
104, 9, 63eqtr4d 2814 . 2 ((𝜑𝑋 = 𝑌) → (𝑇𝑌) = 𝑋)
11 simpr 471 . . . . . . 7 ((𝜑𝑋𝑌) → 𝑋𝑌)
1211neneqd 2947 . . . . . 6 ((𝜑𝑋𝑌) → ¬ 𝑋 = 𝑌)
1312iffalsed 4234 . . . . 5 ((𝜑𝑋𝑌) → if(𝑋 = 𝑌, ( I ↾ 𝐴), ((pmTrsp‘𝐴)‘{𝑋, 𝑌})) = ((pmTrsp‘𝐴)‘{𝑋, 𝑌}))
145, 13syl5eq 2816 . . . 4 ((𝜑𝑋𝑌) → 𝑇 = ((pmTrsp‘𝐴)‘{𝑋, 𝑌}))
1514fveq1d 6334 . . 3 ((𝜑𝑋𝑌) → (𝑇𝑌) = (((pmTrsp‘𝐴)‘{𝑋, 𝑌})‘𝑌))
16 pmtridf1o.a . . . . 5 (𝜑𝐴𝑉)
1716adantr 466 . . . 4 ((𝜑𝑋𝑌) → 𝐴𝑉)
18 pmtridf1o.x . . . . 5 (𝜑𝑋𝐴)
1918adantr 466 . . . 4 ((𝜑𝑋𝑌) → 𝑋𝐴)
201adantr 466 . . . 4 ((𝜑𝑋𝑌) → 𝑌𝐴)
21 eqid 2770 . . . . 5 (pmTrsp‘𝐴) = (pmTrsp‘𝐴)
2221pmtrprfv2 30182 . . . 4 ((𝐴𝑉 ∧ (𝑋𝐴𝑌𝐴𝑋𝑌)) → (((pmTrsp‘𝐴)‘{𝑋, 𝑌})‘𝑌) = 𝑋)
2317, 19, 20, 11, 22syl13anc 1477 . . 3 ((𝜑𝑋𝑌) → (((pmTrsp‘𝐴)‘{𝑋, 𝑌})‘𝑌) = 𝑋)
2415, 23eqtrd 2804 . 2 ((𝜑𝑋𝑌) → (𝑇𝑌) = 𝑋)
2510, 24pm2.61dane 3029 1 (𝜑 → (𝑇𝑌) = 𝑋)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 382 = wceq 1630 ∈ wcel 2144 ≠ wne 2942 ifcif 4223 {cpr 4316 I cid 5156 ↾ cres 5251 ‘cfv 6031 pmTrspcpmtr 18067 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1869 ax-4 1884 ax-5 1990 ax-6 2056 ax-7 2092 ax-8 2146 ax-9 2153 ax-10 2173 ax-11 2189 ax-12 2202 ax-13 2407 ax-ext 2750 ax-rep 4902 ax-sep 4912 ax-nul 4920 ax-pow 4971 ax-pr 5034 ax-un 7095 This theorem depends on definitions: df-bi 197 df-an 383 df-or 827 df-3or 1071 df-3an 1072 df-tru 1633 df-ex 1852 df-nf 1857 df-sb 2049 df-eu 2621 df-mo 2622 df-clab 2757 df-cleq 2763 df-clel 2766 df-nfc 2901 df-ne 2943 df-ral 3065 df-rex 3066 df-reu 3067 df-rab 3069 df-v 3351 df-sbc 3586 df-csb 3681 df-dif 3724 df-un 3726 df-in 3728 df-ss 3735 df-pss 3737 df-nul 4062 df-if 4224 df-pw 4297 df-sn 4315 df-pr 4317 df-tp 4319 df-op 4321 df-uni 4573 df-iun 4654 df-br 4785 df-opab 4845 df-mpt 4862 df-tr 4885 df-id 5157 df-eprel 5162 df-po 5170 df-so 5171 df-fr 5208 df-we 5210 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-ord 5869 df-on 5870 df-lim 5871 df-suc 5872 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-om 7212 df-1o 7712 df-2o 7713 df-er 7895 df-en 8109 df-dom 8110 df-sdom 8111 df-pmtr 18068 This theorem is referenced by: reprpmtf1o 31038
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07 Sep 2012, 12:15
I used to be an old and regular member of the GmatClub. I took the GMAT in 2005 Dec, and got a decent score (730). However, due to personal and professional reasons, never got to do my MBA. 7 years later, I am planning to get back. I guess it is never too late to take it, as long as you have the drive and motivation in place. But, how times have changed. GmatClub has changed it's interface radically, and so has the format of the exam itself (the new IR section). I will be an active participant of this forum, so you will hear more from me.
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07 Sep 2012, 12:17
I used to be an old and regular member of the GmatClub. I took the GMAT in 2005 Dec, and got a decent score (730). However, due to personal and professional reasons, never got to do my MBA. 7 years later, I am planning to get back. I guess it is never too late to take it, as long as you have the drive and motivation in place. But, how times have changed. GmatClub has changed it's interface radically, and so has the format of the exam itself (the new IR section). I will be an active participant of this forum, so you will hear more from me.
Thanks
DMD
Welcome back!
I think your Avatar may have gotten lost over the years so if you have a chance, please upload a new one.
Good luck with the new wave of prep and again, welcome back!
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I found your forum site more user-friendly and fantastic. Keep it up guys.
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Thank you for free grammar book. I am sure it is excellent material.
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What type of work experience do you need for GMAT?
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14 Aug 2013, 16:49
Hello Everyone,
i have just joined the GMAT Club. I am a Software Test Engineer from Pune, India.
All this while i was learning about GMAT. I have collected most of the Material such as MGMAT, OG & PowerScore CR Bible. Also,i have downloaded the GMATClub Math Book.
I am all set to start my Prep from tomorrow morning. I gave my first GMATPrep Mock as a diagnostic and scored a total of 320. Pathetic score but i haven't lost hope.I scored very poorly in Quant(Very Weak here) and Verbal (Don't Even Ask). I made the following mistakes during the test:
1. Lack of concentration: Couldn't stare at the screen for too long and couldn't sit in one place for 4 hours straight
2. I am very poor at both Verbal and Quant(Guess the Indian Engineer Myth is busted). Didn't know what to write for Argument and Screwed up my IR too.
3. So many mistakes in Sentence Correction. Trusted my ear and it all fell apart.
4. Reading speed is slow and so is Mental Calculation speed. Forgot Tables,Squares,Cubes and so on.
I have decided to Complete my Basic Prep by November end and start Extra Practice and Mocks from then on.
I am going to Give GMAT only 1 attempt and i'm going to make it my best .I'm starting prep tomorrow .
Bye for now.
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19 Sep 2013, 00:44
hello all! It will be really interesting to participate in this discussion.
Re: Welcome to the NEW GMAT Club [#permalink] 19 Sep 2013, 00:44
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```YCK2 Form 7 Physics First Assessment 2004-05
Section A (30)
1.
2.
3.
4.
5.
C
B
B
D
C……
11.
12.
13.
14.
15.
6.
7.
8.
B
B
D
16. A
17. B
18. A
26. D
27. D
28. A
9. D
10. D
19. B
20. A
29. D
30. D
Section B
1.
C
B
D
C
A..........
21.
22.
23.
24.
25.
A
A
C
C
C..............
(32)
(a)
(1)
(1)
At coil position shown, the magnetic flux linking each turn of coil is
= B A cos t
( 12 )
By Faraday law of Induction, the induced e.m.f.,
d
V = N
= B A N sin t
dt
or
V = Vo sin t
( 12 )
( 12 + 12 )
where Vo = B A N (the maximum induced e.m.f.)
From diagrams in (b), V is the maximum when d/dt is maximum and the
coil is horizontal;
V is zero when d/dt = 0 and the coil is vertical. (1)
1
(b)
Diagram
Values of L, C and R roughly correct
(c)
(1)
Transformer The primary and the secondary windings are wound
on the same core, thus the rate of change of magnetic flux, d/dt at
any time is the same.
( 12 )
Vp = Np (d/dt)
(equal to back e.m.f.)
( 12 )
Vs = Ns (d/dt)
(when no current is drawn)
( 12 )
Hence
(2)
(2)
(1)
Vs
N
s
Vp N p
( 12 )
Full-wave rectification Rectifiers offer low resistance against
current flow in one direction but high resistance in the opposite
direction.
( 12 )
( 12 )
In the positive half cycle current passes through the rectifier in one
direction; while in the negative half cycle, current passes through the
rectifier in the opposite direction.
( 12 )
The circuit is connected such that current always passes through the
Thus the output voltage becomes d.c. (unidirectional)
( 12 )
(3)
Filter
( 12 )
2
The smoothing of the output voltage is achieved by separating the
d.c. component from the a.c. component of the signal.
( 12 )
A capacitor connected across the load is in fact a storage capacitor
which stores charge and energy when it is charged up and then
releases charge and energy through the load.
( 12 )
( 12 )
An inductor is connected in series with a capacitor act as a potential
divider.
( 12 )
The inductor offers a great impedance to the a.c. component whereas
the capacitor offers a great resistance to the d.c. component.
( 12 )
Thus the a.c. component (unwanted ripples) mostly drops across the
inductor; whereas the d.c. component appears across the capacitor.
( 12 )
For good smoothing action, the time-constant RC must be larger
than 1/2f for the first capacitor.
For 50 Hz, R is about 1 kHz, C is about 10 F.
If L and C series circuit is also used, L is about 10 H, C is about
47F
( 12 )
2.
(a)
(i)
(ii)
(b)
(i)
Electric field, E, in the circuit is the force per coulomb acting on the
free electrons which move in a net direction along the connecting
wires and across the resistor.
(1)
The direction of electric field is along the direction of current (from
higher potential to lower potential).
(0.5)
(suitable figure: 0.5)
Potential difference across the resistor is the energy converted to
some other forms from electric potential energy per coulomb of
charge passing through the resistor.
(1)
According to (a)(ii), the energy converted in R when charge Q passes
through it is given by
E=QV
where V is the potential difference across R (0.5)
The rate of heat conversion is
3
dE d
dQ
(QV) V
VI I 2 R
dt dt
dt
(0.5+0.5)
(ii)
IX
V
IR
In an a.c. circuit with reactance X,
V
I
where I and V are r.m.s. values;
2
2
R X
(I is the current through R and V is the voltage across R)
(c)
(i)
(0.5)
The rate of heat conversion is still I2 R.
There is a time delay in the lighting up of the light bulb
intensity.
It is because a large induced e.m.f. is created in the inductor.
The induced e.m.f. opposes the current through the inductor.
Thus the current rises slowly to its maximum value.
Since
induced e.m.f.
dI
dt
(0.5)
to full
(0.5)
(0.5)
(0.5)
(0.5)
Thus, induced e.m.f. decreases with time and becomes zero when
current reaches its maximum.
(0.5)
(ii)
When the d.c. supply is switched off, there is a large induced e.m.f.
in the inductor.
(0.5)
This opposes the collapse of current.
(0.5)
This is large enough to light up the neon lamp.
(0.5)
(d)
a.c. V
(small R)
Suppose
L
I I o sin t
(0.5)
dI
LI o cos t (0.5)
dt
If R is small, V =
(1)
Thus, V leads I by /2 or T/4 (0.5)
4
Figures: (1+0.5)
X L L
R
R
For small R,
90
Also, Y1 for measuring V
Y2 for measuring IR, or I.
tan
(0.5)
(0.5)
(0.5)
(0.5) //(6)
Section C (40)
1.
By Fleming left hand rule, the charge carriers drift to the top surface. Thus the
charge carrier has positive charge.
(0.5+0.5)
V 6 10 3
6 V m 1
3
d
10
The drift velocity, v , is given by, (at equilibrium)
qvB=qE
The electric field is,
v
or
E
E 6
6 m s 1
B 1
(0.5)
(0.5)
(0.5)
The current is given by
I nAvq
Thus,
2.
n
(0.5)
I
10
1.04 10 25 m 3
19
6
qvA 1.6 10 6 10
(a) The motor does not rotate. Thus there is no back e.m.f.
Current
V 3
0 .6 A
R 5
(1) //(4)
(1)
(1)
(b) Mechanical power output = m g v = 0.05 10 0.3 = 0.15 W
Let I be the current through the coil.
Power lost in resistive heating = I2 R = 5 I2
Power input = I V = 3 I
At constant speed, power is conserved,
(0.5)
3 I = 5 I2 + 0.15
or
5 I2 3 I + 0.15 = 0
Solve for I, we get
I = 0.545 A or
I = 0.055 A
(c) Consider the voltage of the circuit,
(0.5)
(1)
(0.5)
(0.5)
V=IR+
(where is the back e.m.f. )
(1)
For I = 0.545 A, = 0.275 V
(0.5)
For I = 0.055 A, = 2.755 V
(0.5)
Since is proportional to the speed of motor, this shows that the motor
rotates faster when the current is smaller.
(1) //(8)
5
3. (a)
The capacitor blocks all d.c. and
the equivalent circuit is as shown
in the right.
(0.5)
Let R' be the equivalent resistance
of the two shunt resistors, then
3
R'
(0.5)
6 10 6
R' = 0.5 M
Thus,
(b)
10 6 R
10 R
6
R
(0.5)
(0.5)
05 10 6
(0.5)
Hence, R = 1 M
When the input is a.c., the
equivalent circuit is as shown in
(0.5)
the right.
Let Z be the equivalent impedance
of the black box parallel with the
1 M resistor.
Then
3 10
9
3
(0.5)
Z
(1)
10 6 Z
Thus, Z = 0.33 k
(0.5)
This impedance is the result of C in parallel with 0.5 M resistor. Yet
Z << 0.5 M. This means that the 0.5 M resistor can be ignored and Z is
essentially the reactance of C.
(1)
1
Z
Thus,
C
1
1
or
(1) //(7)
C
96 pF
2 f Z 2 (5 10 6 ) (0.33 10 3 )
4.
(a) Voltage across RL = 6 1.5 = 4.5 V
Thus,
RL
4.5
1.5 k
0.003
(1)
(b) Voltage across RB = 6 0.7 = 5.3 V
I
0.003
IB C
1.5 10 5 A
and
200
Thus,
RB
5.3
1.5 10 5
(0.5)
353 k
6
(0.5)
(0.5)
(1)
(c)
5.
IB
5.3
0.106 mA
50 10 3
I C 200 I B 21.2 mA
(a)
Thus,
If
(1)
(0.5) //(5)
VCC Vout I C R L
(0.5)
Vin VBE I B R B
RL
Vout VCC
(Vin VBE )
RB
130 R L
Vout 6
(Vin 0.7)
RB
(0.5)
Vout = 2.5 V ,
Then Vin = 0.88 V
(1)
(1)
(b) The smallest value of Vout is zero. Thus the maximum variation is 2.5 V and
the maximum peak-to-peak voltage of Vout is 5 V.
(1)
Vout
RL
(c) The voltage gain,
(0.5)
Vin
RB
When Vout = 5 V,
Vin = 0.26
(d) At saturation, VCC 0.2 I CS R L
(0.5)
(0.5)
6 0.2 (12 10 3 ) R L
RL = 483
Vout
RL
Vin
RB
Since
10
(1)
(0.5)
130 483
RB
RB = 6279
6.
(a) G
(b)
Rf
100
10
R in
10
(0.5+1)
Saturated at positive side means Vout = 13 V
Thus,
(c)
(1)
I in
Vout
13
1.3 V
G
10
1.3
0.13 mA
10 10 3
Vin
Vin
R in
7
(0.5)
(0.5+1)
(0.5+1)
//(8)
(d)
1.5
1.5
(3) //(8)
8
```
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TEACHING EXPERIENCE
2000 – Present: South Kingstown RI School District Grade 7/8 Mathematics teacher
MY TEACHING STYLE
Fun and engaging that meets the needs of every learner.
HONORS/AWARDS/SHINING TEACHER MOMENT
2011 Presidential Award for Excellence in Mathematics and Science Teaching Winner 2011 National Institute for Summer Learning Award Nominee
MY OWN EDUCATIONAL HISTORY
2008-2014: Walden University Ed.D Teacher Leadership Program Completed all course work with 4.0 GPA. 2004-2006: Walden University Master’s Program – M.S. in Education (Mathematics Grades 6-8) – Graduated with a 4.0 GPA 1995-1999: Bachelors’ Degree from Springfield College (MA) Major – Mathematics, Science and Technology | 1,647 | 7,015 | {"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} | 3.53125 | 4 | CC-MAIN-2017-47 | longest | en | 0.891265 |
https://www.huffpost.com/entry/ukrainian-parliament-fight-painting_n_5659149?ir=Arts | 1,717,067,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00177.warc.gz | 698,303,829 | 64,846 | # That Time A Fight In The Ukrainian Parliament Looked <em>Exactly</em> Like A Renaissance Painting
That Time A Fight In The Ukrainian Parliament LookedLike A Renaissance Painting
Often, art is where you least expect it. Sometimes art is a particularly foggy morning, other times it's a child's drawing made from pure imagination, or it could be, you know, a nasty brawl in the Ukrainian Parliament.
Someone took a candid photo of a fight in Ukranian Parliament that is as well-composed as the best renaissance art pic.twitter.com/BBhw6rdT3l
— James Harvey (@jamesharveytm) August 6, 2014
Yup, that glorious freeze frame of an epic face palm is an example of what's called the golden ratio, a term Renaissance artists used to describe the most aesthetically pleasing -- and therefore obviously divine -- ordering of elements in an artwork. Two quantities are in the golden ratio if their ratio equals the ratio of their sum to the larger of the two quantities.
Confused? Well, you can see the phenomenon in Michelangelo's "The Creation of Adam," and about a thousand other Renaissance masterworks.
According to Mashable, the accidental artwork above was first noticed by Facebook user Manzil Lajura, who highlighted why the photo's composition was so compelling, even placing it inside a fancy frame just for fun. It was later tweeted by artist James Harvey, and retweeted ad infinitum. The Guardian kindly contributed a lovely narrative explanation.
"The Fibonacci spiral has been placed on top of it to show just why its elements cohere so satisfyingly. Starting with one added to one, if you add each number in a series to the one preceding it you create the Fibonacci sequence, whose beautifully exponential growth can be transposed on to everything from the arrangements of petals in flowers to cauliflowers. Here, the violence spirals exponentially outward from the focal point of the fight up to the reddened face of the man at the top of the image; in another brawl image, at the top of this article, the spiral similarly flows around the image to the shiny-suited politico grappling in the middle."
Of course, the Ukrainian Parliament probably wasn't trying to choreograph a perfect flow of movement and drama when they broke into a scuffle over a decree permitting more reservists to defend against Russian troops, but that's what makes the whole thing so magical. Dan Scully then tweeted another example of accidental Renaissance artwork, this time with a soccer moment that looks strangely like Christ's descent from the cross -- sort of.
Just your daily reminder to always keep your eyes open for stealthy artworks lurking in your neighborhood Parliament. Have you noticed any other examples of the golden ratio where you least expected it? Let us know in the comments.
Close | 574 | 2,799 | {"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} | 2.71875 | 3 | CC-MAIN-2024-22 | latest | en | 0.940289 |
http://us.metamath.org/mpeuni/bj-cbvexdv.html | 1,642,478,509,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00350.warc.gz | 67,855,089 | 3,823 | Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-cbvexdv Structured version Visualization version GIF version
Theorem bj-cbvexdv 32431
Description: Version of cbvexd 2277 with a dv condition, which does not require ax-13 2245. (Contributed by BJ, 16-Jun-2019.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
bj-cbvaldv.1 𝑦𝜑
bj-cbvaldv.2 (𝜑 → Ⅎ𝑦𝜓)
bj-cbvaldv.3 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
Assertion
Ref Expression
bj-cbvexdv (𝜑 → (∃𝑥𝜓 ↔ ∃𝑦𝜒))
Distinct variable groups: 𝑥,𝑦 𝜑,𝑥 𝜒,𝑥
Allowed substitution hints: 𝜑(𝑦) 𝜓(𝑥,𝑦) 𝜒(𝑦)
Proof of Theorem bj-cbvexdv
StepHypRef Expression
1 bj-cbvaldv.1 . . . 4 𝑦𝜑
2 bj-cbvaldv.2 . . . . 5 (𝜑 → Ⅎ𝑦𝜓)
32nfnd 1782 . . . 4 (𝜑 → Ⅎ𝑦 ¬ 𝜓)
4 bj-cbvaldv.3 . . . . 5 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
5 notbi 309 . . . . 5 ((𝜓𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
64, 5syl6ib 241 . . . 4 (𝜑 → (𝑥 = 𝑦 → (¬ 𝜓 ↔ ¬ 𝜒)))
71, 3, 6bj-cbvaldv 32430 . . 3 (𝜑 → (∀𝑥 ¬ 𝜓 ↔ ∀𝑦 ¬ 𝜒))
87notbid 308 . 2 (𝜑 → (¬ ∀𝑥 ¬ 𝜓 ↔ ¬ ∀𝑦 ¬ 𝜒))
9 df-ex 1702 . 2 (∃𝑥𝜓 ↔ ¬ ∀𝑥 ¬ 𝜓)
10 df-ex 1702 . 2 (∃𝑦𝜒 ↔ ¬ ∀𝑦 ¬ 𝜒)
118, 9, 103bitr4g 303 1 (𝜑 → (∃𝑥𝜓 ↔ ∃𝑦𝜒))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∀wal 1478 ∃wex 1701 Ⅎwnf 1705 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-10 2016 ax-11 2031 ax-12 2044 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-ex 1702 df-nf 1707 This theorem is referenced by: bj-cbvexdvav 32437
Copyright terms: Public domain W3C validator | 878 | 1,584 | {"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} | 2.921875 | 3 | CC-MAIN-2022-05 | latest | en | 0.238939 |
https://beastpapers.com/marketing-101-if-the-total-costs-are-10000-for-the-production/ | 1,721,399,405,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00021.warc.gz | 103,562,576 | 8,901 | Select Page
if the total costs are \$10,000 for the production of 1,000 units, assuming we completed then all, the cost per unit is simple: \$10,000 divided by 1,000 equals \$10 per unit. But when we didn’t complete all of the units that we worked on, it gets more complicated. So if total costs are \$10,000 for the production of 1,000 units and 800 of them are completed with 200 of them remaining in work-in-process at the end of the month that are 50% complete, we have a new situation. 200 units X 50% = 100 equivalent units that is added to 800 completed units to get 900 equivalent units of production. Now \$10,000/900 E.U. = \$11.11 per E.U.
Now we can use this unit cost to work through reconciliation of the costs transferred out and the cost in ending inventory. So, class, how much cost will remain in the ending inventory, and how much will be transferred out of this department? | 222 | 895 | {"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} | 3.625 | 4 | CC-MAIN-2024-30 | latest | en | 0.946602 |
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