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http://www.jiskha.com/members/profile/posts.cgi?name=Byron | 1,462,425,644,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.19/warc/CC-MAIN-20160428161525-00015-ip-10-239-7-51.ec2.internal.warc.gz | 596,486,022 | 4,798 | Thursday
May 5, 2016
# Posts by Byron
Total # Posts: 21
math
calculate the verticle distance a object dropped from rest covers in 1 seconds
March 1, 2016
math
for a game , three people are chosen in the first round. Rach those people chooses 3 people in the second round and so on. how many people are chosen in the 6th round.
January 11, 2016
Algebra
Factor the Polynomial wv-4v+wk-4k
October 29, 2015
math
Ashton broke apart a number into 30+7. What number did he start with?
September 17, 2014
art
Context Concept Content Iconography Convention Media/Medium
January 27, 2014
physics
Find the magnitude of the linear momentum of (i) a 7.1 kg bowling ball traveling at 12 m/s and (ii) a 200 kg automobile traveling at 90 km/h. (b) A pool player imparts an impulse of 3.2 N-s to a stationary 0.25 kg cue ball with a cue stick. What is the speed of the ball just ...
December 3, 2013
Math Homework, Help
Aaron is 21 years younger than Ryan. In 5 years Ryan will be twice as old as Aaron. How old are they now?
September 7, 2013
To begin you must plot the graph y=2x-1. If you can already do this disregard the next 2 paragraphs. There are two ways you can plot this line. The first and fastest way is to use the gradient intercept method. The basic format of a straight line is y=mx+b where m is the ...
April 10, 2013
math
if the mean is 69 the median is 83 and the mode is 85 and the range is 70 what is the second lowest test score?
December 20, 2011
Physics
The maximum speed of a 3.1-{\rm kg} mass attached to a spring is 0.70 m/s, and the maximum force exerted on the mass is 12 N. What is the amplitude of motion for this mass? Express your answer using two significant figures.
November 20, 2011
College economics
The long-run supply curve for a good is a horizontal line at a price $3 per unit of the good. The demand curve for the good is QD = 50-2P. (a) What is the equilibrium output of the good? (b) A$1 excise tax is imposed on the good. What will be the long-run effect on the ...
November 8, 2011
Economics
The long-run supply curve for a good is a horizontal line at a price $3 per unit of the good. The demand curve for the good is QD = 50-2P. (a) What is the equilibrium output of the good? (b) A$1 excise tax is imposed on the good. What will be the long-run effect on the ...
November 7, 2011
Economics
The long-run supply curve for a good is a horizontal line at a price $3 per unit of the good. The demand curve for the good is QD = 50-2P. (a) What is the equilibrium output of the good? (b) A$1 excise tax is imposed on the good. What will be the long-run effect on the ...
November 7, 2011
Med terms
Chest Computed Tomography I think should be "X-ray Computed Tomography", but I'm not sure.
July 12, 2011
science
how do juries eat
April 13, 2011
physics
A strong child pulls a sled with a force of 108 N at an angle of 33.6° above the horizontal. Find the vertical and horizontal components of this pull.
January 19, 2011
math 156
This week, the Laketown Cruise Company took its boat on 19 cruises, carrying the boat's full capacity of 44 guests on each cruise. If each guest paid \$24 admission, then about how much admission was collected by the company this week? A) Exact calculation B) Estimate
December 19, 2010
socical studies
what mountain rang lies in northeaster afghanistan
March 27, 2009
statistics
sample- a recent study reports that older adults who got regular physical exercise experienced fewer symptoms of depression, even when tested 2 years late. regular excersize scores= 1,3,4,3,5,2,3,4 no regular excersize scores=5,4,6,3,5,7,6,6 a)calculate the mean and the ...
January 30, 2008
math
-0.126/-0.9
August 29, 2007
Math
-0.126/-0.9
August 29, 2007
1. Pages:
2. 1 | 1,082 | 3,731 | {"found_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} | 3.734375 | 4 | CC-MAIN-2016-18 | longest | en | 0.930926 |
https://www.physicsforums.com/threads/two-linked-bodies-via-a-spring.933624/ | 1,679,409,222,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00004.warc.gz | 1,082,410,770 | 18,365 | # Two linked bodies via a spring
inv4lid
## Homework Statement
two linked are attached to each other with a spring. If the second body is placed on a fixed support, the length of the spring is 5 cm. If we fix the first body as in the picture, the length of the spring becomes 15cm. Determine the length of the spring in the non-deformed state.
m1 = 1 kg;
m2 = 4kg;
l1 = 5cm;
l2 = 15cm;
## Homework Equations
This time i have really no idea. The m1 pushes the spring with a force of 10N and m2 should like respond with another force? Quite don't get it.
## The Attempt at a Solution
Any help would be greatly appreciated.
#### Attachments
• _aCHfcg6QW6Hm3xwVqg3mg.png
5 KB · Views: 236
Homework Helper
Gold Member
2022 Award
I suggest using Hooke's law.
inv4lid
inv4lid
I suggest using Hooke's law.
10 N = kΔL?
ΔL1 = L1 - L0 from there
L0 = -ΔL1+L1 -> L0 = -ΔL1+5
ΔL2 = L2-L0 from there
L0 = -ΔL2+L2 -> L0 = -ΔL2+15
Quite still don't get it.
Homework Helper
Gold Member
2022 Award
10 N = kΔL?
ΔL1 = L1 - L0 from there
L0 = -ΔL1+L1 -> L0 = -ΔL1+5
ΔL2 = L2-L0 from there
L0 = -ΔL2+L2 -> L0 = -ΔL2+15
Quite still don't get it.
If we take the first case with the mass ##m_1##. Let ##L## be the natural length of the spring and ##x_1 > 0## be the contraction due to ##m_1##. Then, by Hooke's law:
##m_1g = kx_1##
Where ##k## is the (unknown) sprong constant.
Does that get you started?
inv4lid
inv4lid
If we take the first case with the mass ##m_1##. Let ##L## be the natural length of the spring and ##x_1 > 0## be the contraction due to ##m_1##. Then, by Hooke's law:
##m_1g = kx_1##
Where ##k## is the (unknown) sprong constant.
Does that get you started?
What is x?
I assume it's a different writing form of ΔL? Ok.
Sorry, but i have already tried that above.
mg (which is 10N) = kx, where both k & x are unknown
Homework Helper
Gold Member
2022 Award
What is x?
##x_1## is the distance that the spring is compressed from its natural length under the weight of ##m_1##.
inv4lid
Homework Helper
Gold Member
2022 Award
What is x?
I assume it's a different writing form of ΔL? Ok.
Sorry, but i have already tried that above.
mg (which is 10N) = kx, where both k & x are unknown
Yes, I forgot you had used ##\Delta L_1##.
It doesn't matter how many unknowns you have at this stage. The trick is keep going with the next equation and hope that you can get rid of the unknowns at some stage.
inv4lid
inv4lid
Yes, I forgot you had used ##\Delta L_1##.
It doesn't matter how many unknowns you have at this stage. The trick is keep going with the next equation and hope that you can get rid of the unknowns at some stage.
A question there: why do we need m2 there if it doesn't influence the object?
Homework Helper
Gold Member
2022 Award
A question there: why do we need m2 there if it doesn't influence the object?
We don't. I think the premise is that they are joined together. The mass of ##m_2## doesn't affect the first scenario, nor does ##m_1## affect the second.
You might also ask how they managed to attach ##m_1## to the ceiling, but I wouldn't worry about that either.
inv4lid
inv4lid
Okay.
m1g = k(x1-x0)
m1g = k(x2-x0)
->
10 = k(5-x0)
10 = k(15-x0) that's though quite non-sense
Last edited:
Homework Helper
Gold Member
2022 Award
Okay.
m1g = k(x0-x1)
m1g = k(x0-x2)
->
10 = k(x0-5)
10 = k(x0-15) that's though quite non-sense
##m_1g = kx_1##
That seemed the simplest approach. My thinking was: the spring is being compressed by a certain amount. Let's call that ##x_1##.
But, if you are going to use:
##m_1g = k(x_0 - x_1)##
Then you have to be careful about what ##x_0## and ##x_1## are.
In your second equation things have gone wrong. You've still got ##10## for the force, which can't be right. And, you've got negative numbers creeping in.
Anyway, you need to fix those equations. We can stick with your notation, but you need to be careful about how you are defining things.
inv4lid
inv4lid
##m_1g = kx_1##
That seemed the simplest approach. My thinking was: the spring is being compressed by a certain amount. Let's call that ##x_1##.
But, if you are going to use:
##m_1g = k(x_0 - x_1)##
Then you have to be careful about what ##x_0## and ##x_1## are. ##x_1## can't be "5" here.
In your second equation things have gone wrong. You've still got ##10## for the force, which can't be right. And, you've got negative numbers creeping in.
Anyway, you need to fix those equations. We can stick with your notation, but you need to be careful about how you are defining things.
Okay. Ty for everything, going to solve it. | 1,417 | 4,573 | {"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 | 4 | CC-MAIN-2023-14 | latest | en | 0.914673 |
https://studylib.net/doc/7087684/wind-power-extension | 1,656,542,353,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00306.warc.gz | 597,351,338 | 12,450 | # Wind Power Extension
```Wind Power – A Viable Solution to the Energy Crisis?
For maximum efficiency, wind turbines are built to optimise the wind power they harness. For example, the blades
of a wind turbine must be at least 10m above the ground, as the wind closest to the ground experiences surface drag
as it passes over the Earth’s surface.
Wind turbines take energy from the wind,
creating a “shading” effect upon the land
directly beyond them, where the wind has
less energy (see diagram). Wind turbines are
therefore built with a separation much
greater than this “shading length”, which
ranges from 5-10 times the diameter of the
In the following calculations, you will determine the area required for a wind turbine farm to replace a fossil fuel
power station, and estimate the numbers and area that would be needed to meet the UK’s power requirements.
1. First calculate the area of land required for one wind turbine, so that it is correctly separated from its
neighbouring turbines. Assume that the turbine has a blade diameter of 50 m, and assume a separation
length of 10 blade diameters. Hint: a small “bird’s eye” sketch may help you to visualise this
2. Now calculate the area that would be required by a wind turbine farm with a power output equal to a fossil
fuel power station of 1 GW. Hint: take an average power output of 1 wind turbine as 2 MW
3. Compare the area of this wind farm with the area of the Royal Borough of Kingston upon Thames, 37.25 km2,
and Greater London, 1572 km2.
4. Calculate how many people this wind turbine farm could support, assuming that the average power
consumption per capita in the UK is 631 W. Compare this figure to the UK population of approximately 62
million.
5. From your calculations, do you think wind turbines are a viable solution for all of the UK’s energy
requirements. If so, where would you place the turbines.
``` | 429 | 1,884 | {"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} | 3.296875 | 3 | CC-MAIN-2022-27 | latest | en | 0.917883 |
http://nrich.maths.org/public/leg.php?code=-68&cl=2&cldcmpid=1015 | 1,502,927,602,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00591.warc.gz | 311,612,234 | 10,110 | # Search by Topic
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Can you fit the tangram pieces into the outline of this junk? | 2,205 | 9,624 | {"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-2017-34 | latest | en | 0.937724 |
https://www.mometrix.com/academy/adding-mixed-numbers-with-the-same-denominator/ | 1,566,226,068,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314752.21/warc/CC-MAIN-20190819134354-20190819160354-00189.warc.gz | 915,977,186 | 8,602 | Adding Mixed Numbers with the Same Denominator
There are two primary methods for adding and subtracting mixed numbers. The first method is to convert both terms to an improper fraction before adding. The second method is to add the whole numbers together and the fractions together separately.
Adding Mixed Numbers with the Same Denominator
Adding Mixed Numbers with the Same Denominator
There are two primary methods for adding or subtracting mixed numbers, and I’ll demonstrate both in each of these examples here. The first method is to convert both terms to an improper fraction before adding.
We’ll take 3 times 2 is 6, plus 2 is 8 over 3, plus 3 times 3 is 9, plus 2 is 11 over 3. We’ll add these together 8 plus 11 is 19 over 3, and 3 goes into 19, 6 times with the remainder of 1, so our solution is 6 and 1/ 3.
The second method is to add the whole number parts and the fraction parts separately. 2 plus 3 is 5, and 2/3 plus 2/3 is 4/3. Unfortunately, this is not a proper mixed number because the fractional part is greater than 1, so we’ll have to convert this fraction to a mixed number, 1 and 1/3, and then we can add the 5 back in and we get 6 and 1/3.
Now on the second problem we have 4 and 1/6 minus 2 and 5/6. First we’re going to do the improper fraction method 6 times 4 is 24, plus 1 is 25 over 6, minus, 6 times 2 is 12, plus 5 is 17 also over 6, and we will subtract 25 minus 17 is 8 over 6, or 4 over 3, which is equal to 1 and 1/3.
The second method has us subtracting 2 from 4 to get 2, and subtracting 5/6 from 1/6 to get negative 4/6. Now obviously this doesn’t work because we have a positive and a negative part to this number, so what we have to do is we can convert this whole number to 1 and 6 over 6, and then we can subtract the 4 over 6 to get 1 and 2 over 6, or 1 and 1/3.
396906
by Mometrix Test Preparation | Last Updated: August 15, 2019 | 543 | 1,887 | {"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} | 4.6875 | 5 | CC-MAIN-2019-35 | latest | en | 0.924972 |
https://mathhelpforum.com/threads/finding-a-cartesian-equation-for-an-r-value.221828/ | 1,582,146,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00384.warc.gz | 465,127,752 | 13,953 | # Finding a Cartesian equation for an r value
How would I find a Cartesian equation for r = 4secx?
Thanks.
#### Soroban
MHF Hall of Honor
$$\displaystyle \text{How would I find a Cartesian equation for: }\: r \,=\, 4\sec\theta\,?$$
I will assume you know the conversion identities.
We have: .$$\displaystyle r \:=\:\frac{4}{\cos\theta} \quad\Rightarrow\quad r\cos\theta \:=\:4 \quad\Rightarrow\quad x \,=\,4$$
1 person | 135 | 426 | {"found_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} | 3.015625 | 3 | CC-MAIN-2020-10 | latest | en | 0.703677 |
https://turingbotsoftware.com/blog/category/machine-learning/ | 1,606,150,554,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141163411.0/warc/CC-MAIN-20201123153826-20201123183826-00568.warc.gz | 540,437,910 | 21,872 | ## Decision boundary discovery with symbolic regression
An interesting classification problem is trying to find a decision boundary that separates two categories of points. For instance, consider the following cloud of points:
Clearly, we could hand draw a line that separates the two colors. But can this problem be solved in an automatic way?
Several machine learning methods could be used for this, including for instance a Support Vector Machine or AdaBoost. What all of these methods have in common is that they perform complex calculations under the hood and spill out some number, that is, they are black boxes. An interesting comparison of several of these methods can be found here.
A simpler and more elegant alternative is to try to find an explicit mathematical formula that separates the two categories. Not only would this be easier to compute, but it would also offer some insight into the data. This is where symbolic regression comes in.
### Symbolic regression
The way to solve this problem with symbolic regression is to look for a formula that returns 0 for points of one category and 1 for points of another. That is, a formula for classification = f(x, y).
We can look for that formula by generating a CSV file with our points and loading it into TuringBot. Then we can run the optimization with classification accuracy as the search metric.
If we do that, the program ends up finding a simple formula with an accuracy of 100%:
``classification = ceil(-1*tanh(round(x*y-cos((-2)*(y-x)))))``
To visualize the decision boundary associated with this formula, we can generate some random points and keep track of the ones classified as orange. Then we can find the alpha shape that encompasses those points, which will be the decision boundary:
```import alphashape
from descartes import PolygonPatch
import numpy as np
from math import *
def f(x, y):
return ceil(-1*tanh(round(x*y-cos((-2)*(y-x)))))
pts = []
for i in range(10000):
x = np.random.random()*2-1
y = np.random.random()*2-1
if f(x, y) == 1:
pts.append([x, y])
pts = np.array(pts)
alpha_shape = alphashape.alphashape(pred, 2.)
fig, ax = plt.subplots()
And this is the result:
It is worth noting that even though this was a 2D problem, the same procedure could have been carried out for a classification problem in any number of dimensions.
## How to create an AI trading system
Predicting whether the price of a stock will rise or fall is perhaps one of the most difficult machine learning tasks. Signals must be found on datasets which are dominated by noise, and in a robust way that will not overfit the training data.
In this tutorial, we are going to show how an AI trading system can be created using a technique called symbolic regression. The idea will be to try to find a formula that classifies whether the price of a stock will rise or fall in the following day based on its price candles (open, high, low, close) in the last 14 days.
Our AI trading system will be a classification algorithm: it will take past data as input, and output 0 if the stock is likely to fall in the following day and 1 if it is likely to rise. The first step in generating this model is to prepare a training dataset in which each row contains all the relevant past data and also a 0 or 1 label based on what happened in the following day.
We can be very creative about what past data to use as input while generating the model. For instance, we could include technical indicators such as RSI and MACD, sentiment data, etc. But for the sake of this example, all we are going to use are the OHLC prices of the last 14 candles.
Our training dataset should then contain the following columns:
` open_1,high_1,low_1,close_1,...,open_14,high_14,low_14,close_14,label`
Here the index 1 denotes the last trading day, the index 2 the trading day prior to that, etc.
### Generating the training dataset
To make things interesting, we are going to train our model on data for the S&P 500 index over the last year, as retrieved from Yahoo Finance. The raw dataset can be found here: S&P 500.csv.
To process this CSV file into the format that we need for the training, we have created the following Python script which uses the Pandas library:
```import pandas as pd
training_data = []
for i,row in df.iterrows():
if i < 13 or i+1 >= len(df):
continue
features = []
for j in range(i, i-14, -1):
features.append(df.iloc[j]['Open'])
features.append(df.iloc[j]['High'])
features.append(df.iloc[j]['Low'])
features.append(df.iloc[j]['Close'])
if df.iloc[i+1]['Close'] > row['Close']:
features.append(1)
else:
features.append(0)
training_data.append(features)
columns = []
for i in range(1, 15):
columns.append('open_%d' % i)
columns.append('high_%d' % i)
columns.append('low_%d' % i)
columns.append('close_%d' % i)
columns.append('label')
training_data = pd.DataFrame(training_data, columns=columns)
training_data.to_csv('training.csv', index=False)```
All this script does is iterate through the rows in the Yahoo Finance data and generate rows with the OHLC prices of the last 14 candles, and an additional ‘label’ column based on what happened in the following day. The result can be found here: training.csv.
### Creating a model with symbolic regression
Now that we have the training dataset, we are going to try to find formulas that predict what will happen to the S&P 500 in the following day. For that, we are going to use the desktop symbolic regression software TuringBot. This is what the interface of the program looks like:
The input file is selected from the menu on the upper left. We also select the following settings:
• Search metric: classification accuracy.
• Test/train split: 50/50. This will allow us to easily discard overfit models.
• Test sample: the last points. The other option is “chosen randomly”, which would make it easier to overfit the data due to autocorrelation.
With these settings in place, we can start the search by clicking on the play button at the top of the interface. The best solutions found so far will be shown in real time, ordered by complexity, and their out-of-sample errors can be seen by toggling the “show cross validation” button on the upper right.
After letting the optimization run for a few minutes, these were the models that were encountered:
The one with the best ouf-of-sample accuracy turned out to be the one with size 23. Its win rate in the test domain was 60.5%. This is the model:
`label = 1-floor((open_5-high_4+open_12+tan(-0.541879*low_1-high_1))/high_13)`
It can be seen that it depends on the low and high of the current day, and also on a few key parameters of previous days.
### Conclusion
In this tutorial, we have generated an AI trading signal using symbolic regression. This model had good out-of-sample accuracy in predicting what the S&P 500 would do the next day, using for that nothing but the OHLC prices of the last 14 trading days. Even better models could probably be obtained if more interesting past data was used for the training, such as technical indicators (RSI, MACD, etc).
## How to create an equation for data points
In order to find an equation from a list of values, a special technique called symbolic regression must be used. The idea is to search over the space of all possible mathematical formulas for the ones with the greatest accuracy, while trying to keep those formulas as simple as possible.
In this tutorial, we are going to show how to find formulas using the desktop symbolic regression software TuringBot, which is very easy to use.
### How symbolic regression works
Symbolic regression starts from a set of base functions to be used in the search, such as addition, multiplication, sin(x), exp(x), etc, and then tries to combine those functions in all possible ways with the goal of finding a model that will be as accurate as possible in predicting a target variable. Some examples of base functions used by TuringBot are the following:
As important as the accuracy of a formula is its simplicity. A huge formula can predict with perfect accuracy the data points, but if the number of free parameters in the model is the same as the number of points then this model is not really informative. For this reason, a symbolic regression optimization will discard a larger formula if it finds a smaller one that performs just as well.
### Finding a formula with TuringBot
Finding equations from data points with TuringBot is a simple process. The first step is selecting the input file with the data through the interface. This input file should be in TXT or CSV format. After it has been loaded, the target variable can be selected (by default it will be the last column in the file), and the search can be started. This is what the interface looks like:
Several options are available on the menus on the left, such as setting a test/train split to be able to detect overfit solutions, selecting which base functions should be used, and selecting the search metric, which by default is root-mean-square error, but that can also be set to classification accuracy, mean relative error and others. For this example, we are going to keep it simple and just use the defaults.
The optimization is started by clicking on the play button at the top of the interface. The best formulas found so far will be shown in the solutions box, ordered by complexity:
The software allows the solutions to be exported to common programming languages from the menu, and also to simply be exported as text. Here are the formulas in the example above exported in text format:
```Complexity Error Function
1 1.91399 -0.0967549
3 1.46283 0.384409*x
4 1.362 atan(x)
5 1.18186 0.546317*x-1.00748
6 1.11019 asinh(x)-0.881587
9 1.0365 ceil(asinh(x))-1.4131
13 0.985787 round(tan(floor(0.277692*x)))
15 0.319857 cos(x)*(1.96036-x)*tan(x)
19 0.311375 cos(x)*(1.98862-1.02261*x)*tan(1.00118*x)```
### Conclusion
In this tutorial, we have seen how symbolic regression can be used to find formulas from values. Symbolic regression is very different from regular curve-fitting methods, since no assumption is made about what the shape of the formulas should be. This allows patterns to be found in datasets with an arbitrary number of dimensions, making symbolic regression a general purpose machine learning technique.
## Machine learning black box models: some alternatives
In this article, we will discuss a very basic question regarding machine learning: is every model a black box? Certainly most methods seem to be, but as we will see, there are very interesting exceptions to this.
### What is a black box method?
A method is said to be a black box when it performs complicated computations under the hood that cannot be clearly explained and understood. Data is fed into the model, internal transformations are performed on this data and an output is given, but these transformations are such that basic questions cannot be answered in a straightforward way:
• Which of the input variables contributed the most to generating the output?
• Exactly what features did the model derive from the input data?
• How does the output change as a function of one of the variables?
Not only are black box models hard to understand, they are also hard to move around: since complicated data structures are necessary for the relevant computations, they cannot be readily translated to different programming languages.
### Can there be machine learning without black boxes?
The answer to that question is yes. In the simplest case, a machine learning model can be a linear regression and consist of a line defined by an explicit algebraic equation. This is not a black box method, since it is clear how the variables are being used to compute an output.
But linear models are quite limited and cannot perform the same kinds of tasks that neural networks do, for example. So a more interesting question is: is there a machine learning method capable of finding nonlinear patterns in an explicit and understandable way?
It turns out that such method exists, and is called symbolic regression.
### Symbolic regression as an alternative
The idea of symbolic regression is to find explicit mathematical formulas that connect input variables to an output, while trying to keep those formulas as simple as possible. The resulting models end up being explicit equations that can be written on a sheet of paper, making it apparent how the input variables are being used despite the presence of nonlinear computations.
To give a clearer picture, consider some models found by TuringBot, a symbolic regression software for PC:
In the “Solutions” box above, a typical result of a symbolic regression optimization can be seen. A set of formulas of increasing complexity was found, with more complex formulas only being shown if they perform better than all simpler alternatives. A nonlinearity in the input dataset was successfully recovered through the use of nonlinear base functions like cos(x), atan(x) and multiplication.
Symbolic regression is a very general technique: although the most obvious use case is to solve regression problems, it can also be used to solve classification problems by representing categorical variables as different integer numbers, and running the optimization with classification accuracy as the search metric instead of RMS error. Both of these options are available in TuringBot.
### Conclusion
In this article, we have seen that despite most machine learning methods indeed being black boxes, not all of them are. A simple counterexample are linear models, which are explicit and hence not black boxes. More interestingly, we have seen how symbolic regression is capable of solving machine learning tasks where nonlinear patterns are present, generating models that are mathematical equations that can be analyzed and interpreted.
## A regression model example and how to generate it
Regression models are perhaps the most important class of machine learning models. In this tutorial, we will show how to easily generate a regression model from data values.
### What is regression
The goal of a regression model is to be able to predict a target variable taking as input one or more input variables. The simplest case is that of a linear relationship between the variables, in which case basic methods such as least squares regression can be used.
In real-world datasets, the relationship between the variables is often highly non-linear. This motivates the use of more sophisticated machine learning techniques to solve the regression problems, including for instance neural networks and random forests.
A regression problem example is to predict the value of a house from its characteristics (location, number of bedrooms, total area, etc), using for that information from other houses which are not identical to it but for which the prices are known.
### Regression model example
To give a concrete example, let’s consider the following dataset of house prices: house_prices.txt. It contains the following columns:
```Index;
Local selling prices, in hundreds of dollars;
Number of bathrooms;
Area of the site in thousands of square feet;
Size of the living space in thousands of square feet;
Number of garages;
Number of rooms;
Number of bedrooms;
Age in years;
Construction type (1=brick, 2=brick/wood, 3=aluminum/wood, 4=wood);
Number of fire places;
Selling price.```
The goal is to predict the last column, the selling price, as a function of all the other variables. In order to do that, we are going to use a technique called symbolic regression, which attempts to find explicit mathematical formulas that connect the input variables to the target variable.
We will use the desktop software TuringBot, which can be downloaded for free, to find that regression model. The usage is quite straightforward: you load the input file through the interface, select which variable is the target and which variables should be used as input, and then start the search. This is what its interface looks like with the data loaded in:
We have also enabled the cross validation feature with a 50/50 test/train split (see the “Search options” menu in the image above). This will allow us to easily discard overfit formulas.
After running the optimization for a few minutes, the formulas found by the program and their corresponding out-of-sample errors were the following:
The highlighted one turned out to be the best — more complex solutions did not offer increased out-of-sample accuracy. Its mean relative error in the test dataset was of roughly 8%. Here is that formula:
`price = fire_place+15.5668+(1.66153+bathrooms)*local_pric`
The variables that are present in it are only three: the number of bathrooms, the number of fire places and the local price. It is a completely non-trivial fact that the house price should only depend on these three parameters, but the symbolic regression optimization made this fact evident.
### Conclusion
In this tutorial, we have seen an example of generating a regression model. The technique that we used was symbolic regression, implemented in the desktop software TuringBot. The model that was found had a good out-of-sample accuracy in predicting the prices of houses based on their characteristics, and it allowed us to clearly see the most relevant variables in estimating that price.
## A free AI software for PC
If you are interested in solving AI problems and would like an easy to use desktop software that yields state of the art results, you might like TuringBot. In this article, we will show you how it can be used to easily solve classification and regression problems, and explain the methodology that it uses, which is called symbolic regression.
### The software
TuringBot is a desktop application that runs on both Windows and Linux, and that can be downloaded for free from the official website. This is what its interface looks like:
The usage is simple: you load your data in CSV or TXT format through the interface, select which column should be predicted and which columns should be used as input, and start the search. The program will look for explicit mathematical formulas that predict this target variable, and show the results in the Solutions box.
### Symbolic regression
The name of this technique, which looks for explicit formulas that solve AI problems, is symbolic regression. It is capable of solving the same problems as neural networks, but in an explicit way that does not involve black box computations.
Think of what Kepler did when he extracted his laws of planetary motion from observations. He looked for algebraic equations that could explain this data, and found timeless patterns that are taught to this day in schools. What TuringBot does is something similar to that, but millions of times faster than a human could ever do.
An important point in symbolic regression is that it is not sufficient for a model to be accurate — it also has to be simple. This is why TuringBot’s algorithm tries to find the best formulas of all possible sizes simultaneously, discarding larger formulas that do not perform better than simpler alternatives.
### The problems that it can solve
Some examples of problems that can be solved by the program are the following:
• Regression problems, in which a continuous target variable should be predicted. See here a tutorial in which we use the program to recover a mathematical formula without previous knowledge of what that formula was.
• Classification problems, in which the goal is to classify inputs into two or more different categories. The rationale of solving this kind of problem using symbolic regression is to represent different categorical variables as different integer numbers, and run the optimization with “classification accuracy” as the search metric (this can easily be selected through the interface). In this article, we teach how to use the program to classify the Iris dataset.
• Classification of rare events, in which a classification task must be solved on highly imbalanced datasets. The logic is similar to that of a regular classification problem, but in this case a special metric called F1 score should be used (also available in TuringBot). In this article, we found a formula that successfully classified credit card frauds on a real-world dataset that is highly imbalanced.
### Getting TuringBot
If you liked the concept of TuringBot, you can download it for free from the official website. There you can also find the official documentation, with more information about the search metrics that are available, the input file formats and the various features that the program offers.
## How to find a formula for the nth term of a sequence
Given a sequence of numbers, finding an explicit mathematical formula that computes the nth term of the sequence can be challenging, except in very special cases like arithmetic and geometric sequences.
In the general case, this task involves searching over the space of all mathematical formulas for the most appropriate one. A special technique exists that does just that: symbolic regression. Here we will introduce how it works, and use it to find a formula for the nth term in the Fibonacci sequence (A000045 in the OEIS) as an example.
### What symbolic regression is
Regression is the task of establishing a relationship between an output variable and one or more input variables. Symbolic regression solves this task by searching over the space of all possible mathematical formulas for the ones with the greatest accuracy, while trying to keep those formulas as simple as possible.
The technique starts from a set of base functions — for instance, sin(x), exp(x), addition, multiplication, etc. Then it tries to combine those base functions in various ways using an optimization algorithm, keeping track of the most accurate ones found so far.
An important point in symbolic regression is simplicity. It is easy to find a polynomial that will fit any sequence of numbers with perfect accuracy, but that does not really tell you anything since the number of free parameters in the model is the same as the number of input variables. For this reason, a symbolic regression procedure will discard a larger formula if it finds a smaller one that performs just as well.
### Finding the nth Fibonacci term
Now let’s show how symbolic regression can be used in practice by trying to find a formula for the Fibonacci sequence using the desktop symbolic regression software TuringBot. The first two terms of the sequence are 1 and 1, and every next term is defined as the sum of the previous two terms. Its first terms are the following, where the first column is the index:
```1 1
2 1
3 2
4 3
5 5
6 8
7 13
8 21
9 34
10 55```
A list of the first 30 terms can be found on this file: fibonacci.txt.
TuringBot takes as input TXT or CSV files with one variable per column and efficiently finds formulas that connect those variables. This is how it looks like after we load fibonacci.txt and run the optimization:
The software finds not only a single formula, but the best formulas of all possible complexities. A larger formula is only shown if it performs better than all simpler alternatives. In this case, the last formula turned out to predict with perfect accuracy every single one of the first 30 Fibonacci terms. The formula is the following:
`f(x) = floor(cosh(-0.111572+0.481212*x))`
Clearly a very elegant solution. The same procedure can be used to find a formula for the nth term of any other sequence (if it exists).
### Conclusion
In this tutorial, we have seen how the symbolic regression software TuringBot can be used to find a closed-form expression for the nth term in a sequence of numbers. We found a very short formula for the Fibonacci sequence by simply writing it into a text file with one number per row and loading this file into the software.
If you are interested in trying TuringBot your own data, you can download it from the official website. It is available for both Windows and Linux.
## A machine learning software for data science
Data science is becoming more and more widespread, pushed by companies that are finding that very valuable and actionable information can be extracted from their databases.
It can be challenging to develop useful models from raw data. Here we will introduce a tool that makes it very easy to develop state of the art models from any dataset.
### What is TuringBot
TuringBot is a desktop machine learning software. It runs on both Windows and Linux, and what it does is generate models that predict some target variable taking as input one or more input variables. It does that through a technique called symbolic regression. This is what its interface looks like:
The idea of symbolic regression is to search over the space of all possible mathematical formulas for the ones that best connect the input variables to the target variable, while trying to keep those formulas as simple as possible. The target variable can be anything: for instance, it can represent different categorical variables as different integer numbers, allowing the program to solve classification problems, or it can be a regular continuous variable.
### Machine learning with TuringBot
The usage of TuringBot is very straightforward. All you have to do is save your data in CSV or TXT format, with one variable per column, and load this input file through the program’s interface.
Once the data is loaded, you can select the target variable and which variables should be used as input, as well as the search metric, and then start the search. Several search metrics are available, including RMS error, mean error and classification accuracy. A list of formulas encountered so far will be shown in real time, ordered by complexity. Those formulas can be easily exported as Python, C or text from the interface:
Most machine learning methods are black boxes, which carry out complex computations under the hood before giving a result. This is how neural networks and random forests work, for instance. A great advantage of TuringBot over these methods is that the models that it generates are very explicit, allowing some understanding to be gained into the data. This turns data science into something much more similar to natural science and its search for mathematical laws that explain the world.
### How to get the software
If you are interested in trying TuringBot on your own data, you can download it for free from the official website. There you can also find the official documentation, with detailed information about all the features and parameters of the software. Many engineers and data scientists are already making use of the software to find hidden patterns in their data.
## Symbolic regression tutorial with TuringBot
In this tutorial, we are going to show how you can find a formula from your data using the symbolic regression software TuringBot. It is a desktop software that runs on both Windows and Linux, and as you will see the usage is very simple.
### Preparing the data
TuringBot takes as input files in .txt or CSV format containing one variable per column. The first row may contain the names of the variables, otherwise they will be labelled col1, col2, col3, etc.
For instance, the following is a valid input file:
```x y z w classification
5.20 2.70 3.90 1.40 1
6.50 2.80 4.60 1.50 1
7.70 2.80 6.70 2.00 2
5.90 3.20 4.80 1.80 1
5.00 3.50 1.60 0.60 0
5.10 3.50 1.40 0.20 0
4.60 3.10 1.50 0.20 0
6.90 3.20 5.70 2.30 2```
This is what the program looks like when you open it:
By clicking on the “Input file” button on the upper left, you can select your input file and load it. Different search metrics are available, including for instance classification accuracy, and a handy cross validation feature can also be enabled in the “Search options” box — if enabled, it will automatically create a test/train split and allow you to see the out-of-sample error as the optimization goes on. But in this example we are going to keep things simple and just use the defaults.
### Finding the formulas
After loading the data, you can click on the play button at the top of the interface to start the optimization. The best formulas found so far will be shown in the “Solutions” box, in ascending order of complexity. A formula is only shown if its accuracy is greater than that of all simpler alternatives — in symbolic regression, the goal is not simply to find a formula, but to find the simplest ones possible.
Here are the formulas it found for an example dataset:
The formulas are all written in a format that is compatible out of the box with Python and C. Indeed, the menu on the upper right allows you to export the solutions to these languages:
In this example, the true formula turned out to be sqrt(x), which was recovered in a few seconds. The methodology would be the same for a real-world dataset with many input variables and an unknown dependency between them.
### How to get TuringBot
If you have liked this tutorial, we encourage you to download TuringBot for free from the official website. As we have shown, it is very simple to use, and its powerful mathematical modelling capabilities allow you to find very subtle numerical patterns in your data. Much like a scientist would do from empirical observations, but in an automatic way and millions of times faster.
## Machine learning with symbolic regression
Many machine learning methods are presently available, including for instance neural networks, random forests and support vector machines. In this article, we will talk about a very unexplored algorithm called symbolic regression, and will show how it can be used to solve machine learning problems in a very transparent and explicit way.
### What is machine learning
Machine learning concerns algorithms capable of predicting numerical values (regression) and creating classifications, among other tasks. The real world is messy and randomness appears everywhere, so a major challenge that these algorithms face is being able to discern meaningful signals from the underlying noise contained in the training datasets.
What most machine learning methods have in common is that they are very implicit and resemble black boxes: numbers are fed into the model, and it spits out a result after performing a series of complex computations under the hood. This kind of processing of information is strongly connected to the notion of “artificial intelligence”, since the inner workings of the human brain are also very hard to describe, while it is capable of learning and recognizing patterns across a very wide range of domains.
### Symbolic regression
Symbolic regression is a technique that looks for mathematical formulas that predict some target variable taking as input one or more input variables. Thus, a symbolic model is nothing more than an algebraic formula that can be written on a piece of paper.
A simple case of symbolic model is a polynomial. Any dataset can be represented with perfect accuracy by a polynomial, but that is not very interesting because polynomials quickly diverge outside the train domain, and because they contain as many free parameters as the training dataset itself. So they do not really compress information in any way.
More interesting models are found by combining a set of base functions and trying to find the simplest combinations that predict some target variable. Examples of base functions are trigonometric functions, exponentials, sum, multiplication, division, etc.
For instance, these are some of the base functions used by the symbolic regression software TuringBot:
After the base functions are defined, the task is then to combine them in such way that a target variable is successfully predicted from the input variables. There is more than one way to carry out the optimization — one might be interested in maximizing the classification accuracy, or in recovering the overall shape of a curve without much regard for outliers, etc. For this reason, TuringBot allows many different search metrics to be used:
Some examples of problems that can be solved with symbolic regression include:
Clearly the method is very general, and can be creatively used to solve a variety of problems.
### Conclusion
In this article, we have seen how symbolic regression is an alternative machine learning method capable of generating explicit models and solving various classes of problems in an elegant way. If you are interested in generating symbolic models from your own data and seeing what patterns it can find, you can download the symbolic regression software TuringBot, which works on both Windows and Linux, for free. | 6,771 | 32,890 | {"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-2020-50 | longest | en | 0.909988 |
https://researchoutput.ncku.edu.tw/zh/publications/the-approximability-of-the-p-hub-center-problem-with-parameterize | 1,627,266,909,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00024.warc.gz | 515,130,300 | 12,590 | # The approximability of the p-hub center problem with parameterized triangle inequality
Li Hsuan Chen, Sun Yuan Hsieh, Ling Ju Hung, Ralf Klasing
8 引文 斯高帕斯（Scopus）
## 摘要
A complete weighted graph G = (V, E, w) is called Δβ -metric, for some β ≥ 1/2, if G satisfies the β-triangle inequality, i.e., w(u, v) ≤ β · (w(u, x) + w(x, v)) for all vertices u, v, x ∈ V. Given a Δβ -metric graph G = (V, E, w) and an integer p, the Δβ -pHub Center Problem (Δβ -pHCP) is to find a spanning subgraph H of G such that (i) vertices (hubs) in C ⊂V form a clique of size p in H ; (ii) vertices (non-hubs) in V \C form an independent set in H ; (iii) each non-hub v ∈ V \C is adjacent to exactly one hub in C ; and (iv) the diameter D(H ) is minimized. For β = 1, Δβ -pHCP is NP-hard. (Chen et al., CMCT 2016) proved that for any ε > 0, it is NP-hard to approximate the Δβ -pHCP to within a ratio43 − ε for β = 1. In the same paper, a53 -approximation algorithm was given for Δβ-pHCP for β = 1. In this paper, we study Δβ -pHCP for all β ≥12. We show that for any ε > 0, to approximate the Δβ -pHCP to a ratio g(β) − ε is NP-hard and we give r(β)-approximation algorithms for the same problem where g(β) and r(β) are functions of β. If β ≤3−√23, we have r(β) = g(β) = 1, i.e., Δβ -pHCP is polynomial time solvable. If3−√23 < β ≤23, we have r(β) = g(β) =3β−2β23(1−β). For23 ≤ β ≤5+√105, r(β) = g(β) = β + β2. Moreover, for β ≥ 1, we have g(β) = β ·4β−13β−1 and r(β) = 2β, the approximability of the problem (i.e., upper and lower bound) is linear in β.
原文 English Computing and Combinatorics - 23rd International Conference, COCOON 2017, Proceedings Yixin Cao, Jianer Chen Springer Verlag 112-123 12 9783319623887 https://doi.org/10.1007/978-3-319-62389-4_10 Published - 2017 23rd International Conference on Computing and Combinatorics, COCOON 2017 - Hong Kong, China持續時間: 2017 八月 3 → 2017 八月 5
### 出版系列
名字 Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) 10392 LNCS 0302-9743 1611-3349
### Other
Other 23rd International Conference on Computing and Combinatorics, COCOON 2017 China Hong Kong 17-08-03 → 17-08-05
• 理論電腦科學
• 電腦科學(全部) | 788 | 2,211 | {"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} | 2.6875 | 3 | CC-MAIN-2021-31 | latest | en | 0.793299 |
https://www.coursehero.com/tutors-problems/Computer-Science/11511207-Implement-a-program-that-requests-a-decimal-value-representing-a-resta/ | 1,544,506,435,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823565.27/warc/CC-MAIN-20181211040413-20181211061913-00292.warc.gz | 842,927,709 | 21,968 | View the step-by-step solution to:
# Implement a program that requests a decimal value representing a restaurant bill total and an integer representing a percentage.
Implement a program that requests a decimal value representing a restaurant bill total and an integer representing a percentage. The program should compute the dollar amount of the tip and print all three values to the screen. You may assume that the user enters numbers for the two values and that the numbers are not negative.
Note that although the numbers below format as we would expect dollars and cents to format (mostly), you do not have to concern yourself with formatting to two decimal places .
>>> Enter a restaurant bill (> 0): 45.37
>>> Enter the tip percentage (e.g. 15 = 15%): 20
The tip for a bill of \$45.37 at 20% is 9.074.
Sample output of the program is as follows: Enter a restaurant bill (> 0): 45.37 Enter the tip... View the full answer
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Browse Documents | 282 | 1,319 | {"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} | 3.171875 | 3 | CC-MAIN-2018-51 | latest | en | 0.878792 |
http://mathhelpforum.com/pre-calculus/147735-simultaneous-equation.html | 1,508,301,577,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822739.16/warc/CC-MAIN-20171018032625-20171018052625-00429.warc.gz | 207,855,495 | 11,385 | 1. ## Simultaneous equation
By eliminating y, find the solutions to the simultaneous equations.
x^2+y^2=25
y=x-7
I've tried many different approaches, come up with 4+ different answers so I have no idea what to do.
2. Originally Posted by Mukilab
By eliminating y, find the solutions to the simultaneous equations.
x^2+y^2=25
y=x-7
I've tried many different approaches, come up with 4+ different answers so I have no idea what to do.
Substitute the second equation into the first:
x^2 + (x - 7)^2 = 25.
Expand, simplify, re-arrange to get a quadratic = 0 and solve for x in the usual way.
If you need more help, please show all your work and say where you get stuck.
3. Originally Posted by Mukilab
By eliminating y, find the solutions to the simultaneous equations.
x^2+y^2=25
y=x-7
I've tried many different approaches, come up with 4+ different answers so I have no idea what to do.
You should get two pairs of solutions. It tells you to eliminate y so put x-7 in for y in the first equation
$x^2+(x-7)^2 = 25$ When it's expanded and put into standard form we get ......
Solve using your favourite method although this one factorises to .....
From this find your value of x and for each one put the value into either equation to find y.
4. Got 2x^2-14x+24=0 after expanding and simplifying.
divided by 2
x^2-7x+12=0
put into the quadratic equation got
$
x=\frac{7+\sqrt{1}}{2}$
or $x=\frac{7-\sqrt{1}}{2}$
correct?
5. Sorry how is this calculas.
Actually. What is calculas? All I know that it has some meagre connection to engineering and Newton, I think he created it?
Time for research.
6. Originally Posted by Mukilab
Got 2x^2-14x+24=0 after expanding and simplifying.
divided by 2
x^2-7x+12=0
put into the quadratic equation got
$
x=\frac{7+\sqrt{1}}{2}$
or $x=\frac{7-\sqrt{1}}{2}$
correct?
That is correct but since $\sqrt{1} = 1$ you can simplify your answers
This is pre-calc rather than calculus and although I don't know how the US system works I'd imagine pre-calc sets you up for calculus. By and large calculus is largely about differentiation (rate of change) and integration (area under a graph). Of course it's far more complicated than that.
7. Thanks, got x=3,4
Didn't know about the squareroute of 1 being 1, thanks. I guess I just didn't bother to think about that. How silly of me.
8. Originally Posted by Mukilab
Thanks, got x=3,4
Didn't know about the squareroute of 1 being 1, thanks. I guess I just didn't bother to think about that. How silly of me.
I got those answers too. Now you can use those values to find y using either equation (equation 2 is simplest). You should get two pairs of values | 748 | 2,657 | {"found_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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-43 | longest | en | 0.930068 |
https://inches-to-cm.appspot.com/2/hr/50.4-palaca-u-centimetara.html | 1,721,157,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00749.warc.gz | 284,336,988 | 6,413 | Palaca U Centimetara
# 50.4 in na cm50.4 Palaca na Centimetara
in
=
cm
## Kako to pretvoriti 50.4 palaca na centimetara?
50.4 in * 2.54 cm = 128.016 cm 1 in
## Pretvori 50.4 in na zajedničke duljine
Mjerna jedinicaJedinica za duljinu
Nanometar1280160000.0 nm
mikrometar1280160.0 µm
Milimetar1280.16 mm
Centimetar128.016 cm
Palac50.4 in
Stopa4.2 ft
Jard1.4 yd
Metar1.28016 m
Kilometar0.00128016 km
Milja0.0007954545 mi
Nautička milja0.0006912311 nmi
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https://yasincapar.com/taylor-expansion/ | 1,618,842,740,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00385.warc.gz | 1,196,678,829 | 17,482 | The Taylor expansion is one of the most useful ideas in mathematics. Most functions and polynomials are generally smooth to solve. Polynomials are easier to work than almost any other sort of function, and polynomials are largely able to approximate to functions. The Taylor formula gives us an equation for the polynomial expansion for nearly every smooth function of f.
## Taylor Series
In mathematics, the Taylor series of a function is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor’s series are named after Brook Taylor who introduced them in 1715.
## Taylor’s Theorem
In calculus, Taylor’s theorem gives an approximation of a k-times differentiable function around a given point by a polynomial of degree k, called the kth order Taylor polynomial. For a smooth function, the Taylor polynomial is the truncation at the order k of the Taylor series of the function. The first-order Taylor polynomial is the linear approximation of the function, and the second-order Taylor polynomial is often referred to as the quadratic approximation. There are several versions of Taylor’s theorem, some giving explicit estimates of the approximation error of the function by its Taylor polynomial.
Taylor’s theorem gives simple arithmetic formulas to accurately compute values of many transcendental functions such as the exponential function and trigonometric functions. It is the starting point of the study of analytic functions as well as in numerical analysis and mathematical physics.
The partial sum formed by the first n + 1 terms of a Taylor series is a polynomial of degree n that is called the nth Taylor polynomial of the function. Taylor polynomials are approximations of a function, which become generally better as n increases.
## Taylor Series vs Taylor Theorem
Both are commonly used to describe a sum to n formulated to match up to the nth order derivatives of a function around a point. Taylor series implies that this sum is infinite, while Taylor theorem (polynomial) can take any positive integer value of n.
A Taylor polynomial has a finite number of terms, whereas a Taylor series has infinitely many terms. The Taylor polynomials are the partial sums of the Taylors series. For instance, the Taylor series of ex about x = 0 is → 1 + x + x2/2! + x3/3! + x4/4! + ….
The 3th degree for Taylor polynomial of ex about x = 0 is → 1 + x + x2/2! + x3/3!
### Formulation
The idea behind the Taylor expansion is that we can rewrite every smooth function as an infinite sum of polynomial terms. Let f : R → R is a differentiable function and a R, then a Taylor series of the function f(x) around the point a is:
The Taylor series (also known as Power series), where n! denotes the factorial of n, can be written as;
In particular, if zero is the point, where the derivatives → a = 0, then the expansion is known as the Maclaurin series after Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century, and thus is given by:
The Maclaurin series for 1 / 1 – x is the geometric series of 1 + x + x2 + x3 + ⋯ , so the Taylor series for 1 / x at a = 1 is → 1 − (x − 1) + (x − 1)2 − (x − 1)3 + ⋯ . By integrating the Maclaurin series, we find the Maclaurin series for ln(1 − x), where ln denotes the natural logarithm:
The corresponding Taylor series for ln x at a = 1 is:
and more generally, the corresponding Taylor series for ln x at an arbitrary nonzero point a is:
The Maclaurin series for the exponential function → ex = 1 + x + x2/2! + x3/3! + x4/4! + x5/5! + …
### Exemplary
e2 = 2.71828… × 2.71828… = 7.389056…
let’s try more terms of infinite series:
The higher degree of the Taylor polynomial gives better approximation to the function at x, if the Taylor series converges to the function at x.
#### Newton’s Method and Root Finding
• Newton’s method is an iterative method for approximating solutions (finding roots) to equations. If f is a positive definite quadratic function, Newton’s method can find the minimum of the function directly in practice, this almost never happens. Instead, Newton’s method can be applied when the function f is not truly quadratic but, it can be locally approximated as a positive definite quadratic.
For more detail about The Newton-Raphson Method
## In Conclusion
### Taylor’s Theorem
Taylor’s theorem is used for the expansion of the infinite series such as sing(x), log(x) etc. So that we can approximate the values of these functions or polynomials.
In higher mathematics, Taylor’s theorem gives an approximation of k-times differentiable function around a given point by a with k-nth order Taylor polynomial. For analytic functions, the Taylor polynomials at a given point are finite order truncations of its Taylor series, which completely determines the function in some point. There are several versions of different situations, and some of them contain explicit estimates on the approximation error of the function by its Taylor polynomial.
### Functions
#### 1 – log(1+x)
as a formulation for the convergence of the series → |x| < 1;
#### 2 – ex
For each k-time we have a Taylor polynomial for the function.
as a formulation → |x| < 1;
#### 3 – cos x
The derivative of cos is −sin, and the derivative of sin is cos;
cos(x) = cos(a) − [sin(a)/1!](x-a) − [cos(a)/2!](x-a)2 + [sin(a)/3!](x-a)3 + …
If a = 0, then cos(0) = 1 and sin(0) = 0:
as a formulation → |x| < 1;
#### 4 – sin x
as a formulation → |x| < 1;
#### 5 – 1/(1−x)
as a formulation → |x| < 1;
Sources:
le.ac.uk
brilliant.org
medium.com
feaforall.com
wikipedia.org
physicsforums.com
suzyahyah.github.io
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## 金融代写|投资组合代写Investment Portfolio代考|RISK-NEUTRAL PROBABILITIES
Another way to represent prices is via a risk-neutral probability. To distinguish between a risk-neutral probability and the probability under which we have been taking expectations, it is common to call the latter (whether objective or subjective) the physical probability. ${ }^5$ A risk-neutral probability is defined from the physical probability and a strictly positive SDF. Assume there is a strictly positive $\operatorname{SDF} \tilde{m}$, and, for each event $A$, let $1_A$ denote the indicator function of $A$, that is, for each state of the world $\omega, 1_A(\omega)=1$ if $\omega \in A$ and $1_A(\omega)=0$ if $\omega \notin A$.
Suppose first that there is a risk-free asset. For each event $A$, define
$$\mathbb{Q}(A)=R_f E\left[\tilde{m} 1_A\right] .$$
Then, $\mathbb{Q}$ is the risk-neutral probability associated to $\tilde{m}$. In a finite-state world, (3.21) means that the risk-neutral probability of each state is the product of $R_f$, the physical probability of the state, and the value of $\tilde{m}$ in the state (it is also the product of $R_f$ with the state price). In general, $\mathbb{Q}$ is a probability: $\mathbb{Q}(A) \geq 0, \mathbb{Q}(\Omega)=1$, where $\Omega$ is the set of all states of the world, and if $A_1, A_2, \ldots$ is a sequence of disjoint events, then $\mathbb{Q}\left(\cup A_i\right)=\sum \mathbb{Q}\left(A_i\right)$. As with any probability, there is an expectation operator associated with $\mathbb{Q}$. Denote it by $E^$. The definition of $\mathbb{Q}$ can be restated as $$\mathrm{E}^\left[1_A\right]=R_f \mathrm{E}\left[\tilde{m} 1_A\right],$$
because the expectation of an indicator function is the probability of the event. More generally, the definition of $\mathbb{Q}$ implies that
$$\mathrm{E}^[\tilde{x}]=R_f \mathrm{E}[\tilde{m} \tilde{x}]$$ for every $\tilde{x}$ for which the expectation $\mathrm{E}[\tilde{m} \tilde{x}]$ exists. Because the price of any payoff $\tilde{x}$ is $\mathrm{E}[\tilde{m} \tilde{x}]$, equation (3.22) implies that the price of any payoff $\tilde{x}$ is $$\frac{1}{R_f} \mathrm{E}^[\tilde{x}]$$
## 金融代写|投资组合代写Investment Portfolio代考|HANSEN-JAGANNATHAN BOUNDS
This section derives lower bounds on the standard deviations of SDFs due to Hansen and Jagannathan (1991). The Hansen-Jagannathan bounds have real economic significance. As discussed previously, an asset pricing model is a specification of an SDF $\tilde{m}$. A model can be rejected by the Hansen-Jagannathan bound if $\tilde{m}$ is not sufficiently variable. An illustration of the economic significance of the Hansen-Jagannathan bound with a risk-free asset is given in Exercise 7.2.
Continue to assume that asset payoffs have finite variances and the law of one price holds. Therefore, an SDF exists.
Hansen-Jagannathan Bound with a Risk-Free Asset
Assume there is a risk-free asset. Then, (3.12) holds for any SDF $\tilde{m}$, which we repeat here:
$$\mathrm{E}[\tilde{R}]-R_f=-R_f \operatorname{cov}(\tilde{m}, \tilde{R}) .$$
Letting $\operatorname{corr}(\tilde{m}, \tilde{R})$ denote the correlation of $\tilde{m}$ with $\tilde{R}$, we can write this as
$$\operatorname{corr}(\tilde{m}, \tilde{R}) \times \operatorname{stdcv}(\tilde{m})=\frac{\mathrm{E}[\tilde{R}]-R_f}{R_f \operatorname{stdev}(\tilde{R})} .$$
Because the correlation is between $-1$ and 1 , this implies
$$\operatorname{stdev}(\tilde{m}) \geq \frac{\left|\mathrm{E}[\tilde{R}]-R_f\right|}{R_f \operatorname{stdev}(\tilde{R})} .$$
Recalling that $1 / R_f=\mathrm{E}[\tilde{m}]$, we can rewrite this as
$$\frac{\operatorname{stdev}(\tilde{m})}{\mathrm{E}[\tilde{m}]} \geq \frac{\left|\mathrm{E}[\tilde{R}]-R_f\right|}{\operatorname{stdev}(\tilde{R})} .$$
The Sharpe ratio of a risky asset with return $\tilde{R}$ is the ratio of the risk premium $\mathrm{E}[\tilde{R}]-R_f$ to the risk $\operatorname{stdev}(\tilde{R})$. Thus, the ratio on the right-hand side of (3.35) is the absolute value of the Sharpe ratio of the return $\tilde{R}$. Hence, the ratio of the standard deviation of any SDF to its mean must be at least as large as the maximum absolute Sharpe ratio of all returns. This is one version of the Hansen-Jagannathan (1991) bounds.
## 金融代写|投资组合代写Investment Portfolio代考|RISK-NEUTRAL PROBABILITIES
$$\mathbb{Q}(A)=R_f E\left[\tilde{m} 1_A\right] .$$
$\mathbb{Q}(A) \geq 0, \mathbb{Q}(\Omega)=1$ ， 在哪里 $\Omega$ 是世界上所有状态的集合，如果 $A_1, A_2, \ldots$ 是一系列不相交的事件， 那么 $\mathbb{Q}\left(\cup A_i\right)=\sum \mathbb{Q}\left(A_i\right)$. 与任何概率一样，有一个期望算子与 $\mathbb{Q}$. 表示为 $\mathrm{E}^{\wedge}$. 的定义 $\mathbb{Q}$ 可以重述为
$$\mathrm{E}^{\left[1_A\right]}=R_f \mathrm{E}\left[\tilde{m} 1_A\right]$$
$$\left.\mathrm{E}^{[} \tilde{x}\right]=R_f \mathrm{E}[\tilde{m} \tilde{x}]$$
$$\left.\frac{1}{R_f} \mathrm{E}^{[} \tilde{x}\right]$$
## 金融代写|投资组合代写Investment Portfolio代考|HANSEN-JAGANNATHAN BOUNDS
Hansen-Jagannathan 与无风险资产绑定
$$\mathrm{E}[\tilde{R}]-R_f=-R_f \operatorname{cov}(\tilde{m}, \tilde{R}) .$$
$$\operatorname{corr}(\tilde{m}, \tilde{R}) \times \operatorname{stdcv}(\tilde{m})=\frac{\mathrm{E}[\tilde{R}]-R_f}{R_f \operatorname{stdev}(\tilde{R})} .$$
$$\operatorname{stdev}(\tilde{m}) \geq \frac{\left|\mathrm{E}[\tilde{R}]-R_f\right|}{R_f \operatorname{stdev}(\tilde{R})} .$$
$$\frac{\operatorname{stdev}(\tilde{m})}{\mathrm{E}[\tilde{m}]} \geq \frac{\left|\mathrm{E}[\tilde{R}]-R_f\right|}{\operatorname{stdev}(\tilde{R})} .$$
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2 Nuclear Physics Nuclear Physics comprises the study of: The general properties of nuclei The particles contained in the nucleus The interaction between these particles Radioactivity and nuclear reactions Practical applications of nuclear phenomena: Medical radio-isotopes (imaging & therapy) Magnetic Resonance Imaging (MRI) Identification of materials (NAA, AMS) Dating of materials Power generation (fusion and fission) Weapons of mass destruction (WMD)
3 Properties of Nuclei Every atom contains at its center an extremely dense, positively charged nucleus, which is much smaller than the overall size of the atom, but contains most of its total mass. The nucleus is made of protons and neutrons Protons have positive electric charge Neutrons have no electrical charge 16 8 O A Z O Z = number of protons (atomic number) N = number of neutrons A = Z+N (mass number) Isotopes of an element have the same number of protons but different number of neutrons. Isotopes have similar chemical properties, and different physical properties.
4 Isotope A Z N Atomic Mass (u) Abundance 12 6 C % 13 6 C % [The chemical atomic weight of carbon is u] 14 6 C Radioactive [ 14 6 C decays by β- emission with Half-Life 5730 y]
5 Nuclear Size and Mass The radii of most nuclei is given by the equation: R = R 0 A 1/3 R: radius of the nucleus, A: mass number, R 0 = 1.2x10-15 m The mass of a nucleus is given by: m = A u u = 1.66 x10-27 kg (unified mass unit)
6 Nuclear Density Fe is the most abundant isotope of iron (91.7%) R = 1.2x10-15 m (56) 1/3 = 4.6x10-15 m [Compare with inter-atomic separations 0.1x x10-9 m] V = 4/3 π R 3 = 4.1x10-43 m 3 m = (56) 1.66x10-27 kg = 9.3x10-26 kg ρ = m / V = 2.3x10 17 kg/m 3 [Compare with the density of solid iron 7x10 3 kg/m 3 ] Since: m = A u and R = R 0 A 1/3 ρ = m / V = u / (4/3 π R 03 ) constant All nuclei have approximately the same density
7 Nuclear Binding Energy The total rest energy of the separated nucleons is greater than the rest energy of the nucleus. The difference is called the Binding Energy E B E B is a measure of the energy gained in forming the nucleus from the individual nucleons E B = (Z M H + N m N A ZM) c 2 Note that A ZM is the mass of the neutral atom containing the nucleus, and M H refers to the mass of hydrogen (to balance the Z electrons contained in A Z M)
8 Binding Energy per Nucleon The binding energy per nucleon E B /A is nearly constant as a function of A, suggesting that the nuclear force is saturated (a nucleon interacts only with its nearest neighbors).
9 The Nuclear Force The force that binds together protons and neutrons inside the nucleus is called the Nuclear Force Some characteristics of the nuclear force are: It does not depend on charge It is very short range m It is much stronger than the electric force It is saturated (nucleons interact only with near neighbors) It favors formation of pairs of nucleons with opposite spins We do not have a simple equation to describe the nuclear force
10 Nuclear Stability The energy of a group of nucleons in a square well ( a reasonable first approximation to a nucleus) is smaller if Z N. For large values of A we need N Z to compensate for the electric force
11 Nuclear Stability Plot of N vs. Z for known nuclides. The stable nuclides are indicated by the black dots. Non-stable nuclides decay by emission of particles, or electromagnetic radiation, in a process called radioactivity
12 Radioactivity Alpha Decay: nuclei that are too large to be stable tend to decay by alpha decay, the emission of an alpha particle. [An alpha particle is the 4 He nucleus, two protons and two neutrons] U Th + α Gamma Decay: the energy of internal motion (protons and neutrons in a nucleus is quantized. A nucleus has a set of allowed energy states (ground state and excited states) much like in an atom. Transitions between states lead to the emission of very energetic electromagnetic radiation called γ (gamma) rays U * U + γ
13 Radioactivity Beta Decay: consist in a) the emission of an electron β -, b) the emission of a positron β +, or c) electron capture. a) Nuclides with neutron to proton ratio too large for stability tend to decay by conversion of a neutron into a proton, an electron and an antineutrino: n p + β - + ν b) Nuclides with proton to neutron ratio too large for stability tend to decay by conversion of a proton into a neutron, a positron, and a neutrino: p n + β + + ν c) In certain cases an orbital electron can combine with a proton in the nucleus to form a neutron and a neutrino: p + β - n + ν
14 Radioactivity Radioactive decay is a statistical process. There is now way to predict when an individual nucleus will decay. However, it is possible to predict the decay rate of a group of nuclei. If the initial number of radioactive nuclei at time t = 0 is N 0, then the number of remaining nuclei at time t is given by: N(t) = N 0 exp(-λt) The constant λ is called the decay constant and can be interpreted as the probability per unit time that a nucleus will decay
15 Radioactivity The Half-Life T 1/2 is the time required for the number of radioactive nuclei to decrease to half the initial value N 0 N 0 /2 = N 0 exp(-λt 1/2 ) T 1/2 = ln 2 / λ The mean lifetime T mean, or lifetime, is the time required for the number of radioactive nuclei to decrease to 1/e of the initial value N 0 λt mean = 1 T mean = 1 / λ The number of decays per unit time dn/dt is the activity of a sample, and is measured in Curie (Ci) 1 Ci = 3.7x10 10 decays/second t 1/2 T 1/2 τ T mean
16 238 U decay series
17 14 C Dating Radioactive 14 C is continuously produced in the atmosphere [ 14 N(n,p) 14 C, the neutrons being produced by cosmic rays]. 14 C decays by β - emission with half life T 1/2 = 5730 y 14 C 14 N + β - + ν The chemical activity of 14 C is similar to that of 12 C, so living organisms have the same 14 C / 12 C ratio as in the atmosphere, which is about 1.35x When an organism dies it stops absorbing 14 C, and the ratio 14 C / 12 C decreases. Measurement of this ratio allows the calculation of the time of death of the organism.
18 Nuclear Reactions A nuclear reaction is a rearrangement of nuclear components induced by particle bombardment 4 2 He N 17 8 O H 14 7 N(α,p) 17 8 O Nuclear reactions are subject to the following conservation laws: Charge Momentum and angular momentum Energy Total number of nucleons
19 Nuclear Reactions Some nuclear reactions release energy, while other reactions require input energy to proceed The amount of energy released or absorbed in a nuclear reaction (in the center of mass reference frame) is called the Q value, or reaction energy: In a reaction: M A + M B M C + M D Q = (M A + M B -M C -M D ) c 2 If M A + M B >M C + M D Q > 0 exoergic reaction If M A + M B <M C + M D Q < 0 endoergic reaction An endoergic reaction will not proceed unless the incoming particle provides the reaction energy Q (in CM)
20 Particle Induced Nuclear Reactions: are used mostly for nuclear physics research, and for analytical purposes The reaction 7 Li(p,α) 4 He can be used to detect Li in solids [1-2 MeV protons]
21 Neutron Induced Nuclear Reactions Nuclear reactions induced by neutron bombardment are used: a) In analytical techniques such as neutron activation analysis b) In the generation of energy by Fission or Fusion It is convenient to distinguish between fast and slow neutrons Fast neutron kinetic energy 1 MeV Slow neutron or thermal neutron kinetic energy ~ ev The probability for a reaction to proceed or Cross Section depends strongly on the energy of the neutrons
22 Neutron Activation Analysis Low energy neutrons are likely to be captured by nuclei, with emission of radiation from the excited nucleus, as in X(n,γ)Y. The emitted radiation is a signature for the presence of element X Various layers of a painting are revealed using NAA
23 Fission and Fusion Binding energy per nucleon The binding energy per nucleon peaks for A ~ 70, then: a) splitting a large mass nucleus into two medium mass nuclei (fission), or b) fusing two low mass nuclei into a larger mass nucleus (fusion) result in the release of energy.
24 Fission Uranium has two main isotopes: 238 U with an abundance of 99.3% 235 U with an abundance of 0.7% When 235 U captures a thermal, or low energy, neutron it forms 236 U that decays by undergoing fission (85% of the time) n U 141 Ba + 92 Kr + 3n There is about 1 MeV per nucleon higher binding energy in the products of the reaction than in the Uranium. As a consequence, more that 200 MeV of energy are liberated in this reaction. Furthermore, the emitted neutrons can induce additional fission
25 Chain Fission Reaction To sustain a chain reaction one of the emitted neutrons must be captured by another 235 U nucleus The reproduction constant k is the average number of neutrons from each fission that cause subsequent fission k < 1 the reaction will die out k 1 the reaction will be sustained (nuclear reactor) k > 1 the reaction will run away (nuclear bomb)
26 Nuclear Fission Reactor Fission of 235 U is most likely for low energy neutrons. The neutrons emitted in fission are more energetic so they need to be slowed down by means of a moderator (water, graphite) placed in between the fuel rods. Control rods, of Cadmium or other materials, are further used to regulate the number of fission inducing neutrons
27 Fusion In fusion, two light nuclei such as deuterium 2 H and tritium 3 H fuse together to form a heavier nucleus: 2 H + 3 H 4 He + n MeV The energy released per nucleon is larger than that released in fission The production of power from fusion of light nuclei holds great promise because of the relative abundance of fuel, and the absence of some of the dangers inherent in fission. However, the technology necessary to make fusion practical, has not been developed yet.
28 Fusion The proton proton cycle, resulting in the formation of 4 He, and the release of considerable energy, is the primary source of the sun s energy In practice it is very difficult to generate the necessary density of super-hot gas (T > 10 7 K - plasma), and keep it confined, to facilitate the fusion process.
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### Chapter 7. Electron Structure of the Atom. Chapter 7 Topics
Chapter 7 Electron Structure of the Atom Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Chapter 7 Topics 1. Electromagnetic radiation 2. The Bohr model of | 8,746 | 36,550 | {"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} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.847524 |
https://brainly.com/question/13958 | 1,485,236,651,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284270.95/warc/CC-MAIN-20170116095124-00046-ip-10-171-10-70.ec2.internal.warc.gz | 797,364,628 | 8,778 | # A Businessman assigns five-digits account numbers to each of his clients using only the digits 2,3,4,5,6,7,8, and 9. How many account numbers are available if no digit is repeated ?
1
by jaylac1332
## Answers
• Brainly User
2014-02-26T16:25:50-05:00
### This Is a Certified Answer
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There are 8 digits. The possibilities are 8*7*6*5*4 (do the multiplications yourself) | 169 | 648 | {"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-2017-04 | latest | en | 0.925047 |
https://byjus.com/question-answer/study-the-following-information-carefully-and-answer-the-given-questions-s-t-u-v-w-3/ | 1,643,229,064,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00259.warc.gz | 213,302,873 | 17,011 | Question
# Study the following information carefully and answer the given questions. S, T, U, V, W, X, Y and Z are sitting around a circular table facing the centre but not necessarily in the same order. S sits third to the right of T. Only three people sit between U and Y. U is neither an immediate neighbour of S nor T. Only three people sit between T and W. Z sits to the immediate right of W. V sits third to the left of W. Which amongst the following is true regarding V, as per the given arrangement?
A
None of the given statements are true
B
Only three people sit between V and S
C
U sits second to the right of V
D
V sits second to the right of Y
E
Only two people sit between V and Z
Solution
## The correct option is C None of the given statements are trueNone of the given statements are true.Logical Reasoning
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View More | 213 | 883 | {"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-2022-05 | latest | en | 0.927427 |
http://www.kylesconverter.com/speed-or-velocity/kilometres-per-hour-to-meters-per-millenium | 1,603,367,562,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00412.warc.gz | 149,234,216 | 5,612 | # Convert Kilometres Per Hour to Meters Per Millenium
### Kyle's Converter > Speed Or Velocity > Kilometres Per Hour > Kilometres Per Hour to Meters Per Millenium
Kilometres Per Hour (km/h) Meters Per Millenium Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Meters Per Millenium to Kilometres Per Hour
(or just enter a value in the "to" field)
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Unit Descriptions
1 Kilometre per Hour:
Distance of one kilometer or 1 000 meters travelled in the time span of one hour or exactly 3 600 seconds. 1 Kilometer per hour (kph, km/h) = 0.277 777 778 meters per second (SI base unit). 1 km/h = 0.277777778 m/s.
1 Meter per Millenium:
1 Meter per millenium is approximately 3.168 874 x 10-11 meters per second (SI unit). Using 1000 Gregorian years of 365.2425 days.
Conversions Table
1 Kilometres Per Hour to Meters Per Millenium = 876582000070 Kilometres Per Hour to Meters Per Millenium = 613607400000
2 Kilometres Per Hour to Meters Per Millenium = 1753164000080 Kilometres Per Hour to Meters Per Millenium = 701265600000
3 Kilometres Per Hour to Meters Per Millenium = 2629746000090 Kilometres Per Hour to Meters Per Millenium = 788923800000
4 Kilometres Per Hour to Meters Per Millenium = 35063280000100 Kilometres Per Hour to Meters Per Millenium = 876582000000
5 Kilometres Per Hour to Meters Per Millenium = 43829100000200 Kilometres Per Hour to Meters Per Millenium = 1.753164E+12
6 Kilometres Per Hour to Meters Per Millenium = 52594920000300 Kilometres Per Hour to Meters Per Millenium = 2.629746E+12
7 Kilometres Per Hour to Meters Per Millenium = 61360740000400 Kilometres Per Hour to Meters Per Millenium = 3.506328E+12
8 Kilometres Per Hour to Meters Per Millenium = 70126560000500 Kilometres Per Hour to Meters Per Millenium = 4.38291E+12
9 Kilometres Per Hour to Meters Per Millenium = 78892380000600 Kilometres Per Hour to Meters Per Millenium = 5.259492E+12
10 Kilometres Per Hour to Meters Per Millenium = 87658200000800 Kilometres Per Hour to Meters Per Millenium = 7.012656E+12
20 Kilometres Per Hour to Meters Per Millenium = 175316400000900 Kilometres Per Hour to Meters Per Millenium = 7.889238E+12
30 Kilometres Per Hour to Meters Per Millenium = 2629746000001,000 Kilometres Per Hour to Meters Per Millenium = 8.76582E+12
40 Kilometres Per Hour to Meters Per Millenium = 35063280000010,000 Kilometres Per Hour to Meters Per Millenium = 8.76582E+13
50 Kilometres Per Hour to Meters Per Millenium = 438291000000100,000 Kilometres Per Hour to Meters Per Millenium = 8.76582E+14
60 Kilometres Per Hour to Meters Per Millenium = 5259492000001,000,000 Kilometres Per Hour to Meters Per Millenium = 8.76582E+15 | 885 | 2,681 | {"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-2020-45 | latest | en | 0.520135 |
http://www.jiskha.com/display.cgi?id=1332388888 | 1,493,437,814,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123270.78/warc/CC-MAIN-20170423031203-00272-ip-10-145-167-34.ec2.internal.warc.gz | 564,587,233 | 3,727 | # Math
posted by on .
The dimensions of an aquarium are 20 inches long, 14 inches wide, and 18 inches high. If 231 cubic inches of water fill 1 gallon, how many gallons of water are in the aquarium to the nearest gallon?
• Math - ,
Round to nearest whole numer.
Done!
• Math - correction - ,
almost . . .
volume of tank: 20*14*18 = 5040
volume of 1 gallon: 231
gallons in tank: 5040/231 = 21.8 | 118 | 402 | {"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-2017-17 | latest | en | 0.79659 |
https://byjusexamprep.com/gate-2022-rapid-revision-weekly-quiz-6-i-7c66b9a0-7870-11ec-ada7-94b704be1e31 | 1,669,622,284,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00082.warc.gz | 201,906,614 | 56,666 | Time Left - 25:00 mins
# GATE 2022 Rapid Revision Weekly Quiz 6
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Question 1
The semi-compact section of a laterally unsupported steel beam has an elastic section modulus, plastic section modulus and design bending compressive stress of 500 cm3, 650cm3 and 200MPa, respectively. The design flexural capacity (expressed in kNm) of the section is _________.
Question 2
For a deflection angle of 135°, the maximum efficiency (in %) for a Pelton wheel is:
Question 3
On the apron of spillway model built to scale of 1:25, a hydraulic jump of 90 mm is formed and time required to open the spillway gate is 100 seconds. If the discharge in prototype spillway is 1200 cumecs, then height of the jump in the prototype and discharge over the model are
Question 4
A symmetrical I-section is shown in figure below
The distance between the kerns for the section is _________ cm.
Question 5
The following data were obtained for planning the road development project on a district
a. Total area = 12000 km2
b. Agricultural and developed area = 1700 Km2
c. Existing railway track length = 130 km
d. Existing length of metalled road = 310 km
e. Existing length of unmetalled road = 450 km
f. Number of towns or village in different population ranges are given below
Calculate the additional length of metalled road by Nagpur plan formula for district.
Question 6
A 500 mm diameter pipeline of length 4.5 km connects reservoir A and B whose constant difference of water level is 12m. From a point located at a distance of 1.5km from reservoir A leads to the reservoir C, a branch pipe of length 1.25km such that water level in reservoir C is 15m below that of reservoir A. The diameter of the branch pipe such that the flow into both the reservoirs is same is __________mm [rounded to nearest lower integer]
Take co-efficient of friction = 0.0075
Question 7
Diameter a pipeline PQR suddenly changes from 18cm to 36 cm at B. If the pipe is carrying a discharge of 45 l/sec from P to R, then the head loss at Q is _________ cm [correct to two decimal places]
Question 8
The distance from the pipe wall at which the local velocity is equal to the average velocity for turbulent flow in pipes is given by kR,where R is the radius of the pipe.The value of k will be…(correct to 3 decimal places)
Question 9
A stepped circular shaft is fixed at ‘A’ and ‘C’ as shown in fig. The diameter of the shaft BC is twice of ‘AB’. The torsional – rigidity of AB is ‘GJ’. The stiffness coefficient k11 is —
Question 10
At a site, the subsoil consists 6m thick layer of dry sand (e = 0.85, = 14.3 kN/ , = 0.14mm, = 18.9 kN/ ). The water table is located at 4m depth below G.L what is height of capillary rise (if C= 0.5) and effective stress at the point where dry sand and saturated sand meet?
Question 11
There is a gravity dam of height 60 m and base width 300 m. The length of drainage filter provided is 120 m. Length of the projection of water surface on upstream side is calculated by assuming side slope as 1 in 2.75.
Calculate the seepage through the dam body assuming a free board of 2.5 m and permeability 4 × 10–7 m/sec.
Question 12
The rectangular column shown in the figure below carries 50 kN load having eccentricity 50 mm & 25 mm along x and y axis respectively the stress at point ‘a’ is
• 111 attempts | 873 | 3,370 | {"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-2022-49 | latest | en | 0.899349 |
https://resources.quizalize.com/view/quiz/percent-change-bb0ee703-2b70-4c8b-9e5b-7b4dc52a4b29 | 1,716,346,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00666.warc.gz | 431,244,469 | 14,960 | # Percent Change
## Quiz by Ryan Dagenhart
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If a shirt originally costs \$40 and is on sale for \$32, what is the percent decrease in price?
20%
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A garden originally had 200 flowers. After planting 50 more flowers, what is the percent increase in the number of flowers?
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A cookie recipe called for 200 grams of sugar, but Jane decided to decrease the sugar by 25%. How many grams of sugar did she use?
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150 grams
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A pet store had 80 fish in stock. After receiving a new shipment, they had 120 fish. What was the percent increase in the number of fish?
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60%
40%
30s
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A basketball team won 20 games this season, which was an increase of 25% from the previous season. How many games did they win in the previous season?
15 games
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18 games
30s
Teachers give this quiz to your class | 277 | 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} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.971229 |
https://www.studyhosts.com/consider-a-hydrodynamically-and-thermally-fully-developed-flow-through-a-tube-of-radius-r-the/ | 1,653,080,666,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00256.warc.gz | 1,174,168,977 | 11,185 | Consider a hydrodynamically and thermally fully developed flow through a tube of radius R. The…
Consider a hydrodynamically and thermally fully developed flow through a tube of radius R. The…
Consider a hydrodynamically and thermally fully developed flow through a tube of radius R. The velocity profile is parabolic. The governing equation for the radial variation is given by θ ′′ + (θ ′ /η) + λ 2 (1 − η 2 )θ = 0. A constant temperature boundary condition (T = Tw) is maintained at the wall. The non-dimensional temperature is given by θ = (T − Tw)/(Tc − Tw). Tc is the centreline temperature. Show that the radial variation of the temperature is given by
View complete question » Consider a hydrodynamically and thermally fully developed flow through a tube of radius R. The velocity profile is parabolic. The governing equation for the radial variation is given by θ ′′ + (θ ′ /η) + λ 2 (1 − η 2 )θ = 0. A constant temperature boundary condition (T = Tw) is maintained at the wall. The non-dimensional temperature is given by θ = (T − Tw)/(Tc − Tw). Tc is the centreline temperature. Show that the radial variation of the temperature is given by The parameter, −λ2 is the separation constant, used while applying the separation of variables technique. View less »Dec 08 2021 01:37 PM
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 629 | 2,684 | {"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-2022-21 | latest | en | 0.882742 |
https://csdt.org/culture/adinkra/spirals.html | 1,696,303,434,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00546.warc.gz | 214,998,406 | 16,758 | Adinkra symbols show traditional African ideas about geometry. Here we see two kinds of spirals. A coil of rope is a linear spiral. The space between each revolution is the same. Ropes are dead, they do not grow.
In contrast, the snail's shell is a logarithmic ("log") spiral. As the snail grows, its shell gets larger. Log spirals are growing spirals, so that is why they are used by nature. All living things must grow.
Rope coil as a linear spiral
Snail shell as log spiral
The log spiral is the most common shape in Adinkra symbols. Can you guess why? Click below for some hints:
Animal horns are not the only logarithmic spiral from nature. Many shapes created by living organisms can be modeled using a log spiral. Some have several full revolutions, like the snail shell. Others are just a small section (arc of a spiral), like the chicken's foot.
## Comparing the African and European Discovery of Log Spirals
The adinkra symbol Sankofa ("san-KOH-fah") shows a bird looking backwards. It means “You can always go back.” It is often a symbol for recovering cultural knowledge lost from colonialism or slavery.
The Sankofa symbol comes in two different versions. The first one looks like a natural bird. If we trace the outer edge, we see a log spiral. Below we see the abstract version of the same symbol. The bird is removed, so only the spiral remains in the abstract version. Note that they added reflection symmetry.
Now we know how Ghanaians discovered log spirals: by looking at nature. How did Europeans discover the log spiral? In 1637, Rene Descartes wrote a famous letter in which he describes the log spiral. But he describes it as “mechanical motion” (think of peeling fruit with a knife, starting from the top). Other mathematicians applied the log spirals to mechanical things like springs and ship's anchors. Europeans did not apply log spirals to nature until Sir John Leslie wrote Geometry of Curve Lines in 1813, almost 200 years after Descartes.
Abstract Sankofa
Descartes' spiral
In other words, Africans started with log curves representing nature. Europeans took 200 years to get there. Recall the section on making adinkra ink from tree bark: it does not kill the trees, and the bark collectors protect the forest. In contrast, the mechanical focus of Europeans created factories that destroyed trees and forests. It is not that Africans lacked math and technology. It is that their approach is so different that Europeans did not even recognize it as math. It is based on respectful relations with nature and people. The Europeans called that “primitive”. It did lack equations, but in other ways it was more advanced.
The goal of our website is not to say that one culture's math and science is better than another's. If we do that, we just become the mirror image of the colonial approach. Revenge and tit-for-tat never solves problems. Rather, we want to encourage a future where Indigenous knowledge and modern science can work hand in hand. Innovation needs to be open to contributions from everyone. One name for that approach is “ethnocomputing”. By using the simulation tools on this website, you can create your own ethnocomputing investigation.
## Dwennimmen
Geometry that Reminds Us Not to Bully
This Adinkra symbol, Dwennimmen ("GEN-ah-men") is a good example. It shows four log spirals. The spirals are in pairs because this represents two rams butting heads. Its proverb says: “it is the heart, and not the horns, that leads a ram to bully.” In other words, we have to take responsibility for our actions. Being a big person does not give you an excuse to push people around. Notice that symbol makes use of the log spiral, just like the real ram's horns.
## Gye Nyame
The Curves of Life
Recall that the other adinkra symbols with log spirals were representing shapes of living things: Dwennimmen is based on the ram's horn; sankofa on the bird's neck; Akoko nan on the chicken's foot; and Akoben on the horn of a bull. But what do the logarithmic curves of the Gye Nyame symbol represent?
The knobs down the middle of the symbol represent the knuckles on a fist, a symbol for power. The full meaning of the the Gye Nyame symbol is, "No one except God has the power of life." The curves at each end are not representing any one particular living shape. Rather, they are a general abstraction for life itself. Developing a general abstraction that fits all cases is the fundamental goal of science. Log curves are just one example of power laws in biology. Others include fractals and exponential scaling: these are still active areas of research. For more about the reason we see power laws in biology, click here.
Gye Nyame ("jeh N-yah-mee")
## How Tightly Coiled?
Adinkra Makes Use of Different Spiral Shapes
Adinkra carvers speak Twi. They say that some spirals are aboapua awan or “tightly coiled.” Others are ntitim awan or “loosely coiled”. You will be using the block below to control that variable (“c” for “coil”). See if you can change C to match the spiral shape. After you enter the number, click on the blue button. (Hint: stay between 0.1 and 2.1) | 1,171 | 5,130 | {"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-2023-40 | latest | en | 0.928608 |
https://ask.sagemath.org/questions/7982/revisions/ | 1,606,395,513,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00520.warc.gz | 206,000,320 | 6,580 | # Revision history [back]
### Porting a finite-differences-matrix from Matlab to Numpy
Hi there,
I'm an engineering student plagued by a prof who adores Matlab. For reasons I'm sure you understand, I'd like to avoid Matlab and learn Numpy/Scipy instead. Now I'm given this code, which I'm supposed to adapt to a problem given in class (irrelevant).
If you're somewhat familiar with Matlab and interested in helping me port a weird piece of code, then this question is for you -- read on :)
Short explanation: We're creating a meshgrid of nodes, and assign voltages (matrix Vstart) to some of them. We'd like to find the potentials of the nodes in between. You'd assume we want the voltage drop to follow ?²=0, and indeed I get a fine, smooth potential drop if I loop through the in-between-nodes 50 times and repeatedly assign to them the average value of the nodes around them (that was my approach -- code not shown here).
For starters, here's the Matlab initialization code (my comments):
for m = 1:length(N)
d = a/N(m); #distance between mesh nodes
N1 = N(m)+1; #side length of the central conductor
N2 = b/a*N(m)+1; #side length of meshgrid
Vstart = zeros(N2,N2);
Vstart(1,:) = Vb; #set the outer nodes to Vb
Vstart(:,1) = Vb;
Vstart(N2,:) = Vb;
Vstart(:,N2) = Vb;
lim1=(N2-N1)/2+1;
lim2=(N2+N1)/2;
Vstart(lim1:lim2,lim1:lim2) = Va; #set the inner nodes to Va
Now we're told to use a different approach than the one described above, and we're given a piece of Matlab code that fills a matrix A with a bunch of negative-quarters, and another matrix A with a long line of 0's, Vb's and Va's. It then solves a system and puts the results into V. I'm not quite sure how the matrix works here, although the many 1/4's make me think this is just my previous approach in disguise.
function[A,C] = mACfd(Vstart,N2)
C = zeros(N2^ 2,1);
for i = 1:N2
for j = 1:N2
k = (i-1)*N2+j;
A(k,k) = 1;
if (Vstart(i,j) == 0)
A(k,k-N2) = -1/4;
A(k,k+N2) = -1/4;
A(k,k+1) = -1/4;
A(k,k-1) = -1/4;
else C(k) = Vstart(i,j);
end;
end;
end;
[A,C] = mACfd(Vstart,N2);
V = inv(A)*C;
for i = 1:N2
V2D(i,:) = V((i-1)*N2+1:i*N2);
end;
I don't understand how matrices A and C are filled, what their dimensions are, what they're for, and how I can efficiently reproduce the algorithm in Python. Particularly the weird indexing irks me.
If I didn't make myself clear, please ask in the comments.
Thanks! | 727 | 2,447 | {"found_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} | 3.453125 | 3 | CC-MAIN-2020-50 | longest | en | 0.921356 |
http://mathhelpforum.com/calculus/7350-please-help-me.html | 1,524,218,404,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00498.warc.gz | 196,947,788 | 10,050 | Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.
3. The next one is REALLY confusing.
F(x)=f(x^3) and G(x)=(f(x))^3
a^2=13
f(a)=3
f'(a)=11
f'(a^3)=15
I found F'(a), but I can't figure out G'(a)
I really hate to ask so many questions, but I've been working on this for an hour and I've gotten nowhere on these. Help on ANY of them would be great.
Thank you.
2. Originally Posted by nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.
1. lnx^5 and I need to find the first and second derivatives
Let f(x) = [ln(x)]^5; I am assuming you mean this? Or, do you mean ln(x^5)?
I'll do both; I'll assume for case 1 you mean the first and for case 2 you mean the latter:
Case 1:
f(x) = [ln(x)]^5
f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
f''(x) = Use quotient rule:
Quotient rule states:
[f(x)/g(x)]' = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2
[x*20*[ln(x)]^3*(1/x) - [5*[ln(x)]^4]*1]/[x]^2
= [20*[ln(x)]^3 - 5*[ln(x)]^4]/x^2
Case 2:
f(x) = ln(x^5)
f'(x) = 5/x
f''(x) = -5/(x^2)
I think the above is pretty easy to see, so I think you mean the first one.
3. Originally Posted by nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.
5. find f'(x) when f(x)=-6cos(sin(x^6))
This is actually easier than it looks; we have a composition of functions.
f(x) = -6cos(sin(x^6))
f'(x) = -6*-sin(sin(x^6))*cos(x^6)*6*x^5
Simplify:
Thus, f'(x) = 36*x^5*cos(x^6)*sin(sin(x^6))
4. I was talking about the first way.
I tried putting in the answer you had for that one, and it said that it was incorrect. My question is how did you get the 5ln^4 times 1/x?
5. Originally Posted by AfterShock
This is actually easier than it looks; we have a composition of functions.
f(x) = -6cos(sin(x^6))
f'(x) = -6*-sin(sin(x^6))*cos(x^6)*6*x^5
Simplify:
Thus, f'(x) = 36*x^5*cos(x^6)*sin(sin(x^6))
Ah, thanks, I just made several stupid mistakes in that one.
6. Originally Posted by nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.
4. I need to find the f'(x) when f(x)=8e^(xcosx)
f(x) = 8e^(xcosx)
Another easy one;
f'(x) = 8*e^(x*cos(x))*{derivative of inside of composition here = g(x)}
g(x) = x*cos(x)
g'(x) = 1*cos(x) - x*sin(x)
Thus, f'(x) = 8*e^(x*cos(x))*(cos(x) - x*sin(x))
7. Originally Posted by AfterShock
f(x) = 8e^(xcosx)
Another easy one;
f'(x) = 8*e^(x*cos(x))*{derivative of inside of composition here = g(x)}
g(x) = x*cos(x)
g'(x) = 1*cos(x) - x*sin(x)
Thus, f'(x) = 8*e^(x*cos(x))*(cos(x) - x*sin(x))
I ALWAYS forget product rule, thank you so much, you're a life saver!
8. Originally Posted by nobodygirl
I was talking about the first way.
I tried putting in the answer you had for that one, and it said that it was incorrect. My question is how did you get the 5ln^4 times 1/x?
If you integrate both of them, you will see they are correct. That is, you will end up with f'(x) integrating f''(x) and f(x) integrating f'(x).
f(x) = [ln(x)]^5
f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
(1/x) is taking the derivative of the inside function; do you see why?
5*[ln(x)]^4*{derivative of composition here}, in which case we need to find the derivative of ln(x), which is 1/x, thus,
f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
9. Originally Posted by AfterShock
If you integrate both of them, you will see they are correct. That is, you will end up with f'(x) integrating f''(x) and f(x) integrating f'(x). | 1,448 | 4,470 | {"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} | 3.90625 | 4 | CC-MAIN-2018-17 | latest | en | 0.973437 |
https://www.jiskha.com/questions/1784591/you-deposit-2000-in-an-account-that-pays-7-interest-compounded-semiannually | 1,638,285,468,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00171.warc.gz | 905,537,090 | 4,922 | # math
You deposit \$2000
in an account that pays 7
%
interest compounded semiannually. After 2
years, the interest rate is increased to 7.32
%
compounded quarterly. What will be the value of the account after a total of 4
years?
1. 👍
2. 👎
3. 👁
1. amount after first 2 years = 2000(1.035)^4
then for the next 2 years we have i = .0732/4 = .0183
and n = 8 , number of quarters in 2 years
final amount = 2000(1.035)^4 * (1.0183)^8
= .....
1. 👍
2. 👎
2. You deposit2000 \$ in an account that pays6 % interest compounded semiannually. After years, the interest rate is increased to6.84 % compounded quarterly. What will be the value of the account after a total of 2years?
1. 👍
2. 👎
## Similar Questions
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Find the future value of \$800 deposited at 5% for 2 years if the account pays simple interest, and the account pays interest compounded annually | 734 | 2,710 | {"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} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.924866 |
http://www.jiskha.com/display.cgi?id=1334512208 | 1,455,271,259,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163663.52/warc/CC-MAIN-20160205193923-00173-ip-10-236-182-209.ec2.internal.warc.gz | 495,411,822 | 3,631 | Friday
February 12, 2016
# Homework Help: math
Posted by Gabby ( 5 grade) on Sunday, April 15, 2012 at 1:50pm.
find the area of a rectangle that is 4 meters by 8 meters.
A-12 SQUARE METERS
B-24 SQUARE METERS
C-36 SQUARE METERS
D- 32 SQUARE METERS
I PUT D, IS IT RIGHT? | 96 | 271 | {"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-2016-07 | longest | en | 0.859775 |
http://www.imp.kiev.ua/~kord/wiki/method_of_images.html | 1,511,218,829,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806258.88/warc/CC-MAIN-20171120223020-20171121003020-00705.warc.gz | 442,117,822 | 3,535 | dS Wiki Home | Publications | Talks | Courses
### Method of images
Method of images (or method of mirror images) is used in electrostatics (magnetostatics) to simply calculate or visualize the distribution of the electric (magnetic) field of a charge (magnet) in a vicinity to the conducting (superconducting) surface. It is based on the fact that the tangential (normal) component of the electrical (magnetic) field to the surface of a conductor (superconductor) is zero, and that some field E with rot E = 0 and div E = 0 in some region is uniquely defined by its normal component over the surface which confines this region (the uniqueness theorem).
Fig. 1. The field of a charge near a flat conducting surface, found by the method of images.
Fig. 2. A magnetic dipole over the superconducting surface. The field between the magnet and surface is the same as between this magnet and a symmetric one.
A textbook example [1] is an infinitely flat conducting surface (see Fig. 1). In this case the field distribution between the surface and a charge (right side in Fig. 1) is the same as between this charge and another charge (imaginary charge), which is the mirror image of the real charge in respect to the surface but has the opposite sign. Evidently, in case of an electrical dipole, the vector of the mirrored dipole will have the opposite sign. Therefore, the force between the electrical charge or system of charges and the conducting surface is attractive.
In case of magnet-superconductor pair (the superconductor here is ideal, to which the magnetic field does not penetrate), the mirror image of the magnet will have a magnetization vector which is mirrored but of the same sign (Fig. 2). This can be thought as due to additional sign change upon mirroring of an axial vector which the magnetization is. The force between the magnet and the superconducting surface is therefore repulsive.
Method of frozen mirror images is a modification of the method of images for magnet-superconductor systems that has been introduced [2] to take into account the magnetic flux pinning phenomenon. The method gives a simple representation of the magnetic field distribution generated by a magnet (a system of magnets) outside an infinitely flat surface of a perfectly hard (with infinite pinning force) type-II superconductor in a most general field cooled (FC) case, i.e. when the superconductor goes into superconducting state been already exposed to the magnetic field. The name originates from the replacement of certain elements in the original layout with imaginary magnets, which replicates the boundary conditions of the problem (see Dirichlet boundary conditions). In a simplest case of the magnetic dipole over the flat superconducting surface (see Fig. 3), the magnetic field, generated by a dipole moved from its initial position (at which the superconductor is cooled to the superconducting state) to a final position and by the screening currents at the superconducting surface, is equivalent to the field of three magnetic dipoles: the original one (1), its mirror image (3), and the magnetization inversed mirror image of it in its initial position (2). The method has been shown to work for the melt-textured high temperature superconductors (MT HTSC) [2], which are characterized by a strong pinning and is useful for calculation of the interaction in magnet-HTSC systems such as superconducting magnetic bearings [3], superconducting flywheels [4], MAGLEV [3], etc. [5,6] Fig. 3. Illustration of the frozen mirror image method for a simplest case of the magnetic dipole over a flat superconducting surface. References R. P. Feynman, R. B. Leighton, M. Sands The Feynman Lectures on Physics, Mainly Electromagnetism and Matter. A. A. Kordyuk, Magnetic levitation for hard superconductors - J. Appl. Phys. 83, 610 (1998). J. R. Hull, Superconducting bearings - Supercond. Sci. Technol. 13 (2), R1 (2000). A. V. Filatov, E. H. Maslen, Passive magnetic bearing for flywheel energy storage systems - IEEE Trans. Mag. 37 (6), 3913–3924 (2001). A. A. Kordyuk and V. V. Nemoshkalenko, Study of the dynamics of vortex structures in bulk HTS with levitation techniques - J. Low Temp. Phys. 130, 207-235 (2003). See also Superconducting Levitation Demos. | 989 | 4,267 | {"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-2017-47 | longest | en | 0.947491 |
http://mazentech.com/index.php/books/diagram-genus-generators-and-applications-chapman-hall-crc-monographs-and | 1,580,305,022,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00072.warc.gz | 107,314,270 | 9,175 | # Diagram genus, generators, and applications by Alexander Stoimenow
By Alexander Stoimenow
In knot concept, diagrams of a given canonical genus may be defined through a finite variety of styles ("generators"). Diagram Genus, turbines and Applications provides a self-contained account of the canonical genus: the genus of knot diagrams. the writer explores fresh examine at the combinatorial thought of knots and provides proofs for a few theorems.
The booklet starts off with an advent to the beginning of knot tables and the history information, together with diagrams, surfaces, and invariants. It then derives a brand new description of turbines utilizing Hirasawa’s set of rules and extends this description to push the compilation of knot turbines one genus extra to accomplish their category for genus four. next chapters conceal functions of the genus four class, together with the braid index, polynomial invariants, hyperbolic quantity, and Vassiliev invariants. the ultimate bankruptcy offers extra study with regards to turbines, which is helping readers see functions of turbines in a broader context.
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Extra info for Diagram genus, generators, and applications
Example text
1) Gi , then we call the (special) diagrams Di with Γ(Di ) = Gi the Murasugi atoms of D. Note that each separating Seifert circle of D gives a cut vertex of Γ(D). , [MP2]). 1 below, and [QW]). However, the information of how under a Murasugi sum in D crossings attached to a Seifert circle on either side are arranged relatively to each other is lost in Γ(D). , ind (G1 ∗ G2 ) = ind (G1 ) + ind(G2 ) if G1,2 are bipartite. We will treat below, simultaneously with (the sharpness of) MWF, the following conjecture of theirs.
Consequently, every underpass in D is followed by an overpass, and one easily sees also that every overpass is followed by an underpass. Thus D is alternating. 4 is called special alternating. A knot is special alternating if it has a special alternating diagram. Such knots were introduced and studied by Murasugi [Mu] and have a series of special features. Conversely all knots have a special (not necessarily alternating) diagram. Hirasawa [Hi2] shows how to a modify any knot diagram D into a special diagram D′ so that g(D) = g(D′ ).
2 The checkerboard coloring and checkerboard graph . . . . . Diagrammatic moves . . . . . . . . . . . . . . . . . . . . . . . 1 Flypes and mutations . . . . . . . . . . . . . . . . . . . 2 Bridges and wave moves . . . . . . . . . . . . . . . . . . Braids and braid representations . . . . . . . . . . . . . . . . . . Link polynomials . . . . . . . . . . . . . . . . . . . . . . . . | 795 | 3,574 | {"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} | 3.0625 | 3 | CC-MAIN-2020-05 | latest | en | 0.888793 |
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INSTABILITY IN GEOPHYSICAL FLOWS
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A defining aspect of the course is its emphasis on the numerical solution of boundary value problems; the student learns techniques whose value extends beyond the current topic. Oceanographers think of ocean circulation in terms of a "global conveyor belt," in which cold polar waters sink and then circulate around ocean basins, eventually warming in the tropics.
What Is Instability?
If the equilibrium is stable, disturbances will often take the form of oscillations (eg the car in Figure 1.3a), or waves. For example, the surface of a lake is never perfectly horizontal, but it is usually quite close, because the horizontal equilibrium condition is stable.
Goals
Once we have identified an equilibrium state, the next step is to determine its stability. Forcing by wind, sun, gravity and planetary rotation tends to push the system towards unstable conditions.
Tools
Numerical Solution of a Boundary Value Problem
Since the matrix eigenvalue problem can be easily solved using standard numerical routines (for example, the Matlab function eig5), this approach suggests a convenient way to solve the differential eigenvalue problem. We cannot say the value of the error term in general because it depends on the function f.
The Equations of Motion
This is where viscosity comes in – it models the frictional effect that molecular interactions exert on the macroscopic motions we can observe and measure. As with molecular motions, however, we must account for the effect the gusts have on the larger-scale motions we are trying to understand.
Although this analogy is imperfect,9 the eddy viscosity concept is a useful first step toward understanding the effects of small-scale turbulence. In this book, the quantity we call "viscosity" can refer to either molecular or eddy viscosity.
Appendix: A Closer Look at Perturbation Theory .1 The Parking Problem Revisited
This describes the unrestricted movement of the car away from an unstable equilibrium, i.e. top of a hill. In addition, we have the additional term that arises from the presence of the spring.
The Perturbation Equations
In the simplest convection problems, there is no need to solve differential equations; the problem is purely algebraic. From (2.5) we see that the sum of the first two terms on the right-hand side is equal to zero.1 All remaining terms are proportional to, which therefore cancels out and remains:. 2.14) Again, the equation is linear and does not involve.
Simple Case: Inviscid, Nondiffusive, Unbounded Fluid
Note that the greatest amplitude of the vertical velocity is found at the interface level, and that the motions decrease with distance from the interface over a length scale proportional to the wavelength of the perturbation, k˜−1. The integral on the right is simplified with property 5 of the delta function (Figure 2.5), resulting in . 2.41).
Viscous and Diffusive Effects
A beautiful laboratory demonstration of convective instability arising at the interface between two liquids is shown in Figure 2-6. The removal of this port can be seen to the left of the top panel.
Boundary Effects: the Rayleigh-Benard Problem
As we saw previously in the unbounded cases (section 2.3), the relative strength of gravity and viscosity/diffusion is determined by two factors: the spatial scale of the normal mode, and the orientation of the motions. For a fixed value ofm, as k˜ decreases (i.e. for wider convection cells, see Figure 2.9), the motion becomes mainly horizontal, so the gravitational acceleration is weaker.
Nonlinear Effects
Cells are formed with fluid rising in the center and sinking at the edges (NOAA). b) Convection cells on a 1000 km scale in the solar photosphere. The distance between the strongest plumes is generally of the same order of magnitude as the thickness of the convective layer, reflecting the underlying linear instability.
Summary
In most geophysical examples of convection, Ra is many orders of magnitude larger than the critical value.
Appendix: Waves and Convection in a Compressible Fluid
If the vertical displacement is small, then the buoyant acceleration is −gρ/ρ, where ρ is the density of the parcel minus that of the surrounding fluid. The difference between the density of the parcel and its surroundings is −ρozη, and the parcel oscillates with frequency ω=.
The Perturbation Equations We assume that
Next, assume that the velocity field consists of a parallel shear flow and a disturbance:. where is an arbitrary constant as defined in Section 3.1.1. But the buoyancy and viscosity terms are now neglected and instead we have two new terms that describe the interactions between the turbulence and the parallel shear flow.
Rayleigh’s Equation .1 Normal Modes in a Shear Flow
The tilt angle, ϕ, is the angle between the wave vector and the direction of the background current (x), and is in the range −π/2 ≤ ϕ ≤ π/2. Returning to the perturbation equation (3.12) and replacing it with (3.13), the normal mode form written in terms of σ, we obtain a second-order ordinary differential equation forw(ˆ z):. 3.17) This is called Rayleigh's equation after Lord Rayleigh, the inventor of the normal modes (Rayleigh, 1880).
Analytical Example: the Piecewise-Linear Shear Layer
Based on these results, we can state four rules about the fastest growing instability of a piecewise linear shear layer:. i) The slant angle is zero, i.e. the wave crests are perpendicular to the mean flow. iv). The disturbance moves with the speed of the background flow in the middle of the shear layer.
Solution Types for Rayleigh’s Equation
Numerical Solution of Rayleigh’s Equation
In the ocean example, one could easily imagine that a boundary placed e.g. 1000 m below the shear layer would have a negligible effect on the results and reduce N to ∼1000. This is precisely what we did in the analytical example of the piecewise-linear shear layer (Section 3.3).
Shear Scaling
Now suppose we want to analyze the stability of some class of profiles of the form (3.46) using the Rayleigh equation. Substituting the scaling transformations (3.47) and (3.51) and multiplying by h3/u20 yields. 3.53) The shear scaling is both a labor-saving device (you can analyze a whole class of flows at once) and a source of insight.
Oblique Modes and Squire Transformations
Conversely, for each oblique mode with wavevector(k, ) and growth rateσ, there is a corresponding 2D mode(k˜,0) with higher growth rateσ˜ =σ/cosϕ (Figure 3.10). As a result, the fastest growing mode is always 2D. If we have some arbitrary flow profile U(z), and we want to know the growth rate for allk, we only need to calculate the growth rate for the 2D cases, i.e. = 0, then for any = 0 simply multiply the result by cosϕ.
Rules of Thumb for a General Shear Instability
Numerical Examples
The result shown in Figure 3.12 is comparable to the piecewise linear shear layer: the growth rate rises to a peak around k =0.44, and the maximum value is σ =0.19. This aspect ratio can be compared to Figure 3.13: take the beam width as the width of the island and λ as the wavelength of the instability.
Perturbation Energetics
The energy flow carries kinetic energy in a vertical direction and its convergence (or divergence) at a given z results in an accumulation (or depletion) of energy at that height. Kinetic energy (a) is extracted from the mean flow within the shear layer by shear production (b) and then transported vertically away from the shear layer by energy flux (c).
Necessary Conditions for Instability
Therefore, SP must have at least one positive local maximum somewhere in the interior of the flow (e.g., Figure 3.16b). Therefore, the inflection point is also an extremum of the squared displacement:(Uz2)z =0, and we can rewrite (3.70) as.
The Wave Resonance Mechanism of Shear Instability
Now suppose that the lower edge of the shear layer is also deformed into a stationary sinusoid (Figure 3.20b). Let us next ask what becomes of a sinusoidal disturbance at the upper edge of the shear layer (Figure 3.22b).
Quantitative Analysis of Wave Resonance
Can these discrepancies be explained by the influence of each wave on the phase speed of the other. All of this can be seen quantitatively by writing the rate of rise and the phase velocity in terms of the phase relationship between the upper and lower waves.
Summary
Finally, the resonant interaction of vortex waves provides an explanation of the shear instability mechanism, Rayleigh's and Fjørtoft's theorems, and criteria for critical level and phase tilt. In more complex parallel shear flows, the instability can be understood in terms of the resonance of multiple eddy waves.
Appendix: Classical Proof of the Rayleigh and Fjørtoft Theorems
Characteristics of the fastest growing state:. i) The wave vector is directed parallel to the background current, i.e. there is no variation in the cross-flow direction. ii) If handu0 is defined as length and velocity scales characteristic of the mean flow U(z), then the wavelength is proportional to h and the growth rate is proportional tou0/h. Since dUz2/d z must change sign atzI, (3.95) requires atdUz2/d z to be negative just above zI and positive just below zI instead of the opposite, i.e. the inflection point must be a maximum (not a minimum) of Uz2.
When the tank is tilted, the dense lower layer flows down, forcing the upper layer up, resulting in an accelerated, stratified shear flow. However, in some cases, statically stable stratification can interact with shear to create new instability mechanisms that would not exist in a homogeneous shear flow.
The Richardson Number
Equilibria and Perturbations
The last term on the left describes the advection of the background buoyancy gradient by the vertical velocity perturbation, just as we saw in (2.12). The second term on the left is new; it describes the advection of buoyant disturbances by the background flow (which was zero in the motionless case).
Oblique Modes
The growth rate of the 3D mode is cosϕ times larger than the corresponding 2D mode, which exists in a liquid with stronger stratification. In most circumstances this means that the oblique mode will have a slower growth rate, but if stratification somehow increased the growth rate fast enough to compensate for the skewness factor cosϕ, then the oblique mode may grow faster.
The Taylor-Goldstein Equation
Now suppose that, when the rib is not too large, there is a stationary instability like the one we found in the unstratified case (section 3.9.1). Also, as this critical Richardson number is approached, the wavenumber of the fastest growing mode approaches 1/2, not much different from the value of 0.44 found in the unstratified case.
Application to Internal Wave Phenomena
More precisely, the ratio of the wavelength 2π/k to the shear layer thickness 2h approaches 2π, while in the homogeneous case the value is 7. For example, in the weakly nonlinear theory of solitary waves in a stratified shear flow, the dependence on x and t is described by the Korteweg-De Vries equation, while the vertical structure of the solution of equation TG (4.18) in hydrostasis is the tic limit (Lee and Beardsley, 1974 ).
Analytical Examples of Instability in Stratified Shear Flows Like the Rayleigh equation (3.16–3.19), the TG equation (4.18) is easy to solve
This is due to the delta function behavior of the vorticity and buoyancy gradient profiles viz. As Ribis increased, the growth rate of the instability is reduced and the band of wavenumbers it occupies shrinks to zero.
The Miles-Howard Theorem
It is important to understand what the Miles-Howard theorem says about the stability of certain U(z) and B(z) profiles and what it doesn't. We have already seen the case of a continuously stratified shear layer (4.19), where instability requires that the smallest Ri is less than 0.25, consistent with the Miles-Howard theorem.
Howard’s Semicircle Theorem
The stratification term on the left-hand side of (4.43) becomes insignificant when we take out the imaginary part. Therefore, in the second term on the right-hand side of (4.45), we can change U tocr, which results in
Energetics
By analogy with the development of the kinetic energy equation, we multiply by bybˆ∗, take the real part and divide by 2. Referring to the kinetic energy budget (4.52) we see that, when Bz >0, the momentum flux can only act to reduce growth, as we assumed in Section 41.
Summary
Appendix: Veering Flows
Appendix: Spatial Growth
So the real part of the phase velocity is, not surprisingly, the average of the velocities of the two currents. The momentum equation (1.19), neglecting buoyancy but retaining viscosity, is Du. 5.2) The viscosity term raises the order of the system and therefore requires additional boundary conditions.
Conditions for Equilibrium Consider a parallel shear flow
We now imagine a parallel shear flow in a fluid that is homogeneous but does not have a viscosity ν (e.g. Figure 5.1). For now we will just observe that a viscous fluid must move with the boundary, so for a stationary boundary =v=w=0.
Conditions for Quasi-Equilibrium: the Frozen Flow Approximation
In this case, the boundary atz= H moves with speed u0 and the boundary atz=0 is stationary (Figure 5.2b). For example, suppose the instability is a shear instability, so the growth rate is similar.
The Orr-Sommerfeld Equation We now substitute perturbation forms
Boundary Conditions for Viscous Fluid
Outside the viscous boundary layer, the velocity changes much more slowly; the fluid slides across as if the boundary were frictionless. An equivalent way of stating this is that the viscous flux flux-ν∂u/∂zand−ν∂v/∂zvanises at the boundary.
Numerical Solution of the Orr-Sommerfeld Equation Write the Orr-Sommerfeld equation in the form (5.15)
Oblique Modes
But there is another important difference: the corresponding 2D mode grows on a flow with increased viscosityν˜ =ν/cosϕ. If this were true, and if that viscous destabilization was sufficient to overcome the leading factor cosϕ, then a 3D mode could grow faster than the corresponding 2D mode.
Shear Scaling and the Reynolds Number Here again is the Orr-Sommerfeld equation (3D)
As in the inviscid case, a general 3D mode with φ = 0 and growth rate σ corresponds to a 2D mode (φ =0), whose growth rate reduces to σcosϕ. It is often true that viscosity has the effect of moisture instability, as we have seen in the case of convection (Chapter 2).
Numerical Examples
Frictionless boundaries are placed at z = ±5. a) Fastest growth rate of shear layer instability versus Re. b) Wavenumber of the fastest growing mode. For Re >8×104 (above the line), the growth rate decreases with increasing Reor, in other words, increases with increasing viscosity.
Perturbation Energetics in Viscous Flow
The term νKzz is the convergence of a flux of kinetic energy due to viscosity, namely−νKz. The dissipation term-ε is negative definite and therefore destroys the kinetic energy of the turbulence (turning it into heat).
Summary
In this chapter, we investigate equilibria and disturbances in stratified, parallel shear flow with viscosity and diffusion effects included, effectively unifying Chapters 2, 3, 4, and 5.
Expanding the Basic Equations
The perturbation part of (6.9) is obtained by subtracting (6.10) and omitting terms of order2:. 6.11) Note that (6.11) combines the buoyancy perturbation equations for a motionless, inhomogeneous fluid (2.12) and a stratified, nondiffusive, parallel shear flow (4.10). We eliminate the pressure, as we did before, by combining the divergence of (6.8) with the Laplacian of its vertical component.
Numerical Solution
Solutions of the Taylor-Goldstein equation for incompressible stratified shear flows (Chapter 4) are obtained by settingν =κ = 0. The corresponding 2D mode has a wave vector of the same magnitude,k, but parallel˜ to the thex axis (Figure 3.10).
Shear and Diffusion Scalings
In this case, the velocity gap in the solution process would be filled by U = Re Pr z. Remember that the algorithm for solving F has not changed since it was first defined in (6.16).
Application: Instabilities of a Stably Stratified Shear Layer Back in section 4.6, we studied Kelvin-Helmholtz and Holmboe instabilities using
This suggests that the essential mechanisms of the instabilities are captured in the simple piecewise profiles. Each of the three mode families in Figure 6-9a has its own fastest growing mode and its own critical level.
Summary
The cylindrical (or columnar) vortex is the simplest example of a non-parallel shear flow and is a useful model for tornadoes and other geophysical vortices. Here we will examine two classes of eddy instabilities: (1) barotropic instabilities are closely analogous to the instabilities of a parallel shear flow, while (2) axisymmetric instabilities are similar to convection, but with the centrifugal force playing the role of gravity.
Cyclostrophic Equilibrium
The Perturbation Equations Now imagine a small perturbation to cyclostrophic equilibrium
For barotropic modes (Figures 7.2 and 7.3), the trick is to recognize that the perturbation flow is two-dimensional and non-divergent, and thus can be represented by a stream function. If an impermeable boundary is placed at summer1, then the radial velocity must vanish there, i.e. the boundary condition is justuˆ(r1)=0.
Analytical Example: the Rankine Vortex
The dispersion relation is obtained as in the analysis of both convection at an interface (Section 2.2.4) and the instability of a piecewise linear shear layer (Section 1).
Numerical Example: a Continuous Vortex
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sgpp::base::OperationFirstMomentModBspline Class Reference
FirstMomemnt of sparse grid function, linear grid without boundaries. More...
#include <OperationFirstMomentModBspline.hpp>
Inheritance diagram for sgpp::base::OperationFirstMomentModBspline:
## Public Member Functions
double doQuadrature (const DataVector &alpha, DataMatrix *bounds=nullptr) override
Compute first moment of the function
$\int_{\Omega} x\cdot f(x) dx.$
OperationFirstMomentModBspline (Grid *grid)
Constructor of OperationFirstMomentModBspline. More...
~OperationFirstMomentModBspline () override
Public Member Functions inherited from sgpp::base::OperationFirstMoment
OperationFirstMoment ()
Constructor. More...
virtual ~OperationFirstMoment ()
Destructor. More...
## Protected Attributes
sgpp::base::Gridgrid
## Detailed Description
FirstMomemnt of sparse grid function, linear grid without boundaries.
## Constructor & Destructor Documentation
sgpp::base::OperationFirstMomentModBspline::OperationFirstMomentModBspline ( Grid * grid )
inlineexplicit
Constructor of OperationFirstMomentModBspline.
Parameters
grid Pointer to a sparse grid object
sgpp::base::OperationFirstMomentModBspline::~OperationFirstMomentModBspline ( )
inlineoverride
## Member Function Documentation
double sgpp::base::OperationFirstMomentModBspline::doQuadrature ( const DataVector & alpha, DataMatrix * bounds = nullptr )
overridevirtual
Compute first moment of the function
$\int_{\Omega} x\cdot f(x) dx.$
.
Parameters
alpha Coefficient vector for current grid bounds describes the boundaries of the hypercube of the original function
Implements sgpp::base::OperationFirstMoment.
Referenced by ~OperationFirstMomentModBspline().
## Member Data Documentation
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http://easy-ciphers.com/dominu | 1,560,840,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998690.87/warc/CC-MAIN-20190618063322-20190618085322-00316.warc.gz | 52,815,468 | 9,920 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: dominu
cipher variations: epnjov fqokpw grplqx hsqmry itrnsz jusota kvtpub lwuqvc mxvrwd nywsxe ozxtyf payuzg qbzvah rcawbi sdbxcj tecydk ufdzel vgeafm whfbgn xigcho yjhdip zkiejq aljfkr bmkgls cnlhmt
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: dominu Cipher: wlnrmf
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
Plain: dominu Cipher: AAABB ABBAB ABABB ABAAA ABBAA BAABB
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: dominu
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snvlep
yptbed
ktphef
qvnxet
wxlneh
czjdev
ibhtej
odfjex
vgeafm
bicqfa
hkagfo
nmywfc
towmfq
zqucfe
luqifg
rwoyfu
xymofi
dakefw
jciufk
pegkfy
whfbgn
cjdrgb
ilbhgp
onzxgd
upxngr
arvdgf
mvrjgh
sxpzgv
yznpgj
eblfgx
kdjvgl
qfhlgz
xigcho
dkeshc
jmcihq
poayhe
vqyohs
bswehg
nwskhi
tyqahw
zaoqhk
fcmghy
lekwhm
rgimha
yjhdip
elftid
kndjir
qpbzif
wrzpit
ctxfih
oxtlij
uzrbix
abpril
gdnhiz
mflxin
shjnib
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fmguje
loekjs
rqcajg
xsaqju
duygji
pyumjk
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tikojc
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gnhvkf
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ytbrkv
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qzvnkl
wbtdkz
cdrtkn
ifpjkb
ohnzkp
ujlpkd
bmkgls
hoiwlg
nqgmlu
tsecli
zucslw
fwailk
rawolm
xcuela
desulo
jgqklc
pioalq
vkmqle
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ipjxmh
orhnmv
utfdmj
avdtmx
gxbjml
sbxpmn
ydvfmb
eftvmp
khrlmd
qjpbmr
wlnrmf
dominu
jqkyni
psionw
vugenk
bweuny
hycknm
tcyqno
zewgnc
fguwnq
lismne
rkqcns
xmosng
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: dominu Cipher: qbzvah
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: dominu Cipher: 414323423354
Extended Methods:
Method #1
Plaintext: dominu
method variations: itrosz oywtxe tdbyck yigdhp
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
d o m i n u
4 4 2 4 3 5
1 3 3 2 3 4
They are then read out in rows:
442435133234
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: dominu Cipher: trxlhs
Method #3
Plaintext: dominu
method variations: qhsmxt hsmxtq smxtqh mxtqhs xtqhsm tqhsmx
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: dominu
all 720 cipher variations:
dominu
domiun
domniu
domnui
domuni
domuin
doimnu
doimun
doinmu
doinum
doiunm
doiumn
donimu
donium
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dmioun
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dmiuon
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dmnoiu
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# Important Questions Test: Seating Arrangement- 1
Test Description
## 10 Questions MCQ Test IBPS Clerk Prelims - Study Material, Mock Tests | Important Questions Test: Seating Arrangement- 1
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Important Questions Test: Seating Arrangement- 1 - Question 1
### A, P, R, X, S and Z are sitting in a row. S and Z are in the centre. A and P are at the ends. R is sitting to the left of A. Who is to the right of P ?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 1
The seating arrangement is as follows:
Therefore, right of P is X.
Important Questions Test: Seating Arrangement- 1 - Question 2
### A, B, C, D and E are sitting on a bench. A is sitting next to B, C is sitting next to D, D is not sitting with E who is on the left end of the bench. C is on the second position from the right. A is to the right of B and E. A and C are sitting together. In which position A is sitting?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 2
Therefore, A is sitting in between B and C.
Important Questions Test: Seating Arrangement- 1 - Question 3
### PASSAGE -2 Six friends are sitting in a circle and are facing the centre of the circle. Deepa is between Prakash and Pankaj. Priti is between Mukesh and Lalit. Prakash and Mukesh are opposite to each other. Q. Who is just right to Pankaj?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 3
Hence, Deepa is sitting right to Pankaj
Important Questions Test: Seating Arrangement- 1 - Question 4
PASSAGE -2
Six friends are sitting in a circle and are facing the centre of the circle. Deepa is between Prakash and Pankaj. Priti is between Mukesh and Lalit. Prakash and Mukesh are opposite to each other.
Q. Who is sitting opposite to Priti?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 4
Hence, Deepa is sitting opposite to Priti.
Important Questions Test: Seating Arrangement- 1 - Question 5
PASSAGE -1
Five girls are sitting on a bench to be photographed. Seema is to the left of Rani and to the right of Bindu. Mary is to the right of Rani. Reeta is between Rani and Mary.
Q. Who is second from the left in a photograph?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 5
Seema is sitting second from the left in photograph.
Important Questions Test: Seating Arrangement- 1 - Question 6
PASSAGE -2
Six friends are sitting in a circle and are facing the centre of the circle. Deepa is between Prakash and Pankaj. Priti is between Mukesh and Lalit. Prakash and Mukesh are opposite to each other.
Q. Who is sitting opposite to Prakash?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 6
Important Questions Test: Seating Arrangement- 1 - Question 7
PASSAGE -1
Five girls are sitting on a bench to be photographed. Seema is to the left of Rani and to the right of Bindu. Mary is to the right of Rani. Reeta is between Rani and Mary.
Q. Who is second from the right?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 7
Mary is sitting immediate right to Reeta.
Important Questions Test: Seating Arrangement- 1 - Question 8
PASSAGE -1
Five girls are sitting on a bench to be photographed. Seema is to the left of Rani and to the right of Bindu. Mary is to the right of Rani. Reeta is between Rani and Mary.
Q. Who is sitting immediate right to Reeta?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 8
The order would be..
Bindu Seema Rani Reeta Mary.
Therefore Mary is at the immediate right of reeta.
Important Questions Test: Seating Arrangement- 1 - Question 9
PASSAGE -2
Six friends are sitting in a circle and are facing the centre of the circle. Deepa is between Prakash and Pankaj. Priti is between Mukesh and Lalit. Prakash and Mukesh are opposite to each other. Who is sitting opposite to Prakash?
Q. Who are the neighbours of Mukesh?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 9
Hence, Priti and Pankaj are the neighbours of Mukesh.
Important Questions Test: Seating Arrangement- 1 - Question 10
PASSAGE -1
Five girls are sitting on a bench to be photographed. Seema is to the left of Rani and to the right of Bindu. Mary is to the right of Rani. Reeta is between Rani and Mary.
Q. Who is in the middle of the photograph?
Detailed Solution for Important Questions Test: Seating Arrangement- 1 - Question 10
The number of Girls are 5. They are seema, rani, reeta, mary and bindu
Seema is to the left of rani. So arrangement is:
⇒ Seema, ..., Rani
Seema is to the to the right of bindu. So,arrangement is:
⇒ Bindhu , .. , Seema
Mary is to the right of rani. So arrangement is:
⇒ Rani, ..., Mary
Reeta is between rani and mary. So arrangement is:
⇒ Rani Reeta Mary
So, the arrangement in photograph from left to right side is:-
Bindu , Seema , Rani , reeta and Mary
## IBPS Clerk Prelims - Study Material, Mock Tests
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## IBPS Clerk Prelims - Study Material, Mock Tests
72 videos|42 docs|81 tests | 1,739 | 6,903 | {"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} | 2.796875 | 3 | CC-MAIN-2022-33 | latest | en | 0.892261 |
https://socratic.org/questions/a-little-help-please#637936 | 1,660,873,370,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573540.20/warc/CC-MAIN-20220819005802-20220819035802-00551.warc.gz | 475,911,772 | 6,311 | ## Spotting a police car, you brake your Porsche from a speed of 100 km/h to a speed of 80.0 km/h during a displacement of 88.0 m, at a constant acceleration. a. What is that acceleration? b. How much time is required for the given decrease in speed?
Jul 2, 2018
a. $a = - 1.579 \frac{m}{s} ^ 2$
b. $t = 3.52 s$
#### Explanation:
We have velocity in km/hr and displacement in m. Since we have to make a unit conversion, might as well do it so it will yield the answer in the normal (SI) units for acceleration.
$100 \frac{k m}{h r} \cdot \frac{1000 m}{1 k m} \cdot \frac{1 h r}{3600 s} = 27.78 \frac{m}{s}$
$80.0 \frac{k m}{h r} \cdot \frac{1000 m}{1 k m} \cdot \frac{1 h r}{3600 s} = 22.22 \frac{m}{s}$
a. Using the formula of motion ${v}^{2} = {u}^{2} + 2 \cdot a \cdot d$
#(22.22 m/s)^2 = (27.78 m/s)^2 + 2a88.0 m
$493.7 {m}^{2} / {s}^{2} = 771.7 {m}^{2} / {s}^{2} + 176 m \cdot a$
$a = \frac{493.7 {m}^{2} / {s}^{2} - 771.7 {m}^{2} / {s}^{2}}{176 m} = - 1.579 \frac{m}{s} ^ 2$
b. Since the acceleration was constant, we can use the average velocity, and the distance to determine the time.
$t = \frac{88.0 m}{\frac{1}{2} \cdot \left(27.78 \frac{m}{s} + 22.22 \frac{m}{s}\right)} = 3.52 s$
As a double-check, if we convert $v = u + a \cdot t$ to $t = \frac{v - u}{a}$ and plug in the data, let's see if we get the same time.
$t = \frac{22.22 \frac{m}{s} - 27.78 \frac{m}{s} ^ 2}{- 1.579 \frac{m}{s} ^ 2} = 3.52 s$
I hope this helps,
Steve | 593 | 1,456 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-33 | latest | en | 0.673397 |
tijarohonline.blogspot.com | 1,466,974,434,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395560.14/warc/CC-MAIN-20160624154955-00105-ip-10-164-35-72.ec2.internal.warc.gz | 309,737,274 | 26,723 | ## Tuesday, June 15, 2010
### Analog Signal Transmission for Process Control
Although the microprocessor and digital network technologies have fundamentally reinvented the ways in which today's data acquisition systems handle data, much laboratory and manufacturing information is still communicated the "old" way, via analog electrical signals. And a fundamental understanding of how analog signal transmission works must first begin with a discussion of electrical basics.
To understand the ways in which an analog signal is transmitted over a circuit, it is first important to understand the relationships that make analog signal transmission possible. It is the fundamental relationship between voltage, current, and electrical resistance (Figure 3-1) that allow either a continuously varying current or voltage to represent a continuous process variable.
Figure 3-1:A Basic Electric Circuit
While charge flow is electric current, voltage is the work done in moving a unit of charge (1 coulomb) from one point to another. The unit of voltage is often called the potential difference, or the volt (V). The International System of Units (SI) unit for electrical flow is the ampere (A), defined as one coulomb per second (c/s).
A signal source of voltage, V, will cause a current, I, to flow through a resistor of resistance, R. Ohm's law, which was formulated by the German physicist Georg Simon Ohm (1787-1854), defines the relation:
V=IR
While most single-channel analog signal transmissions use direct current (dc) variations in current or voltage to represent a data value, frequency variations of an alternating current (ac) also can be used to communicate information. In the early 19th century, Jean Baptiste Joseph Fourier, a French mathematician and physicist, discovered that ac signals could be defined in terms of sine waves. A sine wave is described by three quantities: amplitude, period, and frequency. The amplitude is the peak value of the wave in either the positive or negative direction, the period is the time it takes to complete one cycle of the wave, and the frequency is the number of complete cycles per unit of time (the reciprocal of the period).
Analog Signal Types
Most data acquisition signals can be described as analog, digital, or pulse. While analog signals typically vary smoothly and continuously over time, digital signals are present at discrete points in time (Figure 3-2). In most control applications, analog signals range continuously over a specified current or voltage range, such as 4-20 mA dc or 0 to 5 V dc. While digital signals are essentially on/off (the pump is on or off, the bottle is there or isn't), analog signals represent continuously variable entities such as temperatures, pressures, or flow rates. Because computer-based controllers and systems understand only discrete on/off information, conversion of analog signals to digital representations is necessary (and discussed in Chapter 1).
Transduction is the process of changing energy from one form into another. Hence, a transducer is a device that converts physical energy into an electrical voltage or current signal for transmission. There are many different forms of analog electrical transducers. Common transducers include load cells for measuring strain via resistance, and thermocouples and resistance temperature detectors (RTDs) for measuring temperature via voltage and resistance measurement, respectively. Transmission channels may be wires or coaxial cables.
Figure 3-2:Digital and Analog Signal Representations
For noise-resistant transmission over significant distances, the raw transducer signal is often converted to a 4-20 mA signal by a two-wire, loop-powered transmitter. The bottom value of a process variable's range, for example, a temperature, is typically designated as 4 mA, making it easy to distinguish transmitter failure (0 mA) from a valid signal. If the current source is of good quality, current loops tend to be less sensitive to noise pickup by electromagnetic interference than voltage-based signals.
Noise & Grounding
In transmitting analog signals across a process plant or factory floor, one of the most critical requirements is the protection of data integrity. However, when a data acquisition system is transmitting low level analog signals over wires, some signal degradation is unavoidable and will occur due to noise and electrical interference. Noise and signal degradation are two basic problems in analog signal transmission.
Noise is defined as any unwanted electrical or magnetic phenomena that corrupt a message signal. Noise can be categorized into two broad categories based on the source-internal noise and external noise. While internal noise is generated by components associated with the signal itself, external noise results when natural or man-made electrical or magnetic phenomena influence the signal as it is being transmitted. Noise limits the ability to correctly identify the sent message and therefore limits information transfer. Some of the sources of internal and external noise include:
Electromagnetic interference (EMI);
Leakage paths at the input terminals;
Turbulent signals from other instruments;
Electrical charge pickup from power sources;
Switching of high-current loads in nearby wiring;
Self-heating due to resistance changes;
Arcs;
Lightning bolts;
Electrical motors;
High-frequency transients and pulses passing into the equipment;
Improper wiring and installation;
Signal conversion error; and
Uncontrollable process disturbances.
Signal leads can pick up two types of external noise-common mode and normal mode. Normal mode noise enters the signal path as a differential voltage and cannot be distinguished from the transducer signal. Noise picked up on both leads from ground is referred to as common-mode interference.
Figure 3-3:Signal, Noise, and Filtering Frequencies
Typical ranges for data signals and noise are shown in Figure 3-3. Whether the noise is detrimental to the proper performance of the system depends on the ratio of the total signal power to the total noise level. This is referred to as the signal-to-noise ratio. If the signal power is large in comparison to the noise signal, the noise can often times be ignored. However, with long-distance signals operating with limited signal power, the noise may disrupt the signal completely.
Current-driven devices have been most widely accepted in processing plants, with a common current range of 4-20 mA. Low-level current signals are not only safe, but are not as susceptible to noise as voltage signals. If a current is magnetically coupled into the connecting wires in the transmission of the signal from a current source, no significant change in the signal current will result. If the transducer is a voltage-driven device, the error adds directly to the signal. After current transmission, voltage signals can be easily rederived.
Even though internal and external interference can be problematic in sending analog signals, analog signal transmission is widely and successfully used in industry. The effects of noise can be reduced with careful engineering design, proper installation, routing techniques of wires and cables, and shielding and grounding.
One of the ways in which engineers have tried to minimize the effects of noise is to maximize the signal-to-noise ratio. This involves increasing the power of the signal being sent. Although this works in some cases, it has its limitations. By increasing the signal, nonlinear effects become dominant, as the signal amplitude is increased-it enhances the signal and the noise in the same proportion.
Proper grounding also is essential for effective operation of any measurement system. Improper grounding can lead to potentially dangerous ground loops and susceptibility to interference. To understand the principles involved in shielding and grounding, some terms must first be understood. A ground is a conducting flow path for current between an electric circuit and the earth. Ground wires are typically made with materials that have very low resistance. Because current takes the path of least resistance, the ground wires connected from the system provide a stable reference for making voltage measurements. Ground wires also safeguard against unwanted common-mode signals and prevent accidental contact with dangerous voltages. Return lines carry power or signal currents (Figure 3-4). A ground loop is a potentially dangerous loop formed when two or more points in an electrical system are grounded to different potentials.
Figure 3-4:A Ground Conductor
There are many different grounding techniques designed to not only protect the data being transmitted, but to protect employees and equipment. There are two ways in which all systems should be grounded. First of all, all of the measuring equipment and recording systems should be grounded so that measurements can be taken with respect to a zero voltage potential. This not only ensures that potential is not being introduced at the measuring device, but ensures that enclosures or cabinets around equipment do not carry a voltage. To ground an enclosure or cabinet, one or more heavy copper conductors are run from the device to a stable ground rod or a designated ground grid. This system ground provides a base for rejecting common-mode noise signals. It is very important that this ground is kept stable.
The second ground is for the signal ground. This ground is necessary to provide a solid reference for the measurement of all low-level signals. It is very important that this ground is grounded separate and isolated from the system ground. If a signal return line is grounded at the signal source and at the system ground, a difference in potential between the two grounds may cause a circulating current (Figure 3-5). In this case, the circulating current will be in series with the signal leads and will add directly to the signal from the measuring instrument. These ground loops are capable of creating noise signals 100 times the size of the original signal. This current can also be potentially dangerous. In a single-point ground configuration, minimal current can flow in the ground reference. Figure 3-6 shows that by grounding the wire at the signal end only, the current has no path, eliminating the ground loop.
Figure 3-5:Incorrect Grounding of Signal Circuit
For off-ground measurements, the shield or the ground lead is stabilized with respect to either the low-level of the signal or at a point between the two. Because the shield is at a potential above the zero-reference ground, it is necessary to have proper insulation.
Wire & Cable Options
Another important aspect to consider in analog signal transmission is a proper wiring system, which can effectively reduce noise interference. Analog signal transmission typically consists of two-wire signal leads or three-wire signal leads. In systems that require high precision and accuracy, the third signal lead, or shield, is necessary. In the three-wire configuration, the shield is grounded at the signal source to reduce common-mode noise. However, this does not eliminate all of the possibilities of the introduction of noise. It is crucial to prevent the noise pickup by protecting the signal lines. For example, in the case where the noise and signal frequency are the same. In this scenario, the signal cannot be isolated/filtered from the noise at the receiving device.
Figure 3-6: Correct Grounding of Signal Circuit
Generally, two-wire transmission mediums are used to carry an analog signal to or from the field area. A wire carrying an alternating current and voltage may induce noise in a pair of nearby signal leads. A differential voltage/noise will be created since the two wires may be at different distances from the disturbing signal. There are many different wiring options that are available to reduce unwanted noise pickup from entering the line. Four types of wires are fundamental in data acquisition-plain pair, shielded pair, twisted pair, and coaxial cable.
While plain wire can be used, it is generally not very reliable in screening out noise and is not suggested. A shielded pair is a pair of wires surrounded by a conductor that does not carry current. The shield blocks the interfering current and directs it to the ground. When using shielded pair, it is very important to follow the rules in grounding. Again, the shield must only be grounded at one source, eliminating the possibility of ground-loop currents.
Twisted-pairs help in elimination of noise due to electromagnetic fields by twisting the two signal leads at regular intervals. Any induced disturbance in the wire will have the same magnitude and result in error cancellation.
A coaxial cable is another alternative for protecting data from noise. A coaxial cable consists of a central conducting wire separated from an outer conducting cylinder by an insulator. The central conductor is positive with respect to the outer conductor and carries a current (Figure 3-7). Coaxial cables do not produce external electric and magnetic fields and are not affected by them. This makes them ideally suited, although more expensive, for transmitting signals.
Figure 3-7: Coaxial Cable Construction
Although noise and interference cannot be completely removed in the transmission of an analog signal, with good engineering and proper installation, many of the effects of noise and interference can be substantially reduced.
References and Further Reading Analog Signal Processing and Instrumentation, Arie F. Arbel, Cambridge University Press, 1980. Basic Circuit Analysis, David R. Cunningham, and John A. Stuller, Houghton Mifflin Co., 1932. Circuits: Principles, Analysis, and Simulation, Frank P. Yatsko, and David M. Hata, Saunders College Publishing, 1992. Data Acquisition and Control, Microcomputer Applications for Scientists and Engineers, Joseph J. Carr, Tab Books Inc., 1988. Data Communications, A Beginner's Guide to Concepts and Technology, Scott A. Helmers, Prentice-Hall, Inc., 1989. Digital and Analog Communication Systems, K. Sam Shanmugam, John Wiley & Sons, 1979. Fundamentals of Transducers, Stan Gibilisco and R. H. Warring, Tab Books Inc., 1985. Instrument Engineers' Handbook, Third Edition, Bela Liptak, Chilton Book Co., 1995. Introduction to Signal Transmission, Electrical and Electronic Engineering Series, William R. Bennett, McGraw-Hill, 1970. Microprocessors in Industrial Measurement and Control, Marvin D. Weiss, Tab Books Inc., 1987. Modern Digital and Analog Communication Systems, B.P. Lathi, Holt, Rinehart, & Winston, 1983. Process/Industrial Instruments & Controls Handbook, Fourth Edition, Douglas M. Considine, McGraw-Hill Inc., 1993. Signals & Systems Made Ridiculously Simple, Zohey Z. Karu, Zi Zi Press, 1995. Signals, The Telephone and Beyond, John R. Pierce, W. H. Freeman and Co., 1981.
#### 1 comment:
1. The website
www.LoopSlooth.com
has a very extensive, physics-based tutorial on electrical interference. This tutorial covers electrostatic pickup, inductive pickup, radiative pickup, ground loops, and effective inductance or capacitance of mismatched transmission lines. | 2,974 | 15,205 | {"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} | 3.546875 | 4 | CC-MAIN-2016-26 | longest | en | 0.918584 |
http://www.casadosmotores-se.com.br/qtonw/698b8e-fundamental-theorem-of-arithmetic-brainly | 1,627,852,663,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00664.warc.gz | 50,921,889 | 11,976 | It states that any integer greater than 1 can be expressed as the product of prime number s in only one way. Answer: 1 question What type of business organization is owned by a single person, has limited life and unlimited liability? The fundamental theorem of arithmetic is Theorem: Every n∈ N,n>1 has a unique prime factorization. and obviously tru practice problems solutions hw week select (by induction) ≥ 4 5 The Fundamental Theorem of Arithmetic | L. A. Kaluzhnin | download | Z-Library. If A and B are two independent events, prove that A and B' are also independent. It also contains the seeds of the demise of prospects for proving arithmetic is complete and self-consistent because any system rich enough to allow for unique prime factorization is subject to the classical proof by Godel of incompleteness. Please be Join for late night masturbation and sex boys and girls ID - 544 152 4423pass - 1234, The radius of a cylinder is 7cm, while its volume is 1.54L. Use the Fundamental Theorem of Arithmetic to justify that... Get solutions . This means p belongs to p 1 , p 2 , p 3 , . The Fundamental Theorem of Arithmetic for $\mathbb Z[i]$ Ask Question Asked 2 days ago. ОООО If the proposition was false, then no iterative algorithm would produce a counterexample. Thank You for A2A, In a layman term, A rational number is that number that can be expressed in p/q form which makes every integer a rational number. Well, we can also divide polynomials. It’s still true that we’re depending on an interpretation of the integral … So the Assumptions states that : (1) $\sqrt{3}=\frac{a}{b}$ Where a and b are 2 integers This is called the Fundamental Theorem of Arithmetic. Implicit differentiation. You can write a book review and share your experiences. Mathematics College Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainder, all rational roots, all possible roots, and actual roots of the given function. The two legs meet at a 90° angle and the hypotenuse is the longest side of the right triangle and is the side opposite the right angle. (9 Hours) Chapter 8 Binomial Theorem: History, statement and proof of the binomial theorem for positive integral indices. Paul Erdős gave a proof that also relies on the fundamental theorem of arithmetic. There are systems where unique factorization fails to hold. The history of the Fundamental Theorem of Arithmetic is somewhat murky. 5 does not occur in the prime factorization of 4 n for any n. Therefore, 4 n does not end with the digit zero for any natural number n. Question 18. Carl Friedrich Gauss gave in 1798 the first proof in his monograph “Disquisitiones Arithmeticae”. thefundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorizationtheorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. Paul Erdős gave a proof that also relies on the fundamental theorem of arithmetic. 11. A Startling Fact about Brainly Mathematics Uncovered Once the previous reference to interpretation was removed from the proofs of these facts, we’ll have a true proof of the Fundamental Theorem. According to Fundamental theorem of Arithmetic, every composite number can be written (factorised) as the product of primes and this factorization is Unique, apart from the order in which prime factors occur. The course covers several variable calculus, optimization theory and the selected topics drawn from the That course is aimed at teaching students to master comparative statics problems, optimization Fundamental Methods of Mathematical Economics, 3rd edition, McGrow-Hill, 1984. Technology Manual (10th Edition) Edit edition. Exercise 1.2 Class 10 Maths NCERT Solutions were prepared according to … Click now to get the complete list of theorems in mathematics. Applications of the Fundamental Theorem of Arithmetic are finding the LCM and HCF of positive integers. 2 Addition and Subtraction of Polynomials. Simplify: ( 2)! Free Rational Roots Calculator - find roots of polynomials using the rational roots theorem step-by-step This website uses cookies to ensure you get the best experience. Factorial n. Permutations and combinations, derivation of formulae and their connections, simple applications. The same thing applies to any algebraically closed field, … Theorem 6.3.2. For example, 75,600 = 2 4 3 3 5 2 7 1 = 21 ⋅ 60 2. Remainder Theorem and Factor Theorem. Thus 2 j0 but 0 -2. 225 can be expressed as (a) 5 x 3^2 (b) 5^2 x … (・∀・). Fundamental Theorem of Arithmetic. The fourth roots are ±1, ±i, as noted earlier in the section on absolute value. home / study / math / applied mathematics / applied mathematics solutions manuals / Technology Manual / 10th edition / chapter 5.4 / problem 8A. Do you remember doing division in Arithmetic? Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a 2 + b 2 = c 2.Although the theorem has long been associated with Greek mathematician-philosopher Pythagoras (c. 570–500/490 bce), it is actually far older. Every positive integer has a unique factorization into a square-free number and a square number rs 2. This is because we could multiply by 1 as many times as we like in the decomposition. The most important maths theorems are listed here. The number $\sqrt{3}$ is irrational,it cannot be expressed as a ratio of integers a and b.To prove that this statement is true, let us Assume that it is rational and then prove it isn't (Contradiction).. 8.ОТА начало.ogv 9 min 47 s, 854 × 480; 173.24 MB. The following are true: Every integer $$n\gt 1$$ has a prime factorization. (By uniqueness of the Fundamental Theorem of Arithmetic). The Basic Idea is that any integer above 1 is either a Prime Number, or can be made by multiplying Prime Numbers together. Within abstract algebra, the result is the statement that the ring of integers Zis a unique factorization domain. (Q.48) Find the H.C.F and L.C.M. Can two numbers have 15 as their HCF and 175 … . Write the first 5 terms of the sequence whose nth term is ( 3)!! ( )! Proving with the use of contradiction p/q = square root of 6. Or another way of thinking about it, there's exactly 2 values for X that will make F of X equal 0. mitgliedd1 and 110 more users found this answer helpful. Mathway: Scan Photos, Solve Problems (9 Similar Apps, 6 Review Highlights & 480,834 Reviews) vs Cymath - Math Problem Solver (10 Similar Apps, 4 Review Highlights & 40,238 Reviews). * The number 1 is not considered a prime number, being more traditionally referred to … Theorem 1: Equal chords of a circle subtend equal angles, at the centre of the circle. By … of 25152 and 12156 by using the fundamental theorem of Arithmetic 9873444080 (a) 24457576 (b) 25478976 (c) 25478679 (d) 24456567 (Q.49) Find the largest number which divides 245 and 1029 leaving remainder 5 in each case. See answer hifsashehzadi123 is waiting for your help. It provides us with a good reason for defining prime numbers so as to exclude 1. "7 divided by 2 equals 3 with a remainder of 1" Each part of the division has names: Which can be rewritten as a sum like this: Polynomials. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. Converted file can differ from the original. Proof: To prove Quotient Remainder theorem, we have to prove two things: For any integer a … If possible, download the file in its original format. It is used to prove Modular Addition, Modular Multiplication and many more principles in modular arithmetic. You can specify conditions of storing and accessing cookies in your browser. The fundamental theorem of algebra tells us that because this is a second degree polynomial we are going to have exactly 2 roots. In the case of C [ x], this fact, together with the fundamental theorem of Algebra, means what you wrote: every p (x) ∈ C [ x] can be written as the product of a non-zero complex number and first degree polynomials. It may take up to 1-5 minutes before you receive it. Other readers will always be interested in your opinion of the books you've read. corporation partnership sole proprietorship limited liability company - the answers to estudyassistant.com For example, 75,600 = 2 4 3 3 5 2 7 1 = 21 ⋅ 60 2. Use sigma notation to write the sum: 9 14 6 8 5 6 4 4 3 2 5. Book 7 deals strictly with elementary number theory: divisibility, prime numbers, Euclid's algorithm for finding the greatest common divisor, least common multiple. Fundamental principle of counting. Converse of Theorem 1: If two angles subtended at the centre, by two chords are equal, then the chords are of equal length. Find the value of b for which the runk of matrix A=and runk is 2, 1=112=223=334=445=556=667=778=8811=?answer is 1 because if 1=11 then 11=1, Describe in detail how you would create a number line with the following points: 1, 3.25, the opposite of 2, and – (–4fraction of one-half). Real Numbers Class 10 Maths NCERT Solutions are extremely helpful while doing your homework or while preparing for the exam. Viewed 59 times 1. For example, 1200 = 2 4 ⋅ 3 ⋅ 5 2 = ⋅ 3 ⋅ = 5 ⋅ … Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. ivyong22 ivyong22 ... Get the Brainly App Download iOS App Elements of the theorem can be found in the works of Euclid (c. 330–270 BCE), the Persian Kamal al-Din al-Farisi (1267-1319 CE), and others, but the first time it was clearly stated in its entirety, and proved, was in 1801 by Carl Friedrich Gauss (1777–1855). Fundamental theorem of algebra (complex analysis) Fundamental theorem of arbitrage-free pricing (financial mathematics) Fundamental theorem of arithmetic (number theory) Fundamental theorem of calculus ; Fundamental theorem on homomorphisms (abstract algebra) Fundamental theorems of welfare economics Of particular use in this section is the following. NCERT Solutions of all chapters of Class 10 Maths are provided with videos. For example: However, this is not always necessary or even possible to do. Fundamental Theorem of Arithmetic The Basic Idea. Play media. Which of the following is an arithmetic sequence? Media in category "Fundamental theorem of arithmetic" The following 4 files are in this category, out of 4 total. Definition 1.1 The number p2Nis said to be prime if phas just 2 divisors in N, namely 1 and itself. Active 2 days ago. Carl Friedrich Gauss gave in 1798 the first proof in his monograph “Disquisitiones Arithmeticae”. Functions in this section derive their properties from the fundamental theorem of arithmetic, which states that every integer n > 1 can be represented uniquely as a product of prime powers, … (See Gauss ( 1863 , Band II, pp. Get Free NCERT Solutions for Class 10 Maths Chapter 1 ex 1.2 PDF. Within abstract algebra, the result is the statement that the Mathematics College Use the Fundamental Theorem of Calculus to find the "area under curve" of f (x) = 6 x + 19 between x = 12 and x = 15. This article was most recently revised and … Euclid anticipated the result. For example, 252 only has one prime factorization: 252 = 2 2 × 3 2 × 7 1 In this and other related lessons, we will briefly explain basic math operations. Theorem 2: The perpendicular to a chord, bisects the chord if drawn from the centre of the circle. This site is using cookies under cookie policy. It may help for you to draw this number line by hand on a sheet of paper first. The file will be sent to your Kindle account. A right triangle consists of two legs and a hypotenuse. The functions we’ve been dealing with so far have been defined explicitly in terms of the independent variable. The fundamental theorem of arithmetic is Theorem: Every n∈ N,n>1 has a unique prime factorization. Euclid anticipated the result. According to fundamental theorem of arithmetic: Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. The fundamental theorem of arithmetic is truly important and a building block of number theory. Find a formula for the nth term of the sequence: , 24 10, 6 8, 2 6, 1 4, 1 2 4. Any positive integer $$N\gt 1$$ may be written as a product From Fundamental theorem of Arithmetic, we know that every composite number can be expressed as product of unique prime numbers. Also, the important theorems for class 10 maths are given here with proofs. The divergence theorem part of the integral: Here div F = y + z + x. Stokes' theorem is a vast generalization of this theorem in the following sense. The Fundamental Theorem of Arithmetic An integer greater than 1 whose only positive integer divisors… 2 positive integers a and b, GCD (a,b) is the largest positive… If you are considering these as subjects or concepts of Mathematics and not from a biology perspective, then arithmetic represents a constant growth and a geometric growth represents an exponential growth. More formally, we can say the following. The Fundamental Theorem of Arithmetic 1.1 Prime numbers If a;b2Zwe say that adivides b(or is a divisor of b) and we write ajb, if b= ac for some c2Z. This theorem forms the foundation for solving polynomial equations. can be expressed as a unique product of primes and their exponents, in only one way. Every positive integer has a unique factorization into a square-free number and a square number rs 2. The Fundamental theorem of Arithmetic, states that, “Every natural number except 1 can be factorized as a product of primes and this factorization is unique except for the order in which the prime factors are written.” This theorem is also called the unique factorization theorem. If is a differentiable function of and if is a differentiable function, then . Prime numbers are thus the basic building blocks of all numbers. What is the height of the cylinder. If 1 were a prime, then the prime factor decomposition would lose its uniqueness. n n 3. In general, by the Fundamental Theorem of Algebra, the number of n-th roots of unity is n, since there are n roots of the n-th degree equation z u – 1 = 0. Video transcript. Take $$\pi = 22/7$$ Pls dont spam. By the choice of F, dF / dx = f(x).In the parlance of differential forms, this is saying that f(x) dx is the exterior derivative of the 0-form, i.e. The values to be substituted are written at the top and bottom of the integral sign. So, this exercise deals with problems in finding the LCM and HCF by prime factorisation method. Find books It simply says that every positive integer can be written uniquely as a product of primes. The fundamental theorem of arithmetic: For each positive integer n> 1 there is a unique set of primes whose product is n. Which assumption would be a component of a proof by mathematical induction or strong mathematical induction of this theorem? In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to the order of the factors. Problem 8A from Chapter 5.4: a. Quotient remainder theorem is the fundamental theorem in modular arithmetic. Download books for free. So I encourage you to pause this video and try to … The fundamental theorem of arithmetic or the unique-prime-factorization theorem. * The Fundamental Theorem of Arithmetic states that every positive integer/number greater than 1 is either a prime or a composite, i.e. 1 $\begingroup$ I understand how to prove the Fundamental Theory of Arithmetic, but I do not understand how to further articulate it to the point where it applies for $\mathbb Z[I]$ (the Gaussian integers). The square roots of unity are 1 and –1. Thefundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorizationtheorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. Also, the relationship between LCM and HCF is understood in the RD Sharma Solutions Class 10 Exercise 1.4. p n and is one of them. The unique factorization is needed to establish much of what comes later. The fundamental theorem of calculus and accumulation functions. The Fundamental Theorem of Arithmetic is one of the most important results in this chapter. Следствия из ОТА.ogv 10 min 5 s, 854 × 480; 204.8 MB. Suppose f is a polynomial function of degree four, and $f\left(x\right)=0$. Or: how to avoid Polynomial Long Division when finding factors. Play media. Propositions 30 and 32 together are essentially equivalent to the fundamental theorem of arithmetic. 437–477) and Legendre ( 1808 , p. 394) .) Basic math operations include four basic operations: Addition (+) Subtraction (-) Multiplication (* or x) and Division ( : or /) These operations are commonly called arithmetic operations.Arithmetic is the oldest and most elementary branch of mathematics. Using Euclid’s lemma, this theorem states that every integer greater than one is either itself a prime or the product of prime numbers and that there is a definite order to primes. Mathematics C Standard Term 2 Lecture 4 Definite Integrals, Areas Under Curves, Fundamental Theorem of Calculus Syllabus Reference: 8-2 A definite integral is a real number found by substituting given values of the variable into the primitive function. Precalculus – Chapter 8 Test Review 1. It may takes up to 1-5 minutes before you received it. n n a n. 2. The fundamental theorem of arithmetic says that every integer larger than 1 can be written as a product of one or more prime numbers in a way that is unique, except for the order of the prime factors. Add your answer and earn points. All exercise questions, examples and optional exercise questions have been solved with video of each and every question.Topics of each chapter includeChapter 1 Real Numbers- Euclid's Division Lemma, Finding HCF using Euclid' sure to describe on which tick marks each point is plotted and how many tick marks are between each integer. One of the best known mathematical formulas is Pythagorean Theorem, which provides us with the relationship between the sides in a right triangle. The fundamental theorem of arithmetic states that any integer greater than 1 has a unique prime factorization (a representation of a number as the product of prime factors), excluding the order of the factors. function, F: in other words, that dF = f dx. Every such factorization of a given $$n$$ is the same if you put the prime factors in nondecreasing order (uniqueness). ... Get the Brainly App Download iOS App …. We've done several videos already where we're approximating the area under a curve by breaking up that area into rectangles and then finding the sum of the areas of those rectangles as an approximation. Fundamental theorem of arithmetic, Fundamental principle of number theory proved by Carl Friedrich Gauss in 1801. Here is a set of practice problems to accompany the Rational Functions section of the Common Graphs chapter of the notes for Paul Dawkins Algebra course at Lamar University. The file will be sent to your email address. The unique factorization fails to hold specify conditions of storing and accessing in... $\mathbb Z [ i ]$ Ask Question Asked 2 days ago principle of number theory by... Definition 1.1 the number p2Nis said to be prime if phas just 2 divisors in N, N > has... Be interested in your browser you received it to hold each integer number proved... 3 3 5 2 7 1 = 21 ⋅ 60 2 a differentiable function of and if is differentiable! 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It fundamental theorem of arithmetic brainly says that every positive integer/number greater than 1 is not considered a prime or a,. Prime factor decomposition would fundamental theorem of arithmetic brainly its uniqueness factorisation method in his monograph “ Arithmeticae. Another way of thinking about it, there 's exactly 2 values for X will. 10 min 5 s, 854 × 480 ; 173.24 MB as many times we. Section is the following consists of two legs and a square number rs 2 prime... Of thinking about it, there 's exactly 2 values for X that will f! Chapters of Class 10 exercise 1.4 that... Get Solutions first 5 terms the! Always necessary or even possible to do the exam its original format App the theorem. Centre of the integral sign × 480 ; 173.24 MB fundamental theorem of arithmetic brainly number s in only one way were a number... Building block of number theory ). integer has a unique factorization domain 1 is either a or! 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And other related lessons, we will briefly explain basic math operations ( 3 )! four, and latex! | 6,663 | 28,913 | {"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": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-31 | latest | en | 0.906159 |
http://www.milliliter.org/cubic-millimeters-to-quart-dry-conversion | 1,511,548,956,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00389.warc.gz | 433,788,778 | 5,931 | # Cubic Millimeters to Quart (U.S. dry) Conversion
Convert Cubic Millimeters to Quart (U.S. dry) by entering the Cubic Millimeters (mm³) value in the calculator form. mm³ to dry qt conversion.
1 mm³ = 1.0E-6 dry qt
Quart (U.S. dry) to Cubic Millimeters Conversion
## How Many Quart (U.S. dry) in a Cubic Millimeter
There are 1.0E-6 Quart (U.S. dry) in a Cubic Millimeter.
### Conversion Factors for Quart (U.S. dry) and Cubic Millimeters
Volume UnitSymbolFactor
Cubic Millimeters mm³ 1 × 10 -9
Quart (U.S. dry) dry qt 1.101221 × 10 -3
Cubic Millimeters volume unit is equal to 1.0E-6 Quart (U.S. dry).
### Cubic Millimeters to Quart (U.S. dry) Calculation
We calculate the base unit equivalent of Cubic Millimeters and Quart (U.S. dry) with the base unit factor of volume cubic meter.
1 mm³ = 1 * 10-9 m³
1 dry qt = 1.101221 * 10-3 m³
1 dry qt = 0.001101221 m³
1 m³ = (1/0.001101221) dry qt
1 m³ = 908.0829370308 dry qt
1 mm³ = 1 * 10-9 * 908.0829370308 dry qt
1 mm³ = 9.080829370308E-7 dry qt
### Cubic Millimeters to Quart (U.S. dry) Conversion Table
Cubic MillimetersQuart (U.S. dry)
1 mm³1.0E-6 dry qt
1000000 mm³1 dry qt
2000000 mm³2 dry qt
3000000 mm³3 dry qt
4000000 mm³4 dry qt
5000000 mm³5 dry qt
6000000 mm³6 dry qt
7000000 mm³7 dry qt
8000000 mm³8 dry qt
9000000 mm³9 dry qt
10000000 mm³10 dry qt
11000000 mm³11 dry qt
12000000 mm³12 dry qt
13000000 mm³13 dry qt
14000000 mm³14 dry qt
15000000 mm³15 dry qt
16000000 mm³16 dry qt
17000000 mm³17 dry qt
18000000 mm³18 dry qt
19000000 mm³19 dry qt
20000000 mm³20 dry qt
#### Abbreviations
• mm³ : Cubic Millimeters
• dry qt : Quart (U.S. dry)
• m³ : Cubic Meter
### Related Volume Conversions
List all Cubic Millimeters Conversions » | 627 | 1,710 | {"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-2017-47 | latest | en | 0.474214 |
http://mathhelpforum.com/trigonometry/220234-supplementary-exercise-trigo-print.html | 1,516,546,819,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00286.warc.gz | 212,417,640 | 3,240 | # supplementary exercise of trigo
Printable View
• Jun 28th 2013, 08:54 PM
Trefoil2727
supplementary exercise of trigo
If A, B, C are the angles of a triangle, prove that cos2A + cos2B + cos2 C = 1-2cosAcosBcosC
I've tried to use A+B+C=180, but I can't get the cos2 separately..
• Jun 28th 2013, 11:28 PM
mpx86
Re: supplementary exercise of trigo
this will help u
it is done for A+B=C
if A+B=C then prove that cos2A+cos2B + cos2C=1+2cosAcosB cos
it is ur task to do the same for a+b+c=180(method used will be identical)
• Jun 29th 2013, 06:27 AM
Trefoil2727
Re: supplementary exercise of trigo
Quote:
Originally Posted by mpx86
this will help u
it is done for A+B=C
if A+B=C then prove that cos2A+cos2B + cos2C=1+2cosAcosB cos
it is ur task to do the same for a+b+c=180(method used will be identical)
why cos2A+cos2B= 2cos(A+B)cos(A-B) ?
• Jun 29th 2013, 07:10 AM
mpx86
Re: supplementary exercise of trigo
½[sin ( Q+ β) + sin (Q − β)] = sin Q cos β,
so that
sin ( Q+ β) + sin ( Q− β) = 2 sin Q cos β. . . . . . (1)
Now put
Q + β = A
and
Q− β = B. . . . . . . . . . . . . . . .(2)
The left-hand side of line (1) then becomes
sin A + sin B.
This is now the left-hand side of (e), which is what we are trying to prove.
To complete the right−hand side of line (1), solve those simultaneous equations (2) for Q and β.
On adding them, 2Q = A + B,
so that
Q = ½(A + B).
On subtracting those two equations, 2β = A − B,
so that
β = ½(A − B).
On the right−hand side of line (1), substitute those expressions for and β. Line (1) then becomes
sin A + sin B = 2 sin ½(A + B) cos ½(A − B).
HERE A=2C AND B=2D
SUCH THAT sin 2C + sin 2D = 2 sin(C + D) cos (C − D).
IT IS YOUR TASK TO PROVE TH SAME FOR ADDITION OF 2 COSINES.... | 632 | 1,720 | {"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} | 4.28125 | 4 | CC-MAIN-2018-05 | longest | en | 0.763101 |
https://api-project-1022638073839.appspot.com/questions/what-is-the-derivative-of-ln-1-1-x-1-x | 1,726,059,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00676.warc.gz | 79,903,177 | 6,304 | # What is the derivative of ln(1+1/x) / (1/x)?
Nov 22, 2015
I like the form: $\ln \left(\frac{x + 1}{x}\right) - \frac{1}{x + 1}$
#### Explanation:
Before we differentiate, let's get rid of that silly "divided by $\frac{1}{x}$"
$\ln \frac{1 + \frac{1}{x}}{\frac{1}{x}} = x \ln \left(1 + \frac{1}{x}\right)$
We can now use the product rule, but also note that
$1 + \frac{1}{x} = \frac{x + 1}{x}$,
So we have $x \ln \left(\frac{x + 1}{x}\right)$
When we differentiate the second factor, we'll need the chain rule and we'll need $\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x}\right) = \frac{d}{\mathrm{dx}} \left(1 + \frac{1}{x}\right) = 0 - \frac{1}{x} ^ 2 = \frac{- 1}{x} ^ 2$.
So, here we go:
$\frac{d}{\mathrm{dx}} \left(\ln \frac{1 + \frac{1}{x}}{\frac{1}{x}}\right) = \frac{d}{\mathrm{dx}} \left(x \ln \left(\frac{x + 1}{x}\right)\right)$
$= \left[1\right] \ln \left(\frac{x + 1}{x}\right) + x \left[\frac{1}{\frac{x + 1}{x}} \left(\frac{- 1}{x} ^ 2\right)\right]$
$= \ln \left(\frac{x + 1}{x}\right) + {x}^{2} / \left(x + 1\right) \left(\frac{- 1}{x} ^ 2\right)$
$= \ln \left(\frac{x + 1}{x}\right) - \frac{1}{x + 1}$
Nov 22, 2015
Using the property of logs ...
$y = x \ln \left(\frac{x + 1}{x}\right) = x \left[\ln \left(x + 1\right) - \ln \left(x\right)\right]$
#### Explanation:
$y = x \left[\ln \left(x + 1\right) - \ln \left(x\right)\right]$
Using only the product rule :
$y ' = \left(1\right) \left[\ln \left(x + 1\right) - \ln \left(x\right)\right] + x \left[\frac{1}{x + 1} - \frac{1}{x}\right]$
Now simplify ...
$y ' = \ln \left(\frac{x + 1}{x}\right) - \frac{1}{x + 1}$
hope that helped | 683 | 1,622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-38 | latest | en | 0.476951 |
https://www.coursehero.com/file/50219/Homework-4/ | 1,547,733,086,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658981.19/warc/CC-MAIN-20190117123059-20190117145059-00437.warc.gz | 759,631,349 | 53,907 | PHY
Homework 4
# Homework 4 - homework 04 KANUNGO ANIL Latest unpenalized...
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homework 04 – KANUNGO, ANIL 1 Latest unpenalized work: Sep 10 2007 Mon- day 04:00 (after this date you can not make a perfect score). Work cutoff: Sep 11 2007, 4:00 am. Question 1, chap 23, sect 3. part 1 of 4 10 points Consider symmetrically placed rectangular insulators with uniformly charged distribu- tions of equal magnitude as shown in the fig- ures. x y ++++ ---- Find the net electric field at the origin. 1. aligned with the positive y -axis 2. aligned with the positive x -axis 3. aligned with the negative y -axis 4. non-zero and not aligned with either the x - or y -axis 5. aligned with the negative x -axis 6. zero with the direction undefined. Question 2, chap 23, sect 3. part 2 of 4 10 points x y ++++ ++++ In the figure above what is the net field at the origin? 1. aligned with the negative y -axis 2. aligned with the positive y -axis 3. non-zero and not aligned with either the x - or y -axis 4. aligned with the negative x -axis 5. zero and the direction is undefined 6. aligned with the positive x -axis Question 3, chap 23, sect 3.
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https://metanumbers.com/1532927 | 1,642,492,922,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00532.warc.gz | 459,647,499 | 7,497 | # 1532927 (number)
1,532,927 (one million five hundred thirty-two thousand nine hundred twenty-seven) is an odd seven-digits composite number following 1532926 and preceding 1532928. In scientific notation, it is written as 1.532927 × 106. The sum of its digits is 29. It has a total of 4 prime factors and 16 positive divisors. There are 1,298,880 positive integers (up to 1532927) that are relatively prime to 1532927.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 29
• Digital Root 2
## Name
Short name 1 million 532 thousand 927 one million five hundred thirty-two thousand nine hundred twenty-seven
## Notation
Scientific notation 1.532927 × 106 1.532927 × 106
## Prime Factorization of 1532927
Prime Factorization 11 × 23 × 73 × 83
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 1532927 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,532,927 is 11 × 23 × 73 × 83. Since it has a total of 4 prime factors, 1,532,927 is a composite number.
## Divisors of 1532927
16 divisors
Even divisors 0 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 1.79021e+06 Sum of all the positive divisors of n s(n) 257281 Sum of the proper positive divisors of n A(n) 111888 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1238.11 Returns the nth root of the product of n divisors H(n) 13.7005 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,532,927 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 1,532,927) is 1,790,208, the average is 111,888.
## Other Arithmetic Functions (n = 1532927)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1298880 Total number of positive integers not greater than n that are coprime to n λ(n) 162360 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 116200 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 1,298,880 positive integers (less than 1,532,927) that are coprime with 1,532,927. And there are approximately 116,200 prime numbers less than or equal to 1,532,927.
## Divisibility of 1532927
m n mod m 2 3 4 5 6 7 8 9 1 2 3 2 5 4 7 2
1,532,927 is not divisible by any number less than or equal to 9.
## Classification of 1532927
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (1532927)
Base System Value
2 Binary 101110110001111111111
3 Ternary 2212212210002
4 Quaternary 11312033333
5 Quinary 343023202
6 Senary 52504515
8 Octal 5661777
10 Decimal 1532927
12 Duodecimal 61b13b
20 Vigesimal 9bc67
36 Base36 wutb
## Basic calculations (n = 1532927)
### Multiplication
n×y
n×2 3065854 4598781 6131708 7664635
### Division
n÷y
n÷2 766464 510976 383232 306585
### Exponentiation
ny
n2 2349865187329 3602171792016681983 5521866398620756262154241 8464618092838520034675314193407
### Nth Root
y√n
2√n 1238.11 115.303 35.1868 17.2625
## 1532927 as geometric shapes
### Circle
Diameter 3.06585e+06 9.63166e+06 7.38232e+12
### Sphere
Volume 1.50887e+19 2.95293e+13 9.63166e+06
### Square
Length = n
Perimeter 6.13171e+06 2.34987e+12 2.16789e+06
### Cube
Length = n
Surface area 1.40992e+13 3.60217e+18 2.65511e+06
### Equilateral Triangle
Length = n
Perimeter 4.59878e+06 1.01752e+12 1.32755e+06
### Triangular Pyramid
Length = n
Surface area 4.07009e+12 4.2452e+17 1.25163e+06
## Cryptographic Hash Functions
md5 70190f315cd00296a5e1d1a24475c2c9 70a02e5736c38ef1340a60ec00da2dc8e3e3256f a189828aa822a29ccc6feab2f3376b9f0ffb977d548432d30ccd77bd6a50cf13 e41e7d6a087a77bae1b493447993c939314cdf1fc11b5e3affc81b0c558cd7f9a4e7d50183979d54864034c26148c66f5a7b3ed1b93acf54c48689d079438e9a 67dee4b7c9a80a0e47c772bc9a9efb6ffa31dbc0 | 1,565 | 4,400 | {"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} | 3.390625 | 3 | CC-MAIN-2022-05 | latest | en | 0.789501 |
https://www.jiskha.com/display.cgi?id=1361557496 | 1,501,146,800,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00718.warc.gz | 775,270,743 | 4,620 | # math
posted by .
Kevin plotted a linear equation on a graph with 18 as the y-intercept. Which of the following could be the equation plotted by Kevin?
• math -
1/2y = 9x + 5
3x = 6 - 1/3y
• math -
y-intercept is a point where the graph of a function or relation intersects with the y-axis of the coordinate system.
OR
y-intercept is a point where x = 0
1 / 2 y = 9 x + 5 Multiply both sides by 2
y = 18 x + 10
for x = 0
y = 18 * 0 + 10 = 0 + 10 = 10
y-intercept = 10
3 x = 6 - 1 / 3 y Multiply both sides by 3
9 x = 18 - y Add y to both sides
9 x + y = 18 - y + y
9 x + y = 18 Subtract 9 x to both sides
9 x + y - 9 x = 18 - 9 x
y = - 9 x + 18
for x = 0
y = - 9 * 0 + 18 = 0 + 18 = 18
y-intercept = 18
3x = 6 - 1 / 3 y
• math -
That is the correct answer. How did you come up with it?
lol
:)
• math -
lit
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More Similar Questions
Post a New Question | 833 | 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} | 4.03125 | 4 | CC-MAIN-2017-30 | longest | en | 0.906547 |
https://mtp.tools/converters/kilogram-to-dram-calculator | 1,571,417,799,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00357.warc.gz | 606,085,042 | 5,133 | # Kilograms (kg) to Drams (dr) calculator
Input the amount of kilograms you want to convert to drams in the below input field, and then click in the "Convert" button. But if you want to convert from drams to kilograms, please checkout this tool.
## Formula
Formula used to convert kg to dr:
F(x) = x * 564.3833911933
For example, if you want to convert 15 kg to dr, just replace x by 15 [kg]:
15 kg = 15*564.3833911933 = 8465.7508678995 dr
## Steps
1. Multiply the amount of kilograms by 564.3833911933.
2. The result will be expressed in drams.
## Kilogram to Dram Conversion Table
The following table will show the most common conversions for Kilograms (kg) to Drams (dr):
Kilograms (kg) Drams (dr)
0.001 kg 0.5643833912 dr
0.01 kg 5.6438339119 dr
0.1 kg 56.4383391193 dr
1 kg 564.3833911933 dr
2 kg 1128.7667823866 dr
3 kg 1693.1501735799 dr
4 kg 2257.5335647732 dr
5 kg 2821.9169559665 dr
6 kg 3386.3003471598 dr
7 kg 3950.6837383531 dr
8 kg 4515.0671295464 dr
9 kg 5079.4505207397 dr
10 kg 5643.833911933 dr
20 kg 11287.667823866 dr
30 kg 16931.501735799 dr
40 kg 22575.335647732 dr
50 kg 28219.169559665 dr
60 kg 33863.003471598 dr
70 kg 39506.837383531 dr
80 kg 45150.671295464 dr
90 kg 50794.505207397 dr
100 kg 56438.33911933 dr
A kilogram is a widely used unit of weight, defined on the International System of Units (SI). One kilogram is equal to 1000 grams. The symbol used to represent kilograms is kg.
The kilogram is used to measure almost everything, from food, materials such as wood, your weight, among others.
The dram (symbol ʒ or ℨ; or also dr) is a unit of mass in the avoirdupois system, and both a unit of mass and a unit of volume in the apothecaries's system.
## FAQs for Kilogram to Dram calculator
### What is Kilogram to Dram calculator?
Kilogram to Dram is a free and online calculator that converts Kilograms to Drams.
### How do I use Kilogram to Dram?
You just have to insert the amount of Kilograms you want to convert and press the "Convert" button. The amount of Drams will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Kilogram to Dram work on?
Kilogram to Dram calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 730 | 2,447 | {"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} | 3.234375 | 3 | CC-MAIN-2019-43 | latest | en | 0.683303 |
https://meinwords.wordpress.com/2006/01/19/indians-dilemma-part-1/ | 1,531,888,879,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00462.warc.gz | 706,386,311 | 15,328 | Indian’s Dilemma – Part 1
I have always wondered why India always misses the right note. Why we never played it right after 50 years. I cannot criticize by telling how other countries did far better in shorter periods. Because my Indian critics always opposes any such statement by telling me that India is much different from every country. In this post, I have some thoughts which can possibly explain what went wrong using a non-relative approach. I used my feeble human intellect, incomplete knowledge of game theory and some stolen thoughts from Amrtya Sen to organize my beliefs.
Some Game Theory before I continue ( from wikipedia )
• Game theory is a branch of applied mathematics that studies strategic situations where players choose different actions in an attempt to maximize their returns.
• Given a set of alternative allocations and a set of individuals, a movement from one alternative allocation to another that can make at least one individual better off, without making any other individual worse off is called a Pareto improvement or Pareto optimization. An allocation of resources is Pareto efficient or Pareto optimal when no further Pareto improvements can be made.
• Nash equilibrium (named after John Nash who proposed it) is a kind of optimal collective strategy in a game involving two or more players, where no player has anything to gain by changing only his or her own strategy. If each player has chosen a strategy and no player can benefit by changing his or her strategy while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitute a Nash equilibrium.
Classical Prisoner’s Dilemma ( from wikipedia )
The classical prisoner’s dilemma (PD) is as follows:
Two suspects, A and B, are arrested by the police. The police have insufficient evidence for a conviction, and having separated both prisoners, visit each of them and offer the same deal: if one testifies for the prosecution (turns Queen’s Evidence) against the other and the other remains silent, the silent accomplice receives the full 10-year sentence and the betrayer goes free. If both stay silent, the police can only give both prisoners 6 months for a minor charge. If both betray each other, they receive a 2-year sentence each. Each prisoner must make a choice – to betray the other, or to remain silent. However, neither prisoner knows for sure what choice the other prisoner will make. What will happen?
(to be continued…) | 491 | 2,494 | {"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-2018-30 | latest | en | 0.949917 |
https://mathoverflow.net/questions/107782/reference-for-list-of-left-regular-representations-of-real-associative-algebras/107950 | 1,621,107,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00328.warc.gz | 415,296,493 | 35,218 | # reference for list of left-regular representations of real associative algebras
Suppose $\mathcal{A}$ is a unital associative algebra over $\mathbb{R}$. If we identify $\mathcal{A} = \mathbb{R}^n$ then the $\mathcal{A}$ multiplication corresponds to particular linear maps on $\mathbb{R}^n$. Of course any linear map on $\mathbb{R}^n$ corresponds uniquely to its standard matrix hence we obtain a correspondence between vectors in $\mathcal{A}$ and matrices in $\mathbb{R}^{n \times n}$. These square matrices are known as the left regular representation of the algebra. This is not unique unless we add additional data about the correspondence of the $n$-dimensional algebra and its presentation on $\mathbb{R}^n$.
My favorite examples, $\mathbb{C} = \mathbb{R}^2$ is naturally identified with the subalgebra of $2 \times 2$ matrices of the form:
$$\left[\begin{array}{cc} a &-b \\ b &a \end{array} \right]$$
Or the hyperbolic numbers $\mathbb{R}+j\mathbb{R}$ identified with
$$\left[ {\begin{array}{cc} a & b \\ b & a \end{array}} \right]$$
Or the direct product of $\mathbb{R}$ with $\mathbb{R}$ the element $(a,b)$ is identified with
$$\left[ \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right]$$
Up to isomorphism the last two examples are actually the same example. I know of one other two-dimensional algebra up to isomorphism.
Question: where can I find a complete tabulation of the low-dimensional left-regular representations of unital algebras?
I have found many results on google and here, but I can't find one which stands out as a continuation of the list I began at the start of this post. In particular, complete list of complex associative algebra, is great except the base-field is $\mathbb{C}$. If there was a simple theorem that allowed me to extract the list I desire from that list then that would also be a useful answer. But, I'd rather have direct reference for a list of the real associative left regular representations. Ideally this will help me choose a good notation if there already is an agreed notation accepted among those who worked on such classifications.
As always the help of the MO community is greatly appreciated.
• At most, you will find lists of associative algebras up to isomorphism. Their left regular representations can be immediately obtained from a description of the algebras themselves. As for the algebras: there are way too many of them as soon as the dimension is slightly big, so you will not find lists except in the very small cases. People do study the geometry of the variety of algebras of a fixed dimension, though —but that is rather different than making lists of examples. – Mariano Suárez-Álvarez Sep 21 '12 at 19:00
• I'll take a list up to isomorphism. And, I'm happy to have a list to just to 3 or 4 if that is all that is reasonable. – James S. Cook Sep 21 '12 at 19:34
• You can go to dimension 4 by hand with a little work. – Mariano Suárez-Álvarez Sep 24 '12 at 4:30
• Not what you're looking for exactly...but here is a paper classifying associative commutative algebras over an algebraically closed field up to dimension 6: math.mit.edu/~poonen/papers/dimension6.pdf – Bill Cook Sep 24 '12 at 14:45
Let $A$ be an algebra of dimension $d$ over an alg. closed field $k$, and $r\subseteq A$ its radical.
• If $d=1$, then of course $A\cong k$.
• If $d=2$, then either $\dim r=0$ and then $A\cong k^2$ because of Wedderburn's theorem, or $\dim r=1$. In the latter case, we must have $r^2=0$, so the ordinary quiver $Q$ of $A$ is a loop. The only $2$-dimensional admissible quotient of $kQ$ is $k[X]/(X^2)$.
• Suppose $d=3$. Since $A/r$ is semisimple and $k$ algebraically closed, Wedderburn tells us that $\dim A/r$ is a sum of squares; it is at most $3$, the only possible square is which fits is $1$. If $\dim A=\dim A/r=1+1+1$, then $A\cong k^3$; if $\dim A=1+1$, then $\dim r=1$ and $r^2=0$ (because $r$ is nilpotent) and the ordinary quiver $Q$ of $A$ is then $\bullet\to\bullet$ or that of $k[X]/(X^2)\times k$. In the first case, since $kQ$ is $3$-dimensional, we must have $A\cong kQ$, which is in fact the algebra $T$ of $2\times 2$ upper triangular matrices. In the second case, $A\cong k[X]/(X^2)\times k$.
• Finally, suppose $d=4$. Weedderburn's theorem, as above, tells us that either $A/r\cong M_2(k)$, in which case in fact $A\cong M_2(k)$, or $A/r\cong k^s$ with $s\leq 4$.
• If $s=4$, of course $A\cong k^4$.
• If $s=3$, then $\dim s=1$, the ordinary quiver $Q$ has only an arrow, and counting dimensions we see that $A\cong kQ$: if the arrow is a loop, we have $A\cong k[x]/(x^2)\times k^2$, and if it is not a loop, we habe $A\cong T\times k$.
• If $s=2$, the quver $Q$ has two vertices. If $r^2\neq 0$, we have one arrow which does not square to zero, and this is only possible if $A=k[X]/(X^3)\times k$. If instead $r^2=0$, we have two arrows in $Q$ and $A$ is the quotient of $kQ$ by the square of the arrow-ideal: this gives six possibilities which are inbijection with the possible quivers.
• if $s=1$, the quiver has one vertex. It has $\ell=\dim r/r^2$ loops, with $1\leq\ell\leq3$ If $\ell=1$, then there is only one loop and the only admissible quotient is $k[X]/(X^4)$. If $\ell=3$, then we have three loops and all products of arrows vanish, so $A\cong k\langle x,y,z\rangle/(x^2,y^2,z^2,xy,yx,xz,zx,yz,zy)$. If $\ell=2$, we have two arrows in $Q$, call them $x$ and $y$, and the products $x^2$, $xy$, $yx$, $y^2$, being in $r^2$, are all multiples of a fixed non-zero element of $A$. Using this one can complete the list —here, for the first time, we get a non-discrete family.
• Plus/minus a case... – Mariano Suárez-Álvarez Sep 24 '12 at 5:07
• In case $k=\mathbb{C}$ a complete list up to dimension 5 is given in arxiv.org/abs/0910.0932 but according to the heading, the OP seems to be primarily interested in algebras over the reals. – Ralph Sep 24 '12 at 10:01
• @Mariano Suárez-Alvarez this is very interesting, I wish I understood more of it. However, I am for my current purpose primarily interested in algebras over $\mathbb{R}$ as Ralph mentions and I was aware of the tabulation for $k=\mathbb{C}$. – James S. Cook Sep 25 '12 at 3:06
• The same method can be used to deal with the real case, replacing quivers by «modulated quivers» (in which vertices are adorned with division algebras and arrows with bivector spaces) The quivers that can appear are the same as in the complex case, so there are very very few cases to consider apart from those above. (My list is missing the matrix algbera $M_2(k)$, by the way, which is indecomposable but non-basic —the only non-basic example in dimension at most $4$) – Mariano Suárez-Álvarez Sep 25 '12 at 4:14
• @Mariano Suárez-Alvarez Thanks. I would still appreciate a more accessible classification. Hopefully this appears as a reference in a paper primarily intended for an undergraduate audience. I'm surprised the list I seek isn't given in some well-known algebra text. – James S. Cook Sep 27 '12 at 5:20
(Too long for a comment). More (modern and not-so-modern) references, some of them may (partially) contain list(s) you are interested in:
• A.A. Albert, Structure of Algebras, AMS, 1939: on page 172 discusses classification of 4-dimensional algebras.
• S.C. Althoen, K.D. Hansen, L.D. Kugler, A survey of four-dimensional C-associative algebras, Algebras Groups Geom. 21 (2004), N1, 9-27
• R. Ballieu, Anneaux finis; systèmes hypercomplexes de rang trois sur un corps commutatif, Ann. Soc. Sci. Bruxelles Sér. I. 61 (1947), 222-227: presumably contains classification of complex 3-dimensional algebras.
• W.A. de Graaf, Classification of nilpotent associative algebras of small dimension, arXiv:1009.5339.
• D. Happel, Klassifikationstheorie endlich-dimensionaler Algebren in der Zeit von 1880 bis 1920, Enseign. Math. 26 (1980), 91-102 DOI:10.5169/seals-51060 : A nice historical survey from the modern viewpoint, with a large bibliography.
• O.C. Hazlett, On the classification and invariantive characterization of nilpotent algebras Amer. J. Math. 38 (1916), N2, 109-138 http://www.jstor.org/stable/2370262
• G. Pickert, Dreidimensionale assoziative nichtkommutative Algebren, J. Algebra 234 (2000), N2, 280-290 DOI:10.1006/jabr.2000.8550
• Scorza, Atti Acad. Sci. Fis. Mat. Napoli 20 (1935), N13 and N14: Classification of 3- and 4-dimensional algebras.
• D.A. Suprunenko and R.I. Tyshkevich, Commutative Matrices, Acad. Press, 1968 (translation from Russian): On p.61 (of the Russian edition) there is a discussion of commutative nilpotent algebras of dimension 5.
• thanks for the list, after I posted this question it became clear I just needed a tabulation of semi-simple associative algebras over $\mathbb{R}$. The pleathora of nilpotent algebras is outside the scope of my project at the present. On the other hand, the semi-simple algebras are isomorphic to direct sums of matrix algebras over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. – James S. Cook Jul 7 '13 at 20:58
I'm not sure how far he got, but it seems that Benjamin Peirce in his 1882 book "Linear Associative Algebra" made the first attempt at classifying associative algebras of low dimensions. The book is available free at http://archive.org/details/linearassociati00peirgoog
• Thanks. However, my first read was not terribly successful. It's difficult to penetrate the 19-th century terminology. – James S. Cook Sep 25 '12 at 3:02 | 2,821 | 9,413 | {"found_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} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.846608 |
https://www.systutorials.com/docs/linux/man/docs/linux/man/3-cher/ | 1,632,260,008,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00246.warc.gz | 1,011,710,768 | 4,117 | # cher (3) - Linux Man Pages
cher.f -
## SYNOPSIS
### Functions/Subroutines
subroutine cher (UPLO, N, ALPHA, X, INCX, A, LDA)
CHER
## Function/Subroutine Documentation
### subroutine cher (characterUPLO, integerN, realALPHA, complex, dimension(*)X, integerINCX, complex, dimension(lda,*)A, integerLDA)
CHER Purpose:
``` CHER performs the hermitian rank 1 operation
A := alpha*x*x**H + A,
where alpha is a real scalar, x is an n element vector and A is an
n by n hermitian matrix.
```
Parameters:
UPLO
``` UPLO is CHARACTER*1
On entry, UPLO specifies whether the upper or lower
triangular part of the array A is to be referenced as
follows:
UPLO = 'U' or 'u' Only the upper triangular part of A
is to be referenced.
UPLO = 'L' or 'l' Only the lower triangular part of A
is to be referenced.
```
N
``` N is INTEGER
On entry, N specifies the order of the matrix A.
N must be at least zero.
```
ALPHA
``` ALPHA is REAL
On entry, ALPHA specifies the scalar alpha.
```
X
``` X is COMPLEX array of dimension at least
( 1 + ( n - 1 )*abs( INCX ) ).
Before entry, the incremented array X must contain the n
element vector x.
```
INCX
``` INCX is INTEGER
On entry, INCX specifies the increment for the elements of
X. INCX must not be zero.
```
A
``` A is COMPLEX array of DIMENSION ( LDA, n ).
Before entry with UPLO = 'U' or 'u', the leading n by n
upper triangular part of the array A must contain the upper
triangular part of the hermitian matrix and the strictly
lower triangular part of A is not referenced. On exit, the
upper triangular part of the array A is overwritten by the
upper triangular part of the updated matrix.
Before entry with UPLO = 'L' or 'l', the leading n by n
lower triangular part of the array A must contain the lower
triangular part of the hermitian matrix and the strictly
upper triangular part of A is not referenced. On exit, the
lower triangular part of the array A is overwritten by the
lower triangular part of the updated matrix.
Note that the imaginary parts of the diagonal elements need
not be set, they are assumed to be zero, and on exit they
are set to zero.
```
LDA
``` LDA is INTEGER
On entry, LDA specifies the first dimension of A as declared
in the calling (sub) program. LDA must be at least
max( 1, n ).
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Date:
November 2011
Further Details:
``` Level 2 Blas routine.
-- Written on 22-October-1986.
Jack Dongarra, Argonne National Lab.
Jeremy Du Croz, Nag Central Office.
Sven Hammarling, Nag Central Office.
Richard Hanson, Sandia National Labs.
```
Definition at line 136 of file cher.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code. | 741 | 2,784 | {"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-2021-39 | latest | en | 0.781677 |
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# Tag Info
### Can a total programming language be Turing-complete?
I'm pretty sure McBride isn't claiming that a total programming language can be Turing complete, just that it's a useless distinction in practice. You can take any program in a partial language and ...
• 2,157
### Can a total programming language be Turing-complete?
A rhetorical question to keep in mind: would anyone take you seriously if you said ZFC is "not Turing complete" and therefore insufficient to express the algorithms that we write in ...
• 2,697
Accepted
### Does co-inductive and co-recursive types also have their recursors?
To understand coinduction, it helps to understand the categorical presentation of induction (since, as far as I know, coinduction comes from dualizing it there). The idea behind induction is that we ...
• 2,697
### How to write a coterminating, effectful program?
In these sorts of cases, one idiomatic way is to run it with gas, and to create one non-total gas value to let you finish the program. Yes, perhaps Idris should offer a combinator for IO that allows ...
• 326
Accepted
### Proving with co-induction principles
What likely makes it confusing is that you are doing very different things in inductive versus coinductive cases. This is somewhat alluded to by Chlipala's reference to "infinite proofs"1. (Another ...
• 12.1k
### Can a total programming language be Turing-complete?
I'm inclined to disagree with McBride here, for the simple reason that while you can express program semantics totally via coinduction, it is not enough to solve the halting problem for the object ...
Accepted
### When can the coinduction hypothesis be used?
First, let me recall least and greatest fixed points for $\subseteq$. We are working relative to some set $U$, the universe. In the case of (co)inductive definitions, $U$ is the set of all terms. A ...
• 2,254
### Coinduction in mathematical analysis?
The signed digit representation of Real Numbers seems like an efficient approach for: Defining the arithmetic operations and proving their algebraic properties. This was done already in a Master's ...
• 479
Accepted
### Bisimulations: Proof that the following LTS are not bisimilar
The left diagram has a b and the right diagram has a c. Thus, the pair $(2,4')$ does not satisfy the conditions required to be a bisimulation. In particular, the book is correct that those two ...
• 161k
1 vote
### "Largest set" in coinductive definitions
I think the precise definition of a final coalgebra might help. Fix a set of symbols $\Sigma$. For any set $X$ define $F(X) = \Sigma \times X$, and for any $f : X \to Y$ let $F(f) : F(X) \to F(Y)$ be ...
• 30.8k
1 vote
### "Largest set" in coinductive definitions
Here is a tentative self-answer. I think it must just be that there is an additional requirement that wasn't mentioned in the tutorials I read, namely that one of the destructors must apply to each ...
• 976
1 vote
### Definition of M-type in type theory
As far as I can tell, these are the rules: $$\frac{A:Type\quad x:A⊦B:Type}{(M x:A)B(x):Type}-\text{M-Formation}$$ \frac{C:Type\quad t: C\rightarrow \Sigma(a: A)(B[a / a]\rightarrow C)\quad c:C}{...
• 351
Only top scored, non community-wiki answers of a minimum length are eligible | 813 | 3,328 | {"found_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} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.933551 |
http://math.stackexchange.com/questions/91212/is-my-trig-result-unique?answertab=oldest | 1,462,351,350,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122533.7/warc/CC-MAIN-20160428161522-00140-ip-10-239-7-51.ec2.internal.warc.gz | 192,846,375 | 19,863 | # Is my trig result unique?
I recently determined that for all integers $a$ and $b$ such that $a\neq b$ and $b\neq 0$,
$$\arctan\left(\frac{a}{b}\right) + \frac{\pi}{4} = \arctan\left(\frac{b+a}{b-a}\right)$$
This implies that 45 degrees away from any angle with a rational value for tangent lies another angle with a rational value for tangent. The tangent values are related.
If anyone can let me know if this has been done/shown/proven before, please let me know. Thanks!
-
There are other ways to prove that 45 degrees from a rational slope lies a rational slope, and it is not very hard to do, but I must say that I have never seen this identity before. Welcome to MSE! =) +1. – Patrick Da Silva Dec 13 '11 at 20:09
Remember that $\arctan$ (the inverse tangent) always takes values between $-\pi/2$ and $\pi/2$. If you pick a rational that is greater than $\pi/4$ and less than $\pi/2$, then the left hand side of your question cannot be equal to the value of the arctangent at any point, let alone at a rational point. So what you write is not what you meant to write. What you mean, I think, is that if $\alpha$ is an angle such that $\tan(\alpha)=\frac{a}{b}$, then $\tan(\alpha+\frac{\pi}{4}) = \frac{b+a}{b-a}$. – Arturo Magidin Dec 13 '11 at 20:11
@Arturo : Perhaps that is what OP did in his proof? Maybe we should ask him how he did this and help him on his definitions so that such details might not slip his mind again. After all he had an idea in mind. – Patrick Da Silva Dec 13 '11 at 20:14
As written, the formula is not true: the values of $\arctan(x)$ are always between $-\frac{\pi}2$ and $\frac{\pi}{2}$. Pick a rational number $\frac{a}{b}$ with $\frac{\pi}{4}\lt \frac{a}{b}\lt \frac{\pi}{2}$. For example, $a=11$, $b=10$. Then the left hand side, $$\arctan\left(\frac{11}{10}\right)+\frac{\pi}{4}\approx 1.6184$$ whereas the right hand side is negative: $$\arctan\left(\frac{11+10}{10-11}\right) = \arctan(-21) \approx -1.5232.$$
I think that what you mean is that if $\alpha$ is an angle such that $\tan(\alpha)$ is rational, different from $1$, $$\tan(\alpha)=\frac{a}{b}\neq 1,\qquad a,b\text{ integers},$$ then $$\tan\left(\alpha+\frac{\pi}{4}\right) = \frac{b+a}{b-a}.$$
Certainly, well done if you discovered it by yourself! However, it is not new. In fact, the result is true even if $a$ and $b$ are not integers; all you need is for $a$ to be different from $b$, that is, for $\alpha\neq\frac{\pi}{4}$.
There are well-known formulas that express the sine, cosine, and tangent of a sum of angles in terms of the sines, cosines, and tangents of the summands:
\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\ \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\ \tan(\alpha+\beta) &= \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}. \end{align*} Taking $\beta=\frac{\pi}{4}$, since $\tan(\frac{\pi}4) = 1$, we get $$\tan\left(\alpha+\frac{\pi}{4}\right) = \frac{\frac{a}{b}+1}{1-\frac{a}{b}} = \frac{\quad\frac{a+b}{b}\quad}{\frac{b-a}{b}} = \frac{a+b}{b-a},$$ giving your formula.
-
Arturo, thanks for putting what I found into a better format. I discovered this when I noted that $arctan\left(\frac{1}{2}\right)$ had the same decimal as $arctan(3)$ as I was creating an answer key for some right triangle trigonometry problems. About an hour later and an excel spreadsheet of information, I'd determined it. I appreciate you (and Patrick's) help! – Brian Abend Dec 14 '11 at 2:10
@BrianAbend: You do realize that what you see in spreadsheets, etc, are just approximations of the true value of the function, in the vast majority of cases, right? I can find a value $a$ with $\arctan(a)$ that agrees wit $\arctan(3)$ to as many decimal places as you care to specify but is not equal to $\arctan(3)$. – Arturo Magidin Dec 14 '11 at 4:14
I'm well aware of how decimals can be deceiving and also how you can match arctan values to any number of decimals, but I did not make this claim/statement on the basis of those decimals. They piqued my interest, causing me to go searching for a relationship/connection, and I spent 90 minutes trying to figure out the connection. Within reason, I found it, and I'm mostly looking to see if this has been stated elsewhere in either the same or different format. I like how you and others have helped me hone down what I'm looking for, too, so it's another plus towards this discovery :). – Brian Abend Dec 14 '11 at 4:29
@Brian: Good; just double-checking. You'd be surprised how many students today, even in "advanced classes" like calculus, cannot tell the difference between the actual value and a decimal approximation given by a calculator. – Arturo Magidin Dec 14 '11 at 4:37
Quoting from Wikipedia's list of trigonometric identities:
BEGIN QUOTE
$$f(x) = \frac{(\cos\alpha)x - \sin\alpha}{(\sin\alpha)x + \cos\alpha},$$
[$\ldots\ldots$ some material omitted here $\ldots\ldots$]
If $x$ is the slope of a line, then $f(x)$ is the slope of its rotation through an angle of $-\alpha$.
END QUOTE
Dividing the numerator and denominator by $\tan\alpha$ may give the same result as is posted here.
-
If you differentiate the function $$f(t)=\arctan t - \arctan\frac{1 + t}{1 - t},$$ you get zero, so the function is constant in each of the two intervals $(-\infty,1)$ and $(1,+\infty)$ on which it is defined.
• Its value at zero is $\pi/2$, so that $f(t)=-\pi/4$ for all $t<1$, so $$\arctan t + \frac\pi4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t<1.$$
• On the other hand, one easily shows that $\lim_{t\to+\infty}f(t)=\frac{3\pi}{4}$, so $$\arctan t - \frac{3\pi}4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t>1.$$
If $t=a/b$ is a rational number smaller that $1$, then the first point is your identity. If it larger than $1$, we see that you have to change things a bit.
- | 1,802 | 5,840 | {"found_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} | 4.0625 | 4 | CC-MAIN-2016-18 | latest | en | 0.893843 |
https://flatearth.ws/t/hot-air-balloon | 1,716,332,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00439.warc.gz | 225,311,622 | 31,401 | Countering Gravity
Gravity is not the only force. Other forms of force exist. Multiple forces can affect an object at the same time. Forces can counteract gravity, making an object move against the direction of gravity. These cases do not disprove gravity.
Objects can move against the direction of gravity, and flat-Earthers use it as “evidence” that gravity does not exist. In reality, there is at least a force other than gravity affecting the object, in the opposite direction from gravity, with a larger magnitude. Continue reading “Countering Gravity”
Buoyancy
Buoyancy is an upward force exerted by a fluid (liquid or gas) that opposes the weight of an immersed object. Buoyancy happens because the fluid has a pressure gradient. Pressure gradient occurs because the fluid is affected by acceleration, such as the Earth’s gravitational acceleration.
Flat-Earthers makes buoyancy as an “explanations” on how things fall. They are wrong. Without Earth’s gravitational acceleration, buoyancy will not occur.
Hot Air Balloon and the Force of Gravity
A hot air balloon rises because it has buoyancy. Buoyancy is the upward force that is exerted on the trapped hot air because its density is lower than the surrounding air. If buoyancy is greater than the force of gravity — or the weight of the balloon—, then the balloon will rise.
The basic physics escape most flat-Earthers. To them, the fact that a hot-air balloon rises is an ‘evidence’ that gravity does not exist. They believe things go up or down because of density. They are wrong. | 331 | 1,550 | {"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} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.938739 |
http://library.thinkquest.org/C005705/English/Waves/interference.htm | 1,386,211,266,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163038307/warc/CC-MAIN-20131204131718-00071-ip-10-33-133-15.ec2.internal.warc.gz | 107,338,123 | 3,093 | Phase Relationships: Interference Demo2 Interference Interference is when a propagating wave encounters another object that may alter its path or change the amplitude in the medium. There are two types of interference: constructive and destructive. Two or more waves cross paths but do not interrupt the separate wave structure and as a result they either create a larger displacement or cancel each other; this is called the superposition of waves. In constructive interference, waves that cross paths while traveling in the same direction are considered to be in phase. This means the crests are in phase within the medium resulting in an amplitude equal to the sum of the separate amplitudes. If both amplitudes are equal, this doubles the resulting amplitude. Destructive interference is exactly opposite of constructive interference. Assuming that both waves have the same amplitude, when the crest and the trough line up, that is to say, when the two waves are out of phase, the result is an amplitude equal to zero. Demo1 Back to Phase Relationships To Top | 205 | 1,065 | {"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} | 2.953125 | 3 | CC-MAIN-2013-48 | latest | en | 0.899047 |
https://javascripts.com/pad-leading-zeros-in-javascript/ | 1,719,080,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00871.warc.gz | 281,205,075 | 11,726 | Leading zeros: those silent placeholders that help to keep our numbers uniform and orderly. Often overlooked, they play a crucial role in formatting numbers, especially in the world of programming. You might wonder, how can we add these zeros at the start of a number? Well, JavaScript has a few tricks up its sleeve that can help us achieve this.
This task might seem daunting if you’re new to JavaScript, but don’t worry, it’s simpler than you think. We’re going to walk through step by step, using easy-to-understand language. By the end, you’ll be padding leading zeros like a pro!
In this post, we will be looking at the following 3 ways to pad leading zeros in JavaScript:
• By using the padStart() method
• By using a custom function with a while loop
• By using the slice() or substring() method along with the repeat() method
Let’s explore each…
## #1 – By using the padStart() method
The padStart() method in JavaScript is a string function that pads the current string with a given string (repeated, if needed) until the resulting string reaches the provided length. The padding is applied from the start of the current string. Let’s understand this by considering an example where we want to pad a number with leading zeros.
``````
let number = "5";
``````
### How it works
The padStart() method pads the current string with another string (second argument) until the resulting string reaches the given length (first argument). The padding is applied from the start or beginning of the current string. In our example, we padded a string representation of a number with leading zeros until its length is 5.
• Initially, we declare a variable ‘number’ with a value of “5”.
• Then, we use the padStart() method on this string. The first argument is the total length we desire for our string, which is 5 in this case.
• The second argument is the string with which we want to pad our initial string. We use “0” here to add leading zeros.
• The padStart() method therefore adds enough zeros to the beginning of our ‘number’ string to make its length equal to 5. If ‘number’ was already 5 characters long or longer, no padding would occur.
• Finally, we log the result to the console, and we see that our number string is now “00005”, padded with zeros to reach a length of 5 characters.
## #2 – By using a custom function with a while loop
This is a custom JavaScript function that pads leading zeros to a number up to a specified length using a while loop. For example, if we want to pad the number 7 to a length of 3, we get “007”.
``````
num = num.toString();
while(num.length < length) {
num = '0' + num;
}
return num;
}
```
```
### How it works
This function first converts the number to a string. Then, it appends a '0' to the front of the string until it reaches the desired length. The padded string is then returned.
• The function takes two inputs: 'num', the number to be padded, and 'length', the desired length of the string after padding.
• The number 'num' is first converted into a string.
• The function enters a while loop that continues as long as the length of the string is less than the desired length.
• Inside the loop, a '0' is prepended to the start of the string.
• This process is repeated until the string reaches the specified length, effectively padding the number with leading zeros.
• The padded string is finally returned by the function.
## #3 - By using the slice() or substring() method along with the repeat() method
This approach involves creating a string of zeros using the repeat() method, afterwards adding it to the beginning of the input number, and finally using the slice() or substring() method to return the desired length of characters from the end of the string.
For instance, if we need to pad the number '5' to a length of 5, we will first create a string '00000'. Then, we concatenate '00000' with '5' to get '000005'. Finally, we use the slice() method to return the last 5 characters: '00005'.
``````
return ('0'.repeat(length) + num).slice(-length);
}
``````
### How it works
The padLeadingZeros method takes two arguments: the number to be padded (num) and the desired length of the resulting string (length). It first creates a string of zeros equal to the desired length, then adds the number to the end of this string. Finally, it uses the slice() method to return the desired length from the end of the string.
• Repeat '0' for the desired length: '0'.repeat(length) generates a string of zeros.
• Concatenation: ('0'.repeat(length) + num) joins the number to the end of the zeros string.
• Slice: .slice(-length) returns the last 'length' characters from the string, effectively discarding any excess zeros from the start of the string.
Related:
In summary, padding leading zeros in JavaScript is a simple procedure that can be carried out with methods like 'padStart'. Such methods help format numbers and strings with ease, making your code cleaner and more efficient.
Understanding and applying these methods can take your JavaScript coding to the next level. Therefore, never stop learning and experimenting with these useful techniques to enhance your programming skills. | 1,157 | 5,172 | {"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} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.852296 |
http://www.sciforums.com/threads/what-is-magnitive.160029/ | 1,527,258,704,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00206.warc.gz | 457,680,271 | 11,665 | # What is Magnitive
Discussion in 'Pseudoscience' started by Asexperia, Oct 11, 2017.
1. ### AsexperiaRegistered Senior Member
Messages:
987
What is magnitive refers to the property of beings that can be measured, but It is imperceptible.
What is magnitive is objective, but not concrete; for example in Physics: force, gravity and time.
We feel weight, but not gravity.
Magnitive (created by Asexperia) is a modification of the word magnitude.
Last edited: Oct 11, 2017
Messages:
1,478
5. ### AsexperiaRegistered Senior Member
Messages:
987
To stablish the relation between the observer and some physical magnitudes.
Objective and subjective are relations between the observer and reality
7. ### mathmanValued Senior Member
Messages:
1,478
I believe you have the wrong forum. Try philosophy.
Asexperia likes this.
8. ### AsexperiaRegistered Senior Member
Messages:
987
We know by intuition FORCE in: f = m . a
We know by intuition TIME in: s = d / t
What is magnitive is imperceptible, but known
by intuition.
9. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
10,271
Reported so it can be moved out of the science section.
10. ### AsexperiaRegistered Senior Member
Messages:
987
We intuit time in the waiting and in: s = d / t
We intuit gravity in the weight and in: F = G m1 . m2 / d2
11. ### exchemistValued Senior Member
Messages:
7,168
Ballocks. We learn these laws at school.
(Why am I bothering? [rhetorical].)
12. ### AsexperiaRegistered Senior Member
Messages:
987
Yes, but you didn't know that force, gravity and time are MAGNITIVE.
13. ### CounterRegistered Senior Member
Messages:
512
You tell 'im asexperia!
14. ### DaveC426913Valued Senior Member
Messages:
8,378
Dude. You just made the word up. Don't pretend like it's some sort of fact.
15. ### exchemistValued Senior Member
Messages:
7,168
And do you think that enlightens me? At all?
16. ### AsexperiaRegistered Senior Member
Messages:
987
Other concepts created by Asexperia are:
Philochrony, philochron line, Chromnesia and
becoming-duration duality.
17. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
10,271
Asexperia, IMO your ideas are mostly flarctic.
18. ### CounterRegistered Senior Member
Messages:
512
Yes. Inventing words that do not exist? Madness!
19. ### CounterRegistered Senior Member
Messages:
512
"Discombobulate": Third person present.
20. ### DaveC426913Valued Senior Member
Messages:
8,378
Much more doable than inventing words that do exist.
21. ### AsexperiaRegistered Senior Member
Messages:
987
It's never late to learn.
My ideas are on my blog for History.
22. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
10,271
Maintaining Delusions of Grandeur is not a particularly healthy way to live a life.
23. ### AsexperiaRegistered Senior Member
Messages:
987
Out of topic. Focus. | 745 | 2,939 | {"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-2018-22 | latest | en | 0.918888 |
https://www.typingonlinefromhome.com/how-to-write-piecewise-functions-in-desmos/ | 1,713,518,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00734.warc.gz | 958,721,302 | 14,867 | # A Guide To Writing Piecewise Functions in Desmos
Are you ready to take your math skills to the next level? Piecewise functions are a powerful tool that can help you solve complex problems and model real-life scenarios. And with Desmos, an online graphing calculator, writing and graphing piecewise functions has never been easier.
In this article, we’ll guide you through the process of writing piecewise functions in Desmos. We’ll start by explaining what piecewise functions are and why they’re useful. Then, we’ll show you how to get started with Desmos and write your own piecewise functions. Along the way, we’ll share some tips and tricks for graphing these functions effectively.
By the end of this article, you’ll have a solid understanding of how to use piecewise functions in Desmos to tackle even the most challenging math problems. So let’s dive in!
## Key Takeaways
• Piecewise functions use multiple sub-functions for different intervals of the domain.
• Domain and range of each sub-function are important factors to consider when writing piecewise functions.
• Desmos is an online graphing calculator that makes writing and graphing piecewise functions easier.
• Breaking down each piece into smaller segments and using different color schemes or line styles can make it easier to accurately plot and understand piecewise functions.
## Understanding Piecewise Functions
Let’s dive into understanding piecewise functions – it’s gonna be a wild ride! First things first, let’s define what a piecewise function is. In mathematics, a piecewise function is a function that’s defined by multiple sub-functions, each applying to different intervals of the domain. This means that the function changes its behavior depending on which part of the domain you’re looking at.
One way to represent a piecewise function is through its graphical representation. Each sub-function can be plotted separately on the same coordinate plane and connected with open circles or squares at their endpoints. This allows us to see how the function behaves differently in different parts of the domain.
When working with piecewise functions, it’s important to consider the domain and range of each sub-function. The domain refers to all possible input values for which the given expression produces meaningful output values. The range, on the other hand, refers to all possible output values produced by those input values within that particular interval.
By understanding how these factors play into creating and analyzing piecewise functions, we can create more accurate models for real-world situations and better understand mathematical concepts as a whole.
So buckle up and get ready – with this knowledge under your belt, you’ll be able to tackle any problem involving piecewise functions!
## Getting Started with Desmos
You can start exploring Desmos by simply clicking on the colorful icons and watching the graphs come to life before your eyes. This powerful graphing calculator has many features for advanced graphing, including piecewise functions.
To get started, just click on “Functions”in the menu bar and select “Piecewise”from the drop-down menu. Once you’ve created your piecewise function, you can customize it further using Desmos’ other features. For example, you can add sliders to your graph to make it interactive, or use colors and styles to make it more visually appealing.
You can also save your work and share it with others by collaborating on Desmos projects. Collaborating on Desmos projects with others is easy – simply click on “Share”in the top right corner of your screen and choose whether you want to collaborate via a link or email invitation.
With this feature, you can work with classmates or colleagues in real-time, making changes and discussing ideas together. Whether you’re working on a school project or a professional presentation, Desmos makes it easy to create and share complex graphs quickly and efficiently.
## Writing Piecewise Functions in Desmos
Creating complex graphs is made simpler with Desmos’ ability to handle piecewise expressions. Piecewise functions are used in data analysis when different parts of the data follow different mathematical models. You can use Desmos to graph these functions by using the ‘piecewise’ command.
To write a piecewise function in Desmos, start by typing “f(x)=”followed by the first equation for the domain range you want to define. Then add a comma and enter the next equation for another part of the domain range. Repeat this process for each part of your function until all domains have been defined. Make sure that you use parentheses around each equation and separate them with commas.
Common mistakes to avoid when graphing piecewise functions include forgetting parentheses, not defining all parts of the domain, and not checking for continuity at points where two pieces meet. It’s important to check that both sides approach the same value at these points to ensure continuity in your graph.
Writing piecewise functions in Desmos is a powerful tool for analyzing complex data sets. By following a few simple steps and avoiding common mistakes, you can create accurate and visually appealing graphs that help you better understand your data. So, go ahead and experiment with piecewise expressions on Desmos – who knows what insights you might uncover!
## Tips and Tricks for Graphing Piecewise Functions
Maximizing the potential of your data analysis requires mastering tips and tricks for graphing piecewise expressions. Graphing techniques can help you present your data in a visual manner that’s easy to understand, making it easier to draw conclusions and make decisions based on your findings.
When working with piecewise functions, there are some specific graphing techniques that can be particularly helpful. One of the most important things to keep in mind when graphing piecewise functions is domain restrictions. Each piece of the function will have its own domain, which may differ from the overall domain of the function. It’s important to carefully consider these domain restrictions when creating your graph so that you accurately represent all parts of the function without including any irrelevant information.
Another tip for graphing piecewise functions is to break down each piece into smaller segments if needed. This can make it easier to accurately plot points and see how each segment fits together within the overall function. You may also want to experiment with different color schemes or line styles for each segment in order to make it easier to distinguish between them visually.
By using these tips and tricks for graphing piecewise functions, you can create clear, accurate representations of complex data sets that are easy for anyone to understand. Whether you’re working on a research project or simply trying to better understand a particular phenomenon, taking the time to master these skills will pay off in more effective data analysis and decision-making capabilities over time.
## Applying Piecewise Functions in Real-Life Scenarios
Imagine using piecewise functions to model real-life scenarios. Piecewise functions are a powerful tool that can help you analyze and understand complex situations. For example, suppose you’re an urban planner trying to calculate the number of people using a specific bus route at different times of the day. You might use a piecewise function to model the data, with one function for rush hour and another for off-peak times.
Another real-life scenario where piecewise functions come in handy is calculating the cost of a taxi ride with different rates for distance traveled or time spent waiting at lights. By breaking down the trip into segments based on distance or time, you can use separate equations for each segment to create a more accurate representation of the total cost.
While piecewise functions have many advantages when modeling real-life scenarios, there are also some disadvantages compared to other types of functions. They can be more complicated mathematically than simpler equations like linear or quadratic models. Additionally, it may not always be clear how to break down a problem into multiple pieces in order to use piecewise functions effectively. However, if used correctly, piecewise functions offer an efficient and innovative way to tackle complex problems and gain new insights into real-world phenomena.
### Can I use Desmos to graph piecewise functions with more than two conditions?
Yes, Desmos allows you to graph and analyze multi-dimensional piecewise functions with more than two conditions. Use techniques for simplifying complex functions, and enjoy exploring innovative ways to visualize your data.
### How do I find the domain and range of a piecewise function on Desmos?
To find the domain and range of a piecewise function on Desmos, first identify the critical points using techniques like setting each condition equal to zero. Then, adjust the formatting and appearance with strategies like changing colors or adding labels.
### Is it possible to label the different sections of a piecewise function on Desmos?
Labeling different sections of a piecewise function on Desmos is possible. Use the “Note”tool to add labels and explain each condition. To create smooth transitions, use techniques like using parentheses or absolute value functions.
### Can I export the graph of my piecewise function from Desmos to use in other documents or presentations?
Yes, Desmos offers exporting options for piecewise function graphs. You can customize the appearance of your graph by adjusting colors and adding labels. Export as an image or PDF to use in other documents or presentations.
### How do I add a vertical line or point to my graph to represent a discontinuity in the function?
To add vertical lines or points to represent discontinuities in your Desmos graph, simply use the “add item”button and select “vertical line”or “point.”Customize colors as desired. Keep your graphs innovative and efficient with these simple steps!
## Conclusion
Congratulations! You’ve successfully learned how to write piecewise functions in Desmos. With this new skill, you can easily and accurately graph complex functions. By understanding the concept of piecewise functions and following the step-by-step guide on how to write them in Desmos, you now have a valuable tool to help you solve real-life problems.
But don’t stop here! Keep practicing and experimenting with different types of piecewise functions. Try applying them to various scenarios and see how they can help you visualize data or make predictions. Use the tips and tricks provided to enhance your graphing skills and create professional-looking charts.
Writing piecewise functions in Desmos may seem daunting at first, but with practice, it becomes second nature. So keep exploring and pushing yourself to learn more about this powerful tool. Who knows what kind of innovative solutions you may discover along the way? | 2,073 | 11,024 | {"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} | 4.4375 | 4 | CC-MAIN-2024-18 | latest | en | 0.871618 |
https://www.mathway.com/examples/algebra/vectors/finding-an-orthonormal-basis-by-gram-schmidt-method?id=2458 | 1,701,317,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00819.warc.gz | 981,295,627 | 34,824 | # Algebra Examples
Find an Orthonormal Basis by Gram-Schmidt Method
, ,
Step 1
Assign a name for each vector.
Step 2
The first orthogonal vector is the first vector in the given set of vectors.
Step 3
Use the formula to find the other orthogonal vectors.
Step 4
Find the orthogonal vector .
Step 4.1
Use the formula to find .
Step 4.2
Substitute for .
Step 4.3
Find .
Step 4.3.1
Find the dot product.
Step 4.3.1.1
The dot product of two vectors is the sum of the products of the their components.
Step 4.3.1.2
Simplify.
Step 4.3.1.2.1
Simplify each term.
Step 4.3.1.2.1.1
Multiply by .
Step 4.3.1.2.1.2
Multiply by .
Step 4.3.1.2.1.3
Multiply by .
Step 4.3.1.2.2
Step 4.3.1.2.3
Step 4.3.2
Find the norm of .
Step 4.3.2.1
The norm is the square root of the sum of squares of each element in the vector.
Step 4.3.2.2
Simplify.
Step 4.3.2.2.1
One to any power is one.
Step 4.3.2.2.2
One to any power is one.
Step 4.3.2.2.3
One to any power is one.
Step 4.3.2.2.4
Step 4.3.2.2.5
Step 4.3.3
Find the projection of onto using the projection formula.
Step 4.3.4
Substitute for .
Step 4.3.5
Substitute for .
Step 4.3.6
Substitute for .
Step 4.3.7
Simplify the right side.
Step 4.3.7.1
Rewrite as .
Step 4.3.7.1.1
Use to rewrite as .
Step 4.3.7.1.2
Apply the power rule and multiply exponents, .
Step 4.3.7.1.3
Combine and .
Step 4.3.7.1.4
Cancel the common factor of .
Step 4.3.7.1.4.1
Cancel the common factor.
Step 4.3.7.1.4.2
Rewrite the expression.
Step 4.3.7.1.5
Evaluate the exponent.
Step 4.3.7.2
Multiply by each element of the matrix.
Step 4.3.7.3
Simplify each element in the matrix.
Step 4.3.7.3.1
Multiply by .
Step 4.3.7.3.2
Multiply by .
Step 4.3.7.3.3
Multiply by .
Step 4.4
Substitute the projection.
Step 4.5
Simplify.
Step 4.5.1
Combine each component of the vectors.
Step 4.5.2
Subtract from .
Step 4.5.3
Write as a fraction with a common denominator.
Step 4.5.4
Combine the numerators over the common denominator.
Step 4.5.5
Subtract from .
Step 4.5.6
Write as a fraction with a common denominator.
Step 4.5.7
Combine the numerators over the common denominator.
Step 4.5.8
Subtract from .
Step 5
Find the orthogonal vector .
Step 5.1
Use the formula to find .
Step 5.2
Substitute for .
Step 5.3
Find .
Step 5.3.1
Find the dot product.
Step 5.3.1.1
The dot product of two vectors is the sum of the products of the their components.
Step 5.3.1.2
Simplify.
Step 5.3.1.2.1
Simplify each term.
Step 5.3.1.2.1.1
Multiply by .
Step 5.3.1.2.1.2
Multiply by .
Step 5.3.1.2.1.3
Multiply by .
Step 5.3.1.2.2
Step 5.3.1.2.3
Step 5.3.2
Find the norm of .
Step 5.3.2.1
The norm is the square root of the sum of squares of each element in the vector.
Step 5.3.2.2
Simplify.
Step 5.3.2.2.1
One to any power is one.
Step 5.3.2.2.2
One to any power is one.
Step 5.3.2.2.3
One to any power is one.
Step 5.3.2.2.4
Step 5.3.2.2.5
Step 5.3.3
Find the projection of onto using the projection formula.
Step 5.3.4
Substitute for .
Step 5.3.5
Substitute for .
Step 5.3.6
Substitute for .
Step 5.3.7
Simplify the right side.
Step 5.3.7.1
Rewrite as .
Step 5.3.7.1.1
Use to rewrite as .
Step 5.3.7.1.2
Apply the power rule and multiply exponents, .
Step 5.3.7.1.3
Combine and .
Step 5.3.7.1.4
Cancel the common factor of .
Step 5.3.7.1.4.1
Cancel the common factor.
Step 5.3.7.1.4.2
Rewrite the expression.
Step 5.3.7.1.5
Evaluate the exponent.
Step 5.3.7.2
Multiply by each element of the matrix.
Step 5.3.7.3
Simplify each element in the matrix.
Step 5.3.7.3.1
Multiply by .
Step 5.3.7.3.2
Multiply by .
Step 5.3.7.3.3
Multiply by .
Step 5.4
Find .
Step 5.4.1
Find the dot product.
Step 5.4.1.1
The dot product of two vectors is the sum of the products of the their components.
Step 5.4.1.2
Simplify.
Step 5.4.1.2.1
Simplify each term.
Step 5.4.1.2.1.1
Multiply .
Step 5.4.1.2.1.1.1
Multiply by .
Step 5.4.1.2.1.1.2
Multiply by .
Step 5.4.1.2.1.2
Multiply by .
Step 5.4.1.2.1.3
Multiply by .
Step 5.4.1.2.2
Step 5.4.1.2.3
Step 5.4.2
Find the norm of .
Step 5.4.2.1
The norm is the square root of the sum of squares of each element in the vector.
Step 5.4.2.2
Simplify.
Step 5.4.2.2.1
Use the power rule to distribute the exponent.
Step 5.4.2.2.1.1
Apply the product rule to .
Step 5.4.2.2.1.2
Apply the product rule to .
Step 5.4.2.2.2
Raise to the power of .
Step 5.4.2.2.3
Multiply by .
Step 5.4.2.2.4
Raise to the power of .
Step 5.4.2.2.5
Raise to the power of .
Step 5.4.2.2.6
Apply the product rule to .
Step 5.4.2.2.7
One to any power is one.
Step 5.4.2.2.8
Raise to the power of .
Step 5.4.2.2.9
Apply the product rule to .
Step 5.4.2.2.10
One to any power is one.
Step 5.4.2.2.11
Raise to the power of .
Step 5.4.2.2.12
Combine the numerators over the common denominator.
Step 5.4.2.2.13
Step 5.4.2.2.14
Combine the numerators over the common denominator.
Step 5.4.2.2.15
Step 5.4.2.2.16
Cancel the common factor of and .
Step 5.4.2.2.16.1
Factor out of .
Step 5.4.2.2.16.2
Cancel the common factors.
Step 5.4.2.2.16.2.1
Factor out of .
Step 5.4.2.2.16.2.2
Cancel the common factor.
Step 5.4.2.2.16.2.3
Rewrite the expression.
Step 5.4.2.2.17
Rewrite as .
Step 5.4.2.2.18
Multiply by .
Step 5.4.2.2.19
Combine and simplify the denominator.
Step 5.4.2.2.19.1
Multiply by .
Step 5.4.2.2.19.2
Raise to the power of .
Step 5.4.2.2.19.3
Raise to the power of .
Step 5.4.2.2.19.4
Use the power rule to combine exponents.
Step 5.4.2.2.19.5
Step 5.4.2.2.19.6
Rewrite as .
Step 5.4.2.2.19.6.1
Use to rewrite as .
Step 5.4.2.2.19.6.2
Apply the power rule and multiply exponents, .
Step 5.4.2.2.19.6.3
Combine and .
Step 5.4.2.2.19.6.4
Cancel the common factor of .
Step 5.4.2.2.19.6.4.1
Cancel the common factor.
Step 5.4.2.2.19.6.4.2
Rewrite the expression.
Step 5.4.2.2.19.6.5
Evaluate the exponent.
Step 5.4.2.2.20
Simplify the numerator.
Step 5.4.2.2.20.1
Combine using the product rule for radicals.
Step 5.4.2.2.20.2
Multiply by .
Step 5.4.3
Find the projection of onto using the projection formula.
Step 5.4.4
Substitute for .
Step 5.4.5
Substitute for .
Step 5.4.6
Substitute for .
Step 5.4.7
Simplify the right side.
Step 5.4.7.1
Simplify the denominator.
Step 5.4.7.1.1
Apply the product rule to .
Step 5.4.7.1.2
Rewrite as .
Step 5.4.7.1.2.1
Use to rewrite as .
Step 5.4.7.1.2.2
Apply the power rule and multiply exponents, .
Step 5.4.7.1.2.3
Combine and .
Step 5.4.7.1.2.4
Cancel the common factor of .
Step 5.4.7.1.2.4.1
Cancel the common factor.
Step 5.4.7.1.2.4.2
Rewrite the expression.
Step 5.4.7.1.2.5
Evaluate the exponent.
Step 5.4.7.1.3
Raise to the power of .
Step 5.4.7.1.4
Cancel the common factor of and .
Step 5.4.7.1.4.1
Factor out of .
Step 5.4.7.1.4.2
Cancel the common factors.
Step 5.4.7.1.4.2.1
Factor out of .
Step 5.4.7.1.4.2.2
Cancel the common factor.
Step 5.4.7.1.4.2.3
Rewrite the expression.
Step 5.4.7.2
Multiply the numerator by the reciprocal of the denominator.
Step 5.4.7.3
Cancel the common factor of .
Step 5.4.7.3.1
Cancel the common factor.
Step 5.4.7.3.2
Rewrite the expression.
Step 5.4.7.4
Multiply by each element of the matrix.
Step 5.4.7.5
Simplify each element in the matrix.
Step 5.4.7.5.1
Cancel the common factor of .
Step 5.4.7.5.1.1
Move the leading negative in into the numerator.
Step 5.4.7.5.1.2
Factor out of .
Step 5.4.7.5.1.3
Cancel the common factor.
Step 5.4.7.5.1.4
Rewrite the expression.
Step 5.4.7.5.2
Move the negative in front of the fraction.
Step 5.4.7.5.3
Multiply .
Step 5.4.7.5.3.1
Multiply by .
Step 5.4.7.5.3.2
Multiply by .
Step 5.4.7.5.4
Multiply .
Step 5.4.7.5.4.1
Multiply by .
Step 5.4.7.5.4.2
Multiply by .
Step 5.5
Substitute the projections.
Step 5.6
Simplify.
Step 5.6.1
Combine each component of the vectors.
Step 5.6.2
Combine each component of the vectors.
Step 5.6.3
Multiply .
Step 5.6.3.1
Multiply by .
Step 5.6.3.2
Multiply by .
Step 5.6.4
Combine fractions.
Step 5.6.4.1
Combine the numerators over the common denominator.
Step 5.6.4.2
Simplify the expression.
Step 5.6.4.2.1
Step 5.6.4.2.2
Divide by .
Step 5.6.5
Multiply by .
Step 5.6.6
Subtract from .
Step 5.6.7
To write as a fraction with a common denominator, multiply by .
Step 5.6.8
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
Step 5.6.8.1
Multiply by .
Step 5.6.8.2
Multiply by .
Step 5.6.9
Simplify the expression.
Step 5.6.9.1
Combine the numerators over the common denominator.
Step 5.6.9.2
Subtract from .
Step 5.6.10
Cancel the common factor of and .
Step 5.6.10.1
Factor out of .
Step 5.6.10.2
Cancel the common factors.
Step 5.6.10.2.1
Factor out of .
Step 5.6.10.2.2
Cancel the common factor.
Step 5.6.10.2.3
Rewrite the expression.
Step 5.6.11
Move the negative in front of the fraction.
Step 5.6.12
Find the common denominator.
Step 5.6.12.1
Write as a fraction with denominator .
Step 5.6.12.2
Multiply by .
Step 5.6.12.3
Multiply by .
Step 5.6.12.4
Multiply by .
Step 5.6.12.5
Multiply by .
Step 5.6.12.6
Reorder the factors of .
Step 5.6.12.7
Multiply by .
Step 5.6.13
Combine the numerators over the common denominator.
Step 5.6.14
Simplify by subtracting numbers.
Step 5.6.14.1
Subtract from .
Step 5.6.14.2
Subtract from .
Step 5.6.15
Cancel the common factor of and .
Step 5.6.15.1
Factor out of .
Step 5.6.15.2
Cancel the common factors.
Step 5.6.15.2.1
Factor out of .
Step 5.6.15.2.2
Cancel the common factor.
Step 5.6.15.2.3
Rewrite the expression.
Step 6
Find the orthonormal basis by dividing each orthogonal vector by its norm.
Step 7
Find the unit vector where .
Step 7.1
To find a unit vector in the same direction as a vector , divide by the norm of .
Step 7.2
The norm is the square root of the sum of squares of each element in the vector.
Step 7.3
Simplify.
Step 7.3.1
One to any power is one.
Step 7.3.2
One to any power is one.
Step 7.3.3
One to any power is one.
Step 7.3.4
Step 7.3.5
Step 7.4
Divide the vector by its norm.
Step 7.5
Divide each element in the vector by .
Step 8
Find the unit vector where .
Step 8.1
To find a unit vector in the same direction as a vector , divide by the norm of .
Step 8.2
The norm is the square root of the sum of squares of each element in the vector.
Step 8.3
Simplify.
Step 8.3.1
Use the power rule to distribute the exponent.
Step 8.3.1.1
Apply the product rule to .
Step 8.3.1.2
Apply the product rule to .
Step 8.3.2
Raise to the power of .
Step 8.3.3
Multiply by .
Step 8.3.4
Raise to the power of .
Step 8.3.5
Raise to the power of .
Step 8.3.6
Apply the product rule to .
Step 8.3.7
One to any power is one.
Step 8.3.8
Raise to the power of .
Step 8.3.9
Apply the product rule to .
Step 8.3.10
One to any power is one.
Step 8.3.11
Raise to the power of .
Step 8.3.12
Combine the numerators over the common denominator.
Step 8.3.13
Step 8.3.14
Combine the numerators over the common denominator.
Step 8.3.15
Step 8.3.16
Cancel the common factor of and .
Step 8.3.16.1
Factor out of .
Step 8.3.16.2
Cancel the common factors.
Step 8.3.16.2.1
Factor out of .
Step 8.3.16.2.2
Cancel the common factor.
Step 8.3.16.2.3
Rewrite the expression.
Step 8.3.17
Rewrite as .
Step 8.4
Divide the vector by its norm.
Step 8.5
Divide each element in the vector by .
Step 8.6
Simplify.
Step 8.6.1
Multiply the numerator by the reciprocal of the denominator.
Step 8.6.2
Multiply by .
Step 8.6.3
Move to the left of .
Step 8.6.4
Move to the left of .
Step 8.6.5
Multiply the numerator by the reciprocal of the denominator.
Step 8.6.6
Multiply by .
Step 8.6.7
Multiply the numerator by the reciprocal of the denominator.
Step 8.6.8
Multiply by .
Step 9
Find the unit vector where .
Step 9.1
To find a unit vector in the same direction as a vector , divide by the norm of .
Step 9.2
The norm is the square root of the sum of squares of each element in the vector.
Step 9.3
Simplify.
Step 9.3.1
Raising to any positive power yields .
Step 9.3.2
Use the power rule to distribute the exponent.
Step 9.3.2.1
Apply the product rule to .
Step 9.3.2.2
Apply the product rule to .
Step 9.3.3
Raise to the power of .
Step 9.3.4
Multiply by .
Step 9.3.5
One to any power is one.
Step 9.3.6
Raise to the power of .
Step 9.3.7
Apply the product rule to .
Step 9.3.8
One to any power is one.
Step 9.3.9
Raise to the power of .
Step 9.3.10
Step 9.3.11
Combine the numerators over the common denominator.
Step 9.3.12
Step 9.3.13
Cancel the common factor of and .
Step 9.3.13.1
Factor out of .
Step 9.3.13.2
Cancel the common factors.
Step 9.3.13.2.1
Factor out of .
Step 9.3.13.2.2
Cancel the common factor.
Step 9.3.13.2.3
Rewrite the expression.
Step 9.3.14
Rewrite as .
Step 9.3.15
Any root of is .
Step 9.4
Divide the vector by its norm.
Step 9.5
Divide each element in the vector by .
Step 9.6
Simplify.
Step 9.6.1
Multiply the numerator by the reciprocal of the denominator.
Step 9.6.2
Multiply by .
Step 9.6.3
Multiply the numerator by the reciprocal of the denominator.
Step 9.6.4
Combine and .
Step 9.6.5
Multiply the numerator by the reciprocal of the denominator.
Step 9.6.6
Combine and .
Step 10
Substitute the known values. | 5,107 | 12,882 | {"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} | 4.125 | 4 | CC-MAIN-2023-50 | longest | en | 0.721166 |
https://www.calendar-uk.co.uk/frequently-asked-questions/what-is-the-hcf-of-100-and-50 | 1,713,608,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817576.41/warc/CC-MAIN-20240420091126-20240420121126-00465.warc.gz | 613,262,385 | 8,831 | # What is the HCF of 100 and 50?
Step-by-step explanation:
So the greatest common factor 50 and 100 is 50.
## What are factors of 50 and 100?
There are 6 common factors of 50 and 100, that are 1, 2, 5, 10, 50, and 25. Therefore, the greatest common factor of 50 and 100 is 50.
## What is the LCM of 100 and 50?
The LCM of 50 and 100 is 100.
## What are multiples of 50 and 100?
The first few multiples of 50 and 100 are (50, 100, 150, 200, 250, 300, 350, . . . ) and (100, 200, 300, 400, 500, 600, . . . )
## How to calculate hcf?
How to Find HCF?
1. Step 1: Write each number as a product of its prime factors. This method is called here prime factorization.
2. Step 2: Now list the common factors of both the numbers.
3. Step 3: The product of all common prime factors is the HCF ( use the lower power of each common factor)
## Pet Simulator X EGGWARS VS GravyKoalaMan
45 related questions found
### What is example of HCF?
Let us take an example: 1. Problem: Find the H.C.F. of 36 and 45. Solution: 36 = 1,2,3,4,6.9,12,18,3645 =1,3,5,9,45 Since the highest common factor of 36 and 45 is 9, therefore H.C.F of 36 and 45 is 9.
### What is the formula of HCF with example?
HCF = Product of the common prime factors with the lowest powers. Division Method: We use Euclid's Division Lemma( a = bq+ r.) to find the HCF of two numbers a and b. i.e. Dividend = Divisor × Quotient + Remainder. The lemma states when a divides b, q is the quotient and r is the remainder.
### What is the HCF of 100?
What is the HCF of 100 and 190? The HCF of 100 and 190 is 10. To calculate the Highest Common Factor of 100 and 190, we need to factor each number (factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100; factors of 190 = 1, 2, 5, 10, 19, 38, 95, 190) and choose the highest factor that exactly divides both 100 and 190, .
### What is the HCF of 50?
Factors of 50 (Fifty) = 1, 2, 5, 10, 25 and 50. Factors of 70 (Seventy) = 1, 2, 5, 14, 7, 10, 35 and 70. Therefore, common factor of 50 (Fifty) and 70 (Seventy) = 1, 2, 5, 10. Highest common factor (H.C.F) of 50 (Fifty) and 70 (Seventy) = 10.
### How many prime numbers are there between 100 and 50?
All the prime numbers included between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. These all are the fixed 10 prime numbers which are included between 50 and 100. Because these prime numbers can not be divisible with another number, they can be divided by itself.
### What is the HCF of 55 and 100?
The GCF of 55 and 100 is 5. To calculate the GCF (Greatest Common Factor) of 55 and 100, we need to factor each number (factors of 55 = 1, 5, 11, 55; factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100) and choose the greatest factor that exactly divides both 55 and 100, i.e., 5.
### What is the LCM of 100?
LCM of 100 and 200 is 200. By making a note of the multiples of 100 and 200, the number evenly divisible by 100 and 200 is the LCM. Least common multiple of 100 and 200 is the multiple we get commonly using the process of multiplication. (100, 200, 300, 400, 500, 600, 700, ….)
### What are the multiples of 50?
Multiples of 50 are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600, 650, 700,… Multiples of 80 are 80, 160, 240, 320, 400, 480, 560, 640, 720, 800,… Clearly, the lowest common multiple or LCM of 50 and 80 is 400.
### What is the LCM of 50?
LCM of 50 and 60 is 300. In Maths, the value which is divisible by the two numbers indicates the LCM. Least common multiple of 50 and 60 is the smallest number which can be obtained from the common multiples. (50, 100, 150, 200, 250, 300, ….)
### What factors make up 50?
Factors of 50
• Factors of 50: 1, 2, 5, 10, 25, 50.
• Prime Factorization of 50: 2 × 5 × 5.
### What are the common factors of 10 and 50?
There are 4 common factors of 10 and 50, that are 1, 2, 10, and 5. Therefore, the greatest common factor of 10 and 50 is 10.
### What is the HCF of 100 and 52?
Since 4 is the largest of these common factors, the GCF of 52 and 100 would be 4.
### What is the HCF of 110 and 50?
Solution: The GCF of 110 and 50 is 10.
### How do you find the HCF of two numbers?
HCF of two numbers by Division Method
1. First, divide the large number by a small number.
2. If the remainder is left, then divide the first divisor by remainder.
3. If the remainder divides the first divisor completely, then it is the HCF or highest common factor of the given two numbers.
### What is the HCF of 100 and 54?
2 is the hcf of 54 and 100.
### What is the HCF of 100 and 30?
Therefore, the greatest common factor of 30 and 100 is 10.
### What is HCF of 100 and 10?
Hence \[10\] is the HCF of \[10\] and \[100\] . So, the correct answer is “10”.
### How do you find a number if HCF is given?
The formula that shows the relationship between their LCM and HCF is: LCM (a,b) × HCF (a,b) = a × b. For example, let us take two numbers 12 and 8. Let us use the formula: LCM (12,8) × HCF (12,8) = 12 × 8. The LCM of 12 and 8 is 24; and the HCF of 12 and 8 is 4. | 1,712 | 5,004 | {"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} | 4.78125 | 5 | CC-MAIN-2024-18 | latest | en | 0.905165 |
https://profound.academy/algorithms-data-structures/gprNjbcnllOW1Q1LCEJ3 | 1,723,426,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00367.warc.gz | 369,016,145 | 30,137 | # Check if a Graph is Complete
A graph is complete if all the vertices are connected to all the other vertices.
Given an undirected graph with `n` vertices and `m` edges, you are asked to check if itβs a complete graph.
Β
#### Input
The first line of the input contains two integers `n`(1 β€ v β€ 500) and `m` (1 β€ e β€ 100 000).
The following `m` lines contain pairs of integers `v1`, `v2` (1 β€ v1, v2 β€ v) which means that the vertex `v1` is connected to the vertex `v2` and vice versa.
#### Output
The program should print `Yes` if the graph is complete and `No` otherwise.
#### Examples
Input Output 3 2 1 2 2 3 No 3 3 1 2 2 3 3 1 Yes 7 21 1 2 1 3 1 4 1 5 1 6 1 7 2 3 2 4 2 5 2 6 2 7 3 4 3 5 3 6 3 7 4 5 4 6 4 7 5 6 5 7 6 7 Yes
Β
#### Constraints
Time limit: 2 seconds
Memory limit: 512 MB
Output limit: 1 MB | 327 | 835 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.691901 |
http://www.mywordsolution.com/question/an-airplane-must-reach-a-speed-of-189mih-to-take-off-if/9292644 | 1,701,893,325,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00378.warc.gz | 76,678,425 | 7,822 | +61-413 786 465
info@mywordsolution.com
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Q. A car starts from rest on a curve with a radius of 130 and accelerates at 0.200. How many revolutions will the car have gone throughout when the magnitude of its total acceleration is 3?
Q. An airplane must reach a speed of 189mi/h to take off. If the runway is 472m long, what is the minimum value of acceleration that will allow the airplane to take off successfully?
• Category:- Physics
• Reference No.:- M9292644
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https://math.stackexchange.com/questions/4607342/what-kind-of-mathematical-knot-is-the-square-knot | 1,721,268,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00663.warc.gz | 337,644,939 | 35,494 | # What kind of mathematical knot is the square knot?
What kind of knot is the square knot? I made an attempt to calculate the Connway notation of the square knot and it did not match with any of the prime knots. Either I did the calculation wrong or the prime knot is a composition of two primes.
Depending on which ends are fused, I have the following diagrams of a square knot.
The first diagram looks more like a link of two loops than a knot, but I am not sure.
This second diagram came from me tying a square knot and fusing the open end. This is from a single string so I don't think it is possible for it to be a loop.
The second diagram represents a nontrivial knot, but it is not prime (this requires proof, but it's true) so you won't find it in the knot tables. A typical notation for it is $$K_{3_1} \mathbin{\#} \overline{K}_{3_1}$$ to indicate that it is the connected sum of the trefoil knot $$K_{3_1}$$ and its mirror image $$\overline{K}_{3_1}$$. | 243 | 968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.942173 |
http://www.nucksmisconduct.com/2011/1/29/1964004/bettman-points-what-how-and-who-are-they-screwing-over | 1,409,445,103,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500835844.19/warc/CC-MAIN-20140820021355-00341-ip-10-180-136-8.ec2.internal.warc.gz | 528,752,797 | 22,903 | ## Bettman Points. What, how, and who are they screwing over
Bettman points. The shootout. Loser points. Call 'em what you want, but the extra points added to NHL games by Gary Bettman in an attempt to "increase viewership" are probably one of the most oft-discussed and maligned additions to the NHL this side of Sean Avery. While Bettman has claimed the extra points don't affect the actual standings at the end of the year, they have the tendency to create the false sense of parity, or just a sense of parody as I read someone comment somewhere (might have been here, don't remember who). So, you must be saying, do they actually affect the standings to a significant degree? Or is Mr. Bettman correct that things would "all work out in the wash" as I hear people say (my mom). Let's take a look at the situation now...
First off we'll need to define what exactly a "Bettman point" is, and how what kind of system we're going to bring in (or back) to replace it. I'm going to suggest (and do, because this is my post and I'll do whatever I want) that the Shootout is cut, 2 points are awarded for a win, no points are awarded for a loss (even in OT), and 1 point is awarded for a tie after OT. We'll assume there's some amount of overtime - 5 mins, 10 mins, 4 on 4, 5 on 5, whatever.
I've seen other people define the Bettman point differently, like Quisp over at Jewels from the Crown. In his calculations, he just removes the loser point from OT losses and Shootout losses, but keeps shootout wins. I don't think that's a good way to go about it, because the Shootout is pretty much tied to the notion of Bettman points. Also it makes the Kings look really good because they are 5-0 in the shootout. Anyways. No shootout for you.
So, to get the new standings we have to do the following:
Wins = previous Wins - SO wins
Losses = previous losses + OTL - SO losses (OTL = total points from OT and SO losses)
Ties = SO wins + SO losses
Doing this, and projecting out to the end of the year, gives us a very very interesting result (not for us, we're still amazing):
MINNEWHAT?
That's right kids (and senior citizens), Minnesota would be in 4th (!!!!) place with no shootouts and OTL points. I could not believe it, I thought my numbers had to be wrong. But if you look at it, it makes sense. The Wild have gone 0-4 in the shootout, and with only 5 OTLs have only one other OT loss in there. So their wins are not affected at all by converting SO wins to ties, and they're only saddled with one more loss. Whereas most of the teams in front of them lose a large percentage of their points because of the all the SO wins forfeited (cough Nashville cough).
So, to conclude, if the playoffs come around and Minny is on the outside looking in, they have some pretty solid reasons to be pissed about the Shootouts and Bettman points. Nashville, on the other hand, should be sending Gary some nice flowers or chocolates.
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We have our own Community Guidelines at Nucks Misconduct. You should read them. | 935 | 3,789 | {"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} | 2.953125 | 3 | CC-MAIN-2014-35 | latest | en | 0.962744 |
https://www.physicsforums.com/threads/topology-question.189747/ | 1,508,673,654,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825227.80/warc/CC-MAIN-20171022113105-20171022133105-00109.warc.gz | 950,049,098 | 14,145 | # Topology question
1. Oct 7, 2007
### JasonJo
Suppose I have some subset of R, not necessarily an interval, let it be denoted as A. I have some union (might be countable, might be finite, might be uncountable) of sets where each set is an open set of A and the union of the open sets is equal to A. Can I conclude that A is open?
I am not sure because the sets are open in A, not necessarily open in R. I don't know much about A other than it is some subset of R.
Any help?
2. Oct 7, 2007
### morphism
You can't conclude that A is open. For a trivial example, take a non-open subset A of R. Note that A is open in itself.
3. Oct 7, 2007
### Dick
If all that you know is that the sets are open in A this can't tell you anything about whether A is open in R. E.g. A is always open in A. | 225 | 796 | {"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} | 2.6875 | 3 | CC-MAIN-2017-43 | longest | en | 0.972244 |
http://www.fixya.com/support/t15279650-need_help_puzzle_100_professor_layton | 1,529,407,429,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862248.4/warc/CC-MAIN-20180619095641-20180619115641-00186.warc.gz | 410,336,471 | 34,528 | Question about Nintendo Professor Layton & the Curious Village Games for DS
I need help with puzzle 100 on professor layton and the miracle mask
Posted by on
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• Expert
Posted on Dec 06, 2012
Hi there,
Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
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Posted on Jan 02, 2017
first square is the only one that is straight, all the others are at an angle..
1st - top four pins on left hand side
2nd - top row 3rd pin, 3rd row 2nd pin, 4th row 3rd pin (position 4) 2nd row 5th pin
3rd - 1st row 4th pin, 3rd row 1st pin, bottom row 2nd pin, 4th row last pin
4th - 1st row last pin, 2nd row 4th pin, 3rd row 4th pin, 2nd row last pin
5th - 2nd row 3rd pin, 3rd row last pin, bottom row last pin, 5th row 2nd pin
6th - 3rd row 3rd pin, 4th row 2nd pin, 5th row 4th pin, 4th row 3rd pin
7th - 4th row 1st pin, 5th row 1st pin, last row first pin, 5th row 2nd pin.
hope this makes sense!!
Posted on Dec 28, 2008
67 = 9 sweets
a has 3
b has 5
c has 1
Posted on Feb 25, 2009
SOURCE: i need help with puzzle 100
There are 28 PINS -- Count each PIN starting from left to right on each line. (1 -- 28)
Square ( 1) Join together Pin No's ( 1 -- 2 -- 6 -- 7 )
Square ( 2) Join together Pin No's ( 4 -- 12 -- 21 -- 27 )
Square ( 3) Join " " " ( 5 -- 9 -- 11 -- 15 )
Square ( 4) " " " " ( 3 -- 10 -- 13 -- 19 )
Square ( 5) " " " " ( 8 -- 16 -- 23 -- 28 )
Square ( 6) " " " " (14 -- 18-- 20 -- 25 )
Square ( 7) " " " " (17 -- 22 - 24 -- 26 )
GOOD LUCK.........
Posted on Apr 17, 2009
SOURCE: Puzzle 100 line error
I think your game file is corrupted erase it and start all over no other way
Posted on Aug 25, 2009
Testimonial: "Thanks for the answer "
http://professorlaytonwalkthrough.blogspot.com/2008/02/puzzle105.html
please let me know how you get on and if you need any further help
jonniecullen@ireland.com
Posted on Dec 30, 2009
×
my-video-file.mp4
×
Related Questions:
Professor Layton Miracle Mask puzzle no 61 cant solve it doesnt add up
You seem to believe that any person who reads your question has that Professor Layton Miracle Mask Puzzle available. It ain't so. What you can do is spell out the puzzle in your post. Maybe someone will solve it.
Nov 27, 2013 | Nintendo Professor Layton Miracle Mask
Puzzle 100 line error
I think your game file is corrupted erase it and start all over no other way
Aug 23, 2009 | Nintendo Professor Layton & the Curious...
On professer layton cant figure out were the 7 squares are
Go on youtube and type in: professor Layton and the curious village puzzle 100 and click on professor Layton and the curious village puzzle #100 seven...
and that will show you how to do it.
Jul 17, 2009 | Nintendo Professor Layton & the Curious...
Professor Layton & The Curious Village: I have
The Inventor's House (puzzles 121 to 123): Assemble the gizmo.
The Decorator's House (puzzles 124 to 126): Achieve maximum satisfaction by filling the rooms at the inn.
The Art Lover's House (puzzles 127 to 129): Complete the portrait.
The Golden Apple's House (puzzles 130 to 132): Complete the game.
The Puzzle Master's House (puzzles 133 to 135): Complete all the previous 132 puzzles.
Well,Yeah I hope I've helped xD
Apr 11, 2009 | Nintendo Professor Layton & the Curious...
Prof Layton
yes, here is a website that is really helpful as i can't really show you:
http://professorlaytonwalkthrough.blogspot.com/2008/02/puzzle100.html
this can also help with any other puzzle solution
Jan 05, 2009 | Nintendo Professor Layton & the Curious...
Cant solve puzzle 100 in professor layton cant go on im desperate
first square is the only one that is straight, all the others are at an angle..
1st - top four pins on left hand side
2nd - top row 3rd pin, 3rd row 2nd pin, 4th row 3rd pin (position 4) 2nd row 5th pin
3rd - 1st row 4th pin, 3rd row 1st pin, bottom row 2nd pin, 4th row last pin
4th - 1st row last pin, 2nd row 4th pin, 3rd row 4th pin, 2nd row last pin
5th - 2nd row 3rd pin, 3rd row last pin, bottom row last pin, 5th row 2nd pin
6th - 3rd row 3rd pin, 4th row 2nd pin, 5th row 4th pin, 4th row 3rd pin
7th - 4th row 1st pin, 5th row 1st pin, last row first pin, 5th row 2nd pin.
hope this makes sense!!
Dec 06, 2008 | Nintendo Professor Layton & the Curious...
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Level 2 Expert | 1,522 | 4,853 | {"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-2018-26 | latest | en | 0.942615 |
https://git.ucc.asn.au/?p=progcomp2012.git;a=commitdiff;h=9e4bc3c0b49f5e2796a62c8fa91fe0ec78d96af2 | 1,642,817,693,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00656.warc.gz | 331,857,373 | 4,483 | author Sam Moore Fri, 2 Mar 2012 15:47:24 +0000 (23:47 +0800) committer Sam Moore Fri, 2 Mar 2012 15:47:24 +0000 (23:47 +0800)
simulate.py wasn't setting the "gameID" variable correctly,
meaning all games were logged to "red.vs.blue.1.1"
Bug in vixen found by [SLX]; it attempts to move outside the board.
Usually when losing.
Path finding algorithm (path.py) shouldn't move outside the board.
But somehow it is.
Hacky fix at the moment by checking for moves outside the board in
vixen's score calculation, and allocating -100 to these moves.
index 975f6a7..a4010b1 100755 (executable)
@@ -61,7 +61,7 @@ class Vixen(BasicAI):
moveList.append({"unit":unit, "direction":bestScore[0], "score":bestScore[1]})
- if len(moveList) == 0:
+ if len(moveList) <= 0:
print "NO_MOVE"
return True
@@ -86,7 +86,8 @@ class Vixen(BasicAI):
def CalculateScore(self, attacker, defender, path):
p = move(attacker.x, attacker.y, path[0], 1)
-
+ if p[0] < 0 or p[0] >= len(self.board) or p[1] < 0 or p[1] >= len(self.board[p[0]]):
+ return -100.0
total = 0.0
count = 0.0
index 5daafb7..7235624 100755 (executable)
@@ -221,13 +221,16 @@ for roundNumber in range(totalRounds, totalRounds + nRounds):
for blue in agents: #against each other agent, playing as blue
if red == blue:
continue #Exclude battles against self
- gameID = str(roundNumber) + "." + str(gameNumber)
+
+
for i in range(1, nGames/2 + 1): | 597 | 1,476 | {"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-2022-05 | longest | en | 0.077738 |
https://isabelle.in.tum.de/repos/isabelle/rev/b9dd40e2c470 | 1,624,137,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00342.warc.gz | 299,921,298 | 2,406 | author nipkow Sun, 11 Nov 2018 14:34:02 +0100 changeset 69285 b9dd40e2c470 parent 69283 39044da8bb5a (current diff) parent 69284 3273692de24a (diff) child 69286 e4d5a07fecb6
merged
```--- a/src/HOL/Analysis/Sigma_Algebra.thy Sun Nov 11 12:13:24 2018 +0100
+++ b/src/HOL/Analysis/Sigma_Algebra.thy Sun Nov 11 14:34:02 2018 +0100
@@ -1333,7 +1333,7 @@
fix A assume "A \<subseteq> \<Omega> \<and> A \<inter> D \<in> M"
moreover have "(\<Omega> - A) \<inter> D = (\<Omega> - (A \<inter> D)) - (\<Omega> - D)"
by auto
- ultimately show "\<Omega> - A \<subseteq> \<Omega> \<and> (\<Omega> - A) \<inter> D \<in> M"
+ ultimately show "(\<Omega> - A) \<inter> D \<in> M"
using \<open>D \<in> M\<close> by (auto intro: diff)
next
fix A :: "nat \<Rightarrow> 'a set"```
```--- a/src/HOL/Set.thy Sun Nov 11 12:13:24 2018 +0100
+++ b/src/HOL/Set.thy Sun Nov 11 14:34:02 2018 +0100
@@ -1122,7 +1122,7 @@
text \<open>\<^medskip> Set difference.\<close>
-lemma Diff_subset: "A - B \<subseteq> A"
+lemma Diff_subset[simp]: "A - B \<subseteq> A"
by blast
lemma Diff_subset_conv: "A - B \<subseteq> C \<longleftrightarrow> A \<subseteq> B \<union> C"``` | 441 | 1,140 | {"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-2021-25 | latest | en | 0.451527 |
https://en.khanacademy.org/math/early-math/cc-early-math-geometry-topic/cc-early-math-fractions-of-shapes/v/equal-parts-of-circles-and-rectangles | 1,721,601,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00023.warc.gz | 191,202,787 | 89,522 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Early math review>Unit 8
Lesson 4: Fractions of shapes
# Equal parts of circles and rectangles
Sal determines if shapes are divided into 2 or 4 equal sections.
## Want to join the conversation?
• Can a triangle be divided in to 3rds?
• You can achieve this to my knowledge atleast three different ways.
• Can a circle be divided by 6 equally?
• Yes! A circle can be divided into any number of equal parts.
• I already know that circles and rectangles can be equally divided, but how can triangles be divided into more than 2 pieces?
• Assuming you are talking about an equilateral triangle (a triangle with similar side lengths and angles for all three sides), you can split a triangle into three congruent (absolutely equal) pieces by drawing lines from each of the corners until they all reach the middle. To split an equilateral triangle into four pieces, Draw a horizontal line slightly above the middle to make one triangle, then split the lower part into a (going horizontally) regular triangle, upside-down triangle, and then another regular triangle.
Pictures of both are shown here: https://math.stackexchange.com/questions/158290/is-it-possible-to-divide-an-equilateral-triangle-into-12-congruent-triangles | 317 | 1,455 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.897173 |
https://socratic.org/questions/what-is-the-equation-of-the-line-passing-through-60-16-and-18-26 | 1,721,252,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00028.warc.gz | 490,849,976 | 6,167 | # What is the equation of the line passing through (60,16) and (18,26)?
##### 1 Answer
Jun 7, 2018
$\left(y - 16\right) = - \frac{5}{21} \left(x - 60\right)$
#### Explanation:
First you determine the slope:
$\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right) = \left(60 , 16\right)$
$\left(\textcolor{red}{{x}_{2}} , \textcolor{red}{{y}_{2}}\right) = \left(18 , 26\right)$
$\textcolor{g r e e n}{m} = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
$\textcolor{g r e e n}{m} = \frac{\textcolor{red}{26} - \textcolor{b l u e}{16}}{\textcolor{red}{18} - \textcolor{b l u e}{60}} = - \frac{5}{21}$
Now use the Point Slope form of a line:
$\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{g r e e n}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$
$\left(y - \textcolor{b l u e}{16}\right) = \textcolor{g r e e n}{- \frac{5}{21}} \left(x - \textcolor{b l u e}{60}\right)$
graph{(y-16) = -5/21(x-60) [-67, 93, -0.96, 79.04]} | 445 | 1,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.455343 |
https://assignmentpoint.com/nonparametric-method/ | 1,670,444,182,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711218.21/warc/CC-MAIN-20221207185519-20221207215519-00581.warc.gz | 141,792,276 | 9,407 | # Nonparametric Method
A nonparametric method is a mathematical approach to statistical inference that ignores the underlying assumptions about the form of the observed probability distribution. ANOVA, Pearson’s correlation, t-test, and other well-known statistical methods offer reliable information about the data being studied only if the underlying population follows those assumptions. It is a theory test that doesn’t need the populace’s dissemination to be portrayed by specific boundaries. Many hypothesis tests, for example, are predicated on the premise that the population follows a normal distribution with parameters μ and σ. Nonparametric tests, on the other hand, do not make this statement, making them useful when your data is strongly nonnormal and resistant to transformation.
Nonparametric statistics do not depend on the population data meeting the same assumptions as parametric statistics. The brought about estimation of covariate information is a typical issue in measurable information investigation because of the utilization of flawed proxy covariates, mistakes in factors, or the issue of missing information. Nonparametric measurements, along these lines, fall into a class of insights now and then alluded to as conveyance free. Where the population data has an uncertain distribution or the sample size is limited, nonparametric approaches are often used.
Nonparametric tests, on the other hand, are not completely free of assumptions about your results. For example, it’s critical to assume that the samples’ observations are independent and come from the same distribution. Likewise, in two-example plans the supposition of equivalent shape and spread is required. The nonparametric strategy helps in the demonstrating of suitable measurable strategies as a model structure apparatus in monetary time arrangement and econometrics. Although the nonparametric method is less efficient than the parametric approach, it does work under a few assumptions.
On various types of data, both parametric and nonparametric methods are commonly used. In most cases, interval or ratio data are required for parametric statistics. Nonparametric statistics, on the other hand, are usually used on data that are trivial or ordinal. Ostensible factors are factors for which the qualities have not quantitative worth. The nonparametric technique doesn’t need the populace under investigation to meet specific suppositions or explicit boundaries to describe the perceptions, similar to the case with parametric strategies. To provide an example, traditional parametric methods like the t-test and ANOVA only provide true and reliable results if the population under study meets certain assumptions.
Nonparametric tests have the following limitations:
• When the normality assumption is met, nonparametric tests are normally less efficient than parametric tests. If the data comes from a normal distribution, you are less likely to dismiss the null hypothesis when it is incorrect.
• Nonparametric experiments often necessitate the modification of hypotheses. Many nonparametric tests of the population center, for example, are tests of the median rather than the mean. If the population is not symmetric, the test does not answer the same question as to the corresponding parametric approach.
If a statistical approach follows the following criteria, it is considered nonparametric. First, the approach is applied to quantitative data where no population assumptions are made. Second, the strategy utilizes subjective information in a fairly casual manner; consequently, the nonparametric technique is an indicative instrument for a model structure where it tests, checks, appraises and approves information. Non-parametric methods have much broader applicability than parametric methods because they make less assumptions. They can be used in cases where little is understood about the application in question, for example. Non-parametric approaches are often more stable because they depend on fewer assumptions.
Albeit nonparametric measurements have the upside of meeting not many presumptions, they are less incredible than parametric insights. This implies that they may not show a connection between two factors when truth be told one exists. Another reason for using non-parametric methods is their simplicity. Non-parametric methods can be easier to use in some situations, even though parametric methods are justified. Due both to this straightforwardness and to their more prominent heartiness, non-parametric techniques are seen by certain analysts as leaving less space for ill-advised use and misconstruing.
Parametric and nonparametric approaches are used to analyze various types of data. Nonparametric data is concerned with nominal or ordinal data, while parametric data is concerned with interval or ratio data. The more extensive appropriateness and expanded strength of non-parametric tests includes some significant pitfalls: in situations where a parametric test would be proper, non-parametric tests have less force. To put it another way, a greater sample size might be needed to achieve the same level of confidence in conclusions.
Nonparametric methods are also widely used in financial econometrics to estimate returns, bond yields, volatility, return, and state price densities of stock prices. For instance, the technique is favored while looking at the additional time variety of stock costs and bonds. The term non-parametric isn’t intended to suggest that such models totally need boundaries yet that the number and nature of the boundaries are adaptable and not fixed ahead of time.
The curse of dimensionality problem has an effect on density estimation techniques. Using the nonparametric approach to estimate density function tends to be easy on the surface. Chi Square, Wilcoxon rank-sum test, Kruskal-Wallis test, and Spearman’s rank-order correlation are examples of nonparametric tests. A nonparametric strategy is hailed for its benefit of working under a couple of suspicions. Nonetheless, the idea is for the most part viewed as less amazing than the parametric methodology. In light of this, statisticians suggest using parametric methods in cases where both are acceptable.
Information Sources: | 1,154 | 6,266 | {"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-2022-49 | latest | en | 0.906452 |
https://kumonbooks.com/shop/?filter_cat=calculations,focus-on,math-boosters,word-problems&filter_grade=3rd-grade,kindergarten,pre-k | 1,638,384,201,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00472.warc.gz | 424,977,391 | 17,085 | # Kumon Workbooks
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The Kumon Math Boosters series is designed to help children who either need to improve their basic math skills, or who are slightly ahead of | 429 | 1,894 | {"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-2021-49 | latest | en | 0.938136 |
http://cccbdb.nist.gov/comp2vib3x.asp?casno=12597034&charge=-1&method1=1&basis1=1&method2=3&basis2=1&frombad=7 | 1,495,996,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610374.3/warc/CC-MAIN-20170528181608-20170528201608-00485.warc.gz | 78,930,178 | 6,541 | Computational Chemistry Comparison and Benchmark DataBase Release 18 (October 2016) Standard Reference Database 101 National Institute of Standards and Technology Home All data for one species Geometry Experimental Calculated Comparisons Bad Calculations Tutorials and Explanations Vibrations Experimental Calculated Scale factors Reactions Entropies Ions List Ions Energy Electron Affinity Proton Affinity Ionization changes point group Experimental One molecule all properties One property a few molecules Geometry Vibrations Energy Electrostatics Reference Data Calculated Energy Optimized Reaction Internal Rotation Orbital Nuclear repulsion energy Correlation Ion Excited State Basis Set Extrapolation Geometry Vibrations Frequencies Zero point energy (ZPE) Scale Factors Bad Calculations Electrostatics Charges Dipole Quadrupole Polarizability Spin Entropy and Heat Capacity Reaction Lookup by property Comparisons Geometry Vibrations Energy Entropy Electrostatics Ion Resources Info on Results Calculations Done Basis functions used I/O files Glossary Conversion Forms Links NIST Links External links Thermochemistry Tutorials Vibrations Entropy Energy Electrostatics Geometry Cost Bad Calculations FAQ Help Units Choose Units Explanations Credits Just show me Summary Using List Recent molecules Molecules Geometry Vibrations Energy Similar molecules Ions, Dipoles, etc. Index of CCCBDB Feedback You are here: Comparisons > Vibrations > Vibrations > 2 calculations
# Compare vibrational frequencies for two calculations for S3- (Sulfur trimer anion)
### A = HF/6-31G* B = MP2/6-31G*
scale factors=0.8985, 0.943
symmetry frequency (cm-1) reduced mass (amu) IR Intensity (km mol-1)
mode number A B A B diff. ratio A B diff. ratio A B diff. ratio
1 A1 A1 521 526 -5.4 0.990 31.972 31.972 0.000 1.000 3.98 0.88 4.527
2 A1 A1 228 225 2.5 1.011 31.972 31.972 0.000 1.000 3.77 2.71 1.394
3 B2 B2 466 753 -286.6 0.619 31.972 31.972 0.000 1.000 959.81 25.72 37.312
scaled by 0.8985 0.943
See section III.C.1 List or set vibrational scaling factors to change the scale factors used here.
See section III.C.2 Calculate a vibrational scaling factor for a given set of molecules to determine the least squares best scaling factor. | 582 | 2,263 | {"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-2017-22 | longest | en | 0.564914 |
https://engineermind.in/if-somehow-we-supply-high-voltage-current-to-the-railway-track-will-it-travel-throughout-the-network-of-railway-tracks/ | 1,685,643,902,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00089.warc.gz | 271,084,257 | 19,848 | # If somehow we supply high voltage current to the railway track will it travel throughout the network of railway tracks?
## Why can’t we transmit electrical power with transmission voltage more than 1000kv?
Dark Light
This is one of our fantasy about electricity. That if we supply the high voltage current to a track in Mumbai CST, peoples standing on the track in New Delhi [~1400 km far] will get a shock?
No, Current always chooses the shortest path [rather a resistance-free path] in the circuit before to be grounded.
Since the ground is much nearer than New Delhi that too with much less resistance path, the maximum distance traveled by current will be just 17.2 cm [i.e the depth of rail] and not 1400 km [length of rail]. It shall be grounded in no time. You will simply end up with ‘earthing’.
But, Sleeper and ballast crabs are generally hard core insulators. So in this case, the current will find the nearest grounding spots.
which are available at every 100, 300, 500 or 1000 meters interval with a special code of practice for earthing 25kV like this:
Or, manually [for inspection] it can be done anywhere by clippers as:
and it will be earthed.
Let us assume there are not such earthing points available, and the sleepers to are insulators; even then it is practically impossible. You will require astronomical high amperes of current [do also consider the huge ${I}^{2}R$ loss] to circulate it through the whole railway network.
## Why Earth Pin Larger And Longer Than The Phase And Neutral Pin?
We all know that the earth pin is used to protect the user From the short circuit. in…
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Car sensors are electronic devices installed in vehicles to monitor various aspects of their operation and provide data…
## Fleming’s Left-Hand Rule And Right-Hand Rule
John Ambrose Fleming introduced two rules to determine the direction of motion in motors or the direction of induced current…
## What is Distributed Winding And Concentrated Winding?
Imagine a pen and a string. You start winding the string around the pen, and you continue… | 476 | 2,105 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-23 | latest | en | 0.870215 |
https://electronics.stackexchange.com/questions/165884/chosing-the-right-hall-effect-sensor | 1,653,231,558,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00404.warc.gz | 287,416,354 | 67,553 | chosing the right Hall effect sensor
For a student project, I need a Hall Effect current sensor. What I am going to do is to charge 8 rechargeable Li-Ion batteries of 3.7 volts in series with 34 volt supply and the buck circuit frequency is between 5KHz to 10KHz. My concern is finding the right hall-effect sensor for calculation of the input current to adjust the current flowing to the batteries.
The batteries are 4000-6000 mAh and I am going to use AVR to sense the output of current sensor. So, I need a sensor with 5 volt output. I am thinking about finding the right current sensor for this circuit. ACS712ELC-05B is an example however, it senses 5A which is too high for this circuit and reduces the sensitivity of my measurement. I need something in safe range but not too far. Could anybody help me with finding a better option?
• How about a small resistor to measure current? Apr 21, 2015 at 12:12
• @GeorgeHerold not acceptable. btw, on resistor, both pins a floating (and with high voltage). Apr 21, 2015 at 12:25
• You could do a differential voltage measurement of the resistor with an opamp or instrument amp... 34 V is not what I would call high voltage. Apr 21, 2015 at 12:41
• @GeorgeHerold it is still high for damaging AVR. Apr 21, 2015 at 13:05
• Instrument amps have a reference output so you can put the output where ever you like... maybe an ina826? ti.com/lit/ds/symlink/ina826.pdf Apr 21, 2015 at 13:37
You should use a sense resistor and a current sense amplifier -- this will give you far superior accuracy to either a Hall effect sensor or an instrumentation amplifier.
The LTC6102 is an excellent current sense amplifier. Its maximum offset voltage of $10\rm{\mu V}$ permits great accuracy, but it can only sense current in one direction. If you need to sense bidirectional currents and are willing to live with a bit more error, the LT1787 is a good choice, with $75\rm{\mu V}$ of offset.
• One direction is fine. But a few questions: can it be used alone in my case or it needs another amplifier. This image link made me worried to use another amplifier. Are those extra diodes and transistors really necessary? and is it required to remove voltage offset by a potentiometer? I do not like to use any variable resistor solution. Apr 21, 2015 at 23:53
• @barej: A) The image in your link has a single amplifier. The LTC2433-1 is an Analog to Digital converter. You don't have to convert to digital if you don't want to. B) There are no diodes in the diagram, and the only transistor is inside the LTC6102. C) If you don't want to compensate for the offset, then you don't have to. In fact, that's why the offset is engineered to be so low: so that, under most cases, it's small enough that you won't need to correct for it.
– Zulu
Apr 22, 2015 at 1:14
• Thank you very much. For the case of my question, how much ohm resistance do you suggest me to use? Apr 22, 2015 at 12:32
• This element is surface mounted. How to use it on prototype PCB boards? Apr 22, 2015 at 16:01
• @barej, Page 12 of the LTC6102's datasheet provides guidance on selecting the sense resistor. Selection of the other components is covered in the following pages.
– Zulu
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Discussion in 'C++' started by anders.johansen@gmail.com, Apr 1, 2008.
1. ### Guest
Hi,
I have a tight loop where I basically do this over and over (millions
of times):
int cell = <calculation>
const int above = <calculation>
if (cell > above) cell = above;
const int below = <calculation>
if (cell > below) cell = below;
const int trans = <calculation>
if (cell > trans) cell = trans;
Is there a faster way to do this?
Sincerely,
Anders
, Apr 1, 2008
2. ### Guest
On Apr 1, 1:42 pm, wrote:
> Hi,
>
> I have a tight loop where I basically do this over and over (millions
> of times):
>
> int cell = <calculation>
> const int above = <calculation>
> if (cell > above) cell = above;
> const int below = <calculation>
> if (cell > below) cell = below;
> const int trans = <calculation>
> if (cell > trans) cell = trans;
>
> Is there a faster way to do this?
>
> Sincerely,
> Anders
the if could be replaced like
cell=(cell > trans ? trans : cell);
By replacing i cannot say what would be the improvement. There are
other factors to be considered.
moreover, you haven't said anything about the <calculations>. you may
also want to consider running in parallel. all depends on what you are
trying to achieve, about which you haven't said much.
Thanks,
Balaji.
, Apr 1, 2008
3. ### MartinGuest
On Apr 1, 1:42 pm, wrote:
> Hi,
>
> I have a tight loop where I basically do this over and over (millions
> of times):
>
> int cell = <calculation>
> const int above = <calculation>
> if (cell > above) cell = above;
> const int below = <calculation>
> if (cell > below) cell = below;
> const int trans = <calculation>
> if (cell > trans) cell = trans;
>
> Is there a faster way to do this?
>
> Sincerely,
> Anders
One obvious improvement is moving all <calculations> (or at least some
of them) out of the loop, if it's possible. If no - IMO there is no
considerable improvement.
Martin, Apr 1, 2008
4. ### Kai-Uwe BuxGuest
wrote:
> Hi,
>
> I have a tight loop where I basically do this over and over (millions
> of times):
>
> int cell = <calculation>
> const int above = <calculation>
> if (cell > above) cell = above;
> const int below = <calculation>
> if (cell > below) cell = below;
> const int trans = <calculation>
> if (cell > trans) cell = trans;
>
> Is there a faster way to do this?
Not from an algorithmic point of view. To find the minimum of four
values will require computing all of them and three additional comparisons.
On the other hand, there could be bit-fiddling techniques to find the
minimum without using branch operations. This can lead to better pipelining
on some processors. See, for instance:
www.cellperformance.com/articles/2006/04/benefits_to_branch_elimination.html
YMMV
Best
Kai-Uwe Bux
Kai-Uwe Bux, Apr 1, 2008
5. ### Lionel BGuest
On Tue, 01 Apr 2008 01:49:40 -0700, kasthurirangan.balaji wrote:
> On Apr 1, 1:42Â pm, wrote:
>> Hi,
>>
>> I have a tight loop where I basically do this over and over (millions
>> of times):
>>
>> int cell = <calculation>
>> const int above = <calculation>
>> if (cell > above) cell = above;
>> const int below = <calculation>
>> if (cell > below) cell = below;
>> const int trans = <calculation>
>> if (cell > trans) cell = trans;
>>
>> Is there a faster way to do this?
>>
>> Sincerely,
>> Â Anders
>
> the if could be replaced like
> cell=(cell > trans ? trans : cell);
Pointless. That would be no faster, is not (to my mind) any clearer and
potentially involves a redundant self-assignment of cell to itself
(although an optimising compiler would probably not generate code for
this).
[...]
--
Lionel B
Lionel B, Apr 1, 2008
6. ### Yannick TremblayGuest
In article <fssubs\$o9u\$>, Lionel B <> wrote:
>On Tue, 01 Apr 2008 01:49:40 -0700, kasthurirangan.balaji wrote:
>
>>
>> the if could be replaced like
>> cell=(cell > trans ? trans : cell);
>
>Pointless. That would be no faster, is not (to my mind) any clearer and
>potentially involves a redundant self-assignment of cell to itself
>(although an optimising compiler would probably not generate code for
>this).
>
Agree, any decent compiler should generate the exact same machine code for:
cell = ( cell > trans ? trans : cell);
and
if( cell > trans )
cell = trans;
else
cell = cell;
The fact that the first one can be written is C on one line is of no
relevance to the compiler.
Yan
Yannick Tremblay, Apr 1, 2008
7. ### Puppet_SockGuest
On Apr 1, 4:42 am, wrote:
> I have a tight loop where I basically do this over and over (millions
> of times):
>
> int cell = <calculation>
> const int above = <calculation>
> if (cell > above) cell = above;
> const int below = <calculation>
> if (cell > below) cell = below;
> const int trans = <calculation>
> if (cell > trans) cell = trans;
>
> Is there a faster way to do this?
Question: In the above psuedo code, the various <calculation>
entries use a lot of time, yes? That is, the various if-greater
tests are a tiny portion of the overall run-time, yes?
So, if the calcs happen every time through this loop, you
probably will find more opportunity for optimizing in those
calcs rather than the comparatively small potential in
the if-greater tests.
Also: Every time somebody starts asking about "make it faster"
I have a standard response I trot out. Where's your stop-watch?
Where's your performance spec? First measure how much time
various parts of the code take to run. Then look to your spec
to see if that is acceptable. If you've made the spec, relax!
And don't sweat the stuff that takes a small part of the time.
Socks
Puppet_Sock, Apr 1, 2008
8. ### Guest
First of all, thanks to all who answered.
I was kinda hoping somebody had a magic bit-twiddeling trick up their
sleeve for this. Never hurts to ask
More info follows and responses to some concrete questions.
On Apr 1, 4:21 pm, Puppet_Sock <> wrote:
> On Apr 1, 4:42 am, wrote:
>
> > I have a tight loop where I basically do this over and over (millions
> > of times):
>
> > int cell = <calculation>
> > const int above = <calculation>
> > if (cell > above) cell = above;
> > const int below = <calculation>
> > if (cell > below) cell = below;
> > const int trans = <calculation>
> > if (cell > trans) cell = trans;
>
> > Is there a faster way to do this?
>
> Question: In the above psuedo code, the various <calculation>
> entries use a lot of time, yes? That is, the various if-greater
> tests are a tiny portion of the overall run-time, yes?
> So, if the calcs happen every time through this loop, you
> probably will find more opportunity for optimizing in those
> calcs rather than the comparatively small potential in
> the if-greater tests.
Actually the calculations are mainly table look-ups. It looks like
this:
int thismin = 100
<...>
<loop starts>
int cell = (acceptMatrix[false][source[i-1]][target[1]]) ?
distmatrix[i-1][1] : distmatrix[i-1][1]+1 ;
const int above (distmatrix[i-1][2]+1);
if (cell>above) cell=above;
const int left (distmatrix[1]+1);
if (cell>left) cell=left;
const int trans (
distmatrix[i-2][0] + 1 +
(!acceptMatrix[false][source[i-2]][target[1]]) +
(!acceptMatrix[true][source[i-1]][target[0]])
);
if (cell>trans) cell=trans;
distmatrix[2]=cell;
if (cell<thismin) thismin=cell;
<loop ends>
distmatrix is a 2-D array of int.
acceptmatrix is a 3D array of bool.
i and j are unsigned int loop indexes
source and target are const unsigned char * const
An additional question is whether the use of named const temporaries
is detrimental. Should I perhaps instead hand-inline the table
lookups, and let the compiler find the common expressions and
optimize, or...
I don't know how smart the compiler is, but I suspect "not too
bright". It's Borland C++ Builder version 6, so a 6 years old compiler
at best.
> Also: Every time somebody starts asking about "make it faster"
> I have a standard response I trot out. Where's your stop-watch?
> Where's your performance spec? First measure how much time
> various parts of the code take to run. Then look to your spec
> to see if that is acceptable. If you've made the spec, relax!
> And don't sweat the stuff that takes a small part of the time.
> Socks
The "stop-watch" is outside this algorithm, but currently this code
fragment dominates execution time. Changing the calculation of "trans"
from an if/then based structure to the current gobbledygook decreased
run time by approximately 15%, probably by eliminating a few branches
and a temporary or two. I compare the output to the original
implementation for a large dataset, so I am sure that behaviour was
not changed in any obvious way.I am also pursuing other ways to
decrease the data set that the algorithm iterates over, but that's an
entirely different discussion.
Of course I could just get some hotshot to hand-assy it for me, but
then I would be locked-in to that particular "magic" implementation.
Further, I doubt that solution would be as easy to hand-unroll as the
C/C++ version is.
Anders
, Apr 1, 2008
9. ### Michael DOUBEZGuest
a écrit :
> First of all, thanks to all who answered.
>
> I was kinda hoping somebody had a magic bit-twiddeling trick up their
> sleeve for this. Never hurts to ask
>
> More info follows and responses to some concrete questions.
> [snip]
>
> int thismin = 100
>
> <...>
>
> <loop starts>
>
> int cell = (acceptMatrix[false][source[i-1]][target[1]]) ?
> distmatrix[i-1][1] : distmatrix[i-1][1]+1 ;
> const int above (distmatrix[i-1][2]+1);
> if (cell>above) cell=above;
> const int left (distmatrix[1]+1);
> if (cell>left) cell=left;
> const int trans (
> distmatrix[i-2][0] + 1 +
> (!acceptMatrix[false][source[i-2]][target[1]]) +
> (!acceptMatrix[true][source[i-1]][target[0]])
> );
Funny that you add integers and boolean.
What value do you expect from a conversion from bool to int ?
Or is the code incomplete ?
> if (cell>trans) cell=trans;
> distmatrix[2]=cell;
> if (cell<thismin) thismin=cell;
>
> <loop ends>
>
> distmatrix is a 2-D array of int.
>
> acceptmatrix is a 3D array of bool.
>
> i and j are unsigned int loop indexes
>
I see no j variable in your code.
> source and target are const unsigned char * const
>
> An additional question is whether the use of named const temporaries
> is detrimental. Should I perhaps instead hand-inline the table
> lookups, and let the compiler find the common expressions and
> optimize, or...
>
> I don't know how smart the compiler is, but I suspect "not too
> bright". It's Borland C++ Builder version 6, so a 6 years old compiler
> at best.
In this case I suggest that you get a look at the assembly generated by
the compiler. And depending on the operations that seems consuming, you
might want to tinker a bit with compiler's directives.
Here is how I would rewrite the code you gave in order to minimize the
number of operation but I wouldn't expect to outsmart any modern compiler.
<loop starts>
int cell=distmatrix[i-1][1];
const char source_im1(source[i-1]); //reused later
const char target_1 (target[1]); //reused later
if(!acceptMatrix[false][src_im1][target_1])
{ //increment cell only in given case
++cell;
}
//change here to gain additions - might lose in memory load/unload
int cell2 (distmatrix[i-1][2]);
const int left (distmatrix[1]);
if (cell2>left) cell2=left;
int trans (distmatrix[i-2][0]);
if(!acceptMatrix[false][source[i-2]][target[1]])
{
++trans;
}
if(!acceptMatrix[true][source_im1][target_1])
{
++trans;
}
if (cell2>trans) cell2=trans;
//addition we add not done before
++cell2
if (cell>cell2) cell=cell2;
distmatrix[2]=cell;
if (cell<thismin) thismin=cell;
<loop ends>
Michael
Michael DOUBEZ, Apr 2, 2008
10. ### Guest
Hmmm,
How about this:
int cell = (acceptMatrix[false][source[i-1]][target[1]]) ?
distmatrix[i-1][1] : distmatrix[i-1][1]+1 ;
// Newval replaces above. All add one instrs. moved down
int newval (distmatrix[i-1][2]);
const int left (distmatrix[1]);
const int trans (
distmatrix[i-2][0] +
(!acceptMatrix[false][source[i-2]][target[1]]) +
(!acceptMatrix[true][source[i-1]][target[0]])
);
if (left < newval) newval = left;
if (trans < newval) newval = trans;
// Add one moved here
if (++newval < cell) cell=newval;
if (distmatrix[2]=cell < thismin) thismin=cell;
This saves two "+ 1" instructions, and replaces the remaining one with
a preincrement. The ++newval could (should?) be moved out of the if
statement for clearer code.
, Apr 2, 2008
11. ### Daniel PittsGuest
wrote:
> Actually the calculations are mainly table look-ups. It looks like
> this:
>
> int thismin = 100
>
> <...>
>
> <loop starts>
>
> int cell = (acceptMatrix[false][source[i-1]][target[1]]) ?
> distmatrix[i-1][1] : distmatrix[i-1][1]+1 ;
> const int above (distmatrix[i-1][2]+1);
> if (cell>above) cell=above;
> const int left (distmatrix[1]+1);
> if (cell>left) cell=left;
> const int trans (
> distmatrix[i-2][0] + 1 +
> (!acceptMatrix[false][source[i-2]][target[1]]) +
> (!acceptMatrix[true][source[i-1]][target[0]])
> );
> if (cell>trans) cell=trans;
> distmatrix[2]=cell;
> if (cell<thismin) thismin=cell;
>
> <loop ends>
Array lookups aren't terribly expensive, but they do have a cost. if
you can save a reference to distmatrix[i-1] (since its used in a few
places), that might shave a nanosecond or two. If you have nested
loops, look for calculations or references you can do outside of the
inner most loop (or outside of both loops).
For instance, if "i" is from your outer loop, you could "precalculate"
distmatrix, distmatrix[i-1], dismatrix[i-2], source[i-1],
source[i-2], etc...
Chances are, you'll have more luck finding a better algorithm than
you'll have shaving significant time off by optimizing this algorithm.
>
> distmatrix is a 2-D array of int.
>
> acceptmatrix is a 3D array of bool.
>
> i and j are unsigned int loop indexes
>
> source and target are const unsigned char * const
>
> An additional question is whether the use of named const temporaries
> is detrimental. Should I perhaps instead hand-inline the table
> lookups, and let the compiler find the common expressions and
> optimize, or...
It shouldn't really hurt, they would probably compile to the same thing.
>
> I don't know how smart the compiler is, but I suspect "not too
> bright". It's Borland C++ Builder version 6, so a 6 years old compiler
> at best.
>
>> Also: Every time somebody starts asking about "make it faster"
>> I have a standard response I trot out. Where's your stop-watch?
>> Where's your performance spec? First measure how much time
>> various parts of the code take to run. Then look to your spec
>> to see if that is acceptable. If you've made the spec, relax!
>> And don't sweat the stuff that takes a small part of the time.
>> Socks
>
> The "stop-watch" is outside this algorithm, but currently this code
> fragment dominates execution time. Changing the calculation of "trans"
> from an if/then based structure to the current gobbledygook decreased
> run time by approximately 15%, probably by eliminating a few branches
> and a temporary or two. I compare the output to the original
> implementation for a large dataset, so I am sure that behaviour was
> not changed in any obvious way.I am also pursuing other ways to
> decrease the data set that the algorithm iterates over, but that's an
> entirely different discussion.
>
> Of course I could just get some hotshot to hand-assy it for me, but
> then I would be locked-in to that particular "magic" implementation.
> Further, I doubt that solution would be as easy to hand-unroll as the
> C/C++ version is.
>
> Anders
--
Daniel Pitts' Tech Blog: <http://virtualinfinity.net/wordpress/>
Daniel Pitts, Apr 6, 2008
12. ### Martin YorkGuest
On Apr 6, 2:43 pm, Daniel Pitts
<> wrote:
> Array lookups aren't terribly expensive, but they do have a cost. if
> you can save a reference to distmatrix[i-1] (since its used in a few
> places), that might shave a nanosecond or two.
>
> Daniel Pitts' Tech Blog: <http://virtualinfinity.net/wordpress/>
Not sure I buy that. Could be wrong!
Of course we are getting down to machine specific behavior at this
point. But accessing a memory location via an array or reference would
be just as expensive as most memory access instructions (Its been a
while so things may have changed) allow a major pointer and and offset
as part of the same instruction.
Actually the expensive bit is not in the memory retrieval but in
loading the memory from main memory into the local cache. If your code
is getting cache misses the processor stall time for this is many
orders of magnitude greater than any theoretical difference between an
array verses reference read.
As a result I think any changes to code for a perceived code
performance increase must be significant before you consider change to
the code that makes it less readable (read maintainable) as any
performance increases on a particular data size may evaporate when
used on a new data size because of changes in cache traversal/usage.
Martin York, Apr 7, 2008
13. ### ppiGuest
On Apr 1, 4:42 am, wrote:
> Hi,
>
> I have a tight loop where I basically do this over and over (millions
> of times):
>
> int cell = <calculation>
> const int above = <calculation>
> if (cell > above) cell = above;
> const int below = <calculation>
> if (cell > below) cell = below;
> const int trans = <calculation>
> if (cell > trans) cell = trans;
>
> Is there a faster way to do this?
>
> Sincerely,
> Anders
you may get better results by helping the compiler filling up the
pipeline:
int a = cell > above ? above : cell;
int b = below > trans ? trans : below;
cell = a < b ? b : a;
ideally 'a' and can 'b' be computed efficiently: they are both
independent and cached in registers.
By doing so you provide the compiler with more information regarding
the computation, namely that his free to compute a or b whenever it
wants, in this order or not.
Keeping on re-assigning the same variable again and again just make
the program instructions code stuck in the pipeline.
If you can write something like the previous code, you should notice a
speed-up.
This is nice feature off powerpc and AMD/Intel pentium processors.
You may be interested by:
- http://www.agner.org/optimize/optimizing_cpp.pdf
- http://developer.amd.com/TechnicalArticles/Articles/pages/7162004127.aspx
- http://developer.amd.com/TechnicalArticles/Articles/pages/6212004126.aspx
Even if the last 2 links are from the amd website, they are still
relevant for general optimization.
Unfortunately this means that you do have a good compiler and that the
code/algorithm you use can actually make this possible.
Hope it will help,
-- paulo
ppi, Apr 7, 2008
14. ### James KanzeGuest
On Apr 7, 2:15 am, Martin York <> wrote:
> On Apr 6, 2:43 pm, Daniel Pitts
> <> wrote:
> > Array lookups aren't terribly expensive, but they do have a cost. if
> > you can save a reference to distmatrix[i-1] (since its used in a few
> > places), that might shave a nanosecond or two.
> > Daniel Pitts' Tech Blog: <http://virtualinfinity.net/wordpress/>
> Not sure I buy that. Could be wrong!
More to the point (and this applies to the last response by ppi
as well, concerning pipeline stall): the compiler knows this
better than you, and turning on optimization will normally
result in the best solution here. Compilers have been caching
array accesses and common subexpressions for several decades
now, and all modern compilers know about pipelines, memory
caches, etc., and take them into account when optimizing. All
such micro-optimizations do in source code is make it more
convoluted---more difficult for the human reader to understand,
and more difficult for the compiler to optimize.
--
James Kanze (GABI Software) email:
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
James Kanze, Apr 8, 2008
15. ### Guest
Hi,
Well, I can tell you that manual loop unrolling for one special case
gave a massive speedup, and it now runs in about 25% of the time it
used to. I may attempt the approach of helping the compiler bound the
life of temporaries by using curlies, but initial results indicate
that this is detrimental to performance. Next I will probably try to
eliminate named temporaries, on the assumption that the compiler is
able to identify the references easier withour the aliassing
introduced by named temps.
Still, so far I have gotten some nice speedups. They are not related
to the issue I brought up here, but the solutions are in many cases
inspired by the discussion, so you have all helped me in various
ways
Thank you all!
Anders
, Apr 9, 2008
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## #351 2006-01-07 23:18:08
krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905
### Re: Problems and Solutions
x^2+y^2=x^2+y^2+2xy-2xy=(x+y)^2-2xy=1-2xy.
x^4+y^4=17=x^2^2+y^2^2+2x^2y^2-2x^2y^2=(x^2+y^2)^2-2x^2y^2=(1-2xy)^2-2x^2y^2 = 1-4xy+4x^2y^2-2x^2y^2=1-4xy+2x^2y^2 =>
2x^2y^2-4xy=17-1=16
x^2y^2-2xy=16/2=8
IPBLE: Increasing Performance By Lowering Expectations.
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## #352 2006-01-08 04:51:53
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
I might be missing something obvious, but...
1/2 + 1/3 + 1/4 = 13/12? Doesn't one of the partners in k+78 own a 1/12 imaginary share?
I think that this problem could only be solved if we knew who was holding the fake share.
Last edited by irspow (2006-01-08 04:54:55)
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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## #353 2006-01-08 16:22:43
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
krassi_holmz is right!
To irspow : In problem # k + 78, when it is stated that the shares are in the ration a:b:c, it need not be true that a+b+c be equal to 1. a's share of the total would be a/(a+b+c), b's share would be b/(a+b+c) and c's c/(a+b+c).
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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## #354 2006-01-08 17:28:19
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
Problem # k + 83
Four horses are tethered at 4 corners of a square field of side 70 metres so that they just cannot reach one another. What is the area left ungrazed by the horses?
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
## #355 2006-01-08 17:45:49
krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905
### Re: Problems and Solutions
70^2 - 45^2 Pi ?
IPBLE: Increasing Performance By Lowering Expectations.
Offline
## #356 2006-01-08 22:50:51
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
krassi_holmz, you are given another chance.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
## #357 2006-01-09 03:36:33
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: Problems and Solutions
Last edited by Ricky (2006-01-09 03:40:26)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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## #358 2006-01-09 13:06:26
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
Last edited by irspow (2006-01-09 13:39:29)
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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## #359 2006-01-09 16:16:22
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
Well done, irspow
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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## #360 2006-01-09 16:39:06
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
Problem # k + 84
Two spheres of radii 6 cm and 1 cm are inscribed in a right circular cone. The bigger sphere touches the smaller sphere and also the base of the cone. What is the height of the cone?
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
## #361 2006-01-10 11:21:02
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
I'll take an ill-attempted stab.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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## #362 2006-01-10 21:54:50
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
irspow is correct! Although the actual solution is arrived at differently!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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## #363 2006-01-10 21:58:32
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
Probelm # k + 85
If u and v are the roots of the equation x² + ax + b = 0, what are roots of the equation x² -ax + b = 0?
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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## #364 2006-01-11 03:02:37
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: Problems and Solutions
That's an interesting one. I think it's something like this:
Why did the vector cross the road?
It wanted to be normal.
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## #365 2006-01-11 04:16:41
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: Problems and Solutions
I'm not sure, this is what I get:
Last edited by Ricky (2006-01-11 04:16:58)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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## #366 2006-01-11 05:09:16
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: Problems and Solutions
I think our answers are the same, and I've just taken a long way round.
Why did the vector cross the road?
It wanted to be normal.
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## #367 2006-01-11 08:34:24
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: Problems and Solutions
Yea, I guess they are. They looked completely different at first glance.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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## #368 2006-01-11 11:32:20
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
Offline
## #369 2006-01-14 03:44:50
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
Offline
## #370 2006-01-14 06:33:56
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
Offline
## #371 2006-01-14 08:12:05
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
Offline
## #372 2006-01-14 12:15:28
irspow
Member
Registered: 2005-11-24
Posts: 1,055
### Re: Problems and Solutions
darn that k+42, I just cant figure it. Please someone, put me out of my misery. I think that you have to incorporate a geometric series somehow, but everything that I try turns to nonsense.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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## #373 2006-01-14 12:47:28
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: Problems and Solutions
Yes, k+42 is an incredibly difficult one. I think it could probably be solved brutally by getting excel to do all the calculations for you, but it's still tough.
Also, I think your answer to k+40 is wrong. I remember it being much smaller.
Why did the vector cross the road?
It wanted to be normal.
Offline
## #374 2006-01-14 16:12:47
ryos
Member
Registered: 2005-08-04
Posts: 394
### Re: Problems and Solutions
I didn't check if this one has already been solved, but since irspow asked about it, I gave it a go.
Last edited by ryos (2006-01-14 16:15:25)
El que pega primero pega dos veces.
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## #375 2006-01-14 17:13:17
Jai Ganesh
Registered: 2005-06-28
Posts: 47,365
### Re: Problems and Solutions
Four days Pongal break, and so many solutions posted! I shall reply to all of them after I return from Holiday on Jan 16. mathsyperson is right, irspow's solution to problem # k + 40 isn't correct. It is much smaller. . Same about ryos' solution to problem # k + 42.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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# from decimal to hex, the journey?
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## Recommended Posts
Hi folks! I was just wondering, I have always been given the code for decimal/octal/binary to hex converters, never coded them myself. How do I do? I have no clue.. I have serached google, no results... Is it some kind of formula or something? Object Pascal (delphi) and C++ are the languages I use.. Thanks!
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the algo is really easy, just think a bit about it:
- let num be the dec number you want to convert.
- store num % 16 as the last cipher in the hex number
- num = num / 16
- repeat
it's pretty straightforward, just run it trough on paper and you'll see
as you see if you want to convert to another base just replace the 16 in the algo with it.
if you want to see code to convert to/from/between arbitrary bases look here
hope that helps
Edit: just to make that clearer, in the second cycle of the loop num % 16 is stored as the second-to last number and so on, and you use A for 10, B for 11 and so on, but i guess you know that
also, the algo finishes once num is smaller then 16, then num itself is the first cipher of the hex number
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[edited by - burning_ice on April 19, 2003 7:38:28 PM]
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[edited by - mathematix on April 19, 2003 7:37:05 PM]
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Here''s some pseudo code:
num_str is the string with the number you want to convert
base is an integer that says what base num_str is in (binary=2, octal=8, decimal=10)
len is the length of num_str
also for the sake of simplicity, lets have num_str be in order from least-significant to most-significant, so if num_str = "12345", then num_str[0] is ''5'' and num_str[4] is ''1'' (this is the reverse of a normal string)
so,
int answer = 0;for (n = 0; n < len; n++) int this_digit = convertToInteger(num_str[n]) answer = answer + this_digit * (base to the power of n)end for
then you have ''answer'' which contains the value of the input string, so then if you want to convert it to a string with a different base, do this:
again, base is an integer (since you want to convert to hex, use 16)
int current_char = 0string new_string = ""while (answer != 0) int this_digit = answer mod base char this_char = convertToCharacter(this_digit) new_string[current_char] = this_char answer = answer / base current_char++end
which will create a string in the same ''reverse'' order as before.
have fun
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quote:
Original post by Mathematix
umm..isn't that one for finding the gcd of two numbers?
Edit: i think the one for converting bases is called horner's algorithm
My Homepage
[edited by - burning_ice on April 19, 2003 7:47:55 PM]
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Burning_Ice,
Yep, you''re right. But modifying it slightly yields a method for converting a decimal into any of the bases required. A very good universal method. If required, I shall explain in a later post.
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Student would see a visual representation of the "Position Equation" which is a good example of quadratic equations.
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## Question:
I need to build a neural network which is known as learnable parameter or a weight vector. In that way, only one vector will be generated and multiplied with the data. I have created that as following:
``````from tensorflow.keras.layers import Layer
from tensorflow.keras.layers import Input, Dense
from tensorflow.keras.models import Model
class LearnableMultiplier(Layer):
def __init__(self, **kwargs):
super(LearnableMultiplier, self).__init__(**kwargs)
def build(self, input_shape):
shape=(input_shape[-1],),
initializer='glorot_uniform',
trainable=True)
super(LearnableMultiplier, self).build(input_shape)
def call(self, inputs):
return inputs * self.kernel
inputs = Input(shape=(64,))
multiplier = LearnableMultiplier()(inputs)
model = Model(inputs=inputs, outputs=multiplier )
``````
I need the vector learnable parameter defined above to be selected from values of either 1 or -1. I mean each value of that vector which is multiplied with my data can only be 1 or -1. Is that feasible? How can I do it?
Yes, it is feasible to constrain the learnable weight vector to be either 1 or -1. One way to achieve this is by using the sign function to convert the weights to either 1 or -1. You can modify your call function to apply the sign function to the weights, like this:
``````import tensorflow as tf
from tensorflow.keras.layers import Layer
from tensorflow.keras.layers import Input, Dense
from tensorflow.keras.models import Model
class LearnableMultiplier(Layer):
def __init__(self, **kwargs):
super(LearnableMultiplier, self).__init__(**kwargs)
def build(self, input_shape):
shape=(input_shape[-1],),
initializer='glorot_uniform',
trainable=True)
super(LearnableMultiplier, self).build(input_shape)
def call(self, inputs):
return inputs * tf.math.sign(self.kernel)
``````
The `tf.math.sign` function will convert each element of the weight vector to either 1 or -1, depending on whether it is positive or negative. This will ensure that only 1 or -1 are used as the values for the weight vector during the multiplication.
Update: How about if I want to extend it to make every value selected from a list of different values, for example [1, -1, 0.5, -0.5], is that still feasible to be done?
To do this, you can replace the `tf.math.sign` function in the call method with a custom logic that selects the appropriate value from the list based on the sign of the `kernel` weight. Here is an example implementation:
``````import tensorflow as tf
from tensorflow.keras.layers import Layer
class LearnableMultiplier(Layer):
def __init__(self, values, **kwargs):
super(LearnableMultiplier, self).__init__(**kwargs)
self.values = values
def build(self, input_shape):
shape=(input_shape[-1],),
initializer='glorot_uniform',
trainable=True)
super(LearnableMultiplier, self).build(input_shape)
def call(self, inputs):
signs = tf.math.sign(self.kernel)
values = tf.gather(self.values, tf.cast((signs + 1) / 2, tf.int32))
return inputs * values
``````
The modified `LearnableMultiplier` layer takes an additional argument `values`, which is a list of values from which to select the learnable scalar factor. In the `call` method, the `tf.math.sign` function is replaced with a call to `tf.gather` that selects the appropriate value from the `values` list based on the sign of the `kernel` weight. Specifically, we first map the sign values from the range [-1, 1] to the range [0, 1] by adding 1 and dividing by 2, and then use the resulting integer indices to select the corresponding values from the list.
To use the modified layer in a Keras model, you can create an instance of the `LearnableMultiplier` class and pass it the values argument. For example:
``````import tensorflow as tf
from tensorflow.keras.layers import Input, Dense
from tensorflow.keras.models import Model
values = [1, -1, 0.5, -0.5]
inputs = Input(shape=(10,))
x = LearnableMultiplier(values=values)(inputs)
outputs = Dense(1)(x)
model = Model(inputs=inputs, outputs=outputs)
``````
Categories: questions
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner. | 1,003 | 4,239 | {"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} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.65175 |
https://www.semanticscholar.org/paper/On-Sets-Defining-Few-Ordinary-Lines-Green-Tao/8153a4913c80d7223f20cc66174f69a4a1f54f59 | 1,660,922,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573699.52/warc/CC-MAIN-20220819131019-20220819161019-00200.warc.gz | 839,208,394 | 76,644 | # On Sets Defining Few Ordinary Lines
@article{Green2013OnSD,
title={On Sets Defining Few Ordinary Lines},
author={Ben Green and Terence Tao},
journal={Discrete \& Computational Geometry},
year={2013},
volume={50},
pages={409-468}
}
• Published 23 August 2012
• Mathematics
• Discrete & Computational Geometry
Let $$P$$P be a set of $$n$$n points in the plane, not all on a line. We show that if $$n$$n is large then there are at least $$n/2$$n/2ordinary lines, that is to say lines passing through exactly two points of $$P$$P. This confirms, for large $$n$$n, a conjecture of Dirac and Motzkin. In fact we describe the exact extremisers for this problem, as well as all sets having fewer than $$n-C$$n-C ordinary lines for some absolute constant $$C$$C. We also solve, for large $$n$$n, the “orchard…
102 Citations
A new progress on Weak Dirac conjecture
• Mathematics
• 2016
In 2014, Payne-Wood proved that every non-collinear set $P$ of $n$ points in the Euclidean plane contains a point in at least $\dfrac{n}{37}$ lines determined by $P.$ This is a remarkable answer for
On sets of points with few ordinary hyperplanes
Let $S$ be a set of $n$ points in the projective $d$-dimensional real space $\mathbb{RP}^d$ such that not all points of $S$ are contained in a single hyperplane and such that any subset of $d$ points
A finite version of the Kakeya problem
• Mathematics
Australas. J Comb.
• 2016
Following Dvir's proof of the finite field Kakeya conjecture and the idea of using multiplicities of Dvir, Kopparty, Saraf and Sudan, it is proved that a lower bound on the size of S is dependent on the ideal generated by the homogeneous polynomials vanishing on D.
On sets of n points in general position that determine lines that can be pierced by n points
• Mathematics
Discret. Comput. Geom.
• 2020
It is shown that P, the set of points disjoint from P, must be contained in a cubic curve.
Many Collinear $$k$$k-Tuples with no $$k+1$$k+1 Collinear Points
• Mathematics, Computer Science
Discret. Comput. Geom.
• 2013
This work significantly improves the previously best known lower bound for the largest number of collinear k-tuples in a planar point sets, and gets reasonably close to the trivial upper bound O(n^2)$$O(n2). Orchards in elliptic curves over finite fields • Mathematics Finite Fields Their Appl. • 2020 A$$t_k$$tk Inequality for Arrangements of Pseudolines A new combinatorial inequality is presented which holds if no more than n-3 pseudolines intersect at one point and is unrelated to the Hirzebruch inequality for arrangements of complex lines in the complex projective plane. On Sets Defining Few Ordinary Solids • Mathematics Discret. Comput. Geom. • 2021 It is proved that if the number of solids incident with exactly four points of S is less than Kn^3 for some K=o(n^{\frac{1}{7}}) then, for n sufficiently large, all but at most O(K) points of \mathcal{S} are contained in the intersection of five linearly independent quadrics. On the Number of Ordinary Lines Determined by Sets in Complex Space • Mathematics SoCG • 2017 This paper shows that when the points span four or more dimensions, with at most n/2 points contained in any three dimensional affine subspace, it is shown that there must be a quadratic number of ordinary lines. On the Number of Ordinary Conics • Mathematics SIAM J. Discret. Math. • 2016 A lower bound on the number of ordinary conics determined by a finite point set in R is proved, based on the group structure of elliptic curves, that shows that the exponent in the bound is best possible. ## References SHOWING 1-10 OF 58 REFERENCES On a problem of Erdős and Moser • Mathematics • 2015 A set A of vertices in an r-uniform hypergraph$$\mathcal H$$H is covered in$$\mathcal H$$H if there is some vertex$$u\not \in A$$u∉A such that every edge of the form$$\{u\}\cup B$${u}∪B,$$B\in
On a question of Erdős and Moser
• Mathematics
• 2005
For two finite sets of real numbers A and B, one says that B is sum-free with respect to A if the sum set {b + b | b, b ∈ B, b 6= b} is disjoint from A. Forty years ago, Erdős and Moser posed the
Rational points on elliptic curves
• Mathematics
• 2006
We consider the structure of rational points on elliptic curves in Weierstrass form. Let x(P)=A_P/B_P^2 denote the $x$-coordinate of the rational point P then we consider when B_P can be a prime
The orchard problem
• Mathematics
• 1974
Abstract : The geometric version of the problem of Kirkman-Steiner triples may be formulated as follows: What is the maximal possible number t(p) of lines each of which contains precisely three
There exist 6n/13 ordinary points
• Mathematics
Discret. Comput. Geom.
• 1993
One of the main theorems used by Hansen is false, thus leavingn/2 open, and the 3n/7 estimate is improved to 6n/13 (apart from pencils and the Kelly-Moser example), with equality in the McKee example.
On the Number of Ordinary Lines Determined by n Points
• Mathematics
• 1958
More than sixty years ago, Sylvester (13) proposed the following problem: Let n given points have the property that the straight line joining any two of them passes through a third point of the set.
Cayley-Bacharach theorems and conjectures
• Mathematics
• 1996
A theorem of Pappus of Alexandria, proved in the fourth century A.D., began a long development in algebraic geometry. In its changing expressions one can see reflected the changing concerns of the | 1,407 | 5,392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.864203 |
http://bootmath.com/how-to-calculate-the-lens-distortion-coefficients-with-a-known-displacement-vector-field.html | 1,529,396,421,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861981.50/warc/CC-MAIN-20180619080121-20180619100121-00169.warc.gz | 47,044,285 | 6,656 | # How to calculate the lens distortion coefficients with a known displacement vector field?
I have this vector field full of displacement vectors, which indicates radial distortions by a lens system.
(Source)
I know where each of the displacement vectors starts $(x,y)$ and ends $(x’,y’)$ and I know the distortion equations look like
$$x’ = (1 + k_1r^2 + k_2r^4)x\\ y’ = (1 + k_1r^2 + k_2r^4)y$$
where $r^2 = x^2 + y^2$. Since each $(x’,y’)$ was measured, they are most likely biased. How can I estimate coefficients $k_1$ and $k_2$?
#### Solutions Collecting From Web of "How to calculate the lens distortion coefficients with a known displacement vector field?"
Pick two starting points $(x_1,y_1)$ and $(x_2,y_2)$ that are not collinear with the origin, and the corresponding end points. Then you get the two equations
$$x’_1 = (1+k_1r_1^2+k_2r_1^4)x_1\\ x’_2 = (1+k_1r_2^2+k_2r_2^4)x_2.$$
Then rewrite as a linear equation
$$\begin{bmatrix}x’_1/x_1-1\\x’_2/x_2-1\end{bmatrix}= \begin{bmatrix}r_1^2 & r_1^4 \\ r_2^2 & r_2^4 \end{bmatrix} \begin{bmatrix}k_1\\k_2\end{bmatrix}$$
and solve for $k_1$ and $k_2$.
Edit: If you want to include all measured vectors into your estimation of $k_1$ and $k_2$, you can set up a linear system like $Ax=b$ follows:
$$Ax= \begin{bmatrix}r_1^2 & r_1^4 \\r_1^2 & r_1^4 \\ r_2^2 & r_2^4 \\ r_2^2 & r_2^4 \\ \vdots & \vdots \\ r_n^2 & r_n^4 \end{bmatrix} \begin{bmatrix}k_1\\k_2\end{bmatrix}= \begin{bmatrix}x’_{1}/x_{1}-1\\y’_{1}/y_{1}-1\\x’_{2}/x_{2}-1\\y’_{2}/y_{2}-1\\\vdots\\y’_{n}/y_{n}-1\end{bmatrix} =b$$
This system is clearly overdetermined ($A$ is a $2n\times2$ matrix), so you need to employ least squares methods to solve it. Two methods to solve it:
1. Normal equations: premultiply both sides of the equation by $A^T$ and solve $A^TAx=A^Tb$.
2. QR decomposition: find a decomposition $QR=A$ such that $Q\in\mathbb R^{2\times n}$ has orthogonal columns, i.e. $Q^TQ=I$, and $R$ is an upper triangular $2\times2$ matrix. Then solve $Rx=Q^Tb$ (to see this just premultiply $Ax=b$ by $Q^T$).
The second method is numerically more stable, and you’ll find the $QR$ decomposition in nearly all linear algebra packages or mathematical software environments. If you have Matlab, you can directly use b=A\x, since matlab will automatically use a least squares method if $A$ is not square.
For a single point $(x_1,y_1)$ and its conjugate $(x_2,y_2)$:
$$x_2^2 + y_2^2 = (1+k_1 r^2 + k_2 r^4)^2 (x_1^2 + y_1^2) = (1+k_1 r^2 + k_2 r^4)^2 r^2$$
so that
$$1+k_1 r^2 + k_2 r^4 = \pm \sqrt{\frac{x_2^2+y_2^2}{r^2}}$$
The sign of course depends on whether the distortion also inverts. In your case of simple pincushion distortion, I don’t think so, so use the positive solution.
Repeat this for another point and its conjugate and you have two equations and two unknowns which will determine $k_1$ and $k_2$. You may wish to do this over the pupil/field to get a sense of how well your imaging obeys this model. | 1,043 | 2,959 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-26 | latest | en | 0.77065 |
personalnarrativeessay1.blogspot.com | 1,600,614,062,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00023.warc.gz | 587,228,124 | 11,901 | ## Tuesday, June 18, 2019
### Earned Value & Forecasting Essay Example | Topics and Well Written Essays - 750 words
Earned Value & Forecasting - Essay ExampleThe budgeted personify of survive performed at closedown of period 8 comes out to be \$868,000. The same is shown in Table 1. constitute Variance (CV) can be calculated as the difference betwixt budgeted cost and positive cost of work performed which is check to \$(868,000-1,005,000) i.e. - \$137,000. A negative cost variance here implies that the project is over-budget by the oddment of period 8. The Cost Performance indication (CPI) is given by the ratio of budgeted cost of work performed to the actual cost of work performed i.e. (868,000/1,005,000) which is equal to .86. As the Cost Performance Index is less than 1, it again implies that the project is over-budget. Schedule Variance (SV) can be found out as the difference between budgeted cost of work performed and the budgeted cost of work inscriptiond trough the kibosh of period 8 which is equal to \$(868,000-955,000) i.e. - \$87000. A negative schedule variance here implies that the project is behind schedule by the end of period 8. The Schedule Performance Index (SPI) is given by the ratio of budgeted cost of work performed to the budgeted cost of work scheduled i.e. (868,000/955,000) which is equal to .91. As the Schedule Performance Index is less than 1, it gain implies that the project is behind schedule. ... 500000 1.3.2 60000 80 48000 1.4.1 120000 0 0 1.4.2 40000 50 0 1.4.3 75000 100 75000 1.5.2 15000 20 0 1.5.3 30000 0 0 1.6.1 45000 20 0 1.6.2 60000 50 0 1.6.3 30000 0 0 1.6.4 25000 0 0 Total 124500 (BAC) 868000 (BCWP/EV) The planned percentage of work unblemished can be calculated by dividing the planned work completed (in terms of enumerate of periods) by the append work (in terms of number of periods). The total work periods can be calculated by adding the planned durations of each activity. This comes out as 26 periods. The planned work periods can be calculated by multiplying the planned percent of work completed for each activity by its plann ed duration. This is equal to 15. Hence, the planned percent of work completed is equal to 15/26 i.e. 57.7%. Similarly, the actual percent of work completed is computed as the ratio of actual work completed (in terms of number of periods) to the total work (in terms of number of periods). The actual work periods are calculated as 15.2 in the same fashion as for planned work periods. The actual percent of work completed, therefore, comes out as 15.2/26 i.e. 58.46%. The percent cost completed is simply defined as the ratio of actual cost incurred till end of period 8 to the total planned cost of the project i.e. (1,005,000/1,245,000) which comes out as 80.72% Table 2 Calculating percent task outcome Task Duration Actual Percentage of work completed at end of period 8 Actual work completed (in periods) at end of period 8 Planned percentage of work completed at end of period 8 Planned work completed (in periods) at end of period 8 1.2.1 3 100 3 100 3 1.2.2 1 100 1 100 1 1.2.3 2 100 2 1 00 2 1.3.1 3 100 3 100 3 1.3.2 2 80 1.6 100 2 1.4.1 3 0 0 66.66666667 2 1.4.2 1 50 0.5 0 0 1.4.3 3 100 3 66.66666667 2 1.5.2 1 20 0.2 0 0 1.5.3 2 0 0 0 | 999 | 3,252 | {"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} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.916568 |
https://www.tes.com/teaching-resources/shop/jhutcheon | 1,713,104,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00042.warc.gz | 958,076,207 | 17,822 | # J Hutcheon's Shop
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Physic teacher and HOD. I have worked in independent , LA and academy schools. Survived four Ofsted inspections. Still passionate about the curriculum and quality of the education our students receive.
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Physic teacher and HOD. I have worked in independent , LA and academy schools. Survived four Ofsted inspections. Still passionate about the curriculum and quality of the education our students receive.
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AS OCR Physics A Presentation covering charge and current. Worksheets referred to in the presentation are from the IOP teaching advanced Physics site. There is a link to these in the presentation. | 1,113 | 5,424 | {"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-2024-18 | latest | en | 0.907745 |
https://legacy.cs.indiana.edu/classes/a201-dger/spr2002/labs/Four.html | 1,660,198,424,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00535.warc.gz | 353,113,691 | 4,117 | Spring Semester 2002
Lab Notes Four: Classes and objects.
1. Here's a set of practice problems.
2. Here's a set of warm-up questions.
3. Here are the answers to the warm-up questions.
4. Here's a broad overview of the programs to be implemented.
5. Here are the solutions.
Below you have your LAB ASSIGNMENT (#4). Next time you are expected to:
• Have all the steps below done and understood (we develop a program below).
• Look over and solve (or learn the posted solutions to) the review exercises and problems from Chapter Three (pp. 134-137). The texts and the solutions to all of these problems are posted above (steps 1, 2, 3, 5 above).
• Read and understand and be able to explain the crux of the slides presented at step 4 above (the part that talks about Kanamits and the Twilight Zone). Why do you think the slides talk about, open up, and end in reference to Kanamits (whatever they are). Hint: abstraction.
Let's go through the complete and annotated development of a solution to a problem.
(This is problem 3.7 from the book, page 137).
1. Let's implement a class `Student`.
```frilled.cs.indiana.edu%pico Student.java
frilled.cs.indiana.edu%cat Student.java
public class Student {
}
frilled.cs.indiana.edu%javac Student.java
frilled.cs.indiana.edu%ls -ld Student*
-rw------- 1 dgerman 188 Feb 1 08:18 Student.class
-rw------- 1 dgerman 27 Feb 1 08:17 Student.java
frilled.cs.indiana.edu%```
2. Looks like we're done. Can we test it?
3. We need a tester class with a `main` method.
```frilled.cs.indiana.edu%pico StudentTest.java
frilled.cs.indiana.edu%cat StudentTest.java
public class StudentTest {
public static void main(String[] args) {
Student a = new Student();
}
}
frilled.cs.indiana.edu%javac StudentTest.java
frilled.cs.indiana.edu%ls -ld Student*
-rw------- 1 dgerman 188 Feb 1 08:18 Student.class
-rw------- 1 dgerman 27 Feb 1 08:17 Student.java
-rw------- 1 dgerman 297 Feb 1 08:21 StudentTest.class
-rw------- 1 dgerman 110 Feb 1 08:21 StudentTest.java
frilled.cs.indiana.edu%```
4. Can we test it?
5. We can run `StudentTest` but we get no output.
```frilled.cs.indiana.edu%java StudentTest
frilled.cs.indiana.edu%```
6. Does it matter?
7. Do we know what happens inside?
8. The `Student` class is empty. `Student` objects are amorphous.
9. I see... Let's make it such each `Student` have (at least) a name, then.
```frilled.cs.indiana.edu%pico Student.java
frilled.cs.indiana.edu%cat Student.java
public class Student {
private String name;
}
frilled.cs.indiana.edu%```
10. What's the meaning of `private`?
11. It means that to know the name of a `Student` you need to ask the `Student` what its name is.
12. I don't feel very comfortable using he or she for a `Student object. `
``` 13. Fine. How do you inquire about a Student's name? ```
``` 14. We need to add this functionality to class Student first, then make use of it. ```
``` 15. Here's a more comprehensive blueprint of Student objects. ```
``` ```
``````frilled.cs.indiana.edu%pico Student.java
frilled.cs.indiana.edu%cat Student.java
public class Student {
private String name;
}
frilled.cs.indiana.edu%pico Student.java
frilled.cs.indiana.edu%cat Student.java
public class Student {
private String name;
public String whatsYourName () {
return name;
}
}
frilled.cs.indiana.edu%``````
``` 16. What does it mean for the whatsYourName method to be public? 17. It means you can ask a Student "What's your name?" 18. What if we make it private? 19. Then we can never ask. 20. How do we create a Student? 21. Just invoke new the way we did in the tester's main. 22. And if we invoke it, how do things get created, and initialized. 23. Well, a default no-arg constructor is present, but we don't see it. 24. I think we should add it, so that we not forget that it's there. frilled.cs.indiana.edu%pico Student.java frilled.cs.indiana.edu%cat Student.java public class Student { private String name; public String whatsYourName () { return name; } Student() { } } frilled.cs.indiana.edu% 25. It's empty, but it gets called at creation time. 26. Can we create a Student with an initial name? 27. Only if we provide that type of constructor. 28. To be able to create a Student with an initial name we need to be able to initialize the name of any Student (at creation time) with an actual name (to be specified when we create the Student). 29. We're looking for something like this: new Student("Larry Johnson") 30. Let's provide class Student with that capability. frilled.cs.indiana.edu%pico Student.java frilled.cs.indiana.edu%cat Student.java public class Student { private String name; public String whatsYourName () { return name; } Student() { } Student(String givenName) { name = givenName; } } frilled.cs.indiana.edu% 31. Let's enhance our tester's main to exploit the new features. frilled.cs.indiana.edu%pico StudentTest.java frilled.cs.indiana.edu%cat StudentTest.java public class StudentTest { public static void main(String[] args) { Student a = new Student("Larry"); Student b = new Student("Michael"); String answer; System.out.print("Printing the name of the first student: "); answer = a.whatsYourName(); System.out.println(answer); System.out.print("Printing the name of the second student: "); answer = b.whatsYourName(); System.out.println(answer); } } frilled.cs.indiana.edu%javac StudentTest.java frilled.cs.indiana.edu%java StudentTest Printing the name of the first student: Larry Printing the name of the second student: Michael frilled.cs.indiana.edu% 32. Great! What else were we supposed to do? 33. Let's enable the Students to keep track of their scores. frilled.cs.indiana.edu%pico Student.java frilled.cs.indiana.edu%cat Student.java public class Student { private String name; public String whatsYourName () { return name; } Student() { } Student(String givenName) { name = givenName; } void addQuizScore(int newScore) { totalScore = totalScore + newScore; } private int totalScore; } frilled.cs.indiana.edu% 34. I see... If there's a new score to be added to the total score for a student then we just add it to the totalScore as if it were an amount to be placed as deposit over a current, given, existing balance. 35. Yes, so you need to define an instance variable totalScore (which will keep the cumulative score for the Student) and use it as if it were a balance. 36. This way a Student is like a BankAccount with a name. 37. Let's write getBalance, then. frilled.cs.indiana.edu%pico Student.java frilled.cs.indiana.edu%cat Student.java public class Student { private String name; public String whatsYourName () { return name; } Student() { } Student(String givenName) { name = givenName; } void addQuizScore(int newScore) { totalScore = totalScore + newScore; } private int totalScore; int whatsYourTotalScore() { return totalScore; } } frilled.cs.indiana.edu% 38. Let's test it. frilled.cs.indiana.edu%pico StudentTest.java frilled.cs.indiana.edu%cat StudentTest.java public class StudentTest { public static void main(String[] args) { Student a = new Student("Larry"); Student b = new Student("Michael"); String answer; System.out.print("Printing the name of the first student: "); answer = a.whatsYourName(); System.out.println(answer); System.out.print("Printing the name of the second student: "); answer = b.whatsYourName(); System.out.println(answer); a.addQuizScore(100); a.addQuizScore(90); a.addQuizScore(100); System.out.println("Student " + a.whatsYourName() + "reports: "); System.out.println(" cumulative score: " + a.whatsYourTotalScore()); } } frilled.cs.indiana.edu%javac StudentTest.java frilled.cs.indiana.edu%java StudentTest Printing the name of the first student: Larry Printing the name of the second student: Michael Student Larryreports: cumulative score: 290 frilled.cs.indiana.edu% 39. I think you need a space between Larry and reports. 40. I'll let you fix that. But overall we've come a long way, don't you think? 41. I sure do so. What if I want the Students to be able to report the average score in addition to the cumulative score? I don't think this is possible at the moment, because they don't remember how many quizzes they have taken. 42. Indeed, they only keep the cumulative score. 43. To remember how many quizzes they have taken they would need to keep a counter, to be updated (incremented by 1) every time a new score is added to the totalScore. 44. If we kept the number updated we could easily report the average at any time, as follows. frilled.cs.indiana.edu%pico Student.java frilled.cs.indiana.edu%cat Student.java public class Student { private String name; public String whatsYourName () { return name; } Student() { } Student(String givenName) { name = givenName; } void addQuizScore(int newScore) { totalScore = totalScore + newScore; } private int totalScore; int whatsYourTotalScore() { return totalScore; } private int numberOfScores; double reportAverage() { return (double)totalScore / numberOfScores; } } frilled.cs.indiana.edu% 45. I think you forgot to update the counter in addQuizScore, haven't you? 46. Ooops!... frilled.cs.indiana.edu%pico Student.java frilled.cs.indiana.edu%cat Student.java public class Student { private String name; public String whatsYourName () { return name; } Student() { } Student(String givenName) { name = givenName; } void addQuizScore(int newScore) { totalScore = totalScore + newScore; numberOfScores = numberOfScores + 1; } private int totalScore; int whatsYourTotalScore() { return totalScore; } private int numberOfScores; double reportAverage() { return (double)totalScore / numberOfScores; } } frilled.cs.indiana.edu% 47. There you go. 48. Can you test that? 49. Sure, how about this: frilled.cs.indiana.edu%pico StudentTest.java frilled.cs.indiana.edu%cat StudentTest.java public class StudentTest { public static void main(String[] args) { Student a = new Student("Larry"); Student b = new Student("Michael"); String answer; System.out.print("Printing the name of the first student: "); answer = a.whatsYourName(); System.out.println(answer); System.out.print("Printing the name of the second student: "); answer = b.whatsYourName(); System.out.println(answer); a.addQuizScore(100); a.addQuizScore(90); a.addQuizScore(100); System.out.println("Student " + a.whatsYourName() + "reports: "); System.out.println(" cumulative score: " + a.whatsYourTotalScore()); System.out.println(" average score: " + a.reportAverage()); } } 50. Nice. You only changed one line! 51. Indeed. And here's the actual test: frilled.cs.indiana.edu%javac StudentTest.java frilled.cs.indiana.edu%java StudentTest Printing the name of the first student: Larry Printing the name of the second student: Michael Student Larryreports: cumulative score: 290 average score: 96.66666666666667 frilled.cs.indiana.edu% 52. Good Student! 53. Yes. Isn't it time for a break? 54. I sure think so. 55. See you next week! Until then, here's a brief summary of chapter 3: Last updated: Jan 19, 2002 by Adrian German for A201 ``` | 2,810 | 11,003 | {"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} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.796864 |
http://macphilly.com/page/lr-method-eigen-values-30276017.html | 1,600,901,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00028.warc.gz | 77,456,720 | 6,750 | # Lr method eigen values
Note that. View Usage Statistics. The eigenvalues of a triangular matrix are listed on the diagonal, and the eigenvalue problem is solved. Baltimore: Johns Hopkins University Press. This Collection. Namespaces Article Talk. A typical symmetric QR algorithm isolates each eigenvalue then reduces the size of the matrix with only one or two iterations, making it efficient as well as robust.
• Rutishauser's LR Method for finding eigenvalues of matrices
• Complete eigenvalues problem solution using LR method
• ## Rutishauser's LR Method for finding eigenvalues of matrices
matrix whose eigenvalues are the same as those of the original matrix. The problem reduces case the LR algorithm is equivalent to the QD algorithm. It will be. the eigenvalues of A. This then is the LR algorithm. Since the algorithm is based upon the triangular decomposi- tion of a matrix A, we shallintroduce a method.
In numerical linear algebra, the QR algorithm is an eigenvalue algorithm: that is, a procedure to. The QR algorithm was preceded by the LR algorithm, which uses the LU decomposition instead of the QR decomposition.
The QR algorithm is.
In modern computational practice, the QR algorithm is performed in an implicit version which makes the use of multiple shifts easier to introduce.
## Complete eigenvalues problem solution using LR method
However, it represents an important step in the development of the QR algorithm. This operation is known as bulge chasingdue to the peculiar shape of the non-zero entries of the matrix along the steps of the algorithm. The QR algorithm is more stable, so the LR algorithm is rarely used nowadays. Floating point Numerical stability.
PROBLEM SOLVING MODELS IN ELEMENTARY EDUCATION This will allow all visitors to view the contents of the thesis. If the original matrix is symmetricthen the upper Hessenberg matrix is also symmetric and thus tridiagonaland so are all the A k. Login Register.Abstract Not available. Instead, the QR algorithm works with a complete basis of vectors, using QR decomposition to renormalize and orthogonalize.
The LR method (Rutishauser, ) is an iterative method used for solving eigenvalues and eigenvectors of a square matrix. A single step of LR method uses. Introduction. The LR and QR algorithms have proved to be two of the most important general purpose methods for solving the unsymmetric eigenvalue problem.
The LR and QR algorithms, two of the best available iterative methods for finding the eigenvalues of a nonsymmetric matrix associated with a system of linear.
Rutishauser took an algorithm of Alexander Aitken for this task and developed it into the quotient—difference algorithm or qd algorithm.
The QR algorithm is more stable, so the LR algorithm is rarely used nowadays.
Video: Lr method eigen values power method for eigenvalues and eigenvectors examples - part #3
Metadata Show full item record. Numerical linear algebra. Note that.
Video: Lr method eigen values 7: Power Method for Eigenvalues - Learning Linear Algebra
Rutishauser's LR Method for finding eigenvalues of matrices.
HOTEL HAPPY END BRODNICA KAMERA Rutishauser took an algorithm of Alexander Aitken for this task and developed it into the quotient—difference algorithm or qd algorithm. By using this site, you agree to the Terms of Use and Privacy Policy. Moreover, because the Hessenberg form is already nearly upper-triangular it has just one nonzero entry below each diagonalusing it as a starting point reduces the number of steps required for convergence of the QR algorithm.If this is your thesis or dissertation, you can make it open-access. In numerical linear algebrathe QR algorithm is an eigenvalue algorithm : that is, a procedure to calculate the eigenvalues and eigenvectors of a matrix. Applied Numerical Linear Algebra.
If it is unsymmetrical apply LR or QR method. If you are interested in largest or smallest in magnitude eigenvalues then apply Power's method. Methods.
for. Eigenvalue. Problems.
Now we shall discuss some numerical the QR method and the LR method are the most widely used techniques for. Rutishauser's LR Method for finding eigenvalues of matrices. Thumbnail. View/Open. Thesis Fpdf (Mb). Date. Author. Fikes, William Ralph.
Matrix Computations 3rd ed.
Abstract Not available. Moreover, because the Hessenberg form is already nearly upper-triangular it has just one nonzero entry below each diagonalusing it as a starting point reduces the number of steps required for convergence of the QR algorithm.
This will allow all visitors to view the contents of the thesis. If the original matrix is symmetricthen the upper Hessenberg matrix is also symmetric and thus tridiagonaland so are all the A k. However, it represents an important step in the development of the QR algorithm. Metadata Show full item record.
CINTAI AKU KARNA ALLAH KUNCI GITAR DANGDUT This operation is known as bulge chasingdue to the peculiar shape of the non-zero entries of the matrix along the steps of the algorithm. Rutishauser took an algorithm of Alexander Aitken for this task and developed it into the quotient—difference algorithm or qd algorithm. System of linear equations Matrix decompositions Matrix multiplication algorithms Matrix splitting Sparse problems.This will allow all visitors to view the contents of the thesis. One variant of the QR algorithmthe Golub-Kahan-Reinsch algorithm starts with reducing a general matrix into a bidiagonal one. Categories : Numerical linear algebra. Login Register.
## 4 thoughts on “Lr method eigen values”
1. Guzilkree:
A typical symmetric QR algorithm isolates each eigenvalue then reduces the size of the matrix with only one or two iterations, making it efficient as well as robust.
2. Vojinn:
Subject mathematics. Rutishauser's LR Method for finding eigenvalues of matrices.
3. Tolabar:
The eigenvalues of a triangular matrix are listed on the diagonal, and the eigenvalue problem is solved.
4. Zuk:
Non-affiliated individuals should request a copy through their local library's interlibrary loan service. | 1,256 | 6,078 | {"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-2020-40 | latest | en | 0.84473 |
https://degreessymbols.com/degrees-celsius/ | 1,627,710,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00255.warc.gz | 216,689,580 | 11,644 | # Degree Celsius
The degree Celsius (symbol in ° C) is the unit of a temperature measurement scale , named after the Swedish astronomer Anders Celsius (1701 – 1744), who first proposed it in 1742. The Celsius scale is a scale to indicate the temperature at which the temperature ranges are degrees Celsius.
The Celsius scale defines the melting point of ice in a mixture of water saturated with air at 0 ° C and the boiling point at 99.974 ° C under standard pressure conditions (1 bar, slightly less than the atmosphere, pressure to water) boils at 100 degrees Celsius).
Originally conceived by the Celsius scale, it had a boiling point of water at 0 ° C and a freezing point of 100 ° C. After his death, however, the scale was reversed in 1745 by Linnaeus, transforming it into what is now commonly used.
The current official definition of the Celsius scale places 0.01 ° C as a triple point of water and a degree as 1 / 273.16 of the temperature difference between the triple point of water and absolute zero. This definition of degrees Celsius ensures that the temperature difference of one degree Celsius represents the same temperature difference as a Kelvin.
## CONVERSION OF DEGREES CELSIUS TO OTHER TEMPERATURE UNITS
Formulas for converting degrees Celsius to other temperature units.
• Conversion of degrees Celsius to degrees Fahrenheit at ° F = (9/5 × ° C) + 32
• Conversion of degrees Fahrenheit to degrees Celsius at ° C = (5/9) × (° F – 32)
• Conversion of degrees Celsius to kelvin at K = ° C + 273.15
• Conversion Kelvin to degrees Celsius ° K = – 273.15
Anders Celsius originally proposed that the water freezing point be 100 ° C and the boiling point 0 ° C. The ladder was reversed in 1745 at the suggestion of Linnaeus, or perhaps by Daniel Ekström, producer of most of the thermometers used by Celsius.
The Celsius and Fahrenheit thermometric scales cross at the same value of -40 degrees, that is, the value of -40 ° C corresponds to -40 ° F.
One way to convert Celsius to Fahrenheit is to multiply by 1.8 and add 32.
° F = ° C × 1,8 + 32
On the other hand, to convert Fahrenheit to degrees Celsius, subtract 32 and divide by 1.8.
° C = (° F – 32) / 1,8
The Celsius scale is used daily in most parts of the world, although in mass media it is still called centigrade until the 1990s, especially in weather forecasts. In the United States (as well as in non-corporate territories and freely associated states of the Pacific), in Belize, Liberia and some Caribbean islands, the Fahrenheit scale is used, but these countries also use the Celsius or Kelvin scale in Applications. scientific or technological.
Other temperature ranges are: Newton (around 1700), Rømer (1701), Fahrenheit (1724), Réaumur (1731), Delisle or de Lisle (1738), Rankine (1859), Kelvin (1862) and Leyden (by around 1894). Note that “kelvin” is tiny, because it is an SI unit, even though it is derived from the noble title of scientist William Thomson.
## WHAT IS THE DIFFERENCE BETWEEN DEGREES CELSIUS AND DEGREES CELSIUS?
Since there are a hundred divisions between these two landmarks, the original term for this system was either centigrade or centesimal. In 1948, the ninth General Conference of Animals (CR 64) officially changed its name to Celsius to remove several ambiguities:
• the centi prefix is used by the SI system for submultiples;
• a more correct name, at least since Linnaeus was invested, would have been “hectograd”;
• the hectogram is also ambiguous and, in oenology, indicates the alcohol content per hectolitre of wine;
• the term “degrees centigrade” was also used to indicate the hundredth degree used to measure angles;
• Kelvin is also a degree centigrade, in that it is obtained, like Celsius, by dividing the interval between two fundamental fixed points in equal parts.
The decision of the Conférence générale des poids et mesures aimed to remove ambiguities and to unequivocally associate the creation of the scale with Celsius (despite Linnaeus’ important contribution); in addition, the scale was standardized with those of Kelvin (created by Lord Kelvin), Fahrenheit, Réaumur and Rankine , all named after their inventor.
Despite this, there are often people who refer to degrees centigrade as “degrees Celsius”, just as they use the centigrade concept scale.
## WHY IS THE CELSIUS SCALE NOT CONSIDERED AN ABSOLUTE SCALE?
The existence of an absolute temperature scale is a consequence of the second principle of thermodynamics .
According to the second law, the absolute temperature scale starts from the existence of 0º absolute. This condition is met only by two basic units:
• Degree of Kelvin
• Grau Rankine.
Zero degrees Kelvin corresponds to -273.16 degrees Celsius.Author: Oriol Planas – Industrial Technical Engineer, specializing in mechanics | 1,164 | 4,808 | {"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} | 3.5625 | 4 | CC-MAIN-2021-31 | longest | en | 0.810136 |
http://mathisfunforum.com/viewtopic.php?pid=265502 | 1,398,045,682,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00130-ip-10-147-4-33.ec2.internal.warc.gz | 161,083,873 | 4,943 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2013-04-27 15:13:03
Agnishom
Real Member
Online
### Slider Puzzle
How do I devise an algorithm to solve a sliding puzzle?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #2 2013-04-27 15:21:58
bobbym
Offline
### Re: Slider Puzzle
Hi;
Do you mean the 15 puzzle?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #3 2013-04-27 16:04:55
Agnishom
Real Member
Online
### Re: Slider Puzzle
The 15 puzzle is unsolvable.
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #4 2013-04-27 16:08:50
bobbym
Offline
### Re: Slider Puzzle
Some positions are not possible but others are.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #5 2013-04-27 16:25:44
Agnishom
Real Member
Online
### Re: Slider Puzzle
Ok, so how to do it?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #6 2013-04-27 16:27:25
bobbym
Offline
### Re: Slider Puzzle
There is some math that tells you whether it is solveable or not but I have forgotten it.
Programming it, looks tough.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #7 2013-04-27 16:50:13
mathaholic
Star Member
Offline
### Re: Slider Puzzle
Mathematica?
246 pages on Prime Numbers Wiki (+1)
## #8 2013-04-27 17:16:08
Agnishom
Real Member
Online
### Re: Slider Puzzle
But programming it is what I am thinking.
After all, programs are threre which solve them
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #9 2013-04-27 19:10:17
bobbym
Offline
### Re: Slider Puzzle
I will try one and see what comes up.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #10 2013-04-27 21:48:51
Agnishom
Real Member
Online
### Re: Slider Puzzle
Did you?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #11 2013-04-27 21:51:40
bobbym
Offline
### Re: Slider Puzzle
I have not even scratched the surface because I am still wrestling with a representation of the puzzle itself. It is unlikely that I will get far even when I solve that small problem because I do not understand how the program should pick from its many choices.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #12 2013-04-27 21:57:23
Agnishom
Real Member
Online
### Re: Slider Puzzle
Are you on Win 7?
The algorithm isn't simple, it involves a lot of graph theory...
Last edited by Agnishom (2013-04-27 21:58:55)
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #13 2013-04-27 22:11:42
bobbym
Offline
### Re: Slider Puzzle
Heck no!
That is the problem right there. I am not very good with graph theory. As a matter of fact, I stink.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #14 2013-04-27 22:22:16
Agnishom
Real Member
Online
### Re: Slider Puzzle
http://www.brian-borowski.com/Software/Puzzle/
Are they calling each state a node?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #15 2013-04-27 22:34:56
bobbym
Offline
### Re: Slider Puzzle
Probably, that would make sense.
Here is the math behind every solution.
http://mathworld.wolfram.com/15Puzzle.html
Also, there are tons of these programs to download. I need some sleep, see you.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #16 2013-04-27 22:43:30
Agnishom
Real Member
Online
### Re: Slider Puzzle
Okay Sweet Dreams!
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda
## #17 2013-04-27 22:47:24
bobbym
Offline
### Re: Slider Puzzle
Thanks and same to you.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 1,782 | 6,613 | {"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} | 2.9375 | 3 | CC-MAIN-2014-15 | longest | en | 0.885424 |
https://de.mathworks.com/matlabcentral/cody/problems/43077-put-all-numbers-in-a-string-inside-square-brackets/solutions/1980178 | 1,579,824,702,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00332.warc.gz | 388,089,066 | 15,918 | Cody
# Problem 43077. Put all numbers in a string inside square brackets
Solution 1980178
Submitted on 17 Oct 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = 'left3down2'; y_correct = 'left[3]down[2]'; assert(strcmp(nestedNums(x),y_correct))
y = 1×2 cell array {'3'} {'2'}
Error using assert The condition input argument must be a scalar logical. Error in Test1 (line 3) assert(strcmp(nestedNums(x),y_correct))
2 Fail
x = 'fiat500'; y_correct = 'fiat[500]'; assert(strcmp(nestedNums(x),y_correct))
y = 1×1 cell array {'500'}
Assertion failed.
3 Fail
x = 'no numbers here'; y_correct = 'no numbers here'; assert(strcmp(nestedNums(x),y_correct))
y = 0×0 empty cell array
Error using assert The condition input argument must be a scalar logical. Error in Test3 (line 3) assert(strcmp(nestedNums(x),y_correct))
4 Fail
x = '1234567890'; y_correct = '[1234567890]'; assert(strcmp(nestedNums(x),y_correct))
y = 1×1 cell array {'1234567890'}
Assertion failed.
5 Fail
x = 'var1 = 20; var2 = 47; var3 = 59;'; y_correct = 'var[1] = [20]; var[2] = [47]; var[3] = [59];'; assert(strcmp(nestedNums(x),y_correct))
y = 1×6 cell array {'1'} {'20'} {'2'} {'47'} {'3'} {'59'}
Error using assert The condition input argument must be a scalar logical. Error in Test5 (line 3) assert(strcmp(nestedNums(x),y_correct))
6 Fail
x = '14et38z1n541z8ne'; y_correct = '[14]et[38]z[1]n[541]z[8]ne'; assert(strcmp(nestedNums(x),y_correct))
y = 1×5 cell array {'14'} {'38'} {'1'} {'541'} {'8'}
Error using assert The condition input argument must be a scalar logical. Error in Test6 (line 3) assert(strcmp(nestedNums(x),y_correct)) | 555 | 1,757 | {"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-2020-05 | latest | en | 0.216376 |
https://www.dovov.com/date-2.html | 1,653,428,444,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00309.warc.gz | 852,117,689 | 15,017 | # 确定两个date范围是否重叠
（StartA <= EndB）和（EndA> = StartB）
certificate：
`_ |---- DateRange A ------| |---Date Range B -----| _`
（如果`StartA > EndB`真）
`|---- DateRange A -----| _ _ |---Date Range B ----|`
（如果`EndA < StartB`真）
（如果一个范围既不完全相同，
`Not (A Or B)` <=> `Not A And Not B`
`>=`运算符更改为`>` ，将`<=`更改为`<`
{ `endA-startA``endA - startB``endB-startA``endB - startB` }
`(StartA <= EndB) and (EndA >= StartB)` `(StartA <= EndB) and (StartB <= EndA)`
`DateRangesOverlap = max(start1, start2) < min(end1, end2)`
`(StartA <= EndB) and (StartB <= EndA) and (StartA <= EndA) and (StartB <= EndB)`或者：
`(StartA <= EndB) and (StartA <= EndA) and (StartB <= EndA) and (StartB <= EndB)`
`(StartA <= Min(EndA, EndB) and (StartB <= Min(EndA, EndB))`或者：
`(Max(StartA, StartB) <= Min(EndA, EndB)`
`(StartA > StartB? Start A: StartB) <= (EndA < EndB? EndA: EndB)`
` `(StartDate1 <= EndDate2) and (StartDate2 <= EndDate1)` `
` `// ------------------------------------------------------------------------ public enum PeriodRelation { After, StartTouching, StartInside, InsideStartTouching, EnclosingStartTouching, Enclosing, EnclosingEndTouching, ExactMatch, Inside, InsideEndTouching, EndInside, EndTouching, Before, } // enum PeriodRelation` `
` `A.end >= B.start AND A.start <= B.end` `
ErikE评论：
` `----------------------|-------A-------|---------------------- |----B1----| |----B2----| |----B3----| |----------B4----------| |----------------B5----------------| |----B6----| ----------------------|-------A-------|---------------------- |------B7-------| |----------B8-----------| |----B9----| |----B10-----| |--------B11--------| |----B12----| |----B13----| ----------------------|-------A-------|----------------------` `
` `overlap = max(0, min(EndDate1, EndDate2) - max(StartDate1, StartDate2)) if (overlap > 0) { ... }` `
` `|----------------------| range 1 |---> range 2 overlap |---> range 2 overlap |---> range 2 overlap |---> range 2 no overlap` `
` `def doesOverlap (r1, r2): if r1.s > r2.s: swap r1, r2 if r2.s > r1.e: return false return true` `
` `def doesOverlap (r1, r2): if r1.s > r2.s: swap r1, r2 return r2.s <= r1.e` `
` `|----------------------| range 1 |---> range 2 overlap |---> range 2 overlap |---> range 2 no overlap |---> range 2 no overlap` `
• 将空值处理为无穷大
• 假设下界是包含的，上界是排他的。
• 附带一堆testing
testing是基于整数，但由于JavaScript中的date对象是可比较的，所以您可以只input两个date对象。 或者你可以扔在毫秒时间戳。
# 码：
` `/** * Compares to comparable objects to find out whether they overlap. * It is assumed that the interval is in the format [from,to) (read: from is inclusive, to is exclusive). * A null value is interpreted as infinity */ function intervalsOverlap(from1, to1, from2, to2) { return (to2 === null || from1 < to2) && (to1 === null || to1 > from2); }` `
# testing：
` `describe('', function() { function generateTest(firstRange, secondRange, expected) { it(JSON.stringify(firstRange) + ' and ' + JSON.stringify(secondRange), function() { expect(intervalsOverlap(firstRange[0], firstRange[1], secondRange[0], secondRange[1])).toBe(expected); }); } describe('no overlap (touching ends)', function() { generateTest([10,20], [20,30], false); generateTest([20,30], [10,20], false); generateTest([10,20], [20,null], false); generateTest([20,null], [10,20], false); generateTest([null,20], [20,30], false); generateTest([20,30], [null,20], false); }); describe('do overlap (one end overlaps)', function() { generateTest([10,20], [19,30], true); generateTest([19,30], [10,20], true); generateTest([10,20], [null,30], true); generateTest([10,20], [19,null], true); generateTest([null,30], [10,20], true); generateTest([19,null], [10,20], true); }); describe('do overlap (one range included in other range)', function() { generateTest([10,40], [20,30], true); generateTest([20,30], [10,40], true); generateTest([10,40], [null,null], true); generateTest([null,null], [10,40], true); }); describe('do overlap (both ranges equal)', function() { generateTest([10,20], [10,20], true); generateTest([null,20], [null,20], true); generateTest([10,null], [10,null], true); generateTest([null,null], [null,null], true); }); });` `
PhantomJS 1.9.8（Linux）：执行20的20成功（0.003秒/0.004秒）
` `StartDate1.IsBetween(StartDate2, EndDate2) || EndDate1.IsBetween(StartDate2, EndDate2)` `
`IsBetween`是类似的
` ` public static bool IsBetween(this DateTime value, DateTime left, DateTime right) { return (value > left && value < right) || (value < left && value > right); }` `
``` ---------------------- | ------- A ------- | ----------- -----------
| ---- B1 ---- |
| ---- ---- B2 |
| ---- ---- B3 |
| ---------- ---------- B4 |
| ---------------- B5 ---------------- |
| ---- ---- B6 |
---------------------- | ------- A ------- | ----------- -----------
| ------ B7 ------- |
| ---------- ----------- B8 |
| ---- B9 ---- |
| ---- ----- B10 |
| -------- -------- B11 |
| ---- ---- B12 |
| ---- ---- B13 |
---------------------- | ------- A ------- | ----------- ----------- ```
``` AND（
（STARTDATE和ENDDATE之间的'start_date'） - 迎合内部和结束date外部
要么
（'end_date'在STARTDATE和ENDDATE之间） - 迎合内部和开始date外部
要么
（STARTDATE在'start_date'和'end_date'之间） - 只有一个date在里面的外部范围需要。
）
```
` `// Current row dates var dateStart = moment("2014-08-01", "YYYY-MM-DD"); var dateEnd = moment("2014-08-30", "YYYY-MM-DD"); // Check with dates above var rangeUsedStart = moment("2014-08-02", "YYYY-MM-DD"); var rangeUsedEnd = moment("2014-08-015", "YYYY-MM-DD"); // Range covers other ? if((dateStart <= rangeUsedStart) && (rangeUsedEnd <= dateEnd)) { return false; } // Range intersects with other start ? if((dateStart <= rangeUsedStart) && (rangeUsedStart <= dateEnd)) { return false; } // Range intersects with other end ? if((dateStart <= rangeUsedEnd) && (rangeUsedEnd <= dateEnd)) { return false; } // All good return true;` `
` `(Startdate BETWEEN '".\$startdate2."' AND '".\$enddate2."') //overlap: starts between start2/end2 OR (Startdate < '".\$startdate2."' AND (enddate = '0000-00-00' OR enddate >= '".\$startdate2."') ) //overlap: starts before start2 and enddate not set 0000-00-00 (still on going) or if enddate is set but higher then startdate2` `
` `someInterval.overlaps( anotherInterval )` `
# java.time＆ThreeTen-Extra
## `Interval`
`org.threeten.extra.Interval`类是方便的，但需要date – 时间矩（ `java.time.Instant`对象）而不是仅包含date的值。 因此，我们继续使用UTC中的一天中的第一个时刻来表示date。
` `Instant start = Instant.parse( "2016-01-01T00:00:00Z" ); Instant stop = Instant.parse( "2016-02-01T00:00:00Z" );` `
` `Interval interval_A = Interval.of( start , stop );` `
` `Instant start_B = Instant.parse( "2016-01-03T00:00:00Z" ); Interval interval_B = Interval.of( start_B , Duration.of( 3 , ChronoUnit.DAYS ) );` `
` `Boolean overlaps = interval_A.overlaps( interval_B );` `
• `abuts`
• `contains`
• `encloses`
• `equals`
• `isAfter`
• `isBefore`
• `overlaps`
` `A = [StartA, EndA] B = [StartB, EndB] [---- DateRange A ------] (True if StartA > EndB) [--- Date Range B -----] [---- DateRange A -----] (True if EndA < StartB) [--- Date Range B ----]` `
` `A = (StartA, EndA) B = (StartB, EndB) (---- DateRange A ------) (True if StartA >= EndB) (--- Date Range B -----) (---- DateRange A -----) (True if EndA <= StartB) (--- Date Range B ----)` `
` `A = [StartA, EndA) B = [StartB, EndB) [---- DateRange A ------) (True if StartA >= EndB) [--- Date Range B -----) [---- DateRange A -----) (True if EndA <= StartB) [--- Date Range B ----)` `
` `A = (StartA, EndA] B = (StartB, EndB] (---- DateRange A ------] (True if StartA >= EndB) (--- Date Range B -----] (---- DateRange A -----] (True if EndA <= StartB) (--- Date Range B ----]` `
` `A = [StartA, EndA) B = [StartB, EndB] [---- DateRange A ------) (True if StartA > EndB) [--- Date Range B -----] [---- DateRange A -----) (True if EndA <= StartB) [--- Date Range B ----]` `
（StartA <🞐EndB）和（EndA>🞐StartB）
` `private Boolean overlap (Timestamp startA, Timestamp endA, Timestamp startB, Timestamp endB) { return (endB == null || startA == null || !startA.after(endB)) && (endA == null || startB == null || !endA.before(startB)); }` `
` ` // Takes a list and returns all records that have overlapping time ranges. public static IEnumerable<T> GetOverlappedTimes<T>(IEnumerable<T> list, Func<T, bool> filter, Func<T,DateTime> start, Func<T, DateTime> end) { // Selects all records that match filter() on left side and returns all records on right side that overlap. var overlap = from t1 in list where filter(t1) from t2 in list where !object.Equals(t1, t2) // Don't match the same record on right side. let in1 = start(t1) let out1 = end(t1) let in2 = start(t2) let out2 = end(t2) where in1 <= out2 && out1 >= in2 let totover = GetMins(in1, out1, in2, out2) select t2; return overlap; } public static void TestOverlap() { var tl1 = new TempTimeEntry() { ID = 1, Name = "Bill", In = "1/1/08 1:00pm".ToDate(), Out = "1/1/08 4:00pm".ToDate() }; var tl2 = new TempTimeEntry() { ID = 2, Name = "John", In = "1/1/08 5:00pm".ToDate(), Out = "1/1/08 6:00pm".ToDate() }; var tl3 = new TempTimeEntry() { ID = 3, Name = "Lisa", In = "1/1/08 7:00pm".ToDate(), Out = "1/1/08 9:00pm".ToDate() }; var tl4 = new TempTimeEntry() { ID = 4, Name = "Joe", In = "1/1/08 3:00pm".ToDate(), Out = "1/1/08 8:00pm".ToDate() }; var tl5 = new TempTimeEntry() { ID = 1, Name = "Bill", In = "1/1/08 8:01pm".ToDate(), Out = "1/1/08 8:00pm".ToDate() }; var list = new List<TempTimeEntry>() { tl1, tl2, tl3, tl4, tl5 }; var overlap = GetOverlappedTimes(list, (TempTimeEntry t1)=>t1.ID==1, (TempTimeEntry tIn) => tIn.In, (TempTimeEntry tOut) => tOut.Out); Console.WriteLine("\nRecords overlap:"); foreach (var tl in overlap) Console.WriteLine("Name:{0} T1In:{1} T1Out:{2}", tl.Name, tl.In, tl.Out); Console.WriteLine("Done"); /* Output: Records overlap: Name:Joe T1In:1/1/2008 3:00:00 PM T1Out:1/1/2008 8:00:00 PM Name:Lisa T1In:1/1/2008 7:00:00 PM T1Out:1/1/2008 9:00:00 PM Done */ }` `
` `CREATE FUNCTION IsOverlapDates ( @startDate1 as datetime, @endDate1 as datetime, @startDate2 as datetime, @endDate2 as datetime ) RETURNS int AS BEGIN DECLARE @Overlap as int SET @Overlap = (SELECT CASE WHEN ( (@startDate1 BETWEEN @startDate2 AND @endDate2) -- caters for inner and end date outer OR (@endDate1 BETWEEN @startDate2 AND @endDate2) -- caters for inner and start date outer OR (@startDate2 BETWEEN @startDate1 AND @endDate1) -- only one needed for outer range where dates are inside. ) THEN 1 ELSE 0 END ) RETURN @Overlap END GO --Execution of the above code DECLARE @startDate1 as datetime DECLARE @endDate1 as datetime DECLARE @startDate2 as datetime DECLARE @endDate2 as datetime DECLARE @Overlap as int SET @startDate1 = '2014-06-01 01:00:00' SET @endDate1 = '2014-06-01 02:00:00' SET @startDate2 = '2014-06-01 01:00:00' SET @endDate2 = '2014-06-01 01:30:00' SET @Overlap = [dbo].[IsOverlapDates] (@startDate1, @endDate1, @startDate2, @endDate2) SELECT Overlap = @Overlap` `
` `public static class NumberExtensionMethods { public static Boolean IsBetween(this Int64 value, Int64 Min, Int64 Max) { if (value >= Min && value <= Max) return true; else return false; } public static Boolean IsBetween(this DateTime value, DateTime Min, DateTime Max) { Int64 numricValue = value.Ticks; Int64 numericStartDate = Min.Ticks; Int64 numericEndDate = Max.Ticks; if (numricValue.IsBetween(numericStartDate, numericEndDate) ) { return true; } return false; } } public static Boolean IsOverlap(DateTime startDate1, DateTime endDate1, DateTime startDate2, DateTime endDate2) { Int64 numericStartDate1 = startDate1.Ticks; Int64 numericEndDate1 = endDate1.Ticks; Int64 numericStartDate2 = startDate2.Ticks; Int64 numericEndDate2 = endDate2.Ticks; if (numericStartDate2.IsBetween(numericStartDate1, numericEndDate1) || numericEndDate2.IsBetween(numericStartDate1, numericEndDate1) || numericStartDate1.IsBetween(numericStartDate2, numericEndDate2) || numericEndDate1.IsBetween(numericStartDate2, numericEndDate2)) { return true; } return false; } if (IsOverlap(startdate1, enddate1, startdate2, enddate2)) { Console.WriteLine("IsOverlap"); }` `
` ` public static boolean checkTimeOverlaps(Date startDate1, Date endDate1, Date startDate2, Date endDate2) { if (startDate1 == null || endDate1 == null || startDate2 == null || endDate2 == null) return false; if ((startDate1.getTime() <= endDate2.getTime()) && (startDate2.getTime() <= endDate1.getTime())) return true; return false; }` `
@Bretana提供的math解决scheme是好的，但忽略了两个具体细节：
1. closures或半开放的时间段
2. 空的间隔
（StartA <= EndB）和（EndA> = StartB）
（StartA <EndB）和（EndA> StartB）
` `MomentInterval a = MomentInterval.between(Instant.now(), Instant.now().plusSeconds(2)); MomentInterval b = a.collapse(); // make b an empty interval out of a System.out.println(a); // [2017-04-10T05:28:11,909000000Z/2017-04-10T05:28:13,909000000Z) System.out.println(b); // [2017-04-10T05:28:11,909000000Z/2017-04-10T05:28:11,909000000Z)` `
` `System.out.println( "startA < endB: " + a.getStartAsInstant().isBefore(b.getEndAsInstant())); // false System.out.println( "endA > startB: " + a.getEndAsInstant().isAfter(b.getStartAsInstant())); // true System.out.println("a overlaps b: " + a.intersects(b)); // a overlaps b: false` `
` `if (StartDate1 > StartDate2) swap(StartDate, EndDate); (StartDate1 <= EndDate2) and (StartDate2 <= EndDate1);` `
“两个date范围相交”的情况包含两种情况 – 第一个date范围在第二个范围内开始，或者第二个date范围在第一个范围内开始。
` `//custom date for example \$d1 = new DateTime("2012-07-08"); \$d2 = new DateTime("2012-07-11"); \$d3 = new DateTime("2012-07-08"); \$d4 = new DateTime("2012-07-15"); //create a date period object \$interval = new DateInterval('P1D'); \$daterange = iterator_to_array(new DatePeriod(\$d1, \$interval, \$d2)); \$daterange1 = iterator_to_array(new DatePeriod(\$d3, \$interval, \$d4)); array_map(function(\$v) use (\$daterange1) { if(in_array(\$v, \$daterange1)) print "Bingo!";}, \$daterange);` `
This was my solution, it returns true when the values don't overlap:
X START 1 Y END 1
A START 2 B END 2
` `TEST1: (X <= A || X >= B) && TEST2: (Y >= B || Y <= A) && TEST3: (X >= B || Y <= A) X-------------Y A-----B TEST1: TRUE TEST2: TRUE TEST3: FALSE RESULT: FALSE --------------------------------------- X---Y A---B TEST1: TRUE TEST2: TRUE TEST3: TRUE RESULT: TRUE --------------------------------------- X---Y A---B TEST1: TRUE TEST2: TRUE TEST3: TRUE RESULT: TRUE --------------------------------------- X----Y A---------------B TEST1: FALSE TEST2: FALSE TEST3: FALSE RESULT: FALSE` `
For ruby I also found this:
` `class Interval < ActiveRecord::Base validates_presence_of :start_date, :end_date # Check if a given interval overlaps this interval def overlaps?(other) (start_date - other.end_date) * (other.start_date - end_date) >= 0 end # Return a scope for all interval overlapping the given interval, including the given interval itself named_scope :overlapping, lambda { |interval| { :conditions => ["id <> ? AND (DATEDIFF(start_date, ?) * DATEDIFF(?, end_date)) >= 0", interval.id, interval.end_date, interval.start_date] }} end` `
Found it here with nice explaination -> http://makandracards.com/makandra/984-test-if-two-date-ranges-overlap-in-ruby-or-rails
Below query gives me the ids for which the supplied date range (start and end dates overlaps with any of the dates (start and end dates) in my table_name
` `select id from table_name where (START_DT_TM >= 'END_DATE_TIME' OR (END_DT_TM BETWEEN 'START_DATE_TIME' AND 'END_DATE_TIME'))` `
The answer is too simple for me so I have created a more generic dynamic SQL statement which checks to see if a person has any overlapping dates.
` `SELECT DISTINCT T1.EmpID FROM Table1 T1 INNER JOIN Table2 T2 ON T1.EmpID = T2.EmpID AND T1.JobID <> T2.JobID AND ( (T1.DateFrom >= T2.DateFrom AND T1.dateFrom <= T2.DateTo) OR (T1.DateTo >= T2.DateFrom AND T1.DateTo <= T2.DateTo) OR (T1.DateFrom < T2.DateFrom AND T1.DateTo IS NULL) ) AND NOT (T1.DateFrom = T2.DateFrom)` `
I had a situation where we had dates instead of datetimes, and the dates could overlap only on start/end. 示例如下：
(Green is the current interval, blue blocks are valid intervals, red ones are overlapping intervals).
` ` (startB <= startA && endB > startA) || (startB >= startA && startB < endA)` `
` `compare the two dates: A = the one with smaller start date, B = the one with bigger start date if(A.end < B.start) return false return true` ` | 5,084 | 16,248 | {"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-2022-21 | latest | en | 0.314846 |
https://www.convertopedia.com/unit-converters/angle-converter/gradian-to-milliradian-converter/ | 1,726,334,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00455.warc.gz | 663,973,203 | 21,606 | Gradian, also known as gon, grad or grade is a unit of measurement of an angle in which the angle of an entire circle is 400 gradians. A right angle is therefore 100 gradians. One gradian equals 1/400 of a full circle.
A Milliradian, often called as mil or mrad is a unit of measurement of an angle which is defined as a thousanth of the radian (0.001 radian), meaning that a full circle of 360 degree equals to 2000π or approximately 6283 milliradians.
### Example 1:
Therefore, 15g = 15 * 15.708 = 235.62 mrad
### Example 2:
Therefore, 50g = 50 * 15.708 = 785.40 mrad
0g0
1g15.708
2g31.416
3g47.124
4g62.832
5g78.540
6g94.248
7g109.956
8g125.664
9g141.372
10g157.08
11g172.788
12g188.496
13g204.204
14g219.912
15g235.62
20g314.16
50g785.40
100g1570.8 | 283 | 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} | 3.578125 | 4 | CC-MAIN-2024-38 | latest | en | 0.89183 |
https://msd.com.ua/a-companion-to-theoretical-econometrics/sample-attrition-and-sample-selection/ | 1,653,328,233,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00726.warc.gz | 467,140,008 | 17,919 | A COMPANION TO Theoretical Econometrics
Sample Attrition and Sample Selection
Missing observations occur frequently in panel data. If individuals are missing randomly, most estimation methods for the balanced panel can be extended in a straightforward manner to the unbalanced panel (e. g. Hsiao, 1986). For instance, suppose that
ditVit = dit [a, + 7 'zit + uit], (16.34)
where dit is an observable scalar indicator variable which denotes whether information about (yit, z 'it) for the ith individual at fth time period is available or not. The indicator variable dit is assumed to depend on a ^-dimensional variables, wit, individual specific effects Xi and an unobservable error term nit,
dit = I(Xi + §'wu + Чи > 0), (16.35)
where I(-) is the indicator function that takes the value of 1 if Xi + S'wit + nit > 0 and 0 otherwise. In other words, the indicator variable dit determines whether (yit, zit) in (16.34) is observed or not (e. g. Hausman and Wise, 1979).
Without sample selectivity, that is dit = 1 for all i and t, (16.31) is the standard variable intercept (or fixed effects) model for panel data discussed in Section 2. With sample selection and if nit and uit are correlated, E(uit | zit, dit = 1) Ф 0. Let 0( ) denote the conditional expectation of uit conditional on dit = 1 and wit, then (16.31) can be written as
Vit = ai + 7' zit + 0(Xi + §'wit) + £n
where E(eit | zit, dit = 1) = 0. The form of the selection function is derived from the joint distribution of u and n. For instance, if u and n are bivariate normal, then we have the Heckman (1979) sample selection model with 0(X, + §'wit) =
Ф(^, + § Wu), where aun denotes the covariance between u and n, ф( ) and Ф( )
Ф(^і + §'Wit)
are standard normal density and distribution, respectively, and the variance of n is normalized to be 1. Therefore, in the presence of sample attrition or selection, regressing Vu on zit using only the observed information is invalidated by two problems. First, the presence of the unobserved effects a i, and second, the "selection bias" arising from the fact that E(uit| zit, dit = 1) = 0(X,- + §wit).
When individual effects are random and the joint distribution function of (u, n, Y, Xi) is known, both the maximum likelihood and two - or multi-step estimators can be derived (e. g. Heckman, 1979; and Ryu, 1998). The resulting estimators are consistent and asymptotically normally distributed. The speed of convergence is proportional to the square root of the sample size. However, if the joint distribution of u and n is misspecified, then even without the presence of
ai, both the maximum likelihood and Heckman (1979) two-step estimators will be inconsistent. This sensitivity of parameter estimate to the exact specification of the error distribution has motivated the interest in semiparametric methods.
The presence of individual effects is easily solved by pairwise differencing those individuals that are observed for two time periods t and s, i. e. who has dit = dis = 1. However, the sample selectivity factors are not eliminated by pairwise differencing. The expected value of yit - yis given dit = 1 and dis = 1 takes the form
E(Уи - yis I dit = 1, diS = 1) = (§it - ZisYj + E[uu - Uis | du = 1, dis = 1]. (16.37)
In general,
QUs = E(Uit - Uis I du = 1, dis = 1) * 0 (16.38)
and are different from each other. If (uit, nit) are independent, identically distributed (iid) and are independent of ai, Ki, § and w, then
Qu = E(Uit | du = 1, dis = 1) = E(Uit | du = 1)
= E(UU Int > - w'it8 - K) = 9 (8'wit + A,-), (16.39)
where the second equality is due to the independence over time assumption of the error vector and the third equality is due to the independence of the errors to the individual effects and the explanatory variables. The function 9() of the single index, S'wit + K, is the same over i and t because of the iid assumption of (Uit, nit), but in general, 9(8'wit + K) * 9(S'wis + Ki) because of the time variation of the scalar index S'wit. However, for an individual i that has S'wit = S'wis and dit = dis = 1, the sample selection effect 9it will be the same in the two periods. Therefore, for this particular individual, time differencing eliminates both the unobserved individual effect and the sample selection effect,
У it - yis = У' (§it - §is) + (£it - Sis). (16.40)
This suggests estimating у by the least squares from a subsample that consists of those observations that satisfy S'wit = 8'wis and dit = dis = 1,
1
X X (§it - §is)(§it - §is)'1{(Wit - Wis)§ = 0}dudis
i=1 1< s<t < Ti N
X X (§it - §is)(yit - yis)1{(Wit - Wis)'8}dudis
i =1 1<s <t <Ti
where Ti denotes the number of ith individual's time series observations.
The estimator (16.41) cannot be directly implemented because 8 is unknown. Moreover, the scalar index 8'wit will typically be continuous if any of the variables
in wit is continuous. Ahn and Powell (1993) note that if 9 is a sufficiently "smooth" function, and 8 is a consistent estimator of 8, observations for which the difference (wit - wis)'8 is close to zero should have 9it - 9is — 0. They propose a two-step procedure. In the first step, consistent semiparameter estimates of the coefficients of the "selection" equation are obtained. The result is used to obtain estimates of the "single index, wJ8," variables characterizing the selectivity bias in the equation of index. The second step of the approach estimates the parameters of the equation of interest by a weighted instrumental variables regression of pairwise differences in dependent variables in the sample on the corresponding differences in explanatory variables; the weights put more emphasis on pairs with w't8 — wit-18.
Kyriazidou (1997) and Honore and Kyriazidou (1998) generalize this concept and propose to estimate the fixed effects sample selection models in two steps: In the first step, estimate 8 by either the Anderson (1970), Chamberlain (1980) conditional maximum likelihood approach or the Manski (1975) maximum score method. In the second step, the estimated 8 is used to estimate y, based on pairs of observations for which dit = dis = 1 and for which (wit - wis)'8 is "close" to zero. This last requirement is operationalized by weighting each pair of observations with a weight that depends inversely on the magnitude of (wit - wis)'8, so that pairs with larger differences in the selection effects receive less weight in the estimation. The Kyriazidou (1997) estimator takes the form:
where K is a kernel density function which tends to zero as the magnitude of its argument increases and hN is a positive constant that decreases to zero as N ^ ^.
Under appropriate regularity conditions, Kyriazidou (1997) shows that ' (16.42) is consistent and asymptotically normally distributed. However, the rate of convergence is slower than the standard square root of the sample size.
2 Conclusions
There is an explosion of techniques and procedures for the analysis of panel data (e. g. Matyas and Sevestre, 1996). In this chapter we have discussed some popular panel data models. We did not discuss issues of duration and count data models (e. g. Cameron and Trivedi, 1998; Heckman and Singer, 1984; Lancaster, 1990; Lancaster and Intrator, 1998), simulation-based inference (e. g. Gourieroux and Monfort, 1993), specification analysis (e. g. Baltagi and Li, 1995; Lee, 1987; Li and Hsiao, 1998; Maddala, 1995; Wooldridge, 1995), measurement errors (e. g. Biorn, 1992; Griliches and Hausman, 1984; Hsiao, 1991; Hsiao and Taylor, 1991) pseudo panels or matched samples (e. g. Deaton, 1985; Moffit, 1993; Peracchi and Welsch,
1995; Verbeek, 1992), etc. In general, there does not exist a panacea for panel data analysis. It appears more fruitful to explicitly recognize the limitations of the data and focus attention on providing solutions for a specific type of model. A specific model often contains specific structural information that can be exploited. However, the power of panel data depends on the validity of the assumptions upon which the statistical methods have been built (e. g. Griliches, 1979).
Notes
* This work was supported in part by National Science Foundation grant SBR96-19330. I would like to thank two referees for helpful comments.
1 Normality is made for ease of relating sampling approach and Bayesian approach estimators. It is not required.
2 See Chamberlain (1984), Hausman and Taylor (1981) for the approaches of estimating models when u and є are correlated.
3 Under smooth conditions, Horowitz (1992) proposed a smoothed maximum score estimator that has a n~2/5 rate of convergence. With even stronger conditions Lee (1999) is able to propose a root-n consistent semiparametric estimator.
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Lesson 1 A How to Modeling Skills Exercise
My solution:
To start I used a recessed slot and tab arrangement with a 4-40 flat head screw which achieves the objective, however, if this were placed on a shaker table to top would vibrate badly.
To alleviate this problem I put in a series of tongue and groves to pin it together both at the bottom and along the sides. Along the bottom are a set of simple slots with chamfers in which a series of tongue features fit into the slot with a 2 degree draft to ensure the fit without interference.
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I considered using a dove tail type of arrangement along the corners but decided against it because if built such a solution would require exceptionally tight tolerances and most likely result in 4 points of contact which as stated is a problem.
Objective: mechanically assemble a hollow tetrahedron
Requirements and Instructions to get started:
Requirements:
• Utilize the inherit origin plain's and axis that are provided in the native file of your CAD software. Every sketch used to build the part must directly or indirectly reference the origin, plain's or axis. Every part must utilize the origin, plains or axis for the assembly process.
• Robust design: The model must be scale-able. This will demonstrate the robustness of the model when changes are made to the model. Method Suggestions: skeleton modeling, parametric associations.
• Mechanical means of assembly using pins, screws or other means of assembly such that when completed the results would have a high probability of providing a tight fitting assembly that would not fall part and may be constructed in such a way that once completed, might not be able to be disassembled with out being destroyed.
Instructions to start:
• Plan out how you will control the size of the tetrahedron using skeleton techniques or other means as suggested above. For more on skeleton modeling click here.
• Create a hollow tetrahedron and section it into 4 parts. Hint, this is one of the rare times you will need to use an offset work plane from the origin planes to achieve the required geometry.
• Remember, the edges of the resultant parts will need to be trimmed to allow the edges to be re-assembled such that they do not create an interference. See the examples shown | 545 | 2,702 | {"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} | 2.796875 | 3 | CC-MAIN-2021-39 | latest | en | 0.940516 |
https://www.litscape.com/word_analysis/hectares | 1,606,270,501,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00529.warc.gz | 746,721,119 | 11,849 | # Definition of hectares
## "hectares" in the noun sense
### 1. hectare
abbreviated `ha') a unit of surface area equal to 100 ares (or 10,000 square meters)
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WordNet®. Princeton University. 2010.
# hectares in Scrabble®
The word hectares is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.
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## Seven Letter Word Alert: (14 words)
cerates, chaster, cheater, creates, etchers, heaters, hectare, reaches, recheat, reheats, retches, reteach, teacher, teaches
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(89 = 39 + 50)
CHEATER
(89 = 39 + 50)
ETCHERS
(89 = 39 + 50)
ETCHERS
(89 = 39 + 50)
CHEATER
(89 = 39 + 50)
CHEATER
(89 = 39 + 50)
REACHES
(89 = 39 + 50)
CHASTER
(89 = 39 + 50)
TEACHER
(89 = 39 + 50)
HECTARE
(89 = 39 + 50)
RETEACH
(89 = 39 + 50)
RECHEAT
(89 = 39 + 50)
TEACHER
(89 = 39 + 50)
TEACHER
(89 = 39 + 50)
RECHEAT
(89 = 39 + 50)
HECTARE
(89 = 39 + 50)
HECTARE
(89 = 39 + 50)
HECTARE
(89 = 39 + 50)
HECTARE
(89 = 39 + 50)
HECTARE
(89 = 39 + 50)
TEACHER
(89 = 39 + 50)
TEACHER
(89 = 39 + 50)
RETEACH
(89 = 39 + 50)
RETEACH
(89 = 39 + 50)
RECHEAT
(89 = 39 + 50)
RETEACH
(89 = 39 + 50)
RETEACH
(89 = 39 + 50)
REACHES
(89 = 39 + 50)
CHASTER
(89 = 39 + 50)
TEACHES
(89 = 39 + 50)
RETCHES
(89 = 39 + 50)
RETCHES
(89 = 39 + 50)
REACHES
(89 = 39 + 50)
RETCHES
(89 = 39 + 50)
RETCHES
(89 = 39 + 50)
REACHES
(89 = 39 + 50)
RETCHES
(89 = 39 + 50)
REACHES
(89 = 39 + 50)
RECHEAT
(89 = 39 + 50)
RETEACH
(89 = 39 + 50)
HECTARE
(88 = 38 + 50)
REHEATS
(86 = 36 + 50)
ETCHERS
(86 = 36 + 50)
TEACHES
(86 = 36 + 50)
HECTARE
(86 = 36 + 50)
ETCHERS
(86 = 36 + 50)
TEACHER
(86 = 36 + 50)
RECHEAT
(86 = 36 + 50)
RETCHES
(86 = 36 + 50)
CREATES
(86 = 36 + 50)
RECHEAT
(86 = 36 + 50)
RETEACH
(86 = 36 + 50)
REACHES
(86 = 36 + 50)
RETEACH
(86 = 36 + 50)
CHEATER
(86 = 36 + 50)
CHASTER
(86 = 36 + 50)
CREATES
(86 = 36 + 50)
CERATES
(86 = 36 + 50)
CERATES
(86 = 36 + 50)
HECTARE
(86 = 36 + 50)
TEACHER
(84 = 34 + 50)
TEACHES
(84 = 34 + 50)
HECTARES
(84)
RETCHES
(84 = 34 + 50)
HECTARES
(84)
RETEACH
(84 = 34 + 50)
REACHES
(84 = 34 + 50)
HEATERS
(83 = 33 + 50)
HEATERS
(83 = 33 + 50)
HEATERS
(83 = 33 + 50)
HEATERS
(83 = 33 + 50)
HEATERS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
HEATERS
(83 = 33 + 50)
HEATERS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
REHEATS
(83 = 33 + 50)
CHEATER
(82 = 32 + 50)
HECTARE
(82 = 32 + 50)
TEACHES
(82 = 32 + 50)
HECTARE
(82 = 32 + 50)
CHEATER
(82 = 32 + 50)
TEACHER
(82 = 32 + 50)
CHASTER
(82 = 32 + 50)
RECHEAT
(82 = 32 + 50)
RETEACH
(82 = 32 + 50)
CHASTER
(82 = 32 + 50)
RETCHES
(82 = 32 + 50)
REACHES
(82 = 32 + 50)
ETCHERS
(82 = 32 + 50)
RETEACH
(82 = 32 + 50)
CREATES
(80 = 30 + 50)
HEATERS
(80 = 30 + 50)
CREATES
(80 = 30 + 50)
RECHEAT
(80 = 30 + 50)
HEATERS
(80 = 30 + 50)
CREATES
(80 = 30 + 50)
CREATES
(80 = 30 + 50)
REHEATS
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
CERATES
(80 = 30 + 50)
RETEACH
(80 = 30 + 50)
CHASTER
(80 = 30 + 50)
CREATES
(80 = 30 + 50)
CHEATER
(80 = 30 + 50)
CHEATER
(80 = 30 + 50)
REHEATS
(80 = 30 + 50)
ETCHERS
(80 = 30 + 50)
CHASTER
(80 = 30 + 50)
HECTARE
(80 = 30 + 50)
CREATES
(80 = 30 + 50)
CREATES
(80 = 30 + 50)
RETEACH
(78 = 28 + 50)
ETCHERS
(78 = 28 + 50)
RETEACH
(78 = 28 + 50)
CHEATER
(78 = 28 + 50)
CHASTER
(78 = 28 + 50)
CHASTER
(78 = 28 + 50)
HECTARE
(78 = 28 + 50)
HECTARE
(78 = 28 + 50)
TEACHER
(78 = 28 + 50)
CHASTER
(78 = 28 + 50)
CHASTER
(78 = 28 + 50)
TEACHER
(78 = 28 + 50)
TEACHER
(78 = 28 + 50)
RETEACH
(78 = 28 + 50)
RETCHES
(78 = 28 + 50)
REACHES
(78 = 28 + 50)
HEATERS
(78 = 28 + 50)
# hectares in Words With Friends™
The word hectares is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.
HECTARE
(113 = 78 + 35)
## Seven Letter Word Alert: (14 words)
cerates, chaster, cheater, creates, etchers, heaters, hectare, reaches, recheat, reheats, retches, reteach, teacher, teaches
HECTARES
(90)
HECTARES
(90)
HECTARES
(81)
HECTARES
(69)
HECTARES
(64)
HECTARES
(56)
HECTARES
(52)
HECTARES
(52)
HECTARES
(51)
HECTARES
(51)
HECTARES
(51)
HECTARES
(51)
HECTARES
(45)
HECTARES
(45)
HECTARES
(42)
HECTARES
(38)
HECTARES
(34)
HECTARES
(30)
HECTARES
(30)
HECTARES
(30)
HECTARES
(30)
HECTARES
(30)
HECTARES
(30)
HECTARES
(28)
HECTARES
(28)
HECTARES
(28)
HECTARES
(26)
HECTARES
(26)
HECTARES
(26)
HECTARES
(26)
HECTARES
(26)
HECTARES
(26)
HECTARES
(26)
HECTARES
(26)
HECTARES
(23)
HECTARES
(21)
HECTARES
(21)
HECTARES
(20)
HECTARES
(18)
HECTARES
(18)
HECTARES
(18)
HECTARES
(17)
HECTARES
(17)
HECTARES
(17)
HECTARES
(17)
HECTARES
(17)
HECTARES
(16)
HECTARES
(16)
HECTARES
(16)
HECTARES
(15)
HECTARES
(15)
HECTARES
(15)
HECTARES
(15)
HECTARES
(15)
HECTARES
(15)
HECTARES
(14)
HECTARES
(14)
HECTARES
(14)
HECTARE
(113 = 78 + 35)
RETEACH
(101 = 66 + 35)
TEACHER
(101 = 66 + 35)
RECHEAT
(101 = 66 + 35)
REACHES
(101 = 66 + 35)
CHASTER
(101 = 66 + 35)
CHEATER
(101 = 66 + 35)
RETCHES
(101 = 66 + 35)
TEACHER
(101 = 66 + 35)
RETCHES
(101 = 66 + 35)
REACHES
(101 = 66 + 35)
ETCHERS
(101 = 66 + 35)
TEACHES
(101 = 66 + 35)
TEACHES
(101 = 66 + 35)
RECHEAT
(95 = 60 + 35)
REACHES
(95 = 60 + 35)
CHEATER
(95 = 60 + 35)
RECHEAT
(95 = 60 + 35)
REACHES
(95 = 60 + 35)
TEACHES
(95 = 60 + 35)
CREATES
(95 = 60 + 35)
RETCHES
(95 = 60 + 35)
CHEATER
(95 = 60 + 35)
CHASTER
(95 = 60 + 35)
TEACHES
(95 = 60 + 35)
REACHES
(95 = 60 + 35)
CHASTER
(95 = 60 + 35)
RETCHES
(95 = 60 + 35)
ETCHERS
(95 = 60 + 35)
RETEACH
(95 = 60 + 35)
RETEACH
(95 = 60 + 35)
CERATES
(95 = 60 + 35)
TEACHES
(95 = 60 + 35)
TEACHER
(95 = 60 + 35)
TEACHER
(95 = 60 + 35)
ETCHERS
(95 = 60 + 35)
RETCHES
(95 = 60 + 35)
TEACHER
(95 = 60 + 35)
HECTARES
(90)
HECTARES
(90)
HECTARE
(89 = 54 + 35)
ETCHERS
(89 = 54 + 35)
CERATES
(89 = 54 + 35)
CHEATER
(89 = 54 + 35)
RECHEAT
(89 = 54 + 35)
ETCHERS
(89 = 54 + 35)
CREATES
(89 = 54 + 35)
RETEACH
(89 = 54 + 35)
CHASTER
(89 = 54 + 35)
RECHEAT
(89 = 54 + 35)
REHEATS
(86 = 51 + 35)
HEATERS
(86 = 51 + 35)
RECHEAT
(83 = 48 + 35)
HECTARE
(83 = 48 + 35)
HECTARE
(83 = 48 + 35)
RETCHES
(83 = 48 + 35)
TEACHER
(83 = 48 + 35)
RETCHES
(83 = 48 + 35)
TEACHER
(83 = 48 + 35)
TEACHER
(83 = 48 + 35)
RETEACH
(83 = 48 + 35)
RETEACH
(83 = 48 + 35)
CHASTER
(83 = 48 + 35)
RETEACH
(83 = 48 + 35)
ETCHERS
(83 = 48 + 35)
HECTARE
(83 = 48 + 35)
ETCHERS
(83 = 48 + 35)
ETCHERS
(83 = 48 + 35)
TEACHES
(83 = 48 + 35)
ETCHERS
(83 = 48 + 35)
RETEACH
(83 = 48 + 35)
TEACHES
(83 = 48 + 35)
RETEACH
(83 = 48 + 35)
CHASTER
(83 = 48 + 35)
TEACHES
(83 = 48 + 35)
TEACHES
(83 = 48 + 35)
CHASTER
(83 = 48 + 35)
HECTARE
(83 = 48 + 35)
TEACHER
(83 = 48 + 35)
RETCHES
(83 = 48 + 35)
CHEATER
(83 = 48 + 35)
CHEATER
(83 = 48 + 35)
RECHEAT
(83 = 48 + 35)
RECHEAT
(83 = 48 + 35)
REACHES
(83 = 48 + 35)
REACHES
(83 = 48 + 35)
RECHEAT
(83 = 48 + 35)
CHASTER
(83 = 48 + 35)
REACHES
(83 = 48 + 35)
CHEATER
(83 = 48 + 35)
CHASTER
(83 = 48 + 35)
REACHES
(83 = 48 + 35)
RETCHES
(83 = 48 + 35)
CHEATER
(83 = 48 + 35)
HECTARE
(83 = 48 + 35)
HECTARE
(83 = 48 + 35)
CHEATER
(83 = 48 + 35)
HECTARES
(81)
HEATERS
(80 = 45 + 35)
TEACHER
(77 = 42 + 35)
TEACHES
(77 = 42 + 35)
RECHEAT
(77 = 42 + 35)
TEACHES
(77 = 42 + 35)
HECTARE
(77 = 42 + 35)
ETCHERS
(77 = 42 + 35)
TEACHES
(77 = 42 + 35)
HECTARE
(77 = 42 + 35)
HECTARE
(77 = 42 + 35)
RECHEAT
(77 = 42 + 35)
RETEACH
(77 = 42 + 35)
RECHEAT
(77 = 42 + 35)
CREATES
(77 = 42 + 35)
ETCHERS
(77 = 42 + 35)
REACHES
(77 = 42 + 35)
ETCHERS
(77 = 42 + 35)
TEACHER
(77 = 42 + 35)
REACHES
(77 = 42 + 35)
TEACHER
(77 = 42 + 35)
RETCHES
(77 = 42 + 35)
RETEACH
(77 = 42 + 35)
REACHES
(77 = 42 + 35)
RETEACH
(77 = 42 + 35)
RETCHES
(77 = 42 + 35)
RETCHES
(77 = 42 + 35)
RECHEAT
(77 = 42 + 35)
TEACHER
(77 = 42 + 35)
ETCHERS
(77 = 42 + 35)
RETEACH
(77 = 42 + 35)
TEACHES
(77 = 42 + 35)
RETCHES
(77 = 42 + 35)
HECTARE
(77 = 42 + 35)
REACHES
(77 = 42 + 35)
HECTARE
(77 = 42 + 35)
CERATES
(77 = 42 + 35)
CHASTER
(77 = 42 + 35)
CHASTER
(77 = 42 + 35)
CHASTER
(77 = 42 + 35)
CHASTER
(77 = 42 + 35)
CERATES
(77 = 42 + 35)
CHEATER
(77 = 42 + 35)
CREATES
(77 = 42 + 35)
CHEATER
(77 = 42 + 35)
CREATES
(77 = 42 + 35)
CERATES
(77 = 42 + 35)
CHEATER
(77 = 42 + 35)
CHEATER
(77 = 42 + 35)
RETEACH
(75 = 40 + 35)
CREATES
(75 = 40 + 35)
CERATES
(75 = 40 + 35)
CERATES
(75 = 40 + 35)
CREATES
(75 = 40 + 35)
CHEATER
(75 = 40 + 35)
CREATES
(75 = 40 + 35)
HECTARE
(75 = 40 + 35)
CERATES
(75 = 40 + 35)
RECHEAT
(75 = 40 + 35)
CHASTER
(75 = 40 + 35)
ETCHERS
(75 = 40 + 35)
REHEATS
(74 = 39 + 35)
REHEATS
(74 = 39 + 35)
REHEATS
(74 = 39 + 35)
HEATERS
(74 = 39 + 35)
HEATERS
(74 = 39 + 35)
HEATERS
(74 = 39 + 35)
REHEATS
(71 = 36 + 35)
REHEATS
(71 = 36 + 35)
REACHES
(71 = 36 + 35)
REHEATS
(71 = 36 + 35)
CHEATER
(71 = 36 + 35)
TEACHES
(71 = 36 + 35)
CERATES
(71 = 36 + 35)
RETCHES
(71 = 36 + 35)
CERATES
(71 = 36 + 35)
TEACHER
(71 = 36 + 35)
CERATES
(71 = 36 + 35)
HECTARE
(71 = 36 + 35)
CREATES
(71 = 36 + 35)
CERATES
(71 = 36 + 35)
CREATES
(71 = 36 + 35)
RETEACH
(71 = 36 + 35)
HEATERS
(71 = 36 + 35)
CHASTER
(71 = 36 + 35)
HEATERS
(71 = 36 + 35)
CREATES
(71 = 36 + 35)
CERATES
(71 = 36 + 35)
HEATERS
(71 = 36 + 35)
CERATES
(71 = 36 + 35)
CREATES
(71 = 36 + 35)
CREATES
(71 = 36 + 35)
CREATES
(71 = 36 + 35)
HECTARES
(69)
REHEATS
(68 = 33 + 35)
HEATERS
(68 = 33 + 35)
HEATERS
(68 = 33 + 35)
HEATERS
(68 = 33 + 35)
REHEATS
(68 = 33 + 35)
REHEATS
(68 = 33 + 35)
REHEATS
(68 = 33 + 35)
REHEATS
(68 = 33 + 35)
HEATERS
(68 = 33 + 35)
# Word Growth involving hectares
## Shorter words in hectares
he hectare
ar are tare hectare
re are tare hectare
ar tar tare hectare
ta tar tare hectare
ar are ares tares
re are ares tares
ar are tare tares
re are tare tares
ar tar tare tares
ta tar tare tares
## Longer words containing hectares
(No longer words found) | 6,240 | 12,003 | {"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} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.631913 |
https://nrich.maths.org/public/leg.php?code=-68&cl=2&cldcmpid=2103 | 1,508,455,702,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823482.25/warc/CC-MAIN-20171019231858-20171020011858-00201.warc.gz | 749,987,364 | 9,223 | # Search by Topic
#### Resources tagged with Visualising similar to Taking Steps:
Filter by: Content type:
Stage:
Challenge level:
### Taking Steps
##### Stage: 2 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
### Circles, Circles
##### Stage: 1 and 2 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Dicey
##### Stage: 2 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### Green Cube, Yellow Cube
##### Stage: 2 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
### Midpoint Triangle
##### Stage: 2 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Hexagon Transformations
##### Stage: 2 Challenge Level:
Can you cut a regular hexagon into two pieces to make a parallelogram? Try cutting it into three pieces to make a rhombus!
### Three Cubed
##### Stage: 2 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Endless Noughts and Crosses
##### Stage: 2 Challenge Level:
An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5.
### A Chain of Eight Polyhedra
##### Stage: 2 Challenge Level:
Can you arrange the shapes in a chain so that each one shares a face (or faces) that are the same shape as the one that follows it?
### More Building with Cubes
##### Stage: 2 Challenge Level:
Here are more buildings to picture in your mind's eye. Watch out - they become quite complicated!
### Counter Roundup
##### Stage: 2 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
### A Puzzling Cube
##### Stage: 2 Challenge Level:
Here are the six faces of a cube - in no particular order. Here are three views of the cube. Can you deduce where the faces are in relation to each other and record them on the net of this cube?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### Move Those Halves
##### Stage: 2 Challenge Level:
For this task, you'll need an A4 sheet and two A5 transparent sheets. Decide on a way of arranging the A5 sheets on top of the A4 sheet and explore ...
### Peg Rotation
##### Stage: 2 Challenge Level:
Can you work out what kind of rotation produced this pattern of pegs in our pegboard?
### Matchsticks
##### Stage: 2 Challenge Level:
Reasoning about the number of matches needed to build squares that share their sides.
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### The Path of the Dice
##### Stage: 2 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
### Wrapping Presents
##### Stage: 2 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
### Three Squares
##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### Making Maths: Rolypoly
##### Stage: 1 and 2 Challenge Level:
Paint a stripe on a cardboard roll. Can you predict what will happen when it is rolled across a sheet of paper?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 25 - Pentominoes
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 26 - Old Chestnut
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 19 - Working Men
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### Folding, Cutting and Punching
##### Stage: 2 Challenge Level:
Exploring and predicting folding, cutting and punching holes and making spirals.
### World of Tan 20 - Fractions
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### World of Tan 28 - Concentrating on Coordinates
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### Holes
##### Stage: 1 and 2 Challenge Level:
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
##### Stage: 2 Challenge Level:
This practical problem challenges you to make quadrilaterals with a loop of string. You'll need some friends to help!
### Square to L
##### Stage: 2 Challenge Level:
Find a way to cut a 4 by 4 square into only two pieces, then rejoin the two pieces to make an L shape 6 units high.
### Fractional Triangles
##### Stage: 2 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Coin Cogs
##### Stage: 2 Challenge Level:
Can you work out what is wrong with the cogs on a UK 2 pound coin?
### Trace the Edges
##### Stage: 2 Challenge Level:
On which of these shapes can you trace a path along all of its edges, without going over any edge twice?
### A Square in a Circle
##### Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
### Jomista Mat
##### Stage: 2 Challenge Level:
Looking at the picture of this Jomista Mat, can you decribe what you see? Why not try and make one yourself?
### Right or Left?
##### Stage: 2 Challenge Level:
Which of these dice are right-handed and which are left-handed?
### Triangular Faces
##### Stage: 2 Challenge Level:
This problem invites you to build 3D shapes using two different triangles. Can you make the shapes from the pictures? | 1,987 | 8,453 | {"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} | 4.1875 | 4 | CC-MAIN-2017-43 | latest | en | 0.905766 |
https://discourse.julialang.org/t/use-linear-interpolation-between-two-lla-coordinate-system-points/28296 | 1,652,829,762,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00097.warc.gz | 278,513,467 | 7,423 | # Use linear interpolation between two LLA coordinate system points
I have the following simple piece of code to find the position between two coordinate points using linear interpolation. I’m not sure if this is a good / feasible solution or not. So what do you think?
``````mutable struct LLA
lat::Float64
lon::Float64
alt::Float64
end
LLA(lat,lon) = LLA(lat,lon,0.0)
function linearInterpolation ( startLoc::LLA, endLoc::LLA, fraction::Float )
startLat = startLoc.lat
startLon = startLoc.lon
endLat = endLoc.lat
endLon = endLoc.lon
return LLA( startLat + fraction * (endLat - startLat), startLon + fraction * (endLon - startLon) )
end
``````
Linear interpolation of the coordinates will only be approximately correct for points that are near each other, and it fails completely if you cross the date line. For larger distances you need to account for the curvature of the earth by calculating the great circle distance - and for that you need to use spherical trigonometry (instead of ordinary planar). The standard approach for lat-lon coordinates is to use the haversine formula, or alternatively n-vectors. Then I guess you could just interpolate the two altitudes if that makes sense for your application.
5 Likes
The most accurate way to do this would be to compute a geodesic line via the inverse geodesic problem, and use this object to interpolate between the lat-lon endpoints. This is the method used in the GeographicLib waypoints example in python (see also the underlying C++ documentation).
I don’t think we have a registered julia package which can do this in the most natural way yet. The obvious candidates would be Proj4.jl (wraps the C proj.4 library) and Geodesy.jl (pure julia, with some algorithms ported from GeographicLib). This may eventually be in Geodesy.jl (see https://github.com/JuliaGeo/Geodesy.jl/issues/40) but in the meantime you could try out https://github.com/anowacki/GeographicLib.jl or `Proj4.jl` with a similar two-step procedure:
• Compute the azimuth and great circle distance along the geodesic from the first to second point using `GeographicLib.inverse` or `Proj4.geod_inverse`
• Interpolate along this geodesic from the first point toward the second using `GeographicLib.forward` or `Proj4.geod_destination`.
@visr @anowacki is there a better alternative?
All the above is for interpolating the lat-lon components. For the altitude you can probably just use linear interpolation based on the fractional distance along the geodesic.
5 Likes
@NiclasMattsson, Is there a great circle route?
GMT.jl can do it.
2 Likes
Thanks for giving that explanation, @Chris_Foster. That is indeed how I would do this. GeographicLib.jl now also exposes the waypoint functionality of the original library with `waypoints`, though this doesn’t include fractional distances. (Perhaps an additional option could be added.)
As Chris says, this should be added to Geodesy if I can get round to it at some point this coming semester.
2 Likes
The Haversine method suggested by Niclas is for great circles on the sphere, so if the accuracy you need is not high, or you are only interested in spheres, that will do the job fine. (Feel free to steal the implementation of `delta` and `step` in https://github.com/anowacki/assorted-julia-modules/blob/master/SphericalGeom.jl). If accuracy on the real Earth is required, the other routes explained above with Proj or GeographicLib are needed.
1 Like
Here’s a solution which uses the original `GeographicLib` C++ library via Cxx.jl:
``````using Cxx
using Libdl
# Workaround weird issue with dynamic linking
Libdl.dlopen("/usr/lib/x86_64-linux-gnu/libGeographic.so", Libdl.RTLD_GLOBAL)
cxx"""
#include <GeographicLib/Geodesic.hpp>
#include <GeographicLib/Constants.hpp>
#include <GeographicLib/GeodesicLine.hpp>
using namespace GeographicLib;
"""
# Get `(lat,lon)` position `d` meters along a GeodesicLine
function geoline_position(line, d)
lat = Ref(0.0)
lon = Ref(0.0)
@cxx line->Position(d, lat, lon)
lat[], lon[]
end
latlon1 = (40.1, 116.6)
latlon2 = (37.6, -122.4)
geod = icxx"Geodesic(Constants::WGS84_a(), Constants::WGS84_f());"
line = @cxx geod->InverseLine(latlon1..., latlon2...)
dist = @cxx line->Distance()
numpoints = ceil(Int, dist/1000e3)
positions = geoline_position.(Ref(line), LinRange(0, dist, numpoints))
``````
2 Likes
I was not sure of this but checked, GMT only does this on the sphere. If that’s enough, with the master version, to interpolate at 5000 km steps along a line from long, lat (0,0) to (45,45)
``````using GMT
julia> project(start_pt=(0,0), end_pt=(45,45), step=5000, km=true)
1-element Array{GMT.GMTdataset,1}:
GMT.GMTdataset([-3.1805546814635168e-15 3.1805546814635168e-15 0.0; 29.970589500178583 35.24036543346766 5000.0; 44.99999999999997 44.999999999999986 6671.703118513764], String[], "", String[], "", "")
``````
with GMT.jl 0.11 version the syntax will have to be a bit different
``````julia> project(start_pt=(0,0), end_pt=(45,45), G=5000, Q=true)
``````
1 Like | 1,331 | 5,007 | {"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} | 3.40625 | 3 | CC-MAIN-2022-21 | latest | en | 0.843338 |
https://mathoverflow.net/questions/306896/why-are-modular-curves-non-trivial-covers-of-the-j-line | 1,606,719,515,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00105.warc.gz | 378,890,057 | 28,764 | # Why are modular curves non-trivial covers of the $j$-line
This is a very soft question.
Let $n\geq 1$ and let $Y(n)$ be the (open) modular curve associated to $\Gamma(n)\subset SL_2(\mathbb{Z})$. Interpreted correctly, $Y(n)\to Y(d)$ is finite etale, whenever $d$ is a positive integer dividing $n$.
Now, it is a priori possible that $Y(n)\to Y(d)$ is a "trivial" finite etale cover, i.e., it has a section.
But, of course, it isn't. With a view towards higher-dimensional analogues of the modular curves I would like to know:
Why is $Y(n)\to Y(d)$ a non-trivial finite etale cover?
I'm looking for an argument which is as "soft" as possible, and could be applied to higher-dimensional situations (or even situations different from the moduli of abelian varieties).
For instance, the genus of $Y(n)$ grows with $n$, so these covers can't be trivial. But is there a "softer", slightly more direct, argument?
• Maybe I am missing something: a morphism of irreducible curves cannot have a section unless it's an isomorphism. – Pop Jul 26 '18 at 12:02
• Oops, I should have said "birational" instead of "isomorphism". (But the point remains the same.) – Pop Jul 26 '18 at 12:27
• Yes, but let's not use that $Y(n)$ is irreducible. (In fact, if one defines level $n$ structure without demanding compatibility with the Weil pairing, then the resulting modular curve is not necessarily irreducible.) Basically, I am looking for an argument which only uses the moduli-interpretation somehow. – Dasz Jul 26 '18 at 13:05
Since you are looking for moduli interpretations, we may either work with stacks, or assume $d$ is large enough that $Y(d)$ is representable (i.e., at least 3).
Perhaps the easiest answer is that in a small neighborhood of the infinite cusp, $Y(n)$ is ramified over $Y(1)$ with degree $n$, while $Y(d)$ is ramified with degree $d$. In other words, there is a $\mathbb{C}((q))$-point of $Y(n)$ whose image in $Y(d)$ factors through a map from the subfield $\mathbb{C}((q^{n/d}))$, and the point from this subfield does not lift to $Y(n)$. If there were a section, all points of $Y(d)$ would have lifts.
A moduli-theoretic interpretation of this ramification is that if we are given an elliptic curve with level $n$ structure $(E,a,b)$, monodromy around the infinite cusp induces a shearing automorphism $(E,a,b) \mapsto (E,a,a+b)$ of order $n$, and this automorphism induces an automorphism $(E,n\frac{a}{d},n\frac{b}{d}) \mapsto (E,n\frac{a}{d},n\frac{a+b}{d})$ of order $d$ on the corresponding curve with level $d$ structure. | 719 | 2,549 | {"found_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} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.903829 |
https://www.physicsforums.com/threads/cmos-circuit-jsim-solving-coding-help.709401/ | 1,716,623,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00462.warc.gz | 791,080,163 | 16,353 | # CMOS Circuit JSIM: Solving & Coding Help
• Engineering
• DODGEVIPER13
In summary, CMOS Circuit JSIM is a software tool that helps simulate and analyze digital circuits using CMOS technology. It uses advanced algorithms and simulations to solve circuit problems and offers coding help for debugging and optimization. It can be used for both analog and digital circuits and is suitable for beginners with some basic knowledge.
DODGEVIPER13
## The Attempt at a Solution
I have Jsim installed and working and have the circuit correctly done. I however don't understand how the coding works in Jsim. I essentially need the equivalent circuit in Jsim code. I need help doing this as I don't understand Jsim at all.
#### Attachments
• EPSON025.jpg
7.4 KB · Views: 428
• physicsforumshelponlab1.jpg
6 KB · Views: 516
Do you have to do this in Jsim, or just work out the output for different values of A, B and C?
Yah we have to do this in Jsim
## 1. What is CMOS Circuit JSIM?
CMOS Circuit JSIM is a software tool used for simulating and analyzing digital circuits, specifically those using Complementary Metal-Oxide-Semiconductor (CMOS) technology. It allows users to create and test complex circuit designs, and provides coding help for debugging and optimization.
## 2. How does CMOS Circuit JSIM solve circuit problems?
CMOS Circuit JSIM uses advanced algorithms and simulation techniques to solve circuit problems. It models the behavior of the circuit using mathematical equations and simulations, allowing users to test and analyze different scenarios and identify potential issues.
## 3. Can CMOS Circuit JSIM be used for both analog and digital circuits?
Yes, CMOS Circuit JSIM can be used for both analog and digital circuits. It is particularly useful for digital circuits that use CMOS technology, but it also has features for modeling and analyzing analog circuits.
## 4. How can I get coding help while using CMOS Circuit JSIM?
CMOS Circuit JSIM has a built-in coding help feature that provides suggestions and tips for debugging and optimizing your circuit design. Additionally, there are online resources and forums where you can ask for help from other users and experts in the field.
## 5. Is CMOS Circuit JSIM suitable for beginners?
CMOS Circuit JSIM can be used by beginners, but it does require some basic knowledge of digital circuits and coding. The user interface is designed to be user-friendly, and there are tutorials and guides available to help beginners get started.
• Engineering and Comp Sci Homework Help
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1K | 719 | 3,113 | {"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-2024-22 | latest | en | 0.856037 |
https://estebantorreshighschool.com/equation-help/how-to-solve-a-multi-step-equation.html | 1,611,826,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704839214.97/warc/CC-MAIN-20210128071759-20210128101759-00197.warc.gz | 324,853,402 | 29,781 | ## What are the steps to solving multi step equations?
Step-by-Step Solution: Combine like terms on both sides. Subtract 6 y 6y 6y on both sides to keep the variable y to the left side only. Add 11 to both sides of the equation. Finally, divide both sides by −10 to get the solution.
## What are the 4 steps to solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal.
## How do you solve multi step equations with like terms?
Multi-step Equations with Like TermsProcedure to Solve Equations:Step 1: Remove any parentheses by using the Distributive Property or the Multiplication Property of Equality.Step 2: Simplify each side of the equation by combining like terms.Step 3: Isolate the begin{align*}axend{align*} term. Step 4: Isolate the variable.
## How do you solve an equation with multiple variables?
Section 2-4 : Equations With More Than One VariableMultiply both sides by the LCD to clear out any fractions.Simplify both sides as much as possible. Move all terms containing the variable we’re solving for to one side and all terms that don’t contain the variable to the opposite side.
## What is the golden rule for solving equations?
Do unto one side of the equation, what you do to the other! When solving math equations, we must always keep the ‘scale’ (or equation) balanced so that both sides are ALWAYS equal.
## What are the rules for solving equations?
A General Rule for Solving EquationsSimplify each side of the equation by removing parentheses and combining like terms.Use addition or subtraction to isolate the variable term on one side of the equation.Use multiplication or division to solve for the variable.
You might be interested: Equation for spring force
## What is an example of a multi step equation?
If the equation is in the form, ax + b = c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division. Solve 3y + 2 = 11. Subtract 2 from both sides of the equation to get the term with the variable by itself.
## How do you solve multi step word problems?
Here are steps to solving a multi-step problem: Step 1: Circle and underline. Circle only the necessary information and underline what ultimately needs to be figured out. Step 2: Figure out the first step/problem in the paragraph and solve it. Last step: Find the answer by using the information from Steps 1 and 2.
## How do you solve 3 equations with 3 variables?
Here, in step format, is how to solve a system with three equations and three variables:Pick any two pairs of equations from the system.Eliminate the same variable from each pair using the Addition/Subtraction method.Solve the system of the two new equations using the Addition/Subtraction method.
### Releated
#### Equation for decibels
How do you calculate dB? Find the logarithm of the power ratio. log (100) = log (102) = 2 Multiply this result by 10 to find the number of decibels. decibels = 10 × 2 = 20 dB If we put all these steps together into a single equation, we once again have the definition […]
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […] | 806 | 3,534 | {"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} | 4.78125 | 5 | CC-MAIN-2021-04 | longest | en | 0.903598 |
https://calculat.io/en/number/percent-of/30--50 | 1,721,789,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00434.warc.gz | 125,503,548 | 24,969 | # 30 percent of 50
## What's 30 percent of 50?
Answer: 30 percent of 50 it is 15
(Fifteen)
15 is 30% of 50
## 30 percent of 50 calculation explanation
In order to calculate 30% of 50 let's write it as fractional equation.
We have 50 = 100% and X = 30%. So our fraction will look like:
50
/
X
=
100%
/
30%
Now we can solve our fraction by writing it as an equation:
X = (50 × 30) ÷ 100
=
X = 1500 ÷ 100
=
X = 15
Therefore, 30% of 50 is 15
Another way to solve our problem is to find the value of 1% of the number and then multiply it by the number of percent (30). To find 1% of a number 50 you need to divide it by 100:
X = (50 ÷ 100) × 30
=
X = 0.5 × 30
=
X = 15
So we got the same result again: 15
PercentValue
7.5
8
8.5
9
19% of 509.5
10
10.5
11
11.5
12
12.5
13
13.5
28% of 5014
29% of 5014.5
15
15.5
16
16.5
34% of 5017
17.5
18
18.5
19
39% of 5019.5
20
20.5
21
43% of 5021.5
22
## About "Percent of Number" Calculator
This calculator will help you to calculate percent of a given number. For example, it can help you find out what's 30 percent of 50? (The answer is: 15). Enter the percent (e.g. '30') and the number (e.g. '50'). Then hit the 'Calculate' button.
## FAQ
### What's 30 percent of 50?
30 percent of 50 it is 15 | 470 | 1,256 | {"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} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.880101 |
https://math.answers.com/math-and-arithmetic/What_is_25_minus_45_percent | 1,726,099,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00649.warc.gz | 354,962,157 | 47,403 | 0
# What is 25 minus 45 percent?
Updated: 9/22/2023
Wiki User
11y ago
25 minus 45 percent = 13.75
25 - 45% of 25
= 25 - (0.45 * 25)
= 25 - 11.25
= 13.75
Wiki User
11y ago | 83 | 181 | {"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} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.714173 |
http://physics.stackexchange.com/questions/16631/what-are-the-polarization-states-of-the-photons-in-a-polarized-and-unpolarized-l/16632 | 1,464,112,240,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049272823.52/warc/CC-MAIN-20160524002112-00144-ip-10-185-217-139.ec2.internal.warc.gz | 237,704,071 | 17,743 | # What are the polarization states of the photons in a polarized and unpolarized light?
The photons are completely polarized, i.e their polarization states can be expressed as $a|R\rangle+b|L\rangle$, where $|R\rangle$ and $|L\rangle$ are two helicity eigenstates of the photon. For example, the $|R\rangle$ photon is right circularly polarized and the $|H\rangle=\frac{1}{\sqrt{2}}(|R\rangle+|L\rangle)$ is horizontally linearly polarized photon.
Is the polarized light a pure state and the unpolarized light a statistical mixture of photons with different polarizations?
-
Yes, a photon in a polarized light is found in a pure state such as $|H\rangle$, $|V\rangle$, $|L\rangle$, $|R\rangle$, or any complex linear combination of them. A photon in (completely) unpolarized light is described by the density matrix $$\rho = \frac{1}{2} \left( |L\rangle \langle L| + |R\rangle \langle R| \right) = \frac{1}{2} \left( |H\rangle \langle H| + |V\rangle \langle V| \right)$$ Note that you omitted the relationship for the vertically polarized state, $|V\rangle = i(|R\rangle - |L\rangle)/\sqrt{2}$, up to an overall sign which is a convention (well, the whole phase including $i$ is physically inconsequential, so it doesn't matter at all but one must be self-consistent with the conventions).
Your density matrix corresponds to a completely unpolarized light. Partially unpolarized light density matrix is expressed via projection operators with different "weights" $w_n$ (probabilities), not 1/2. – Vladimir Kalitvianski Nov 7 '11 at 10:37
Dear @ANKU, to discuss frequency (and/or direction), you need to extend the Hilbert space by tensor-multiplying it with the space of different $\vec k$. Again, one may have pure states and mixed states with respect to frequencies, too. The totally non-monochromatic, unpolarized light is given by the density matrix like mine, but it would also have extra labels $\vec k$ in all the bra,ket states and one would integrated over some interval of $\vec k$. However, one may also have pure states relatively to the position/frequencies. There are many states, pure and mixed; what's your exact question? – Luboš Motl Nov 7 '11 at 22:24 | 571 | 2,174 | {"found_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} | 3.09375 | 3 | CC-MAIN-2016-22 | latest | en | 0.891372 |
http://www.mathworks.com/help/images/ref/edge.html?nocookie=true | 1,430,873,803,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430457655609.94/warc/CC-MAIN-20150501052055-00008-ip-10-235-10-82.ec2.internal.warc.gz | 540,262,200 | 12,937 | # edge
Find edges in intensity image
## Syntax
• `BW = edge(I)`
• `BW = edge(I,method)` example
• `BW = edge(I,method,threshold)`
• `BW = edge(I,method,threshold,direction)`
• `BW = edge(I,method,threshold,direction,'nothinning')`
• `BW = edge(I,method,threshold,direction,sigma)`
• ```[BW,threshOut] = edge(___)```
• ```[gpuarrayBW,threshOut] = edge(gpuarrayI,___)``` example
## Description
````BW = edge(I)` returns a binary image `BW` containing `1`s where the function finds edges in the input image `I` and `0`s elsewhere. By default, `edge` uses the Sobel edge detection method.This function supports code generation (see Tips).```
example
````BW = edge(I,method)` detects edges in image `I`, where `method` specifies the edge detection method used.```
````BW = edge(I,method,threshold)` detects edges in image `I`, where `threshold` specifies the sensitivity threshold. `edge` ignores all edges that are not stronger than `threshold`.```
````BW = edge(I,method,threshold,direction)` detects edges in image `I`, where `direction` specifies the direction in which the function looks for edges in the image: horizontally, vertically, or in both directions. Used only with the Sobel and Prewitt methods.```
````BW = edge(I,method,threshold,direction,'nothinning')` detects edges in image `I`, where `‘nothinning' ` speeds up the operation of the algorithm by skipping the additional edge-thinning stage. By default, the algorithm applies edge thinning. Used only with the Sobel, Prewitt, and Roberts methods. ```
````BW = edge(I,method,threshold,direction,sigma)` detects edges in image `I`, where `sigma` specifies the standard deviation of the filter. Used only with the Laplacian of Gaussian and Canny methods. ```
``````[BW,threshOut] = edge(___)``` returns the threshold value.```
example
``````[gpuarrayBW,threshOut] = edge(gpuarrayI,___)``` performs the edge detection operation on a GPU. The input image and the output image are gpuArrays. This syntax requires Parallel Computing Toolbox™. ```
## Examples
collapse all
### Compare Edge Detection Using Canny and Prewitt Methods
Read a grayscale image into the workspace and display it.
```I = imread('circuit.tif'); imshow(I) ```
Find edges using the Canny method.
```BW1 = edge(I,'Canny'); ```
Find edges using the Prewitt method.
```BW2 = edge(I,'Prewitt'); ```
Display both results side-by-side.
```imshowpair(BW1,BW2,'montage') ```
### Find Edges Using Prewitt Method on a GPU
Read grayscale image, creating a gpuArray.
`I = gpuArray(imread('circuit.tif'));`
Find edges using the Prewitt method.
```BW = edge(I,'prewitt'); ```
Display results.
`figure, imshow(BW)`
## Input Arguments
collapse all
### `I` — Input intensity or binary image2-D, real, nonsparse numeric or logical array
Input intensity or binary image, specified as a 2-D, real, nonsparse, numeric, or logical array.
Example: `I = imread('circuit.tif'); BW = edge(I);`
Data Types: `single` | `double` | `int16` | `uint8` | `uint16` | `logical`
### `method` — Edge detection method`‘Sobel'` (default) | string
Edge detection method, specified as one of the following strings.
Method Description
`'Canny'`Finds edges by looking for local maxima of the gradient of `I`. The `edge` function calculates the gradient using the derivative of a Gaussian filter. This method uses two thresholds to detect strong and weak edges, including weak edges in the output if they are connected to strong edges. By using two thresholds, the Canny method is less likely than the other methods to be fooled by noise, and more likely to detect true weak edges.
Not supported on a GPU.
`‘log'` (Laplacian of Gaussian)Finds edges by looking for zero-crossings after filtering `I` with a Laplacian of Gaussian filter.
`'Prewitt`'Finds edges using the Prewitt approximation to the derivative. It returns edges at those points where the gradient of `I` is maximum.
`'Roberts'`Finds edges using the Roberts approximation to the derivative. It returns edges at those points where the gradient of `I` is maximum.
`'Sobel'`Finds edges using the Sobel approximation to the derivative. It returns edges at those points where the gradient of I is maximum.
`'zerocross'`Finds edges by looking for zero-crossings after filtering I with a filter that you specify.
Example: `BW = edge(I,'Canny');`
Data Types: `char`
### `threshold` — Sensitivity thresholdscalar | two-element vector (Canny method only)
Sensitivity threshold, specified as a numeric scalar or, for the Canny method only, a two-element vector. `edge` ignores all edges that are not stronger than `threshold`. If you do not specify `threshold`, or if you specify an empty array (`[]`), `edge` chooses the value automatically. For information about this parameter, see Tips.
For the `‘log'` (Laplacian of Gaussian) and `'zerocross'` methods, if you specify the threshold value `0`, the output image has closed contours because it includes all of the zero-crossings in the input image.
Example: `BW = edge(I,'Prewitt',3);`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`
### `direction` — Direction of detection`‘both'` (default) | `'horizontal'` | `'vertical'`
Direction of detection, specified as the string `'horizontal'`, `'vertical'`, or `'both'`. Supported only by the Sobel and Prewitt methods.
Example: `BW = edge(I,'Prewitt',3,'horizontal');`
Data Types: `char`
### `sigma` — Standard deviation of the filter`2` (default) | scalar
Standard deviation of the filter, specified as a scalar. The following table describes this option for each method that supports it.
MethodDescription
Canny
Scalar value that specifies the standard deviation of the Gaussian filter. The default `sigma` is `sqrt(2)`; the size of the filter is chosen automatically, based on `sigma`.
Laplacian of Gaussian (log)
Scalar value that specifies the standard deviation of the Laplacian of Gaussian filter. The default `sigma` is 2. The size of the filter is `n`-by-`n`, where `n` `=` `ceil(sigma*3)*2+1`.
Scalar value that specifies the standard deviation of the Gaussian filter. The default `sigma` is `sqrt(2)`. The size of the filter is chosen automatically, based on `sigma`.
Example: `BW = edge(I,'Canny',3,5);`
Data Types: `double`
### `gpuarrayI` — Input imagegpuArray
Input image, specified as a gpuArray.
Example: ```gpuarrayI = gpuArray(imread('circuit.tif')); gpuarrayBW = edge(gpuarrayI,'prewitt');```
## Output Arguments
collapse all
### `BW` — Output binary imagelogical array
Output binary image, returned as a logical array, the same size as `I`, with `1`s where the function finds edges in `I` and `0`s elsewhere.
### `threshOut` — Threshold value used in the computationnumeric scalar
Threshold value used in the computation, returned as a numeric scalar.
### `gpuarrayBW` — Output binary image when run on a GPUgpuArray
Output binary image when run on a GPU, returned as a gpuArray.
collapse all
### Tips
• This function supports the generation of C code using MATLAB® Coder™. Note that if you choose the generic `MATLAB Host Computer` target platform, the function generates code that uses a precompiled, platform-specific shared library. Use of a shared library preserves performance optimizations but limits the target platforms for which code can be generated. For more information, see Understanding Code Generation with Image Processing Toolbox.
When generating code, note the following:
• The `method`, `direction`, and `sigma` arguments must be compile-time constants.
• Nonprogrammatic syntaxes are not supported. For example, if you do not specify a return value,`edge` displays an image. This syntax is not supported.
• Notes about the `threshold` parameter:
• For the gradient-magnitude edge detection methods (Sobel, Prewitt, Roberts), `edge` uses `threshold` to threshold the calculated gradient magnitude. For the zero-crossing methods, including Laplacian of Gaussian, `edge` uses `threshold` as a threshold for the zero-crossings. In other words, a large jump across zero is an edge, while a small jump isn't.
• The Canny method applies two thresholds to the gradient: a high threshold for low edge sensitivity and a low threshold for high edge sensitivity. `edge` starts with the low sensitivity result and then grows it to include connected edge pixels from the high sensitivity result. This helps fill in gaps in the detected edges.
• In all cases, `edge` chooses the default threshold heuristically, depending on the input data. The best way to vary the threshold is to run `edge` once, capturing the calculated threshold as the second output argument. Then, starting from the value calculated by `edge`, adjust the threshold higher (fewer edge pixels) or lower (more edge pixels).
• The function `edge` changed in Version 7.2 (R2011a). Previous versions of the Image Processing Toolbox™ used a different algorithm for computing the Canny method. If you need the same results produced by the previous implementation, use the following syntax:
`BW = edge(I,'canny_old',...)`
• The syntax `BW = edge(... ,K)` has been removed. Use the `BW = edge(... ,direction)` syntax instead.
• The syntax `edge(I,'marr-hildreth',...)` has been removed. Use the `edge(I,'log',...)` syntax instead.
## References
[1] Canny, John, "A Computational Approach to Edge Detection," IEEE Transactions on Pattern Analysis and Machine Intelligence,Vol. PAMI-8, No. 6, 1986, pp. 679-698.
[2] Lim, Jae S., Two-Dimensional Signal and Image Processing, Englewood Cliffs, NJ, Prentice Hall, 1990, pp. 478-488.
[3] Parker, James R., Algorithms for Image Processing and Computer Vision, New York, John Wiley & Sons, Inc., 1997, pp. 23-29. | 2,417 | 9,733 | {"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-2015-18 | latest | en | 0.680584 |
https://www.coursehero.com/file/6005346/week-6/ | 1,496,032,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612013.52/warc/CC-MAIN-20170529034049-20170529054049-00326.warc.gz | 1,063,814,400 | 32,270 | # week 6 - SOLUTIONS TO EXERCISES EXERCISE 10-1(1520 minutes...
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SOLUTIONS TO EXERCISES EXERCISE 10-1 (15–20 minutes) Item Land Land Improvements Building Other Accounts (a) (\$275,000) Notes Payable (b) \$275,000 (c) \$ 10,000 (d) 7,000 (e) 6,000 (f) (1,000) (g) 25,000 (h) 250,000 (i) 9,000 (j) \$ 4,000 (k) 11,000 (l) (5,000) (m) 13,000 (n) 19,000 (o) 14,000 (p) 3,000
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EXERCISE 10-3 (10–15 minutes) 1. Truck #1. ............................................................... 13,900 Cash. ............................................................ 13,900 2. Truck #2. ............................................................... 18,364* Discount on Notes Payable. ............................. 1,636 Cash. ............................................................ 2,000 Notes Payable. ........................................... 18,000 *PV of \$18,000 @ 10% for 1 year = \$18,000 X .90909 = \$16,364 \$16,364 + \$2,000 = \$18,364 3. Truck #3. ............................................................... 15,200 Cost of Goods Sold. ........................................... 12,000 Inventory. .................................................... 12,000 Sales. ........................................................... 15,200 [Note to instructor : The selling (retail) price of the computer system appears to be a better gauge of the fair value of the consideration given than is the list price of the truck as a gauge of the fair value of the consideration received (truck). Vehicles are very often sold at a
price below the list price.] 4. Truck #4. ............................................................... 13,000 Common Stock. ......................................... 10,000 Paid-in Capital in Excess of Par (1,000 shares X \$13 = \$13,000; \$13,000 less \$10,000 par value). ........ 3,000
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EXERCISE 10-4 (20–25 minutes) Purchase Cash paid for equipment, including sales tax of \$5,000. \$105,000 Freight and insurance while in transit. ............................... 2,000 Cost of moving equipment into place at factory. .............. 3,100 Wage cost for technicians to test equipment. ................... 6,000 Special plumbing fixtures required for new equipment. 8,000 Total cost. ................................................................................. \$124,100 The insurance premium paid during the first year of operation on this equipment should be reported as insurance expense, and not be capitalized. Repair cost incurred in the first year of operations related to this equipment should be reported as repair and maintenance expense, and not be capitalized. Both these costs relate to periods subsequent to purchase. Construction Material and purchased parts (\$200,000 X .99). ............... \$198,000 Labor costs. ............................................................................. 190,000 Overhead costs. ...................................................................... 50,000 Cost of installing equipment. ................................................ 4,400 Total cost. ................................................................................. \$442,400
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## This note was uploaded on 11/03/2010 for the course AC550 505 taught by Professor Weiss during the Summer '09 term at Keller Graduate School of Management.
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week 6 - SOLUTIONS TO EXERCISES EXERCISE 10-1(1520 minutes...
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Ask a homework question - tutors are online | 868 | 3,863 | {"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} | 3.0625 | 3 | CC-MAIN-2017-22 | longest | en | 0.637649 |
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