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# Stats Notes: Week 3 STAT 3332-0U1
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Everything from the continuation of normal distribution through randomization and gambling probability.
COURSE
Statistics for Life Sciences
PROF.
Dr. Yuly Koshevnik
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Stats, life, sciences, normal, distribution, Randomization, plot, curves, Probability
KARMA
25 ?
## Popular in Statistics
This 9 page Class Notes was uploaded by rkl130030 on Sunday June 12, 2016. The Class Notes belongs to STAT 3332-0U1 at University of Texas at Dallas taught by Dr. Yuly Koshevnik in Summer 2016. Since its upload, it has received 14 views. For similar materials see Statistics for Life Sciences in Statistics at University of Texas at Dallas.
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Associated Topics || Dr. Math Home || Search Dr. Math
### Converting Logs to Another Base
```
Date: 04/27/98 at 01:50:52
From: Ashraf Suleman
Subject: Finding the nth log
Hello,
I'm taking a computer programming course in Grade 12 (Pascal), and on
the side I've decided to write a calculator program for use in DOS.
The compiler I'm using, however, only performs logarithms to the
base e. How can I use it to find log to the base 10, or log to the
base n (if possible)? I don't need the specific programming code, just
the method I would use is sufficient. I've searched through the site
here but I haven't found anything pertaining to this (I've yet to take
algebra, calculus, or finite, so I don't have a grasp of many of the
terms). Any help would be greatly appreciated. Thanks :)
```
```
Date: 04/27/98 at 07:41:37
From: Doctor Jerry
Subject: Re: Finding the nth log
Hi Ashraf,
If you want y = log_a(w), then:
a^y = w
ln(a^y) = ln(w)
y*ln(a) = ln(w)
y = log_a(w) = ln(w)/ln(a)
-Doctor Jerry, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
```
Date: 04/29/98 at 07:44:14
Subject: Re: Finding the nth log
There is a nice way of converting logarithms from any base to another.
This is very useful, as you have found in trying to write your
program.
Before showing you the rule for conversion, let me show you one way to
obtain it so that you can see how and why it works. Let's suppose you
want to find the log base n of x; call this y:
y = log x
n
What this really means is that x = n^y. Now suppose I am your
computer and can only take logarithms using the base e. I could take
the base e logarithm of both sides of the equation x = n^y; this would
give me:
ln x = ln (n^y)
Writing "ln" means "log base e". Next we make use of the property of
logarithms which says that ln (n^y) = y * ln (n), so our equation
becomes:
ln x = y ln n
Now we divide both sides by ln n to get:
(ln x)/(ln n) = y
Remember that y is the original logarithm we wanted to find using
base n, so all we have to do is take the natural (base e) log of x
and divide by the natural log of n. This conversion rule works for any
base, not just e; the general rule to convert from base n to any
base b looks like this:
log x = (log x)/(log n)
n b b
Hope this helps. Good luck with your programming!
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Logs
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
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# discrete series
discrete series
дискретная серия
The New English-Russian Dictionary of Radio-electronics. . 2005.
### Смотреть что такое "discrete series" в других словарях:
• Discrete series representation — In mathematics, a discrete series representation is an irreducible unitary representation of a locally compact topological group G that is a subrepresentation of the left regular representation of G on L²(G). In the Plancherel measure, such… … Wikipedia
• Discrete signal — Discrete sampled signal Digital signal (Discrete here … Wikipedia
• Discrete — in science is the opposite of continuous: something that is separate; distinct; individual. This article is about the possible uses of the word discrete . For a definition of the word discreet , see the Wiktionary entry discreet. Discrete may… … Wikipedia
• Discrete logarithm records — are the best results achieved to date in solving the discrete logarithm problem, which is the problem of finding solutions x to the equation gx = h given elements g and h of a finite cyclic group G. The difficulty of this problem… … Wikipedia
• Discrete cosine transform — A discrete cosine transform (DCT) expresses a sequence of finitely many data points in terms of a sum of cosine functions oscillating at different frequencies. DCTs are important to numerous applications in science and engineering, from lossy… … Wikipedia
• Discrete Fourier transform — Fourier transforms Continuous Fourier transform Fourier series Discrete Fourier transform Discrete time Fourier transform Related transforms In mathematics, the discrete Fourier transform (DFT) is a specific kind of discrete transform, used in… … Wikipedia
• Discrete wavelet transform — An example of the 2D discrete wavelet transform that is used in JPEG2000. The original image is high pass filtered, yielding the three large images, each describing local changes in brightness (details) in the original image. It is then low pass… … Wikipedia
• Discrete valuation ring — In abstract algebra, a discrete valuation ring (DVR) is a principal ideal domain (PID) with exactly one non zero maximal ideal. This means a DVR is an integral domain R which satisfies any one of the following equivalent conditions: R is a local… … Wikipedia
• Discrete-time Fourier transform — In mathematics, the discrete time Fourier transform (DTFT) is one of the specific forms of Fourier analysis. As such, it transforms one function into another, which is called the frequency domain representation, or simply the DTFT , of the… … Wikipedia
• Discrete mathematics — For the mathematics journal, see Discrete Mathematics (journal). Graphs like this are among the objects studied by discrete mathematics, for their interesting mathematical properties, their usefulness as models of real world problems, and their… … Wikipedia
• Discrete Fourier series — A Fourier series is a representation of a function in terms of a summation of an infinite number of harmonically related sinusoids with different amplitudes and phases. The amplitude and phase of a sinusoid can be combined into a single complex… … Wikipedia
### Книги
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CC-MAIN-2019-51
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latest
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http://www.buenastareas.com/ensayos/Estadisticas/479890.html
| 1,485,176,644,000,000,000
|
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| 375,153,025
| 14,785
|
Solo disponible en BuenasTareas
• Páginas : 2 (331 palabras )
• Descarga(s) : 7
• Publicado : 1 de julio de 2010
Vista previa del texto
Coffee is a leading export from several developing countries. When coffee prices are high, farmers often clear forest to plant more cofee trees. Here are data onprices paid to coffe growers in Indonesia and the rate of deforestation in a national park that lies in a coffee-producing region for five years:
Price (cents perpound) | Deforestation(percent) |
29 | 0,49 |
40 | 1,59 |
54 | 1,69 |
55 | 1,82 |
72 | 3,1 |
x | y | (xi-x)/Sx | (yi-y)/Sy | (xi-x)/Sx*(yi-y)/Sy |
29 |0,49 | -1,2863842 | -1,3450728 | 1,73028039 |
40 | 1,59 | -0,6125639 | -0,1595118 | 0,09771119 |
54 | 1,69 | 0,24502556 | -0,0517336 | -0,012676 |
55 | 1,82 |0,30628195 | 0,08837818 | 0,02706864 |
72 | 3,1 | 1,34764056 | 1,46794005 | 1,97825556 |
| | | Total | 3,82063974 |
r = 3,82063974/4
r= 0,95515994
yʹ= a +bx
y’= -0.9763527 + 0.05428705x
α=0.05
Ho: ρ=0
H1: ρ>0
t= r√n-2 / √1-r² ; gl= n-2
gl= 3
Si t>2.353 se rechaza Ho.Si t<2.353 no se rechaza Ho.
t= 0.95519√5-2/√1-0.95519²
t= 5.589441
t es mayor que 2.353, por lo tanto se rechaza Ho, lo que comprueba que hay evidencia queel coeficiente de correlacion poblacional es positivo.
Ho: β=0
H1: β>0
α= 0.05
t= b/SEb
SEb= s / √Σ(xi-x̃)²
S= √Σei²/n-2
e= yi-ỹ
x | y | x̃| ỹ | xi-x̃ | yi-ỹ |
29 | 0,49 | 50 | 1.738 | -21 | -1.248 |
40 | 1,59 | 50 | 1.738 | -10 | -0.148 |
54 | 1,69 | 50 | 1.738 | 4 | -0.048 |
55 | 1,82 | 50 |1.738 | 5 | 0.082 |
72 | 3,1 | 50 | 1.738 | 22 | 1.362 |
| | | total | 0 | 0 |
Si t>2.353 se rechaza Ho
Si t<2.353 no se rechaza Ho.
S=0
SE=0
t=0
| 763
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https://gmatclub.com/forum/in-a-certain-game-a-large-container-is-filled-with-red-yel-144902.html
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| 542,687,706
| 149,170
|
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# In a certain game, a large container is filled with red, yel
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In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0
Math Expert
Joined: 02 Sep 2009
Posts: 58416
Re: In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
29 Dec 2012, 06:14
13
4
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0
$$147,000=2^3*3*5^3*7^2$$.
Since no other beads value is a multiple of 7, then there must be two red beads removed.
Similar question to practice: in-a-certain-game-a-large-bag-is-filled-with-blue-green-126425.html
Hope it helps.
_________________
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Joined: 13 May 2014
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Concentration: General Management, Strategy
Re: In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
22 May 2014, 13:01
5
MavenQ wrote:
Dear amarmeena1992
This is not a Combinations question. (And it cannot be, since you are not even told what the total number of balls is.)
You are told that the product of the scores of all the balls that were taken out is 147000.
Now, $$147000 = 7^2*5*3*4*2^3$$
From this expression, you can easily deduce that:
2 Red balls were taken out, along with
3 Yellow balls, 1 Green ball and 3 Blue balls.
Hope this helps.
Hi MavenQ,
Kindly correct the factors :
$$147,000 = 7^2*5^3*3*2^3$$
which would mean
2 R, 3 Y , 1 G and 3 B beads
Please press kudos if it helps you
Kudos is the best form of appreciation
Dear amarmeena1992,
The beads R,Y,G,B are allotted some numbers which are prime.
So now since the product of the values of removed beads are known,
we can decompose or factorize the product into unique combination/product of the bead values
as shown :
147,000 = 7*7*3*5*5*5*2*2*2 = $$7^2*5^3*3*2^3$$
that is 2 R, 3 Y , 1 G and 3 B
Kudos is the best form of appreciation
##### General Discussion
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In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
22 May 2014, 11:18
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
A. 5
B. 4
C. 3
D. 2
E. 0
I am not able to comprehend the question correctly. There could be a lot of combinations for removing the beads. How could we find any particular answer?
Math Expert
Joined: 02 Sep 2009
Posts: 58416
Re: In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
22 May 2014, 11:29
amarmeena1992 wrote:
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
A. 5
B. 4
C. 3
D. 2
E. 0
I am not able to comprehend the question correctly. There could be a lot of combinations for removing the beads. How could we find any particular answer?
Merging similar topics. please refer to the discussion above and ask if anything remains unclear.
Similar questions to practice:
in-a-certain-game-a-large-bag-is-filled-with-blue-green-126425.html
a-cargo-ship-carrying-four-kinds-of-items-doohickies-geega-167523.html
All OG13 questions with solutions are here: the-official-guide-quantitative-question-directory-143450.html
Hope this helps.
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Re: In a certain game, a large container is filled with red, yel [#permalink]
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Updated on: 22 May 2014, 14:37
2
Dear amarmeena1992
This is not a Combinations question. (And it cannot be, since you are not even told what the total number of balls is.)
You are told that the product of the scores of all the balls that were taken out is 147000.
Now, $$147000 = 7^2*5^3*3*4*2^3$$
From this expression, you can easily deduce that:
2 Red balls were taken out (since each Red ball has a score of 7. So, number of Red balls = power of 7 in the expression), along with
3 Yellow balls, 1 Green ball and 3 Blue balls.
Hope this helps.
_________________
Please press Kudos if you were helped by my post!
Originally posted by JapinderKaur on 22 May 2014, 12:15.
Last edited by JapinderKaur on 22 May 2014, 14:37, edited 2 times in total.
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Re: In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
15 Jul 2016, 04:43
3
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0
The first thing we want to do is to decipher the problem. The key word here is “product;" we are told that the product of the point values of the removed beads is 147,000.
Let's look at an example. Let’s say 2 red beads, 2 yellow beads, 2 green beads, and 2 blue beads were removed. The product of the point values would be:
7 x 7 x 5 x 5 x 3 x 3 x 2 x 2 = 44,100
Conversely, if we were to take 44,100 and break it down into its prime factors, we would get:
44,100 = 7 x 7 x 5 x 5 x 3 x 3 x 2 x 2
Note that this prime factorization of 44,100 corresponds exactly to the point values of the 2 red beads, the 2 yellow beads, the 2 green beads, and the 2 blue beads from the example.
This example is important because we can now use the same approach for answering the actual question. We are given that the product of the point values of the removed beads is 147,000. Thus, some number of 7’s, 5’s, 3’s, and 2’s is multiplied together to equal 147,000. Since we only care about the red beads, we only care about the number of 7’s in the product. Let’s keep this in mind as we break down 147,000.
147,000 = 147 x 1,000 = 49 x 3 x 1,000 = 7 x 7 x 3 x 1,000
Since 1,000 does not have 7 as a factor, we see that there are two 7’s in the product of 147,000; thus, 2 red beads were removed.
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Re: In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
18 May 2017, 11:26
2
Red – 7
Yellow – 5
Green – 3
Blue – 2
How many read beads have been removed, when the product of point values is 147 000.
Here we have a problem with prime factorization.
In other words, the problem is asking how many 7 factors are in 147 000
Using factor tree =>
147 000 = 147 * 1000
1000 = 10*10*10 =>2*5 * 2*5 *2*5 => no factors of 7 here
147 = 7*21
21=3*7 here we have 2 factors of 7.
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Re: In a certain game, a large container is filled with red, yel [#permalink]
### Show Tags
10 Aug 2019, 02:51
147000=7^2 ∗ 5^3 ∗ 3 ∗ 4 ∗2^3
From this expression, you can easily deduce that:
2 Red balls were taken out (since each Red ball has a score of 7. So, number of Red balls = power of 7 in the expression), along with
3 Yellow balls, 1 Green ball and 3 Blue balls.
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Joined: 09 Sep 2013
Posts: 13234
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# Class 4
Class 4. Hold onto your hypothetical executive summaries. You should have 3 hard copies & 1 electronic copy. Please turn in your code book, spreadsheets and Work Logs at the front table. Class Agenda. Dale Carnegie Speeches A Few Final Report Formatting Tips Research Methods Paper
Télécharger la présentation
## Class 4
E N D
### Presentation Transcript
1. Class 4 Hold onto your hypothetical executive summaries. You should have 3 hard copies & 1 electronic copy. Please turn in your code book, spreadsheets and Work Logs at the front table.
2. Class Agenda • Dale Carnegie Speeches • A Few Final Report Formatting Tips • Research Methods Paper • Meet with Supervisors on Ex. Summaries
3. Conor Sullivan Jack Donisch Nicholas Impson Arianna Rogers Celina Bridges Enith Sanchez Asha Shirwa Catherine Warren Dale Carnegie Presentations Today
4. Research Methods • Any questions on questions 1-5? • Review of questions 6-8
5. Question 6- Misleading Graphs • Some statisticians manipulate the way data are presented • Depending on the y-axis scale you use, you can make the difference between 2 bars or a spike in a line graph appear larger/smaller to the reader • This effect is influenced by the following • Whether or not the y-axis begins at 0 • The size of the intervals • Any “jumps” in the y-axis scale
6. Question 6- Misleading Graphs • These 2 graphs present the same data but their scales are different.
7. Question 7- Scaling Numbers • A percentage is a way of expressing a number as a fraction out of 100 • Scaling numbers involves converting raw numbers into rates • A rate is the one number divided by the total of cases. • Example, 5 out of 10 students are silly. It would read. 5 per 10 students are silly.
8. Total Traffic Deaths, Selected States, 1985 Source: National Safety Council. The World Almanac and Book of Facts. New York: Newspaper Enterprise Association, 1986. p. 781.
9. Scaling Numbers (not %) (ch.16) • Allows you to compare items of different size • Example • Raw values make NY appear more dangerous to drive in than NM. • But when shown as deaths per 100 million miles driven NM has a considerably higher rate. • You may NOT use percentages.
10. Problem with Percentages • % of accidents per total miles • New Mexico • .000000043 • New York • .000000024%
11. Question 8- Percentage Change • Pgs. 88-89 & 198-200 in Maxwell Manual • In order to see if a pattern exists in your data, you need to determine the percent change between years • It gives you a precise indicator of the amount of change from one time period to the next Percent change = [(New figure – Old figure) / Old figure] x 100
12. Differences between two periods using percent change (show calculations) (ch.16) • Percent Difference shows change between time periods • Example: Number of felonies in NYC in 1981 was 637,451. There were 538,051 in 1984. HOW TO CALCULATE PERCENT CHANGE: • New Figure – Original Figure * 100 = Percent Difference Original Figure INCLUDE IN YOUR PAPER: • 1984 felonies – 1981 felonies * 100 = Percent Difference 1981 felonies • 538,051 – 637,451 * 100 = -99,400 * 100 = -15.99% = -16% 637,451 637,451
13. Dale Carnegie 2/13 • Coralis Rivera-Aponte • Brenda Palacios • Adam Pelner • Allison Adams • Edward Palmer • Iasiah Jones
14. Meet with Supervisors • Review of Executive Summaries
15. We’ll break into groups of two Circle any mistakes you find Choose best one to present to class Hypothetical Executive Summary
16. Hypothetical Executive Summary • Go to MaxPal if you need to make major revisions to your executive summary • Minor corrections can be written in by hand • Turn in your paper today in the PAF office in the PAF INBOX (to the right of the desk) by 4:00PM!
17. Reminders • Read Chapters 5-7 in “Is That a Fact?” • If it’s ready, turn in your Exec Summary as you leave • Due next class: • Research Methods Paper (Complete using template) • Due in 2 weeks 2/20 • Revised/Updated Agenda & Contact Log
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469,963 Members | 1,943 Online
# Partitions of an integer
Wrote a python script to find the partitions of an integer (a list of
all the ways you can express n as a sum of integers). For example, the
partitions of 5 are
5
4+1
3+2
2+2+1
3+1+1
2+1+1+1
1+1+1+1+1
My method, however, generates plenty of duplicates (note the if
statement to catch them). Generating the 15 partitions of 7, for
example, also generates 19 duplicates. Can someone spot a way to improve
or eliminate the duplicates?
Nick.
def partitions(n):
"""Create a list of all of n's partitions"""
allpartitions = []
onepartition = [n]
allpartitions.append(onepartition)
# p steps through allpartitions, looking for those which are not all 1s
p = 0
while p != len(allpartitions):
onepartition = allpartitions[p]
# check each element of this partition
i = 0
while i != len(onepartition):
# if there is a non-1, then create a new partition
if onepartition[i] > 1:
hi = onepartition[i] - 1
lo = 1
while hi >= lo:
newpartition = onepartition[:i] + [hi] + [lo] +
onepartition[i+1:]
newpartition.sort()
newpartition.reverse()
# newpartition may be a copy of an existing one!!!
if not newpartition in allpartitions:
allpartitions.append(newpartition)
hi -= 1
lo += 1
i += 1
p += 1
print dups
return allpartitions
print partitions(7)
Jul 18 '05 #1
5 4064
Am Samstag, 24. Juli 2004 18:08 schrieb Nick J Chackowsky:
My method, however, generates plenty of duplicates (note the if
statement to catch them). Generating the 15 partitions of 7, for
example, also generates 19 duplicates. Can someone spot a way to improve
or eliminate the duplicates?
def _yieldParts(num,lt):
if not num:
yield []
for i in range(min(num,lt),0,-1):
for parts in yieldParts(num-i,i):
yield [i]+parts
def yieldParts(num):
for part in _yieldParts(num,num):
yield part
parts = list(yieldParts(7,7))
print len(parts)
print parts
will only yield non-duplicate partitions, and functions as a generator, so you
don't need to store all partitions in memory. It's not iterative as your
example, but you could implement caching so that its runtime decreases to
iterative.
Heiko.
Jul 18 '05 #2
Nick J Chackowsky wrote:
Wrote a python script to find the partitions of an integer (a list of
all the ways you can express n as a sum of integers). For example, the
partitions of 5 are
5
4+1
3+2
2+2+1
3+1+1
2+1+1+1
1+1+1+1+1
My method, however, generates plenty of duplicates (note the if
statement to catch them). Generating the 15 partitions of 7, for
example, also generates 19 duplicates. Can someone spot a way to improve
or eliminate the duplicates?
Nick.
You should check this recipe:
Generator for integer partitions
http://aspn.activestate.com/ASPN/Coo.../Recipe/218332
This is a cool example of how to use generators with recursion.
--
George
Jul 18 '05 #3
Am Samstag, 24. Juli 2004 18:25 schrieb Heiko Wundram:
[snip]
Ahh, slight error crept in:
def _yieldParts(num,lt):
if not num:
yield []
for i in range(min(num,lt),0,-1): - for parts in yieldParts(num-i,i):
+ for parts in _yieldParts(num-i,i): yield [i]+parts
def yieldParts(num):
for part in _yieldParts(num,num):
yield part
parts = list(yieldParts(7,7))
print len(parts)
print parts
Heiko.
Jul 18 '05 #4
In article <pN**************@animal.nntpserver.com>,
Nick J Chackowsky <me*************@hotmail.com> wrote:
Wrote a python script to find the partitions of an integer (a list of
all the ways you can express n as a sum of integers). For example, the
partitions of 5 are
5
4+1
3+2
2+2+1
3+1+1
2+1+1+1
1+1+1+1+1
My method, however, generates plenty of duplicates (note the if
statement to catch them). Generating the 15 partitions of 7, for
example, also generates 19 duplicates. Can someone spot a way to improve
or eliminate the duplicates?
Nick.
I propose the following recursive solution:
def sorted(tpl):
copy = list(tpl)[:] ; copy.sort()
return tuple(copy)
def nat_partitions(n, mem = {}):
assert n > 0
if mem.has_key(n):
return mem[n]
parts = {}
for b in range(1, n):
for p in nat_partitions(n - b):
parts[sorted(p + (b,))] = True
mem[n] = parts.keys() + [(n,)]
return mem[n]
from pprint import pprint
pprint(nat_partitions(7))
[(1, 1, 1, 1, 3),
(1, 1, 2, 3),
(3, 4),
(1, 3, 3),
(1, 2, 4),
(1, 1, 1, 2, 2),
(1, 1, 1, 1, 1, 2),
(1, 1, 1, 1, 1, 1, 1),
(1, 1, 1, 4),
(1, 1, 5),
(2, 2, 3),
(2, 5),
(1, 2, 2, 2),
(1, 6),
(7,)]
Certainly the algorithm is simple enough, though there is a certain
amount of thrashing between lists and tuples doing it this way. But,
with the memoization, it performs reasonably well.
-M
--
Michael J. Fromberger | Lecturer, Dept. of Computer Science
http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA
Jul 18 '05 #5
* Nick J Chackowsky <me*************@hotmail.com> in comp.lang.python:
Wrote a python script to find the partitions of an integer (a list of
all the ways you can express n as a sum of integers). For example, the
partitions of 5 are
5
4+1
3+2
2+2+1
3+1+1
2+1+1+1
1+1+1+1+1
give more Python code here. I just wanted to point out this draft from
Knuth where many details about the theory of partitions and the related
algorithms can be found. I guess this will be of interest for all
contributors of the thread if they do not know this document already.
<http://www-cs-faculty.Stanford.EDU/~knuth/fasc3b.ps.gz>
--
DW
Jul 18 '05 #6
### This discussion thread is closed
Replies have been disabled for this discussion.
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Module Mlpost.Point
`module Point: `sig` .. `end``
Points in the plane
`type t = `Signature.point` `
The abstract type for points
`val pt : `Num.t * Num.t -> t``
Construct a point from two numeric values
The following functions create points of length 1. They are especially useful to specify directions with `Path.Vec`
`val dir : `float -> t``
`dir f` is the point at angle `f` on the unit circle. `f` shall be given in degrees
The unitary vectors pointing up, down, left and right
`val up : `t``
`val down : `t``
`val left : `t``
`val right : `t``
`val origin : `t``
`val length : `t -> Num.t``
`length p` is the length of vector from the origin to `p`
`val xpart : `t -> Num.t``
`xpart p` is the x coordinate of point `p`
`val ypart : `t -> Num.t``
`ypart p` is the y coordinate of point `p`
Operations on points
`val transform : `Transform.t -> t -> t``
Apply a transformation to a point
`val segment : `float -> t -> t -> t``
`segment f p1 p2` is the point `(1-f)p1 + fp2`. Stated otherwise, if `p1` is at `0.` and `p2` is at `1.`, return the point that lies at `f`
`val add : `t -> t -> t``
`val shift : `t -> t -> t``
Sum two points
`val sub : `t -> t -> t``
Substract two points
`val mult : `Num.t -> t -> t``
`val scale : `Num.t -> t -> t``
Multiply a point by a scalar
`val rotate : `float -> t -> t``
Rotate a point by an angle in degrees
`val rotate_around : `t -> float -> t -> t``
`rotate_around p1 f p2` rotates `p2` around `p1` by an angle `f` in degrees
`val xscale : `Num.t -> t -> t``
Scales the X coordinate of a point by a scalar
`val yscale : `Num.t -> t -> t``
Scales the Y coordinate of a point by a scalar
`val normalize : `t -> t``
Normalize the vector represented by the point. The origin becomes the origin
Convenient constructors
The following functions build a point at a given scale (see `Num.t` for scales)
`val bpp : `float * float -> t``
`val inp : `float * float -> t``
`val cmp : `float * float -> t``
`val mmp : `float * float -> t``
`val ptp : `float * float -> t``
Same as the previous functions but build list of points
`val map_bp : `(float * float) list -> t list``
`val map_in : `(float * float) list -> t list``
`val map_cm : `(float * float) list -> t list``
`val map_mm : `(float * float) list -> t list``
`val map_pt : `(float * float) list -> t list``
`val p : `?scale:(float -> Num.t) -> float * float -> t``
Builds a point from a pair of floats
`scale` : a scaling function to be applied to each float; see `Num.t` for scaling functions for usual units
`val ptlist : `?scale:(float -> Num.t) -> (float * float) list -> t list``
Same as `p`, but builds a list of points
`val draw : `?brush:Brush.t -> ?color:Color.t -> ?pen:Pen.t -> t -> Command.t``
Draw a point
`color` : the color of the point; default is black
`pen` : the pen used to draw the pen; default is `Brush.Pen.default`
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Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
# Amortization calculator
Amortization is the process of repaying a loan in equal, monthly payments. This calculator lets you estimate your monthly loan repayments. The calculator will produce a full explanation of how the computation was carried out.
Amortization calculator
How much will your monthly loan payment be?
show help ↓↓ examples ↓↓
Hide steps
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Examples
example 1:ex 1:
What is the monthly payment on a mortgage of $12,000 with an annual interest rate of 5.5% that runs for 10 years? example 2:ex 2: What is the original value of a mortgage if it is amortized over 10 months at 7% interest and$25.3 monthly payments?
example 3:ex 3:
You wish to take out a $50000 mortgage with monthly payments of 4.5%, and you can afford$550 per month. How long would it take for you to pay off the mortgage?
example 4:ex 4:
What is the interest rate on a $23,000 mortgage with$350 monthly payments for ten years?
Search our database of more than 200 calculators
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What does the time complexity O(log n) actually mean? by@maazrk
# What does the time complexity O(log n) actually mean?
### @maazrkMaaz
Knowing the complexity of algorithms beforehand is one thing, and other thing is knowing the reason behind it being like that.
Whether you’re a CS graduate or someone who wants to deal with optimization problems effectively, this is something that you must understand if you want to use your knowledge for solving actual problems.
Complexities like O(1) and O(n) are simple and straightforward. O(1) means an operation which is done to reach an element directly (like a dictionary or hash table), O(n) means first we would have to search it by checking n elements, but what could O(log n) possibly mean?
One place where you might have heard about O(log n) time complexity the first time is Binary search algorithm. So there must be some type of behavior that algorithm is showing to be given a complexity of log n. Let us see how it works.
Since binary search has a best case efficiency of O(1) and worst case (average case) efficiency of O(log n), we will look at an example of the worst case. Consider a sorted array of 16 elements.
For the worst case, let us say we want to search for the the number 13.
A sorted array of 16 elements
Selecting the middle element as pivot (length / 2)
Since 13 is less than pivot, we remove the other half of the array
Repeating the process for finding the middle element for every sub-array
You can see that after every comparison with the middle term, our searching range gets divided into half of the current range.
So, for reaching one element from a set of 16 elements, we had to divide the array 4 times,
We can say that,
Simplified Formula
Similarly, for n elements,
Generalization
Separating the power for the numerator and denominator
Multiplying both sides by 2^k
Final result
Now, let us look at the definition of logarithm, it says that
A quantity representing the power to which a fixed number (the base) must be raised to produce a given number.
Which makes our equation into
Logarithmic form
So the log n actually means something doesn’t it? A type of behavior nothing else can represent.
Well, i hope the idea of it is clear in your mind. When working in the field of computer science, it is always helpful to know such stuff (and is quite interesting too). Who knows, maybe you’re the one in your team who is able to find an optimized solution for a problem, just because you know what you’re dealing with. Good luck!
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Cody
# Problem 42706. Step up
Solution 806384
Submitted on 14 Jan 2016 by Moe_2015
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 Pass
%% x = [1 8 9 2]; y_correct = [3 10 11 4]; assert(isequal(your_fcn_name(x),y_correct))
y = 3 10 11 4
2 Pass
%% x = [1 11 19 21]; y_correct = [3 13 21 23]; assert(isequal(your_fcn_name(x),y_correct))%% x = [30 63 12 2]; y_correct = [32 65 14 4]; assert(isequal(your_fcn_name(x),y_correct))
y = 3 13 21 23 y = 32 65 14 4
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Find the treasures in MATLAB Central and discover how the community can help you!
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# Maths - Real Normed Algebras
In the set of real numbers the square root of a minus number is a problem because it does not exist.
However, √-1can be used to generate algebras which use higher dimensional numbers, or to put it the other way round, algebras with higher dimensional numbers may have one or more solutions to the square root of a minus number.
By 'higher dimensional numbers' we mean that each element in the algebra, each number, may contain multiple real values similar to vectors. However there are a lot of possible algebras using multidimensional elements, when we are talking about real normed algebras then we define multiplication so that:
• Inverse multiplication (division) always exists, i.e. we can always find the solution to a/b
• norms are preserved by multiplication |a * b| = |a| |b|.
Where |a| = the distance from origin.
It turns out that these requirements can only be met by algebras which have 1,2,4 or 8 dimensional elements. These 4 algebras, which satisfy this condition, have the following properties:
dimension √-1 exists Multiplication commutative Multiplication associative Real Numbers R 1 no yes yes Complex Numbers C 2 yes yes yes Quaternions Q 4 yes no yes Octonions O 8 yes no no
Complex Numbers contain a copy of the real numbers plus a copy of the real numbers multiplied by the square root of -1.
Quaternions contain a copy of the Complex Numbers plus another copy of the real numbers multiplied by another square root of -1 (going via a different dimension at 90 degrees to the first).
Octonions contain a copy of the Quaternions and double up again.
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## Mathematics of the Environment (Part 7)
13 November, 2012
Last time we saw how the ice albedo effect could, in theory, make the Earth’s climate have two stable states. In a very simple model, we saw that a hot Earth might stay hot since it’s dark and absorbs lots of sunlight, while a cold Earth might stay cold—since it’s icy and white and reflects most of the sunlight.
If you haven’t tried it yet, make sure to play around with this program pioneered by Lesley de Cruz and then implemented in Java by Allan Erskine:
The explanations were all given in Part 6 so I won’t repeat them here!
This week, we’ll see how noise affects this simple climate model. We’ll borrow lots of material from here:
And we’ll use software written by these guys together with Allan Erskine and Jim Stuttard. The power of the Azimuth Project knows no bounds!
### Stochastic differential equations
The Milankovich cycles are periodic changes in how the Earth orbits the Sun. One question is: can these changes can be responsible for the ice ages? On the first sight it seems impossible, because the changes are simply too small. But it turns out that we can find a dynamical system where a small periodic external force is actually strengthened by random ‘noise’ in the system. This phenomenon has been dubbed ‘stochastic resonance’ and has been proposed as an explanation for the ice ages. It also shows up in many other phenomena:
• Roberto Benzi, Stochastic resonance: from climate to biology.
But to understand it, we need to think a little about stochastic differential equations.
A lot of systems can be described by ordinary differential equations:
$\displaystyle{ \frac{d x}{d t} = f(x, t)}$
If $f$ is nice, the time evolution of the system will be a nice smooth function $x(t),$ like the trajectory of a thrown stone. But there are situations where we have some kind of noise, a chaotic, fluctuating influence, that we would like to take into account. This could be, for example, turbulence in the air flow around a rocket. Or, in our case, short term fluctuations of the weather of the earth. If we take this into account, we get an equation of the form
$\displaystyle{ \frac{d x}{d t} = f(x, t) + w(t) }$
where the $w$ is a ‘random function’ which models the noise. Typically this noise is just a simplified way to take into account rapidly changing fine-grained aspects of the system at hand. This way we do not have to explicitly model these aspects, which is often impossible.
### White noise
We’ll look at a model of this sort:
$\displaystyle{ \frac{d x}{d t} = f(x, t) + w(t) }$
where $w(t)$ is ‘white noise’. But what’s that?
Very naively speaking, white noise is a random function that typically looks very wild, like this:
White noise actually has a sound, too: it sounds like this! The idea is that you can take a random function like the one graphed above, and use it to drive the speakers of your computer, to produce sound waves of an equally wild sort. And it sounds like static.
However, all this is naive. Why? The concept of a ‘random function’ is not terribly hard to define, at least if you’ve taken the usual year-long course on real analysis that we force on our grad students at U.C. Riverside: it’s a probability measure on some space of functions. But white noise is a bit too spiky and singular too be a random function: it’s a random distribution.
Distributions were envisioned by Dirac but formalized later by Laurent Schwarz and others. A distribution $D(t)$ is a bit like a function, but often distributions are too nasty to have well-defined values at points! Instead, all that makes sense are expressions like this:
$\displaystyle{ \int_\infty^\infty D(t) f(t) \, d t}$
where we multiply the distribution by any compactly supported smooth function $f(t)$ and then integrate it. Indeed, we specify a distribution just by saying what all these integrals equal. For example, the Dirac delta $\delta(t)$ is a distribution defined by
$\displaystyle{ \int_\infty^\infty \delta(t) f(t) \, d t = f(0) }$
If you try to imagine the Dirac delta as a function, you run into a paradox: it should be zero everywhere except at $t = 0$, but its integral should equal 1. So, if you try to graph it, the region under the graph should be an infinitely skinny infinitely tall spike whose area is 1:
But that’s impossible, at least in the standard framework of mathematics—so such a function does not really exist!
Similarly, white noise $w(t)$ is too spiky to be an honest random function, but if we multiply it by any compactly supported smooth function $f(t)$ and integrate, we get a random variable
$\displaystyle{ \int_\infty^\infty w(t) f(t) \, d t }$
whose probability distribution is a Gaussian with mean zero and standard deviation equal to
$\displaystyle{ \sqrt{ \int_\infty^\infty f(t)^2 \, d t} }$
(Note: the word ‘distribution’ has a completely different meaning when it shows up in the phrase probability distribution! I’m assuming you’re comfortable with that meaning already.)
Indeed, the above formulas make sense and are true, not just when $f$ is compactly supported and smooth, but whenever
$\displaystyle{ \sqrt{ \int_\infty^\infty f(t)^2 \, d t} } < \infty$
If you know about Gaussians and you know about this sort of integral, which shows up all over math and physics, you’ll realize that white noise is an extremely natural concept!
### Brownian motion
While white noise is not a random function, its integral
$\displaystyle{ W(s) = \int_0^s w(s) \, ds }$
turns out to be a well-defined random function. It has a lot of names, including: the Wiener process, Brownian motion, red noise and—as a kind of off-color joke—brown noise.
The capital W here stands for Wiener: that is, Norbert Wiener, the famously absent-minded MIT professor who studied this random function… and also invented cybernetics:
Brownian noise sounds like this.
Puzzle: How does it sound different from white noise, and why?
It looks like this:
Here we are zooming in on closer and closer, while rescaling the vertical axis as well. We see that Brownian noise is self-similar: if $W(t)$ is Brownian noise, so is $W(c t)/\sqrt{c}$ for all $c > 0$.
More importantly for us, Brownian noise is the solution of this differential equation:
$\displaystyle{ \frac{d W}{d t} = w(t) }$
where $w(t)$ is white noise. This is essentially true by definition, but making it rigorous takes some work. More fancy stochastic differential equations
$\displaystyle{ \frac{d x}{d t} = f(x, t) + w(t) }$
take even more work to rigorously formulate and solve. You can read about them here:
Stochastic differential equation, Azimuth Library.
It’s actually much easier to explain the difference equations we use to approximately solve these stochastic differential equation on the computer. Suppose we discretize time into steps like this:
$t_i = i \Delta t$
where $\Delta t > 0$ is some fixed number, our ‘time step’. Then we can define
$x(t_{i+1}) = x(t_i) + f(x(t_i), t_i) \; \Delta t + w_i$
where $w_i$ are independent Gaussian random variables with mean zero and standard deviation
$\sqrt{ \Delta t}$
The square root in this formula comes from the definition I gave of white noise.
If we use a random number generator to crank out random numbers $w_i$ distributed in this way, we can write a program to work out the numbers $x(t_i)$ if we are given some initial value $x(0)$. And if the time step $\Delta t$ is small enough, we can hope to get a ‘good approximation to the true solution’. Of course defining what we mean by a ‘good approximation’ is tricky here… but I think it’s more important to just plunge in and see what happens, to get a feel for what’s going on here.
### Stochastic resonance
Let’s do an example of this equation:
$\displaystyle{ \frac{d x}{d t} = f(x, t) + w(t) }$
which exhibits ‘stochastic resonance’. Namely:
$\displaystyle{ \frac{d x}{d t} = x(t) - x(t)^3 + A \; \sin(t) + \sqrt{2D} w(t) }$
Here $A$ and $D$ are constants we get to choose. If we set them both to zero we get:
$\displaystyle{ \frac{d x}{d t} = x(t) - x(t)^3 }$
This has stable equilibrium solutions at $x = \pm 1$ and an unstable equilibrium in between at $x = 0$. So, this is a bistable model similar to the one we’ve been studying, but mathematically simpler!
Then we can add an oscillating time-dependent term:
$\displaystyle{ \frac{d x}{d t} = x(t) - x(t)^3 + A \; \sin(t) }$
which wiggles the system back and forth. This can make it jump from one equilibrium to another.
And then we can add on noise:
$\displaystyle{ \frac{d x}{d t} = x(t) - x(t)^3 + A \; \sin(t) + \sqrt{2D} w(t) }$
Let’s see what the solutions look like!
In the following graphs, the green curve is $A \sin(t),$, while the red curve is $x(t)$. Here is a simulation with a low level of noise:
As you can see, within the time of the simulation there is no transition from the stable state at 1 to the one at -1. If we were doing a climate model, this would be like the Earth staying in the warm state.
Here is a simulation with a high noise level:
The solution $x(t)$ jumps around wildly. By inspecting the graph with your eyes only, you don’t see any pattern in it, do you?
But finally, here is a simulation where the noise level is not too small, and not too big:
Here we see the noise helping the solution $x(t)$ hops from around $x = -1$ to $x = 1$, or vice versa. The solution $x(t)$ is not at all ‘periodic’—it’s quite random. But still, it tends to hop back and forth thanks to the combination of the sinusoidal term $A \sin(t)$ and the noise.
Glyn Adgie, Allan Erskine, Jim Stuttard and Tim van Beek have created an online model that lets you solve this stochastic differential equation for different values of $A$ and $D$. You can try it here:
You can change the values using the sliders under the graphic and see what happens. You can also choose different ‘random seeds’, which means that the random numbers used in the simulation will be different.
To read more about stochastic resonance, go here:
Stochastic resonance, Azimuth Library.
In future weeks I hope to say more about the actual evidence that stochastic resonance plays a role in our glacial cycles! It would also be great to go back to our climate model from last time and add noise. We’re working on that.
## Mathematics of the Environment (Part 6)
10 November, 2012
Last time we saw a ‘bistable’ climate model, where the temperatures compatible with a given amount of sunshine can form an S-shaped curve like this:
The horizontal axis is insolation, the vertical is temperature. Between the green and the red lines the Earth can have 3 temperatures compatible with a given insolation. For example, the black vertical line intersects the S-shaped curve in three points. So we get three possible solutions: a hot Earth, a cold Earth, and an intermediate Earth.
But last time I claimed the intermediate Earth was unstable, so there are just two stable solutions. So, we say this model is bistable. This is like a simple light switch, which has two stable positions but also an intermediate unstable position halfway in between.
(Have you ever enjoyed putting a light switch into this intermediate position? If not, you must not be a physicist.)
Why is the intermediate equilibrium unstable? It seems plausible from the light switch example, but to be sure, we need to go back and study the original equation:
$\displaystyle{ C \frac{d T}{d t} = - A - B T + Q c(T(t)) }$
We need see what happens when we push $T$ slightly away from one of its equilibrium values. We could do this analytically or numerically.
Luckily, Allan Erskine has made a wonderful program that lets us study it numerically: check it out!
Here’s you’ll see a bunch of graphs of temperature as a function of time, $T(t).$ To spice things up, Allan has made the insolation a function of time, which starts out big for some interval $[0,\tau]$ and then drops to its usual value $Q = 341.5.$ So, these graphs are solutions of
$\displaystyle{ C \frac{d T}{d t} = - A - B T + Q(t) c(T(t)) }$
where $Q(t)$ is a step function with
$Q(t) = \left\{ \begin{array}{ccl} Q + X & \mathrm{for} & 0 \le t \le \tau \\ Q & \mathrm{for} & t > \tau \end{array} \right.$
The different graphs show solutions with different initial conditions, ranging from hot to cold. Using sliders on the bottom, you can adjust:
• the coalbedo transition rate $\gamma,$
• the amount $X$ of extra insolation,
• the time $\tau$ at which the extra insolation ends.
I urge you to start by setting $\tau$ to its maximum value. That will make the insolation be constant as a function of time. Then you if $\gamma$ and $X$ are big enough, you’ll get bistability. For example:
I get this with $\gamma$ about 0.08, $X$ about 28.5. You can see a hot stable equilibrium, a cold one, and a solution that hesitates between the two for quite a while before going up to the hot one. This intermediate solution must be starting out very slightly above the unstable equilibrium.
When $X$ is zero, there’s only one equilibrium solution: the cold Earth.
I can’t make $X$ so big that the hot Earth is the only equilibrium, but it’s possible according to our model: I’ll need to change the software a bit to let us make the insolation bigger.
All sorts of more interesting things happen when we move $\tau$ down from its maximum value. I hope you play with the parameters and see what happens. But essentially, what happens is that the hot Earth is only stable before $t = \tau,$ since we need the extra insolation to make that happen. After that, the Earth is fated to go to a cold state.
Needless to say, these results should not be trusted when it comes to the actual climate of our actual planet! More about that later.
We can also check the bistability in a more analytical way. We get an equilibrium solution of
$\displaystyle{ C \frac{d T}{d t} = - A - B T + Q c(T(t)) }$
whenever we find a number $T$ obeying this equation:
$- A - B T + Q c(T) = 0$
We can show that for certain values of $\gamma$ and $Q,$ we get solutions for three different temperatures $T.$ It’s easy to see that $- A - B T + Q c(T)$ is positive for very small $T$: if the Earth were extremely cold, the Sun would warm it up. Similarly, this quantity is negative for very large $T$: the Earth would cool down if it were very hot. So, the reason
$- A - B T + Q c(T) = 0$
has three solutions is that it starts out positive, then goes down below zero, then goes up above zero, and then goes down below zero again. So, for the intermediate point at which it’s zero, we have
$\displaystyle{\frac{d}{dT}( -A - B T + Q c(T)) > 0 }$
That means that if it starts out slightly warmer than this value of $T,$ the temperature will increase—so this solution is unstable. For the hot and cold solutions, we get
$\displaystyle{ \frac{d}{dT}(-A - B T + Q c(T)) < 0 }$
so these equilibria are stable.
### A moral
What morals can we extract from this model?
As far as climate science goes, one moral is that it pays to spend some time making sure we understand simple models before we dive into more complicated ones. Right now we’re looking at a very simple one, but we’re already seeing some interesting phenomena. The kind of model we’re looking at now is called a Budyko-Sellers model. These have been studied since the late 1960′s:
• M. I. Budyko, On the origin of glacial epochs (in Russian), Meteor. Gidrol. 2 (1968), 3-8.
• M. I. Budyko, The effect of solar radiation variations on the climate of the earth, Tellus 21 (1969), 611-619.
• William D. Sellers, A global climatic model based on the energy balance of the earth-atmosphere system, J. Appl. Meteor. 8 (1969), 392-400.
• Carl Crafoord and Erland Källén, A note on the condition for existence of more than one steady state solution in Budyko-Sellers type models, J. Atmos. Sci. 35 (1978), 1123-1125.
• Gerald R. North, David Pollard and Bruce Wielicki, Variational formulation of Budyko-Sellers climate models, J. Atmos. Sci. 36 (1979), 255-259.
I should talk more about some slightly more complex models someday.
It also pays to compare our models to reality! For example, the graphs we’ve seen show some remarkably hot and cold temperatures for the Earth. That’s a bit unnerving. Let’s investigate. Suppose we set $\gamma = 0$ on our slider. Then the coalbedo of the Earth becomes independent of temperature: it’s 0.525, halfway between its icy and ice-free values. Then, when the insolation takes its actual value of 342.5 watts per square meter, the model says the Earth’s temperature is very chilly: about -20 °C!
Does that mean the model is fundamentally flawed? Maybe not! After all, it’s based on very light-colored Earth. Suppose we use the actual albedo of the Earth. Of course that’s hard to define, much less determine. But let’s just look up some average value of the Earth’s albedo: supposedly it’s about 0.3. That gives a coalbedo of $c = 0.7.$ If we plug that in our formula:
$\displaystyle{ Q = \frac{ A + B T } {c} }$
we get 11 °C. That’s not too far from the Earth’s actual average temperature, namely about 15 °C. So the chilly temperature of -20 °C seems to come from an Earth that’s a lot lighter in color than ours.
Our model includes the greenhouse effect, since the coeficients $A$ and $B$ were determined by satellite measurements of how much radiation actually escapes the Earth’s atmosphere and shoots out into space. As a further check to our model, we can look at an even simpler zero-dimensional energy balance model: a completely black Earth with no greenhouse effect. We discussed that earlier.
As he explains, this model gives the Earth a temperature of 6 °C. He also shows that in this model, lowering the albedo to a realistic value of 0.3 lowers the temperature to a chilly -18 ° C. To get from that to something like our Earth, we must take the greenhouse effect into account.
This sort of fiddling around is the sort of thing we must do to study the flaws and virtues of a climate model. Of course, any realistic climate model is vastly more sophisticated than the little toy we’ve been looking at, so the ‘fiddling around’ must also be more sophisticated. With a more sophisticated model, we can also be more demanding. For example, when I said 11 °C is “is not too far from the Earth’s actual average temperature, namely about 15 °C”, I was being very blasé about what’s actually a big discrepancy. I only took that attitude because the calculations we’re doing now are very preliminary.
## Mathematics of the Environment (Part 5)
6 November, 2012
We saw last time that the Earth’s temperature seems to have been getting colder but also more erratic for the last 30 million years or so. Here’s the last 5 million again:
People think these glacial cycles are due to variations in the Earth’s orbit, but as we’ll see later, those cause quite small changes in ‘insolation’—roughly, the amount of sunshine hitting the Earth (as a function of time and location). So, M. I. Budyko, an expert on the glacial cycles, wrote:
Thus, the present thermal regime and glaciations of the Earth prove to be characterized by high instability. Comparatively small changes of radiation—only by 1.0-1.5%—are sufficient for the development of ice cover on the land and oceans that reaches temperate latitudes.
How can small changes in the amount of sunlight hitting the Earth, or other parameters, create big changes in the Earth’s temperature? The obvious answer is positive feedback: some sort of amplifying effect.
But what could it be? Do we know feedback mechanisms that can amplify small changes in temperature? Yes. Here are a few obvious ones:
Water vapor feedback. When it gets warmer, more water evaporates, and the air becomes more humid. But water vapor is a greenhouse gas, which causes additional warming. Conversely, when the Earth cools down, the air becomes drier, so the greenhouse effect becomes weaker, which tends to cool things down.
Ice albedo feedback. Snow and ice reflect more light than liquid oceans or soil. When the Earth warms up, snow and ice melt, so the Earth becomes darker, absorbs more light, and tends to get get even warmer. Conversely, when the Earth cools down, more snow and ice form, so the Earth becomes lighter, absorbs less light, and tends to get even cooler.
Carbon dioxide solubility feedback. Cold water can hold more carbon dioxide than warm water: that’s why opening a warm can of soda can be so explosive. So, when the Earth’s oceans warm up, they release carbon dioxide. But carbon dioxide is a greenhouse gas, which causes additional warming. Conversely, when the oceaans cool down, they absorb more carbon dioxide, so the greenhouse effect becomes weaker, which tends to cool things down.
Of course, there are also negative feedbacks: otherwise the climate would be utterly unstable! There are also complicated feedbacks whose overall effect is harder to evaluate:
Planck feedback. A hotter world radiates more heat, which cools it down. This is the big negative feedback that keeps all the positive feedbacks from making the Earth insanely hot or insanely cold.
Cloud feedback. A warmer Earth has more clouds, which reflect more light but also increase the greenhouse effect.
Lapse rate feedback. An increased greenhouse effect changes the vertical temperature profile of the atmosphere, which has effects of its own—but this works differently near the poles and near the equator.
See also “week302″ of This Week’s Finds, where Nathan Urban tells us more about feedbacks and how big they’re likely to be.
Understanding all these feedbacks, and which ones are important for the glacial cycles we see, is a complicated business. Instead of diving straight into this, let’s try something much simpler. Let’s just think about how the ice albedo effect could, in theory, make the Earth bistable.
To do this, let’s look at the very simplest model in this great not-yet-published book:
• Gerald R. North, Simple Models of Global Climate.
This is a zero-dimensional energy balance model, meaning that it only involves the average temperature of the earth, the average solar radiation coming in, and the average infrared radiation going out.
The average temperature will be $T,$ measured in Celsius. We’ll assume the Earth radiates power square meter equal to
$\displaystyle{ A + B T }$
where $A = 218$ watts/meter2 and $B = 1.90$ watts/meter2 per degree Celsius. This is a linear approximation taken from satellite data on our Earth. In reality, the power emitted grows faster than linearly with temperature.
We’ll assume the Earth absorbs solar energy power per square meter equal to
$Q c(T)$
Here:
$Q$ is the average insolation: that is, the amount of solar power per square meter hitting the top of the Earth’s atmosphere, averaged over location and time of year. In reality $Q$ is about 341.5 watts/meter2. This is one quarter of the solar constant, meaning the solar power per square meter that would hit a panel hovering in space above the Earth’s atmosphere and facing directly at the Sun. (Why a quarter? We’ve seen why: it’s because the area of a sphere is $4 \pi r^2$ while the area of a circle is just $\pi r^2$.)
$c(T)$ is the coalbedo: the fraction of solar power that gets absorbed. The coalbedo depends on the temperature; we’ll have to say how.
Given all this, we get
$\displaystyle{ C \frac{d T}{d t} = - A - B T + Q c(T(t)) }$
where $C$ is Earth’s heat capacity in joules per degree per square meter. Of course this is a funny thing, because heat energy is stored not only at the surface but also in the air and/or water, and the details vary a lot depending on where we are. But if we consider a uniform planet with dry air and no ocean, North says we may roughly take $C$ equal to about half the heat capacity at constant pressure of the column of dry air over a square meter, namely 5 million joules per degree Celsius.
The easiest thing to do is find equilibrium solutions, meaning solutions where $\frac{d T}{d t} = 0,$ so that
$A + B T = Q c(T)$
Now $C$ doesn’t matter anymore! We’d like to solve for $T$ as a function of the insolation $Q,$ but it’s easier to solve for $Q$ as a function of $T$:
$\displaystyle{ Q = \frac{ A + B T } {c(T)} }$
To go further, we need to guess some formula for the coalbedo $c(T).$ The coalbedo, remember, is the fraction of sunlight that gets absorbed when it hits the Earth. It’s 1 minus the albedo, which is the fraction that gets reflected. Here’s a little chart of albedos:
If you get mixed up between albedo and coalbedo, just remember: coal has a high coalbedo.
Since we’re trying to keep things very simple right not, not model nature in all its glorious complexity, let’s just say the average albedo of the Earth is 0.65 when it’s very cold and there’s lots of snow. So, let
$c_i = 1 - 0.65 = 0.35$
be the ‘icy’ coalbedo, good for very low temperatures. Similarly, let’s say the average albedo drops to 0.3 when its very hot and the Earth is darker. So, let
$c_f = 1 - 0.3 = 0.7$
be the ‘ice-free’ coalbedo, good for high temperatures when the Earth is darker.
Then, we need a function of temperature that interpolates between $c_i$ and $c_f.$ Let’s try this:
$c(T) = c_i + \frac{1}{2} (c_f-c_i) (1 + \tanh(\gamma T))$
If you’re not a fan of the hyperbolic tangent function $\tanh,$ this may seem scary. But don’t be intimidated!
The function $\frac{1}{2}(1 + \tanh(\gamma T))$ is just a function that goes smoothly from 0 at low temperatures to 1 at high temperatures. This ensures that the coalbedo is near its icy value $c_i$ at low temperatures, and near its ice-free value $c_f$ at high temperatures. But the fun part here is $\gamma,$ a parameter that says how rapidly the coalbedo rises as the Earth gets warmer. Depending on this, we’ll get different effects!
The function $c(T)$ rises fastest at $T = 0,$ since that’s where $\tanh (\gamma T)$ has the biggest slope. We’re just lucky that in Celsius $T = 0$ is the melting point of ice, so this makes a bit of sense.
Now Allan Erskine‘s programming magic comes into play! I’m very fortunate that the Azimuth Project has attracted some programmers who can make nice software for me to show you. Unfortunately his software doesn’t work on this blog—yet!—so please hop over here to see it in action:
You can slide a slider to adjust the parameter $\gamma$ to various values between 0 and 1.
In the little graph at right, you can see how the coalbedo $c(T)$ changes as a function of the temperature $T.$ In this graph the temperature ranges from -50 °C and 50 °C; the graph depends on what value of $\gamma$ you choose with slider.
In the big graph at left, you can see how the insolation $Q$ required to yield a given temperature $T$ between -50 °C and 50 °C. As we’ve seen, it’s easiest to graph $Q$ as a function of $T:$
$\displaystyle{ Q = \frac{ A + B T } {c_i + \frac{1}{2} (c_f-c_i) (1 + \tanh(\gamma T))} }$
Solving for $T$ here is hard, but we can just flip the graph over to see what equilibrium temperatures $T$ are allowed for a given insolation $Q$ between 200 and 500 watts per square mater.
The exciting thing is that when $\gamma$ gets big enough, three different temperatures are compatible with the same amount of insolation! This means the Earth can be hot, cold or something intermediate even when the amount of sunlight hitting it is fixed. The intermediate state is unstable, it turns out—we’ll see why later. Only the hot and cold states are stable. So, we say the Earth is bistable in this simplified model.
Can you see how big $\gamma$ needs to be for this bistability to kick in? It’s certainly there when $\gamma = 0.05,$ since then we get a graph like this:
When the insolation is less than about 385 W/m2 there’s only a cold state. When it hits 385 W/m2, as shown by the green line, suddenly there are two possible temperatures: a cold one and a much hotter one. When the insolation is higher, as shown by the black line, there are three possible temperatures: a cold one, and unstable intermediate one, and a hot one. And when the insolation gets above 465 W/m2, as shown by the red line, there’s only a hot state!
Mathematically, this model illustrates catastrophe theory. As we slowly turn up $\gamma,$ we get different curves showing how temperature is a function of insolation… until suddenly the curve isn’t the graph of a function anymore: it becomes infinitely steep at one point! After that, we get bistability:
$\gamma = 0.00$
$\gamma = 0.01$
$\gamma = 0.02$
$\gamma = 0.03$
$\gamma = 0.04$
$\gamma = 0.05$
This is called a cusp catastrophe, and you can visualize these curves as slices of a surface in 3d, which looks roughly like this picture:
from here:
• Wolfram Mathworld, Cusp catastrophe. (Includes Mathematica package.)
The cusp catastrophe is ‘structurally stable’, meaning that small perturbations don’t change its qualitative behavior. In other words, whenever you have a smooth graph of a function that gets steeper and steeper until it ‘overhangs’ and ceases to be the graph of a function, it looks like this cusp catastrophe. This statement is quite vague as I’ve just said it— but it’s made 100% precise in catastrophe theory.
Structural stability is a useful concept, because it focuses our attention on robust features of models: features that don’t go away if the model is slightly wrong, as it always is.
There are lots more things to say, but the most urgent question to answer is this: why is the intermediate state unstable when it exists? Why are the other two equilibria stable? We’ll talk about that next time!
## Mathematics of the Environment (Part 3)
13 October, 2012
This week I’ll release these notes before my seminar, so my students (and all of you, too) can read them ahead of time. The reason is that I’m pushing into topics I don’t understand as well as I’d like. So, my notes wrestle with some ideas in too much detail to cover in class—and I’m hoping some students will look at these notes ahead of time to prepare. Also, I’d appreciate your comments!
This week I’ll borrow material shamelessly from here:
• Seymour L. Hess, Introduction to Theoretical Meteorology, Henry Holt and Company, New York, 1959.
It’s an old book: for example, it talks about working out the area under a curve using a gadget called a planimeter, which is what people did before computers.
It also talks about how people measured the solar constant (roughly, the brightness of the Sun) before we could easily put satellites up above the Earth’s atmosphere! And it doesn’t mention global warming.
But despite or perhaps even because of these quaint features, it’s simple and clear. In case it’s not obvious yet, I’m teaching this quarter’s seminar in order to learn stuff. So, I’ll sometimes talk about old work… but if you catch me saying things that are seriously wrong (as opposed to merely primitive), please let me know.
### The plan
Last time we considered a simple model Earth, a blackbody at uniform constant temperature absorbing sunlight and re-emitting the same amount of power in the form of blackbody radiation. We worked out that its temperature would be 6 °C, which is not bad. But then we took into account the fact that the Earth is not black. We got a temperature of -18 °C, which is too cold. The reason is that we haven’t yet equipped our model Earth with an atmosphere! So today let’s try that.
At this point things get a lot more complicated, even if we try a 1-dimensional model where the temperature, pressure and other features of the atmosphere only depend on altitude. So, I’ll only do what I can easily do. I’ll explain some basic laws governing radiation, and then sketch how people applied them to the Earth.
It’ll be good to start with a comment about what we did last time.
### Kirchoff’s law of radiation
When we admitted the Earth wasn’t black, we said that it absorbed only about 70% of the radiation hitting it… but we still modeled it as emitting radiation just like a blackbody! Isn’t there something fishy about this?
Well, no. The Earth is mainly absorbing sunlight at visible frequencies, and at these frequencies it only absorbs about 70% of the radiation that hits it. But it mainly emits infrared light, and at these frequencies it acts like it’s almost black. These frequencies are almost completely different from those where absorption occurs.
But still, this issue is worth thinking about.
After all, emission and absorption are flip sides of the same coin. There’s a deep principle in physics, called reciprocity, which says that how X affects Y is not a separate question from how Y affects X. In fact, if you know the answer to one of these questions, you can figure out the answer to the other!
The first place most people see this principle is Newton’s third law of classical mechanics, saying that if X exerts a force on Y, Y exerts an equal and opposite force on X.
For example: if I punched your nose, your nose punched my fist just as hard, so you have no right to complain.
This law is still often stated in its old-fashioned form:
For every action there is an equal and opposite reaction.
I found this confusing as a student, because ‘force’ was part of the formal terminology of classical mechanics, but not ‘action’—at least not as used in this sentence!—and certainly not ‘reaction’. But as a statement of the basic intuition behind reciprocity, it’s got a certain charm.
In engineering, the principle of reciprocity is sometimes stated like this:
Reciprocity in linear systems is the principle that the response $R_{ab}$ measured at a location $a$ when the system is excited at a location $b$ is exactly equal to $R_{ba},$ which is the response at location $b$ when that same excitation is applied at $a.$ This applies for all frequencies of the excitation.
Again this is a bit confusing, at least if you’re a mathematician who would like to know exactly how a ‘response’ or an ‘excitation’ is defined. It’s also disappointing to see the principle stated in a way that limits it to linear systems. Nonetheless it’s tremendously inspiring. What’s really going on here?
I don’t claim to have gotten to the bottom of it. My hunch is that to a large extent it will come down to the fact that mixed partial derivatives commute. If we’ve got a smooth function $f$ of a bunch of variables $x_1, \dots, x_n,$ and we set
$\displaystyle{ R_{ab} = \frac{\partial^2 f}{\partial x_a \partial x_b} }$
then
$R_{ab} = R_{ba}$
However, I haven’t gotten around to showing that reciprocity boils down to this in all the examples yet. Yet another unification project to add to my list!
Anyway: reciprocity has lots of interesting applications to electromagnetism. And that’s what we’re really talking about now. After all, light is electromagnetic radiation!
The simplest application is one we learn as children:
If I can see you, then you can see me.
or at least:
If light can go from X to Y in a static environment, it can also go from Y to X.
But we want something that sounds a bit different. Namely:
The tendency of a substance to absorb light at some frequency equals its tendency to emit light at that frequency.
This is too vague. We should make it precise, and in a minute I’ll try, but first let me motivate this idea with a thought experiment. Suppose we have a black rock and a white rock in a sealed mirrored container. Suppose they’re in thermal equilibrium at a very high temperature, so they’re glowing red-hot. So, there’s red light bouncing around the container. The black rock will absorb more of this light. But since they’re in thermal equilibrium, the black rock must also be emitting more of this light, or it would gain energy and get hotter than the white one. That would violate the zeroth law of thermodynamics, which implies that in thermal equilibrium, all the parts of a system must be at the same temperature.
More precisely, we have:
Kirchoff’s Law of Thermal Radiation. For any body in thermal equilibrium, its emissivity equals its absorptivity.
Let me explain. Suppose we have a surface made of some homogeneous isotropic material in thermal equilibrium at temperature $T.$ If it’s perfectly black, we saw last time that it emits light with a monochromatic energy flux given by the Planck distribution:
$\displaystyle{ f_{\lambda}(T) = \frac{2 hc^2}{\lambda^5} \frac{1}{ e^{\frac{hc}{\lambda k T}} - 1 } }$
Here $\lambda$ is the wavelength of light and the ‘monochromatic energy flux’ has units of power per area per wavelength.
But if our surface is not perfectly black, we have to multiply this by a fudge factor between 0 and 1 to get the right answer. This factor is called the emissivity of the substance. It can depend on the wavelength of the light quite a lot, and also on the surface’s temperature (since for example ice melts at high temperatures and gets darker). So, let’s call it $e_\lambda(T).$
We can also talk about the absorptivity of our surface, which is the fraction of light it absorbs. Again this depends on the wavelength of the light and the temperature of our surface. So, let’s call it $a_\lambda(T).$
Then Kirchoff’s law of thermal radiation says
$e_\lambda(T) = a_\lambda(T)$
So, for each frequency the emissivity must equal the absorptivity… but it’s still possible for the Earth to have an average emissivity near 1 at the wavelengths of infrared light and near 0.7 at the wavelengths of visible light. So there’s no paradox.
Puzzle 1. Is this law named after the same guy who discovered Kirchhoff’s laws governing electrical circuits?
### Schwarzschild’s equation
Now let’s talk about light shining through the Earth’s atmosphere. Or more generally, light shining through a medium. What happens? It can get absorbed. It can get scattered, bouncing off in different directions. Light can also get emitted, especially if the medium is hot. The air in our atmosphere isn’t hot enough to emit a lot of visible light, but it definitely emits infrared light and microwaves.
It sounds complicated, and it is, but there are things we can say about it. Let me tell you about Schwarzschild’s equation.
Light comes in different wavelengths. So, can ask how much power per square meter this light carries per wavelength. We call this the monochromatic energy flux $I_{\lambda},$ since it depends on the wavelength $\lambda.$ As mentioned last time, this has units W/m2μm, where μm stands for micrometers, a unit of wavelength.
However, because light gets absorbed, scattered and emitted the monochromatic energy flux is really a function $I_{\lambda}(s),$ where $s$ is the distance through the medium. Here I’m imagining an essentially one-dimensional situation, like a beam of sunlight coming down through the air when the Sun is directly overhead. We can generalize this later.
Let’s figure out the basic equation describing how $I_{\lambda}(s)$ changes as a function of $s.$ This is called the equation of radiative transfer, or Schwarzschild’s equation. It won’t tell us how different gases absorb different amounts of light of different frequencies—for that we need to do hard calculations, or experiments. But we can feed the results of these calculations into Schwarzschild’s equation.
For starters, let’s assume that light only gets absorbed but not emitted or scattered. Later we’ll include emission, which is very important for what we’re doing: the Earth’s atmosphere is warm enough to emit significant amounts of infrared light (though not hot enough to emit much visible light). Scattering is also very important, but it can be treated as a combination of absorption and emission.
For absorption only, we have the Beer–Lambert law:
$\displaystyle{ \frac{d I_\lambda(s)}{d s} = - a_\lambda(s) I_\lambda(s) }$
In other words, the amount of radiation that gets absorbed per distance is proportional to the amount of radiation. However, the constant of proportionality $a_\lambda (s)$ can depend on the frequency and the details of our medium at the position $s.$ I don’t know the standard name for this constant $a_\lambda (s),$ so let’s call it the absorption rate.
Puzzle 2. Assuming the Beer–Lambert law, show that the intensity of light at two positions $s_1$ and $s_2$ is related by
$\displaystyle{ I_\lambda(s_2) = e^{-\tau} \; I_\lambda(s_1) }$
where the optical depth $\tau$ of the intervening medium is defined by
$\displaystyle{ \tau = \int_{s_1}^{s_2} a_\lambda(s) \, ds }$
So, a layer of stuff has optical depth equal to 1 if light shining through it has its intensity reduced by a factor of $1/e.$
We can go a bit further if our medium is a rather thin gas like the air in our atmosphere. Then the absorption rate is given by
$a_\lambda(s) = k_\lambda(s) \, \rho(s)$
where $\rho(s)$ is the density of the air at the position $s$ and $k_\lambda(s)$ is its absorption coefficient.
In other words, air absorbs light at a rate proportional to its density, but also depending on what it’s made of, which may vary with position. For example, both the density and the humidity of the atmosphere can depend on its altitude.
What about emission? Air doesn’t just absorb infrared light, it also emits significant amounts of it! As mentioned, a blackbody at temperature $T$ emits light with a monochromatic energy flux given by the Planck distribution:
$\displaystyle{ f_{\lambda}(T) = \frac{2 hc^2}{\lambda^5} \frac{1}{ e^{\frac{hc}{\lambda k T}} - 1 } }$
But a gas like air is far from a blackbody, so we have to multiply this by a fudge factor. Luckily, thanks to Kirchoff’s law of radiation, this factor isn’t so fudgy: it’s just the absorption rate $a_\lambda(s).$
Here are we generalizing Kirchoff’s law from a surface to a column of air, but that’s okay because we can treat a column as a stack of surfaces; letting these become very thin we arrive at a differential formulation of the law that applies to absorption and emission rates instead of absorptivity and emissivity. (If you’re very sharp, you’ll remember that Kirchoff’s law applies to thermal equilibrium, and wonder about that. Air in the atmosphere isn’t in perfect thermal equilibrium, but it’s close enough for what we’re doing here.)
So, when we take absorption and also emission into account, Beer’s law gets another term:
$\displaystyle{ \frac{d I_\lambda(s)}{d s} = - a_\lambda(s) I_\lambda(s) + a_\lambda(s) f_\lambda(T(s)) }$
where $T$ is the temperature of our gas at the position $s.$ In other words:
$\displaystyle{ \frac{d I_\lambda(s)}{d s} = a_\lambda(s) ( f_\lambda(T) - I_\lambda(s))}$
This is Schwarzschild’s equation.
Puzzle 3. Is this equation named after the same guy who discovered the Schwarzschild metric in general relativity, describing a spherically symmetric black hole?
### Application to the atmosphere
In principle, we can use Schwarzschild’s equation to help work out how much sunlight of any frequency actually makes it through the atmosphere down to the Earth, and also how much infrared radiation makes it through the atmosphere out to space. But this is not a calculation I can do here today, because it’s very complicated.
If we actually measure what fraction of radiation of different frequencies makes it through the atmosphere, you’ll see why:
Everything here is a function of the wavelength, measured in micrometers. The smooth red curve is the Planck distribution for light coming from the Sun at a temperature of 5325 K. Most of it is visible light, with a wavelength between 0.4 and 0.7 micrometers. The jagged red region shows how much of this gets through—on a clear day, I assume—and you can see that most of it gets through. The smooth bluish curves are the Planck distributions for light coming from the Earth at various temperatures between 210 K and 310 K. Most of it is infrared light, and not much of it gets through.
This, in a nutshell, is what keeps the Earth warmer than the chilly -18 °C we got last time for an Earth with no atmosphere!
This is the greenhouse effect. As you can see, the absorption of infrared light is mainly due to water vapor, and then carbon dioxide, and then other lesser greenhouse gases, mainly methane, nitrous oxide. Oxygen and ozone also play a minor role, but ozone is more important in blocking ultraviolet light. Rayleigh scattering—the scattering of light by small particles, including molecules and atoms—is also important at short wavelengths, because its strength is proportional to $1/\lambda^4.$ This is why the sky is blue!
Here the wavelengths are measured in nanometers; there are 1000 nanometers in a micrometer. Rayleigh scattering continues to become more important in the ultraviolet.
But right now I want to talk about the infrared. As you can see, the all-important absorption of infrared radiation by water vapor and carbon dioxide is quite complicated. You need quantum mechanics to predict how this works from first principles. Tim van Beek gave a gentle introduction to some of the key ideas here:
• Tim van Beek, A quantum of warmth, Azimuth, 2 July 2011.
Someday it would be fun to get into the details. Not today, though!
You can see what’s going on a bit more clearly here:
The key fact is that infrared is almost completely absorbed for wavelengths between 5.5 and 7 micrometers, or over 14 micrometers. (A ‘micron’ is just an old name for a micrometer.)
### The work of Simpson
The first person to give a reasonably successful explanation of how the power of radiation emitted by the Earth balances the power of the sunlight it absorbs was George Simpson. He did it in 1928:
• George C. Simpson, Further studies in terrestrial radiation, Mem. Roy. Meteor. Soc. 3 (1928), 1–26.
One year earlier, he had tried and failed to understand this problem using a ‘gray atmosphere’ model where the fraction of light that gets through was independent of its wavelength. If you’ve been paying attention, I think you can see why that didn’t work.
In 1928, since he didn’t have a computer, he made a simple model that treated emission of infrared radiation as follows.
He treated the atmosphere as made of layers of varying thickness, each layer containing 0.03 grams per centimeter2 of water vapor. The Earth’s surface radiates infrared almost as a black body. Part of the power is absorbed by the first layer above the surface, while some makes it through. The first layer then re-radiates at the same wavelengths at a rate determined by its temperature. Half this goes downward, while half goes up. Of the part going upward, some is absorbed by the next layer… and so on, up to the top layer. He took this top layer to end at the stratosphere, since the atmosphere is much drier in the stratosphere.
He did this all in a way that depends on the wavelength, but using a simplified model of how each of these layers absorbs infrared light. He assumed it was:
• completely opaque from 5.5 to 7 micrometers (due to water vapor),
• partly transparent from 7 to 8.5 micrometers (interpolating between opaque and transparent),
• completely transparent from 8.5 to 11 micrometers,
• partly transparent from 11 to 14 micrometers (interpolating between transparent and opaque),
• completely opaque above 14 micrometers (due to carbon dioxide and water vapor).
He got this result, at the latitude 50° on a clear day:
The upper smooth curve is the Planck distribution for a temperature of 280 K, corresponding to the ground. The lower smooth curve is the Planck distribution at 218 K, corresponding to the stratosphere. The shaded region is his calculation of the monochromatic flux emitted into space by the Earth. As you can see, it matches the Planck distribution for the stratosphere where the lower atmosphere is completely opaque in his model—between 5.5 and 7 micrometers, and over 14 micrometers. It matches the Planck distribution for the stratosphere where the lower atmosphere is completely transparent. Elsewhere, it interpolates between the two.
The area of this shaded region—calculated with a planimeter, perhaps?—is the total flux emitted into space.
This is just part of the story: he also took clouds into account, and he did different calculations at different latitudes. He got a reasonably good balance between the incoming and outgoing power. In short, he showed that an Earth with its observed temperatures is roughly compatible with his model of how the Earth absorbs and emits radiation. Note that this is just another way to tackle the problem of predicting the temperature given a model.
Also note that Simpson didn’t quite use the Schwarzschild equation. But I guess that in some sense he discretized it—right?
### And so on
This was just the beginning of a series of more and more sophisticated models. I’m too tired to go on right now.
You’ll note one big thing we’ve omitted: any sort of calculation of how the pressure, temperature and humidity of the air varies with altitude! To the extent we talked about those at all, we treated them as inputs. But for a full-fledged one-dimensional model of the Earth’s atmosphere, we’d want to derive them from some principles. There are, after all, some important puzzles:
Puzzle 4. If hot air rises, why does the atmosphere generally get colder as you go upward, at least until you reach the stratosphere?
Puzzle 5. Why is there a tropopause? In other words, why is there a fairly sudden transition 10 kilometers up between the troposphere, where the air is moist, cooler the higher you go, and turbulent, and the stratosphere, where the air is drier, warmer the higher you go, and not turbulent?
There’s a limit to how much we can understand these puzzles using a 1-dimensional model, but we should at least try to make a model of a thin column of air with pressure, temperature and humidity varying as a function of altitude, with sunlight streaming downward and infrared radiation generally going up. If we can’t do that, we’ll never understand more complicated things, like the actual atmosphere.
## Mathematics of the Environment (Part 2)
11 October, 2012
Here are some notes for the second session of my seminar. They are shamelessly borrowed from these sources:
• Tim van Beek, Putting the Earth In a Box, Azimuth, 19 June 2011.
Climate model, Azimuth Library.
### Climate models
Though it’s not my central concern in this class, we should talk a little about climate models.
There are many levels of sophistication when it comes to climate models. It is wise to start with simple, not very realistic models before ascending to complicated, supposedly more realistic ones. This is true in every branch of math or physics: working with simple models gives you insights that are crucial for correctly handling more complicated models. You shouldn’t fly a fighter jet if you haven’t tried something simpler yet, like a bicycle: you’ll probably crash and burn.
As I mentioned last time, models in biology, ecology and climate science pose new challenges compared to models of the simpler systems that physicists like best. As Chris Lee emphasizes, biology inherently deals with ‘high data’ systems where the relevant information can rarely be captured in a few variables, or even a few field equations.
(Field theories involve infinitely many variables, but somehow the ones physicists like best allow us to make a small finite number of measurements and extract a prediction from them! It would be nice to understand this more formally. In quantum field theory, the ‘nice’ field theories are called ‘renormalizable’, but a similar issue shows up classically, as we’ll see in a second.)
The climate system is in part a system that feels like ‘physics’: the flow of air in the atmosphere and water in the ocean. But some of the equations here, for example the Navier–Stokes equations, are already ‘nasty’ by the standards of mathematical physics, since the existence of solutions over long periods of time has not been proved. This is related to ‘turbulence’, a process where information at one length scale can significantly affect information at another dramatically different length scale, making precise predictions difficult.
Climate prediction is, we hope and believe, somewhat insulated from the challenges of weather prediction: we can hope to know the average temperature of the Earth within a degree or two in 5 years even though we don’t know whether it will rain in Manhattan on October 8, 2017. But this hope is something that needs to be studied, not something we can take for granted.
On top of this, the climate is, quite crucially, a biological system. Plant and animal life really affects the climate, as well as being affected by it. So, for example, a really detailed climate model may have a portion specially devoted to the behavior of plankton in the Mediterranean. This means that climate models will never be as ‘neat and clean’ as physicists and mathematicians tend to want—at least, not if these models are trying to be truly realistic. And as I suggested last time, this general type of challenge—the challenge posed by biosystems too complex to precisely model—may ultimately push mathematics in very new directions.
I call this green mathematics, without claiming I know what it will be like. The term is mainly an incitement to think big. I wrote a little about it here.
However, being a bit of an old-fashioned mathematician myself, I’ll start by talking about some very simple climate models, gradually leading up to some interesting puzzles about the ‘ice ages’ or, more properly, ‘glacial cycles’ that have been pestering the Earth for the last 20 million years or so. First, though, let’s take a quick look at the hierarchy of different climate models.
### Different kinds of climate models
Zero-dimensional models are like theories of classical mechanics instead of classical field theory. In other words, they only consider with globally averaged quantities, like the average temperature of the Earth, or perhaps regionally averaged quantities, like the average temperature of each ocean and each continent. This sounds silly, but it’s a great place to start. It amounts to dealing with finitely many variables depending on time:
$(x_1(t), \dots x_n(t))$
We might assume these obey a differential equation, which we can always make first-order by introducing extra variables:
$\displaystyle{ \frac{d x_i}{d t} = f_i(t, x_1(t), \dots, x_n(t)) }$
This kind of model is studied quite generally in the subject of dynamical systems theory.
In particular, energy balance models try to predict the average surface temperature of the Earth depending on the energy flow. Energy comes in from the Sun and is radiated to outer space by the Earth. What happens in between is modeled by averaged feedback equations.
The Earth has various approximately conserved quantities like the total amount of carbon, or oxygen, or nitrogen—radioactive decay creates and destroys these elements, but it’s pretty negligible in climate physics. So, these things move around from one form to another. We can imagine a model where some of our variables $x_i(t)$ are the amounts of carbon in the air, or in the soil, or in the ocean—different ‘boxes’, abstractly speaking. It will flow from one box to another in a way that depends on various other variables in our model. This idea gives class of models called box models.
Here’s one described by Nathan Urban in “week304″ of This Week’s Finds:
I’m interested in box models because they’re a simple example of ‘networked systems’: we’ve got boxes hooked up by wires, or pipes, and we can imagine a big complicated model formed by gluing together smaller models, attaching the wires from one to the wires of another. We can use category theory to formalize this. In category theory we’d call these smaller models ‘morphisms’, and the process of gluing them together is called ‘composing’ them. I’ll talk about this a lot more someday.
One-dimensional models treat temperature and perhaps other quantities as a function of one spatial coordinate (in addition to time): for example, the altitude. This lets us include one dimensional processes of heat transport in the model, like radiation and (a very simplified model of) convection.
Two-dimensional models treat temperature and other quantities as a function of two spatial coordinates (and time): for example, altitude and latitude. Alternatively, we could treat the atmosphere as a thin layer and think of temperature at some fixed altitude as a function of latitude and longitude!
Three-dimensional models treat temperature and other quantities as a function of all three spatial coordinates. At this point we can, if we like, use the full-fledged Navier–Stokes equations to describe the motion of air in the atmosphere and water in the ocean. Needless to say, these models can become very complex and computation-intensive, depending on how many effects we want to take into account and at what resolution we wish to model the atmosphere and ocean.
General circulation models or GCMs try to model the circulation of the atmosphere and/or ocean.
Atmospheric GCMs or AGCMs model the atmosphere and typically contain a land-surface model, while imposing some boundary conditions describing sea surface temperatures. Oceanic GCMs or OGCMs model the ocean (with fluxes from the atmosphere imposed) and may or may not contain a sea ice model. Coupled atmosphere–ocean GCMs or AOGCMs do both atmosphere and ocean. These the basis for detailed predictions of future climate, such as are discussed by the Intergovernmental Panel on Climate Change, or IPCC.
• Backing down a bit, we can consider Earth models of intermediate complexity or EMICs. These might have a 3-dimensional atmosphere and a 2-dimensional ‘slab ocean’, or a 3d ocean and an energy-moisture balance atmosphere.
• Alternatively, we can consider regional circulation models or RCMs. These are limited-area models that can be run at higher resolution than the GCMs and are thus able to better represent fine-grained phenomena, including processes resulting from finer-scale topographic and land-surface features. Typically the regional atmospheric model is run while receiving lateral boundary condition inputs from a relatively-coarse resolution atmospheric analysis model or from the output of a GCM. As Michael Knap pointed out in class, there’s again something from network theory going on here: we are ‘gluing in’ the RCM into a ‘hole’ cut out of a GCM.
Modern GCMs as used in the 2007 IPCC report tended to run around 100-kilometer resolution. Individual clouds can only start to be resolved at about 10 kilometers or below. One way to deal with this is to take the output of higher resolution regional climate models and use it to adjust parameters, etcetera, in GCMs.
### The hierarchy of climate models
The climate scientist Isaac Held has a great article about the hierarchy of climate models:
• Isaac Held, The gap between simulation and understanding in climate modeling, Bulletin of the American Meteorological Society (November 2005), 1609–1614.
In it, he writes:
The importance of such a hierarchy for climate modeling and studies of atmospheric and oceanic dynamics has often been emphasized. See, for example, Schneider and Dickinson (1974), and, especially, Hoskins (1983). But, despite notable exceptions in a few subfields, climate theory has not, in my opinion, been very successful at hierarchy construction. I do not mean to imply that important work has not been performed, of course, but only that the gap between comprehensive climate models and more idealized models has not been successfully closed.
Consider, by analogy, another field that must deal with exceedingly complex systems—molecular biology. How is it that biologists have made such dramatic and steady progress in sorting out the human genome and the interactions of the thousands of proteins of which we are constructed? Without doubt, one key has been that nature has provided us with a hierarchy of biological systems of increasing complexity that are amenable to experimental manipulation, ranging from bacteria to fruit fly to mouse to man. Furthermore, the nature of evolution assures us that much of what we learn from simpler organisms is directly relevant to deciphering the workings of their more complex relatives. What good fortune for biologists to be presented with precisely the kind of hierarchy needed to understand a complex system! Imagine how much progress would have been made if they were limited to studying man alone.
Unfortunately, Nature has not provided us with simpler climate systems that form such a beautiful hierarchy. Planetary atmospheres provide insights into the range of behaviors that are possible, but the known planetary atmospheres are few, and each has its own idiosyncrasies. Their study has connected to terrestrial climate theory on occasion, but the influence has not been systematic. Laboratory simulations of rotating and/or convecting fluids remain valuable and underutilized, but they cannot address our most complex problems. We are left with the necessity of constructing our own hierarchies of climate models.
Because nature has provided the biological hierarchy, it is much easier to focus the attention of biologists on a few representatives of the key evolutionary steps toward greater complexity. And, such a focus is central to success. If every molecular biologist had simply studied his or her own favorite bacterium or insect, rather than focusing so intensively on E. coli or Drosophila melanogaster, it is safe to assume that progress would have been far less rapid.
It is emblematic of our problem that studying the biological hierarchy is experimental science, while constructing and studying climate hierarchies is theoretical science. A biologist need not convince her colleagues that the model organism she is advocating for intensive study is well designed or well posed, but only that it fills an important niche in the hierarchy of complexity and that it is convenient for study. Climate theorists are faced with the difficult task of both constructing a hierarchy of models and somehow focusing the attention of the community on a few of these models so that our efforts accumulate efficiently. Even if one believes that one has defined the E. coli of climate models, it is difficult to energize (and fund) a significant number of researchers to take this model seriously and devote years to its study.
And yet, despite the extra burden of trying to create a consensus as to what the appropriate climate model hierarchies are, the construction of such hierarchies must, I believe, be a central goal of climate theory in the twenty-first century. There are no alternatives if we want to understand the climate system and our
comprehensive climate models. Our understanding will be embedded within these hierarchies.
It is possible that mathematicians, with a lot of training from climate scientists, have the sort of patience and delight in ‘study for study’s sake’ to study this hierarchy of models. Here’s one that Held calls ‘the fruit fly of climate models’:
For more, see:
• Isaac Held, The fruit fly of climate models.
### The very simplest model
The very simplest model is a zero-dimensional energy balance model. In this model we treat the Earth as having just one degree of freedom—its temperature—and we treat it as a blackbody in equilibrium with the radiation coming from the Sun.
A black body is an object that perfectly absorbs and therefore also perfectly emits all electromagnetic radiation at all frequencies. Real bodies don’t have this property; instead, they absorb radiation at certain frequencies better than others, and some not at all. But there are materials that do come rather close to a black body. Usually one adds another assumption to the characterization of an ideal black body: namely, that the radiation is independent of the direction.
When the black body has a certain temperature $T,$ it will emit electromagnetic radiation, so it will send out a certain amount of energy per second for every square meter of surface area. We will call this the energy flux and denote this as $f.$ The SI unit for $f$ is W/m2: that is, watts per square meter. Here the watt is a unit of energy per time.
Electromagnetic radiation comes in different wavelengths. So, can ask how much energy flux our black body emits per change in wavelength. This depends on the wavelength. We will call this the monochromatic energy flux $f_{\lambda}.$ The SI unit for $f_{\lambda}$ is W/m2μm, where μm stands for micrometer: a millionth of a meter, which is a unit of wavelength. We call $f_\lambda$ the ‘monochromatic’ energy flux because it gives a number for any fixed wavelength $\lambda.$ When we integrate the monochromatic energy flux over all wavelengths, we get the energy flux $f.$
Max Planck was able to calculate $f_{\lambda}$ for a blackbody at temperature $T,$ but only by inventing a bit of quantum mechanics. His result is called the Planck distribution: if
$\displaystyle{ f_{\lambda}(T) = \frac{2 hc^2}{\lambda^5} \frac{1}{ e^{\frac{hc}{\lambda k T}} - 1 } }$
where $h$ is Planck’s constant, $c$ is the speed of light, and $k$ is Boltzmann’s constant. Deriving this would be tons of fun, but also a huge digression from the point of this class.
You can integrate $f_\lambda$ over all wavelengths $\lambda$ to get the total energy flux—that is, the total power per square meter emitted by a blackbody. The answer is surprisingly simple: if the total energy flux is defined by
$\displaystyle{f = \int_0^\infty f_{\lambda}(T) \, d \lambda }$
then in fact we can do the integral and get
$f = \sigma \; T^4$
for some constant $\sigma.$ This fact is called the Stefan–Boltzmann law, and $\sigma$ is called the Stefan-Boltzmann constant:
$\displaystyle{ \sigma=\frac{2\pi^5 k^4}{15c^2h^3} \approx 5.67 \times 10^{-8}\, \frac{\mathrm{W}}{\mathrm{m}^2 \mathrm{K}^4} }$
Using this formula, we can assign to every energy flux $f$ a black body temperature $T,$ which is the temperature that an ideal black body would need to have to emit $f.$
Let’s use this to calculate the temperature of the Earth in this simple model! A planet like Earth gets energy from the Sun and loses energy by radiating to space. Since the Earth sits in empty space, these two processes are the only relevant ones that describe the energy flow.
The sunshine near Earth carries an energy flux of about 1370 watts per square meter. If the temperature of the Earth is constant, as much energy is coming in as going out. So, we might try to balance the incoming energy flux with the outgoing flux of a blackbody at temperature $T$:
$\displaystyle{ 1370 \, \textrm{W}/\textrm{m}^2 = \sigma T^4 }$
and then solve for $T$:
$\displaystyle{ T = \left(\frac{1370 \textrm{W}/\textrm{m}^2}{\sigma}\right)^{1/4} }$
We’re making a big mistake here. Do you see what it is? But let’s go ahead and see what we get. As mentioned, the Stefan–Boltzmann constant has a value of
$\displaystyle{ \sigma \approx 5.67 \times 10^{-8} \, \frac{\mathrm{W}}{\mathrm{m}^2 \mathrm{K}^4} }$
so we get
$\displaystyle{ T = \left(\frac{1370}{5.67 \times 10^{-8}} \right)^{1/4} \mathrm{K} } \approx (2.4 \cdot 10^9)^{1/4} \mathrm{K} \approx 394 \mathrm{K}$
This is much too hot! Remember, this temperature is in kelvin, so we need to subtract 273 to get Celsius. Doing so, we get a temperature of 121 °C. This is above the boiling point of water!
Do you see what we did wrong? We neglected a phenomenon known as night. The Earth emits infrared radiation in all directions, but it only absorbs sunlight on the daytime side. Our calculation would be correct if the Earth were a flat disk of perfectly black stuff facing the Sun and perfectly insulated on the back so that it could only emit infrared radiation over the same area that absorbs sunlight! But in fact emission takes place over a larger area than absorption. This makes the Earth cooler.
To get the right answer, we need to take into account the fact that the Earth is round. But just for fun, let’s see how well a flat Earth theory does. A few climate skeptics may even believe this theory. Suppose the Earth were a flat disk of radius $r,$ made of black stuff facing the Sun but not insulated on back. Then it would absorb power equal to
$1370 \cdot \pi r^2$
since the area of the disk is $\pi r^2,$ but it would emit power equal to
$\sigma T^4 \cdot 2 \pi r^2$
since it emits from both the front and back. Setting these equal, we now get
$\displaystyle{ \frac{1370}{2} \textrm{W}/\textrm{m}^2 = \sigma T^4 }$
or
$\displaystyle{ T = \left(\frac{1370 \textrm{W}/\textrm{m}^2}{2 \sigma}\right)^{1/4} }$
This reduces the temperature by a factor of $2^{-1/4} \approx 0.84$ from our previous estimate. So now the temperature works out to be less:
$0.84 \cdot 394 \mathrm{K} \approx 331 \mathrm{K}$
But this is still too hot! It’s 58 °C, or 136 °F for you Americans out there who don’t have a good intuition for Celsius.
So, a flat black Earth facing the Sun would be a very hot Earth.
But now let’s stop goofing around and do the calculation with a round Earth. Now it absorbs a beam of sunlight with area equal to its cross-section, a circle of area $\pi r^2.$ But it emits infrared over its whole area of $4 \pi r^2$: four times as much. So now we get
$\displaystyle{ T = \left(\frac{1370 \textrm{W}/\textrm{m}^2}{4 \sigma}\right)^{1/4} }$
so the temperature is reduced by a further factor of $2^{-1/4}.$ We get
$0.84 \cdot 331 \mathrm{K} \approx 279 \mathrm{K}$
That’s 6 °C. Not bad for a crude approximation! Amusingly, it’s crucial that the area of a sphere is 4 times the area of a circle of the same radius. The question if there is some deeper reason for this simple relation was posed as a geometry puzzle here on Azimuth.
I hope my clowning around hasn’t distracted you from the main point. On average our simplified blackbody Earth absorbs 1370/4 = 342.5 watts of solar power per square meter. So, that’s how much infrared radiation it has to emit. If you can imagine how much heat a 60-watt bulb puts out when it’s surrounded by black paper, we’re saying our simplified Earth emits about 6 times that heat per square meter.
### The second simplest climate model
The next step is to take into account the ‘albedo’ of the Earth. The albedo is the fraction of radiation that is instantly reflected without being absorbed. The albedo of a surface does depend on the material of the surface, and in particular on the wavelength of the radiation, of course. But in a first approximation for the average albedo of earth we can take:
$\mathrm{albedo}_{\mathrm{Earth}} = 0.3$
This means that 30% of the radiation is instantly reflected and only 70% contributes to heating earth. So, instead of getting heated by an average of 342.5 watts per square meter of sunlight, let’s assume it’s heated by
$0.7 \times 342.5 \approx 240$
watts per square meter. Now we get a temperature of
$\displaystyle{ T = \left(\frac{240}{5.67 \times 10^{-8}} \right)^{1/4} K } \approx (4.2 \cdot 10^9)^{1/4} K \approx 255 K$
This is -18 °C. The average temperature of earth is actually estimated to be considerably warmer: about +15 °C. This should not be a surprise: after all, 70% of the planet is covered by liquid water! This is an indication that the average temperature is most probably not below the freezing point of water.
So, our new ‘improved’ calculation gives a worse agreement with reality. The actual Earth is roughly 33 kelvin warmer than our model Earth! What’s wrong?
The main explanation for the discrepancy seems to be: our model Earth doesn’t have an atmosphere yet! Thanks in part to greenhouse gases like water vapor and carbon dioxide, sunlight at visible frequencies can get into the atmosphere more easily than infrared radiation can get out. This warms the Earth. This, in a nutshell, is why dumping a lot of extra carbon dioxide into the air can change our climate. But of course we’ll need to turn to more detailed models, or experimental data, to see how strong this effect is.
Besides the greenhouse effect, there are many other things our ultra-simplified model leaves out: everything associated to the atmosphere and oceans, such as weather, clouds, the altitude-dependence of the temperature of the atmosphere… and also the way the albedo of the Earth depends on location and even on temperature and other factors. There is much much more to say about all this… but not today!
## The Mathematical Origin of Irreversibility
8 October, 2012
guest post by Matteo Smerlak
### Introduction
Thermodynamical dissipation and adaptive evolution are two faces of the same Markovian coin!
Consider this. The Second Law of Thermodynamics states that the entropy of an isolated thermodynamic system can never decrease; Landauer’s principle maintains that the erasure of information inevitably causes dissipation; Fisher’s fundamental theorem of natural selection asserts that any fitness difference within a population leads to adaptation in an evolution process governed by natural selection. Diverse as they are, these statements have two common characteristics:
1. they express the irreversibility of certain natural phenomena, and
2. the dynamical processes underlying these phenomena involve an element of randomness.
Doesn’t this suggest to you the following question: Could it be that thermal phenomena, forgetful information processing and adaptive evolution are governed by the same stochastic mechanism?
The answer is—yes! The key to this rather profound connection resides in a universal property of Markov processes discovered recently in the context of non-equilibrium statistical mechanics, and known as the ‘fluctuation theorem’. Typically stated in terms of ‘dissipated work’ or ‘entropy production’, this result can be seen as an extension of the Second Law of Thermodynamics to small systems, where thermal fluctuations cannot be neglected. But it is actually much more than this: it is the mathematical underpinning of irreversibility itself, be it thermodynamical, evolutionary, or else. To make this point clear, let me start by giving a general formulation of the fluctuation theorem that makes no reference to physics concepts such as ‘heat’ or ‘work’.
### The mathematical fact
Consider a system randomly jumping between states $a, b,\dots$ with (possibly time-dependent) transition rates $\gamma_{a b}(t)$ where $a$ is the state prior to the jump, while $b$ is the state after the jump. I’ll assume that this dynamics defines a (continuous-time) Markov process, namely that the numbers $\gamma_{a b}$ are the matrix entries of an infinitesimal stochastic matrix, which means that its off-diagonal entries are non-negative and that its columns sum up to zero.
Now, each possible history $\omega=(\omega_t)_{0\leq t\leq T}$ of this process can be characterized by the sequence of occupied states $a_{j}$ and by the times $\tau_{j}$ at which the transitions $a_{j-1}\longrightarrow a_{j}$ occur $(0\leq j\leq N)$:
$\omega=(\omega_{0}=a_{0}\overset{\tau_{0}}{\longrightarrow} a_{1} \overset{\tau_{1}}{\longrightarrow}\cdots \overset{\tau_{N}}{\longrightarrow} a_{N}=\omega_{T}).$
Define the skewness $\sigma_{j}(\tau_{j})$ of each of these transitions to be the logarithmic ratio of transition rates:
$\displaystyle{\sigma_{j}(\tau_{j}):=\ln\frac{\gamma_{a_{j}a_{j-1}}(\tau_{j})}{\gamma_{a_{j-1}a_{j}}(\tau_{j})}}$
Also define the self-information of the system in state $a$ at time $t$ by:
$i_a(t):= -\ln\pi_{a}(t)$
where $\pi_{a}(t)$ is the probability that the system is in state $a$ at time $t$, given some prescribed initial distribution $\pi_{a}(0)$. This quantity is also sometimes called the surprisal, as it measures the ‘surprise’ of finding out that the system is in state $a$ at time $t$.
Then the following identity—the detailed fluctuation theorem—holds:
$\mathrm{Prob}[\Delta i-\Sigma=-A] = e^{-A}\;\mathrm{Prob}[\Delta i-\Sigma=A]$
where
$\displaystyle{\Sigma:=\sum_{j}\sigma_{j}(\tau_{j})}$
is the cumulative skewness along a trajectory of the system, and
$\Delta i= i_{a_N}(T)-i_{a_0}(0)$
is the variation of self-information between the end points of this trajectory.
This identity has an immediate consequence: if $\langle\,\cdot\,\rangle$ denotes the average over all realizations of the process, then we have the integral fluctuation theorem:
$\langle e^{-\Delta i+\Sigma}\rangle=1,$
which, by the convexity of the exponential and Jensen’s inequality, implies:
$\langle \Delta i\rangle=\Delta S\geq\langle\Sigma\rangle.$
In short: the mean variation of self-information, aka the variation of Shannon entropy
$\displaystyle{ S(t):= \sum_{a}\pi_{a}(t)i_a(t) }$
is bounded from below by the mean cumulative skewness of the underlying stochastic trajectory.
This is the fundamental mathematical fact underlying irreversibility. To unravel its physical and biological consequences, it suffices to consider the origin and interpretation of the ‘skewness’ term in different contexts. (By the way, people usually call $\Sigma$ the ‘entropy production’ or ‘dissipation function’—but how tautological is that?)
### The physical and biological consequences
Consider first the standard stochastic-thermodynamic scenario where a physical system is kept in contact with a thermal reservoir at inverse temperature $\beta$ and undergoes thermally induced transitions between states $a, b,\dots$. By virtue of the detailed balance condition:
$\displaystyle{ e^{-\beta E_{a}(t)}\gamma_{a b}(t)=e^{-\beta E_{b}(t)}\gamma_{b a}(t),}$
the skewness $\sigma_{j}(\tau_{j})$ of each such transition is $\beta$ times the energy difference between the states $a_{j}$ and $a_{j-1}$, namely the heat received from the reservoir during the transition. Hence, the mean cumulative skewness $\langle \Sigma\rangle$ is nothing but $\beta\langle Q\rangle,$ with $Q$ the total heat received by the system along the process. It follows from the detailed fluctuation theorem that
$\langle e^{-\Delta i+\beta Q}\rangle=1$
and therefore
$\Delta S\geq\beta\langle Q\rangle$
which is of course Clausius’ inequality. In a computational context where the control parameter is the entropy variation itself (such as in a bit-erasure protocol, where $\Delta S=-\ln 2$), this inequality in turn expresses Landauer’s principle: it impossible to decrease the self-information of the system’s state without dissipating a minimal amount of heat into the environment (in this case $-Q \geq k T\ln2$, the ‘Landauer bound’). More general situations (several types of reservoirs, Maxwell-demon-like feedback controls) can be treated along the same lines, and the various forms of the Second Law derived from the detailed fluctuation theorem.
Now, many would agree that evolutionary dynamics is a wholly different business from thermodynamics; in particular, notions such as ‘heat’ or ‘temperature’ are clearly irrelevant to Darwinian evolution. However, the stochastic framework of Markov processes is relevant to describe the genetic evolution of a population, and this fact alone has important consequences. As a simple example, consider the time evolution of mutant fixations $x_{a}$ in a population, with $a$ ranging over the possible genotypes. In a ‘symmetric mutation scheme’, which I understand is biological parlance for ‘reversible Markov process’, meaning one that obeys detailed balance, the ratio between the $a\mapsto b$ and $b\mapsto a$ transition rates is completely determined by the fitnesses $f_{a}$ and $f_b$ of $a$ and $b$, according to
$\displaystyle{\frac{\gamma_{a b}}{\gamma_{b a}} =\left(\frac{f_{b}}{f_{a}}\right)^{\nu} }$
where $\nu$ is a model-dependent function of the effective population size [Sella2005]. Along a given history of mutant fixations, the cumulated skewness $\Sigma$ is therefore given by minus the fitness flux:
$\displaystyle{\Phi=\nu\sum_{j}(\ln f_{a_j}-\ln f_{a_{j-1}}).}$
The integral fluctuation theorem then becomes the fitness flux theorem:
$\displaystyle{ \langle e^{-\Delta i -\Phi}\rangle=1}$
discussed recently by Mustonen and Lässig [Mustonen2010] and implying Fisher’s fundamental theorem of natural selection as a special case. (Incidentally, the ‘fitness flux theorem’ derived in this reference is more general than this; for instance, it does not rely on the ‘symmetric mutation scheme’ assumption above.) The ensuing inequality
$\langle \Phi\rangle\geq-\Delta S$
shows that a positive fitness flux is “an almost universal evolutionary principle of biological systems” [Mustonen2010], with negative contributions limited to time intervals with a systematic loss of adaptation ($\Delta S > 0$). This statement may well be the closest thing to a version of the Second Law of Thermodynamics applying to evolutionary dynamics.
It is really quite remarkable that thermodynamical dissipation and Darwinian evolution can be reduced to the same stochastic mechanism, and that notions such as ‘fitness flux’ and ‘heat’ can arise as two faces of the same mathematical coin, namely the ‘skewness’ of Markovian transitions. After all, the phenomenon of life is in itself a direct challenge to thermodynamics, isn’t it? When thermal phenomena tend to increase the world’s disorder, life strives to bring about and maintain exquisitely fine spatial and chemical structures—which is why Schrödinger famously proposed to define life as negative entropy. Could there be a more striking confirmation of his intuition—and a reconciliation of evolution and thermodynamics in the same go—than the fundamental inequality of adaptive evolution $\langle\Phi\rangle\geq-\Delta S$?
Surely the detailed fluctuation theorem for Markov processes has other applications, pertaining neither to thermodynamics nor adaptive evolution. Can you think of any?
### Proof of the fluctuation theorem
I am a physicist, but knowing that many readers of John’s blog are mathematicians, I’ll do my best to frame—and prove—the FT as an actual theorem.
Let $(\Omega,\mathcal{T},p)$ be a probability space and $(\,\cdot\,)^{\dagger}=\Omega\to \Omega$ a measurable involution of $\Omega$. Denote $p^{\dagger}$ the pushforward probability measure through this involution, and
$\displaystyle{ R=\ln \frac{d p}{d p^\dagger} }$
the logarithm of the corresponding Radon-Nikodym derivative (we assume $p^\dagger$ and $p$ are mutually absolutely continuous). Then the following lemmas are true, with $(1)\Rightarrow(2)\Rightarrow(3)$:
Lemma 1. The detailed fluctuation relation:
$\forall A\in\mathbb{R} \quad p\big(R^{-1}(-A) \big)=e^{-A}p \big(R^{-1}(A) \big)$
Lemma 2. The integral fluctuation relation:
$\displaystyle{\int_{\Omega} d p(\omega)\,e^{-R(\omega)}=1 }$
Lemma 3. The positivity of the Kullback-Leibler divergence:
$D(p\,\Vert\, p^{\dagger}):=\int_{\Omega} d p(\omega)\,R(\omega)\geq 0.$
These are basic facts which anyone can show: $(2)\Rightarrow(3)$ by Jensen’s inequality, $(1)\Rightarrow(2)$ trivially, and $(1)$ follows from $R(\omega^{\dagger})=-R(\omega)$ and the change of variables theorem, as follows,
$\begin{array}{ccl} \displaystyle{ \int_{R^{-1}(-A)} d p(\omega)} &=& \displaystyle{ \int_{R^{-1}(A)}d p^{\dagger}(\omega) } \\ \\ &=& \displaystyle{ \int_{R^{-1}(A)} d p(\omega)\, e^{-R(\omega)} } \\ \\ &=& \displaystyle{ e^{-A} \int_{R^{-1}(A)} d p(\omega)} .\end{array}$
But here is the beauty: if
$(\Omega,\mathcal{T},p)$ is actually a Markov process defined over some time interval $[0,T]$ and valued in some (say discrete) state space $\Sigma$, with the instantaneous probability $\pi_{a}(t)=p\big(\{\omega_{t}=a\} \big)$ of each state $a\in\Sigma$ satisfying the master equation (aka Kolmogorov equation)
$\displaystyle{ \frac{d\pi_{a}(t)}{dt}=\sum_{b\neq a}\Big(\gamma_{b a}(t)\pi_{a}(t)-\gamma_{a b}(t)\pi_{b}(t)\Big),}$
and
• the dagger involution is time-reversal, that is $\omega^{\dagger}_{t}:=\omega_{T-t},$
then for a given path
$\displaystyle{\omega=(\omega_{0}=a_{0}\overset{\tau_{0}}{\longrightarrow} a_{1} \overset{\tau_{1}}{\longrightarrow}\cdots \overset{\tau_{N}}{\longrightarrow} a_{N}=\omega_{T})\in\Omega}$
the logarithmic ratio $R(\omega)$ decomposes into ‘variation of self-information’ and ‘cumulative skewness’ along $\omega$:
$\displaystyle{ R(\omega)=\underbrace{\Big(\ln\pi_{a_0}(0)-\ln\pi_{a_N}(T) \Big)}_{\Delta i(\omega)}-\underbrace{\sum_{j=1}^{N}\ln\frac{\gamma_{a_{j}a_{j-1}}(\tau_{j})}{\gamma_{a_{j-1}a_{j}}(\tau_{j})}}_{\Sigma(\omega)}.}$
This is easy to see if one writes the probability of a path explicitly as
$\displaystyle{p(\omega)=\pi_{a_{0}}(0)\left[\prod_{j=1}^{N}\phi_{a_{j-1}}(\tau_{j-1},\tau_{j})\gamma_{a_{j-1}a_{j}}(\tau_{j})\right]\phi_{a_{N}}(\tau_{N},T)}$
where
$\displaystyle{ \phi_{a}(\tau,\tau')=\phi_{a}(\tau',\tau)=\exp\Big(-\sum_{b\neq a}\int_{\tau}^{\tau'}dt\, \gamma_{a b}(t)\Big)}$
is the probability that the process remains in the state $a$ between the times $\tau$ and $\tau'$. It follows from the above lemma that
Theorem. Let $(\Omega,\mathcal{T},p)$ be a Markov process and let $i,\Sigma:\Omega\rightarrow \mathbb{R}$ be defined as above. Then we have
1. The detailed fluctuation theorem:
$\forall A\in\mathbb{R}, p\big((\Delta i-\Sigma)^{-1}(-A) \big)=e^{-A}p \big((\Delta i-\Sigma)^{-1}(A) \big)$
2. The integral fluctuation theorem:
$\int_{\Omega} d p(\omega)\,e^{-\Delta i(\omega)+\Sigma(\omega)}=1$
3. The ‘Second Law’ inequality:
$\displaystyle{ \Delta S:=\int_{\Omega} d p(\omega)\,\Delta i(\omega)\geq \int_{\Omega} d p(\omega)\,\Sigma(\omega)}$
The same theorem can be formulated for other kinds of Markov processes as well, including diffusion processes (in which case it follows from the Girsanov theorem).
### References
Landauer’s principle was introduced here:
• [Landauer1961] R. Landauer, Irreversibility and heat generation in the computing process}, IBM Journal of Research and Development 5, (1961) 183–191.
and is now being verified experimentally by various groups worldwide.
The ‘fundamental theorem of natural selection’ was derived by Fisher in his book:
• [Fisher1930] R. Fisher, The Genetical Theory of Natural Selection, Clarendon Press, Oxford, 1930.
His derivation has long been considered obscure, even perhaps wrong, but apparently the theorem is now well accepted. I believe the first Markovian models of genetic evolution appeared here:
• [Fisher1922] R. A. Fisher, On the dominance ratio, Proc. Roy. Soc. Edinb. 42 (1922), 321–341.
• [Wright1931] S. Wright, Evolution in Mendelian populations, Genetics 16 (1931), 97–159.
Fluctuation theorems are reviewed here:
• [Sevick2008] E. Sevick, R. Prabhakar, S. R. Williams, and D. J. Searles, Fluctuation theorems, Ann. Rev. Phys. Chem. 59 (2008), 603–633.
Two of the key ideas for the ‘detailed fluctuation theorem’ discussed here are due to Crooks:
• [Crooks1999] Gavin Crooks, The entropy production fluctuation theorem and the nonequilibrium work relation for free energy differences, Phys. Rev. E 60 (1999), 2721–2726.
who identified $(E_{a}(\tau_{j})-E_{a}(\tau_{j-1}))$ as heat, and Seifert:
• [Seifert2005] Udo Seifert, Entropy production along a stochastic trajectory and an integral fluctuation theorem, Phys. Rev. Lett. 95 (2005), 4.
who understood the relevance of the self-information in this context.
The connection between statistical physics and evolutionary biology is discussed here:
• [Sella2005] G. Sella and A.E. Hirsh, The application of statistical physics to evolutionary biology, Proc. Nat. Acad. Sci. USA 102 (2005), 9541–9546.
and the ‘fitness flux theorem’ is derived in
• [Mustonen2010] V. Mustonen and M. Lässig, Fitness flux and ubiquity of adaptive evolution, Proc. Nat. Acad. Sci. USA 107 (2010), 4248–4253.
Schrödinger’s famous discussion of the physical nature of life was published here:
• [Schrödinger1944] E. Schrödinger, What is Life?, Cambridge University Press, Cambridge, 1944.
## Time Crystals
26 September, 2012
When water freezes and forms a crystal, it creates a periodic pattern in space. Could there be something that crystallizes to form a periodic pattern in time? Frank Wilczek, who won the Nobel Prize for helping explain why quarks and gluons trapped inside a proton or neutron act like freely moving particles when you examine them very close up, dreamt up this idea and called it a time crystal:
• Frank Wilczek, Classical time crystals.
• Frank Wilczek, Quantum time crystals.
‘Time crystals’ sound like something from Greg Egan’s Orthogonal trilogy, set in a universe where there’s no fundamental distinction between time and space. But Wilczek wanted to realize these in our universe.
Of course, it’s easy to make a system that behaves in an approximately periodic way while it slowly runs down: that’s how a clock works: tick tock, tick tock, tick tock… But a system that keeps ‘ticking away’ without using up any resource or running down would be a strange new thing. There’s no telling what weird stuff we might do with it.
It’s also interesting because physicists love symmetry. In ordinary physics there are two very important symmetries: spatial translation symmetry, and time translation symmetry. Spatial translation symmetry says that if you move an experiment any amount to the left or right, it works the same way. Time translation symmetry says that if you do an experiment any amount of time earlier or later, it works the same way.
Crystals are fascinating because they ‘spontaneously break’ spatial translation symmetry. Take a liquid, cool it until it freezes, and it forms a crystal which does not look the same if you move it any amount to the right or left. It only looks the same if you move it certain discrete steps to the right or left!
The idea of a ‘time crystal’ is that it’s a system that spontaneously breaks time translation symmetry.
Given how much physicists have studied time translation symmetry and spontaneous symmetry breaking, it’s sort of shocking that nobody before 2012 wrote about this possibility. Or maybe someone did—but I haven’t heard about it.
It takes real creativity to think of an idea so radical yet so simple. But Wilczek is famously creative. For example, he came up with anyons: a new kind of particle, neither boson nor fermion, that lives in a universe where space is 2-dimensional. And now we can make those in the lab.
Unfortunately, Wilczek didn’t know how to make a time crystal. But now a team including Xiang Zhang (seated) and Tongcang Li (standing) at U.C. Berkeley have a plan for how to do it.
Actually they propose a ring-shaped system that’s periodic in time and also in space, as shown in the picture. They call it a space-time crystal:
Here we propose a space-time crystal of trapped ions and a method to realize it experimentally by confining ions in a ring-shaped trapping potential with a static magnetic field. The ions spontaneously form a spatial ring crystal due to Coulomb repulsion. This ion crystal can rotate persistently at the lowest quantum energy state in magnetic fields with fractional fluxes. The persistent rotation of trapped ions produces the temporal order, leading to the formation of a space-time crystal. We show that these space-time crystals are robust for direct experimental observation. The proposed space-time crystals of trapped ions provide a new dimension for exploring many-body physics and emerging properties of matter.
The new paper is here:
• Tongcang Li, Zhe-Xuan Gong, Zhang-Qi Yin, H. T. Quan, Xiaobo Yin, Peng Zhang, L.-M. Duan and Xiang Zhang, Space-time crystals of trapped ions.
Alas, the press release put out by Lawrence Berkeley National Laboratory is very misleading. It describes the space-time crystal as a clock that
will theoretically persist even after the rest of our universe reaches entropy, thermodynamic equilibrium or “heat-death”.
NO!
First of all, ‘reaching entropy’ doesn’t mean anything. More importantly, as time goes by and things fall apart, this space-time crystal, assuming anyone can actually make it, will also fall apart.
I know what the person talking to the reporter was trying to say: the cool thing about this setup is that it gives a system that’s truly time-periodic, not gradually using up some resource and running down like an ordinary clock. But nonphysicist readers, seeing an article entitled ‘A Clock that Will Last Forever’, may be fooled into thinking this setup is, umm, a clock that will last forever. It’s not.
If this setup were the whole universe, it might keep ticking away forever. But in fact it’s just a small, carefully crafted portion of our universe, and it interacts with the rest of our universe, so it will gradually fall apart when everything else does… or in fact much sooner: as soon as the scientists running it turn off the experiment.
### Classifying space-time crystals
What could we do with space-time crystals? It’s way too early to tell, at least for me. But since I’m a mathematician, I’d be happy to classify them. Over on Google+, William Rutiser asked if there are 4d analogs of the 3d crystallographic groups. And the answer is yes! Mathematicians with too much time on their hands have classified the analogues of crystallographic groups in 4 dimensions:
Space group: classification in small dimensions, Wikipedia.
In general these groups are called space groups (see the article for the definition). In 1 dimension there are just two, namely the symmetry groups of this:
— o — o — o — o — o — o —
and this:
— > — > — > — > — > — > —
In 2 dimensions there are 17 and they’re called wallpaper groups. In 3 dimensions there are 230 and they are called crystallographic groups. In 4 dimensions there are 4894, in 5 dimensions there are… hey, Wikipedia leaves this space blank in their table!… and in 6 dimensions there are 28,934,974.
So, there is in principle quite a large subject to study here, if people can figure out how to build a variety of space-time crystals.
There’s already book on this, if you’re interested:
• Harold Brown, Rolf Bulow, Joachim Neubuser, Hans Wondratschek and Hans Zassenhaus, Crystallographic Groups of Four-Dimensional Space, Wiley Monographs in Crystallography, 1978.
## A Course on Quantum Techniques for Stochastic Mechanics
18 September, 2012
Jacob Biamonte and I have come out with a draft of a book!
It’s based on the first 24 network theory posts on this blog. It owes a lot to everyone here, and the acknowledgements just scratch the surface of that indebtedness. At some later time I’d like to go through the posts and find the top twenty people who need to be thanked. But I’m leaving Singapore on Friday, going back to California to teach at U.C. Riverside, so I’ve been rushing to get something out before then.
If you see typos or other problems, please let us know!
We’ve reorganized the original blog articles and polished them up a bit, but we plan to do more before publishing these notes as a book.
I’m looking forward to teaching a seminar called Mathematics of the Environment when I get back to U.C. Riverside, and with luck I’ll put some notes from that on the blog here. I will also be trying to round up a team of grad students to work on network theory.
The next big topics in the network theory series will be electrical circuits and Bayesian networks. I’m beginning to see how these fit together with stochastic Petri nets in a unified framework, but I’ll need to talk and write about it to fill in all the details.
You can get a sense of what this course is about by reading this:
### Foreword
This course is about a curious relation between two ways of describing situations that change randomly with the passage of time. The old way is probability theory and the new way is quantum theory
Quantum theory is based, not on probabilities, but on amplitudes. We can use amplitudes to compute probabilities. However, the relation between them is nonlinear: we take the absolute value of an amplitude and square it to get a probability. It thus seems odd to treat amplitudes as directly analogous to probabilities. Nonetheless, if we do this, some good things happen. In particular, we can take techniques devised in quantum theory and apply them to probability theory. This gives new insights into old problems.
There is, in fact, a subject eager to be born, which is mathematically very much like quantum mechanics, but which features probabilities in the same equations where quantum mechanics features amplitudes. We call this subject stochastic mechanics
#### Plan of the course
In Section 1 we introduce the basic object of study here: a ‘stochastic Petri net’. A stochastic Petri net describes in a very general way how collections of things of different kinds can randomly interact and turn into other things. If we consider large numbers of things, we obtain a simplified deterministic model called the ‘rate equation’, discussed in Section 2. More fundamental, however, is the ‘master equation’, introduced in Section 3. This describes how the probability of having various numbers of things of various kinds changes with time.
In Section 4 we consider a very simple stochastic Petri net and notice that in this case, we can solve the master equation using techniques taken from quantum mechanics. In Section 5 we sketch how to generalize this: for any stochastic Petri net, we can write down an operator called a ‘Hamiltonian’ built from ‘creation and annihilation operators’, which describes the rate of change of the probability of having various numbers of things. In Section 6 we illustrate this with an example taken from population biology. In this example the rate equation is just the logistic equation, one of the simplest models in population biology. The master equation describes reproduction and competition of organisms in a stochastic way.
In Section 7 we sketch how time evolution as described by the master equation can be written as a sum over Feynman diagrams. We do not develop this in detail, but illustrate it with a predator–prey model from population biology. In the process, we give a slicker way of writing down the Hamiltonian for any stochastic Petri net.
In Section 8 we enter into a main theme of this course: the study of equilibrium solutions of the master and rate equations. We present the Anderson–Craciun–Kurtz theorem, which shows how to get equilibrium solutions of the master equation from equilibrium solutions of the rate equation, at least if a certain technical condition holds. Brendan Fong has translated Anderson, Craciun and Kurtz’s original proof into the language of annihilation and creation operators, and we give Fong’s proof here. In this language, it turns out that the equilibrium solutions are mathematically just like ‘coherent states’ in quantum mechanics.
In Section 9 we give an example of the Anderson–Craciun–Kurtz theorem coming from a simple reversible reaction in chemistry. This example leads to a puzzle that is resolved by discovering that the presence of ‘conserved quantities’—quantities that do not change with time—let us construct many equilibrium solutions of the rate equation other than those given by the Anderson–Craciun–Kurtz theorem.
Conserved quantities are very important in quantum mechanics, and they are related to symmetries by a result called Noether’s theorem. In Section 10 we describe a version of Noether’s theorem for stochastic mechanics, which we proved with the help of Brendan Fong. This applies, not just to systems described by stochastic Petri nets, but a much more general class of processes called ‘Markov processes’. In the analogy to quantum mechanics, Markov processes are analogous to arbitrary quantum systems whose time evolution is given by a Hamiltonian. Stochastic Petri nets are analogous to a special case of these: the case where the Hamiltonian is built from annihilation and creation operators. In Section 11 we state the analogy between quantum mechanics and stochastic mechanics more precisely, and with more attention to mathematical rigor. This allows us to set the quantum and stochastic versions of Noether’s theorem side by side and compare them in Section 12.
In Section 13 we take a break from the heavy abstractions and look at a fun example from chemistry, in which a highly symmetrical molecule randomly hops between states. These states can be seen as vertices of a graph, with the transitions as edges. In this particular example we get a famous graph with 20 vertices and 30 edges, called the ‘Desargues graph’.
In Section 14 we note that the Hamiltonian in this example is a ‘graph Laplacian’, and, following a computation done by Greg Egan, we work out the eigenvectors and eigenvalues of this Hamiltonian explicitly. One reason graph Laplacians are interesting is that we can use them as Hamiltonians to describe time evolution in both stochastic and quantum mechanics. Operators with this special property are called ‘Dirichlet operators’, and we discuss them in Section 15. As we explain, they also describe electrical circuits made of resistors. Thus, in a peculiar way, the intersection of quantum mechanics and stochastic mechanics is the study of electrical circuits made of resistors!
In Section 16, we study the eigenvectors and eigenvalues of an arbitrary Dirichlet operator. We introduce a famous result called the Perron–Frobenius theorem for this purpose. However, we also see that the Perron–Frobenius theorem is important for understanding the equilibria of Markov processes. This becomes important later when we prove the ‘deficiency zero theorem’.
We introduce the deficiency zero theorem in Section 17. This result, proved by the chemists Feinberg, Horn and Jackson, gives equilibrium solutions for the rate equation for a large class of stochastic Petri nets. Moreover, these equilibria obey the extra condition that lets us apply the Anderson–Craciun–Kurtz theorem and obtain equlibrium solutions of the master equations as well. However, the deficiency zero theorem is best stated, not in terms of stochastic Petri nets, but in terms of another, equivalent, formalism: ‘chemical reaction networks’. So, we explain chemical reaction networks here, and use them heavily throughout the rest of the course. However, because they are applicable to such a large range of problems, we call them simply ‘reaction networks’. Like stochastic Petri nets, they describe how collections of things of different kinds randomly interact and turn into other things.
In Section 18 we consider a simple example of the deficiency zero theorem taken from chemistry: a diatomic gas. In Section 19 we apply the Anderson–Craciun–Kurtz theorem to the same example.
In Section 20 we begin the final phase of the course: proving the deficiency zero theorem, or at least a portion of it. In this section we discuss the concept of ‘deficiency’, which had been introduced before, but not really explained: the definition that makes the deficiency easy to compute is not the one that says what this concept really means. In Section 21 we show how to rewrite the rate equation of a stochastic Petri net—or equivalently, of a reaction network—in terms of a Markov process. This is surprising because the rate equation is nonlinear, while the equation describing a Markov process is linear in the probabilities involved. The trick is to use a nonlinear operation called ‘matrix exponentiation’. In Section 22 we study equilibria for Markov processes. Then, finally, in Section 23, we use these equilbria to obtain equilibrium solutions of the rate equation, completing our treatment of the deficiency zero theorem.
## More Second Laws of Thermodynamics
24 August, 2012
Oscar Dahlsten is visiting the Centre for Quantum Technologies, so we’re continuing some conversations about entropy that we started last year, back when the Entropy Club was active. But now Jamie Vicary and Brendan Fong are involved in the conversations.
I was surprised when Oscar told me that for a large class of random processes, the usual second law of thermodynamics is just one of infinitely many laws saying that various kinds of disorder increase. I’m annoyed that nobody ever told me about this before! It’s as if they told me about conservation of energy but not conservation of schmenergy, and phlenergy, and zenergy
So I need to tell you about this. You may not understand it, but at least I can say I tried. I don’t want you blaming me for concealing all these extra second laws of thermodynamics!
Here’s the basic idea. Not all random processes are guaranteed to make entropy increase. But a bunch of them always make probability distributions flatter in a certain precise sense. This makes the entropy of the probability distribution increase. But when you make a probability distribution flatter in this sense, a bunch of other quantities increase too! For example, besides the usual entropy, there are infinitely many other kinds of entropy, called ‘Rényi entropies’, one for each number between 0 and ∞. And a doubly stochastic operator makes all the Rényi entropies increase! This fact is a special case of Theorem 10 here:
• Tim van Erven and Peter Harremoës, Rényi divergence and majorization.
Let me state this fact precisely, and then say a word about how this is related to quantum theory and ‘the collapse of the wavefunction’.
To keep things simple let’s talk about probability distributions on a finite set, though Erven and Harremoës generalize it all to a measure space.
How do we make precise the concept that one probability distribution is flatter than another? You know it when you see it, at least some of the time. For example, suppose I have some system in thermal equilibrium at some temperature, and the probabilities of it being in various states look like this:
Then say I triple the temperature. The probabilities flatten out:
But how can we make this concept precise in a completely general way? We can do it using the concept of ‘majorization’. If one probability distribution is less flat than another, people say it ‘majorizes’ that other one.
Here’s the definition. Say we have two probability distributions $p$ and $q$ on the same set. For each one, list the probabilities in decreasing order:
$p_1 \ge p_2 \ge \cdots \ge p_n$
$q_1 \ge q_2 \ge \cdots \ge q_n$
Then we say $p$ majorizes $q$ if
$p_1 + \cdots + p_k \ge q_1 + \cdots + q_k$
for all $1 \le k \le n.$ So, the idea is that the biggest probabilities in the distribution $p$ add up to more than the corresponding biggest ones in $q.$
In 1960, Alfred Rényi defined a generalization of the usual Shannon entropy that depends on a parameter $\beta.$ If $p$ is a probability distribution on a finite set, its Rényi entropy of order $\beta$ is defined to be
$\displaystyle{ H_\beta(p) = \frac{1}{1 - \beta} \ln \sum_i p_i^\beta }$
where $0 \le \beta < \infty.$ Well, to be honest: if $\beta$ is 0, 1, or $\infty$ we have to define this by taking a limit where we let $\beta$ creep up to that value. But the limit exists, and when $\beta = 1$ we get the usual Shannon entropy
$\displaystyle{ H_1(p) = - \sum_i p_i \ln(p_i) }$
As I explained a while ago, Rényi entropies are important ways of measuring biodiversity. But here’s what I learned just now, from the paper by Erven and Harremoës:
Theorem 1. If a probability distribution $p$ majorizes a probability distribution $q,$ its Rényi entropies are smaller:
$\displaystyle{ H_\beta(p) \le H_\beta(q) }$
for all $0 \le \beta < \infty.$
And here’s what makes this fact so nice. If you do something to a classical system in a way that might involve some randomness, we can describe your action using a stochastic matrix. An $n \times n$ matrix $T$ is called stochastic if whenever $p \in \mathbb{R}^n$ is a probability distribution, so is $T p.$ This is equivalent to saying:
• the matrix entries of $T$ are all $\ge 0,$ and
• each column of $T$ sums to 1.
If $T$ is stochastic, it’s not necessarily true that the entropy of $T p$ is greater than or equal to that of $p,$ not even for the Shannon entropy.
Puzzle 1. Find a counterexample.
However, entropy does increase if we use specially nice stochastic matrices called ‘doubly stochastic’ matrices. People say a matrix $T$ doubly stochastic if it’s stochastic and it maps the probability distribution
$\displaystyle{ p_0 = (\frac{1}{n}, \dots, \frac{1}{n}) }$
to itself. This is the most spread-out probability distribution of all: every other probability distribution majorizes this one.
Why do they call such matrices ‘doubly’ stochastic? Well, if you’ve got a stochastic matrix, each column sums to one. But a stochastic operator is doubly stochastic if and only if each row sums to 1 as well.
Here’s a really cool fact:
Theorem 2. If $T$ is doubly stochastic, $p$ majorizes $T p$ for any probability distribution $p \in \mathbb{R}^n.$ Conversely, if a probability distribution $p$ majorizes a probability distribution $q,$ then $q = T p$ for some doubly stochastic matrix $T$.
Taken together, Theorems 1 and 2 say that doubly stochastic transformations increase entropy… but not just Shannon entropy! They increase all the different Rényi entropies, as well. So if time evolution is described by a doubly stochastic matrix, we get lots of ‘second laws of thermodynamics’, saying that all these different kinds of entropy increase!
Finally, what does all this have to do with quantum mechanics, and collapsing the wavefunction? There are different things to say, but this is the simplest:
Theorem 3. Given two probability distributions $p, q \in \mathbb{R}^n$, then $p$ majorizes $q$ if and only there exists a self-adjoint matrix $D$ with eigenvalues $p_i$ and diagonal entries $q_i.$
The matrix $D$ will be a density matrix: a self-adjoint matrix with positive eigenvalues and trace equal to 1. We use such matrices to describe mixed states in quantum mechanics.
Theorem 3 gives a precise sense in which preparing a quantum system in some state, letting time evolve, and then measuring it ‘increases randomness’.
How? Well, suppose we have a quantum system whose Hilbert space is $\mathbb{C}^n.$ If we prepare the system in a mixture of the standard basis states with probabilities $p_i,$ we can describe it with a diagonal density matrix $D_0.$ Then suppose we wait a while and some unitary time evolution occurs. The system is now described by a new density matrix
$D = U D_0 \, U^{-1}$
where $U$ is some unitary operator. If we then do a measurement to see which of the standard basis states our system now lies in, we’ll get the different possible results with probabilities $q_i,$ the diagonal entries of $D.$ But the eigenvalues of $D$ will still be the numbers $p_i.$ So, by the theorem, $p$ majorizes $q$!
So, not only Shannon entropy but also all the Rényi entropies will increase!
Of course, there are some big physics questions lurking here. Like: what about the real world? In the real world, do lots of different kinds of entropy tend to increase, or just some?
Of course, there’s a huge famous old problem about how reversible time evolution can be compatible with any sort of law saying that entropy must always increase! Still, there are some arguments, going back to Boltzmann’s H-theorem, which show entropy increases under some extra conditions. So then we can ask if other kinds of entropy, like Rényi entropy, increase as well. This will be true whenever we can argue that time evolution is described by doubly stochastic matrices. Theorem 3 gives a partial answer, but there’s probably much more to say.
I don’t have much more to say right now, though. I’ll just point out that while doubly stochastic matrices map the ‘maximally smeared-out’ probability distribution
$\displaystyle{ p_0 = (\frac{1}{n}, \dots, \frac{1}{n}) }$
to itself, a lot of this theory generalizes to stochastic matrices that map exactly one other probability distribution to itself. We need to work with relative Rényi entropy instead of Rényi entropy, and so on, but I don’t think these adjustments are really a big deal. And there are nice theorems that let you know when a stochastic matrix maps exactly one probability distribution to itself, based on the Perron–Frobenius theorem.
### References
I already gave you a reference for Theorem 1, namely the paper by Erven and Harremoës, though I don’t think they were the first to prove this particular result: they generalize it quite a lot.
What about Theorem 2? It goes back at least to here:
• Barry C. Arnold, Majorization and the Lorenz Order: A Brief Introduction, Springer Lecture Notes in Statistics 43, Springer, Berlin, 1987.
The partial order on probability distributions given by majorization is also called the ‘Lorenz order’, but mainly when we consider probability distributions on infinite sets. This name presumably comes from the Lorenz curve, a measure of income inequality. This curve shows for the bottom x% of households, what percentage y% of the total income they have:
Puzzle 2. If you’ve got two different probability distributions of incomes, and one majorizes the other, how are their Lorenz curves related?
When we generalize majorization by letting some other probability distribution take the place of
$\displaystyle{ p_0 = (\frac{1}{n}, \dots, \frac{1}{n}) }$
it seems people call it the ‘Markov order’. Here’s a really fascinating paper on that, which I’m just barely beginning to understand:
• A. N. Gorban, P. A. Gorban and G. Judge, Entropy: the Markov ordering approach, Entropy 12 (2010), 1145–1193.
What about Theorem 3? Apparently it goes back to here:
• A. Uhlmann, Wiss. Z. Karl-Marx-Univ. Leipzig 20 (1971), 633.
though I only know this thanks to a more recent paper:
• Michael A. Nielsen, Conditions for a class of entanglement transformations, Phys. Rev. Lett. 83 (1999), 436–439.
By the way, Nielsen’s paper contains another very nice result about majorization! Suppose you have states $\psi$ and $\phi$ of a 2-part quantum system. You can trace out one part and get density matrices describing mixed states of the other part, say $D_\psi$ and $D_\phi$. Then Nielsen shows you can get from $\psi$ to $\phi$ using ‘local operations and classical communication’ if and only if $D_\phi$ majorizes $D_\psi$. Note that things are going backwards here compared to how they’ve been going in the rest of this post: if we can get from $\psi$ to $\phi$, then all forms of entropy go down when we go from $D_\psi$ to $D_\phi$! This ‘anti-second-law’ behavior is confusing at first, but familiar to me by now.
When I first learned all this stuff, I naturally thought of the following question—maybe you did too, just now. If $p, q \in \mathbb{R}^n$ are probability distributions and
$\displaystyle{ H_\beta(p) \le H_\beta(q) }$
for all $0 \le \beta < \infty$, is it true that $p$ majorizes $q$?
Apparently the answer must be no, because Klimesh has gone to quite a bit of work to obtain a weaker conclusion: not that $p$ majorizes $q$, but that $p \otimes r$ majorizes $q \otimes r$ for some probability distribution $r \in \mathbb{R}^m.$ He calls this catalytic majorization, with $r$ serving as a ‘catalyst’:
I thank Vlatko Vedral here at the CQT for pointing this out!
Finally, here is a good general introduction to majorization, pointed out by Vasileios Anagnostopoulos:
• T. Ando, Majorization, doubly stochastic matrices, and comparison of eigenvalues, Linear Algebra and its Applications 118 (1989), 163-–248.
## Network Theory (Part 20)
6 August, 2012
guest post by Jacob Biamonte
We’re in the middle of a battle: in addition to our typical man versus equation scenario, it’s a battle between two theories. For those good patrons following the network theory series, you know the two opposing forces well. It’s our old friends, at it again:
Stochastic Mechanics vs Quantum Mechanics!
Today we’re reporting live from a crossroads, and we’re facing a skirmish that gives rise to what some might consider a paradox. Let me sketch the main thesis before we get our hands dirty with the gory details.
First I need to tell you that the battle takes place at the intersection of stochastic and quantum mechanics. We recall from Part 16 that there is a class of operators called ‘Dirichlet operators’ that are valid Hamiltonians for both stochastic and quantum mechanics. In other words, you can use them to generate time evolution both for old-fashioned random processes and for quantum processes!
Staying inside this class allows the theories to fight it out on the same turf. We will be considering a special subclass of Dirichlet operators, which we call ‘irreducible Dirichlet operators’. These are the ones where starting in any state in our favorite basis of states, we have a nonzero chance of winding up in any other. When considering this subclass, we found something interesting:
Thesis. Let $H$ be an irreducible Dirichlet operator with $n$ eigenstates. In stochastic mechanics, there is only one valid state that is an eigenvector of $H$: the unique so-called ‘Perron–Frobenius state’. The other $n-1$ eigenvectors are forbidden states of a stochastic system: the stochastic system is either in the Perron–Frobenius state, or in a superposition of at least two eigensvectors. In quantum mechanics, all $n$ eigenstates of $H$ are valid states.
This might sound like a riddle, but today as we’ll prove, riddle or not, it’s a fact. If it makes sense, well that’s another issue. As John might have said, it’s like a bone kicked down from the gods up above: we can either choose to chew on it, or let it be. Today we are going to do a bit of chewing.
One of the many problems with this post is that John had a nut loose on his keyboard. It was not broken! I’m saying he wrote enough blog posts on this stuff to turn them into a book. I’m supposed to be compiling the blog articles into a massive LaTeX file, but I wrote this instead.
Another problem is that this post somehow seems to use just about everything said before, so I’m going to have to do my best to make things self-contained. Please bear with me as I try to recap what’s been done. For those of you familiar with the series, a good portion of the background for what we’ll cover today can be found in Part 12 and Part 16.
### At the intersection of two theories
As John has mentioned in his recent talks, the typical view of how quantum mechanics and probability theory come into contact looks like this:
The idea is that quantum theory generalizes classical probability theory by considering observables that don’t commute.
That’s perfectly valid, but we’ve been exploring an alternative view in this series. Here quantum theory doesn’t subsume probability theory, but they intersect:
What goes in the middle you might ask? As odd as it might sound at first, John showed in Part 16 that electrical circuits made of resistors constitute the intersection!
For example, a circuit like this:
gives rise to a Hamiltonian $H$ that’s good both for stochastic mechanics and quantum mechanics. Indeed, he found that the power dissipated by a circuit made of resistors is related to the familiar quantum theory concept known as the expectation value of the Hamiltonian!
$\textrm{power} = -2 \langle \psi, H \psi \rangle$
Oh—and you might think we made a mistake and wrote our Ω (ohm) symbols upside down. We didn’t. It happens that ℧ is the symbol for a ‘mho’—a unit of conductance that’s the reciprocal of an ohm. Check out Part 16 for the details.
### Stochastic mechanics versus quantum mechanics
Let’s recall how states, time evolution, symmetries and observables work in the two theories. Today we’ll fix a basis for our vector space of states, and we’ll assume it’s finite-dimensional so that all vectors have $n$ components over either the complex numbers $\mathbb{C}$ or the reals $\mathbb{R}.$ In other words, we’ll treat our space as either $\mathbb{C}^n$ or $\mathbb{R}^n.$ In this fashion, linear operators that map such spaces to themselves will be represented as square matrices.
Vectors will be written as $\psi_i$ where the index $i$ runs from 1 to $n,$ and we think of each choice of the index as a state of our system—but since we’ll be using that word in other ways too, let’s call it a configuration. It’s just a basic way our system can be.
#### States
Besides the configurations $i = 1,\dots, n,$ we have more general states that tell us the probability or amplitude of finding our system in one of these configurations:
Stochastic states are $n$-tuples of nonnegative real numbers:
$\psi_i \in \mathbb{R}^+$
The probability of finding the system in the $i$th configuration is defined to be $\psi_i.$ For these probabilities to sum to one, $\psi_i$ needs to be normalized like this:
$\displaystyle{ \sum_i \psi_i = 1 }$
or in the notation we’re using in these articles:
$\displaystyle{ \langle \psi \rangle = 1 }$
where we define
$\displaystyle{ \langle \psi \rangle = \sum_i \psi_i }$
Quantum states are $n$-tuples of complex numbers:
$\psi_i \in \mathbb{C}$
The probability of finding a state in the $i$th configuration is defined to be $|\psi(x)|^2.$ For these probabilities to sum to one, $\psi$ needs to be normalized like this:
$\displaystyle{ \sum_i |\psi_i|^2 = 1 }$
or in other words
$\langle \psi, \psi \rangle = 1$
where the inner product of two vectors $\psi$ and $\phi$ is defined by
$\displaystyle{ \langle \psi, \phi \rangle = \sum_i \overline{\psi}_i \phi_i }$
Now, the usual way to turn a quantum state $\psi$ into a stochastic state is to take the absolute value of each number $\psi_i$ and then square it. However, if the numbers $\psi_i$ happen to be nonnegative, we can also turn $\psi$ into a stochastic state simply by multiplying it by a number to ensure $\langle \psi \rangle = 1.$
This is very unorthodox, but it lets us evolve the same vector $\psi$ either stochastically or quantum-mechanically, using the recipes I’ll describe next. In physics jargon these correspond to evolution in ‘real time’ and ‘imaginary time’. But don’t ask me which is which: from a quantum viewpoint stochastic mechanics uses imaginary time, but from a stochastic viewpoint it’s the other way around!
#### Time evolution
Time evolution works similarly in stochastic and quantum mechanics, but with a few big differences:
• In stochastic mechanics the state changes in time according to the master equation:
$\displaystyle{ \frac{d}{d t} \psi(t) = H \psi(t) }$
which has the solution
$\psi(t) = \exp(t H) \psi(0)$
• In quantum mechanics the state changes in time according to Schrödinger’s equation:
$\displaystyle{ \frac{d}{d t} \psi(t) = -i H \psi(t) }$
which has the solution
$\psi(t) = \exp(-i t H) \psi(0)$
The operator $H$ is called the Hamiltonian. The properties it must have depend on whether we’re doing stochastic mechanics or quantum mechanics:
• We need $H$ to be infinitesimal stochastic for time evolution given by $\exp(-i t H)$ to send stochastic states to stochastic states. In other words, we need that (i) its columns sum to zero and (ii) its off-diagonal entries are real and nonnegative:
$\displaystyle{ \sum_i H_{i j}=0 }$
$i\neq j \Rightarrow H_{i j}\geq 0$
• We need $H$ to be self-adjoint for time evolution given by $\exp(-t H)$ to send quantum states to quantum states. So, we need
$H = H^\dagger$
where we recall that the adjoint of a matrix is the conjugate of its transpose:
$\displaystyle{ (H^\dagger)_{i j} := \overline{H}_{j i} }$
We are concerned with the case where the operator $H$ generates both a valid quantum evolution and also a valid stochastic one:
$H$ is a Dirichlet operator if it’s both self-adjoint and infinitesimal stochastic.
We will soon go further and zoom in on this intersection! But first let’s finish our review.
#### Symmetries
As John explained in Part 12, besides states and observables we need symmetries, which are transformations that map states to states. These include the evolution operators which we only briefly discussed in the preceding subsection.
• A linear map $U$ that sends quantum states to quantum states is called an isometry, and isometries are characterized by this property:
$U^\dagger U = 1$
• A linear map $U$ that sends stochastic states to stochastic states is called a stochastic operator, and stochastic operators are characterized by these properties:
$\displaystyle{ \sum_i U_{i j} = 1 }$
and
$U_{i j} \geq 0$
A notable difference here is that in our finite-dimensional situation, isometries are always invertible, but stochastic operators may not be! If $U$ is an $n \times n$ matrix that’s an isometry, $U^\dagger$ is its inverse. So, we also have
$U U^\dagger = 1$
and we say $U$ is unitary. But if $U$ is stochastic, it may not have an inverse—and even if it does, its inverse is rarely stochastic. This explains why in stochastic mechanics time evolution is often not reversible, while in quantum mechanics it always is.
Puzzle 1. Suppose $U$ is a stochastic $n \times n$ matrix whose inverse is stochastic. What are the possibilities for $U$?
It is quite hard for an operator to be a symmetry in both stochastic and quantum mechanics, especially in our finite-dimensional situation:
Puzzle 2. Suppose $U$ is an $n \times n$ matrix that is both stochastic and unitary. What are the possibilities for $U$?
#### Observables
‘Observables’ are real-valued quantities that can be measured, or predicted, given a specific theory.
• In quantum mechanics, an observable is given by a self-adjoint matrix $O,$ and the expected value of the observable $O$ in the quantum state $\psi$ is
$\displaystyle{ \langle \psi , O \psi \rangle = \sum_{i,j} \overline{\psi}_i O_{i j} \psi_j }$
• In stochastic mechanics, an observable $O$ has a value $O_i$ in each configuration $i,$ and the expected value of the observable $O$ in the stochastic state $\psi$ is
$\displaystyle{ \langle O \psi \rangle = \sum_i O_i \psi_i }$
We can turn an observable in stochastic mechanics into an observable in quantum mechanics by making a diagonal matrix whose diagonal entries are the numbers $O_i.$
### From graphs to matrices
Back in Part 16, John explained how a graph with positive numbers on its edges gives rise to a Hamiltonian in both quantum and stochastic mechanics—in other words, a Dirichlet operator.
Here’s how this works. We’ll consider simple graphs: graphs without arrows on their edges, with at most one edge from one vertex to another, with no edges from a vertex to itself. And we’ll only look at graphs with finitely many vertices and edges. We’ll assume each edge is labelled by a positive number, like this:
If our graph has $n$ vertices, we can create an $n \times n$ matrix $A$ where $A_{i j}$ is the number labelling the edge from $i$ to $j,$ if there is such an edge, and 0 if there’s not. This matrix is symmetric, with real entries, so it’s self-adjoint. So $A$ is a valid Hamiltonian in quantum mechanics.
How about stochastic mechanics? Remember that a Hamiltonian $H$ in stochastic mechanics needs to be ‘infinitesimal stochastic’. So, its off-diagonal entries must be nonnegative, which is indeed true for our $A,$ but also the sums of its columns must be zero, which is not true when our $A$ is nonzero.
But now comes the best news you’ve heard all day: we can improve $A$ to a stochastic operator in a way that is completely determined by $A$ itself! This is done by subtracting a diagonal matrix $L$ whose entries are the sums of the columns of $A$:
$L_{i i} = \sum_i A_{i j}$
$i \ne j \Rightarrow L_{i j} = 0$
It’s easy to check that
$H = A - L$
is still self-adjoint, but now also infinitesimal stochastic. So, it’s a Dirichlet operator: a good Hamiltonian for both stochastic and quantum mechanics!
In Part 16, we saw a bit more: every Dirichlet operator arises this way. It’s easy to see. You just take your Dirichlet operator and make a graph with one edge for each nonzero off-diagonal entry. Then you label the edge with this entry.
So, Dirichlet operators are essentially the same as finite simple graphs with edges labelled by positive numbers.
Now, a simple graph can consist of many separate ‘pieces’, called components. Then there’s no way for a particle hopping along the edges to get from one component to another, either in stochastic or quantum mechanics. So we might as well focus our attention on graphs with just one component. These graphs are called ‘connected’. In other words:
Definition. A simple graph is connected if it is nonempty and there is a path of edges connecting any vertex to any other.
Our goal today is to understand more about Dirichlet operators coming from connected graphs. For this we need to learn the Perron–Frobenius theorem. But let’s start with something easier.
### Perron’s theorem
In quantum mechanics it’s good to think about observables that have positive expected values:
$\langle \psi, O \psi \rangle > 0$
for every quantum state $\psi \in \mathbb{C}^n.$ These are called positive definite. But in stochastic mechanics it’s good to think about matrices that are positive in a more naive sense:
Definition. An $n \times n$ real matrix $T$ is positive if all its entries are positive:
$T_{i j} > 0$
for all $1 \le i, j \le n.$
Similarly:
Definition. A vector $\psi \in \mathbb{R}^n$ is positive if all its components are positive:
$\psi_i > 0$
for all $1 \le i \le n.$
We’ll also define nonnegative matrices and vectors in the same way, replacing $> 0$ by $\ge 0.$ A good example of a nonnegative vector is a stochastic state.
In 1907, Perron proved the following fundamental result about positive matrices:
Perron’s Theorem. Given a positive square matrix $T,$ there is a positive real number $r,$ called the Perron–Frobenius eigenvalue of $T,$ such that $r$ is an eigenvalue of $T$ and any other eigenvalue $\lambda$ of $T$ has $|\lambda| < r.$ Moreover, there is a positive vector $\psi \in \mathbb{R}^n$ with $T \psi = r \psi.$ Any other vector with this property is a scalar multiple of $\psi.$ Furthermore, any nonnegative vector that is an eigenvector of $T$ must be a scalar multiple of $\psi.$
In other words, if $T$ is positive, it has a unique eigenvalue with the largest absolute value. This eigenvalue is positive. Up to a constant factor, it has an unique eigenvector. We can choose this eigenvector to be positive. And then, up to a constant factor, it’s the only nonnegative eigenvector of $T.$
### From matrices to graphs
The conclusions of Perron’s theorem don’t hold for matrices that are merely nonnegative. For example, these matrices
$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) , \qquad \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$
are nonnegative, but they violate lots of the conclusions of Perron’s theorem.
Nonetheless, in 1912 Frobenius published an impressive generalization of Perron’s result. In its strongest form, it doesn’t apply to all nonnegative matrices; only to those that are ‘irreducible’. So, let us define those.
We’ve seen how to build a matrix from a graph. Now we need to build a graph from a matrix! Suppose we have an $n \times n$ matrix $T.$ Then we can build a graph $G_T$ with $n$ vertices where there is an edge from the $i$th vertex to the $j$th vertex if and only if $T_{i j} \ne 0.$
But watch out: this is a different kind of graph! It’s a directed graph, meaning the edges have directions, there’s at most one edge going from any vertex to any vertex, and we do allow an edge going from a vertex to itself. There’s a stronger concept of ‘connectivity’ for these graphs:
Definition. A directed graph is strongly connected if there is a directed path of edges going from any vertex to any other vertex.
So, you have to be able to walk along edges from any vertex to any other vertex, but always following the direction of the edges! Using this idea we define irreducible matrices:
Definition. A nonnegative square matrix $T$ is irreducible if its graph $G_T$ is strongly connected.
### The Perron–Frobenius theorem
Now we are ready to state:
The Perron-Frobenius Theorem. Given an irreducible nonnegative square matrix $T,$ there is a positive real number $r,$ called the Perron–Frobenius eigenvalue of $T,$ such that $r$ is an eigenvalue of $T$ and any other eigenvalue $\lambda$ of $T$ has $|\lambda| \le r.$ Moreover, there is a positive vector $\psi \in \mathbb{R}^n$ with $T\psi = r \psi.$ Any other vector with this property is a scalar multiple of $\psi.$ Furthermore, any nonnegative vector that is an eigenvector of $T$ must be a scalar multiple of $\psi.$
The only conclusion of this theorem that’s weaker than those of Perron’s theorem is that there may be other eigenvalues with $|\lambda| = r.$ For example, this matrix is irreducible and nonnegative:
$\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$
Its Perron–Frobenius eigenvalue is 1, but it also has -1 as an eigenvalue. In general, Perron-Frobenius theory says quite a lot about the other eigenvalues on the circle $|\lambda| = r,$ but we won’t need that fancy stuff here.
Perron–Frobenius theory is useful in many ways, from highbrow math to ranking football teams. We’ll need it not just today but also later in this series. There are many books and other sources of information for those that want to take a closer look at this subject. If you’re interested, you can search online or take a look at these:
• Dimitrious Noutsos, Perron Frobenius theory and some extensions, 2008. (Includes proofs of the basic theorems.)
• V. S. Sunder, Perron Frobenius theory, 18 December 2009. (Includes applications to graph theory, Markov chains and von Neumann algebras.)
• Stephen Boyd, Lecture 17: Perron Frobenius theory, Winter 2008-2009. (Includes a max-min characterization of the Perron–Frobenius eigenvalue and applications to Markov chains, economics, population growth and power control.)
I have not taken a look myself, but if anyone is interested and can read German, the original work appears here:
• Oskar Perron, Zur Theorie der Matrizen, Math. Ann. 64 (1907), 248–263.
• Georg Frobenius, Über Matrizen aus nicht negativen Elementen, S.-B. Preuss Acad. Wiss. Berlin (1912), 456–477.
And, of course, there’s this:
• Wikipedia, Perron–Frobenius theorem.
It’s got a lot of information.
### Irreducible Dirichlet operators
Now comes the payoff. We saw how to get a Dirichlet operator $H$ from any finite simple graph with edges labelled by positive numbers. Now let’s apply Perron–Frobenius theory to prove our thesis.
Unfortunately, the matrix $H$ is rarely nonnegative. If you remember how we built it, you’ll see its off-diagonal entries will always be nonnegative… but its diagonal entries can be negative.
Luckily, we can fix this just by adding a big enough multiple of the identity matrix to $H$! The result is a nonnegative matrix
$T = H + c I$
where $c > 0$ is some large number. This matrix $T$ has the same eigenvectors as $H.$ The off-diagonal matrix entries of $T$ are the same as those of $A,$ so $T_{i j}$ is nonzero for $i \ne j$ exactly when the graph we started with has an edge from $i$ to $j.$ So, for $i \ne j,$ the graph $G_T$ will have an directed edge going from $i$ to $j$ precisely when our original graph had an edge from $i$ to $j.$ And that means that if our original graph was connected, $G_T$ will be strongly connected. Thus, by definition, the matrix $T$ is irreducible!
Since $T$ is nonnegative and irreducible, the Perron–Frobenius theorem swings into action and we conclude:
Lemma. Suppose $H$ is the Dirichlet operator coming from a connected finite simple graph with edges labelled by positive numbers. Then the eigenvalues of $H$ are real. Let $\lambda$ be the largest eigenvalue. Then there is a positive vector $\psi \in \mathbb{R}^n$ with $H\psi = \lambda \psi.$ Any other vector with this property is a scalar multiple of $\psi.$ Furthermore, any nonnegative vector that is an eigenvector of $H$ must be a scalar multiple of $\psi.$
Proof. The eigenvalues of $H$ are real since $H$ is self-adjoint. Notice that if $r$ is the Perron–Frobenius eigenvalue of $T = H + c I$ and
$T \psi = r \psi$
then
$H \psi = (r - c)\psi$
By the Perron–Frobenius theorem the number $r$ is positive, and it has the largest absolute value of any eigenvalue of $T.$ Thanks to the subtraction, the eigenvalue $r - c$ may not have the largest absolute value of any eigenvalue of $H.$ It is, however, the largest eigenvalue of $H,$ so we take this as our $\lambda.$ The rest follows from the Perron–Frobenius theorem. █
But in fact we can improve this result, since the largest eigenvalue $\lambda$ is just zero. Let’s also make up a definition, to make our result sound more slick:
Definition. A Dirichlet operator is irreducible if it comes from a connected finite simple graph with edges labelled by positive numbers.
This meshes nicely with our earlier definition of irreducibility for nonnegative matrices. Now:
Theorem. Suppose $H$ is an irreducible Dirichlet operator. Then $H$ has zero as its largest real eigenvalue. There is a positive vector $\psi \in \mathbb{R}^n$ with $H\psi = 0.$ Any other vector with this property is a scalar multiple of $\psi.$ Furthermore, any nonnegative vector that is an eigenvector of $H$ must be a scalar multiple of $\psi.$
Proof. Choose $\lambda$ as in the Lemma, so that $H\psi = \lambda \psi.$ Since $\psi$ is positive we can normalize it to be a stochastic state:
$\displaystyle{ \sum_i \psi_i = 1 }$
Since $H$ is a Dirichlet operator, $\exp(t H)$ sends stochastic states to stochastic states, so
$\displaystyle{ \sum_i (\exp(t H) \psi)_i = 1 }$
for all $t \ge 0.$ On the other hand,
$\displaystyle{ \sum_i (\exp(t H)\psi)_i = \sum_i e^{t \lambda} \psi_i = e^{t \lambda} }$
so we must have $\lambda = 0.$ █
What’s the point of all this? One point is that there’s a unique stochastic state $\psi$ that’s an equilibrium state: since $H \psi = 0,$ it doesn’t change with time. It’s also globally stable: since all the other eigenvalues of $H$ are negative, all other stochastic states converge to this one as time goes forward.
### An example
There are many examples of irreducible Dirichlet operators. For instance, in Part 15 we talked about graph Laplacians. The Laplacian of a connected simple graph is always irreducible. But let us try a different sort of example, coming from the picture of the resistors we saw earlier:
Let’s create a matrix $A$ whose entry $A_{i j}$ is the number labelling the edge from $i$ to $j$ if there is such an edge, and zero otherwise:
$A = \left( \begin{array}{ccccc} 0 & 2 & 1 & 0 & 1 \\ 2 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 2 & 1 \\ 0 & 1 & 2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array} \right)$
Remember how the game works. The matrix $A$ is already a valid Hamiltonian for quantum mechanics, since it’s self adjoint. However, to get a valid Hamiltonian for both stochastic and quantum mechanics—in other words, a Dirichlet operator—we subtract the diagonal matrix $L$ whose entries are the sums of the columns of $A.$ In this example it just so happens that the column sums are all 4, so $L = 4 I,$ and our Dirichlet operator is
$H = A - 4 I = \left( \begin{array}{rrrrr} -4 & 2 & 1 & 0 & 1 \\ 2 & -4 & 0 & 1 & 1 \\ 1 & 0 & -4 & 2 & 1 \\ 0 & 1 & 2 & -4 & 1 \\ 1 & 1 & 1 & 1 & -4 \end{array} \right)$
We’ve set up this example so it’s easy to see that the vector $\psi = (1,1,1,1,1)$ has
$H \psi = 0$
So, this is the unique eigenvector for the eigenvalue 0. We can use Mathematica to calculate the remaining eigenvalues of $H.$ The set of eigenvalues is
$\{0, -7, -8, -8, -3 \}$
As we expect from our theorem, the largest real eigenvalue is 0. By design, the eigenstate associated to this eigenvalue is
$| v_0 \rangle = (1, 1, 1, 1, 1)$
(This funny notation for vectors is common in quantum mechanics, so don’t worry about it.) All the other eigenvectors fail to be nonnegative, as predicted by the theorem. They are:
$| v_1 \rangle = (1, -1, -1, 1, 0)$
$| v_2 \rangle = (-1, 0, -1, 0, 2)$
$| v_3 \rangle = (-1, 1, -1, 1, 0)$
$| v_4 \rangle = (-1, -1, 1, 1, 0)$
To compare the quantum and stochastic states, consider first $|v_0\rangle.$ This is the only eigenvector that can be normalized to a stochastic state. Remember, a stochastic state must have nonnegative components. This rules out $|v_1\rangle$ through to $|v_4\rangle$ as valid stochastic states, no matter how we normalize them! However, these are allowed as states in quantum mechanics, once we normalize them correctly. For a stochastic system to be in a state other than the Perron–Frobenius state, it must be a linear combination of least two eigenstates. For instance,
$\psi_a = (1-a)|v_0\rangle + a |v_1\rangle$
can be normalized to give stochastic state only if $0 \leq a \leq \frac{1}{2}.$
And, it’s easy to see that it works this way for any irreducible Dirichlet operator, thanks to our theorem. So, our thesis has been proved true!
### Puzzles on irreducibility
Let us conclude with a couple more puzzles. There are lots of ways to characterize irreducible nonnegative matrices; we don’t need to mention graphs. Here’s one:
Puzzle 3. Let $T$ be a nonnegative $n \times n$ matrix. Show that $T$ is irreducible if and only if for all $i,j \ge 0,$ $(A^m)_{i j} > 0$ for some natural number $m.$
You may be confused because today we explained the usual concept of irreducibility for nonnegative matrices, but also defined a concept of irreducibility for Dirichlet operators. Luckily there’s no conflict: Dirichlet operators aren’t nonnegative matrices, but if we add a big multiple of the identity to a Dirichlet operator it becomes a nonnegative matrix, and then:
Puzzle 4. Show that a Dirichlet operator $H$ is irreducible if and only if the nonnegative operator $H + c I$ (where $c$ is any sufficiently large constant) is irreducible.
Irreducibility is also related to the nonexistence of interesting conserved quantities. In Part 11 we saw a version of Noether’s Theorem for stochastic mechanics. Remember that an observable $O$ in stochastic mechanics assigns a number $O_i$ to each configuration $i = 1, \dots, n.$ We can make a diagonal matrix with $O_i$ as its diagonal entries, and by abuse of language we call this $O$ as well. Then we say $O$ is a conserved quantity for the Hamiltonian $H$ if the commutator $[O,H] = O H - H O$ vanishes.
Puzzle 5. Let $H$ be a Dirichlet operator. Show that $H$ is irreducible if and only if every conserved quantity $O$ for $H$ is a constant, meaning that for some $c \in \mathbb{R}$ we have $O_i = c$ for all $i.$ (Hint: examine the proof of Noether’s theorem.)
In fact this works more generally:
Puzzle 6. Let $H$ be an infinitesimal stochastic matrix. Show that $H + c I$ is an irreducible nonnegative matrix for all sufficiently large $c$ if and only if every conserved quantity $O$ for $H$ is a constant.
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Aim at and aim to..which one is right : GMAT Verbal Section
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# Aim at and aim to..which one is right
Author Message
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Aim at and aim to..which one is right [#permalink]
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29 May 2004, 12:26
Violence in the stands at soccer matches has gotten so pronounced in several European countries that some stadiums have adopted new rules that aim to identify fans of visiting teams and that seat them in a separate area.
(A) to identify fans of visiting teams and that seat them
(B) wrong
(C) wrong
(D) at identifying fans of visiting teams so as to seat them
(E) wrong
I have a confusion with aim at and aim to. Which one to use and when.
Satya
If you have any questions
New!
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### Show Tags
29 May 2004, 13:02
Hello Sathya
To Aim at is the right Idiom
Example :- You have to aim at 780+ to score around 750 in gmat.
You have to aim at the red circle to get full points.
Example of Aim at (just for your reference).
One has to have an Aim to suceed in life. You would see that the Aim in here is not a verb. When Aim is used as a Verb, then Aim to is the right one.
I hope this helped.
Regards
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### Show Tags
29 May 2004, 13:35
carsen wrote:
Hello Sathya
To Aim at is the right Idiom
Example :- You have to aim at 780+ to score around 750 in gmat.
You have to aim at the red circle to get full points.
Example of Aim at (just for your reference).
One has to have an Aim to suceed in life. You would see that the Aim in here is not a verb. When Aim is used as a Verb, then Aim to is the right one.
I hope this helped.
Regards
Carsen,
Did you mean "at" follows "Aim" when "Aim" is used as Verb?
I just got confused. Thanks for your help.
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29 May 2004, 17:13
so which one is right. From ur explaination it sounds aim at is right and so the option D seems right. Is it?
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### Show Tags
29 May 2004, 17:26
so which one is right. From ur explaination it sounds aim at is right and so the option D seems right. Is it?
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### Show Tags
29 May 2004, 17:51
D is better because it tells the intension behind identifying the fans.
A makes it look like
that aim at fans
and
that aim at seating fans (awkward)
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29 May 2004, 18:07
But the answer is A. It is a question in 885SC SC question bank by bigB.
Any comment from SC gurus
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29 May 2004, 21:09
Hello Sathya
Pardon me, I did not indicate the right answer, rather I gave you the explaination on where to use AIM AT and AIM TO.
As said before, AIM AT (when used as a Verb, an action). In the original sentence it is not a Verb. When we read the sentence...
Violence in the stands at soccer matches has gotten so pronounced in several European countries that some stadiums have adopted new rules that aim to identify (this is the verb) fans of visiting teams and that seat them in a separate area.
Aim, in the above sentence is like goal, a purpose. Adopted is a verb and aim goes with this verb. Hence in the above aim to, is correct form. when you negate the words "AIM TO" from the original sentence, it still is valid anc clear. Let us negate the words AIM TO from the original and see how it fits.
Violence in the stands at soccer matches has gotten so pronounced in several European countries that some stadiums have adopted new rules to identify fans of visiting teams and that seat them in a separate area.
I hope it is clear. If need be for any clarifiation, feel free.
Hence the original sentence is the best choice.
I hope this helps.
Regards buddy.
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29 May 2004, 21:09
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For any e-curious person, some of the basic questions that one might have are
• Who came up with this constant ?
• Where did it first appear ?
• Why is this number so important ?
• Why should one make a function out of this constant ( exp(x) ) ?
• What is its relation to complex numbers ?
• What is its connection with hyperbola , as it appears in coshx , sinhx etc ?
This book is a collection of stories about various developments around e. However there is a common thread which runs across all the stories. Each of the story has in its essence, a development which lead to the world we are living in, where we take e for granted in most of the applications we deal with.
It all began with Napier’s 20 year effort. John Napier worked on a single idea for 20 years. What was his idea ? Multiplication and Division of large numbers are difficult operations than addition and subtraction. If numbers could be represented as an exponent of some number, then multiplication will turn in to addition and division would turn in to subtraction. Napier thought of a base and created elaborate tables where by you can multiply and divide by looking up relevant numbers in the tables and performing addition and subtraction instead. Henry Briggs, a geometry professor from Gresham college suggested a few improvements like having a base of 10. Thus the concept of logarithm was born.
What’s logarithm got to do with e ?
Finding area under the curve y = 1/x was a problem which racked quite a few mathematicians. Fermat had developed a way to find out the area under the curve y = x^n , but for n=-1 , he said it did not work. The crucial breakthrough took place when Anton de Sarasa, a Jesuit mathematician noted that area under the curve y = 1/x behaved like logarithms. For lengths which grow geometrically on the x axis, the area grew arithmetically. Thus he was able to connect Napier logarithms with the behavior of area under 1/x and thus was born, logarithmic function.
Subsequent to this, there were developments in the area of calculus that brought e closer to discovery. One of the fundamental theorems of calculus states that rate of change of area function with respect to t is equal , at every point x=t, to the value of the original function at that point.
Simply put, for y = 1/x kind of situation , d(A)/d(t) = -1/t …It was already established that A was log(t) though base was something which was not yet fixed. base of the logarithm could have been anything. If we let the base be some number b, then we see that the equation turns in to a problem where we need to find the base such that derivative of a function is the function itself. Thus the base of logarithm e was born. ln(x) and exp(x) thus become known to be inverse functions of each other.
So, as one can see that exp(x) is closely related to hyperbola ( y = 1/x) and this spawned a new set of trigonometric entities called sinhx, coshx, tanhx etc. These are the analogs of sinx , cosx, tanx relevant to the circle. Thus e became to Hyperbola what pi became to a circle.
The author subsequently takes the reader through multiple facets of e, like the use of logarithm/e in representation on the music scale, its presence in the eulers formula , its presence in the nature ( sunflower spiral, nautilus shell ) . its presence in ancient and modern architecture etc
e was first noticed in continuous compounding of money. The expression (1+1/n)^n as n tends to infinity was known to converge to a number 2.7182.. But it took a century for the math to develop that so called number 2.7182 in to a tractable exponential function, e^x and finally stamping the property of transcendental on e.
Today , one cannot think of doing math/fin with out exp(x) function , for this makes a whole set of problems analytically tractable. This book makes one pause and look at the massive efforts of various mathematicians / scientists who contributed to the understanding of e.
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# A 3.0 L solution contains 73.5 g of H_2SO_4. What is the molar concentration of the solution?
Approx. $0.250 \cdot m o l \cdot {L}^{-} 1$.
$\text{Concentration} = \frac{\frac{73.5 \cdot g}{98.08 \cdot g \cdot m o {l}^{-} 1}}{3.0 \cdot L} = 0.250 \cdot m o l \cdot {L}^{-} 1$
The concentration is quoted with respect to ${H}_{2} S {O}_{4} \left(a q\right)$. What is the concentration with respect to ${H}_{3} {O}^{+}$, and why do I make this distinction?
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# Math Problem Solving Number of the Day
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1. This is the Math Problem Solving Mega Bundle! There are lots of themed worksheets in this pack so you can run Math Problem Solving in your class every day as a warm up activity before you start a math lesson! I find doing a 30 minutes math problem solving session each day really helps your students
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Product Description
These worksheets make a great warm up for your students. Select a number of the day and work with students through the activities and word problems. The first sheet in this set is a blank sheet for you to use with any number you would like.
Number 1 – allows students to demonstrate their understanding of the number one through drawing.
Numbers 2 – 5 introduces the part part whole strategy.
Numbers 6 – 20 moves on to maths word problems and applying the part part whole strategy to work out the answer. Some of the word problems throw in a simple ‘red-herring’ to introduce students to more complex maths problems.
This pack allows for differentiation with students and the ability to return again and again to early number work problems.
Aligns with Foundation Year and Year 1 Australian Curriculum (ACARA):
-sharing a collection of readily available materials into two equal portions
-splitting an object into two equal pieces and describing how the pieces are equal
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# Easter and Maths Questions — Perfect!
Primary teachers are Geniuses — or is that “Genii”?
They are able to take the traditional view of Easter — church attendance, hot cross buns, Easter Eggs, holidays — as an ideal structure in which to invent heaps of maths questions that have a natural interest — a natural intrigue — to the students in their care.
And that’s what WE do, here at The Education Shop.
Make maths meaningful! (Just dig that alliteration!)
But if teachers would like a shortcut, they might like to check out our 3-week freebie trial worksheet offer. Just go to http://www.EdShop.net.au and enjoy, enjoy, enjoy.
Go on! Do yourself a favour! That’s http://www.EdShop.net.au
# Primary Teachers, Shane Warne, Baked Beans & Maths. REALLY??
Who would have thought that Shane Warne’s baked beans habits would make a great Maths question for primary students?
But they did!
And they do!
Here it is . . . the BIG QUESTION . . .
“Aussie cricketer Shane Warne had 1,900 tins of baked beans flown in and addressed to him personally in India in 1998. (But they weren’t ALL just for him!) If each tin cost 80 cents, what was the total value of all those cans of baked beans?”
And there are HEAPS MORE topical, FUN maths questions where that one came from.
For an OBLIGATION-FREE 3-WEEK TRIAL SUBS., go to http://www.EdShop.net.au and ENJOY the fun, and the maths success, that we bring to that all-important subject — maths!
Go on, give it a go. www.Edshop.net.au
# Hey, Primary Teacher! Good questions — for your next Maths Lesson!
Well, these 4 questions are from this week’s (late January, 2016) EdShop Maths Word Problem Worksheets. They’re AUSSIE, and THEY’RE FUN!
Loads more where these came from.
MIDDLE-PRIMARY QUESTIONS:
1. In the Australian Open tennis on TV, in one hour of viewing there were 24 minutes of commercials. The rest was tennis. How many minutes of actual tennis were on TV in that hour?
2. The series of Star Wars movies feature the robots C-3PO, R2-D2, and BB-8. What total number do you get by adding all the numbers in these robot names together?
3. Spongebob Squarepants can jog 50 metres in 30 seconds. At that rate, how long will it take him to jog 200 metres?
4. Taylor Swift was born on December 13, 1989. How old (in years) will she be on 1st February 2016?
These questions are from the middle-primary level. We also prepare Upper-Primary and Extension levels of worksheets. ALL OF THEM ARE PREPARED FRESHLY EACH SCHOOL WEEK, AND EMAILED OUT ON SATURDAY AFTERNOONS.
They bring an exciting new, novel, fun, realistic, TOPICAL range of Maths worded questions that the students — and teachers — ABSOLUTELY LOVE!
Try them yourself! For use in your class. Or to share at your school.
FREE for the 3-week trial period. Absolutely no cost! Absolutely no obligation! (Not a single credit card number need change hands!!)
To receive this week’s worksheets TODAY (Sunday!), simply go to http://www.EdShop.net.au and complete the “3-week trial” coupon. We’ll do the rest! We’ll respond STRAIGHT AWAY!
Go on! Check it out NOW! What is there to lose? http://www.EdShop.net.au and enjoy the fruits of this terrific teaching resource!
# Sometimes Primary-age students just need something to make them LOVE maths . . . and here it is!
Yep! Some kids HATE maths.
No surprise there!
Every primary teacher has seen it! THAT LOOK that covers the face when it’s Maths Time!
(There should be an EMOTICON for that look!)
Anyway, the time has come to replace THAT look with a look of POSITIVE INTEREST and OPTIMISM and ENTHUSIASM in maths lessons.
And . . . here’s one way. Just one! But it works! FRESHLY EVERY SCHOOL WEEK!
It’s not JUST maths, either! It involves reading comprehension, usually a little discussion with a classmate or teacher, and an interest-packed Maths exercise that’s FUN, topical, full of interest, and teaches HEAPS about how useful Maths is in everyday life.
AND . . . you can try it out for 3 weeks — totally free of cost or obligation.
MANY teachers have visited! As have many schools! And they’re ecstatic!
Oh, yes, of what am I speaking? (I almost forgot to mention it!)
Visit http://www.EdShop.net.au to find out. Go right NOW, and we’ll respond within the hour!
# Teachers — the oldest trick in the book!
Back in the “olden days”, the Teachers’ Union used to give some sage advice to teachers just starting a new year in the classroom.
The advice was especially pertinent to teachers of Infant classes, but in reality applies to all primary school teachers everywhere.
And the advice is simple . . .
When a child gets home at the end of that first school day, the first question a parent will ask is, “What did you learn today?”
So . . .
EVERY TEACHER should ensure that the first day at school includes a song, a poem, a joke, a limerick, a simple maths fact — ANYTHING that will PROVE to a loving parent that school is a great place of learning.
And if you happen to be a bit short of ideas, then go to our website at http://www.Edshop.net.au and request our FREE 3-week maths worksheet trial, which we’ll deliver pronto to you via email.
Remember, first impressions are crucial! Make them GREAT impressions.
Have a GREAT first day at school this year — 2016!
# What ALL Primary Teachers Want — a GREAT start to 2016
And what better way to do it than have a worthwhile, interest-packed Maths lesson first up!
How?
Well, EVERY teacher brings to the classroom their own approaches, their own skills, their own personality.
But sometimes, just sometimes, the teacher needs a bit of a new idea, just to get the class “ticking all the boxes”, as it were.
So, here’s the place.
A TOTALLY FREE teaching resource for 3 weeks that can set a positive tone for maths lessons for the rest of the school year. How’s THAT for an idea?
If you’re at all interested, you owe it to yourself to check out the http://www.Edshop.net.au website, and trial the unique AUSSIE primary maths worksheets there.
Thousands of students across Australia are using these fascinating maths word problem worksheets every week — all 3 levels of them.
Many, many SCHOOLS across Australia are subscribing to them, and distributing them to their middle-primary and upper-primary students each week. All at a TINY cost.
Oh, yes. And they make GREAT Homework Sheets. Fresh every week! With Answers!
Go on, check it out! No tricks. No obligation. Just a great teaching tool to help you get those positive maths vibes going in your classroom, right from Week 1 of the 2016 school year.
You owe it to yourself! As Molly would say, “Do yourself a favour!”
# Primary Teacher Week 1 Maths Success!
EVERY primary teacher wants a first week success with his / her class.
So, for a FUN — but worthwhile — foray into Maths for teachers of Years 3, 4, 5 and 6, you can do no better than to go to the Aussie Ninja of WORD PROBLEM maths fun worksheets.
Totally free for 3 weeks. And worth EVERY CENT of it!
3 levels: Middle-primary, Upper-primary, Extension
Students love them! Teachers love them! Parents love them!
They’re GREAT for use as ready-made HOMEWORK SHEETS. They come with ANSWERS, which adds to their practicability and ease of use, from the teacher’s perspective.
And there’s nothing to pay. Give them a go. Right now!
Didn’t someone say, “The best things in life are FREE?” Well, for 3 weeks these are free, anyway!
The 3-week FREEBIE is available at http://www.EdShop.net.au and it’s absolutely true that there are NO STRINGS ATTACHED!
After the 3 weeks, you and your teaching colleagues can consider the TRUE value of these worksheets. And take the appropriate action. Or not!
All we can do is offer the freebie of an excellent, fun, Aussie, worthwhile Maths product.
Don’t ponder it! JUST DO IT!!
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https://community.intel.com/t5/Intel-oneAPI-Math-Kernel-Library/PARDISO-Solving-a-large-matrix-containing-many-identical-nonzero/td-p/1048590
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Intel® oneAPI Math Kernel Library
Ask questions and share information with other developers who use Intel® Math Kernel Library.
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6590 Discussions
## PARDISO-Solving a large matrix containing many identical nonzero sparsity structure submatrices
Beginner
121 Views
Hi friends,
I am solving coupled problem analysis in FEM. I want to solve a matrix A.x = b, A could be
A = [A11, A12; A21, A22] and B has a size of [A11, A12]
A could contain 3x3, 4x4...nxn submatrices. The submatrices have the same nonzero sparsity structure (IA and JA). At the moment the solution may be reordering IA and JA from submatrices to a large matrix A to input into Pardiso Solver. However it will increase the storage memory for IA and JA.
It there any way to solve matrix A without reodering IA and JA? As I know multi right hand side method solves the only A matrix with different rhs, it can not apply for this case. Can anyone give me a hint?
2 Replies
Employee
121 Views
Hi,
Probably i didn't get your question clearly... You wrote that matrix 2x2 blocked but further said that it is not generally 2x2 blocked, but number of blocks could be bigger - am i correct?
Thanks,
Alex
Beginner
121 Views
Hi Alex,
2x2 blocked matrix is an example for my case, let us solve the problem of matrix A has 2x2 block. If we solve this broblem, we can also solve nxn blocked matrix in the same way.
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# Question: Weapons damage (in General)
## FlameyMarch 22 2006 11:33 PM EST
i know your all thinking not Flamey again.
im gonna use an example here VB [75x40] (+26)
it says this in wiki: Weapon [base x damage modifier] (+ to-hit),
i know what base damage is, i know what PTH is, but damage modifier, i always thought it was enchantment modifier.
how does it modify your damage? is it spells or a formula or something?
Thanks, Flamey
## LayWaste-[bAMarch 22 2006 11:35 PM EST
I always thought it was a simple multiplication problem.. 75 times 40 = damage Obviously there are many factors that affect that.
## StephenMarch 22 2006 11:38 PM EST
Don't assume that x means multiply in this case
## FlameyMarch 22 2006 11:44 PM EST
yes i got told not to use the actual math symbols. so i know it isn't 75x40+26=total damage.
this just came to me, would damage modifier, be strength?
## AdminQBnovice[Cult of the Valaraukar]March 23 2006 12:06 AM EST
I'm sure there are better explanations somewhere, but essentially you're trying to simplify a big math problem with many variables. Besides the base and multiplier, you also have STR and the nw of your weapons X.
## QBRangerMarch 23 2006 12:10 AM EST
Also, there is an addition to your damage using the nw of your weapon that is via the x of your weapon. That is to compensate for the fact DD spells ramp up linearly while it takes more and more NW to increase your weapons x. The NW applied to the + of the weapon is not used in damage calculations.
## AdminQBGentlemanLoser[{END}]March 23 2006 3:16 AM EST
Ranger, are you sure the NW cost of the x value is also used to calculate damage?
I always thought this was a misnomer, with only the x contributing to damage (and therefore the only NW of the weapon that contributes to damage is the NW invested in the x, but x40 is the same x40 no matter how much or little it cost you to get there).
## MrC[DodgingTheEvilForgeFees]March 23 2006 3:37 AM EST
I'm with GL on this one.
I've never seen any evidence that NW has any relation to damage at all.
In fact, quite the opposite.
What makes you think that it's any more than a rumour, Ranger?
And Flamey, "this just came to me, would damage modifier, be strength? " you wont get a straight answer on this. Jon's done a great job of keeping us all in the dark on weapon damage. Although I've heard from credible people that the modifier *is* a multiplier of sorts... then again, maths never was a strong suite of mine, so I'll leave that to the CB math nutters to figure out.
This thread is closed to new posts. However, you are welcome to reference it from a new thread; link this with the html <a href="/bboard/q-and-a-fetch-msg.tcl?msg_id=001kbd">Question: Weapons damage</a>
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## 38.8 Completion and Mittag-Leffler modules
Lemma 38.8.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $A$ be a set. Assume $R$ is Noetherian and complete with respect to $I$. The completion $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge$ is flat and Mittag-Leffler.
Proof. By More on Algebra, Lemma 15.27.1 the map $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge \to \prod _{\alpha \in A} R$ is universally injective. Thus, by Algebra, Lemmas 10.82.7 and 10.89.7 it suffices to show that $\prod _{\alpha \in A} R$ is flat and Mittag-Leffler. By Algebra, Proposition 10.90.6 (and Algebra, Lemma 10.90.5) we see that $\prod _{\alpha \in A} R$ is flat. Thus we conclude because a product of copies of $R$ is Mittag-Leffler, see Algebra, Lemma 10.91.3. $\square$
Lemma 38.8.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume
1. $R$ is Noetherian and $I$-adically complete,
2. $M$ is flat over $R$, and
3. $M/IM$ is a projective $R/I$-module.
Then the $I$-adic completion $M^\wedge$ is a flat Mittag-Leffler $R$-module.
Proof. Choose a surjection $F \to M$ where $F$ is a free $R$-module. By Algebra, Lemma 10.97.9 the module $M^\wedge$ is a direct summand of the module $F^\wedge$. Hence it suffices to prove the lemma for $F$. In this case the lemma follows from Lemma 38.8.1. $\square$
In Lemmas 38.8.3 and 38.8.4 the assumption that $S$ be Noetherian holds if $R \to S$ is of finite type, see Algebra, Lemma 10.31.1.
Lemma 38.8.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map, and $N$ an $S$-module. Assume
1. $R$ is a Noetherian ring,
2. $S$ is a Noetherian ring,
3. $N$ is a finite $S$-module, and
4. for any finite $R$-module $Q$, any $\mathfrak q \in \text{Ass}_ S(Q \otimes _ R N)$ satisfies $IS + \mathfrak q \not= S$.
Then the map $N \to N^\wedge$ of $N$ into the $I$-adic completion of $N$ is universally injective as a map of $R$-modules.
Proof. We have to show that for any finite $R$-module $Q$ the map $Q \otimes _ R N \to Q \otimes _ R N^\wedge$ is injective, see Algebra, Theorem 10.82.3. As there is a canonical map $Q \otimes _ R N^\wedge \to (Q \otimes _ R N)^\wedge$ it suffices to prove that the canonical map $Q \otimes _ R N \to (Q \otimes _ R N)^\wedge$ is injective. Hence we may replace $N$ by $Q \otimes _ R N$ and it suffices to prove the injectivity for the map $N \to N^\wedge$.
Let $K = \mathop{\mathrm{Ker}}(N \to N^\wedge )$. It suffices to show that $K_{\mathfrak q} = 0$ for $\mathfrak q \in \text{Ass}(N)$ as $N$ is a submodule of $\prod _{\mathfrak q \in \text{Ass}(N)} N_{\mathfrak q}$, see Algebra, Lemma 10.63.19. Pick $\mathfrak q \in \text{Ass}(N)$. By the last assumption we see that there exists a prime $\mathfrak q' \supset IS + \mathfrak q$. Since $K_{\mathfrak q}$ is a localization of $K_{\mathfrak q'}$ it suffices to prove the vanishing of $K_{\mathfrak q'}$. Note that $K = \bigcap I^ nN$, hence $K_{\mathfrak q'} \subset \bigcap I^ nN_{\mathfrak q'}$. Hence $K_{\mathfrak q'} = 0$ by Algebra, Lemma 10.51.4. $\square$
Lemma 38.8.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map, and $N$ an $S$-module. Assume
1. $R$ is a Noetherian ring,
2. $S$ is a Noetherian ring,
3. $N$ is a finite $S$-module,
4. $N$ is flat over $R$, and
5. for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ where $\mathfrak p = R \cap \mathfrak q$ we have $IS + \mathfrak q \not= S$.
Then the map $N \to N^\wedge$ of $N$ into the $I$-adic completion of $N$ is universally injective as a map of $R$-modules.
Proof. This follows from Lemma 38.8.3 because Algebra, Lemma 10.65.5 and Remark 10.65.6 guarantee that the set of associated primes of tensor products $N \otimes _ R Q$ are contained in the set of associated primes of the modules $N \otimes _ R \kappa (\mathfrak p)$. $\square$
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Determination of angle of contact
It should be noted that when two pulleys are connected by a belt drive and they are unequal size, the angle q must be taken as minimum angle of lap or contact i.e., the angel of contact on the smaller pulley.
For open belt:
cos (θ / 2) = r2 – r1 / x
where θ (in radians) is the angle of lap of smaller pulley = π – 2 α
θ = 2 cos–1 (r2 – r1 / x)
For cross belt:
cos (πθ/2) = r1 + r2 / x
θ = 2 [π – cos–1 (r1 + r2 / x)]
where θ (in radians) is the angle of lap on any one of the pulleys and θ = (π + 2α).
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Hints:
b.1)
If you cannot solve this by calculus, the area is:
A • L ,
p
where A is the amplitude of the wave and L is the wavelength.
b.2) Calculate the volume of this 1 m wide section, and use the formula for kinetic energy:
m • v2 ,
2
where v is the speed of the wave. Note that 1 m3 of water contains 1000 kg of water!
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Review question
# When is $6 \times 7 = 42$ a counter-example? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R6993
## Solution
The fact that $6 \times 7 = 42,$ is a counter-example to which of the following statements?
1. the product of any two odd integers is odd;
In order to be a counter-example to this statement, a fact would have to involve two odd integers whose product is even.
Since $6$ is not an odd integer, the fact is not a counter-example to this statement.
In fact, the statement is true, so there are no counter-examples.
1. if the product of two integers is not a multiple of $4$ then the integers are not consecutive;
The number $42$ is not a multiple of $4$, but it is the product of two consecutive integers, $6$ and $7$, so we have a counter-example, and the statement is not true.
1. if the product of two integers is a multiple of $4$ then the integers are not consecutive;
Since $42$ is not a multiple of $4$, the fact that $6 \times 7 = 42$ has no bearing on the truth of this statement.
The statement, however, is clearly false — try $4 \times 5$.
1. any even integer can be written as the product of two even integers.
This statement is clearly false, but to provide a counter-example we would have to show that every factorisation of some even integer includes an odd number.
The fact $6 \times 7 = 42$ alone is not enough.
But if we listed all the factorisations of $42$, that’s $1 \times 42, 2 \times 21, 3 \times 14, 6 \times 7,$ and noted that $42$ is never the product of two even numbers, this would be a counter-example to the statement.
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Concentration Statistics - BQL substitution [General Statistics]
Hi Sara,
» What is your opinion on below quantification limit (BQL) substitution for concentration data summary statistics, by timepoint?
You can use the median and quartiles or $$\small{\bar{x}_{geo}\mp SD_{geo}}$$ if a certain percentage of samples are measurable (I have seen SOPs with 50%, 67%, and 75%) and nothing (‘not reportable’) otherwise. At the end of the day it’s not important at all (not relevant for the BE assessment). Use whatever you like. 1
» Zero substitution is the one I have seen the most, …
To quote Harold Boxenbaum (Crystal City workshop about bioanalytical method validation, Arlington 1990):
After a dose we know only one thing for sure: The concentration is not zero.
Ended in shouting matches.
» … I don't think it is suitable for the calculation of geometric means.
Correct, since$$\lim_{x \to 0} \log x=-\infty.$$For simplicity we can say that $$\small{\log 0}$$ is undefined. It is reasonable to assume that concentrations ($$\small{x \in \mathbb{R}^+}$$) follow a lognormal distribution, and the geometric mean would be the best estimator of location. Some people chicken out, set BQLs to zero, and present arithmetic means. This leads to funny plots with $$\small{\bar{x}\mp SD,}$$ where the lower whisker reaches far below zero. 2 Phew, negative concentrations? Not in this universe.
1. Phoenix/WinNonlin by default calculates descriptive statistics only for numeric values. This can lead to strange results. Say, we have $$n-2$$ values which are BQL and two $$\small{C\geq LLOQ}$$. Then we end up with $$\small{\bar{C}_{ar}=\tilde{C}=\tfrac{C_1+C_2}{2},\bar{C}_{geo}=\sqrt{C_1\times C_2}}$$. Doesn’t make sense. However, we can specify different rule sets for descriptive statistics and plots.
2. A goody from the FDA’s NDA 204-412 (mesalamine delayed release capsules, n = 238, sampling times: pre-dose, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 24, 30, 36, and 48 h post-dose). BQLs were imputed as LLOQ/2.
Splendid, $$\small{\bar{x}\mp SD}$$ in bloody Excel. Hey, wait a minute, that’s a fucking line plot… Oh dear! One hour intervals in the beginning are as wide as the 12 hours at the end.
Let’s see the XY-plot:
Do these guys and dolls really believe that at seven hours there’s a ~16% chance that concentrations are 232 and a ~1% chance that concentrations are 731‽ Any statistic implies an underlying distribution. The arithmetic mean implies a normal distribution with $$\small{x \in \mathbb{R}\:\vee\:x \in \left \{-\infty, +\infty\right \}}$$. Fantastic.
Which cult of Pastafarianism do they belong to?
The one holding that negative mass exist or the one believing in negative lengths?
$$\small{\bar{x}_{geo}\mp SD_{geo}}$$ reflects the terrible variability of this drug much better and shows that high concentrations are more likely than low ones.
Dif-tor heh smusma 🖖
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
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# Continuity and differentiability of a function :$F(r,t) = (r\cos f(t), r\sin f(t))$
I have two question concerning continuity and differentiability, because something is unclear to me.
Let’s say I have a function : $f:\mathbb{R} \rightarrow \mathbb{R}$ (this function is not necessarily continuous), and now let’s defined the function $F$ as follow :
$$\forall (r,t) \in \mathbb{R}^+\times \mathbb{R} F(r,t)=(r \cos f(t), r \sin f(t))$$
Now my questions are :
Is $F$ continuous on a neighborhood of $(0,0)$ even-if $f$ is not continuous ?
According to me it’s most likely true because : $$\lim_(r,t \rightarrow (0,0)) F(r,t) = 0$$
If it wasn’t continuous on a neighborhood of $(0,0)$ we can’t talk about the limit of such function at the point $(0,0)$, so the function is continuous at $(0,0)$.
Now I would like to know why I can’t say that $F$ is différentiable at the point $(0,0)$ using the following argument :
If $F$ is differentiable at the point $(0,0)$ then there is a function $u$ such that :
$$F(h,h) = F(0,0) + u(0)(h) + o(||h||)$$
So by taking the limit as $h \rightarrow 0$ we have : $$u(0)(h) = 0$$
And hence $F$ is differentiable at the point $(0,0)$, and $\mathrm{d}F(0) = x \mapsto 0$.
Yet this argument is actually incorrect, but why ?
Thank you, for taking your time !
• You haven't proved that such an $u$ exists. And the condition you get for $u$ is false. Taking the limit you get the identity $F(0,0)=F(0,0)$. By the way, $h$ can have different components $h_1$ and $h_2$. – Dog_69 Jun 18 '18 at 23:50
We have that $F$ is continuous, because $\sin$ and $\cos$ are bounded and $r\to 0$. The function $f$ has no say in what happens here. But if ${\rm D}F(0,0)$ is the zero map, we would have to check that $$\lim_{(r,t)\to (0,0)} \frac{F(r,t)}{\sqrt{r^2+t^2}} =(0,0),$$since $F(0,0)=(0,0)$. This is equivalent to $$\lim_{(r,t)\to (0,0)} \frac{r}{\sqrt{r^2+t^2}}\cos f(t) =\lim_{(r,t)\to (0,0)} \frac{r}{\sqrt{r^2+t^2}}\sin f(t) =0.$$ Although $r/\sqrt{r^2+t^2}$ is bounded, the terms $\cos f(t)$ and $\sin f(t)$ need not go to zero (take $f=1$ or whatever).
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Converter
Converting hexadecimal C6 to binary? Hex c6 to Bin = 11000110
Ultimate Binary to Decimal to Hexadecimal Converter. The binary-hexadecimal-decimal converter is capable of instantly converting back and forth between binary, decimal, and hexadecimal numbers.
N2
N10
N16
## How to use the Binary-Decimal-Hexadecimal Converter?
Our Bin-to-Dec-to-Hex converter is a very easy-to-use online tool that will enable you to perform a conversion of any binary, decimal, or hexadecimal number quickly. You can enter a number in any field and the conversion is immediate. So for example, if you start to write a binary number in the binary number field, you can see the result of the decimal and hexadecimal conversion in real-time. Our free tool can perform the following operations:
• Binary to Decimal Conversion (as a Bin to Dec Converter)
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• Hexadecimal to Decimal Conversion (as a Hex to Dec Converter)
• Hexadecimal to Binary Conversion (as a Hex to Bin Converter)
## Example calculations: Binary to Decimal Converter
10101100 Binary to Decimal = 172 11111010 Binary to Decimal = 250 11000 Binary to Decimal = 24 10110111 Binary to Decimal = 183 1011001 Binary to Decimal = 89 10111111 Binary to Decimal = 191
## Example calculations: Binary to Hexadecimal Converter
10011110 Binary to Hexadecimal = 9e 11101011 Binary to Hexadecimal = eb 1101100 Binary to Hexadecimal = 6c 11111110 Binary to Hexadecimal = fe 10000111 Binary to Hexadecimal = 87 1110111 Binary to Hexadecimal = 77
## Example calculations: Decimal to Binary Converter
38 Decimal to Binary = 100110 240 Decimal to Binary = 11110000 256 Decimal to Binary = 100000000 185 Decimal to Binary = 10111001 111 Decimal to Binary = 1101111 233 Decimal to Binary = 11101001
## Example calculations: Decimal to Hexadecimal Converter
45 Decimal to Hexadecimal = 2d 241 Decimal to Hexadecimal = f1 179 Decimal to Hexadecimal = b3 152 Decimal to Hexadecimal = 98 221 Decimal to Hexadecimal = dd 186 Decimal to Hexadecimal = ba
## Example calculations: Hexadecimal to Binary Converter
26 Hexadecimal to Binary = 100110 b0 Hexadecimal to Binary = 10110000 fe Hexadecimal to Binary = 11111110 76 Hexadecimal to Binary = 1110110 9f Hexadecimal to Binary = 10011111 9 Hexadecimal to Binary = 1001
## Example calculations: Hexadecimal to Decimal Converter
9 Hexadecimal to Decimal = 9 1 Hexadecimal to Decimal = 1 34 Hexadecimal to Decimal = 52 fe Hexadecimal to Decimal = 254 db Hexadecimal to Decimal = 219 dd Hexadecimal to Decimal = 221
Q&A
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# Explaining the reason for the delay in the commencement
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Explaining the reason for the delay in the commencement [#permalink]
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Explaining the reason for the delay in the commencement of the current academic session, the students’ council informed that while 50 percent of the undergrad students had chosen their courses, the remaining students explored suitable disciplines for specialization having been included in the curriculum introduced by the education board.
A. had chosen their courses, the remaining students explored suitable disciplines for specialization having been included in the curriculum
B. had chosen their courses, the remaining students explored suitable disciplines to specialize on included in the curriculum
C. had chosen their courses, the remaining students were exploring suitable disciplines for specialization included in the curriculum
D. chose their courses, the remaining students were exploring suitable disciplines to specialize on included in the curriculum that
E. chose their courses, the remaining explored suitable disciplines for specialization included in the curriculum and
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Re: Explaining the reason for the delay in the commencement [#permalink]
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02 Feb 2019, 01:18
Explaining the reason for the delay in the commencement of the current academic session, the students’ council informed that while 50 percent of the undergrad students had chosen their courses, the remaining students explored suitable disciplines for specialization having been included in the curriculum introduced by the education board.
A. had chosen their courses, the remaining students explored suitable disciplines for specialization having been included in the curriculum
B. had chosen their courses, the remaining students explored suitable disciplines to specialize on included in the curriculum
C. had chosen their courses, the remaining students were exploring suitable disciplines for specialization included in the curriculum
D. chose their courses, the remaining students were exploring suitable disciplines to specialize on included in the curriculum that
E. chose their courses, the remaining explored suitable disciplines for specialization included in the curriculum and
Re: Explaining the reason for the delay in the commencement [#permalink] 02 Feb 2019, 01:18
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# Search results
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Doping (semiconductor)
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Heat-affected zone - heat input for arc welding
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# Slicing the Democratic Primary Pies with Three Leading Candidates
Slicing the Democratic Primary Pies with Three Leading Candidates.
Previously, we showed how the Democratic district vote would be split between two candidates, because only candidates with more than 15% of the vote were to be counted, and only two existed at that time. Now, however, the Quinnipiac University poll shows three candidates with 15% or over. So we analyze this particular case for 3, 4, 5, 6, 7, or 8 delegates per district.
To start with, the poll has not only Sanders beating Trump, with 53% to 40%, but so do the next five Democratic candidates. In the Democratic Primary country-wide, we have Biden at 30%, Sanders at 19%, Warren at 15%, Buttigieg at 8%, Harris at 7%, and others 3% or lower. 13% are undecided or not available. The accuracy was pm 3.5%.
The first Democratic rule is to drop everybody below 15%. The three above 15% add up to 64%. The other 36% may end up voting for the top 3 by the time they get to vote. Anyway, for now, they are dropped.
The second Democratic rule is to rescale those above 15% to 100% by multiplying by 100/64 = 1.5625.
This gives Biden 46.875%, Sanders 29.685%, and Warren 23.4375%.
These will be used for all districts, regardless of the number of delegates.
The third rule is to multiply this times the number of delegates for the district, which is mostly 4 or 5, but also 3 or 6.
Biden has 2.34, Sanders has 1.48, and Warren has 1.17.
The fourth rule is to first look at the whole numbers, giving Biden 2 delegates, Sanders 1 delegate, and Warren 1 delegate.
There is still 1 delegate left. This is given to the leading fractional number, which is the 0.48 to Sanders.
So we have Biden 2, Sanders 2, and Warren 1 for five total delegates.
Next, we do the case of 4 total delegates.
Multiplying by 4 gives Biden 1.875, Sanders 1.1874, and Warren 0.9375. Truncating, Biden gets 1 and Sanders 1, with Warren 0. The rest are to be awarded in order of the highest fractional remainder. Since Warren wins the fraction, she almost has 1, so she gets 1. Biden has the next highest fraction, so he should get another delegate. Thus, Biden has 2, Sanders 1, and Warren 1.
For 6 delegates, Biden has 2.8125, Sanders 1.7811, and Warren 1.406. So Biden gets 2, Sanders 2, and Warren 1. The sixth delegate goes to the largest fraction, which is Biden. Finally, Biden has 3, Sanders 2, and Warren 1.
For 3 delegates, Biden has 1.406, Sanders 0.891, and Warren 0.703. Clearly, Biden gets 1, and 2 are left over. Sanders is highest, and Warren also close to 1. Biden would round off lower to 1. So Biden 1, Sanders 1, and Warren 1. Equal. Despite Biden actually having twice the vote of Warren.
For 7 delegates, Biden has 3.281, Sanders has 2.078, and Warren has 1.640. The whole numbers give Biden 3, Sanders 2, and Warren 1, adding to 6. Warren has the largest fraction, and gets the last one. The final number is Biden 3, Sanders 2, and Warren 2.
For 8 delegates, Biden has 3.75, Sanders has 2.37, and Warren has 1.875. The whole numbers give Biden 3, Sanders 2, and Warren 1. There are two left over delegates, which by fractional ordering gives Warren 1 and Biden 1. The end result is Biden 4, Sanders 2, and Warren 2. These breakups get closer to the vote percentage as the number of delegates increases.
In California’s 53 House districts, there are 1 with 3 delegates, 16 with 4, 18 with 5, 13 with 6, 3 with 7, and 2 with 8.
We make a table of the results for B, S, and W delegates for districts with a given number of delegates:
Del. B. S. W
3. 1. 1. 1
4. 2. 1. 1
5. 2. 2. 1
6. 3. 2. 1
7. 3. 2. 2
8. 4. 2. 2
Applying this to the number of districts for each number of delegates, gives a total number of delegates for
B. S. W
125. 89. 58
These total 272 delegates.
The percentage of delegates for each candidate is now compared to their relative percentages of the top three candidates, which are over 15%.
B has 0.460, S has 0.327, and W has 0.213. Compare this to their vote ratios:
B with 0.469, S with 0.297, and W with 0.234. The law of large numbers to the rescue.
Since the apportionment of delegates is really a round off to the nearest whole delegates, when you put enough together, it gives a convergent approximation. It’s like pixelating a picture, still leaves an approximation of the picture. Three about equal pixelizations of 4, 5, and 6 delegates are being merged here.
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A013939 Partial sums of sequence A001221 (number of distinct primes dividing n). 33
0, 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 14, 15, 17, 19, 20, 21, 23, 24, 26, 28, 30, 31, 33, 34, 36, 37, 39, 40, 43, 44, 45, 47, 49, 51, 53, 54, 56, 58, 60, 61, 64, 65, 67, 69, 71, 72, 74, 75, 77, 79, 81, 82, 84, 86, 88, 90, 92, 93, 96, 97, 99, 101, 102, 104, 107, 108, 110, 112 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS a(n) = A093614(n) - A048865(n); see also A006218. A027748(a(A000040(n))+1) = A000040(n), A082287(a(n)+1) = n. LINKS Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 Eric Weisstein's World of Mathematics, Distinct Prime Factors FORMULA a(n) = Sum_{k <= n} omega(k). a(n) = Sum_{k = 1..n} floor( n/prime(k) ). a(n) = a(n-1) + A001221(n). a(n) = Sum_{k=1..n} pi(floor(n/k)). - Vladeta Jovovic, Jun 18 2006 a(n) = n log log n + O(n). - Charles R Greathouse IV, Jan 11 2012 a(n) = n*(log log n + B) + o(n), where B = 0.261497... is the Mertens constant A077761. - Arkadiusz Wesolowski, Oct 18 2013 G.f.: (1/(1 - x))*Sum_{k>=1} x^prime(k)/(1 - x^prime(k)). - Ilya Gutkovskiy, Jan 02 2017 a(n) = Sum_{k=1..floor(sqrt(n))} k * (pi(floor(n/k)) - pi(floor(n/(k+1)))) + Sum_{p prime <= floor(n/(1+floor(sqrt(n))))} floor(n/p). - Daniel Suteu, Nov 24 2018 MAPLE A013939 := proc(n) option remember; `if`(n = 1, 0, a(n) + iquo(n+1, ithprime(n+1))) end: seq(A013939(i), i = 1..69); # Peter Luschny, Jul 16 2011 MATHEMATICA a[n_] := Sum[Floor[n/Prime[k]], {k, 1, n}]; Table[a[n], {n, 1, 69}] (* Jean-François Alcover, Jun 11 2012, from 2nd formula *) Accumulate[PrimeNu[Range[120]] (* Harvey P. Dale, Jun 05 2015 *) PROG (PARI) t=0; vector(99, n, t+=omega(n)) \\ Charles R Greathouse IV, Jan 11 2012 (PARI) a(n)=my(s); forprime(p=2, n, s+=n\p); s \\ Charles R Greathouse IV, Jan 11 2012 (PARI) a(n) = sum(k=1, sqrtint(n), k * (primepi(n\k) - primepi(n\(k+1)))) + sum(k=1, n\(sqrtint(n)+1), if(isprime(k), n\k, 0)); \\ Daniel Suteu, Nov 24 2018 (Haskell) a013939 n = a013939_list !! (n-1) a013939_list = scanl1 (+) \$ map a001221 [1..] -- Reinhard Zumkeller, Feb 16 2012 (Python) from sympy.ntheory import primefactors print([sum(len(primefactors(k)) for k in range(1, n+1)) for n in range(1, 121)]) # Indranil Ghosh, Mar 19 2017 (MAGMA) [(&+[Floor(n/NthPrime(k)): k in [1..n]]): n in [1..70]]; // G. C. Greubel, Nov 24 2018 (Sage) [sum(floor(n/nth_prime(k)) for k in (1..n)) for n in (1..70)] # G. C. Greubel, Nov 24 2018 CROSSREFS Cf. A005187, A006218, A011371, A013936. Cf. A022559. Cf. A077761. Sequence in context: A008320 A004439 A050126 * A209921 A268377 A201010 Adjacent sequences: A013936 A013937 A013938 * A013940 A013941 A013942 KEYWORD nonn,easy,nice AUTHOR EXTENSIONS More terms from Henry Bottomley, Jul 03 2001 STATUS approved
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# Decision Support Framework and Parameters for Dynamic Part-Time Shoulder Use: Considerations for Opening Freeway Shoulders for Travel as a Traffic Management Strategy
## Appendix C. Decision Parameter Development Methods
This appendix provides technical details to supplement chapter 4 and provides additional information on two decision parameter methods discussed in chapter 4.
### B1. Breakdown Probability Estimation with Method II — Emperical Performance Data
An empirical approach is adapted from the literature, (Brilon, Geistefeldt, & Regler, 2005) which identifies the probability of traffic flow breakdown based on the assumption that capacity is intrinsically stochastic. The method is based on the statistics of censored data. It delivers an estimation of the capacity distribution function Fc(q), representing the probability of a traffic breakdown in dependence on the flow rate q:
Figure 54. Equation. The capacity distribution function.
where:
Fc(q) is the capacity distribution function, representing the breakdown probability at traffic volume q.
q is the traffic volume in vehicles per hour per lane (veh/h/ln).
c is the capacity (veh/h/ln), i.e., the traffic volume beyond which traffic flow will break down into congested conditions.
P(c>q) is the probability that the capacity is greater than the observed volume.
In the light of the difficulty in selecting an appropriate fixed value for capacity, one could select values based on a "tolerable probability of breakdown." (Elefteriadou, 2014)
Traffic flow observations deliver pairs of average speeds and volumes in selected time intervals (e.g., 5 minutes). In intervals prior to a traffic breakdown that results in a speed drop below a specified threshold speed in the next time interval, capacity can be measured directly. In intervals not followed by a breakdown, capacity must have been greater than the observed volume. These observations are called "censored" observations. To estimate distribution functions based on samples that include censored data, both non-parametric and parametric methods can be used.
A non-parametric method to estimate the distribution function of lifetime variables is the product limit method (PLM). (Elefteriadou, et al., 2009) The PLM is based on work by Kaplan and Meier, which uses lifetime data analysis techniques for estimating the time until failure of mechanical parts or the duration of human life (Kaplan & Meier, 1958). Brilon et al. (2005) used this method in the context of freeway breakdown to estimate the capacity in a true stochastic sense. The productlimit estimator for the capacity or breakdown probability distribution is given by:
Figure 55. Equation. The product-limit estimator for the capacity or breakdown probability distribution. (Elefteriadou, 2014)
where:
q is the traffic volume (veh/h/ln).
qi is the traffic volume in interval i.
P(qi>q) is the probability that the observed breakdown volume is greater than the observed volume.
The product-limit estimator for the probability of observed breakdown volume being greater than the observed volume is given by:
Figure 56. Equation. The product-limit estimator for the probability of observed breakdown volume being greater than the observed volume.
where:
q is the observed traffic volume (veh/h/ln).
qi is the observed traffic volume at interval i, which is the one prior to the drop in speeds; i.e., defined as the observed breakdown flow (veh/h/ln).
ki is the number of intervals with a traffic volume of q≥qi.
di is the number of breakdowns at a volume of qi.
B is the set of breakdown intervals {B1,B2,…}.
However, if each observed volume that causes a breakdown is considered separately; i.e., only one observation of breakdown for every volume qi ; di = 1, then the product-limit estimator is given by:
Figure 57. Equation. Product-limit estimator for observed volume that causes a breakdown but is considered separately.
with all the terms defined as previously.
For a parametric estimation of the breakdown probability distribution, the function type of the distribution must be predetermined. The distribution parameters can be estimated by applying the maximum likelihood technique. For capacity analysis, the likelihood function is:
Figure 58. Equation. The likelihood function for capacity analysis.
where:
fc(qi) is the statistical density function of the capacity c.
Fc(qi) is the cumulative distribution function of the capacity c.
n is the number of intervals.
δI = 1, if interval i contains an uncensored value.
δI = 0, if interval i contains a censored value.
To simplify the computation, it is useful to maximize the log-likelihood function instead of the likelihood function L.
Calculating the breakdown probability distribution will provide agencies with guidance on observing and measuring (1) maximum pre-breakdown throughput, and (2) breakdown flow at the agencies' desirable probability of breakdown value.
Agencies with new dynamic part-time shoulder use (D-PTSU) facilities, including conversions from static part-time shoulder use (S-PTSU), may want to open the shoulder less frequently and be more tolerant of congestion. Agencies more experienced with D-PTSU may want to be more aggressive opening the shoulder and do so even with a lower probability of congestion if it were not opened.
### B2. FREEVAL Modifications to Support D-PTSU Analysis with Method III — Macroscopic Decision Parameter Optimization
The computational procedure in the Highway Capacity Manual (HCM) freeway facilities methodology is defined in terms of two distinct operational regimes. The first regime handles conditions where all segments are operating under capacity and is referred to as the "undersaturated" method. The second applies when at least one segment is operating over capacity or at level of service (LOS) F and is referred to as the "oversaturated" method.
The undersaturated method provides operational analysis in 15-minute increments. This increment is fixed as required by the set of underlying regressions on which the method is based. Alternatively, the oversaturated method is based on the cell transmission model, an approach which allows time steps of any length. The HCM fixes the oversaturated computational time step length at 15 seconds in accordance with certain assumptions of the methodology. Further, in order not to overwhelm users with extensive outputs, but to provide consistency within the undersaturated approach, the analysis using the oversaturated method is always aggregated up to the same 15-minute resolution as the undersaturated methods.
In the context of this project, there are two primary considerations relating to the use of the HCM method. First, the analysis is focused on operational conditions where demand is likely to exceed capacity and result in a breakdown that the PTSU strategy will attempt to mitigate or even eliminate. Since congested conditions are those of primary interest, it is assumed that the oversaturated method will be used for the entirety of the analysis. While it does not provide the exact same operational results as the individual segment methodologies during undersaturated time periods, the oversaturated approach does adequately approximate the method to the extent that it will not demonstrably affect the results of the experiment.
The second consideration is that the default 15-minute time step provides an analysis resolution that is too coarse to capture the necessary responsiveness of a D-PTSU system. However, as mentioned previously, the oversaturated approach actually updates operational conditions at a 15-second resolution before being aggregated to 15-minute results for consistency. By overriding the HCM's default 15-minute aggregation of results and replacing it with a reduced increment, such as a oneminute resolution, this issue can be circumvented in a straightforward manner without any true modifications to the methodology.
Bypassing the 15-minute aggregation of results does not require changing any underlying methodological assumptions or modifying any specific computational steps. Rather, it is accomplished by changing a single global variable of the methodology as defined in chapter 25 of the HCM: S — the number of computational time steps in an analysis period.
Reducing this from the default value of 60 (corresponding to a 15-minute analysis period) to a value of 4 effectively sets the length of an analysis period of the methodology to one minute. This required modification to the aggregation procedure as well as corresponding updates to the interface, which were made directly within the open-source FREEVAL engine to support the computational details needed for this project and the effective modeling of D-PTSU.
United States Department of Transportation - Federal Highway Administration
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It is currently 17 Oct 2017, 16:06
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# Magazine Publisher: Our magazine does not have a liberal
Author Message
Director
Joined: 17 Oct 2005
Posts: 924
Kudos [?]: 267 [0], given: 0
Magazine Publisher: Our magazine does not have a liberal [#permalink]
### Show Tags
17 Apr 2006, 03:53
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
Magazine Publisher: Our magazine does not have a liberal bias. It is true that when a book review we had commissioned last year turned out to express distinctly conservative views, we did not publish it until we had also obtained a second review that took a strongly liberal position. Clearly, however, our actions demonstrate not a bias in favor of liberal views but rather a commitment to a balanced presentation of diverse opinions.
Determining which of the following would be most useful in evaluating the cogency of the magazine publisher’s response?
A. Whether any other magazines in which the book was reviewed carried more than one review of the book
B. Whether the magazine publishes unsolicited book reviews as well as those that it has commissioned
C. Whether in the event that a first review commissioned by the magazine takes a clearly liberal position the magazine would make any efforts to obtain further reviews
D. Whether the book that was the subject of the two reviews was itself written from a clearly conservative or a clearly liberal point of view
E. Whether most of the readers of the magazine regularly read the book reviews that the magazine publishes
Kudos [?]: 267 [0], given: 0
Director
Joined: 13 Nov 2003
Posts: 788
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Location: BULGARIA
### Show Tags
17 Apr 2006, 04:03
Think that C is best option. Since the author claims that they look for balance of opinions, the argument would be cogent in case C is correct
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Manager
Joined: 28 Mar 2006
Posts: 157
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### Show Tags
17 Apr 2006, 08:25
IMO C and E are out..coz in C...it says..that the first review is liberal..hwever in the question the condition present is vice versa
Though confused b/w rest of d choices...D seems better than others..
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Senior Manager
Joined: 24 Jan 2006
Posts: 251
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### Show Tags
17 Apr 2006, 08:47
C seems correct
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Senior Manager
Joined: 07 Mar 2006
Posts: 351
Kudos [?]: 37 [0], given: 1
### Show Tags
17 Apr 2006, 08:53
Even i'll go with C.
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Manager
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Posts: 157
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### Show Tags
17 Apr 2006, 10:19
Hey please notice that in the paragraph...the first review was conservative and thus not liberal..hwever choice C..talks about the first review being liberal n then too going for a second review....
Am I thinking in the right direction?
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Senior Manager
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### Show Tags
17 Apr 2006, 10:50
I go for C too.
- Vipin
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Intern
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### Show Tags
17 Apr 2006, 10:56
C.
Others seem out of scope.
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Director
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### Show Tags
17 Apr 2006, 15:09
C is oa,
I hardly understand this CR, can you guys give explanations to your answers? thanks
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VP
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### Show Tags
17 Apr 2006, 16:25
late but C!
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### Show Tags
17 Apr 2006, 16:30
joemama142000 wrote:
C is oa,
I hardly understand this CR, can you guys give explanations to your answers? thanks
"we did not publish it until we had also obtained a second review that took a strongly liberal position" was given in the stem
In the question he asks us the genuinity of the claim made
Choice C tells us clearly that if the first reading has taken a liberal position which the mag claims do they go fora second review which gives them a MORE liberal view.
Kudos [?]: 38 [0], given: 0
17 Apr 2006, 16:30
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Lemma 87.20.8. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $X_{/T}$ be the formal completion of $X$ along $T$.
1. If $X \setminus T \to X$ is quasi-compact, then $X_{/T}$ is locally adic*.
2. If $X$ is locally Noetherian, then $X_{/T}$ is locally Noetherian.
Proof. Choose a surjective étale morphism $U \to X$ with $U = \coprod U_ i$ a disjoint union of affine schemes, see Properties of Spaces, Lemma 66.6.1. Let $T_ i \subset U_ i$ be the inverse image of $T$. We have $X_{/T} \times _ X U_ i = (U_ i)_{/T_ i}$ (Lemma 87.14.4). Hence $\{ (U_ i)_{/T_ i} \to X_{/T}\}$ is a covering as in Definition 87.11.1. Moreover, if $X \setminus T \to X$ is quasi-compact, so is $U_ i \setminus T_ i \to U_ i$ and if $X$ is locally Noetherian, so is $U_ i$. Thus the lemma follows from the affine case which is Lemma 87.14.6. $\square$
Comment #2029 by Brian Conrad on
Maybe it is worthwhile to include a Remark warning that if $X = {\rm{Spec}}(A)$ is affine and $T$ is the zero locus of a finitely generated ideal $I$ (so $X-T$ is quasi-compact) with radical $J$ then from the definitions $X_{/T} = {\rm{Spf}}(A^{\wedge})$ for the $I$-adic completion $A^{\wedge}$ of $A$ whereas the natural map from $A^{\wedge}$ to the $J$-adic completion of $A$ can fail to be a ring isomorphism.
As an example, to illustrate the issues, one can mention the old standby $A = O_K$ for an algebraically closed field $K$ equipped with a rank-1 valuation, $I = (\pi)$ for a nonzero nonunit $\pi \in A$, and $J = \mathfrak{m}$ the maximal ideal. Since $J^2=J$ the $J$-adic completion of $A$ is the residue field, whereas the $I$-adic completion of $A$ is the valuation ring of the completion of $K$ for the given rank-1 valuation).
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# Interpretation of negative binomial GLM
In this model, the interpretation of the continuous variable tmax for an example would be: a increase 1 unit of tmax (exp(coef)=1.06) increases in 6% the incidence of disease in a month. Considering that casos = monthly number of cases of the disease, and populacao variable being used as offset representing the population in each city (municipio).
Is this interpretation correct?
summary(m1<- glm.nb(casos ~ 0 + municipio + precip_ant + tmax + tmax_ant + umid + umid_ant + enxu_2 + offset(log(populacao)), data = dataset))
Call:
glm.nb(formula = casos ~ 0 + municipio + precip_ant + tmax +
tmax_ant + umid + umid_ant + enxu_2 + offset(log(populacao)),
data = dataset, init.theta = 2.105944887, link = log)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.1763 -1.0107 -0.6286 0.3874 4.2286
Coefficients:
Estimate Std. Error z value Pr(>|z|)
municipio6 -2.566e+01 1.698e+00 -15.110 < 2e-16 ***
municipio1 -2.406e+01 1.706e+00 -14.108 < 2e-16 ***
municipio2 -2.424e+01 1.707e+00 -14.205 < 2e-16 ***
municipio3 -2.530e+01 1.696e+00 -14.914 < 2e-16 ***
municipio4 -2.525e+01 1.701e+00 -14.846 < 2e-16 ***
municipio5 -2.524e+01 1.702e+00 -14.829 < 2e-16 ***
precip_ant 1.750e-03 6.414e-04 2.728 0.006373 **
tmax 6.291e-02 1.922e-02 3.273 0.001066 **
tmax_ant 1.600e-01 1.995e-02 8.020 1.06e-15 ***
umid 2.665e-02 1.230e-02 2.166 0.030297 *
umid_ant 5.555e-02 1.454e-02 3.820 0.000134 ***
enxu_2 3.154e-01 2.074e-01 1.521 0.128384
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for Negative Binomial(2.1059) family taken to be 1)
Null deviance: 41285.6 on 1008 degrees of freedom
Residual deviance: 1002.5 on 996 degrees of freedom
AIC: 2702.5
Number of Fisher Scoring iterations: 1
Theta: 2.106
Std. Err.: 0.308
2 x log-likelihood: -2676.499
> exp(coef(m1))
municipio6 municipio1 municipio2 municipio3 municipio4 municipio5 precip_ant tmax tmax_ant
7.171204e-12 3.553106e-11 2.963580e-11 1.033429e-11 1.082516e-11 1.089931e-11 1.001751e+00 1.064932e+00 1.173543e+00
umid umid_ant enxu_2
1.027009e+00 1.057120e+00 1.370790e+00
In a Negative Binomial (NB) regression model with no offset, the 6.291e-02 coefficient would represent the change in the log of the expected count of the monthly number of cases of the disease for a one-unit increase in the corresponding predictor variable tmax, while holding all other predictor variables constant. i.e. the expected count would be multiplied by ($$\exp(6.291\text{E-02})$$=) 1.06 for a one-unit increase in tmax. Because though our NB regression model has an offset this increase is against the expected rate. The post uses the term: "incidence of disease" which I think is somewhat open to interpretation, it is probably to explicitly say: "incidence rate of disease" but aside from that the interpretation is good to go.
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# Subtraction 3 (ages 7 to 12)
• Sale
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Tax included at checkout (when applicable)
Follow Us Follow us on Twitter, Instagram (@ttl_cic) and Pinterest (we aren't on Facebook), and remember to buy the resources that you viewed! Every worksheet, template, book, toy or service purchased funds our work empowering lifelong learners, whether in the homeschool, classroom or studying independently. 2020 marks the 14th year of Talk Together London's work as a registered ‘Community Interest Company’ - a 'cic' with charitable objectives, supporting those who are socially or economically disadvantaged. Contribute to community projects and subscribe to our newsletter to get valuable resources delivered right to your mailbox.
With the method used in this worksheet, learners can learn to subtract two digit numbers easily by:
• partitioning 2-digit numbers and recognising their place values in subtraction
• regrouping the numbers to be subtracted in columns, in order to calculate their difference in writing
• subtracting with column maths and decomposition or borrowing.
In a nutshell: Confidence breeds more confidence! And, hands on practice in subtracting does wonders for confidence and gradual skills development. Whether homeschooling, working independently or in classroom teaching, this worksheet will make learners win and brighten the day.
When kids are learning two digit addition and subtraction, one of the concepts they'll encounter is regrouping, which is also known as carrying over or column maths.
Carrying over is an important concept to learn, because it makes working with large numbers manageable. For example, when subtracting 17 from 42 in writing, we must carry over because we can not subtract 7 from 2.
Carrying over in subtraction, or as some people refer to it 'borrowing', can be confusing to young learners. Children know how to subtract 20 - 10 and get the correct answer. But, when the problem is 30 - 18, subtraction becomes more challenging! Because, to be able to take away 18, they must understand how to carry over.
Many times, we teach students the process of working a math problem – but not the 'why' behind it. And if students don’t understand how place value works in subtraction, then they don't understand why we 'borrow'.
Best get started with it!
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We are offering for sale as part of our current fundraising drive, because we do not use them anymore or need them any longer and there is limited room for storage, a collection of 'like new' and 'pre-owned' toys.
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Grams Pounds Europe World Zone-1 World Zone-2 100g 0.22 5.65 6.30 6.75 250g 0.55 5.85 7.15 7.69 500g 1.1 7.75 10.60 11.35 750g 1.65 8.99 12.75 13.55 1kg 2.2 10.05 14.99 15.79 1.25kg 2.75 10.89 16.85 17.9 1.5kg 3.3 11.89 18.55 19.89 1.75kg 3.85 12.65 18.90 20.45 2kg 4.4 13.05 19.75 21.55
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Your Order
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All items are dispatched using a recorded service. If you have not received your parcel within the advertised period from receiving your dispatch email, please contact us via:
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or
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#### FREE RESOURCES
DOWNLOAD: Addition Table
DOWNLOAD: Multiplication Square
Note: Both PDF documents contain forms. You may download, print and complete or learners may complete the exercises on a computer.
There is a FREE version of the 7-sentence story template available online here:
https://talktogetherlondon.org/blogs/resources-and-services/story-maker
#### BLOG
Together: News n'Info
A chance to catch up with news, gain insight on issues and share views.
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# 1.if n is an odd number and p is an even then p+p+n=2.what will you get if you add three odd numbers and an even number?3.give the difference between the two odd numbers right after 20.is the difference odd or even?4.add the consecutive even and odd numbers between 5 and 10.is the total odd or even?
1
by ariza
2014-06-25T23:41:38+08:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Remember: even±even/odd±odd = even & odd±even = odd
(or para mas madali, pag same=even at pag magkaiba=odd
1) p (even) + p (even) = even + n (odd) = odd
2) odd+odd = even + odd = odd + even = odd
3) 19-17 = 2 = even
4) 6+7+8+9 = 30 = even
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| 262,011,989
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# Challenge 22 - Largest Pandigital Prime
Welcome, everyone to the discussion thread for the 22nd Go challenge posted to the site
Feel free to show us how you solved the challenge by posting your solutions below!
I dont know why my solution is wrong yet. The number is correct according to my research.
I know it is not the best solution using go. Iโm new on this language.
I also tried not to apply big maths algorithm. Used only simple logic and solutions applied in previous challenges.
According to my research of Problem 41 of Project Euler, the The largest Pandigital prime is 7652413.
My code returns this number, but it does not pass in the test
package main
import (
"fmt"
"math"
"strconv"
)
func LargestPandigitalPrime() int {
// finds the greater possible sequence eliminating divisibles by 3
// I mean, 1..(greater n), where 1+2+3+..+n is not divisble by 3
sequence := findGreaterSequence()
if len(sequence) == 0 {
fmt.Println("It is not possible")
return 0
}
// using the sequence get the upper and lower boundaries
// 1..(greater n) and (greater n) to 1
// like 123456789 and 987654321
lowerBound, upperBound := getLowerAndUpperBoundaries(sequence)
// decreasing from upper to lower, finding primes and pandigital ones
// the bigger found is returned
greater := findGreaterNumber(lowerBound, upperBound)
return greater
}
func runeToInt(r rune) int {
return int(r) - '0'
}
func intToRune(n int) rune {
return rune(n) + '0'
}
func isPrime(n int) bool {
return len(CheckFactors(n)) == 2
}
// used int exercide Even and Odds factors
// could be optimized just returning false when found a factor
func CheckFactors(num int) []int {
factors := []int{1, num}
limit := int(math.Ceil(float64(num) / 2))
for i := 2; i <= limit; i++ {
if num%i == 0 {
factors = append(factors, i)
}
}
return factors
}
func isEven(n int) bool {
return n%2 == 0
}
// Pandigital must have only one digit from 1..(size(n))
func isPandigital(n int) bool {
str := fmt.Sprint(n)
size := len(str)
charMap := make(map[rune]int)
for _, char := range str {
// does have digit out of range from 1..(size(n))
if runeToInt(char) > size {
return false
}
// does have repeated digit
if _, ok := charMap[char]; ok {
return false
}
charMap[char] = 1
}
return true
}
func findGreaterNumber(lower int, upper int) int {
for i := upper; i > lower; i-- {
if !isEven(i) && isPandigital(i) && isPrime(i) {
return i
}
}
return 0
}
// eliminates number divisible by 3
func findGreaterSequence() []int {
bigger := "123456789"
for len(bigger) > 0 {
sum, sequence := sumNumbers(bigger)
if sum%3 != 0 {
return sequence
}
// removes the bigger digit from sequence
bigger = bigger[0 : len(bigger)-1]
}
return []int{}
}
// sum the digits of a string returning the sum and all digits
func sumNumbers(s string) (int, []int) {
sum := 0
var sequence []int
for _, r := range s {
number := runeToInt(r)
sum += number
sequence = append(sequence, number)
}
return sum, sequence
}
func getLowerAndUpperBoundaries(seq []int) (int, int) {
size := len(seq)
min := make([]rune, size, size)
max := make([]rune, size, size)
for i := 0; i < size; i++ {
r := intToRune(seq[i])
min[i] = r
max[size-i-1] = r
}
minValue, _ := strconv.ParseInt(string(min), 10, 0)
maxValue, _ := strconv.ParseInt(string(max), 10, 0)
return int(minValue), int(maxValue)
}
func main() {
fmt.Println("Pandigital Primes")
pandigitalPrime := LargestPandigitalPrime()
fmt.Println(pandigitalPrime)
//7652413
}
Iโve just taken a look, there was a typo in the challenge tests which Iโve just pushed a fix for to the main branch for the site!
I seriously appreciate you highlighting this as it helps the site and the community!
One slight improvement I would make is handling the ParseInt error value as itโs good practice to keep checking these errors. Other than that it is a solid implementation!
1 Like
this is No.41 problem in Project Euler: https://projecteuler.net , I reference one solution from jmoiron
package main
import (
"fmt"
"math"
"strconv"
)
func notPrime(n int) bool {
if n&2 == 2 {
return false
}
if (n-1)%6 == 0 {
return false
}
if (n+1)%6 == 0 {
return false
}
return true
}
func isPrime(n int) bool {
if notPrime(n) {
return false
}
max := int(math.Ceil(math.Sqrt(float64(n))))
for i := int(3); i < max; i += 2 {
if n%i == 0 {
return false
}
}
return true
}
func atoi(s string) int {
n, _ := strconv.Atoi(s)
return n
}
func permute(s string) []string {
if len(s) == 1 {
return []string{s}
}
perms := []string{}
tail := s[1:]
for _, perm := range permute(tail) {
for i := 0; i < len(s); i++ {
newperm := perm[:i] + head + perm[i:]
perms = append(perms, newperm)
}
}
return perms
}
func LargestPandigitalPrime() int {
max := 0
digits := "7654321"
for i := 0; i < len(digits); i++ {
for _, perm := range permute(digits[i:]) {
i := atoi(perm)
if isPrime(i) {
if i > max {
max = i
}
}
}
if max > 0 {
break
}
}
return max
}
func main() {
fmt.Println("Pandigital Primes")
pandigitalPrime := LargestPandigitalPrime()
fmt.Println(pandigitalPrime)
}
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|
# دوره راهنمای مطالعه و تمرین- تست GRE ، فصل 19 : GRE Quantitative Reasoning- Exponents
## دربارهی این فصل:
Review the elements of exponents to prepare for the Graduate Record Exam (GRE). Video lessons and quizzes incorporate these elements of the quantitative reasoning component in a simple and clear manner.
## این شامل 6 زیر است:
In this video, learn how to go from a rational exponent to a radical expression and back. No tricks or magic, just good math! We'll review the basics and look at a few examples.
Simplifying expressions with rational exponents is so easy. In fact, you already know how to do it! We simply use the exponent properties but with fractions as the exponent!
Brand new technologies don't always catch on right away because they can be expensive and don't always work as well as they should. But once the price comes down, and they start to work better, it doesn't take very long before it seems like everyone has one. Learn about the numbers behind this, exponential functions!
We'll look at the five important exponent properties and an example of each. You can think of them as the order of operations for exponents. Learn how to handle math problems with exponents here!
In this video, we will put all of the exponent properties we have learned together. Don't let lots of numbers and letters confuse you! Simplifying exponents can be easy and fun!
The zero and negative exponent properties are two you will use quite a lot in mathematics. The negative exponent property can be confusing, but when you remember a couple fun ideas, you will get it right every time!
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https://exercism.io/tracks/javascript/exercises/grains/solutions/a8e484158ff94fa9a1e77eed88e98271
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# PatrickMcSweeny's solution
## to Grains in the JavaScript Track
Published at Aug 21 2019 · 0 comments
Instructions
Test suite
Solution
#### Note:
This exercise has changed since this solution was written.
Calculate the number of grains of wheat on a chessboard given that the number on each square doubles.
There once was a wise servant who saved the life of a prince. The king promised to pay whatever the servant could dream up. Knowing that the king loved chess, the servant told the king he would like to have grains of wheat. One grain on the first square of a chess board, with the number of grains doubling on each successive square.
There are 64 squares on a chessboard (where square 1 has one grain, square 2 has two grains, and so on).
Write code that shows:
• how many grains were on a given square, and
• the total number of grains on the chessboard
## For bonus points
Did you get the tests passing and the code clean? If you want to, these are some additional things you could try:
• Optimize for speed.
Then please share your thoughts in a comment on the submission. Did this experiment make the code better? Worse? Did you learn anything from it?
## Setup
Go through the setup instructions for Javascript to install the necessary dependencies:
https://exercism.io/tracks/javascript/installation
## Requirements
Install assignment dependencies:
``````\$ npm install
``````
## Making the test suite pass
Execute the tests with:
``````\$ npm test
``````
In the test suites all tests but the first have been skipped.
Once you get a test passing, you can enable the next one by changing `xtest` to `test`.
## Source
JavaRanch Cattle Drive, exercise 6 http://www.javaranch.com/grains.jsp
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### big-integer.spec.js
``````import BigInt from './lib/big-integer';
describe('The big-integer module\'s returned object', () => {
let bigI;
beforeEach(() => {
bigI = BigInt(42);
});
afterEach(() => {
bigI = null;
});
test('is not a number', () => {
expect(typeof 42).toBe('number');
expect(typeof bigI).not.toBe('number');
expect(typeof bigI).toBe('object');
});
test('can be compared to a stringified number by calling \'.toString()\'', () => {
expect(bigI).not.toBe(42);
expect(bigI).not.toBe('42');
expect(bigI.toString()).toBe('42');
// NOTE:
// The '==' operator calls '.toString()' here in order to compare.
expect(bigI == '42').toBe(true);
// While the line above is easier to write and read, we will use the
// 'expect(bigI.toString()).toBe(expected)' way so that test failure
// "Expected '84' to be '42'." instead of
// "Expected false to be true."
});
test('is immutable', () => {
expect(bigI.toString()).toBe('42');
bigI.subtract(10);
expect(bigI.toString()).toBe('42');
});
expect(bigI.toString()).toBe('84');
});
test('can perform power operations', () => {
bigI = BigInt(10);
bigI = bigI.pow(2);
expect(bigI.toString()).toBe('100');
});
// The "Methods" section of the README is especially useful:
//
// https://github.com/peterolson/BigInteger.js#methods
});``````
### grains.spec.js
``````/**
* In JavaScript, integers beyond +/- 9007199254740991 cannot be accurately
* represented. To see this in action, console.log() out the expected number
* of grains on square #64:
*
* console.log(9223372036854775808);
* // => 9223372036854776000
* // ^^^^
*
* This is because, in JavaScript, integers are represented as 64-bit floating
*
* https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/MAX_SAFE_INTEGER
* http://stackoverflow.com/questions/307179/what-is-javascripts-highest-integer-value-that-a-number-can-go-to-without-losin
*
* So, an accurate solution to this problem requires the use of a
* "big integer" type. There are multiple ways to use big integer types.
* We have provided you with BigInteger.js. You can read more about it here:
*
* https://github.com/peterolson/BigInteger.js
* ^--- The "Methods" section of the README will be especially helpful.
*
* https://github.com/peterolson/BigInteger.js/blob/master/spec/spec.js
* ^--- Tests are a good way to understand, in addition to the README.
*
* To get you started, this folder has a file of the big-integer module.
* See its tests in this folder for a quick primer on how to use it! ( :
*/
import { Grains } from './grains';
describe('Grains', () => {
const grains = new Grains();
test('square 1', () => {
expect(grains.square(1)).toBe('1');
});
xtest('square 2', () => {
expect(grains.square(2)).toBe('2');
});
xtest('square 3', () => {
expect(grains.square(3)).toBe('4');
});
xtest('square 4', () => {
expect(grains.square(4)).toBe('8');
});
xtest('square 16', () => {
expect(grains.square(16)).toBe('32768');
});
xtest('square 32', () => {
expect(grains.square(32)).toBe('2147483648');
});
xtest('square 64', () => {
expect(grains.square(64)).toBe('9223372036854775808');
});
xtest('total', () => {
expect(grains.total()).toBe('18446744073709551615');
});
});``````
``````import BigInt from './lib/big-integer';
const NUMBER_OF_SQUARES = 64;
export class Grains {
square(number) {
return BigInt(2).pow(number - 1).toString();
}
total() {
let total = BigInt(0);
for(let i = 1; i <= NUMBER_OF_SQUARES; i++){
}
}
}``````
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https://quizizz.com/admin/quiz/5bbd46b0050b46001bfbda1d
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Multiplication and Division Unknowns
a year ago
sarahivey
Save
Edit
Host a game
Live GameLive
Homework
Solo Practice
Practice
• Question 1
120 seconds
Q.
Solve for the unknown
24÷_____=4
3
2
6
5
• Question 2
120 seconds
Q.
Solve for the unknown
4x____=32
7
8
4
6
• Question 3
120 seconds
Q.
Solve for the unknown
27÷___=9
2
4
5
3
• Question 4
120 seconds
Q.
Solve for the unknown
___x6=18
3
2
1
4
• Question 5
120 seconds
Q.
Solve for the unknown
3x___=30
10
9
7
8
• Question 6
120 seconds
Q.
Solve for the unknown
36÷___=4
9
6
5
3
• Question 7
120 seconds
Q.
Solve for the unknown
7x___=21
2
4
3
1
• Question 8
120 seconds
Q.
Solve for the unknown
56÷___=8
7
6
5
8
• Question 9
120 seconds
Q.
Solve for the unknown
___x5=35
5
3
4
7
• Question 10
120 seconds
Q.
Solve for the unknown
42÷___=7
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https://www.physicsoverflow.org/40786/pseudotensor-in-different-dimensions
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# pseudotensor in different dimensions
+ 0 like - 0 dislike
427 views
In this topic https://physics.stackexchange.com/questions/129417/what-is-pseudo-tensor one answer was the next:
The action of parity on a tensor or pseudotensor depends on the number of indices it has (i.e. its tensor rank):
- Tensors of odd rank (e.g. vectors) reverse sign under parity.
- Tensors of even rank (e.g. scalars, linear transformations, bivectors, metrics) retain their sign under parity.
- Pseudotensors of odd rank (e.g. pseudovectors) retain their sign under parity.
- Pseudotensors of even rank (e.g. pseudoscalars) reverse sign under parity.
But I don't understand one thing. Is that statement only for Euclidean three dimensions? I attempted understand it myself. And it is my thoughts. Pseudotensor is determined as:
$\hat{P}^{i_1\ldots i_q}_{\,j_1\ldots j_p} = (-1)^A A^{i_1} {}_{k_1}\cdots A^{i_q} {}_{k_q} B^{l_1} {}_{j_1}\cdots B^{l_p} {}_{j_p} P^{k_1\ldots k_q}_{l_1\ldots l_p}$
where $(-1)^A = \mathrm{sign}(\det(A^{i_q} {}_{k_q})) = \pm{1}$.
Let's consider a pseudovector in Euclidean three dimensions. Then $\det(A^{i_q} {}_{k_q})$ is
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{pmatrix}
And $(-1)^A=-1$
Let's consider a pseudovector in Euclidean three dimensions. Then $\det(A^{i_q} {}_{k_q})$ is
\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 &0 \\
0 & 0 & 0 & -1
\end{pmatrix}
And $(-1)^A=1$
Let's consider a pseudovector in Minkovski space. Then $\det(A^{i_q} {}_{k_q})$ is
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 &0 \\
0 & 0 & 0 & -1
\end{pmatrix}
And $(-1)^A=-1$
am I right?
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A numeric keypad, numpad or ten key,[1][2][3] is the small, palm-sized, seventeen key section of a computer keyboard, usually on the very far right. The numeric keypad features digits 0 to 9, addition (+), subtraction (-), multiplication (*) and division (/) symbols, a decimal point (.) and Num Lock and Enter keys.[4] Laptop keyboards often do not have a numpad, but may provide numpad input by holding a modifier key (typically labelled "Fn") and operating keys on the standard keyboard. Particularly large laptops (typically those with a 17 inch screen or larger) may have space for a real numpad, and many companies sell separate numpads which connect to the host laptop by a USB connection (many of these also add an additional spacebar off to the side of the number zero where the thumb is located, as well as an additional 00 key typical of modern adding machines and cash registers). It also provides a calculator-style keyboard for efficient entering of numbers.
Numeric keypads usually operate in two modes: when Num Lock is off, keys 8, 6, 2, 4 act like arrow keys and 7, 9, 3, 1 act like Home, PgUp, PgDn and End; when Num Lock is on, digits keys produce corresponding digits. On Apple Macintosh computers, which lack a Num Lock key, the numeric keypad always produces only numbers. The Num Lock key is replaced by the Clear key.
The arrangement of digits on numeric keypads is different from that of telephone Touch-Tone keypads, which have the 1-2-3 keys on top and 7-8-9 keys on the third row, instead of the reverse used on a numeric keypad. It is thought that the layout was arranged this way for the early telephone keypads so that the addition of the alphabetical characters found on most telephones followed the digits numerically.[citation needed] This layout, which matches most modern calculators and cash registers, may be confusing for those who use one of these arrangements more often.
Numeric keypads are useful for entering long sequences of numbers quickly, for example in spreadsheets, financial/accounting programs, and calculators. Input in this style is similar to that of a calculator or adding machine.
## Chinese input methods
The numeric keypad is used by some systems for input of Chinese characters, for example CKC Chinese Input System and Q9 input method.
## Computer games
A 104-key PC US English QWERTY keyboard layout with the numeric keypad at the far right.
Numeric keypads are also used for playing some computer games where the player must control a character, for example roguelikes. Unlike arrow keys, the numeric keypad allows diagonal movement. For keyboards without a numeric keypad, some games provide alternative movement keys, such as classic Rogue's HJKL keys.
The numeric keypad can also be an alternative for navigation in computer gaming from the WASD keys. The mouse would be used in the left hand instead of the right.
## References
1. ^ "What Is a Ten Key? (with pictures)". WiseGeek. Retrieved 8 January 2014.
2. ^ Pasewark, William Robert; Knowlton, Todd (1995-01-01). Ten-Key Skill Builder for Computers. South-Western Educational Pub. ISBN 9780538629195.
3. ^ Stroman, James; Wilson, Kevin; Wauson, Jennifer (2011-09-01). Administrative Assistant's and Secretary's Handbook. AMACOM Div American Mgmt Assn. p. 224. ISBN 9780814417607.
4. ^
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https://photo.stackexchange.com/questions/98650/how-is-focal-length-facial-distortion-different-on-an-aps-c-sensor-compared-to-a?noredirect=1
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# How is focal length facial distortion different on an APS-C sensor compared to a full frame? [duplicate]
From my understanding (and research) focal length affects how a face will end up looking on a photograph.
Here's an example of focal length distortion on a full-frame sensor:
My question is: will a photo taken with a 28mm focal length on an APS-C sensor look identical to a photo taken with a 40mm (28mm equivalent on a full frame sensor) on a full frame sensor? or will the levels of distortion remain the same as with a 28mm - full-frame sensor combo?
## 5 Answers
The facial deformations you are worried about are due to perspective distortion. Perspective is determined by one thing and one thing only: subject distance.
Here's an example of focal length distortion on a full-frame sensor:
That really is an example of how changing the subject distance to frame the same subject with different focal length lenses affects the perspective. To get the subject (roughly) the same size in each image required very different shooting distances for the 20mm lens and the 200mm lens. The 20mm shot was probably taken less that a foot away from the tip of the subject's nose. The 200mm shot was probably taken from around 10 feet (if the lens in question is an "honest" 200mm at short focus distances).
In the first case (20mm) the subject's ears were nearly twice as far from the camera as the subject's nose. In the last case (200mm), the distances between the camera and the nose on the one hand and the camera and the ears on the other was probably less than 10 percent.
If you take a picture with a 50mm lens on a FF camera from a distance of 10 feet and also take a picture with a 30-35mm lens on an APS-C camera from a distance of 10 feet both pictures will have the same perspective and framing (allowing for the rounding of focal lengths - a 50mm lens may be anywhere from about 46-53mm in actual focal length and most other lenses are also usually rounded to the nearest common focal length).
The focal length of a 35mm lens is always 35mm. A 35mm FX lens and a 35mm DX lens will both provide the same field of view on a camera with DX sensor (Again, allowing for that pesky rounding - all 35mm lenses aren't exactly 35mm. Some are 37mm, others might be 33mm).
The only reason we use crop factors is to compare the angle of view (AoV) of lenses used on cameras with smaller sensors to the focal length needed to obtain the same angle of view on a full frame body. When comparing the AoV of a 35mm DX lens or 35mm FX lens used on a DX body to the focal length needed for the same FoV on an FX camera both lenses need to be multiplied by the DX crop factor.
YES, you are correct. A photo taken with a 28mm lens on an APS-C will have the same level of distortion as a 40mm lens on a Full Frame camera provided you shoot the subject from the same distance.
It is the distance from the subject that determines the facial “perspective” distortion, not the focal length.
What actually affects the way a face will look is perspective - how close you are to the subject. The difference in the look between the first image example and the last is entirely driven by how far the camera is from the subject.
The focal length and sensor size will determine the size of the head in the final image but have no effect on the "look" of the facial features.
So your subject's facial features taken with a 28mm lens on an APS-C will look identical to a picture taken with a 28mm lens on a full frame camera and identical to a picture taken with a 28mm lens taken with a medium format camera - taken from the same spot. The difference will be that the head in the shot taken with the APS-C sensor will take up more of the captured pixels than the full frame sensor and much more than the medium format.
The "look" of the face will be the same because all three shots were taken from the same distance.
Now we normally think of a head and shoulders portrait of a person having the head take up the majority of the height of the final image and we usually like to have a certain look to our portraits. The portrait photographer makes the choice of where to stand to get the look they want. But that look is driven by the distance you are standing from your subject. It is the perspective given by that subject distance that you want. Then to achieve the head and shoulders filling the frame you need to choose a different focal length lens on our three example formats.
Say for example you want to choose a look like the one labelled "50mm" in the example above. To get that look you have to choose the right subject distance and then to get the subject's face to fill the frame you need to choose a focal length to get that. If you are using an APS-C camera you will want to choose a 33mm lens. If you are using a full frame 35mm camera you will want a 50mm lens. If you are using a 645 medium format camera you will want an 80mm lens.
But if all you have is a 28mm lens you can still get the look you want by standing in the appropriate place and cropping the APS-C image a little, cropping the full frame a more and cropping the 645 image a larger amount. They will all give you the same final look given that you took the image from the same location.
Facial distortion is evident when features like nose size and ear size are perceived as incorrect. Our view of what we look like is derived from our dressing mirror. If the photographer can replicate this mind’s eye view, facial distortion becomes moot. In most cases, if the photograph just stepped back, the problem is vanished.
OK a more scientific approach. Every image has a “correct” viewing distance associated with it. If the image is viewed from such a distance, most all problems related to perspective distortion are moot. Fortunately, most pictures are immune from perspective distortion so any viewing distance is OK. The human face is a major exception.
What comprises the correct viewing distance? This distance intertwines focal length and the magnification applied to make the final display. Suppose we make 8 by 12 inch print or display same on a monitor / TV with the same dimensions. A full frame 24 x 36mm must the enlarged 8 ½ X to obtain this size. A compact digital frame is 16 X 24mm and we must enlarge 12 ¾ X to get to the same size.
From a scientific point of view, we should use a lens with a focal length of about 2.5X normal when doing portraiture. OK to round most of the values. At this point let me add that in art, there are no rules so you are free to follow your heart. For the FX that’s about 105mm focal length. For the DX that’s 75mm. These values are approximately the corner to corner measurement of the frame size time 2.5.
The correct viewing distance minimizing distortion is focal length multiplied by magnification. We observe an image 8x12 inch shot on FX with 105mm enlarged 8.5x. The viewing distance is 105 X 8.5 = 890mm = 35 inches.
Using this same formula for a DX with a 70mm making a 8x12 display, the focal length is 75mm X 12.75 = 950mm = 37 inches.
My conclusion -- the FX with a 105 will display about the same degree of facial distortion as the DX with a 75mm mounted.
“My question is: will a photo taken with a 28mm focal length on an APS-C sensor look identical to a photo taken with a 40mm (28mm equivalent on a full frame sensor) on a full frame sensor?”
Answer DX 28mm X 12.75 = 350mm = 13.7 inch viewing distance. FX 40mm X 8.5 = 340mm = 14 inch viewing distance. Conclusion – They will look about the same.
Perspective is by capture ratio, not focal length per se, so you are exactly right. If a 40mm lens has the same capture size as a 28mm lens on a larger format of capture, then the perspective will be identical. Assuming the cameras have the same x/y ratio (e.g. 35mm is 24x36mm, i.e. 1:1.5 or 2:3), a good way of determining the proportional focal length is to take the size of the diagonal. You can figure this out using the Pythagorean theorem, since it's a triangle:
x^2 + y^2 = z^2 (z is the diagonal)
24^2 + 36^2 = 1872
z^2 = 1872
z = sqr(1872) = 43.3 mm
If your crop sensor camera has a diagonal of, say, 30mm, then 30/43.3 = 0.69 and if you multiply that by a focal length on 35mm, you'll get the equivalent focal length. For example, a 100mm lens on full frame or 35mm film capture is the equivalent of a 69mm lens on this fictitious crop-sensor camera and will have the same perspective.
Now, you didn't ask about this, but depth of field is another animal. Depth of field is fixed by focal length. A 28mm lens has the same depth of field on all formats. This means, with smaller sensors or film, depth of field is greater at a given capture perspective than you get on larger formats. This means very deep depth of field is easy to get on very tiny sensors, but isolating the background is very difficult. On very large formats, the opposite occurs.
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## Wednesday, May 6, 2015
### PARCC Practice Test Question 17 (Day 162)
Chapter 8 of Mario Livio's The Equation That Couldn't Be Solved is titled, "Who's the Most Symmetrical of Them All." In this chapter, Livio shows how symmetry can be applied to other fields beyond mathematics and physics.
The title of the chapter is an allusion to the question asked by Snow White's stepmother, "Who's the fairest of them all." The point Livio makes here is that "fairest" -- in the sense of "most attractive" -- really means "most symmetrical." According to him, a preference for symmetry has been hard-wired into humans by the forces of evolution.
I once read about a math teacher who wanted to motivate her high school students to learn geometry. I must admit that this was over a year ago, and I don't remember who this teacher was -- otherwise I would provide a link. Anyway, when her class reached the lesson on symmetry and reflections, this teacher motivated her students by discussing the link between symmetry and attractiveness. The point was that these are high school students, at an age where one is confused about one's own sexuality. So these students may be interested in whether or not they are attractive -- and the teacher took advantage by telling them how attractiveness related to symmetry, and therefore to geometry. This was especially effective with the female students. (Still, it's a shame that I couldn't find the link.)
Livio moves on to symmetry in music. We've already discussed the relationship between music and mathematics after I watched the NOVA episode -- the same episode in which Livio appeared (and ultimately led me to reading his book). The author repeats how Pythagoras is credited with the discovery that notes that sound good together have frequencies in simple ratios -- 3:2 for the perfect fifth, 4:3 for the perfect fourth, and so on. Anyway, Livio links this back to group theory, in that if we represent the notes by numbers, so that C is 0, C# is 1, D is 2, and so on up to B as 11, we repeat the first note, C, an octave higher -- so after 11 is 0. He points out that the same think happens on the clock, where 12 o'clock is also equal to 0 (to be followed by 1 o'clock). Here is a link to a Square One Video that discusses this clock arithmetic:
Both musical arithmetic and clock arithmetic are one and the same -- and Livio refers to both of these as addition modulo 12. This is an example of what Livio calls a cyclic group -- and for good reason, since we can see easily how a clock cycles around the hours. Musical notes cycle as well -- I had "discovered" of the G major scale at age 7 simply by playing by ear, but in reality, the G major scale is formed by adding 1 or 2 notes (semitones and tones) in a specific pattern modulo 12, beginning with the note G or 7. Groups theorists often symbolize the additive group modulo 12 as Z/12Z, where Z stands for the usual additive group of integers.
Livio also refers to atonal music, or music based on a twelve-tone row. I've already mentioned the mathematician/musician Vi Hart, who complained when someone tried to copyright the number pi (converted to music). Well, she has produced a 30-minute video based on the twelve-tone row that Livio mentions in this chapter:
(Note: the conversion of pi to music is not generally based on a twelve-tone row. But there was a video that I had posted earlier that used the digits of pi in base 12 so that all twelve tones are used.)
The octave is divided into twelve tones because so many powers of the twelfth root of two are equal to Pythagorean ratios. As Livio points out, 2^(7/12) is 1.498 (close to 1.5, or 3:2) and 2^(3/12) is 1.189 (close to 1.2, or 6:5). A question that sometimes comes up is, is there any other division of the octave other than 12 that produces ratios that are near the simple ratios of Pythagoras? In other words, can music be based on a group other than Z/12Z?
Some musicians believe that the next natural group to consider is Z/19Z. Just as some people have an esoteric interest in calendar reform, others have an esoteric interest in musical reform. According to the 20th century musician Joseph Yasser, 19 is the next natural step after 12:
http://www.bikexprt.com/tunings/fibonaci.htm
"A 19-tone keyboard based on the traditional keyboard would have pairs of black keys (Db and C#, Eb and D#, etc.), and an additional black key between the semitone intervals E-F and B-C."
Another musician, Jonathan Glasier, has produced three YouTube videos on 19-tone music. I don't want to link to all three of these, but a search on YouTube for "Jonathan Glasier 19" suffices. We see that 2^(11/19) = 1.494, which isn't as close to 3:2 as 2^(7/12) is. So, as Glasier points out, we no longer have a perfect fifth, but just a "fifth." But 2^(5/19) = 1.2001, which is much closer to 6:5 than 2^(3/12) is, so we now have what Glasier calls a "perfect minor third." Its inverse ("inversion," according to Livio), the major sixth, also becomes a "perfect major sixth."
Some of the new notes correspond to new intervals. The next ratio to consider after 6:5 is 7:6. There is no note in 12-tone that gives the ratio 7:6, but in 19-tone, we have that 2^(4/19) = 1.157 which, while still a bit off from 7:6 = 1.167, is still better than anything in 12-tone. Ratios based on seven in music are often called "septimal" and sometimes sound good in blues music. So we call the new ratio a "septimal minor third" or "blue third" and is denotes by the interval from C to D#, where D#, as is written above, is not the same note as Eb.
Here's a link to the flutist Anne La Berge, who plays a song based on 19 tones. Let's find out whether this sounds more "bluesy" when one uses 19 tones instead of 12:
With some teenage girls in our classes, we can start talking about attractiveness and symmetry and they will keep discussing it forever. Well, I can keep talking about mathematics and music forever -- and I can't let music take over this post when I need to get back to the PARCC. Question 17 of the PARCC Practice Exam is on trigonometry:
An archaeological team is excavating artifacts from a sunken merchant vessel on the ocean floor. To assist the team, a robotic probe is used remotely. The probe travels approximately 3,900 meters at an angle of depression of 67.4 degrees from the team's ship on the ocean surface down to the sunken vessel on the ocean floor. The figure shows a representation of the team's ship and the probe.
How many meters below the surface of the ocean will the probe be when it reaches the ocean floor?
In this question, the figure helps us immensely. We can form a right triangle where the angle of depression is 67.4 degrees, and the hypotenuse is 3,900 meters, and we wish to find the side opposite the given angle. This is a job for the sine function:
sin(67.4) = x/3900
x = 3900 sin(67.4)
x = 3600.5
Rounded off to the nearest hundred meters, the answer is 3,600 meters. Notice that the answer is already fairly close to a multiple of 100 meters. Indeed, the test writers probably had a 5-12-13 right triangle in mind when coming up with this problem.
The trig ratios appear in Sections 13-3 and 13-4 of the U of Chicago text. I've stated earlier that most angle of elevation problems require the tangent since typically either a horizontal or vertical distance is given and we wish to find the other. But here, 3,900 is clearly the hypotenuse, and so we must use the sine, not the tangent. Common errors include using the wrong trig function as well as having the calculator in radian mode rather than degree mode.
Livio wraps up the chapter by returning to group theory. He describes that some groups are simple -- that is, they can't be broken up into smaller subgroups. All cyclic groups containing a prime number of elements, such as the musical Z/19Z, are simple. But there exists groups that contain a composite number of elements yet aren't simple. One of these contains 60 elements and is the group that Galois discovered as blocking the quintic from being solved. But it wasn't until 1980 that one of the largest simple groups was discovered. It contains a mind-boggling 8 * 10^53 elements -- so many that it has been dubbed the "monster." Yet the "monster," as large as it is, is just simple at heart.
PARCC Practice EOY Exam Question 17
U of Chicago Correspondence: Section 14-4, The Sine and Cosine Ratios
Key Theorem: Definition of sine and cosine
In right triangle ABC with right angle C,
the sine of Angle A, written sin A, is leg opposite Angle A / hypotenuse
the cosine of Angle A, written cos A, is leg adjacent to A / hypotenuse
Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.
Commentary: Most angle of elevation and depression problems require the use of the tangent, but interestingly enough, Question 14 in Section 14-4 -- the only appearance of the phrase "angle of depression" in the U of Chicago text -- requires the use of sine instead. If we change Question 14 so that it asks for the height of your eyes above the ground and not the distance between you and the object, the question becomes analogous to the PARCC problem.
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https://bigpel66.oopy.io/library/ps/boj-platinum/2842
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# 집배원 한상덕
Created
2021/03/23 01:58
문제 번호
2842
카테고리
그래프
이분탐색
DFS
BFS
투포인터
### Code
@12/9/2020
2152 KB
#### 시간
276 ms
#include <iostream> #include <vector> #include <string> #include <algorithm> int g_row_size; int g_matrix_size; int g_search_count; int g_result; std::string g_map; std::vector<int> g_alt; std::vector<int> g_unique_alt; std::pair<int, int> g_post_office; std::pair<int, std::vector<std::pair<int, int> > > g_houses; std::vector<std::pair<int, int> > g_displacement; std::vector<bool> g_visited; void pre_setting(void) { std::ios_base::sync_with_stdio(false); std::cin.tie(NULL); std::cout.tie(NULL); } void input_action(void) { int index; std::cin >> g_row_size; g_matrix_size = g_row_size * g_row_size; g_map.resize(g_matrix_size); g_alt.resize(g_matrix_size); g_unique_alt.resize(g_matrix_size); g_displacement = { {0, 1}, {0, -1}, {1, 0}, {-1 ,0}, {1, 1}, {1, -1}, {-1, -1}, {-1, 1} }; g_result = 1e6; g_houses.first = 0; index = -1; while (++index < g_matrix_size) { std::cin >> g_map[index]; if (g_map[index] == 'P') g_post_office = std::make_pair(index / g_row_size, index % g_row_size); else if (g_map[index] == 'K') { ++g_houses.first; g_houses.second.push_back(std::make_pair(index / g_row_size, index % g_row_size)); } } index = -1; while (++index < g_matrix_size) { std::cin >> g_alt[index]; g_unique_alt[index] = g_alt[index]; } std::sort(g_unique_alt.begin(), g_unique_alt.end()); g_unique_alt.erase(std::unique(g_unique_alt.begin(), g_unique_alt.end()), g_unique_alt.end()); } void depth_search(int x, int y, int min_alt, int max_alt) { int index; index = -1; if (x < 0 || y < 0 || x >= g_row_size || y >= g_row_size || g_visited[x * g_row_size + y] || min_alt > g_alt[x * g_row_size + y] || max_alt < g_alt[x * g_row_size + y]) return ; g_visited[x * g_row_size + y] = true; if (g_map[x * g_row_size + y] == 'K') ++g_search_count; while (++index < 8) depth_search(g_displacement[index].first + x, g_displacement[index].second + y, min_alt, max_alt); } void two_pointer_search(void) { int search_size; int s_index; int e_index; search_size = static_cast<int>(g_unique_alt.size()); s_index = 0; e_index = 0; while (e_index < search_size) { while (s_index < search_size) { g_search_count = 0; g_visited = std::vector<bool>(g_matrix_size, false); depth_search(g_post_office.first, g_post_office.second, g_unique_alt[s_index], g_unique_alt[e_index]); if (g_houses.first != g_search_count) break ; if (g_result > g_unique_alt[e_index] - g_unique_alt[s_index]) g_result = g_unique_alt[e_index] - g_unique_alt[s_index]; ++s_index; } ++e_index; } } void output_action(void) { std::cout << g_result; } void solution(void) { input_action(); two_pointer_search(); output_action(); } int main(void) { pre_setting(); solution(); return (0); }
C++
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https://math.stackexchange.com/questions/3816217/how-do-i-prove-negp-land-q-to-neg-p-lor-neg-p-lor-q-iff-neg-p-lor-q-w
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# How do I prove $\neg(P\land Q)\to(\neg P\lor(\neg P\lor Q))\iff(\neg P\lor Q)$ without using a truth table?
Without using truth tables prove that $$\neg(P\land Q)\to(\neg P\lor(\neg P\lor Q))\iff(\neg P\lor Q)$$
This is a question I've encountered during my examination.
So basically what I did was, I took the RHS of the equation which is $$\neg(P\land Q)\to(\neg P\lor(\neg P\lor Q))$$ and hoped I could equate it to $$(\neg P\lor Q)$$ even though I was sure it was not enough to prove this bi - conditional statement.
Following on my assumption, this is what I've reached
$$\begin{array}{rl} & & \neg(P\land Q)\to(\neg P\lor(\neg P\lor Q)) \iff\\ & \iff & (P\land Q)\lor(\neg P\lor Q) \iff\\ & \iff & (\neg P\lor Q\lor P) \land(\neg P\lor Q\lor Q) \iff\\ & \iff & Q \land(\neg P\lor Q) & \end{array}$$
I'm kinda stuck on what to do after this, I'm not even sure whether this approach is even right. So some help is appreciated on how to prove this bi - conditional statement.
• Note that $(\neg P \lor Q \lor P)$ is a tautology. Sep 6, 2020 at 13:37
• This also depends on what rules you are allowed to use. Sep 6, 2020 at 13:37
• Well, yes indeed it's a tautology and I'd get the LHS, but how can I prove that it's a bi-conditional statement? Sep 6, 2020 at 13:42
• The rules you employed, e.g. material implication, distribution etc. goes both ways. Sep 6, 2020 at 13:45
• So this is what I just have to do? Sep 6, 2020 at 13:45
$$\begin{array}{rl} & & \neg(P\land Q)\to(\neg P\lor(\neg P\lor Q)) \iff\\ & \iff & (P\land Q)\lor(\neg P\lor Q) \iff & \quad & \text{Material Implication}\\ & \iff & (\neg P\lor Q\lor P) \land(\neg P\lor Q\lor Q) & \quad & \text{Distributive Law}\\ \end{array}$$
As you can verify, $$\neg P \vee Q \vee P$$ is a tautology.
$$\neg P \vee Q \vee P \iff \neg P \vee P \vee Q \iff \top \vee Q \iff \top$$
Hence, we have the following.
$$\begin{array}{rl} & & \top \wedge (\neg P \vee Q \vee Q) \iff\\ & \iff & \neg P \vee Q \vee Q \iff\\ & \iff & \neg P \vee Q & \square\\ \end{array}$$
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CC-MAIN-2023-50
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latest
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https://math.stackexchange.com/questions/2242615/metrizability-of-infinite-product-space-under-box-topology
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# Metrizability of infinite product space under box topology
Let $X=[0,1]^\omega$. We know that under the standard product topology, $X$ is metrizable. But is it metrizable under the box topology? My guess is not, and the intuition is that Urysohn's metrization theorem fails because the basis for $X$ under the box topology is uncountable. But that's not sufficient to disprove metrizability.
How can one show that this is indeed the case? (or prove otherwise)
$X$ with the box topology is not even first-countable, which immediately implies it's not metrizable. To see this, suppose we had a countable neighborhood basis $U_1, U_2, \ldots$ at $(0, 0, \ldots)$. Then for each $n$, suppose $[0, \epsilon_{n,1}) \times [0, \epsilon_{n,2}) \times \cdots \subseteq U_n$. Then $V := [0, \frac{1}{2} \epsilon_{1,1}) \times [0, \frac{1}{2} \epsilon_{2,2}) \times \cdots$ is an open neighborhood of $(0, 0, \ldots)$ in the box topology. However, for each $n$, $(0, 0, \ldots, \frac{3}{4} \epsilon_{n,n}, 0, \ldots)$ (with the nonzero entry at position $n$) is in $U_n$ but not in $V$; so no $U_n$ is contained in $V$.
• Well, just in case, here's the one-line proof of that fact: if $X$ is a metric space and $x \in X$ then the collection of open balls $B_{1/n}(x)$ is a neighborhood basis at $x$. – Daniel Schepler Apr 20 '17 at 16:54
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CC-MAIN-2019-39
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latest
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en
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https://www.codeflu.com/post/validate-binary-search-tree
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crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00304.warc.gz
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• Time:O(n)
• Space:O(h)
## C++
``````class Solution {
public:
bool isValidBST(TreeNode* root) {
return isValidBST(root, nullptr, nullptr);
}
private:
bool isValidBST(TreeNode* root, TreeNode* minNode, TreeNode* maxNode) {
if (!root)
return true;
if (minNode && root->val <= minNode->val)
return false;
if (maxNode && root->val >= maxNode->val)
return false;
return isValidBST(root->left, minNode, root) &&
isValidBST(root->right, root, maxNode);
}
};
``````
## JAVA
``````class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, null, null);
}
private boolean isValidBST(TreeNode root, TreeNode minNode, TreeNode maxNode) {
if (root == null)
return true;
if (minNode != null && root.val <= minNode.val)
return false;
if (maxNode != null && root.val >= maxNode.val)
return false;
return isValidBST(root.left, minNode, root) &&
isValidBST(root.right, root, maxNode);
}
}
``````
## Python
``````class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
def isValidBST(root: Optional[TreeNode],
minNode: Optional[TreeNode], maxNode: Optional[TreeNode]) -> bool:
if not root:
return True
if minNode and root.val <= minNode.val:
return False
if maxNode and root.val >= maxNode.val:
return False
return isValidBST(root.left, minNode, root) and \
isValidBST(root.right, root, maxNode)
return isValidBST(root, None, None)
``````
• Time:O(n)
• Space:O(h)
## C++
``````class Solution {
public:
bool isValidBST(TreeNode* root) {
stack<TreeNode*> stack;
TreeNode* pred = nullptr;
while (root || !stack.empty()) {
while (root) {
stack.push(root);
root = root->left;
}
root = stack.top(), stack.pop();
if (pred && pred->val >= root->val)
return false;
pred = root;
root = root->right;
}
return true;
}
};
``````
## JAVA
``````class Solution {
public boolean isValidBST(TreeNode root) {
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode pred = null;
while (root != null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if (pred != null && pred.val >= root.val)
return false;
pred = root;
root = root.right;
}
return true;
}
}
``````
## Python
``````class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
stack = []
pred = None
while root or stack:
while root:
stack.append(root)
root = root.left
root = stack.pop()
if pred and pred.val >= root.val:
return False
pred = root
root = root.right
return True
``````
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https://doubtnut.com/question-answer/in-how-many-of-the-distinct-permutations-of-the-letters-in-mississippi-do-the-four-is-not-come-toget-484
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Class 11
MATHS
Permutations And Combinations
# In how many of the distinct permutations of the letters in MISSISSIPPI do the four Is not come together?
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 6-10-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
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as we know that <br> M= 1 <br> I = 4 <br> S= 4 <br> P=2 <br> total = 11!/(4!*4!*2!) = (11*10*9*8*7*6*5)/(4*3*2*2)= 34650 <br> when Is are together = 8!/(4!*2!) <br> =(8*7*6*5)/2= 28*30= 840 <br> required = 34650-840 <br> = 33810ways <br> answer
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CC-MAIN-2021-17
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latest
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orion.cs.purdue.edu
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## Correlations and missing data
The simple pdf distributions discussed above can be used to represent 1-D pdfs. But in many cases, there exist intra-tuple correlations between the attributes. Table 2 is an example where the uncertain attributes and in the same tuple are correlated. These more complex distributions are supported in Orion 2.0 using joint probability distributions across attributes.
The information about intra-tuple dependencies is captured by the schema dependency information, which is a partition of all the uncertain attributes present in the table . It consists of multiple sets of attributes that are correlated within a tuple, i.e. dependency sets. It also contains singleton sets containing attributes that are uncertain but are not dependent on any other attributes. The attributes not listed in the dependency sets are assumed to be certain.
To illustrate, let us consider a table with schema , where represents the data type of attribute . If all the attributes in the table are certain, then the dependency sets are empty. On the other hand, if and are uncertain and are correlated, this information is captured by the dependency sets: .
In our first sensor data example (Table 1), attribute Sensor ID is a certain attribute while Location is uncertain. Its dependency information is . Note that there is only one dependency set, since there is only one uncertain attribute. However, in the case when there are multiple uncertain attributes, there may be still one dependency set. For example, in Table 2 where the location is 2-D, the only dependency set is because are jointly distributed, hence dependent on each other.
Consider the special case when all the uncertain attributes in a table are jointly distributed (i.e. there is one dependency set). This extreme case captures tuple uncertainty as the complete value of the tuple is uncertain [18], as is illustrated in Table 2. The probability of the presence of a tuple (we call it tuple probability) is then the total probability of the joint distribution. When there are multiple dependency sets within a tuple, we can compute the joint pdf of the tuple by multiplying the individual pdfs of the dependency sets (the justification is that different dependency sets are independent from each other). If the probabilities over the joint pdf sum up (integrate) to , then is the probability that the tuple does not exist in the database. When , we call such pdfs partial pdfs. An important feature of Orion is its support for partial pdfs to ensure that database operations such as selection are consistent with PWS [18].
Rohit Jain 2011-08-02
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https://short-facts.com/is-weight-measured-in-newtons-in-physics/
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# Is weight measured in Newtons in physics?
## Is weight measured in Newtons in physics?
In physics, the term weight has a specific meaning – which is the force that acts on a mass due to gravity. Weight is measured in newtons. The mass of a given object is the same everywhere, but its weight can change. We use balances to measure weights and masses.
### What is something that weighs 1 Newton?
One Newton on the surface of the Earth is equal to 101.972 grams, 0.224809 lb, or 3.59694 oz. Objects that weigh one Newton on the Earth’s surface include a quarter-pound burger, a stick of margarine, and coincidentally a medium size apple given the alleged story of how Newton discovered gravity.
#### What is weight measured in physics?
Weight is a measure ofthe force of gravity on a physical object and is measured in newtons. The weight of a bird of mass 15 g varies with the magnitude ofthe gravitational force acting on it and would be considerably different if measured on the Moon, for example, instead of on Earth.
Is mass weight in physics?
The terms “mass” and “weight” are used interchangeably in ordinary conversation, but the two words don’t mean the same thing. The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass.
What is formula of mass?
Mass is defined as the amount of matter present in a body. The SI unit of mass is the kilogram (kg). The formula of mass can be written as: Mass = Density × Volume.
## How do you convert kg to Newtons?
The relation between 1kg-wt and Newton is given by 1kg-wt =9.81N.
### What is the force of 1 Newton equivalent to?
100,000 dynes
One newton is equal to a force of 100,000 dynes in the centimetre-gram-second (CGS) system, or a force of about 0.2248 pound in the foot-pound-second (English, or customary) system.
#### What is work done formula?
Mathematically, the concept of work done W equals the force f times the distance (d), that is W = f. d and if the force is exerted at an angle θ to the displacement, then work done is calculated as W = f . d cos θ.
How do you find weight in physics?
Summary
1. Weight is a measure of the force of gravity pulling down on an object. It depends on the object’s mass and the acceleration due to gravity, which is 9.8 m/s2 on Earth.
2. The formula for calculating weight is F = m × 9.8 m/s2, where F is the object’s weight in Newtons (N) and m is the object’s mass in kilograms.
What is the weight of 1 kg of mass?
On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.
## What is the SI unit of mass?
kilogram (kg)
The SI unit of mass is the kilogram (kg).
### When was a Newton adopted as a measurement of weight?
Besides being a unit of force it was also adopted as a measurement of weight by the General Conference on Weights and Measures (CGPM) in 1960. A Newton is the force acting on a 1 kg object that would cause an acceleration of 1 m/s 2.
#### What are some examples of things that weigh one newton?
There are many things that weigh one Newton in fast food restaurants. Some examples are an order of chicken fingers from Arby’s or an onion bagel from Bruegger’s. If you eat scrambled eggs at McDonald’s then you are eating one Newton worth of food. There are countless things in this world that are considered to weigh one Newton.
What’s the difference between mass and weight in physics?
In physics, however, there is an important distinction. Weight is the pull of Earth on an object. It depends on the distance from the center of Earth. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon.
What is the force needed to accelerate one newton?
Newtons are a derived unit, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.
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https://forums.reiclub.com/t/tax-foreclosure-short-sale-question/85289
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# tax foreclosure short sale question
Hey all wanted to run something buy you guys. I have a client I am working with who owes like a \$1700 on a tax lien. He also owes about 45K on his mortgage. The property is scheduled to go to sheriffs sale in December. Do you think the bank will pay his tax payment so they wont lose the balance on the note? After they do that will they be more willing to negotitate a short sale payoff. His interest rate is about to go up 13%. Thanks for your opinions!
The bank probably will not consider a short sale unless you can prove that the home owner is having severe financial hardship. The other option is to have him start missing payments.
These days we have been able to get the banks to consider short sales before the home owner is an arrears if we are able to show he can no longer make the payments.
He is already 60 days behind as of November 1st. I have a bunch of medical bills to help prove the financial hardship.
I would say it is highly likely that the bank would pay of the tax lien.
How much is the house worth? Let’s say the house is worth \$50K, if the house were sold via tax foreclosure, the bank would lose \$50K.
If the bank pays the taxes, they would usually work out some type of escrow shortage spread, where the taxes would be paid off over a period of time as an addition to the mortgage payment.
Sounds like it’s in the best interest of the bank to pay the tax lien, then negotiate a short sale.
If he’s behind on his mortgage, negotiate a short sale and add in the tax lien to your purchase cost. It’s possible you could get the tax lien discounted, too. You’ll have to ask if this is customary in your area.
You said the mortgage balance is \$45,000, and the tax lien is \$1700.(in real life, the tax lien could be \$1700 but the back taxes not paid could be considerably more than that.) But let’s take the simple example \$45k and \$1.7k. The bank has to offer \$46,700 because any more than that, and they will have to give the excess to the owner. Any less than that will be a loss on the mortgage. Remember, they hold the \$45k mortgage. One second after the foreclosure auction starts, they will bid \$46,700. They will want to go through the foreclosure process to make sure that all junior liens are wiped out.
Now after foreclosure, you could ask for a short sale. The bank could be getting presure from the bank examiners to clear their books.
The bank is not going to pay the back taxes then let the house go via a short sale. There is no benefit to them in this. They will either foreclose then take care of the back taxes to have a clear title for themselves or they will accept a short sale and you will be left to deal with the back taxes and any other liens.
it could go either way. They could pay the taxes or they could not. Ask the person at the bank who is handling the case. If it is not in default yet, then you will probably end up speaking to someone in their legal department. As long as you have a letter of authorization to discuss the account you should be able to find out what their plans are.
Then again, they could be stubborn and refuse to tell you what they intend to do, in which case you may want to ask what their options are then take an educated guess as to what you think they will most likely do.
I have done 2 SS this past month that I included on the HUD1 delinquent taxes AND utilities and the bank paid all of them off without even asking a quesiton besides who the checks go to. We just over estimated what the actual payoffs were for the delinquent bills. Why not try to do this for the homeowner that you are helping, they will love you for this.
JD
The bank will pay the taxes or lose their 1st lien position. They could wait for the tax auction and pay the taxes from their bid, or do a short sale and pay the taxes from the proceeds of the short sale… but one way or another, they will pay the taxes if the owner does not.
Tax liens are usually recorded on a property shortly after they are delinquent but (depending on your state) rarely enforced through foreclosure unless it’s been an issue for years. Some states will file a lien within a few months but give you plenty of time to catch up. You really need to check the laws in your state, find out if is it a state or federal lien, why is the lien on this property, etc…ask these questions (which no one yet in this forum has addressed). Is it going to sale for the tax lien or a mortgage default? Do you truly have a bargain? Ask yourself why the lender isn’t paying the back taxes? Do they even know there is a tax lien? Is this an IRS lien? If so you have much more work to do. Also, most lenders will require you to keep property taxes current or they can foreclose even if you’re current on the mortgage. I have seen this on every mortgage note I’ve ever reviewed in California. It seems to me there isn’t enough information about this scenario to make an educated decision, yet options are being handed out here.
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https://intelligencemission.com/free-electricity-from-gravity-free-electricity-discovery.html
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##### In most cases of interest there are internal degrees of freedom and processes, such as chemical reactions and phase transitions, which create entropy. Even for homogeneous “bulk” materials, the free energy functions depend on the (often suppressed) composition, as do all proper thermodynamic potentials (extensive functions), including the internal energy.
Research in the real sense is unheard of to these folks. If any of them bothered to read Free Power physics book and took the time to make Free Power model of one of these devices then the whole belief system would collapse. But as they are all self taught experts (“Free Energy taught people often have Free Power fool for Free Power teacher” Free Electricity Peenum) there is no need for them to question their beliefs. I had Free Power long laugh at that one. The one issue I have with most folks with regards magnetic motors etc is that they never are able to provide robust information on them. Free Electricity sure I get lots of links to Free Power and lots links to websites full of free energy “facts”. But do I get anything useful? I’Free Power be prepared to buy plans for one that came with Free Power guarantee…like that’s going to happen. Has anyone who proclaimed magnetic motors work actually got one? I don’t believe so. Where, I ask, is the evidence? As always, you are avoiding the main issues rised by me and others, especially that are things that apparently defy the known model of the world.
Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.
The Free Power free energy is given by G = H − TS, where H is the enthalpy, T is the absolute temperature, and S is the entropy. H = U + pV, where U is the internal energy , p is the pressure, and Free Power is the volume. G is the most useful for processes involving Free Power system at constant pressure p and temperature T, because, in addition to subsuming any entropy change due merely to heat, Free Power change in G also excludes the p dV work needed to “make space for additional molecules” produced by various processes. Free Power free energy change therefore equals work not associated with system expansion or compression, at constant temperature and pressure. (Hence its utility to solution-phase chemists, including biochemists.)
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http://compucolor.org/vmedia.html?diskname=vmedia%2Ffinancial_calc-sector.ccvf&filename=REMBAL.BAS%3B01
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# Compucolor.org – Virtual Media
Listing of file='REMBAL.BAS;01' on disk='vmedia/financial_calc-sector.ccvf'
```10 REM PROGRAM TO COMPUT THE REMAINING BALANCE ON A LOAN
20 PLOT 27,11,14,12,6,1
30 PRINT TAB( 20)"REMAINING BALANCE ON A LOAN"
40 PLOT 15,6,2
50 PRINT "THIS PROGRAM COMPUTES THE BALANCE REMAINING ON A LOAN"
55 PRINT "AFTER A SPECIFIED NUMBER OF PAYMENTS HAVE BEEN MADE."
60 PRINT "IT IS NECCESSARY FOR YOU TO PROVIDE THE AMOUNT OF THE"
70 PRINT "REGULAR PAYMENT, THE NUMBER OF PAYMENTS PER YEAR, THE AMOUNT"
75 PRINT "OF THE PRINCIPAL,THE ANNUAL INTEREST RATE, AND THE PAYMENT"
80 PRINT "NUMBER AFTER WHICH TO CALCULATE THE REMAINING BALANCE..."
90 INPUT "HIT -RETURN- TO BEGIN...";G
100 IF G\$= "GO"THEN 130
130 PLOT 14,10,6,6
140 INPUT "REGULAR PAYMENT = ";R
150 INPUT "PRINCIPAL = ";P
160 INPUT "# OF PAYMENTS PER YEAR = ";N
170 INPUT "ANNUAL INTEREST RATE = ";I
180 I= I/ 100
190 INPUT "# OF PAYMENTS MADE = ";N1
200 B0= P
210 FOR J1= 1TO N1
220 I1= INT ((B0* I/ N)* 100+ .5)/ 100
230 A= R- I1
240 B0= B0- A
250 NEXT J1
260 PLOT 6,3,10
270 PRINT TAB( 20)"REMAINING BALANCE = \$";INT (B0* 100+ .5)/ 100
280 PLOT 15,6,2,10
290 INPUT "HAVE YOU A DIFFERENT PLAN IN MIND ?...(1=YES, 0=N0) ";X
300 IF X= 1THEN RUN
310 PLOT 1
```
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CC-MAIN-2024-10
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https://mail.scipy.org/pipermail/numpy-discussion/2009-July/043961.html
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# [Numpy-discussion] mean of two or more arrays while ignoring a specific value
David Warde-Farley dwf@cs.toronto....
Tue Jul 14 14:59:01 CDT 2009
```On 14-Jul-09, at 3:33 PM, Greg Fiske wrote:
> Dear list,
>
>
>
> I'm learning to work with numpy arrays. Can somebody explain how to
> get the
> average of two separate arrays while ignoring a user defined value
> in one
> array?
>
>
>
> For example:
>
>>>> a = numpy.array([1,5,4,99])
>
>>>> b = numpy.array([3,7,2,8])
>
>
>
> Ignoring the value 99, the result should be an array like c=
> ([2,6,3,8])
The simplest way I can think of doing it:
In [60]: c = concatenate((a[newaxis,:], b[newaxis,:]),axis=0)
In [61]: c = where(c == 99, nan, float64(c))
In [62]: nansum(c,axis=0) / (~isnan(c)).sum(axis=0)
Out[62]: array([ 2., 6., 3., 8.])
The last line could be replaced with scipy.stats.nanmean(c, axis=0) if
you have scipy installed.
David
```
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https://www.numberproperties.com/515101/
| 1,713,953,429,000,000,000
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| 4,120
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# Number 515101
## Properties of the number 515101
Prime Factorization 467 x 1103 Divisors 1, 467, 1103, 515101 Count of divisors 4 Sum of divisors 516672 Previous integer 515100 Next integer 515102 Is prime? NO Previous prime 515089 Next prime 515111 515101st prime number ↻ calculating, please wait Is a Fibonacci number? NO Zeckendorf representation 514229 + 610 + 233 + 21 + 8 Is a Pell number? NO Is a regular number? NO Is a perfect number? NO Is a perfect square number? NO Is a perfect cube number? NO Is power of 2? NO Is power of 3? NO Square 5151012 265329040201 Square root √515101 717.70537130497 Cube 5151013 136671253936575301 Cubic root ∛515101 80.161185439234 Natural logarithm 13.152118276922 Decimal logarithm 5.7118923930063
## Trigonometry of the number 515101
515101 modulo 360° 301° Sine of 515101 radians -0.72749777449833 Cosine of 515101 radians 0.68611004080977 Tangent of 515101 radians -1.0603222970469 Sine of 515101 degrees -0.85716730070217 Cosine of 515101 degrees 0.51503807490996 Tangent of 515101 degrees -1.6642794823509 515101 degrees in radiants 8990.2084303153 515101 radiants in degrees 29513113.322968
## Base conversion of the number 515101
Binary 1111101110000011101 Octal 1756035 Duodecimal 20a111 Hexadecimal 7dc1d
## Recommended Books
Looking for good books to read? Take a look at these books if you want to know more about the theory of numbers
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https://leanprover-community.github.io/archive/stream/252551-graph-theory/topic/k-factors.html
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## Stream: graph theory
### Topic: k-factors
#### Yaël Dillies (Nov 02 2021 at 11:06):
@Kyle Miller, how hard do you think it is to prove that the number of k-factors of a graph is the product over its connected components of their number of k-factors?
#### Yaël Dillies (Nov 02 2021 at 11:07):
And is there a higher generality I'm missing?
#### Yaël Dillies (Nov 02 2021 at 11:09):
In particular, should we define k-factors of a graph or of a subgraph? And how do you plan on modelling connected components? For this latter question, I think defining them as a set of subgraphs is the way to go.
#### Kyle Miller (Nov 02 2021 at 23:38):
Connected components are not very developed in the walks_and_trees branch yet. So far it's just the quotient of V by G.reachable, so it's more like $\pi_0(G)$.
The set of connected components as subgraphs I think is
def connected_components : set (subgraph G) :=
{G' | G'.coe.connected ∧ ∀ G'', G' ≤ G'' → ¬ G'.coe.connected}
and it would be good having a theorem that says these correspond to quot G.reachable, with each connected component being the induced subgraph from an equivalence class of vertices. (And that the union of these graphs is G and that they're disjoint.)
There also ought to be facts like every G.walk u v can be written as a walk in an element of connected_components. The code for reinterpreting walks as walks in subgraphs is very underdeveloped at the moment, though there's not nothing.
#### Kyle Miller (Nov 02 2021 at 23:40):
Another aspect of formalization is writing G specifically as a disjoint union of connected subgraphs. In a way, the stuff about quot G.reachable and induced subgraphs is just a way to show it's possible.
#### Kyle Miller (Nov 02 2021 at 23:42):
I don't know very much about k-factors. My ignorant answer is that it seems ok defining them for graphs, then for subgraphs you have k-factors of G'.coe.
#### Yaël Dillies (Nov 02 2021 at 23:43):
So you mean I should rather prove that the number of k-factors of a sum of graphs is their product? Then the connected component version will be a special case.
#### Yaël Dillies (Nov 02 2021 at 23:43):
Do we have the sum of simple_graphs?
#### Kyle Miller (Nov 02 2021 at 23:47):
I mean write a graph as a disjoint union of connected subgraphs, then relate k-factors of subgraphs (as k-factors of subgraphs coerced to simple graphs) to the original graph.
#### Kyle Miller (Nov 02 2021 at 23:50):
Then again, it's probably fine defining k-factors with respect to a subgraph, since k-factors of the top subgraph is a special case.
#### Yaël Dillies (Nov 02 2021 at 23:51):
Ah okay, so what about (warning, doesn't compile as is) simple_graph.pi (s : set ι) (G : Π i, simple_graph (α i)) : simple_graph (Π i, α i) := s.pi G? And then you can show number_k_factors (imple_graph.pi s G) = ∏ i, number_k_factors (G i)?
#### Yaël Dillies (Nov 02 2021 at 23:52):
Then you can transfer this result to connected components using the equivalence between a graph and the sum of its connected compoents.
#### Kyle Miller (Nov 02 2021 at 23:53):
We'll want that equivalence eventually anyway.
#### Kyle Miller (Nov 02 2021 at 23:54):
I'm not sure it'll make the proof easier though. I think it might be less awkward staying in the realm of subgraphs of a given graph, though it'd take setting things up right.
#### Kyle Miller (Nov 02 2021 at 23:55):
(I'm imagining wrangling all the vertex types, versus having just one vertex type.)
#### Yaël Dillies (Nov 02 2021 at 23:57):
Yeah, I totally get that. Which way sounds the more mathlib-like to you?
#### Kyle Miller (Nov 03 2021 at 00:10):
I think the statements in the end will be the same (and will involve creating similar theory), so whichever way to get there that's simpler :smile:
Here's some start for some pseudocode. The idea is that mk_k_factor is actually defining an equivalence between the product of the sets of k-factors of the connected components and the k-factors of G, which would be explicitly constructed next from it, and then you apply the lemmas for fintype.card of an equivalence and extract the result you want.
def connected_components : set (subgraph G) :=
{G' | G'.coe.connected ∧ ∀ G'', G' ≤ G'' → ¬ G'.coe.connected}
def subgraph.k_factors [locally_finite G] (G' : subgraph G) (k : ℕ) : set (subgraph G) :=
{G'' | G'.verts = G''.verts ∧ G''.coe.is_regular_of_degree k}
def mk_k_factor [locally_finite G] (k : ℕ)
(f : Π G' (hG' : G' ∈ G.connected_components), subgraph G)
(hf : ∀ G' hG', f G' hG' ∈ G'.k_factors k) :
{G' : subgraph G | G' ∈ (⊤ : simple_graph V).k_factors k} := sorry
#### Kyle Miller (Nov 03 2021 at 00:16):
(The inverse of mk_k_factor would be from intersecting the k-factor G' with each connected component.)
#### Yaël Dillies (Nov 03 2021 at 00:16):
I like the idea! But I'm not convinced we should rely on connected_components to even define mk_k_factor
#### Kyle Miller (Nov 03 2021 at 00:30):
You could develop a theory of disjoint unions of "locally complete" subgraphs (maybe subgraphs that contain their incidence sets) and define mk_k_factor in terms of choosing a k-factor for each subgraph in the disjoint union.
Then as a special case, the connected components (as defined above) form such a system.
#### Yaël Dillies (Nov 03 2021 at 00:31):
Yeah that's what I was describing above.
Where?
#### Yaël Dillies (Nov 03 2021 at 00:32):
Here
Yaël Dillies said:
So you mean I should rather prove that the number of k-factors of a sum of graphs is their product? Then the connected component version will be a special case.
and there
Yaël Dillies said:
Ah okay, so what about (warning, doesn't compile as is) simple_graph.pi (s : set ι) (G : Π i, simple_graph (α i)) : simple_graph (Π i, α i) := s.pi G? And then you can show number_k_factors (imple_graph.pi s G) = ∏ i, number_k_factors (G i)?
#### Yaël Dillies (Nov 03 2021 at 00:33):
Well, I wasn't talking about the ismorphism, but counting kind of implies having the isomorphism under the rug. Sorry if that was unclear.
#### Kyle Miller (Nov 03 2021 at 00:34):
I for some reason thought you meant for the components of the sum to be connected, so I missed that.
I guess it's the difference between doing in synthetically (like yours) or analytically (like mine). The only way to know which is "better" is to implement it both ways.
#### Kyle Miller (Nov 03 2021 at 00:35):
But we still would want both points of view in the end.
#### Kyle Miller (Nov 03 2021 at 00:35):
It's like the whole thing about direct sums of vector spaces vs internal direct sums of subspaces.
#### Yaël Dillies (Nov 03 2021 at 00:37):
Oh yeah, I guess.
#### Yaël Dillies (Nov 03 2021 at 00:37):
So k-factors over subgraphs but isomorphism over disjoint graphs sounds like the most general setting?
#### Kyle Miller (Nov 03 2021 at 00:39):
What's the application? (Also, I'm having trouble parsing your suggestion.)
#### Yaël Dillies (Nov 03 2021 at 00:40):
Mario's formalization project. In particular, the bit that asks that the graph is connected if there are at least two perfect matchings and each edge is forced.
#### Yaël Dillies (Nov 03 2021 at 00:41):
Yes, what I'm asking for is overpowered.
Last updated: Dec 20 2023 at 11:08 UTC
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[Agda] Modelling term-rewriting systems in Agda
Twan van Laarhoven twanvl at gmail.com
Wed Oct 23 12:08:30 CEST 2013
```I threw it up on github, https://gist.github.com/twanvl/7115932
I have some more proofs, such as
confluent-by-sn : LocalConfluent R → StronglyNormalizing R → Confluent (* R)
but these are scattered over several files, so I didn't include them.
Twan
On 22/10/13 16:22, Jacques Carette wrote:
> Thank you, this was exactly the kind of example I was looking for. Is your
> complete example available somewhere? Or would you mind sharing it?
>
> Jacques
>
> On 2013-10-22 11:13 AM, Twan van Laarhoven wrote:
>>
>> As a somewhat related data point, a while ago I proved confluence of beta
>> reduction for untyped lambda calculus (as I am sure many others have as well).
>> To encode the rewrite rules I used a data type to represent the reduction
>> relation. Here is single step beta reduction
>>
>> data _-B->_ {V} : Term V → Term V → Set where
>> beta : ∀ {x y} → (ap (lam x) y) -B-> (subst y x)
>> under-ap₁ : ∀ {x y x'} → (x -B-> x') → (ap x y -B-> ap x' y)
>> under-ap₂ : ∀ {x y y'} → (y -B-> y') → (ap x y -B-> ap x y')
>> under-lam : ∀ {x x'} → (x -B-> x') → (lam x -B-> lam x')
>>
>> And here is the parallel beta reduction:
>>
>> data _-PB->_ {V} : Term V → Term V → Set where
>> var : ∀ {v} → var v -PB-> var v -- no reduction in leaf
>> beta : ∀ {x y x' y'} → (x -PB-> x') → (y -PB-> y')
>> → (ap (lam x) y) -PB-> (subst y x)
>> under-ap : ∀ {x y x' y'} → (x -PB-> x') → (y -PB-> y')
>> → (ap x y -PB-> ap x' y')
>> under-lam : ∀ {x x'} → (x -PB-> x') → (lam x -PB-> lam x')
>>
>> In the generic case you might try to split the `under` constructors into
>> another data type.
>>
>> For confluence it turned out to be useful to define a datatype:
>>
>> data CommonReduct {a r s} {A : Set a} (R : Rel A r) (S : Rel A s)
>> (x y : A) : Set (a ⊔ r ⊔ s) where
>> _||_ : {z : A} → R x z → S y z → CommonReduct R S x y
>>
>> then confluence can be expressed as
>>
>> Confluent : ∀ {a r A} → (R : Rel A r) → Set (a ⊔ r)
>> Confluent R = ∀ {x y z} → R x y → R x z → CommonReduct R R y z
>>
>> And using Data.Star you can prove lemma's like strong normalization & local
>> confluence imply global confluence.
>>
>>
>> Twan
>>
>> _______________________________________________
>> Agda mailing list
>> Agda at lists.chalmers.se
>> https://lists.chalmers.se/mailman/listinfo/agda
>
```
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# Equation Radius Of An Arc
By | April 18, 2017
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Radius and central angle calculator arc length formula to finding the given gcse maths steps 4 2 mathematics libretexts of diffe bend angles in radian measure a circle you from for an arch
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# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## AP Board Class 7 Maths Chapter 3 Simple Equations Ex 3.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 3 Simple Equations Ex 3.2 Book Answers
AP Board Class 7 Maths Chapter 3 Simple Equations Ex 3.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 3 Simple Equations Ex 3.2 Book Answers
## Andhra Pradesh State Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Books Solutions
Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 7th Subject Maths Chapters Maths Chapter 3 Simple Equations Ex 3.2 Provider Hsslive
2. Click on the Andhra Pradesh Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Answers.
3. Look for your Andhra Pradesh Board STD 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbooks PDF.
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## AP Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
Question 1.
Give the steps you will use to separate the variables and then solve the equation.
(i) 5𝑚3 = 10
Given 5𝑚3 = 10
⇒ 5𝑚3 × 3 = 10 × 3 (Multiply with 3 on both sides)
⇒ 5m = 30
⇒ 5𝑚5 = 305 (Divide both sides by 5)
⇒ m = 6
Check:
Substitute m = 6 in the given equation,
LHS = 5𝑚3 = 5(6)3 = 303 = 10 = RHS.
Hence verified.
(ii) 4M – 23 = 13
Given 4n – 23 = 13
⇒ 4n = 36
⇒ 4𝑛4 = 364 (Divide both sides by 4)
⇒ n = 9
Check:
Substitute n = 9 in the given equation.
LHS = 4n – 23
= 4(9) – 23
= 36 – 23 = 13 = RHS
Hence verified.
(iii) – 5 + 3x = 16
Given – 5 + 3x = 16
⇒ – 5 + 3x + 5 = 16 + 5 (Add 5 on both sides)
⇒ 3x = 21
⇒ 3𝑥3 = 213 (Divide by 3 on both sides)
⇒ x = 7
Check:
Substitute x = 7 in the given equation
LHS = – 5 + 3x
= – 5 + 3(7)
= – 5 + 21 = 16 = RHS
Hence verified.
(iv) 2(y – 1) =8
Given 2(y – 1) = 8
⇒ 2y – 2 = 8 (Distributive property)
⇒ 2y – 2 + 2 = 8 + 2 (Add 2 on both sides)
⇒ 2y = 10
⇒ 2𝑦2 = 102 (Divide by 2 on both sides)
⇒ y = 5
Check: Substitute y = 5 in the given equation
LHS = 2(y – 1)
= 2(5 – 1)
= 2 × 4 = 8 = RHS
Hence verified.
Question 2.
Solve the following simple equations and check the results.
(i) 3x = 18
Given 3x = 18
⇒ 3𝑥3 = 183 (Divide by 3 on both sides)
⇒ x = 6
Check: Substitute x = 6 in 3x = 18
LHS ⇒ 3x = 3(6) = 18 = RHS
Hence verified.
(ii) 𝑏7 = – 2
Sol.
Given 𝑏7 = – 2
⇒ 𝑏7 × 7 = – 2 × 7 (Multiply by 7 on both sides)
⇒ b = – 14
Check: Substitute b = – 14 in
𝑏7 = – 2
LHS = 𝑏7 = 147 = – 2 = RHS
Hence verified
(iii) – 2x = – 10
Given – 2x = – 10
⇒ 2𝑥2 = 102 (Divide by – 2 on both sides)
⇒ x = 5
Check:
Substitute x = 5 in
– 2x = – 10
LHS = – 2x
= – 2(5) = – 10 = RHS
Hence verified.
(iv) 10 + 6a = 40
Given 10 + 6a = 40
⇒ 10 + 6a – 10 = 40 – 10 (Subtract 10 on both sides)
⇒ 6a = 30
⇒ 6𝑎6 = 306 (Divide by 6 on both sides)
⇒ a = 5
Check: Substitute a = 5 in 10 + 6a = 40
LHS = 10 + 6a
= 10 + 6(5)
= 10 + 30
= 40 = RHS
Hence verified.
(v) – 7m = 21
Given – 7m = 21
⇒ 7𝑚7 = 217 (Divide by – 7 on both sides)
⇒ m = – 3
Check:
Substitute m = in – 7m = 21
LHS = – 7m
= – 7 (- 3)
= 21 = RHS
Hence verified.
(iv) 4p + 7 = – 21
Given 4p + 7 = – 21
⇒ 4p + 7 – 7 = – 21 – 7 (Subtract 7 on both sides)
⇒ 4p = – 28
⇒ 4𝑝4 = 284 (Divide by 4 on both sides)
⇒ p = – 7
Check:
Substitute p = – 7 in 4p + 7 = – 21
LHS = 4p + 7
= 4(- 7) + 7
= – 28 + 7
= – 21 = RHS
Hence verified.
(vii) 3x – 13 = 5
Check: Substitute x = 169 in 3x – 13 = 5
LHS = 3x – 13
= 3(169) – 13
16313
1613 = 153
= 5 = RHS
Hence verified.
(viii) 18 – 7n = – 3
Given 18 – 7n = – 3
⇒ 18 – 7n – 18 = – 3 – 18 (Subtract 18 on both sides)
⇒ – 7n = – 21
⇒ 7𝑛7𝑛 = 217 (Divide by – 7 on both sides)
⇒ n = 3
Check:
Substitute n = 3 in 18 – 7n = – 3
LHS = 18 – 7n
= 18 – 7(3)
= 18 – 21
= – 3 = RHS
Hence verified.
(ix) 3(k + 4) = 21
Sol.
Given 3(k + 4) = 21
⇒ 3k + 12 = 21 (Distributive property)
⇒ 3k + 12 – 12 = 21 – 12 (Subtract 12 on both sides)
⇒ 3k = 9
⇒ 3k3 = 93 (Divide by 3 on both sides)
⇒ k = 3
Check:
Substitute k = 3 in 3(k + 4) = 21
LHS = 3(k + 4)
= 3(3 + 4)
= 3 × 7 = 21= RHS
Hence verified.
(x) 9 (a + 1) + 2 = 11
Given 9(a + 1) + 2 = 11
⇒ 9(a + 1) + 2 – 2 = 11 – 2 (Subtract 2 on both sides)
⇒ 9(a + 1) = 9
⇒ 9(𝑎+1)9 = 99 (Divide by 9 on both sides)
⇒ a + 1 = 1
⇒ a + 1 – 1 = 1 – 1 (Subtract 1 on both sides)
⇒ a = 0
Check:
Substitute a = 0 in 9(a + 1) + 2 = 11
LHS = 9(a + 1) + 2
= 9(0 + 1) + 2
= 9(1) + 2
= 9 + 2 = 11 = RHS
Hence verified
## Andhra Pradesh Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbooks for Exam Preparations
Andhra Pradesh Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 exam preparation. The AP Board STD 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Books State Board syllabus with maximum efficiency.
## FAQs Regarding Andhra Pradesh Board Class 7th Maths Chapter 3 Simple Equations Ex 3.2 Textbook Solutions
#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?
Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.
## Important Terms
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1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# A body of mass m is moved to a height equal to the radius of the earth R. The increase in its potential energy is:
A
GMmR
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B
2GMmR
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C
12GMmR
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D
14GMmR
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Open in App
Solution
## The correct option is C 12GMmRWhen the object is placed at surface of earth, potential energy is given as, Ui=−GMmR where m is mass of the object, M is mass of the earth and R is the radius of the earth. when the object is placed at height R above the surface of earth. Potential energy is given by, Uf=−GMmR+R=−GMm2R Now, change in potential energy, ΔU=Uf−Ui or, ΔU=−GMm2R+GMmR or, ΔU=12GMmR
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# Predict the behavior of a time series (P&L trading desk)
I work at the trading desk P&L department at a large bank. The trading desk has positions in almost all sorts of derivatives (options, futures) over a long list of stocks, currencies, commodities... My boss asked me to do the following:
1) Show that it is not possible to predict next years P&L
2) Build a working econometric model to predict next year's P&L
The daily P&L time series is really not stationary.
Can I tell my boss that:
Non-stationarity $\Rightarrow$Cannot be predicted based exclusively on the time series. Is it true? At least to start and show him that the time series is not enough to make predictions and that we will need, at least, a complex machine learning model to do so.
• I think your boss should be more concerned about P&L variations, and those could be stationary. However - before using heavy artillery such as machine learning - visual inspection and simple models are highly advisable: pick good predictors! – Lisa Ann Jul 20 '17 at 21:48
• suppose it is not stationary neither. Is it true that: Non-stationarity ⇒Cannot be predicted based exclusively on the time series? – John Jul 20 '17 at 21:57
• No, that's not true. – Chris Taylor Jul 20 '17 at 22:01
## 1 Answer
Without seeing your trading desk's P&L it's impossible to say whether it is predictable or not. But here are a few thoughts -
1. There's no reason to think that it isn't predictable. In general, financial time series are hardest to predict when the represent the return stream of an investible asset. A trading desk's P&L isn't really investible, so there is no reason to think that it will be unpredictable.
2. You certainly won't be able to predict the P&L with a large amount of certainty, but you may be able to make ballpark estimates.
3. Have you graphed the P&L to check if there is an obvious trend? For example, if markets are getting more efficient, you would expect P&L to be decreasing over time.
4. Is P&L correlated with other variables that you may be able to predict? For example, if the desk is engaged in market making, it will probably make more money when volatility is high and liquidity is low (as the desk is selling liquidity, and the price of liquidity is high when markets are more volatile).
5. If the desk is trading illiquid instruments (e.g. exotic derivatives) the P&L series may be autocorrelated. This is especially true if P&L is correlated with volatility or liquidity, since vol and liquidity measures are autocorrelated.
6. Are the markets the desk is engaged in becoming more heavily traded, or less? P&L from market making is correlated with volume, and you may be able to make a reasonable guess about whether volume will be higher or lower next year.
What you certainly do not need is a complex machine learning model. As Lisa Ann said in a comment, you will do much better with visualisations, simple linear models, and a healthy dose of insight.
• YOur answer makes total sense. But there is no market making activity. It is all concentrated on speculation. he bank is located in a country unstable politically, hence the volatilities are much function of what is on the news. Hence, the source of future variability in the time series is not contained in the own time series as it will depend on the news. The future news are impossible to model. That is why I think it is unpredictable. The past tells nothing about the future. One clue about this is that the time series is definetly not stationary. Do you agree? – John Jul 20 '17 at 23:41
• Chris Taylor, could you give an example of liquidity index? – John Jul 21 '17 at 12:29
• @John You can't draw any conclusion about the predictability of the time series based on what you have said. It may be predictable, it may not. To know for sure we would need to see the data (and try to make some out of sample predictions). I don't understand what you mean when you say "the time series is not stationary". First, stationarity is a property of a time series generating process, not of any particular time series. Do you mean that the P&L is steadily increasing or decreasing over time? Or something else? – Chris Taylor Jul 21 '17 at 12:57
• @John A commonly used liquidity index is the TED Spread, i.e. the excess yield of 3M LIBOR over 3M treasuries. You might also look at cost-based measures (e.g. average bid-offer spread), volume-based measures (e.g. ratio of turnover to market size) or impact-based measures (e.g. typical market move per unit of volume traded). – Chris Taylor Jul 21 '17 at 13:03
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# calculator conveyor tonnes
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### how many tons can a conveyor hold
How Many Tons Can A 30 Conveyor Hold > How Many Tons Can A 30 Conveyor Hold; For a 30 yard dumpster this can range from 2 to 8 tons 400016000 lbs dependi
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### Screw Conveyor Capacity | Engineering Guide
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Calculate the belt power for a belt conveyor with this calculator to assist Engineers and Designers. Sorry - there is a problem with the database. W
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# What fractions are equivalent to 7/8
Decimal and Fraction Conversion Chart
Fraction Equivalent Fractions
7/8 14/16 49/56
1/9 2/18 7/63
2/9 4/18 14/63
4/9 8/18 28/63
23 more rows
## What is 7/8 as a common fraction?
Fractions and decimals are two common ways to write out partial numbers….Some common decimals and fractions.FractionDecimalPercent1/80.12512.5%3/80.37537.5%5/80.62562.5%7/80.87587.5%21 more rows•Apr 13, 2021
## What is 7 9 equivalent to as a fraction?
Hence, the five fractions equivalent to 7/9 are 14/18,21/27,28/36,35/45 and 42/54.
## What is 4 and 7/8 as a fraction?
Answer and Explanation: As an improper fraction, the mixed number 4 7/8 would be 39/8.
## What is 7 5 equivalent to as a fraction?
So two equivalent-fractions of 7/ 5 are 14/10 and 21/15.
## How do you find equivalent fractions?
To find the equivalent fractions for any given fraction, multiply the numerator and the denominator by the same number. For example, to find an equivalent fraction of 3/4, multiply the numerator 3 and the denominator 4 by the same number, say, 2. Thus, 6/8 is an equivalent fraction of 3/4.
## What is equivalent calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## What is 7 over 8 written in a decimal?
0.8757 divided by 8 or 7/8 is equal to 7 divided by 8, which is equal to 0.875.
## Is a equivalent fraction?
Equivalent fractions are two or more fractions that are all equal even though they different numerators and denominators. For example the fraction 1/2 is equivalent to (or the same as) 25/50 or 500/1000. Each time the fraction in its simplest form is ‘one half’.
## What is the improper fraction of a mixed fraction 7 and 7 8?
2. What is mixed number 7 78 in improper fraction? Mixed Number 7 78 in the improper fraction is 638 .
## What is the equivalent to 7 8?
14/16Decimal and Fraction Conversion ChartFractionEquivalent Fractions7/814/1656/641/92/188/722/94/1816/724/98/1832/7223 more rows
## What is 1/3 the same as?
Answer: The fractions equivalent to 1/3 are 2/6, 3/9, 4/12, etc. Equivalent fractions have the same value in the reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number.
## What is a equivalent to?
1 : equal in force, amount, or value also : equal in area or volume but not superposable a square equivalent to a triangle. 2a : like in signification or import. b : having logical equivalence equivalent statements.
## What is 7 9 as a mixed fraction?
Since 79 is a proper fraction, it cannot be written as a mixed number.
## What is 7 out of 9 as a percent?
The percentage value for 7/9 is 77.7778%.
## What is 7 9 as a percentage?
77.777778%Fraction to percent conversion tableFractionPercent5/955.555556%6/966.666667%7/977.777778%8/988.888889%41 more rows
## Which is greater 2/3 or 7 9 by how much?
7/9 is greater by 1/9.
## How to find equivalent fractions?
To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 78) by the same natural number, ie, multiply by 2, 3, 4, 5, 6
## Is 7 8 a fraction?
Important: 7 8 looks like a fraction, but it is actually an improper fraction.
## Can you convert fractions to decimals?
This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.
## What are Equivalent Fractions?
Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same.
## How to make a fraction equivalent?
Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction.
## What is half of a fraction?
For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100
## How to find equivalent fractions?
To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 87) by the same natural number, ie, multiply by 2, 3, 4, 5, 6
## Is 8 7 a fraction?
Important: 8 7 looks like a fraction, but it is actually an improper fraction.
## Can you convert fractions to decimals?
This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.
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## Differential Equation - help
Forum for the GRE subject test in mathematics.
boo78
Posts: 36
Joined: Tue Oct 15, 2013 3:53 pm
### Differential Equation - help
This is an example from the PR Math GRE Review: dy/dx = x(x-2)/e^y. We tried to do it in the class I teach and ran into a problem. PR solves it by sep of var (duh), giving the solution e^y = x^3/3 - x^2 + c. Fine, but this would mean the domain is restricted to values of x for which the cubic is positive. The slope field, however, doesn't indicate any problems in these regions and seems to suggest there should still be solutions there. PR apparently attempts to solve this issue by placing an absolute value around the cubic, giving the general solution of y = log|x^3/3 - x^2 + c|, but when you graph this function it doesn't agree with the slope field in the regions where the cubic is negative. My class and I are thinking 1) that this is yet another typo in PR, and 2) that there must be some correct way to represent the full family of solutions indicated by the slope field. Thoughts? It's been way too many years since I took DiffEq and I can't find my text book.
redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am
### Re: Differential Equation - help
I think its sort of a trick specific to log. Note that for real x != 0 (including x<0), the derivative of log( |x| ) is 1/x
boo78
Posts: 36
Joined: Tue Oct 15, 2013 3:53 pm
### Re: Differential Equation - help
Yeah, that's why I kept thinking the solution had something to do with absolute values, but it was obvious PR didn't do it right. Since posting, I've explored the slope field and overlaying solutions in more detail, and I'm now convinced that the solution does NOT involve absolute values at all. If you vary the value of c enough, you can see solutions that cover the entire slope field. Some of the things that make it tricky are that the solutions have vertical asymptotes that vary in position depending on c, which isn't obvious from either the differential equation or initial examination of the slope field, but is obvious when you look at the cubic argument of the log. Also, the solutions about the negative y-axis only appear if you graph c-values between 0 and 4/3, when the relative maximum of the cubic is positive but the relative minimum is still negative. So the answer is that y = ln ((1/3)x^3-x^2+c) is the full family of solutions, no absolute values needed. YES!
davidthedavid
Posts: 3
Joined: Wed Jan 08, 2014 11:50 pm
### Re: Differential Equation - help
I don't understand your reasoning, boo78. I think the answer y = ln|x^3/3 - x^2 + c| is actually correct.
Taking into account that d( ln|x| )/dx = 1/x, take dy/dx for y:
dy/dx = [ d(x^3/3 - x^2) + c)/dx ] / (x^3/3 - x^2 + c) [chain rule]
= (x^2 - 2x) / e^( ln|x^3/3-x^2 + c| )
= x(x-2) / e^y
I don't think you are correct that there exist points on the slope field where this isn't true.
boo78
Posts: 36
Joined: Tue Oct 15, 2013 3:53 pm
### Re: Differential Equation - help
Your error is in your substitution of e^y for x^3/3 - x^2 + c. e^y > 0, so this substitution is only valid when x^3/3 - x^2 + c > 0. Therefore the differential equation only has solutions for values of x and c that satisfy this inequality. Using the absolute value allows for x^3/3 - x^2 + c to be negative, and thus is not a valid solution in those regions. If you graph the slope field and y = ln |x^3/3 - x^2 + c|, you will observe this.
davidthedavid
Posts: 3
Joined: Wed Jan 08, 2014 11:50 pm
### Re: Differential Equation - help
Your statement that "e^y > 0" is only true on the set of real numbers. On the complex plane there are plenty of numbers y such that e^y < 0.
Does the problem state explicitly whether they are working in real or complex numbers?
Edit: I just plugged it in at "http://www.wolframalpha.com/input/?i=dy%2Fdx+%3D+x%28x-2%29%2Fe^y" , and I think I might be wrong. The answer might not have the "absolute value" in it for the solution on the complex plane. As far as the real solution goes, you either need the absolute value or to restrict y to be above zero.
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# A producer finds that demand for his commodity obeys a linear demand equation p+0.85x=8.5 where p is in thousands of dollars and x is in thousands of units. If the producer's cost are given by...
A producer finds that demand for his commodity obeys a linear demand equation p+0.85x=8.5 where p is in thousands of dollars and x is in thousands of units. If the producer's cost are given by C(x)=1+4x, what should his level of production be to maximize profits?
The level of production is= ( ) in thousands of units
violy | High School Teacher | (Level 1) Associate Educator
Posted on
Solve for p first on the demand function to know the representation of the price.
Subtract both sides by 0.85x on both sides.
`p = 8.5 - 0.85x`
We know the revenue = price * quantity.
So, we will have:
`r = (8.5 - 0.85x)*x = 8.5x - 0.85x^2`
We can now write the profit function by taking note that:
Profit = Revenue - Cost
`P = 8.5x - 0.85x^2 - (1 + 4x) = 8.5x - 0.85x^2 - 1 - 4x`
Combinbe like terms.
`P = -0.85x^2 + 4x - 1`
Take the derivative an equate to zero.
`-1.7x + 4 = 0`
Subtract both sides by 4.
`-1.7x = -4`
Divide both sides by -1.7.
`x = 2.35 or 3` (round up).
Hence, the level of production that will maximize the profit is 3000.
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# Find the sum of series $\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$
Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2} {6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $$\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ and then expanding or using known sum types as $$\displaystyle \sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ or $$\displaystyle \sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$$ but nothing seems to lead to anything!
• What have you tried? Sep 18, 2021 at 21:27
• Please, follow this link math.meta.stackexchange.com/questions/9959/… The current form of your question lacks a whole swath of things. Do consider improvements, and at the very least, show attempts, this is not a "do-it-for-me" site.
– user956717
Sep 18, 2021 at 21:36
• i have tried using the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$ Sep 18, 2021 at 21:37
• @Math3147 you are right.Sorry for that. Sep 18, 2021 at 21:41
• Retracted my downvote, and upvoted. :)
– user956717
Sep 18, 2021 at 21:44
HINT:
Use the partial fraction decomposition $$\frac{1}{n^3(n+1)^3} = \left(\frac{1}{n^3} - \frac{1}{(n + 1)^3}+ \frac{6}{n} - \frac{6}{n+1}\right) - \left(\frac{3}{n^2} + \frac{3}{(n + 1)^2}\right)$$
The sum equals $$(1+6) -( 3 \zeta(2) +3 \zeta(2)- 3)= 10 - \pi^2$$.
I'm basically just going to use the identity you noted in the comments $$\frac 1n-\frac 1{n+1}=\frac 1{n(n+1)}$$ repeatedly and then the closed form for $$\zeta(2)$$.
$$\sum_{n=1}^\infty \frac 1{n^3(n+1)^3}=\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^3=\sum_{n=1}^\infty \frac 1{n^3}-\frac{3}{n^2(n+1)}+\frac{3}{n(n+1)^2}-\frac 1{(n+1)^3}$$
The first and last terms telescope so you get
$$=1-3\sum_{n=1}^\infty \frac 1{n(n+1)}\left(\frac 1n-\frac 1{n+1}\right)=1-3\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^2$$
$$=1-3\sum_{n=1}^\infty \frac 1{n^2}-\frac{2}{n(n+1)}+\frac 1{(n+1)^2}$$
The first and last terms are $$\zeta(2)$$ and $$\zeta(2)-1$$ respectively
$$=4-6\zeta(2)+6\sum_{n=1}^\infty \frac 1{n(n+1)}=4-6\zeta(2)+6\sum_{n=1}^\infty\frac 1n-\frac 1{n+1}=10-6\zeta(2)=10-\pi^2$$
Hint: $$\frac{1}{n^3} - \frac{1}{{(n+1)}^3} \;=\; \frac{3n(n+1) + 1}{n^3{(n+1)}^3}$$
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Updated on 03.13.13
# The \$21 Food Week: Is It Possible? Is It Healthy?
A while back, I read a fascinating article about the governor of Oregon and his attempt at eating for a week for just \$21. This concept wormed around in my mind for months until recently, when I realized that this provided a pretty sharp frugality challenge.
Can I feed myself for one week on just \$21? Since I have a wife at home and also a toddler (whose food costs would be roughly half of an adult), could I actually spend just \$52.50 on food for the three of us for an entire week?
I didn’t want to just load up on carbs and fats, either – I wanted healthy stuff. So I headed over to the USDA’s interactive food pyramid to find out what I should be eating in a given day, along with what my wife and I should be eating.
My nutritional intake for a day should be:
Grains – 9 ounces
Vegetables – 3.5 cups
Fruits – 2 cups
Milk – 3 cups
Meat & Beans – 6.5 ounces
For my wife:
Grains – 7 ounces
Vegetables – 3 cups
Fruits – 2 cups
Milk – 3 cups
Meat & Beans – 6 ounces
For my son:
Grains – 3 ounces
Vegetables – 1 cup
Fruits – 1 cup
Milk – 2 cups
Meat & Beans – 2 ounces
I decided that over the course of a week, it was fine if I exceeded the recommended amount of food one day and dipped under it the next as long as it averaged out to the recommended amount.
Thus, over the course of a week, I need to serve my family:
Grains – 133 ounces
Vegetables – 52.5 cups
Fruits – 42 cups
Milk – 56 cups
Meat & Beans – 101.5 ounces
Can I buy all of that for \$52.50? That’s what this challenge really boils down to. Let’s break it down piece by piece.
First, the milk. There are 16 cups in a gallon, so to get 56 cups of milk, I would need 3.5 gallons of milk. Three gallons of skim (for myself and my wife) and one half gallon of whole milk (for my son – whole milk is what is doc recommends). An average gallon of milk costs \$3.79 and a half-gallon is \$2.09, so we’re spending \$13.46 on milk for the week, leaving us with \$39.04 for the other stuff.
Next, the grains. You can buy three loaves of whole wheat bread for \$0.99 a piece to get 48 ounces of the grains. The rest can be picked up by two containers of Quaker Old Fashioned Oats at 42 ounces at \$5.39 each. Thus, we’re spending \$13.75 on grains for the week, leaving us with \$25.29 for the other stuff.
What about meat and beans? You can buy two pounds of hamburger at \$1.99 a pound (80/20 beef, just fine for grilling), plus get two 32 ounce bags of mixed beans (for burritos and soup) for just \$3.99 a bag. Thus, we’re spending \$11.96 on meats and beans, leaving us just \$13.13 to spend on fruits and vegetables.
The fruits and vegetables are the trickiest part. If you stick with what’s in season, you can get many different kinds of fruits and vegetables for well under a dollar a pound, and on average a pound of fruit or vegetable adds up to about three cups. The USDA recommends that our family eats 94.5 cups of fruits and vegetables in a week, which is roughly equal to 31 pounds of fruits and vegetables (remember, a family of three for a whole week). We have \$13.13 to spend, but even if we found fruits and vegetables for \$0.50 a pound, we wouldn’t quite make it.
Even given this, I don’t think the experiment is entirely a failure.
First of all, one can take advantage of sales, bulk buying, and comparison shopping. I used some average prices in these calculations – and I know from experience that the milk, at the very least, can be found cheaper, particularly in dairy-rich regions. I also didn’t factor in sales or smart shopping, both of which can trim more off of the cost. Also, you can buy some of the items in bulk, like the beans and oatmeal.
Secondly, I focused strongly on healthy options, since I prefer to put only healthy foods on the table. If you open the door to less healthy substitutes, it’s easy to drive the cost down much further than what’s presented here.
Those two factors alone could potentially drive down the cost of these meals in practice to below that magic \$1 per meal number.
What can we really learn here? The biggest thing I learned is that one can eat healthy at a very inexpensive price, and thus it often leaves me with questionable feelings when I find that I’ve spent astonishing amounts on other meals. It also shows me that if you stick with the nutritional staples and basic ingredients, cooking at home can be incredibly cheap. If you’ve dropped \$20 or more at a meal recently, recognize that you could have fed yourself for a week on that bill, including breakfasts, lunches, and dinners.
Of course, you could just go for the “three \$1 double cheeseburgers from McDonald’s a day” plan (which adds up to \$21 in a week), but that might not be the best choice either.
1. Laura says:
Wow, a good article and thank you for the pyramid link. my husband and I would need more than \$42 to eat well. We both enjoy a variety of vegtables and my husband loves fruit. It’s definately a good exercise to do. I’ll be using the charts when planning next month’s grocery list.
2. Alexis says:
Just a couple of problems. Look a the list of food you bought. (Milk, bread, beans, veggies/fruits, oats and hamburger) These items alone do not make up meals! What are you going to put on the bread, or will your family just each milk and plain bread for lunch with some fruit on the side? You could probably get away with some meals, like oats with milk/water for breakfast, but how will you make a weeks worth of meals out of these items? Thus it is less likely that you will feed yourself, even on these items, for \$21/week. Not to say that it should take a \$100’s more, but you are not truly getting meals out of your shopping list.
Secondly, if your wife is nursing the newborn, her food intake needs to be more than what you have listed. Her body needs more in calories and nutrition to be able to support the baby. I understand that this is just an exercise in capability, and not practicality, but still – something to consider.
3. Jean says:
Since you are focusing on eating healthy, I hope you are buying organic milk, especially for your son.
4. Starting with an empty pantry I doubt the \$21 a week concept would work well, but assuming you have a nice collection of spices, condiments, and other base recipe ingredients I imagine it could happen.
I managed to spend less than \$200 for two people on groceries in the month of September, but I was starting with a good bit of food already on hand and a full pantry. I am working at cutting that cost down ever further so it will be interesting to see if I can get close to the \$21 number.
5. Oh, and the Double Cheeseburger plan will work until the ol’ heart gives out. :)
6. !wanda says:
I’m suspicious of the USDA guidelines, since they are heavily influenced by the food industries.
1) Milk: Most of the world’s adult population is lactose intolerant (this is our species’s ancestral condition). You can be perfectly healthy without milk. If milk were cheap I wouldn’t be picking on you, but milk is taking up a significant proportion of your budget. Cut it out, particularly for the adults.
2) Meat: Is hamburger the cheapest kind of meat? Also, meat of any sort is not cheap. You can cut this out entirely and replace it with beans. You do have to be a little more careful with your meal planning- meat provides all the essential amino acids, while you have to eat rice and beans to get them in a vegetarian meal. But rice and beans are dirt cheap.
With those two changes, you can go a long way towards buying those fruits and vegetables.
7. Daria says:
Does making your own bread/soaking and cooking your own beans help? Likely that amount was negligible given that you were over the number, but I was just curious, especially since I know you make your own bread.
An interesting article, even I would cut back a lot of things before I would cut back food to that level.
8. Trent says:
Wanda, so who should you believe when it comes to a healthy diet? There is no universally accepted definition of healthy out there – every single diet you propose will have some critics.
9. Johanna says:
I agree with !wanda – it is not necessary for an adult to drink 1.5 gallons of milk a week. The dairy industry lobbies the USDA to say that it is.
I think you probably ought to be counting servings of beans and grains by their cooked weight, not their dried weight, which means you can buy less of those items. Also, beans, oats and other grains are available for much, much cheaper than the prices you’ve listed here, if you get them from the bulk bins at a health food store. Not everybody lives near a health food store, but if you can drive to one, you can get away with making a trip every six months or so, since dried beans and grains keep for a long time.
10. Amy K. says:
Worth a read: the Emergency \$45 Menu at Hillbilly Housewife. The author includes a shopping list with prices for her area, a daily prep plan, and all the menus.
It’s a well thought out meal plan, starting from a bare cupboard, and though I would not follow it myself (from a personal taste perspective, not nutrition) it did get me interested in United States food policy and keeping a price book. And brought to light exactly how much work I could be doing to make my food, when instead I buy my bread/tortillas/canned beans.
11. kevin says:
Trent,
Did you write this post before your most recent baby was born?
from the article=============
Since I have a wife at home and also a toddler (whose food costs would be roughly half of an adult), could I actually spend just \$52.50 on food for the three of us for an entire week?
=============================
12. Rob in Madrid says:
Interesting, I was about to link to the hillbilly housewife as well. Best part is she starts the week off with pancakes!
13. sean98125 says:
You can save a lot by getting your oats in the bulk food aisle. The local grocery here in Seattle has oats in the bulk bin for 79 cents a pound, which would only be about \$2.00 for 42 ounces.
You can dump the meat entirely if you are on this strict of a budget. The amount of money isn’t meant to give people enough to have meat every night, but enough to enable them to survive.
14. When I was in grad school (which was only about 6 years ago), I fed myself on an average of \$29 a week without even trying. And I love to cook. I did not eat Ramen noodles. I had healthy, nutritious meals with a balanced and varied diet. Like Trent said, the key is buying sales. I also bought 100% of my meat from the “reduced for quick sale” section of the supermarket. At the time, the market I lived next to had great deals there. Shopping at night helps you find those deals easier since they are usually put out just after the meat counter closes.
15. Lana says:
Going back to the frugal vs. cheap conversation, I think if a person can afford it, they shouldn’t skimp on healthy foods. Buying an abundance of fresh fruits and vegetables, low-fat dairy, lean meats and whole grains is an investment in health.
Of course, there are always situations where expenses are tight, and food shopping may have to be done scrupulously. But somehow I don’t think the majority of Americans are overspending on spinach and carrots.
To me, being frugal means spending more at the grocery store to feed myself well and according to the health guidelines I follow, and eating out less. Eating fresh, healthy foods also affords me the energy to get exercise daily, so it’s a win-win situation.
16. 42 says:
second the notion on milk. it is unnecessary (and a bit unnatural) for adults to drink the milk of another species. it is not a significant source of nutrition, and no other mammal consumes milk past the age of weaning, except for cats and that’s only because we give it to them ;)
as has been pointed out, the USDA food pyramid thing was largely paid for by Big Ag and there isn’t a person on earth who follows it to the letter.
another place to save is finding a store that has a large bulk-foods dept. you can get organic rolled oats (my usual breakfast, with raisins) for 99 cents/lb and a huge range of dried legumes for cheap compared to packaged versions. even expensive yuppie-chow palaces like Whole Foods have cheap bulk depts.
17. Carlos says:
I usually spend around \$15 for my breakfasts and lunches (which I eat at work) for the week. Of course, I make the same sandwich every day for lunch, drink the water from the water cooler (lots of it), and my breakfasts consist of one thing of yogurt and a banana.
Tack on the dinners and I’m doing pretty well. The trick, of course, is to get through the boredom of it and actually stick to it for a month straight—not an easy task.
18. !wanda says:
@Trent: You’re right; there are many sources of nutritional information. I suspect that people can live under many nutritional regimes and be healthy; just compare a traditional Inuit diet vs. an agricultural, grain-based diet. My goal was to highlight two pieces of your plan that are both expensive and not clearly necessary for a healthy diet. Milk *cannot* be necessary for adult health if most adults can’t drink it! That doesn’t mean it isn’t good for you in moderation- it just means that you might not want it if you have \$21/week for food.
Meat is probably not absolutely necessary either- I personally know many healthy vegetarians, and from a theoretical standpoint it’s possible to get all the protein you need from vegetarian sources if you’re careful. It’s also cheaper.
19. sfgal says:
Wanda makes a good point about the USDA allowances. I don’t eat by those guidelines – I try to eat a little bit of every food group. And I’ve done it on \$30 a month. Granted I did have the basic staples already. The key to saving money on groceries is to only eat until you’re reasonably satisfied. You don’t need to feel full or stuffed after a meal. I eat just enough and it works wonders.
I think each person should go by their own guidelines and lifestyles. They should match their meat, dairy and cheese consumption based on their own health and dietary needs.
20. Susy says:
I think eating healthy is actually cheaper than unhealthfully. The key is to buy things that are minimally processed. Have you ever compared the cost of the big cans of oatmeal per serving compared to cereal? Also not eating meat is a huge moneysaver in the grocery department and in the health department. I can buy a bag of dried beans for \$1 and it will feed my husband and I for many meals.
My husband and I eat incredibly healthy, including organic milk and organic eggs for around \$40 a month. We eat a lot of fresh produce and a lot of beans & brown rice.
A great read for a healthier food pyramid and for cheaper better ways to eat healthier read, “Eat, Drink, and be Healthy” by Walt Willett.
21. Sora says:
My monthly food budget for my family of 7 (two adults, one of whom is pregnant, 5 children between 12 and 21 month, all but the youngest eat as much or more than the adults) is \$600 / month, which is very close to \$21 per week per child.
Milk: With 5 kids, we go through a lot of it. Instant powdered milk at the grocery store mixes up at a cost of \$2 / gallon. Whenever milk is being used in a recipe (ie. pancakes, muffins, cheese sauce) you can’t tell the difference between powdered and fresh. (It also never goes bad). In our area, major grocery chains offer milk as a loss-leader for \$1 / half gallon about every 2-3 months. When this happens, we put as much as we can fit in our deep freezer. A local gas station (which coincidentally usually has very competitive gas prices) offers up to 2 gallons of milk for \$2.39 / gallon with the purchase of 8 gallons of gas. We buy our gas 8 gallons at a time and get the milk every time.
Beans and whole grains: find and join a food coop. I buy oatmeal in 50 lb bags for \$23.29 a bag. This savings (2 cents per ounce vs. your 7 cents per ounce) allows us to serve the kids oatmeal with raisins, brown sugar and cinnamon, making it much more palatable. :-) A 50 lb bag of wheat is \$21.61. The electric grain mill we bought 5 years ago has paid for itself many times over in savings on our whole wheat bread: I figure my cost per loaf with yeast, honey, electricity, etc. is about \$.35 — but my loaf of bread is about twice as heavy, and twice as filling, as whole wheat bread from the grocery store. Savings on dry beans and lentils (very easy to cook in the crockpot) are significant when bought in 25 lb bags from the food coop instead of 1 lb bags at the grocery store.
We rarely find fruits and veggies other than bananas, carrots, and potatoes for less than \$.50 a lb, but the savings on staples makes it possible to spend more here.
22. Susan says:
I used to live on a plan I called the “\$1 per meal plan” when I first started my job. All my coworkers made fun of me, but those early years of savings helped me fund my Roth. And as we all know, saving wisely for the first 10 years of your career results in more money than the next 25. Now that I’ve sacrificed during those 10 years, I can now eat at \$20 per meal if I wanted without worrying about retirement.
23. Andrew says:
I agree that the USDA guidelines are suspect, at best. Out of all the federal bureaucracies (including the Food & Drug Admin., the Dept. of Education, and Health & Human Services) why should the Dept. of Agriculture determine a “healthy” diet? By definition, they should be regulating the production of food, not the consumption of it.
24. Great article! Be careful of the \$0.99 “whole wheat” bread. Check the ingredients. Chances are you are just getting enriched white flour instead of whole wheat flour. Real whole wheat bread should have 2-3 grams of fiber per slice.
25. st says:
I’m with the others who question the wisdom of adults drinking (or at least, needing) milk. No other mammal in the animal kingdom drinks the milk of some other mammal, and only juvenile mammals drink milk (except when it comes to humans). Think about that for a minute. Milk just isn’t meant for adults.
26. Alex Knapp says:
Here’s a really good article by two people who lived off of less than \$3 a day, and it looks like they did pretty well. (For better clarity, it’s best to start from the bottom and work your way up.)
27. D. A. says:
I liked the reference to the governor who tried to eat decently on food stamps, and for your research as well, Trent. It would be a tremendous challenge, however, for spouse & I to live on so little. The spouse has diabetes, and I’m wheat intolerant, so finding foods to replace the usual inexpensive fillers (pasta, bread, white rice, other easy starches) would take much effort. Regardless, we continue to explore how to make our dollar stretch further, and I’m grateful to this site and the commentors here for the inspiration!
28. suz says:
I just came here to say that seems like a LOT of milk to have daily but I can see lots of people have said this already. There are other foods with plenty of calcium out there for less \$.
29. vh says:
What an interesting post! Gotta try this out.
Did you check out frozen veggies and fruits? Sometimes they’re cheaper than fresh, depending on where you buy and what store has them on sale.
Agreed, with other commenters: If milk tastes good to you and doesn’t upset your stomach, then by all means enjoy it…but you sure don’t need that much of the stuff.
Hamburger is cheaper & far tastier if you watch for chuck and round roasts on sale & have the butcher grind it for you. Today Safeway again had chuck roast for significantly less than hamburger. Grabbed two packages, “saved” (hah) \$12, and now have meat for many, many days of grilled burger, tacos, curry, and chili. Yum!
30. !wanda says:
@st: It’s not that simple. Certain populations of humans have the mutations necessary to digest milk as adults. These populations were the ones that raised dairy cows. In those populations, drinking milk as adults was so favorable to survival and reproduction that the mutations that allowed milk-drinking rapidly ended up in every member of the population. One of those populations was northern Europeans; that’s why the US has a dairy industry and USDA guidelines that promote milk consumption.
Trent’s ancestors depended on dairy for calories and protein. But Trent is not a wandering cow-herder. If he wants to really save money at the grocery store, he can replace milk with cheaper forms of protein and calories that are still healthy.
31. Mariette says:
Cutting back on the foodstuffs is hard – for so many people food=comfort. Now we know that isn’t true really, but it’s a very difficult habit pattern to break, both for your wallet and your waistline. I know I could spend a lot less money on food and still eat well – even organic – and these posts and comments are a great reminder of that and an encouragement for me to continue working harder on it. So thanks everyone!
32. Jen says:
While this is a pretty good post, this can certainly be done. I have a family of 3 (my husband, my 2 year old son and myself) and I’m expecting my second child next March. My grocery budget is \$50 per week and I stick to it. OK, I mostly stick to it. Last week I went over a few bucks because I was craving chips and salsa. :)
I do have a pantry full of staples and that helps considerably. We eat a fairly healthy diet. My husband and I aren’t big on drinking milk, so 1 gallon easily gets us through the week. I make my own whole wheat bread and, when he ate baby food, I made that for my toddler as well. While I do shop sales, I also do a lot of my shopping at a local Aldi, and then pick up whatever they don’t have at the regular grocery store. When there is a sale on meat I will buy more then I need then freeze it servings for later meals. The key for me is sticking to my list and not shopping hungry (which I’ll admit is harder these days).
33. Steve W says:
This is the BEST article I’ve read on the subject of diets, food and human health (“Unhappy Meals” by Michael Pollan):
http://www.nytimes.com/2007/01/28/magazine/28nutritionism.t.html
Here’s the opening:
“Eat food. Not too much. Mostly plants.
That, more or less, is the short answer to the supposedly incredibly complicated and confusing question of what we humans should eat in order to be maximally healthy. I hate to give away the game right here at the beginning of a long essay, and I confess that I’m tempted to complicate matters in the interest of keeping things going for a few thousand more words.”
34. Sharon says:
Okay, This is a small issue with me. \$21 a week is an AVERAGE. If you have such a low income that you have no other money to spend on groceries then you will receive more. So the likelihood of you only having \$21 a week is slim. I suppose that if you have some income and a large lease/mortgage because you are in a temporary situation, it is possible. But then you more likely to have a few things in the pantry, which hopefully frees a few pennies up for sales.
Check out this randomly picked website with the maximum amounts of food stamps for PA
http://www.dpw.state.pa.us/LowInc/FoodStamps/003670291.htm
35. Monica says:
My family and I live off with less than that. It can be done. Having a pantry and buying in bulk helps a lot. And like Trent said, sales really do make a difference. I shop at three different stores buying only sale items from two of them if I NEED them. I think most of our money savings comes from buying from sales, in bulk and cooking from scratch. Most of peoples grocery money goes toward convenience foods like already made and prepared burritos, waffles, breakfast sandwiches, already cooked bacon, chicken tenders…etc etc. Those are super expensive when you look at how much it costs to make it yourself.
I’d recommend reading Miserly Moms which focuses a lot on how to save money with your grocery shopping. There is even another recipe book called Miserly Meals where everything is less than 75 cents per serving. These are healthy meals!
36. Mary says:
From my price book, let’s see how I can fare using the lowest prices:
I would probably follow Trent’s wife’s plan:
Weekly:
Grains -49 oz
Vegetables – 21 cups
Fruits – 14 cups
Milk – 21 cups (168 oz)
Meat & Beans – 42 ounces (appx 21 servings)
My best price for whole grains is about \$.06 cents/oz for oatmeal (dry), so 49*\$.06=\$2.94
Milk is something I don’t really drink or need in my diet, but for the sake of comparison I’ll add it in anyway. I can generally find a sale on a gallon of milk for about \$3.10 a gallon (128 oz), which is about \$.024/oz, so .024*168=\$4.08
Similarly, if I use dry milk instead (\$.020/reconstituted oz)is \$3.36.
As for fruit and vegetables, it’s a combined total of 35 cups, which I’ll estimate is about 2 cups per pound. If I can find bulk carrots or other similar veggies for \$.50/lb, that totals \$8.75.
As for meat and beans, I don’t generally like to cook meat so I eat more legumes, generally canned (I don’t have a kitchen). I’m gonna go by servings on this one, since otherwise we’d be comparing dry to cooked beans, so I estimate about 21 servings per week. Each can of beans costs \$.69 and has 3.5 servings, so I’d be eating roughly 6 cans of beans per week. So, 6*.69=4.14
Now add it up:
Grain: \$2.94
Milk: \$4.08 (fresh milk)
Produce:\$8.75
Beans: \$4.14
=19.91
Plus 8.5% sales tax (it’s California, what can I say)= \$21.60
With dry milk, I’d come in just under the cutoff at \$20.70
I guess it can be done, but there’s no way i’d be eating a meal of oatmeal and canned beans for days on end. But it does go to show how much money goes towards produce, and how much you can really save if you grow your own.
37. kazari says:
wow,
i didn’t know the governor did it, but the idea of a \$21 challenge has been floating around at Simplesavings.com for a little while. You can hear a bit about it here:
http://www.moneytv.co.nz
The concept is slightly different though – can you eat out of the pantry for a week, using only the supplies you have? we all stockpile food (to a lesser or greater extent) and the challenge is to make as many of your meals from your stockpile as you can. then use the \$21 for milk, bread and other fresh stuff.
it’s a great tactic when an emergency happens and you really need to tighten your belt for a week or so to get past it.
38. s says:
This is a great post, but as with the other people, I see some flaws in the presentation.
First, you only include half the money for your son, because he’s young, but the government bases the amount you receive on the number of people, regardless of age. Also, the \$21/week amount was based on the “average benefit received”, which is based on a maximum amount minus a portion of your income. The max amount is what they actually expect eating to cost. That is \$426/month for a family of three. My point here is they don’t expect you to eat for \$21 — the \$21 is just subsidizing your actual cost.
Food stamp issue aside (since that isn’t really the “point” of the article), there are still some issues with your “plan”.
The milk issue: This used to be called “milk and dairy” and would be more appropriate names “calcium sources” or “calcium and extra protein” since that’s the real point. Milk is a reasonable way to get this, but not the only way. Other people have addressed this.
Prices: A lot of your prices seem high (this has also been noted). The oatmeal seems the most ridiculous. But you did note it was for name brand oatmeal. I would have liked to see some brown rice on there. It’s super cheap, and a healthy addition to the beans (which also sounded expensive).
The “ground beef only” meat plan is silly. Not healthy if nothing else. I’ll leave it at that. There are other meats for the same or less. And alternate protein sources for the same or less. But for an “example” it’s fine.
And whoever commented on the spices and staples thing: most of those things can be bought “in bulk” if you want, in which case you can really just buy what you need. I can buy enough spices for a batch of soup for pennies. No need to buy 5 years worth.
While \$21/week isn’t really what food stamps imply you should do, I think it’s doable. Doable while eating “enough” and still eating healthy. Just stick with the food buying rules you always preach: buy seasonal fruits and veggies, avoid processed foods, and looks for generics and sale items.
39. s says:
PS. This was the site with the info on food stamps:
http://www.fns.usda.gov/fsp/faqs.htm
40. Jim says:
In grad school (10 years ago) I ate for \$20 or less per week. nowadays it is more like \$50 per week, but I eat out once or twice a week, and buy more convenience foods. I estimate that I could cut it down to about \$30 a week and still eat healthy.
I never drink milk (no longer like the taste), I only eat meat one or two times a week. I tend to splurge on fresh fruits and vegetables.
\$20 a week is probably doable, but like most americans, I am spoiled by the variety and convenience in my diet. Eating oats and beans all the time would get old quick.
41. Marcus Murphy says:
I have personally studied and trained with doctors, nutritionists, dieticians, etc. on the subject of nutrition which is fascinating. I have a 8′ by 5′ bookshelf dedicated to just nutrition (roughly 120 books most of them thick). I have read about every diet you can think of. There are diets based on blood types, and metabolism types, raw food diets, vegan diets, vegetarian diets, indigenous diets, asian diets, european diets, low-carb diets, low fat diets, etc.
I’ve found in studies (personal, controlled ones), that the best diet is a combination of eating for your blood type and metabolism and keeping your food selection simple based on the foods available from your region. when I say simple, pretend you were a caveman living in the region where you are now before civilization. What grows there and what game resides there. 75% of your diet would probably consist of fruits, vegetables, nuts, and seeds. The rest would be delicacies as you would be able to hunt them: meats, eggs, milk, and all other forms of dairy. I also found that a balanced amount of carbs, proteins, and fats give the right mix of nutrients your body needs to stay in tip-top shape. That means 1/3 of the pie goes to each category of healthy carbs (fruits, vegetables, grains) healthy proteins (vegetables, beans, grains), healthy fats (nuts, seeds, olive oil, coconut oil, beans). Proportionally balancing those categories and avoiding foods in your do not eat lists from respective blood type, metabolism type diets will result in sound health.
For one week I would need 2 lbs vegetables, 2.5 lbs fruit (non-dried), 2 pound beans, 2 pounds rice, 1 pounds trail mix (raw), 1 lb quinoa.
My farmers market will have sales during the last half an hour of the market.
I can fill the fruit and veggies requirement for about \$9.
Beans and rice are available from bulk bins at my local grocery store for about \$1.50/lb.
Quinoa is \$2.50/lb. Trail mix is \$3.50/lb.
Grand Total =\$21
Ofcourse this does assume a full rack of spices, oils, and the like already in stock.
Mission complete.
42. Lauren says:
Last week I bought a week’s worth of meals for three people for \$50, at Whole Foods no less. Granted, breakfast was not included (we buy Cheerios in bulk at Costco), and lunch will simply be leftovers, but it wasn’t difficult at all.
Our two biggest savers – none of us drinks much milk, so we only had one gallon for the week, and one of us is a vegetarian, so we tend to cook veggie meals most of the time — things like chick peas, black beans, and lentils can be REALLY cheap and are incredibly delicious and healthy. For those who say they need variety, there are so many different grains and legumes, it’s hard to get bored (try quinoa, it’s great).
43. Mary McK. says:
This is a very interesting discussion with a lot of great ideas. I will be keeping all of this in mind at the grocery store (or rather, at the time of meal planning, since most of the decisions should be made at home).
One semi-relevant point: I was an evolutionary biologist in my previous career and I don’t buy the “people aren’t meant to drink milk” argument. While it may be true that lactose intolerance is the ancestral condition for adult humans, there’s absolutely no reason why an adult should not drink milk if they want to and tolerate it(most mammals are not monogamous but that’s not an argument for us to model our relationships on the rest of the Mammalia. I’m just sayin’). Milk is a great source of calcium and protein.
No need to overdo the stuff though as it is indeed expensive (why IS that??).
44. sunny says:
Great post, we may try this next week, I’ve been looking to trim some fat in our budget and this may be the place! One trick I learned years ago is to use TVP (textured vegetable protein) crumbles in place of hamburger. I used this stuff for years in spaghetti sauce and my kids were never the wiser. You can find in in bulk bins at health food stores or Bob’s Red Mill makes it prepackaged.
Also buy your oatmeal from the bulk bins – I pay 39 cents a pound. Or 89 cents for organic.
45. Steve W says:
Bulk, dried beans/peas/lentils etc. are a great, healthy value. Here’s a tip (OK, several):
1. The Slow-cooker is invaluable for cooking dried beans.
2. Try to cook Indian. I.e., learn from a culture that has spent centuries (millennia?) perfecting the cooking of vegetables.
3. I’ve mentioned this before: The Joyce Chen Microwave Vegetable streamer is a \$12 miracle (and no, I’m not on Joyce’s payroll, but I should be).
46. Mrs. Micah says:
Mr. Micah and I eat on less than \$35/ea per week. We budget about that for food and also drugstore stuff.
With the milk, see if you can find less expensive powdered. Then mix half hydrated powdered and half normal, using your previous plastic milk bottle. It’s lowered our bill a bit.
47. Erin says:
It doesn’t look like anyone has yet mentioned buying a CSA (Community Supported Agriculture)share for your fruits and veggies. My husband and I get a box of organic produce from a local farm every week from mid-May to the end of October, and the price averages out to about \$10/week. The variety is great; this week we got melons, radishes, broccoli, romaine, baby greens, collars, kale, basil, eggplant, kohlrabi, and a pumpkin. It’s usually more than we can eat in a week, but it keeps well since it’s so fresh.
48. Marsha says:
Wow, what a great discussion!
I feel so blessed that (a) I am not lactose intolerant and (b) I love beans and rice. :D
\$12 seems like a lot for a vegetable steamer. I got one recently at Ross for Less for about \$3.
Thanks to everyone who provided links – LOTS of interesting reading here!!
49. Raymond says:
I’ve tried saving through strategic grocery shopping and cooking, but being cooking-challenged, I end up eating out too often. Food – my biggest savings challenge :(
50. Sharon says:
My husband and I happily eat on \$40 per week with no extra money spent on lunches out or coffee on the go and including our spices and oil, etc. It is really quite easy and I would attribute it to being vegetarian and living near Trader Joe’s which sells very healthy food for very reasonable prices. The only time we spend more on food is with our “fun money” which we occasionally use to supplement meals for dinner parties. I’ve also gotten good at cooking Indian and Mexican dishes, they are hearty and healthy (when eaten in proper serving sizes). That said, without Trader Joe’s and bulk bins, our manner of eating would likely be impossible on this budget. I couldn’t do it in Chicago, that much is certain.
51. Siena says:
I think this is totally possible if you know how to cook, which I think you do. If you buy convenience foods and ramen, you will fail. I’d recommend whole chicken (on sale), roast it (eat as meals), then you can make soup with the leftover carcass, and eat that with rice and veggies. Also just buy whatever fruits and veggies are on sale. I can get fresh apples or nectarines for as low as 50 cents a pound and recently got whole eggplants for 50 cents each. I can make an eggplant stirfry from one eggplant that’ll last 2-3 meals. Also consider using tofu–cheap protein. And eggs are also great cheap protein. Dairy is tough–I would cut back on dairy/cheese for this experiment. You can get protein and calcium from other foods besides milk. And I’d also keep using coupons if you can. Finally, try and find a store that sells spices/dry goods in bulk. That way you can only buy what you need and not a whole bottle (at an expensive price) for this experiment.
One of my pet peevs on other experiments of this nature is that the focus is on how difficult it is to eat on this amount without resorting to beans and rice or ramen. I’d like to see someone do this experiment and show how well one can eat on so little. If one cooks and buys fresh food I think this is very easy to acheive.
52. Interesting post and comments. I notice that the phrase “bulk bins” kept popping up. Am I the only one that is concerned about food contamination under these conditions? I recently visited a Henry’s which has good prices and an interesting selection on items which they sell in bulk from semi covered bins. (The customers lift the hinged lids and scoop what they want into plastic bags.) So dirty hands are probably going in there, including those of little kids who aren’t too fussy about what they touch. And since the lids are not tight, I’m thinking insects pests too.
My prejudice against bulk bins is perhaps colored by an incident early in my housewifely career when I bought a lot of oatmeal at a health food store with bulk bins. I stored it in an air-tight, bug proof container. Despite my good intentions, I didn’t get around to using it for some time. When I finally opened the container again — you guessed it: bugs! I had to throw the stuff out, so it was no bargain.
53. annie says:
Bananas are usually a good source of potassium and are found relatively cheap 3/\$1.00, bread store brand 2@\$1.00, instead of ground beef, try whole chicken legs at 2/\$1.00, and fresh vegetables bunch of broccoli at .99 cents a pound. how about making your tortias a whole 5 pounds of mesa for 1.99 cents and rice is always cheap, I buy the parboiled store brand at 5 lbs for \$1.99.This usually will last for a week. Eggs are a good source of protein as well as dried beans for a buck a pound. I usually only shop the outside perimeter of the store. follow the walls, produce, dairy and meats are all on the outside walls of the supermarkets. Leave the pre-boxed stuff alone, I feed a family of 44 for less than \$75.00 each week. I also buy in bulk and split up meats into portion sizes. Bulk rice’s, bulk frozen vegetables and bulk dairy/cheeses and butter. And sales, make use of online shopping lists that you can put together online from most of the supermarket circulars
54. Margaret says:
Speaking of bulk bins — I was at a store and saw a little kid standing by himself. I thought he might be lost, so I kept an eye on him. Well, he was standing by the bulk candy and eating the smarties. His mom came by a minute or two later, and I thought he was going to get a scolding for stealing. Nope, turns out that is where his mom had him stay while she shopped so he would snack and be out of her way.
55. Johanna says:
I’ve bought lots of things from the bulk bins at lots of different stores, and I’ve never had a problem with bugs. What’s more, there was a story on the news a few weeks ago in my area about a regular large grocery store that was stocking packaged rice with live bugs in it. Buying your food in sealed plastic packages is not a 100% guarantee against contamination.
The one type of bulk-bin contamination I do see sometimes is cross-contamination from one bin to another – something gets spilled, somebody uses the wrong scoop, etc. If you have food allergies so severe that you’ll get really sick if there’s a wheat-flour residue in your brown rice, then maybe you should avoid the bulk bins. Fortunately, that’s not something I have to worry about.
56. Louise says:
Start looking towards ethnic cuisines such as mediterranean, asian and indian to learn how to add flavour and variety to your meals. After all there’s nothing boring about minestrone, falafel with tahini, moroccan chickpea and pumpkin stew or a thai green curry made with whatever veggies are cheapest (use powdered coconut milk, cheaper and lower in fat). Buy pasta – preferably wholemeal – to replace part of the bread and provide variety. This is normally a very cheap item to buy.
No one mentioned just how cheap, delicious and filling soups and stews can be, plus you can make them in bulk and freeze them saving time and money by taking advantage of bulk produce and using less electricity or gas per meal. If you’re doing this you’re less likely to be stuck with just sandwiches for lunch. Freeze the stews/soups in individual serves to take with you to work. If you don’t have a microwave at work to reheat them, you can simply heat them at home and take them in wide mouth thermos.
Also you don’t have to be stuck with just oatmeal for breakfast. When on sale, other grain cereals can work out almost as cheap per serve, plus creamed rice makes a tasty and very healthy breakfast when made with skim milk (powdered of course), brown rice and just a small amount of sugar and vanilla essence. Once again, make it in bulk and freeze individual or family sized portions. Or simply make extra when you have it for dessert so there is left over for the next mornings breakfast. Some mornings have a smoothie and toast instead of cereal, and on weekends make your own pancakes. Baked beans on toast for a savoury breakfast (make your own baked beans in bulk or buy ultra cheap tinned at supermarket). Scrambled eggs with stewed tomatos when you have the time. Just saute some chopped onion, add either fresh or tinned tomatos (pick whichever fits your budget) and cook until thick. Variety is the spice of life so make sure you add to the number of dishes you know how to cook so that you won’t get bored. Try different grains and beans for variety and also extra nutritional benefits.
Budget cooking doesn’t just mean ramen noodles and oatmeal.
57. Frugal Goose says:
I think putting such financial limits on one’s nutrition is more a matter of being cheap not frugal. Theres really no reason you can’t spend an additional \$10-15 to get most of the nutrition you really need. I guess its upto the person but diet is one area I think one should not skimp on.
58. Johanna says:
There are “good” and “bad” reasons to increase your food budget.
On the good side, you can pay more for:
– higher quality
– greater variety (including things like heirloom varieties that don’t take well to large-scale industrial agriculture)
– environmentally friendly agriculture (including organic)
– a decent standard of animal welfare, if you choose to eat animal products
– supporting small farms or businesses that treat their workers well (including fairtrade products)
On the bad side, you can pay more for:
– extra processing
– extra packaging
– extra advertising for name brands versus unbranded
– extra transportation for out-of-season produce flown in from South America or wherever
– wasting food or eating too much
So it’s not really a matter of saying “the cheapest diet is the best” or “the most expensive diet is the best.”
59. Jon says:
I’m also against the argument that drinking milk is bad because “other mammals don’t do it.” It’s obvious why adult animals don’t drink milk — it requires too much technology. Domesticating other animals, maintaining grazing land, building fences or otherwise controlling movement of predators and the milk producers, etc. Good luck getting a lion to protect and tame a buffalo rather than just eating it.
Biologically, they don’t produce their own milk because they don’t have enough surplus energy. It would take a heck of a lot of extra energy for a species to produce its own milk in a large enough quantity to provide substantial nutrition. Imagine if adult cows drank milk. Well an adult cow could drink a lot more milk than a baby cow, so cows would have to start producing 10 or more times as much milk as they currently do in order to feed themselves — impossible.
60. Lisa Knight says:
Great article!!!
I spend an average of \$500 a month on food items. That includes the bulk buying & pantry stocking I do so really I feed us for longer than a month on that amount. And I can go a few weeks only needing fresh stuff like milk, eggs & fruit. I cut a lot of milk out of my diet, instead I eat yogurt to get dairy. I also add milled flaxseed to my baked goods, it’s worth the extra \$. BTW there are 4 of us, so my per person per week is \$31.25. Best I can figure the only nutrient we are lacking is fiber (besides what we get from veggies & bread), I just saw something that said kids should get 7-10g of fiber in the morning, even the “whole wheat” kids cereals don’t have that much!
61. Steve W says:
Marsha — the \$12 Joyce Chen vegetable steamer (and rice cooker) is specifically for a microwave. It’s very different from the inexpensive stove-top steamer. And tremendously faster and no loss in taste or texture. Trust me, I’m an good amateur cook & baker and avoid the microwave as much as possible (I didn’t have experience with them until my son was born and my wife insisted on having one), but the microwave steamer works great.
62. DD says:
I order several different sprout seed mixes online, and have a big jar of them growing on my kitchen window shelf at all times. They’re delicious, packed with vitamins, and couldn’t be fresher! I use them in salads, sandwiches, tuna mixes, omeletes, etc.
63. Andrew Stevens says:
I’ll just add my two cents to the milk debate. If you believe that it is not part of God’s plan for humans to drink milk, then I’m fine with that. If you’re lactose-intolerant (as most people are), then of course you shouldn’t drink milk. If you’re an animal rights activist, then I sympathize with your anti-milk opinions.
However, I am quite annoyed at pulling evolutionary biology in as a justification for the anti-milk cause. Evolution has been very clear. In those societies which bred dairy animals, the ability to drink milk constituted such a survival advantage that the lactose-intolerant were quickly eliminated from the gene pool. How does this happen if milk is such a bad thing? In fact, milk is very close to Nature’s perfect food. The reason humans (and most other mammals) evolved lactose-intolerance was to aid weaning them off their mother’s nipple so she could bear more children and not have so many demands placed on her energy. (By the way, virtually all cats are also lactose-intolerant. Milk should be given to them sparingly as it gives them diarrhea.)
Of course, the anti-milk people are quite correct that we don’t *need* milk and if it’s straining your budget, you can certainly cut it out. Other than milk, the best source of calcium and vitamin D is fish, though you can also pick it up from other sources. By the way, most Westerners don’t eat enough fish.
64. Mr. Nickle says:
There’s a great book on this subject called “Eat Well for 99 Cents a Meal”. Some of it is definitely a bit out there, but there are some very interesting tips/ideas in it, such as buying grains and oats in bulk from a feed and tack store (which are often higher quality than what is packaged and sold for humans). I don’t practice most of what is in this book, but I found it a very entertaining read.
65. Mr. Nickle says:
Also, consider powdered milk. I switched to it recently and haven’t really noticed much of a difference. Plus if you make it in small batches, spoiling never becomes an issue.
66. Vicki Davidson says:
Thank you, Andrew and Jon, for your “cow milk support.” I was beginning to feel like a freak of homo sapien nature. I love milk and have never had an allergic reaction to it (although some shellfish gives me hives and my daughter is allergic to fresh cherries). I drink at least 8-16 oz daily, and at age 50, still have strong teeth and fingernails and have never had a broken bone. Coincidence or genetics, perhaps, but my daily balance of food intake has always included dairy products PLUS milk. Firmly stating that “other mammals don’t drink milk” as the definitive argument against drinking it seems to be fallacious argumentation. In the “they-don’t-do-it-universe,” other mammals also don’t cook and refrigerate fish and other meats, don’t wash fruits and vegetables, don’t simmer beans and rice, don’t husk corn, don’t have clambakes, and don’t mill grains – this “they don’t do it” list can go on and on. Carnivores and herbivores have distinctive nutritional requirements, whereas humans and a handle of other mammals are omnivores – ne’er the twain shall meet. While milk may be a bit costly compared to other sources of calcium, it’s one area I choose to not eliminate. I will continue to make my own bread and shop “bargain stores” and sales to afford my naughty indulgence. Love that milk!
Fabulous article and equally great posts. A wealth of information is here, and while I know that spending just \$21 per person per week to feed my family of four will be a challenge (my athletic son is 14, so does he count as two or three people? LOL), I am excited to give it a try.
67. Siena says:
I buy bulk a lot and usually it’s cheaper–especially spices cause I can just buy what I need for a recipe (vs. a whole bottle). My grocery store (Winco on the west coast) has a large bulk section and it looks very clean and neat. It has the overhead spout bins that you pull a lever and food falls into a bag. It also has a place where you can churn peanuts or almonds into butter. So far I’ve had no problem with bulk buying even from bins. There are “no sampling” signs everywhere and it’s right next to the deli section so employees from the deli area have a clear view of us bulk shoppers.
68. Andrew Stevens says:
I’m actually taking no stand on whether milk is good or bad for us in the long term; evolution only selects for us to get old enough to breed. I’m just pointing out that the evolutionary arguments *against* milk make no sense. If evolution is telling us anything at all, it’s quite the opposite, given the elimination of lactose-intolerance in many sub-groups of humanity.
69. Margaret says:
Put me in the pro-milk camp — LOVE IT. I took an anthropology course once, and the prof actually discussed lactose intolerance in adults. He used to say the “northern european pastoralists” were okay with milk. On the other hand, if you have a limited budget, then milk is definitely one place you can cut down, as long as you are getting your calcium elsewhere. I personally would cut out a lot of other things before I would cut out milk. And yes, I am totally grateful that I have the ability to make the choice. As I tell my 4 year old, we are SUPER RICH compared to most people in the world.
70. Lori says:
I love bulk bins and save a lot of money buying from them. All packaged and bulk dry goods have the potential to be infested with bugs. Most of the grain you buy has larvae in it- that’s just how it is.
If you put your grains and other dry goods in the freezer for a couple of days after you buy them, then you kill the larvae. Then store them in airtight containers and you won’t get bugs. I highly recommend doing this for all flours, grains, cereals, etc., whether you buy in bulk or packaged.
As for germs, well, there probably is a greater potential for germs in bulk items, but not more than you would encounter everyday on doorknobs, grocery cart handles, handrails, phones, keyboards, etc. Germs are everywhere. If you’re cooking it anyway, what’s the big deal?
The best part about bulk-bin buying, apart from the price, is that you can buy a tiny portion of something you are buying for a special recipe or experimentation, that way you aren’t stuck with a pound of an exotic grain that you will never use again.
71. Robin says:
My husband and I have a food budget of \$160 a month for food. That coincidentally works out to be \$20 per week per person. We don’t buy shrimp or expensive fresh fruit and vegetables, but I love to cook and we eat pretty comfortably.
72. Rena says:
Wow this is a great challenge. I have a copy of the Bare Bones Grocery Challenge book (I’m a member of the Cheapskates Club) that gives ideas and recipes to feed a family of four for under \$20 a week which is a lot less than \$21 per person. The thing to remember is that most households have some basic pantry and fridge/freezer staples on hand to help stretch the \$20.
I have followed it with my family as a part of a pantry challenge for two weeks and it was good. I didn’t actually have to spend the \$20 a week, most weeks I only spent about \$15 to top up the meat and fruit/veg.
You can also grow your own vegetables which saves heaps and even if you don’t have a yard for a garden a balcony with pots will do the job.
Milk powder is cheaper than fresh milk too, and that cuts the cost down considerably. Bulk grains and legumes are cheap and of course can be substituted for meat. I serve 4 meatless meals a week and it really helps to keep our grocery budget low.
Menu planning on such a limited budget is fun and you can serve some really creative meals. Eating well and healthily doesn’t have to cost a fortune, it’s just what we have been conditioned to believe. Most of us over-eat and eat all the wrong foods anyway.
73. LawVibe says:
Wendy’s dollar menu is always a good choice. Chicken nuggets are the cornerstone of any nutritious meal.
74. yvie says:
Well I’m grew up on a farm with Bessie the cow, and that’s where we got our milk.
Today I love not having milk products. My stomach is sure happier and so is my budget.
Kudos to those who know how to take inexpensive, otherwise boring veggies and beans and, through the use of spices and sauces, make them very tasty. You can eat very well and very cheaply if you have the skills.
75. danielle says:
I think all of these ideas are great IF: you live NEAR a Whole Foods (which I don’t)…there IS a farmers market nearby (which there isn’t)….if milk did NOT cost 4.50 to 5.25 a gallon, which it does…..and if I COULD cost comparison shop at other grocery stores…not an option. I have two option for cheap grocerys, an HEB and Walmart. There are other local mom and pop grocery stores….cheap veggies and fruits…expensive everything else. And I personally don’t like beans…but that’s me. I have cut down on costs of our grocery bill, but keep in mind be/c of the TWO options for fruits and veggies….I tend to pay too much money for those items…I refuse to take a toddler or myself that matter to 3-4 different stores….but in general, where I live grocery shopping IS expensive! No matter how you cut corners. Yes I do buy in bulk, if I can and it makes sense, b/c not all bulk saves you money! The only bulk stores I live within a 45 minute drive is Sams and Costco! Oh…and I rarely buy convenience foods, I cook dinner every night! I don’t buy pre-packed/pre-cooked meat like chicken tenders or anything like that either! So any other ideas?
76. eric says:
I am the father of a family of five. My wife buys in bulk for the most of our shopping. We have a budget set of 80/wk without taking into account the USDA recommend servings. Using just a calculator (80 divided by 5 =\$16/week/person) that’s just \$2.29/person/day. This includes buying diapers, personal care, and household items. We make almost everything from scratch, buying for convince is expensive. We all eat well with complete meals and snacks.
77. SunshineD says:
I think it’s odd that some people think it’s crazy to live on a diet of basics, like beans and various grains and vegetables and fruit. That’s what we are supposed to be eating! In fact my boyfriend and I eat a pretty basic diet. It only gets boring when the food is cooked into a dish or spiced properly. We do that out of laziness However, it is edible. Vegetables and fruits provide their own flavoring for dishes. We also buy organic, but it does not cost \$21 a week pp! I wish! But it’s not that expensive. There is so much to distract when trying to eat healthy: fast food, restaurants, unhealthy processed foods. And some of them taste good too. I believe though that home cooked food is superior in taste, nutrition and cost! So it’s a win-win!
I have a tip for Danielle: learn to like beans! They are very cheap and nutricious. There are many types and they don’t taste the same: chickpeas, black beans, pinto beans, kidney beans, etc…
78. Xias says:
Great Article, I always love to read about eating on the cheap!
79. efficacyman says:
Note: You could use dried powdered non-fat milk and cut your milk bill in about half, which would give you plenty of money to spend on vegetables (at around 80-90/c a pound). Additionally you could buy canned vegetables or can your own to significantly save on cost.
80. Terese says:
We are the ONLY species who drinks off of another animal’s teet after we are babies. How nasty. http://www.milksucks.com Milk is bad for you and actually makes your bones weak and give osteoperosis to people. Research it. Meat… we don’t even mess with it period. Red dyes they put in it give kids ADD/ADHD symptoms and the dye has also given lab rats cancer. Nobody needs meat or the antibiotics the government allows in it– http://www.goveg.com or http://www.allcreatures.org and http://www.spice-of-life.com
Think of the animal suffering you will stop.
81. Jeremy says:
This is insane. My food bill is easily \$50 a week if not more. Then again, most of you are couch potatoes and can live off low caloric intake. Milk is not bad for you. If you think its bad for you, you are retarded.
Animal suffering? I support PETA.
People for the Eating of Tasty Animals.
82. Artemis says:
I enjoyed the article but would like some sample menus from the author and some of the posters who have low weekly food budgets.
One gap I noticed was in bread. Our day old bread store has two 10 percent off days a week as well as reduced cost for higher quality breads and other items. They also have a further reduced cost shelf. Then there is home made.
Staple items I have to reduce serving cost are old-fashioned oatmeal, whole grain bread with high fiber content, peanut butter, honey, spaghetti, and tuna fish and noodles.
Stock up purchases on reduced prices. Every items seems to have increased in price by 30 cents from fuel price increases getting goods to markets.
83. Annie says:
Okay I just stumbled onto this article today while researching ideas for frugal ways of living. My husband and I are both retired with health problems, I am 57 yr, he is 64 yr, we have our 22 yr son still living at home and working at a part-time job, we live in Michigan which is currently one of worst states for employment and foreclosures.
We live on a fixed income and yes I can feed our family of three adults on \$63.00 per week and eat good meals with a variety each day. In fact I actually do feed us on around \$250 to \$275 per month, I am fortunate to live near Aldi’s and can save 40 to 60% on groceries, we have a small upright freezer and I buy bulk meat deals from a local butcher, this includes pork and beef cuts and depending on the bundle I choose it can also have some luncheon meats and/or hot dogs. Depending on my decision to buy 43 lbs or 62 lbs bundles it can range from \$1.45 per lb to \$1.54 per lb., or I can go higher too all the way to the side of beef bundle at \$1.59 per lb and spending \$509.00…………that I can never afford. But my point is by shopping around, avoiding name brands, eating healthy and good meals does not have to cost a small fortune. I cook from scratch, I do not buy pre-package foods, such as Rice-A-Roni when I can make it myself cheaper and better. I make my scalloped potatoes from scratch, not a box mix. We don’t have to eat meat at every meal three times a day.
Remember back to how your grandmother cooked and how good it was. My grandparents who lived on farms, very rarely bought anything at a grocery store, but both my grandmothers could whip up some delicious meals from their home canned veggies, smoked meats and neither one owned a freezer!
With a freezer or a pantry area to store canned foods you can save money by shopping local farm stands in summertime and freezing or canning your own veggies, jams and fruits for the year.
I make a big turkey dinner every November, a big ham dinner every December 25th and again on Easter. Those are when our daughter, son in law and 3 teenage grandchildren come and eat with us. Leftover ham bones and cut up ham makes wonderful seasoning for good old fashion bean soup the week after the holiday dinner.
I also use my crock pot often to cook, you would be surprised at how a cheap tough cut of meat can become so tender it is falling off the bone when cooked in a crock pot. I have never paid for “stewing beef” at the store for soups or stews, I cut up my own from cuts of beef I have purchased for a much cheaper price. Couple times a month I toss pinto beans into the crock pot and a ham bone for seasoning, we have bean soup, use some of the beans for burritos, I never buy a can of refried beans when I can make my own with some of the pinto beans I have already cooked myself.
Then another day I cook a beef roast or maybe pork steak, take some of the leftover meat, add some pinto beans and we have meat and bean burritos, if I have gravy, I add some salsa, hot sauce, little cheddar cheese and we have wet burritos. It just takes a little planning, but no food goes to waste or gets tossed out. I learned from spending my childhood summers with my grandparents on their farm how to be a spendthrift and it is serving me well in today’s high cost of living.
And eggs are just not for breakfast folks………….we have breakfast supper sometimes, eggs, fried potatoes, or pancakes and if the budget permits a little bacon, sausage or ham slices also. Summertime omelets with fresh veggies are wonderful!
Hillbilly Housewife is a wonderful website for ideas for cheap meals, the prices have gone up since she put the menu online but you can still see how to feed your family cheaply and serve a variety of foods.
Oatmeal for breakfast is cheap, quick to fix and can have many different flavors depending on what you add to it, it is filling, besides it is high in fiber, and good for lowering cholesterol.
It amazes me how so many folks today have no idea how to budget their food bill and cook inexpensive meals for their families……….they all seem to think inexpensive meals means hamburger, macaroni with a can of tomatoes tossed in………..nothing wrong with that meal, but there are so many other meals to prepare on a small budget.
Fact is if we would all eat less, cook more of our foods from scratch, stop spending on high fat convience foods we would be healthier, skinny and have more energy and most of all less likely to have some of the health problems we have in todays society. Learn how to store your produce properly, wrapping your celery in foil will keep it fresh longer, drink more water, use more storage containers instead of plastic storage bags and wraps. Figure out how many different ways you can serve one large cut of meat to your family so you can have more than one meal from a roast, ham, or whole chicken. And trust me if you do your homework and research they won’t even realize they are eating leftovers.
84. Sam says:
To cut down on the high cost of meat, we recently bought a quarter (side of beef). In addition to the cost savings, which added up to approx. \$1.50 per pound (yes you read right), it is organic beef without the hormones and antibiotics and is grass fed. How healthy is that?! We were able to specify how much hamburger, roasts, steaks etc. we wanted. And it came individually wrapped and frozen, freezer ready. Imagine having T-bone steaks for that price! Well worth looking into, even if the cost is somewhat higher in your area.
85. steve says:
Always put newly bought containers of grains or beans into the freezer for a couple of days before moving them to the pantry, unless you will be using them soon. This will kill any grain moth larvae that are in there. And there almost always are grain moth larvae in grains.
As to contamination in bulk bins, well I cook most of my food anyway.
86. J.O. says:
The calcium in milk is necessary for strong bones and teeth, and to help muscles and nerves work properly. Milk is one of the few cheap sources, and provides vitamin D at the same time (for folks who live in northerly climates and can’t get enough D from the sun). Yes, there are other sources of calcium, but they are pretty expensive (salmon, tofu, fortified rice or soy milk, for example). Use all or half powdered milk and the cost goes down. I suspect those who are saying milk is not necessary are too young to have reached the years of osteoporosis. Bone loss starts around the age of 35, so build your bones while you can.
87. Brittany says:
For \$20 a week, I eat at least two servings of fruit, 3-4 servings of veggies, 2-4 servings of milk, 1-2 servings of protein (tends to be cheese, tofu, beans, nuts… very occasionally meat), and 3-4 servings of rice or pasta. Since I got a new roommate who eats similarly, it’s dropped to \$12-\$15. What’s unhealthy about that?
88. Elaine says:
When my husband and I were first married we had a food budget of 10.00 a week and we survived. We ate a lot of spaghetti, macaroni and cheese, stews, soups, casseroles and stir fries-anything that could maximize the ingredients we had. We bought bread from an outlet store or made it from scratch. We drink a lot of milk but found the cheapest store to buy it from. Cookies and sweets were bought on sale or not at all and most were made from scratch. You can have a healthy diet on a surprisingly small amount of money if you do your research and use a little imagination.
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CONFIDENCE Function
### Compatibility Function in Excel (34 Practical Examples)
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### How to Calculate Confidence Interval for Slope in Excel
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### How to Calculate 95 Percent Confidence Interval in Excel (4 Ways)
Excel is a powerful tool for analyzing data and handling large datasets. It is sometimes necessary to perform statistical calculations in order to ...
### How to Find Confidence Interval in Excel for Two Samples
Are you a business analyst or a data science enthusiast? Do you work with a large amount of data? Then, you’ve come across a statistical term ...
### How to Calculate Confidence Interval in Excel (4 Easy Methods)
When you want to do research with a large number of people, it is impossible to take all people individually at a time. That’s why we can take a ...
### How to Calculate Confidence Interval for Population Mean in Excel
Confidence interval or CI is the measurement of the range of estimates for a particular parameter. This is an important concept in data analysis. As ...
### How to Calculate 90 Percent Confidence Interval in Excel
Excel is extremely powerful for analyzing data and handling large datasets. Sometimes, statistical calculations are necessary for interpreting a ...
### How to Find Upper and Lower Limits of Confidence Interval in Excel
If you are looking for ways to find the upper and lower limits of a confidence interval in Excel, then this article will help you with 5 different ...
5 Excel Hacks You Never Knew
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Business mathematics MCQs, business mathematics quiz answers 6 to learn secondary education online courses. Discount formula multiple choice questions (MCQs), business mathematics quiz questions and answers for for online secondary education degree. Discount formula, profit and loss, free online math test for secondary school teaching certification.
Learn high school math multiple choice questions (MCQs): Discount formula, profit and loss, free online math, with choices selling price, discount, loss percentage, and profit percentage for online secondary education degree. Free math study guide for online learning discount formula quiz questions to attempt multiple choice questions based test.
MCQ: (Discount ⁄ mp) × 100 is equal to
1. sales price
2. profit
3. discount percentage
4. none of above
C
MCQ: (Loss⁄Cost Price) × 100 is equal to
1. discount
2. selling price
3. loss percentage
4. profit percentage
C
MCQ: If selling price of an item is greater than its cost price, then we earn
1. nothing
2. profit
3. loss
4. discount
B
MCQ: (100 × Selling price)⁄100 + profit % is called
1. cost price
2. marked price
3. profit
4. discount
A
MCQ: (Profit + cost) price is equal to
1. selling price
2. loss
3. marked price
4. discount
A
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# Computing force over a range
I'm building a murphy bed. I've build a box out of plywood that holds the mattress (the box is shaped like a sandbox. See image below). I plan on putting 5 different 25lb springs across the long side of the bed frame, but want to know if the frame can withstand 125 lbs of constant force.
So, what I did was build three different L-shaped test pieces, joined the same way I did the box (rabbit joints + dowels + screws + glue), to test their breaking force. (I'm going to use a similar setup to http://woodgears.ca/joint_strength/).
Here's my question. Does each of those L-shaped test pieces need to withstand 125 pounds of force? Or do they need to withstand only 25 pounds (assuming the L-shaped pieces are the same length as the average distance between springs)?
• A diagram of the planned bed frame would be very helpful here. I have some concerns based on what you described.
– mac
Feb 13, 2014 at 20:34
• Good idea, mac. I've included a quick sketchup image and edited the post. (Sorry the "springs" don't look like springs. I'm not too savvy at sketchup). Feb 13, 2014 at 21:08
• thanks for the diagram. I still have some questions/concerns. 1) What is the purpose of the springs? 2) When you say "25 pound spring" do you really mean a 25 pound-per-inch spring, or are you saying that each spring will be guaranteed (by some other design feature) to exert 25 pounds of force on the bed frame?
– mac
Feb 14, 2014 at 17:12
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# Polarisation of reflected light
• sperryrand
In summary, the polarity of the reflected light is a function of the angle of incidence, the angle of reflectance to the eye, and the difference between v and r.
sperryrand
Hi,
I have a question regarding the projection of polarised light onto a screen, such as commonly used in 3D or VR cinemas. I'm only concerned with linearly polarised light for the moment.
I understand that when light undergoes reflection the polarity is altered.
I've attached a diagram. Light from projector is typically polarised at 45 degrees to the horizontal.
My question is, is the polarity of the reflected light a function of:
a) angle of incidence θi,
b) the angle of reflectance to the eye θv,
c) or the difference between v and r: θvr, or
d) none of the above?
Is there a simple expression, or does it depend greatly on the optical properties of the screen material? Does the Brewster angle have anything to do with this?
M
#### Attachments
• ReflectedPolarLight.png
5.8 KB · Views: 515
Hi,
Thanks. A few things are unclear to me..
a) How do I use these equations for an opaque material, where there is no transmittance or refraction? (What is the refractive index of an opaque material?)
b) These equations seem to tell me the amount of light reflected or transmitted along r - the reflection of i. But I need the ratio of the "p" component to the "s" component for light scattered in all directions.
Sorry if I am missing something obvious.
M
sperryrand said:
Hi,
I've attached a diagram. Light from projector is typically polarised at 45 degrees to the horizontal.
It's not clear if you are really interested in 3D projection or just in the phenomenon itself.
But I thought that in 3D projection they use circular polarization.
Hi,
I'm interested in the degradation of stereo contrast (i.e. increase in cross-talk) caused by high angles of incidence and reflection.
Yes, in some theatres it is circular, but in others linear. Here, I have a linear system.
M
## 1. What is polarisation of reflected light?
Polarisation of reflected light refers to the orientation of the electric field in a light wave that is reflected off a surface. When light waves bounce off a surface, they can become polarised in a specific direction, which affects how the light is perceived by our eyes.
## 2. How does polarisation occur in reflected light?
Polarisation occurs when unpolarised light, which has electric fields oscillating in all directions, hits a surface and the reflected light is filtered to only allow electric fields oscillating in a specific direction to pass through. This results in the light becoming polarised in that direction.
## 3. What are some examples of polarisation in everyday life?
Polarisation can be observed in various aspects of everyday life, such as the glare on a car windshield or the reflection of light off of a body of water. It is also used in polarised sunglasses to reduce glare and in 3D movie technology to create depth perception.
## 4. How does polarisation affect the appearance of objects?
When light is polarised, it can change how objects appear to our eyes. For example, polarised light can make shiny or reflective surfaces appear darker or more muted, while non-reflective surfaces may appear to have more contrast and detail.
## 5. Can polarisation of reflected light be manipulated?
Yes, the polarisation of reflected light can be manipulated using polarising filters, which are materials that only allow light waves of a specific orientation to pass through. They can be used to control the amount and direction of polarisation in reflected light, making them useful in photography and other scientific applications.
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# Verilog for beginers
1. Write a Verilog model of a synchronous finite state machine whose output is the sequence 0, 2, 4, 6, 8 10, 12, 14, 0 . . . . The machine is controlled by a single input, Run, so that counting occurs while Run is asserted, suspends while Run is de-asserted, and resumes the count when Run is re-asserted. Clearly state any assumptions that you make.
2. Write a Verilog model of the Mealy FSM described by the state diagram in Fig. P5.48. Develop a test bench and demonstrate that the machine state transitions and output correspond to its state diagram.
3. Draw the state diagram of the machine described by the Verilog model given below.
module Prob_5_52 ( output reg y_out, input x_in, clk, reset);
parameter s0 = 2'b00, s1 = 2'b01, s2 = 2'b10, s3 = 2'b11;
reg [1:0] state, next_state;
always @ ( posedge clk, negedge reset) begin
if (reset == 1'b0) state <= s0; else state <= next_state;
always @(state, x_in) begin
y_out = 0; next_state = s0;
case (state)
s0: if x_in = 1 begin y_out = 0; if (x_in) next_state = s1; else next_state = s0; end
s1: if x_in = 1 begin y_out = 0; if (x_in) next_state = s2; else next_state = s1; end
s2: if x_in = 1 if (x_in) begin next_state = s3; y_out = 0; else begin next_state = s2; y_out = 1; end
s3: if x_in = 1 begin y_out = 1; if (x_in) next_state = s0; else next_state = s3; end
default : next_state = s0;
endcase
end
endmodule
4. Develop the state diagram for a Mealy state machine that detects a sequence of three or more consecutive 1's in a string of bits coming through an input line. Write and verify a Verilog behavioral model of the counter designed
Skills: Verilog / VHDL
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Why does AES encryption take more time than decryption?
While I was studying the time consumed by some methods I found that encryption take more time than decryption. Is this right? I am using AES (the same steps will be taken during the enc and dec)
• Fun fact: when using AES-NI for accelerating AES, decryption is actually slower because an additional instruction, `AESIMC`, must be used every round. Commented Mar 7, 2018 at 10:09
• @forest `aesimc` is used on each round key, not every round. The decryption keys can be expanded before decrypting. Commented Jun 7 at 16:50
Several symmetric block ciphers (specifically ones like AES, DES, Blowfish, RC5) will take the same amount of time (within measurement error) for encryption and decryption, when operating on a single block (e.g., 128-bits for AES).
However, there are a couple reasons why it appears different when encrypting/decrypting multiple blocks. For example, with cipher-block chaining (CBC), encryption must be done sequentially (encrypt block 0 before you can encrypt block 1 before you can encrypt block 2 ...), while decryption can be parallelized as the XOR step (with the previous block of ciphertext) is done after the block cipher is applied.
See the diagrams below (XOR is denoted with a circled plus ⊕). To encrypt the second block of plaintext `p[1]`, you use `c[1] = AES_Encrypt(p[1] XOR c[0])`, this means you can't generate `c[1]` until you are finished generating `c[0]`.
Meanwhile to decrypt the second block of ciphertext, you use `p[1] = c[0] XOR AES_Decrypt(c[1])`. This is not dependent on the previous block of plaintext `p[0]` so can be completely parallelized (and can run much faster on multi-core systems).
If you want fast decryption and encryption, you may consider using CTR as it can be parallelized on both encryption and decryption.
You also potentially should be wary of tests that are skewed because of caches on your system. If you randomly decide to read a file from disk and encrypt it, it will take several milliseconds to read each chunk of the file; however the next time you access it the file will generally be cached in memory and be accessed much quicker. Meanwhile for your decryption test that encrypted file that was recently written to disk will likely still be in a cache in memory and will not suffer the penalty from reading from disk.
CTR Timing
``````\$ time openssl aes-256-ctr -e -salt -pass pass:passwd -in Fedora_64-bit.vmdk -out vmdk.encrypted
real 0m58.217s
user 0m5.788s
sys 0m8.493s
\$ time openssl aes-256-ctr -e -salt -pass pass:passwd -in Fedora_64-bit.vmdk -out vmdk.encrypted
real 0m34.780s
user 0m4.800s
sys 0m7.748s
\$ time openssl aes-256-ctr -e -salt -pass pass:passwd -in Fedora_64-bit.vmdk -out vmdk.encrypted
real 0m34.989s
user 0m4.120s
sys 0m6.444s
\$ time openssl aes-256-ctr -d -salt -pass pass:passwd -in vmdk.encrypted -out decrypted
real 0m35.944s
user 0m4.140s
sys 0m7.008s
``````
The first three encryptions show the effect of the disk cache, where the first access of a file was much slower than subsequent accesses. It also shows that for CTR that both encryption / decryption of this 7.4 G virtual machine took roughly 35 seconds. (On repetition occasionally, decryption would be faster or slower).
• There are block ciphers where encrypt/decrypt have slightly different cost. These aren't just based on external effects like mode, or caches. You could see that difference with ECB mode. Threefish is one of them, but I don't know if this applies to AES as well. Commented Jun 26, 2013 at 17:30
• @CodesInChaos - Good point. I was trying to speak concretely about AES where the steps for encryption / decryption have exact analogs of each other in a decent implementation (Round Key generation / SubBytes vs InvSubBytes / ShiftRows vs InvShiftRows / MixColumns vs InvMixColumns). The similar block ciphers was specifically referring to Luby-Rackoff block ciphers (which AES is not an example of) built from Feistel network should have this property as encryption decryption are the same applications of the same pseudo-random function (just in opposite order to opposite halves of the block). Commented Jun 26, 2013 at 18:13
• Just wanted to say that this is a great answer, and I wish there were more like this on the SE network! Commented Jun 26, 2013 at 19:15
• If you use ram (e.g. `/tmp`) to store the files shouldn't you be able to test without being affected by disk io? Commented Jun 27, 2013 at 6:21
• @CodesInChaos - While I can't demonstrate that all symmetric key ciphers take same time on one block for enc/decrypting. However, looking at threefish's implementation (section 2.2 ), it should take the same time to encrypt/decrypt. Tests in python3 for threefish show the same running time; e.g., encrypting 50000 random 64-byte blocks takes 23.134 +/- 0.183 ms, while decrypting 50000 random 64-byte ASCII blocks takes 23.135 +/- 0.095 ms. Here's a pastebin of the python3 tests. Commented Jun 28, 2013 at 16:26
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# BEST ROULETTE STRATEGIES TO HAVE FUN AND RETAIN A FAIR CHANCE OF WINNING BIG
Difficult
A lot of of the casinos had mechanical roulette devices not controlled by computers after that some casinos could be a a small amount sloppy with checking how good the machines were working. All bets are identified using poker chips of varying amounts.
### Strategy Rating
At the outset you have to understand the alacrity of the ball as it plunges into the pockets. The good gossip is advanced techniques can spot a bias in far fewer spins. You need the right combination of broker and wheel. It offers a circle that has the numbers 1 all the way through 36 and, depending on the account, a single zero or a alter ego zero. It is also why a good number players will spread their bets athwart multiple betting selections such as combining a straight bet with a break bet, square bet and street anticipate. Players should simply memorize the four numbers and follow them when assembly their bets. Some were actually defeat the game; meaning they were defeat the mechanism itself. However, for those who take the game more acutely, applying strategy can make the alteration as roulette is a game of chance, and therefore applying strategy addicted to your roulette sessions allow you en route for see small gains versus large ones.
### How to Use the Five Base Strategy in Roulette?
This was seen in casinos in France that had a roulette wheel so as to included the numbers but showed the single zero in the colour burgundy and the double zero in black. Roulette Strategy Strategy in roulette is critical for long-term success. This anticipate is often made to in coincidence with single, split and street bets as part of bankroll management. Roulette is one that does not call for the person playing it to allow any specific mathematical skills as it is one that requires you insertion bets on single numbers or combinations of numbers and if the globe lands on those, a win is guaranteed.
### Progressive Strategies in Roulette
As a replacement for the question is this: can a dealer hit a specific section of a wheel so that betting arrange the numbers in that section bidding give the player an overall advantage at the game. How are you supposed to do this? Should the ball land on any of the six numbers it will pay At the outset you have to understand the alacrity of the ball as it plunges into the pockets. Please Note: By the end of their spins, a good number balls fall at the same agreed speed as the physics of affecting objects and friction take control. The game might be unbeatable but a defective wheel was not. But this tends to slow down the amusement and make spins less frequent.
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What is compound interest?
Compound interest builds your money at an accelerated rate. It's interest on top of interest: Your deposits pile up interest, and the interest piles up interest, too.
This can lead to incredible growth over time.
In contrast, simple interest amounts to interest on only the intial principal year-by-year. Previous interest is not taken into account.
Let's say — hypothetically — that you deposit \$20,000 into a high-interest savings account that compounds interest at 2.80%. If the account stayed at that rate for five years, your savings would grow like this:
Compound interest over 5 years at 2.80%
Year Initial Balance Interest Gained Final Balance
1 \$20,000 \$560 \$20,560
2 \$20,650 \$576 \$21,136
3 \$21,136 \$591 \$21,727
4 \$21,727 \$609 \$22,336
5 \$22,336 \$625 \$22,961
Now let's see what simple interest would look like at the same rate, over the same period, for comparison's sake.
Simple interest over 5 years at 2.80%
Year Initial Balance Interest Gained Final Balance
1 \$20,000 \$560 \$20,560
2 \$20,560 \$560 \$21,120
3 \$21,120 \$560 \$21,680
4 \$21,680 \$560 \$22,240
5 \$22,240 \$560 \$22,800
The difference between the \$22,961 with compound interest doesn't seem like that much more than \$22,800, with simple interest. But flash forward to the end of 23 years under this scenario.
With compound interest, you will have nearly doubled your initial investment to \$37,746, while with simple interest you're looking at just \$32,880. At the end of the 50th year, compound interest will have amassed \$79,558, versus \$48,000 with simple interest over the same period.
Now just imagine the power of compounding interest, if you were making regular contributions to that savings account.
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Compound interest and saving
Here's the good news: Compound interest is far more common than simple interest when you're talking about saving. There are a couple types of savings vehicles you can open to make compound interest work for you.
Guaranteed investment certificates (GICs) are probably one of the safest places for you to keep your money and watch it grow. GICs offer guaranteed high interest rates but lock your funds up for a fixed term.
These terms vary anywhere from 90 days to five years or more, and there are penalties for withdrawing early.
Once the term has ended, you have the option to renew and keep your money growing in the GIC or take it out. All previous accumulations will roll over into the next fixed term.
The locked nature of GICs can be restrictive, especially if you are planning to open an emergency savings account you'll need quick access to.
You also won't be able to deposit more funds into the GIC until it matures and is up for renewal.
High-yield savings accounts will give you flexibility to withdraw and deposit funds whenever you wish, and the accounts provide solid returns over time.
To keep your short-term savings more agile, a high-yield savings account might work best.
Be like Buffett
When you talk about successful investing, U.S. investing giant Warren Buffett is a shining example of how you can grow your money through compounding.
"My wealth has come from a combination of living in America, some lucky genes, and compound interest," he once wrote (CNN).
And if you take a look at examples of Buffett's portfolios, you'll notice he has mostly invested in stocks that pay dividends, which offer another form of compounding.
My wealth has come from a combination of living in America, some lucky genes, and compound interest." — Warren Buffett
One of the best things you can do, if you're OK with a little risk, is invest in a combination of stocks, mutual funds, and ETFs that pay dividends. A popular ETF, or exchange-traded fund, in Canada is the Vanguard VFV, which tracks the U.S. S&P 500 Index.
Reinvested dividends that you earn will generate further dividends, and the compounding dance continues.
Let's take a look at the historical data for the S&P 500 as an example. Say you invested \$100,000 from 2013 to 2017 in an index fund that mimics the S&P 500.
The following table displays what that would look like without (w/o) , and with (w) dividend reinvestment:
S&P 500 Index Dividend Reinvestment Gains
Year % Gain w/o % Gain w Final Balance w/o Final Balance w
2013 27.54% 30.50% \$127,540 \$130,500
2014 10.68% 12.94% \$141,161 \$147,387
2015 -1.44% 0.58% \$139,129 \$148,242
2016 7.32% 9.66% \$149,313 \$162,562
2017 16.95% 19.42% \$174,621 \$194,131
Reinvesting dividends cushioned your loss in 2015, and led to you nearly doubling your initial investment over the course of five years.
If you are regularly contributing to your stock portfolio in a well-thought-out approach, and reinvesting your dividends, the growth on average of your portfolio has the potential to look very green.
In the long term, maybe you won't see same the sort of growth that Buffett has over his storied career. But he can attribute his great success to smart investing, and relying on the power of compounding. And you'll be able to do the same.
If investing on your own intimidates you, there are many new technologies that do the heavy lifting for you and cut the cost of investment fees.
Wealthsimple is an automated investment service that makes automatic adjustments in your portfolio in response to risk, so you don't have to worry. But if you'd still like to discuss your investment mix, there are portfolio managers on hand to address your concerns.
Whether you're saving in the short term, or investing for the long haul, compound interest is your best friend. So give it a hug, and let it work for you.
With CIBC Investor's Edge, kick-start your portfolio with 100 free trades and up to \$4,500 cash back.
Rudro is an editor with Money.ca. Rudro had previously served as Managing Editor of Oola, and as the Content Lead of Tickld before that. Rudro holds a Bachelor of Science in Psychology from the University of Toronto.
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# Thread: Are These 2 Linear Equations Correct?
1. ## Are These 2 Linear Equations Correct?
The First One: 8(5-3x)-4(2+3x)=3
Expansion: -24x -12x 40-8 =3
Combine:-36x 32 =3
Shift: -36x = 3-32= -29
Solution: x= -29/-36
The Second One: 9(1+x)-8(x+2)= 2x
Expansion: 9x-8x 9-16 =2x
Combine: x -7 =2x
Shift: (x-2x) -7= 2x -2x
-x = -7
Solution: x= -7/-1
Is my formula for solving these equations correct (there are other ways but I thought my method is the easiest, or am I wrong?)? And are my solutions correct? If they're wrong, could you show me where I went wrong? Thanks guys!
2. ## Re: Are These 2 Linear Equations Correct?
These are very hard to read as you keep forgetting plus and minus signs! But your logic is fine. The first is correct. The second is not.
You are correct up to \displaystyle \displaystyle \begin{align*} x - 7 = 2x \end{align*}, now subtract \displaystyle \displaystyle \begin{align*} x \end{align*} from both sides to find \displaystyle \displaystyle \begin{align*} -7 = x \end{align*}.
3. ## Re: Are These 2 Linear Equations Correct?
Originally Posted by Prove It
These are very hard to read as you keep forgetting plus and minus signs! But your logic is fine. The first is correct. The second is not.
You are correct up to \displaystyle \displaystyle \begin{align*} x - 7 = 2x \end{align*}, now subtract \displaystyle \displaystyle \begin{align*} x \end{align*} from both sides to find \displaystyle \displaystyle \begin{align*} -7 = x \end{align*}.
For the second one, WolframAlpha says the answer is x=7/-1???
4. ## Re: Are These 2 Linear Equations Correct?
Originally Posted by MathClown
For the second one, WolframAlpha says the answer is x=7/-1???
Which is equal to -7...
5. ## Re: Are These 2 Linear Equations Correct?
Originally Posted by Prove It
Which is equal to -7...
So WolframAlpha got it wrong? The final simplified answer is just x=-7? Is that what you're getting at? Or simply -7 since you're subtracting both x variables.
6. ## Re: Are These 2 Linear Equations Correct?
Originally Posted by MathClown
So WolframAlpha got it wrong? The final simplified answer is just x=-7? Is that what you're getting at? Or simply -7 since you're subtracting both x variables.
Remember that when positive number is divided or multiplied by negative number, answer is negative. Wolfram alpha gave you $\displaystyle \frac{7}{-1}=-7$. So both Wolfram Alpha and Prove It are correct.
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HW_4_Solutions_F05_1
# HW_4_Solutions_F05_1 - Physics 112 HW#4 Solutions 4-26[Step...
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- 1 - Physics 112 - HW #4 Solutions Fall 2005 4-26 [Step on it!] Assuming the truck is accelerating to the right ( ), as shown: (a) Forces on Box: Forces on Truck: f 1 N 1 w = m g B B (normal contact force due to Truck) (friction contact force due to Truck) (weight due to Earth) f 2 N 2 N R f R w = m g T T (normal contact force due to Box) (friction contact force due to Box) (friction contact force due to Road) (drag force due to Air and Road) f d (weight due to Earth) (normal contact force due to Road) NOTE CAREFULLY!: * The kinetic (sliding) friction forces f 1 and f 2 are in the directions shown because the bottom surface of the Box is sliding to the left ( ) relative to the surface of the Truck with which it is in contact. Remember, the accelerating truck is not an inertial reference frame. * N 2 , the normal contact force acting on the Truck due to the Box, is NOT the same thing as the weight of the Box, which is the gravitational force acting on the Box due to the Earth. These are conceptually different forces for different interactions and must be treated as such! (b) There are two 3rd Law interaction pairs among the forces shown above: (i) N 1 and N 2 (normal contact interaction between the Box and Truck) (ii) f 1 and f 2 (frictional contact interaction between the Box and Truck) a
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- 2 - 4-31 [Force is a Vector] We need to sum vectors F 1 and F 2 to form the vector R . F 1 + F 2 + R x components: F 2 cos θ = R (1) y components: F 2 sin θ – F 1 = 0, or F 2 sin θ = F 1 (2) Divide eq (2) by eq (1): tan θ = F 1 R F 1 /R = 1300 N 1300 N = 1 So θ = 45 ° and the angle between F 1 and F 2 is 135 ° Now use eq (1): F 2 cos θ = R F 2 = R cos θ = 1300 N cos 45 ° = 1840 N (Note: drawing a good vector diagram is the key to solving this type of problem easily.) 4-39 [Pulling Crates] (a) 2.50 m/s 2 (Both crates move together.) T N 1 m g 1 (normal contact force due to surface) (tension force due to rope (weight due to Earth) +x +y T N
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# Trying to understand relativity
1. Jul 26, 2012
### kr75
I am a novice in relativity as of now and I'm still learning about it. Currently reading Feynman's lectures and an example has confused me. So here it is,
Consider a spaceship moving at a velocity u. If a light source is placed equidistant from two detectors at opposite ends, one in front of it and the other behind it (with respect to the direction of motion). Then for a person inside the spaceship, a ray of light reaches both detectors at the same time but for a person observing from outside the spaceship the ray of light takes more time to reach the detector in front of it than the one behind it (as the detector in front is seemingly moving away from the source while the one behind is moving towards it).
I do not fully understand this concept. I'd be glad if someone could explain it to me
2. Jul 26, 2012
### PAllen
It is called relativity of simultaneity. The concept of 'at the same time' is not a universal truth, and cannot even be perceived over any significant distance without some measurement convention. For a reasonable convention, we find, that in our universe, if two observers are in relative motion, they will disagree on which 'events' are 'at the same time'. Thus, using normal measurement conventions, both rocket observer and home observer see light as traveling the same speed, but they disagree on which events are 'at the same time'. For the home observer, the 'arrival at back' event occurs before the 'arrival at front event. Inside the rocket, these same two events are considered simultaneous.
3. Jul 26, 2012
### kr75
Ok but i'm still confused as to why for a person outside the spaceship, the ray of light reaches the detector behind it before it reaches the detector in front of it. Since the spaceship is moving at a velocity 'u' wouldn't the light source and the detectors be moving with the same velocity too? Or is my approach to understanding this wrong fundamentally?
4. Jul 26, 2012
### Staff: Mentor
- The spaceship is moving at a velocity u relative to the observer outside the ship. That's a critical detail, so important that it cannot be left unsaid.
- The speed of the source is irrelevant to the speed of the ray of light and to the speed of the detector relative to the observer outside the ship. The different-time effect is caused by the fact that both observers see the ray of light traveling at the same speed (speed of light in vacuum is the same for all observers) but covering different distances. Both agree that the ray of light was emitted at the same place and time, right under the noses of both observers - but because one of the observers sees the detectors changing position while the ray is flight, and the other doesn't, they measure different distances traversed. Same speed, different distance... Different travel time.
5. Jul 26, 2012
### Jimmy
Relativity of Simultaneity:
Last edited by a moderator: Sep 25, 2014
6. Jul 26, 2012
### kr75
ahh now I understand better. I was making a rudimentary mistake while thinking about it. Thanks
7. Jul 26, 2012
### kr75
8. Jul 26, 2012
### ghwellsjr
For the person inside the spaceship, the reason the ray of light reaches both detectors at the same time is purely and only because the person has defined the same time to be when the two rays hit the two detectors. Let's back up a bit. In order for the spaceman to determine when the light rays hit the two detectors, he must have timing devices installed in the two detectors that record the times when the rays hit them. He performs his experiment and then he goes to one detector and sees what time is recorded on it. He goes to the other detector and sees what time is recorded on it. Chances are, the first time he does this, they will not record exactly the same time. So what he does is tweak the time on one of them by whatever amount the difference in the two readings was and then he repeats the experiment, and whatayahknow, the times are now the same.
Now what about the guy outside the spaceship. He can't do the same experiment, because he's not on the spaceship so he sets up his own detectors with timers that he adjusts by the same technique. The only problem is that he needs a bunch more detectors because he doesn't know where the spaceship will be when it flies by. And he also needs to set up all the timers so that no matter where the light rays start from, each pair of detectors at equal distances from the light source will have to read the same time.
Now there is one more added complication. He needs to have camcorders placed at every detector because he doesn't just want to know when the light rays get to each detector, he also wants to know which two of his detectors were adjacent to the two detectors on the spaceship when the rays hit them.
It will take him awhile to do this but when he gets done he is ready for the spaceship to come flying by. When it does, and the light source emits the two rays in opposite directions, this will be recorded by whichever camcorder is closest to the light source. We'll call this camcorder 'S' (for source). Then as the rays travel outwards in both directions, each pair of his detectors that are equal distant from the first one will record equal times, all the way out to the end. However, since the spaceship is moving while this is happening, the detector at the rear of the spaceship will see the light ray first and it will be recorded on a stationary camcorder nearby. We'll call this camcorder 'R' (for rear). Later on, the other ray will eventually catch up to the detector at the front of the spaceship and this will be recorded by another camcorder which we will call 'F' (for front).
Now it should be obvious that the time on 'R' will be earlier than the time on 'F' even though the times on the spaceships two detectors will read the same time as each other, but not the same times as what the stationary detectors record. Each camcorder records both the time on the stationary detector and the time on the adjacent spaceship detector but only two camcorders capture the events of the light rays hitting the spaceships detectors. And although they will record the spaceship's two detectors having the same time on them, the two stationary detectors will have different times on them.
9. Jul 27, 2012
### kr75
yes that does help. Of course i will have to go over it again till I understand perfectly by myself, but this helps me put it into perspective in a better way. Thanks George
10. Jul 27, 2012
### phyti
No one needs clocks.
Using a basic space-time diagram, the outsider concludes a long path, short path to the front and the reverse to the back, in equal times.
The anaut is only coincident with the emission and detection of the return signals which are simultaneous for him. Not knowing his speed in space, he cannot determine when or where the reflections occurred. Using the SR clock convention, he assumes the paths to be equal.
11. Jul 27, 2012
### ghwellsjr
The scenario under consideration in this thread does not have any return signals and there are no reflections.
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# If the plane of the rectangular coil is parallel to the magnetic field (i.e. radial magnetic field), the torque on the coil is?
A. τ = NIAB cosα
B. τ = NIAB sinα
C. Both a and b
D. τ = NIAB
Right Answer: D. τ = NIAB
December 27, 2017 Hassan Ali Physics
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# Decision Mathematics D1 Advanced/Advanced Subsidiary. Tuesday 5 June 2007 Afternoon Time: 1 hour 30 minutes
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## Transcription
2 Write your answers in the D1 answer book for this paper. 1. Explain what is meant by a planar graph. (Total 2 marks) 2. A 1a A 1a E 1b E 1b H 2 H 2 J 3 J 3 L 4 L 4 M 5 M 5 Figure 1 Figure 2 Six workers, Annie, Emma, Hannah, Jerry, Louis and Morand, are to be assigned to five tasks, 1,2,3,4 and 5. For safety reasons, task 1 must be done by two people working together. A bipartite graph showing the possible allocations of the workers is given in Figure 1 and an initial matching is given in Figure 2. The maximum matching algorithm will be used to obtain a complete matching. (a) Although there are five tasks, six vertices have been created on the right hand side of each bipartite graph. Explain why this is necessary when applying this algorithm. (b) Find an alternating path and the complete matching it gives. (3) Hannah is now unable to do task 5 due to health reasons. (c) Explain why a complete matching is no longer possible. (Total 7 marks) N26122A 2
3 3. Start Let A = 0 Input x, y Is x even? No A = A + y Yes x = x - 1 x = x 2 No Is x = 0? Yes y = 2y Output A Stop Figure 3 An algorithm is described by the flow chart shown in Figure 3. (a) Given that x = 54 and y = 63, complete the table in the answer book to show the results obtained at each step when the algorithm is applied. (7) (b) State what the algorithm achieves. (Total 9 marks) N26122A 3 Turn over
4 4. C F 11 I 12 8 B 6 11 E 12 H 11 A 15 D 9 G Figure 4 Figure 4 models a network of underground tunnels that have to be inspected. The number on each arc represents the length, in km, of each tunnel. Joe must travel along each tunnel at least once and the length of his inspection route must be minimised. The total weight of the network is 125 km. The inspection route must start and finish at A. (a) Use an appropriate algorithm to find the length of the shortest inspection route. You should make your method and working clear. (5) Given that it is now permitted to start and finish the inspection at two distinct vertices, (b) state which two vertices should be chosen to minimise the length of the new route. Give a reason for your answer. (Total 7 marks) N26122A 4
5 5. M A B C D E M A B C D E The table shows the cost, in pounds, of linking five automatic alarm sensors, A,B,C,D and E, and the main reception, M. (a) Use Prim s algorithm, starting from M, to find a minimum spanning tree for this table of costs. You must list the arcs that form your tree in the order that they are selected. (3) (b) Draw your tree using the vertices given in Diagram 1 in the answer book. (c) Find the total weight of your tree. (d) Explain why it is not necessary to check for cycles when using Prim s algorithm. (1) (1) (Total 7 marks) N26122A 5 Turn over
6 6. D(11) A() E(4) H(9) K(7) 0 0 B(9) F I(15) J(15) C(11) L(16) 11 G(1) 14 KEY Early event time Late event time Figure 5 The network in Figure 5 shows the activities that need to be undertaken to complete a project. Each activity is represented by an arc. The number in brackets is the duration of the activity in days. The early and late event times are to be shown at each vertex and some have been completed for you. (a) Calculate the missing early and late times and hence complete Diagram 2 in your answer book. (3) (b) List the two critical paths for this network. (c) Explain what is meant by a critical path. N26122A 6
7 The sum of all the activity times is 1 days and each activity requires just one worker. The project must be completed in the minimum time. (d) Calculate a lower bound for the number of workers needed to complete the project in the minimum time. You must show your working. (e) List the activities that must be happening on day 20. (f) Comment on your answer to part (e) with regard to the lower bound you found in part (d). (1) (g) Schedule the activities, using the minimum number of workers, so that the project is completed in 30 days. (3) (Total 15 marks) N26122A 7 Turn over
8 7. The tableau below is the initial tableau for a linear programming problem in x, y and z. The objective is to maximise the profit, P. basic variable x y z r s t Value r s t P Using the information in the tableau, write down (a) the objective function, (b) the three constraints as inequalities with integer coefficients. (3) Taking the most negative number in the profit row to indicate the pivot column at each stage, (c) solve this linear programming problem. Make your method clear by stating the row operations you use. (9) (d) State the final values of the objective function and each variable. (3) One of the constraints is not at capacity. (e) Explain how it can be identified. (1) (Total 18 marks) N26122A 8
9 8. C 1 C H C I S A D F J T B G E C 2 C 1 Figure 6 Figure 6 shows a capacitated, directed network. The number on each arc represents the capacity of that arc. The numbers in circles represent an initial flow. (a) State the value of the initial flow. (b) State the capacities of cuts C 1 and C 2. (1) Diagram 3 in the answer book shows the labelling procedure applied to the above network. (c) Using Diagram 3, increase the flow by a further 19 units. You must list each flow-augmenting path you use, together with its flow. (5) (d) Prove that the flow is now maximal. END (Total marks) TOTAL FOR PAPER: 75 MARKS N26122A 9
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Write your name here Surname Other names Centre Number Candidate Number Edexcel GCE Drama and Theatre Studies Advanced Unit 4: Theatre Text in Context Wednesday 9 June 2010 Morning Time: 2 hours 30 minutes
### Sport Timetabling. Outline DM87 SCHEDULING, TIMETABLING AND ROUTING. Outline. Lecture 15. 1. Problem Definitions
Outline DM87 SCHEDULING, TIMETABLING AND ROUTING Lecture 15 Sport Timetabling 1. Problem Definitions Marco Chiarandini DM87 Scheduling, Timetabling and Routing 2 Problems we treat: single and double round-robin
### Wednesday 5 November 2014 Morning
H Wednesday 5 November 2014 Morning GCSE MATHEMATICS B J567/03 Paper 3 (Higher Tier) * 1 1 8 3 2 9 5 6 3 5 * Candidates answer on the Question Paper. OCR supplied materials: None Other materials required:
### Paper 2 Revision. (compiled in light of the contents of paper1) Higher Tier Edexcel
Paper 2 Revision (compiled in light of the contents of paper1) Higher Tier Edexcel 1 Topic Areas 1. Data Handling 2. Number 3. Shape, Space and Measure 4. Algebra 2 Data Handling Averages Two-way table
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# How you calculate the temperature scaling factor?
Used to obtain the corresponding pressure at ambient temperature. Example when testing Proof or burst pressure to a given value at 300°F, but you will test with corresponding pressure (higher) to ambient temperature.
Example. Proof Supply Pressure @ 300°F = 1161 psig Proof Supply Pressure @ AMBIENT = 1161 X Temperature Scaling Factor = 1161 X 1.15 = 1335 psig
Burst Supply pressure @ 300°F = 1251 psig
Burst Supply pressure @ AMBIENT = 1251 X Temperature Scaling Factor
= 1251 X 1.24
= 1551 psig
Please note different values for TSF for Proof and Burst pressures. Are these values obtained from the material the part is made off? Is there is a calculation, what it is?
• This could be a straightforward application of the ideal gas law or somehting else entirely. Can you tell us more about the problem and possibly the relevant codes etc. you work with? The question was migrated to a site where not everyone is an aerospace person!
– mart
Aug 12, 2020 at 8:53
• Material properties depend on temperature. The proof strength and ultimate tensile strength of the material at room temperature will be higher than at 300F, but they probably don't scale in the same proportion. I don't think this has anything to do with gas laws. We don't know the material specification so there is no way to tell how the 15% and 24% factors were derived. Aug 12, 2020 at 12:28
This is the essence of the ASME Boiler Code ;allowable stresses for materials at various temperatures. It is NOT a calculation . It is based on a great many creep and stress rupture tests which have been analyzed and reviewed for over a hundred years.If you are building a pressure vessel in the civilized world ,it is required to conform to "code" , or a local national interpretation. The code addresses all aspects of the pressure vessels such as welding, inspection,quality control , piping , nuclear pressure vessels, hazardous materials, etc,etc,etc. Some aspects of quality are now handled through ISO 9000. The code roots go back well past 100 years.
• Wikipedia list the first issue of code in 1914 but that is misleading. Steam boiler explosions were unfortunately common and in the 1880's efforts began to write a document to define safe (steam) pressure vessels , particularly on locomotives. Apparently it took about 20 years to finish the first "code". The ASTM material specifications initially were developed for the ASME code. Aug 12, 2020 at 19:47
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# [Get Solution] Linear Regression
Using stocks.csvPreview the document from the previous assignment:Required:Define your dependent and independent variables (pick the column you want to predict and a column for your feature(X))This should be something that has a hope of being linear, from the last assignment we saw not much correlated with tomorrow’s percent change. So that would be a bad Y. Try predicting low,high,volume, etcSplit your data into training and testing sets. Use 20% of the data for testingTrain(fit) a Linear Regression modelReport the square errorFor the heck of it if you want:Use pyplot to plot your X and Y values, along with your predictionPreviousNext
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# Scintillating Snippets: Reconstructing Position From Depth
There are times I wish I’d never responded to this thread over at GDnet, simply because of the constant stream of PM’s that I still get about it. Wouldn’t it be nice if I could just pull out all the important bits, stick it on some blog, and then link everyone to it? You’re right, it would be!
First things first: what am I talking about? I’m talking about something that finds great use for deferred rendering: reconstructing the 3D position of a previously-rendered pixel (either in view-space or world-space) from a single depth value. In practice, it’s really not terribly complicated. You intrinsically know (or can figure out) the 2D position of any pixel when you’re shading it, which means that if you can sample a depth value you can get the whole 3D position. However it’s still easy to get tripped up due to the fact that there’s several ways to go about it, coupled with the fact that many beginners aren’t very proficient at debugging their shaders.
Let’s talk about the first way to do it: storing post-projection z/w, combining it with x/w and y/w, transforming by the inverse of the projection matrix, and dividing by w. In HLSL it looks something like this…
```// Depth pass vertex shader
output.vPositionCS = mul(input.vPositionOS, g_matWorldViewProj);
output.vDepthCS.xy = output.vPositionCS.zw;
// Depth pass pixel shader (output z/w)
return input.vDepthCS.x / input.vDepthVS.y;
// Function for converting depth to view-space position
// in deferred pixel shader pass. vTexCoord is a texture
// coordinate for a full-screen quad, such that x=0 is the
// left of the screen, and y=0 is the top of the screen.
float3 VSPositionFromDepth(float2 vTexCoord)
{
// Get the depth value for this pixel
float z = tex2D(DepthSampler, vTexCoord);
// Get x/w and y/w from the viewport position
float x = vTexCoord.x * 2 - 1;
float y = (1 - vTexCoord.y) * 2 - 1;
float4 vProjectedPos = float4(x, y, z, 1.0f);
// Transform by the inverse projection matrix
float4 vPositionVS = mul(vProjectedPos, g_matInvProjection);
// Divide by w to get the view-space position
return vPositionVS.xyz / vPositionVS.w;
}```
For many this is the preferred approach since it works with hardware depth buffers. It also may seem natural to some: we get depth by projection, we get position by un-projecting. But what if we don’t have access to a hardware depth buffer? If you’re targeting the PC and D3D9, sampling from a depth buffer as if it were a texture is not straightforward since it requires driver hacks. If you’re using XNA, it’s not possible at all since the framework generally attempts to main cross-plaftorm compatibility between the PC and the Xbox 360. In these cases, we can simply render out a depth buffer ourselves using the vertex and pixel shader bits I posted above. But is this really a good idea? z/w is non-linear, and most of the precision will be dedicated to areas very close to the near-clip plane.
A different approach would be to render out normalized view-space z as our depth. Since it’s view-space it’s linear which means we get uniform precision distribution, and this also means we don’t need to bother with projection or unprojection to reconstruct position. Instead we can take the approach of CryTek and multiply the depth value with a ray pointing from the camera to the far-clip plane. In HLSL it goes something like this:
```// Shaders for rendering linear depth
void DepthVS( in float4 in_vPositionOS : POSITION,
out float4 out_vPositionCS : POSITION,
out float out_fDepthVS : TEXCOORD0 )
{
// Figure out the position of the vertex in
// view space and clip space
float4x4 matWorldView = mul(g_matWorld, g_matView);
float4 vPositionVS = mul(in_vPositionOS, matWorldView);
out_vPositionCS = mul(vPositionVS, g_matProj);
out_fDepthVS = vPositionVS.z;
}
float4 DepthPS(in float in_fDepthVS : TEXCOORD0) : COLOR0
{
// Negate and divide by distance to far-clip plane
// (so that depth is in range [0,1])
// This is for right-handed coordinate system,
// for left-handed negating is not necessary.
float fDepth = -in_fDepthVS/g_fFarClip;
return float4(fDepth, 1.0f, 1.0f, 1.0f);
}
// Shaders for deferred pass where position is reconstructed
void QuadVS ( in float3 in_vPositionOS : POSITION,
in float3 in_vTexCoordAndCornerIndex : TEXCOORD0,
out float4 out_vPositionCS : POSITION,
out float2 out_vTexCoord : TEXCOORD0,
out float3 out_vFrustumCornerVS : TEXCOORD1 )
{
// Offset the position by half a pixel to correctly
// align texels to pixels. Only necessary for D3D9 or XNA
out_vPositionCS.x = in_vPositionOS.x - (1.0f/g_vOcclusionTextureSize.x);
out_vPositionCS.y = in_vPositionOS.y + (1.0f/g_vOcclusionTextureSize.y);
out_vPositionCS.z = in_vPositionOS.z;
out_vPositionCS.w = 1.0f;
// Pass along the texture coordinate and the position
// of the frustum corner in view-space. This frustum corner
// position is interpolated so that the pixel shader always
// has a ray from camera->far-clip plane
out_vTexCoord = in_vTexCoordAndCornerIndex.xy;
out_vFrustumCornerVS = g_vFrustumCornersVS[in_vTexCoordAndCornerIndex.z];
}
// Pixel shader function for reconstructing view-space position
float3 VSPositionFromDepth(float2 vTexCoord, float3 vFrustumRayVS)
{
float fPixelDepth = tex2D(DepthSampler, vTexCoord).r;
return fPixelDepth * vFrustumRayVS;
}```
As you can see the reconstruction is quite nice with linear depth, we only need a single multiply instead of the 4 MADD’s and a divide needed for unprojection. If you’re curious on how to get the frustum corner position I use, it’s rather easy with a little trig. This tutorial walks you through it. Or if you’re using XNA, there’s a super-convient BoundingFrustum class that can take care of it for you. My code for getting the positions looks something like this:
```Matrix viewProjMatrix = viewMatrix * projMatrix;
BoundingFrustum frustum = new BoundingFrustum(viewProjMatrix);
frustum.GetCorners(frustumCornersWS);
Vector3.Transform(frustumCornersWS, ref viewMatrix, frustumCornersVS);
for (int i = 0; i < 4; i++)
farFrustumCornersVS[i] = frustumCornersVS[i + 4];```
The farFrustumCornersVS array is what I send to my vertex shader as shader constants. Then you just need to have an index in your quad vertices that tells you which vertex belongs to which corner (which you could also do with shader math, if you want). Another approach would be to simply store the corner positions directly in the vertices as texCoord’s.
Extra Credit: this technique can also be used to to reconstruct world-space position, if that’s what you’re after. All you need to do is rotate (not translate) your frustum corner positions by the inverse of your view matrix to get them back into world space. Then when you multiply the interpolated ray with your depth value, you simply add the camera position to the value (ends up being a single MADD).
Extra-Extra Credit: you can use this technique with arbitrary geometry too, not just quads. You just need to figure out a texture coordinate for each pixel, which you can do by either interpolating the clip-space position and dividing x and y by w, or by using the VPOS semantic. Then for your frustum ray you just calculate the eye->vertex vector and scale it so that it points all the way back to the far-clip plane.
UPDATE: Answers to extra credit questions here
1. Dave says:
Thank you for your beautiful article, it’s exactly what I needed!
Just 2 things:
1) I need to reconstruct the world space position… can you explain a bit more the process? What do you mean for “rotate the frustm corners by the matrix” ?
2) For point lights I use a sphere mesh to approximate the light volume. How can I build the world position from that? I mean, using a full screen quad , I can easily compute the current frustum corner..but with a sphere?
Again, thank you my friend!
2. Thank you for posting this all together! I can delete my 6 book marks to different parts of it now :)
Cheers,
Greg
3. Very interesting, thanks.
All that’s missing to make this perfect is an explanation of how to map a position in view space back into screen/texture space again.
My brain melts each time I try to do this and it’s stopping me from implementing a better SSAO algorithm than I have now.
4. Hi Paul. I replied in that thread on the XNA forums, but I’ll reply here too.
In my SSAO shader I just take the view-space position, transform it by the projection matrix, and then use this function to get a texture coordinate:
// Gets the screen-space texel coord from clip-space position
float2 CalcSSTexCoord (float4 vPositionCS)
{
float2 vSSTexCoord = vPositionCS.xy / vPositionCS.w;
vSSTexCoord = vSSTexCoord * 0.5f + 0.5f;
vSSTexCoord.y = 1.0f – vSSTexCoord.y;
vSSTexCoord += 0.5f / g_vTexDimensions;
return vSSTexCoord;
}
g_vTexDimensions would be the dimensions of the texture you’re sampling. There might be a cheaper way of doing this, but projecting definitely works.
5. :) Thanks. My email’s been off over the weekend.
6. Nik says:
Sorry for doubting your article, but I think that the projective transform is not inversible in general – you loose information when you project. Having depth buffer values helps to restore some of the information, but since depth map has a certain resolution then this approach will not work if the camera is far away from the projected object. Most of the object may have very similar depth value in the depth map and so it will look flat after “unprojecting”. I think you should point this out, since it is misleading to say that projective transforms can be undone in general…
7. Right, if you store perspective z/w in most cases you’ll have inadequate precision in areas closer to the far clip plane. But it’s really a distribution of precision issue more than anything, due to the non-linear curve of z/w. My latest blog (http://mynameismjp.wordpress.com/2010/03/22/attack-of-the-depth-buffer/) demonstrates the sort of error you can expect for different depth formats…perhaps I’ll link it here so that people are aware of this issue. Thanks for bringing it up. :)
8. Phil says:
It looks like this technique of using the far bounding frustum coordinates will only work if you are using a perspective projection, is that right?
I have a parallel projection, and multiplying the depth by the frustum ray just doesn’t seem to make sense.
9. thank you for your post.
can i translate into korean and put on my blog?
10. Phil: indeed the technique is meant for a perspective projection. For an ortho projection it’s unnecessary since you can directly calculate view-space X and Y coordinates based on your clip-space XY and your projection parameters. You can do the same for view-space Z once you’ve sampled it from your depth buffer.
ozael: absolutely, that’s not problem at all.
11. mystran says:
The way I wrote this yesterday, instead of using constants for the frustum corners, I just did mul(float4(x,y,1,1), mInvProj) per vertex to get far-plane positions in view-space. Then in pixel shader multiply with stored (floating point) view-z/farPlane.
At least visually it looks like it should and saves you from having to work out the frustum coordinates on CPU, so you could draw large number of screen-aligned triangles together, yet still do the unprojection work only once-per-vertex.
12. Orcun says:
Hi MPJ,
I have spend nearly a week while reading your thread in gamedev and trying several ssao implamentations.
I’m trying to make a ssao protetype in render monkey before integrating it in to the engine that I working on. Currently my engine does not have developer friendly shader management facilities so I’m doing my best to make the effect as good as posible bofore integration.
I haver read and understood all of the methods that you mentioned hoverever the biggest piece is missing. I can not calculate and feed view frustum corners to vertex shader because of rendermonkey. I’m trying to generate them by using screen space quad mesh (which is a simple 3d rectangle with dimensions) vertices and I can not get result. Can you write how to calculate view vectors by using only screen aligned quad in vertex shader.
Here is my code which does not work;
VS_OUTPUT vs_main(VS_INPUT Input)
{
VS_OUTPUT Output;
Output.Position = mul(Input.Position + float4(ModelPosition.xyz, 0.0f), ViewProjectionMatrix);
float3 ObjectPosition = mul(Input.Position, ViewMatrix);
Output._Position = ObjectPosition;
Output.NormalDepth.xyz = mul(Input.Normal, ViewMatrix);
Output.NormalDepth.w = ObjectPosition.z;
return Output;
}
PS_OUTPUT ps_main( PS_INPUT Input )
{
PS_OUTPUT Output;
Output.Position = float4(Input.Position, 1.0f);
Output.NormalDepth.xyz = Input.NormalDepth.xyz;
Output.NormalDepth.w = Input.NormalDepth.w / FarZ;
return Output;
}
VS_OUTPUT vs_main(float4 Position : POSITION, float2 Texcoord : TEXCOORD0)
{
VS_OUTPUT Output;
Output.Position = float4(Position.xy, 0.0f, 1.0f) + float4(-PixelSize.x, PixelSize.y, 0.0f, 0.0f);
Output.ScreenTexcoord = Texcoord;
Output.ViewVector.x = Position.x * tan(FOV / 2) * (ScreenSize.x / ScreenSize.y);
Output.ViewVector.y = Position.y * tan(FOV / 2);
Output.ViewVector.z = 1;
return Output;
}
float3 GetViewPosition(in float2 Texcoord, in float3 ViewVector)
{
return ViewVector * tex2D(NormalDepthInput, Texcoord).w;
}
Thanks a lot.
13. Camus says:
Hi, the second method of render z on view space as depth, can be used to store shadow depth?
Because I think it will solve in part the bias problem, since it is lineal the distribution, a constant bias will be enough to fix it, seems too good to be true, I’m missing something?
14. If you’re manually writing out shadow depth to a render target, then you can use whatever depth metric you’d like. Linear Z/FarClip definitely works for that purpose. It’s only an issue if you only want to render to a depth buffer, in which case you don’t have a choice but to use z/w.
15. Camus says:
Thank you so much, I will run some tests :)
16. Camus says:
It works:
Thanks for open my eyes, behold the linearity.
17. Camus says:
Me again, I figured out how to render to a depth buffer without using z/w. The pixel fragment should write zero as always, the vertex fragment change from this:
Out.position = mul(In.position, WorldViewProj);
to this:
float4 vpos = mul(In.position, WorldViewProj);
vpos.z = (vpos.z*vpos.w)/FarPlane;
Out.position = vpos;
It allows me to perform the depth comparison by hardware, I still don’t understand why everybody, even the SDK shadow map sample uses z/w, there is even papers written with the purpose of fixing the z-fighting in shadow map.
18. Messing with your z value like that can (and will) screw up rasterization, early z-cull, and z compression (which is why people don’t do it).
One workable alternative is to use a floating point depth buffer, and flip the near and far planes of your projection (and also flip your depth test directions). When you do that, the non-linear distribution of precision in a floating point value *mostly* cancels out the non-linearity of z/w. That helps with precision, but not with the issue of applying a non-uniform bias.
19. Camus says:
Hi mpettineo, again, I just want to point out about the frustum corners method, it is valid only for the current pixel, can’t be used directly to read from neighbor samples, because of the interpolation, it can be modified doing a manual lerp:
lerp(-corner.x,corner.x,uv.x);
lerp(-corner.y,corner.y,uv.y);
I was into it the whole week trying to figure out that. Because many people are getting confused with this.
Also, as you said, Z/FarPlane, can be used only if it’s done in the pixel program, interpolated by the vertex program falls into a nonlinear function with werid results, I don’t know why this page is still alive:
http://www.mvps.org/directx/articles/linear_z/linearz.htm
IT IS WRONG!, thanks for your time anyway, has been very helpful :)
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# Find the speed at the bottom of it's swing
• Dark Visitor
In summary, a simple pendulum with a length of 2 meters is released with an initial velocity from an angle of 25 degrees from the vertical. Using conservation of energy, the speed at the bottom of the swing can be found by calculating the change in height from the geometry and the potential energy. The final speed is found to be 2.3 m/s.
Dark Visitor
A simple pendulum, 2 m in length, is released with a push (i.e. - it has an initial velocity) when the support string is at an angle of 25 degrees from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, use conservation of energy to find it's speed at the bottom of the swing.
* 2.3 m/s
* 2.6 m/s
* 2 m/s
* .5 m/s
I don't really understand how to use conservation of energy that well, so I need help. How would I apply that for this problem? And how do I solve from there? Any help would be much appreciated. I need it by tomorrow night. Thanks.
Do you understand what would happen if there were no push? Energy is continually transferred back and forth between kinetic and potential, but the sum is constant at all times.
First, find the height of the bob above its position at the bottom from the geometr, ie the vertical displacement between 25 degrees and vertical.
Keep track of what the item's total energy is (PE+KE) and this cannot change. At the bottom of the swing all the energy is kinetic (since it's convenient to define PE=0 at the bottom of the swing in this case). :)
What do you mean by the bob and geometer? And I would think the height would always remain 2 m due to the length.
To Matterwave: You're saying that at the bottom when PE = 0, that KE will still equal all of the energy that was there when PE did not equal 0?
Dark Visitor said:
What do you mean by the bob and geometer? And I would think the height would always remain 2 m due to the length.
To Matterwave: You're saying that at the bottom when PE = 0, that KE will still equal all of the energy that was there when PE did not equal 0?
er sorry, that was meant to be geometry and the bob is just the mass on the end. The change in height is given by L(1-cos(25)). Can you see why?
The kinetic energy at the bottom will be the sum of the kinetic energy from the push plus the potential energy from it being displaced by 25 degrees, see above.
It's okay. Well, I see that the cos25 gives me the length on the "x axis" of the pendulum when it is at the 25 degree angle, but I don't know where the 1 comes in, or what L stands for, unless it is length.
Dark Visitor said:
It's okay. Well, I see that the cos25 gives me the length on the "x axis" of the pendulum when it is at the 25 degree angle, but I don't know where the 1 comes in, or what L stands for, unless it is length.
Right L is length = 2m. I'll post a pic here in a minute.
Okay, I think I understand a little more. So the length won't be 2 meters anymore. It will be:
2 m(1 - cos25), which makes it a lot smaller.
My question now is how does this help me get the final speed?
The displacement of the bob vertically is given by L - L(cos(25))=L(1-cos(25))
The potential energy is mgh where h is described as L(1-cos(25)). The total energy is the initial kinetic from the push (1/2mV^2) + the potential energy (mgh). That do it?
Yes, quite a bit. One thing I still don't get is how we can use either of those when we don't have the mass (m) for the equations. Do we assume anything, or do we need to find it?
OOps, I forgot what we we're trying to calculate. Well since all masses fall at the same rate, it doesn't matter. Just add the velocity it would have attained if dropped from that height. Different ways to get this, but V'^2=V^2+2(g*h) is the most straight froward at this point. V' is the final velocity, v is the initial.
Okay:
vf2 = vi2 + 2(g)(h)
= (1.2)2 + 2(9.8)(.1874)
= 5.11304
Then I square-rooted that and got 2.2612, which 2.3 m/s is an answer.
Good job.
Thanks, and thanks for all the help. I really appreciate it. If you have some free time to spare, please help me on some of my other posts. I got a lot of them, and not many people helping me. If you can't, it's fine. Thanks.
Dark Visitor said:
Thanks, and thanks for all the help. I really appreciate it. If you have some free time to spare, please help me on some of my other posts. I got a lot of them, and not many people helping me. If you can't, it's fine. Thanks.
Sure I'll look for them--must be a sunday, things are humming here. ;-D
Yeah. I posted most yesterday to allow time for people to see them and respond, but didn't happen too much. Fortunately, I still have time to complete them all.
## 1. What does "Find the speed at the bottom of its swing" mean?
"Find the speed at the bottom of its swing" refers to determining the velocity of an object at the lowest point of its swinging motion. This can be calculated using the object's mass, length of the swing, and the force of gravity.
## 2. Why is it important to find the speed at the bottom of its swing?
Knowing the speed at the bottom of an object's swing can provide important information about its energy, momentum, and trajectory. This can be useful in understanding the behavior of pendulums, as well as other swinging objects.
## 3. How do you calculate the speed at the bottom of its swing?
The speed at the bottom of an object's swing can be calculated using the formula v = √(2gh), where v is the velocity, g is the acceleration due to gravity, and h is the height of the swing at its lowest point.
## 4. Can the speed at the bottom of its swing be different from the speed at the top?
Yes, the speed at the bottom of its swing can be different from the speed at the top. This is because the object's speed is affected by the force of gravity, which increases as the object moves closer to the ground. Therefore, the object will have a higher speed at the bottom of its swing compared to the top.
## 5. How does the length of the swing affect the speed at the bottom of its swing?
The length of the swing affects the speed at the bottom of its swing by changing the object's potential and kinetic energy. A longer swing will have a higher potential energy at the top, resulting in a higher speed at the bottom. However, the effect of length on speed is not linear, and other factors such as mass and air resistance also play a role.
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This site uses cookies. By continuing to browse the ConceptDraw site you are agreeing to our Use of Site Cookies.
# Euclidean algorithm - Flowchart
## Euclidean algorithm - Flowchart
"In mathematics, the Euclidean algorithm, or Euclid's algorithm, is a method for computing the greatest common divisor (GCD) of two (usually positive) integers, also known as the greatest common factor (GCF) or highest common factor (HCF). ...
The GCD of two positive integers is the largest integer that divides both of them without leaving a remainder (the GCD of two integers in general is defined in a more subtle way).
In its simplest form, Euclid's algorithm starts with a pair of positive integers, and forms a new pair that consists of the smaller number and the difference between the larger and smaller numbers. The process repeats until the numbers in the pair are equal. That number then is the greatest common divisor of the original pair of integers.
The main principle is that the GCD does not change if the smaller number is subtracted from the larger number. ... Since the larger of the two numbers is reduced, repeating this process gives successively smaller numbers, so this repetition will necessarily stop sooner or later - when the numbers are equal (if the process is attempted once more, one of the numbers will become 0)." [Euclidean algorithm. Wikipedia]
The flowchart example "Euclidean algorithm" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Mathematics solution from the Science and Education area of ConceptDraw Solution Park. Read more
Euclid's algorithm flow chart
Used Solutions
## Top 5 Android Flow Chart Apps
For many years CS Odessa corporation has been developing the multifunctional vector diagramming software with supporting sophisticated drawing tools for vector diagramming and design on Macintosh and PC with Windows installed. We feel that our considerable experience gives us an opportunity to look objectively at the current offerings and to offer you our opinion about the main contenders and pretenders. Would you like to know the details about what is going on in the world of Flow Chart drawing applications that support Android? The Top 5 of drawing applications in this space rapidly changes, consequently it is significant understatement to call this application space dynamic. Now, the list of Top 5 popular flowchart makers for Android includes LLNL Flow Charts, Army Flow Charts, DroidDia prime, Note Droid, DroidDia PRO unlocker. These listed business graphics applications are recognized by many experts the most convenient, powerful, and successful for use when drawing on Android devices. Read more
## Solving quadratic equation algorithm - Flowchart
"In elementary algebra, a quadratic equation (from the Latin quadratus for "square") is any equation having the form
ax^2+bx+c=0
where x represents an unknown, and a, b, and c are constants with a not equal to 0. If a = 0, then the equation is linear, not quadratic. The constants a, b, and c are called, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.
Because the quadratic equation involves only one unknown, it is called "univariate". The quadratic equation only contains powers of x that are non-negative integers, and therefore it is a polynomial equation, and in particular it is a second degree polynomial equation since the greatest power is two.
Quadratic equations can be solved by a process known in American English as factoring and in other varieties of English as factorising, by completing the square, by using the quadratic formula, or by graphing." [Quadratic equation. Wikipedia]
The flowchart example "Solving quadratic equation algorithm" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Mathematics solution from the Science and Education area of ConceptDraw Solution Park. Read more
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### Archive
Posts Tagged ‘random numbers’
## Calculating power using Monte Carlo simulations, part 5: Structural equation models
In our last four posts in this series, we showed you how to calculate power for a t test using Monte Carlo simulations, how to integrate your simulations into Stata’s power command, and how to do this for linear and logistic regression models and multilevel models. In today’s post, I’m going to show you how to estimate power for structural equation models (SEM) using simulations.
Our goal is to write a program that will calculate power for a given SEM at different sample sizes. We’ll follow the same general procedure as the previous two posts, but the way we’ll go about simulating data is a bit different. Rather than individually simulating each variable for our specified model, we’ll be simulating all our variables simultaneously from a given covariance matrix. Means for each of the variables can also be used to simulate the data if your SEM has a mean structure, such as in group analysis or growth curve analysis. Read more…
Categories: Statistics Tags:
## Calculating power using Monte Carlo simulations, part 4: Multilevel/longitudinal models
In my last three posts, I showed you how to calculate power for a t test using Monte Carlo simulations, how to integrate your simulations into Stata’s power command, and how to do this for linear and logistic regression models. In today’s post, I’m going to show you how to estimate power for multilevel/longitudinal models using simulations. You may want to review my earlier post titled “How to simulate multilevel/longitudinal data” before you read this post. Read more…
Categories: Statistics Tags:
## Calculating power using Monte Carlo simulations, part 3: Linear and logistic regression
In my last two posts, I showed you how to calculate power for a t test using Monte Carlo simulations and how to integrate your simulations into Stata’s power command. In today’s post, I’m going to show you how to do these tasks for linear and logistic regression models. The strategy and overall structure of the programs for linear and logistic regression are similar to the t test examples. The parts that will change are the simulation of the data and the models used to test the null hypothesis. Read more…
Categories: Statistics Tags:
## Calculating power using Monte Carlo simulations, part 2: Running your simulation using power
In my last post, I showed you how to calculate power for a t test using Monte Carlo simulations. In this post, I will show you how to integrate your simulations into Stata’s power command so that you can easily create custom tables and graphs for a range of parameter values. Read more…
Categories: Statistics Tags:
## Calculating power using Monte Carlo simulations, part 1: The basics
Power and sample-size calculations are an important part of planning a scientific study. You can use Stata’s power commands to calculate power and sample-size requirements for dozens of commonly used statistical tests. But there are no simple formulas for more complex models such as multilevel/longitudinal models and structural equation models (SEMs). Monte Carlo simulations are one way to calculate power and sample-size requirements for complex models, and Stata provides all the tools you need to do this. You can even integrate your simulations into Stata’s power commands so that you can easily create custom tables and graphs for a range of parameter values. Read more…
Categories: Statistics Tags:
## How to generate random numbers in Stata
Overview
I describe how to generate random numbers and discuss some features added in Stata 14. In particular, Stata 14 includes a new default random-number generator (RNG) called the Mersenne Twister (Matsumoto and Nishimura 1998), a new function that generates random integers, the ability to generate random numbers from an interval, and several new functions that generate random variates from nonuniform distributions.
Random numbers from the uniform distribution
In the example below, we use runiform() to create Read more…
Categories: Statistics Tags:
## Using Stata’s random-number generators, part 4, details
For those interested in how pseudo random number generators work, I just wrote something on Statalist which you can see in the Statalist archives by clicking the link even if you do not subscribe:
http://www.stata.com/statalist/archive/2012-10/msg01129.html
To remind you, I’ve been writing about how to use random-number generators in parts 1, 2, and 3, and I still have one more posting I want to write on the subject. What I just wrote on Statalist, however, is about how random-number generators work, and I think you will find it interesting.
To find out more about Statalist, see
Statalist
How to successfully ask a question on Statalist
Categories: Numerical Analysis Tags:
## Using Stata’s random-number generators, part 3, drawing with replacement
The topic for today is drawing random samples with replacement. If you haven’t read part 1 and part 2 of this series on random numbers, do so. In the series we’ve discussed that Read more…
Categories: Numerical Analysis Tags:
## Using Stata’s random-number generators, part 2, drawing without replacement
Last time I told you that Stata’s runiform() function generates rectangularly (uniformly) distributed random numbers over [0, 1), from 0 to nearly 1, and to be precise, over [0, 0.999999999767169356]. And I gave you two formulas,
1. To generate continuous random numbers between a and b, use
generate double u = (ba)*runiform() + a
The random numbers will not actually be between a and b: they will be between a and nearly b, but the top will be so close to b, namely 0.999999999767169356*b, that it will not matter.
2. To generate integer random numbers between a and b, use Read more…
Categories: Numerical Analysis Tags:
## Using Stata’s random-number generators, part 1
I want to start a series on using Stata’s random-number function. Stata in fact has ten random-number functions: Read more…
Categories: Numerical Analysis Tags:
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Thursday, November 23, 2006
Simple Pregnancy Testing (revisited)
A little more detailed description of the simple pregnancy testing example described here. This is a description of the network, that can be downloaded as the HUGIN netfile DRGTTEST.NET
The standard management procedure in sow production is to make a pregnancy test four weeks after mating in order to establish, whether a sow is pregnant or not. Often the farmer will know the expected pregnancy rate in the herd. The test precision will depend on the skill of the farmer. This can be modelled in a Bayesian Network (BN) as shown in the figure
The graph consists of three nodes:
• S, Pregnancy state. Indicates the true state of the sow. The node has two states: (No, Yes). The á priori probabilities are (0.3, 0.7).
• M, Method for pregnancy testing. Indicates the quality of the pregnancy tester. The node has two states (Good, Bad) with á priori probabilities (0.5, 0.5).
• T, Outcome of Pregnancy Test Indicates the outcome of the pregnancy test. The node has two states (Neg, Pos) indicating a negative and a positive outcome respectively.
Method Good Bad State no yes no yes neg 0.85 0.15 0.65 0.15 pos 0.15 0.95 0.35 0.85
Using GRAPPA
The network is available as a Hugin netfile here, but there are several options for handling the network.
One possibility is Peter Greens GRAPPA program available for use in R, which is a free software environment for statistical computing and graphics. R can be downloaded from a CRAN mirror.
Unfortunately, GRAPPA is not available as a standard R-package, which would make it easier to install and use. But if you visit the home page you simple need to download the R source code, the Windows DLL (or the Fortran source code if you're not using Windows), and the user guide in PDF format.
Then you can proceed as follows (currently R produces some warnings, which you may ignorere).
``` # load grappa source code
source("grappa.r")
# define the three nodes
query('S',c(0.3,0.7))
query('M',c(0.5,0.5))
tab(c('T','S','M'),,
c(0.85,0.15,
0.05,0.95,
0.65,0.35,
0.15,0.85),c("neg","pos"))
vs('S',c('no','yes'))
vs('M',c('Good','Bad'))
# compile, initialise and equilibrate
compile()
initcliqs()
trav()
```
Now you may use the network for inference
``` # remove previous evidence
equil()
# add new evidence and
# read the posterior probabilities
prop.evid('T','neg')
pnmarg('S')
pnmarg('M')
# If you know you are good at testing
prop.evid('M','Good')
pnmarg('S')
```
Thus, if you observe a negative test result, the probability of being non-pregnant changes from 0.3 to 0.76, and there is a slight increase in the probability that the method M is a bad method (posterior 50.8 vs the prior 50 %). If you know that the method is good, the probability of no pregnancy increases to 88 %.
Monday, November 20, 2006
Links to other BN repositories
I have just checked my del.icio.us links for other BN-repositories and have added them to the links in the left column. Currently, I have only listed three.
I will follow up with similar links to software.
Thursday, November 16, 2006
Decisions
These are examples used in the 2003 Ph.D. course Reasoning under Uncertainty in Agriculture: Bayesian Networks and Graphical Models organised by the Nordic Informatics Network in the Agricultural Sciences (Nina)
Building of expert systems.
These are examples used in the 2003 Ph.D. course Reasoning under Uncertainty in Agriculture: Bayesian Networks and Graphical Models organised by the Nordic Informatics Network in the Agricultural Sciences (Nina)
Monitoring of pregnancy rate.
This is an example used in the 2003 Ph.D. course Reasoning under Uncertainty in Agriculture: Bayesian Networks and Graphical Models organised by the Nordic Informatics Network in the Agricultural Sciences (Nina)
Monday, November 06, 2006
Simple Pregnancy Testing
• Simple pregnancy testing (DRGTTEST.net) The standard management procedure is to make a pregnancy test four weeks after mating in order to establish whether a sow is pregnant or not. This can be modelled in a Bayesian Network (BN).
• Enhanced pregnancy testing ( DRGTTST1.net) A more realistic example includes the heat detection at three weeks after mating, and allows for the possibility of different herd levels for the pregnancy rate.
Friday, November 03, 2006
Why Blog about Bayesian Networks in agriculture
I have gathered a lot of examples about the use of Bayesian Networks and Markov Processes during the last 15 years. They have been used for different lectures, presentations and workshops. At the moment the examples are more or less scattered on different obsolete servers and some of them on my own hard disk. My plan is to add the examples to this blog, when I have some spare time.
I realise that a wiki may be a better format for this exercise, but it was easier to set up as a blog.
My original intention was to use the name BayesianNetworks.blogspot.com but this address was already in use. As most of my examples are mainly within the agricultural domain, the addition of agriculture is not a big problem.
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# Sample Space
A complete list of all possible outcomes of a random experiment is called sample space or possibility space and is denoted by $S$. Each outcome is called an element of the sample space. A sample space may contain any number of outcomes. If it contains a finite number of outcomes, it is called a finite or discrete sample space. When two bulbs are selected from a lot, the possible outcomes are four, which can be counted as:
1. Both bulbs are defective
2. The first is defective and the second is good
3. The first is good and the second is defective
4. Both are good
Here the sample space is discrete.
When the possibilities of the sample space cannot be contained, it is called continuous. The number of possible readings of temperature from ${45^ \circ }C$ to ${46^ \circ }C$ will make a continuous sample space.
The sample space is a basic term in the theory of probability. We shall discuss some sample spaces in this tutorial. It is not always possible to make a sample space. If it contains a very large number of points, we cannot register all the outcomes, but we must understand how to make the sample space. The outcomes of the sample space are written within the $\left\{ {} \right\}$. Some simple sample spaces are discussed below.
A coin is tossed
When a coin is tossed, it has two possible outcomes: heads and tails. To be brief, heads is denoted by $H$ and tails is denoted by $T$. Thus the sample space consists of heads and tails. In set theory notation, we can write $S$ as:
$S = \left\{ {{\text{head, tail}}} \right\}or\,\,S = \left\{ {H,T} \right\}$
Two coins are tossed
When two coins are tossed, there are four possible outcomes. Let ${H_1}$and ${T_1}$ denote the head and tail on the first coin and ${H_2}$ and ${T_2}$ denote the head and tail on the second coin. The sample space $S$ can be written as
$S = \left\{ {\left( {{H_1},{H_2}} \right),\left( {{H_1},{T_2}} \right),\left( {{T_1},{H_2}} \right),\left( {{T_1},{T_2}} \right)} \right\}$
It may be noted that the sample space of the throw of two coins has $4$ possible points. A sample space of $3$ coins will have ${2^3} = 8$ possible points and for $n$ coins, the number of possible points will be${2^n}$.
A die is thrown
An ordinary die has six faces. These six faces contain $1,2,3,4,5,6$ dots on them. Thus for a single throw of a die, the sample space has $6$ possible outcomes, which are:
$S = \left\{ {1,2,3,4,5,6} \right\}$
Two dice are thrown
A die has six faces. Each face of the first die can occur with all six faces of the second die. Thus there are $6 \times 6 = 36$ possible pairs or points when two dice are tossed together. These $36$ pairs are written as:
$S = \left\{ {\begin{array}{*{20}{c}} {\left( {1,1} \right)}&{\left( {1,2} \right)}&{\left( {1,3} \right)}&{\left( {1,4} \right)}&{\left( {1,5} \right)}&{\left( {1,6} \right)}\\ {\left( {2,1} \right)}&{\left( {2,2} \right)}&{\left( {2,3} \right)}&{\left( {2,4} \right)}&{\left( {2,5} \right)}&{\left( {2,6} \right)}\\ {\left( {3,1} \right)}&{\left( {3,2} \right)}&{\left( {3,3} \right)}&{\left( {3,4} \right)}&{\left( {3,5} \right)}&{\left( {3,6} \right)}\\ {\left( {4,1} \right)}&{\left( {4,2} \right)}&{\left( {4,3} \right)}&{\left( {4,4} \right)}&{\left( {4,5} \right)}&{\left( {4,6} \right)}\\ {\left( {5,1} \right)}&{\left( {5,2} \right)}&{\left( {5,3} \right)}&{\left( {5,4} \right)}&{\left( {5,5} \right)}&{\left( {5,6} \right)}\\ {\left( {6,1} \right)}&{\left( {6,2} \right)}&{\left( {6,3} \right)}&{\left( {6,4} \right)}&{\left( {6,5} \right)}&{\left( {6,6} \right)} \end{array}} \right\}$
If $3$ dice are thrown, the sample space will have ${6^3} = 216$ possible points, with each point being a triplet of $3$ digits.
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# Multivariate Kolmogorov-Smirnov for two samlpes
Let’s say we have data of some fruits and we want to consider different statistics for each fruit (there are various samples for each statistic). For example:
Watermelon:
• Sugar: 50, 57, 36…
• Water: 101, 143, 128…
Melon:
• Sugar: …
• Water: …
(…More fruits…)
We want to know if, based on this statistics, fruits are the same. I thought about using Kolmogorov-Smirnov test in order to check it. So we have a CDF for each statistic (sugar, water, etc.) for each fruit. If we want to consider all statistics for each fruit, it would be nice to use the multivariate Kolmogorov-Smirnov test for two samples and compare every pair of fruits.
The problem is that I cannot find information about the multivariate test with two samples and I read somewhere that the multivariate test is not that good. Could you tell me a different test I could use? Or the Kolmogorov-Smirnov is a good idea (where can I read about the multivariate for two-samples)?
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1. Which rectangular prism is 2 units high? A. B. C. D. Hint 2. Find the surface area of the rectangular prism whose net is shown below. A. 40 units2 B. 20 units2 C. 13 units2 D. 32 units2 Hint 3. What is the surface area of the solid below? A. about 37.8 units2 B. about 52.8 units2 C. about 36.2 units2 D. about 48.9 units2 Hint 4. Gift Wrapping Teisha is wrapping a box that 10 inches long, 12 inches wide, and 2 inches high. What is the surface area of the box? A. 328 in2 B. 164 in2 C. 280 in2 D. 352 in2 Hint 5. Draw a net for this figure. A. B. C. D. Hint
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2 Atoms Molecules & Ions
# 2 Atoms Molecules & Ions - Chapter 2 Atoms Molecules...
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1 Atoms, Molecules, and Ions Chapter 2 2 S t r u c t u r e o f t h e a t o m According to Dalton’s theory, an atom is the basic unit of an element that can enter into chemical combination A t o m Nucleus Electrons (-1) Protons (+1) Neutrons Number of protons = number of electrons. An atom has n o net electrical charge
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2 How to represent an element? X A Z symbol of the element mass number atomic number Atomic Number, Mass Number and Isotopes Atomic number Z: number of protons in the nucleus Mass number A: the sum of the number of protons and neutrons 4 What is the number of protons, neutrons and electrons for the fluorine atom? F 19 9 The mass of an electron is so small we ignore it. Example: Z= number of protons =9 = number of electrons A= mass number = protons + neutrons = 19 neutrons = 19 - 9 = 10
3 Atomic Number, Mass Number and Isotopes (cont) Definition : Two Isotopes are two atoms that have the same atomic number (Z) but different mass numbers (A). E x a m p l e : Hydrogen has three isotopes H 1 1 1 2 1 3 H H Hydrogen Deuterium Tritium (0 neutrons) (1 neutron) (2 neutrons) 6 Example: What is the number of protons, neutrons and electrons for the isotopes of carbon below? 11 14 6 6 C C Atomic Number, Mass Number and Isotopes protons electrons neutrons 6 6 5 11-6= 6 6 8 14-6=
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4 Average Mass E x a m p l e : Carbon is naturally composed of 98.93% 12 C and 1.07% 13 C Averare atomic weight =(0.9893)(12 amu) + (0.0107)(13.00335)=12.01amu Atomic weight of the elements are listed in the periodic table 1 amu=1.66054 x 10 -24 g
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# 2D mesh generation!! HELP...!
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October 23, 2006, 03:21 2D mesh generation!! HELP...! #1 Jesper Sørensen Guest Posts: n/a Hey I am a construction-engineer student from Denmark, and I have a qustion regarding: "How to make a 2D-simulation of windflow around i.e an airfoile profile??" The problem is how to generate a 2D-mesh. This problem i solved in: Ansys CFX release 10 tutorials (Fluid Structure Interaction and Mesh deformation) about a Valve - it is the generation of such a mesh i would like to do! The way too proceed it NOT by using 3D whit changede Boundery Con. while the length-scale in the 3D-mesh gives problems!! If anyone can help me/tell me (step-by-step??) how to solve this problem i would be very thankfull!! Thanks in advance!! \\Jesper
October 24, 2006, 17:58 Re: 2D mesh generation!! HELP...! #2 M. Lemieux Guest Posts: n/a CFX question? Wrong forum. But CFX cannot accept 2D meshes, it is only a 3D solver. So in CFX, you need to make your mesh one element thick. You can make a 2D mesh in Gambit or ICEM if you are running the case in FLUENT.
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# The equation for the motion of an object with constant acceleration is d=vt+0.5at2 where d is distance traveled in ft, v
The equation for the motion of an object with constant acceleration is d=vt+0.5at2
where d is distance traveled in ft, v is starting velocity in fts2, and t is time in seconds. A racecar begins with a velocity of 200 fts2)
Write an expression for the distance the car travels after t seconds.
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# Ch3-6-DNFCriteria.ppt
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### Ch3-6-DNFCriteria.ppt
1. 1. Introduction to Software Testing Chapter 3.6 Disjunctive Normal Form Criteria Paul Ammann & Jeff Offutt http://www.cs.gmu.edu/~offutt/softwaretest/
2. 2. Disjunctive Normal Form <ul><li>Common Representation for Boolean Functions </li></ul><ul><ul><li>Slightly Different Notation for Operators </li></ul></ul><ul><ul><li>Slightly Different Terminology </li></ul></ul><ul><li>Basics: </li></ul><ul><ul><li>A literal is a clause or the negation (overstrike) of a clause </li></ul></ul><ul><ul><ul><li>Examples: a, a </li></ul></ul></ul><ul><ul><li>A term is a set of literals connected by logical “and” </li></ul></ul><ul><ul><ul><li>“ and” is denoted by adjacency instead of </li></ul></ul></ul><ul><ul><ul><li>Examples: ab, ab, ab for a b , a ¬ b , ¬ a ¬ b </li></ul></ul></ul><ul><ul><li>A (disjunctive normal form) predicate is a set of terms connected by “or” </li></ul></ul><ul><ul><ul><li>“ or” is denoted by + instead of </li></ul></ul></ul><ul><ul><ul><li>Examples: abc + ab + ac </li></ul></ul></ul><ul><ul><ul><li>Terms are also called “implicants” </li></ul></ul></ul><ul><ul><ul><ul><li>If a term is true, that implies the predicate is true </li></ul></ul></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
3. 3. Implicant Coverage <ul><li>Obvious coverage idea: Make each implicant evaluate to “true”. </li></ul><ul><ul><li>Problem: Only tests “true” cases for the predicate. </li></ul></ul><ul><ul><li>Solution: Include DNF representations for negation. </li></ul></ul>Implicant Coverage (IC) : Given DNF representations of a predicate f and its negation f , for each implicant in f and f , TR contains the requirement that the implicant evaluate to true. <ul><li>Example: f = ab + bc f = b + ac </li></ul><ul><ul><li>Implicants: { ab, bc, b, ac } </li></ul></ul><ul><ul><li>Possible test set: {TTF, FFT} </li></ul></ul><ul><li>Observation: IC is relatively weak </li></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
4. 4. Improving on Implicant Coverage <ul><li>Additional Definitions: </li></ul><ul><ul><li>A proper subterm is a term with one or more clauses removed </li></ul></ul><ul><ul><ul><li>Example: abc has 6 proper subterms: a, b, c, ab, ac , bc </li></ul></ul></ul><ul><ul><li>A prime implicant is an implicant such that no proper subterm is also an implicant. </li></ul></ul><ul><ul><ul><li>Example: f = ab + abc </li></ul></ul></ul><ul><ul><ul><li>Implicant ab is a prime implicant </li></ul></ul></ul><ul><ul><ul><li>Implicant abc is not a prime implicant (due to proper subterm ac ) </li></ul></ul></ul><ul><ul><li>A redundant implicant is an implicant that can be removed without changing the value of the predicate </li></ul></ul><ul><ul><ul><li>Example: f = ab + ac + bc </li></ul></ul></ul><ul><ul><ul><li>ab is redundant </li></ul></ul></ul><ul><ul><ul><li>Predicate can be written: ac + bc </li></ul></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
5. 5. Unique True Points <ul><li>A minimal DNF representation is one with only prime, nonredundant implicants. </li></ul><ul><li>A unique true point with respect to a given implicant is an assignment of truth values so that </li></ul><ul><ul><li>the given implicant is true, and </li></ul></ul><ul><ul><li>all other implicants are false </li></ul></ul><ul><li>Hence a unique true point test focuses on just one implicant </li></ul><ul><li>A minimal representation guarantees the existence of at least one unique true point for each implicant </li></ul>Unique True Point Coverage (UTPC) : Given minimal DNF representations of a predicate f and its negation f , TR contains a unique true point for each implicant in f and f . Introduction to Software Testing (Ch 3) © Ammann & Offutt
6. 6. Unique True Point Example <ul><li>Consider again: f = ab + bc f = b + ac </li></ul><ul><ul><li>Implicants: { ab, bc, b, ac } </li></ul></ul><ul><ul><li>Each of these implicants is prime </li></ul></ul><ul><ul><li>None of these implicants is redundant </li></ul></ul><ul><li>Unique true points: </li></ul><ul><ul><li>ab: {TTT} </li></ul></ul><ul><ul><li>bc: {FTF} </li></ul></ul><ul><ul><li>b: {FFF, TFF, TFT} </li></ul></ul><ul><ul><li>ac: {FTT} </li></ul></ul><ul><li>Note that there are three possible (minimal) tests satisfying UTPC </li></ul><ul><li>UTPC is fairly powerful </li></ul><ul><ul><li>Exponential in general, but reasonable cost for many common functions </li></ul></ul><ul><ul><li>No subsumption relation wrt any of the ACC or ICC Criteria </li></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
7. 7. Near False Points <ul><li>A near false point with respect to a clause c in implicant i is an assignment of truth values such that f is false, but if c is negated (and all other clauses left as is), i (and hence f ) evaluates to true. </li></ul><ul><li>Relation to determination : at a near false point, c determines f </li></ul><ul><ul><li>Hence we should expect relationship to ACC criteria </li></ul></ul><ul><li>Note that definition only mentions f , and not f . </li></ul><ul><li>Clearly, CUTPNFP subsumes RACC </li></ul>Unique True Point and Near False Point Pair Coverage (CUTPNFP) : Given a minimal DNF representation of a predicate f , for each clause c in each implicant i , TR contains a unique true point for i and a near false point for c such that the points differ only in the truth value of c . Introduction to Software Testing (Ch 3) © Ammann & Offutt
8. 8. CUTPNFP Example <ul><li>Consider f = ab + cd </li></ul><ul><ul><li>For implicant ab , we have 3 unique true points: {TTFF, TTFT, TTTF} </li></ul></ul><ul><ul><ul><li>For clause a , we can pair unique true point T TFF with near false point F TFF </li></ul></ul></ul><ul><ul><ul><li>For clause b, we can pair unique true point T T FF with near false point T F FF </li></ul></ul></ul><ul><ul><li>For implicant cd , we have 3 unique true points: {FFTT, FTTT, TFTT} </li></ul></ul><ul><ul><ul><li>For clause c , we can pair unique true point FF T T with near false point FF F T </li></ul></ul></ul><ul><ul><ul><li>For clause d , we can pair unique true point FFT T with near false point FFT F </li></ul></ul></ul><ul><li>CUTPNFP set: {TTFF, FFTT, TFFF, FTFF, FFTF, FFFT} </li></ul><ul><ul><li>First two tests are unique true points; others are near false points </li></ul></ul><ul><li>Rough number of tests required: # implicants * # literals </li></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
9. 9. DNF Fault Classes <ul><li>ENF: Expression Negation Fault f = ab+c f’ = ab+c </li></ul><ul><li>TNF: Term Negation Fault f = ab+c f’ = ab+c </li></ul><ul><li>TOF: Term Omission Fault f = ab+c f’ = ab </li></ul><ul><li>LNF: Literal Negation Fault f = ab+c f’ = ab+c </li></ul><ul><li>LRF: Literal Reference Fault f = ab + bcd f’ = ad + bcd </li></ul><ul><li>LOF: Literal Omission Fault f = ab + c f’ = a + c </li></ul><ul><li>LIF: Literal Insertion Fault f = ab + c f’ = ab + bc </li></ul><ul><li>ORF+: Operator Reference Fault f = ab + c f’ = abc </li></ul><ul><li>ORF*: Operator Reference Fault f = ab + c f’ = a + b + c </li></ul><ul><li>Key idea is that fault classes are related with respect to testing: </li></ul><ul><li>Test sets guaranteed to detect certain faults are also </li></ul><ul><li>guaranteed to detect additional faults. </li></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
10. 10. Fault Detection Relationships Introduction to Software Testing (Ch 3) © Ammann & Offutt Expression Negation Fault ENF Literal Insertion Fault LIF Term Omission Fault TOF Literal Reference Fault LRF Literal Negation Fault LNF Operator Reference Fault ORF+ Literal Omission Fault LOF Term Negation Fault TNF Operator Reference Fault ORF*
11. 11. Understanding The Detection Relationships <ul><li>Consider the TOF (Term Omission Fault) class </li></ul><ul><ul><li>UTPC requires a unique true point for every implicant (term) </li></ul></ul><ul><ul><li>Hence UTPC detects all TOF faults </li></ul></ul><ul><ul><li>From the diagram, UTPC also detects: </li></ul></ul><ul><ul><ul><li>All LNF faults (Unique true point for implicant now false) </li></ul></ul></ul><ul><ul><ul><li>All TNF faults (All true points for implicant are now false points) </li></ul></ul></ul><ul><ul><ul><li>All ORF+ faults (Unique true points for joined terms now false) </li></ul></ul></ul><ul><ul><ul><li>All ENF faults (Any single test detects this…) </li></ul></ul></ul><ul><li>Although CUTPNFP does not subsume UTPC, CUTPNFP detects all fault classes that UTPC detects (Converse is false) </li></ul><ul><li>Consider what this says about the notions of subsumption vs. fault detection </li></ul><ul><li>Literature has many more powerful (and more expensive) DNF criteria </li></ul><ul><ul><li>In particular, possible to detect entire fault hierarchy (MUMCUT) </li></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
12. 12. Karnaugh Maps for Testing Logic Expressions <ul><li>Fair Warning </li></ul><ul><ul><li>We use , rather than present , Karnaugh Maps </li></ul></ul><ul><ul><li>Newcomer to Karnaugh Maps probably needs a tutorial </li></ul></ul><ul><ul><ul><li>Suggestion: Google “Karnaugh Map Tutorial” </li></ul></ul></ul><ul><li>Our goal: Apply Karnaugh Maps to concepts used to test logic expressions </li></ul><ul><ul><li>Identify when a clause determines a predicate </li></ul></ul><ul><ul><li>Identify the negation of a predicate </li></ul></ul><ul><ul><li>Identify prime implicants and redundant implicants </li></ul></ul><ul><ul><li>Identify unique true points </li></ul></ul><ul><ul><li>Identify unique true point / near false point pairs </li></ul></ul><ul><li>No new material here on testing </li></ul><ul><ul><li>Just fast shortcuts for concepts already presented </li></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt
13. 13. K-Map: A clause determines a predicate <ul><li>Consider the predicate: f = b + ac + ac </li></ul><ul><li>Suppose we want to identify when b determines f </li></ul><ul><li>The dashed line highlights where b changes value </li></ul><ul><ul><li>If two cells joined by the dashed line have different values for f , then b determines f for those two cells. </li></ul></ul><ul><ul><li>b determines f : ac + ac (but NOT at ac or ac ) </li></ul></ul><ul><li>Repeat for clauses a and c </li></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt t t t 1 t t t 0 10 11 01 00 ab c
14. 14. K-Map: Negation of a predicate <ul><li>Consider the predicate: f = ab + bc </li></ul><ul><li>Draw the Karnaugh Map for the negation </li></ul><ul><ul><li>Identify groups </li></ul></ul><ul><ul><li>Write down negation: f = b + a c </li></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt t t 1 t 0 10 11 01 00 ab c t t 1 t t t 0 10 11 01 00 ab c
15. 15. K-Map: Prime and redundant implicants <ul><li>Consider the predicate: f = abc + abd + abcd + abcd + acd </li></ul><ul><li>Draw the Karnaugh Map </li></ul><ul><li>Implicants that are not prime: abd, abcd, abcd, acd </li></ul><ul><li>Redundant implicant: abd </li></ul><ul><li>Prime implicants </li></ul><ul><ul><li>Three: ad, bcd, abc </li></ul></ul><ul><ul><li>The last is redundant </li></ul></ul><ul><ul><li>Minimal DNF representation </li></ul></ul><ul><ul><ul><li>f = ad + bcd </li></ul></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt 01 00 10 11 01 00 ab cd t t t t t t 11 10
16. 16. K-Map: Unique True Points <ul><li>Consider the predicate: f = ab + cd </li></ul><ul><li>Three unique true points for ab </li></ul><ul><ul><li>TTFF, TTFT, TTTF </li></ul></ul><ul><ul><li>TTTT is a true point, but not a unique true point </li></ul></ul><ul><li>Three unique true points for cd </li></ul><ul><ul><li>FFTT, FTTT, TFTT </li></ul></ul><ul><li>Unique true points for f </li></ul><ul><ul><li>f = ac + bc + ad + bd </li></ul></ul><ul><ul><li>FTFT,TFFT, FTTF, TFTF </li></ul></ul><ul><li>Possible UTPC test set </li></ul><ul><ul><li>f : {TTFT, FFTT} </li></ul></ul><ul><ul><li>f : {FTFT, TFFT, FTTF, TFTF} </li></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt 01 00 10 11 01 00 ab cd t t t t 11 10 t t t
17. 17. K-Map: Unique True Point/ Near False Point Pairs <ul><li>Consider the predicate: f = ab + cd </li></ul><ul><li>For implicant ab </li></ul><ul><ul><li>For a , choose UTP, NFP pair </li></ul></ul><ul><ul><ul><li>TTFF, FTFF </li></ul></ul></ul><ul><ul><li>For b , choose UTP, NFP pair </li></ul></ul><ul><ul><ul><li>TTFT, TFFT </li></ul></ul></ul><ul><li>For implicant cd </li></ul><ul><ul><li>For c , choose UTP, NFP pair </li></ul></ul><ul><ul><ul><li>FFTT, FFFT </li></ul></ul></ul><ul><ul><li>For d , choose UTP, NFP pair </li></ul></ul><ul><ul><ul><li>FFTT, FFTF </li></ul></ul></ul><ul><li>Possible CUTPNFP test set </li></ul><ul><ul><li>{TTFF, TTFT, FFTT //UTPs </li></ul></ul><ul><ul><li>FTFF, TFFT, FFFT, FFTF} //NFPs </li></ul></ul>Introduction to Software Testing (Ch 3) © Ammann & Offutt 01 00 10 11 01 00 ab cd t t t t 11 10 t t t
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Copyright © 2010 Pearson Education, Inc. Slide 10 - 1 A least squares regression line was fitted to the weights (in pounds) versus age (in months) of a.
Presentation on theme: "Copyright © 2010 Pearson Education, Inc. Slide 10 - 1 A least squares regression line was fitted to the weights (in pounds) versus age (in months) of a."— Presentation transcript:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 1 A least squares regression line was fitted to the weights (in pounds) versus age (in months) of a group of many young children. The equation of the line is Where is the predicted weight and t is the age of the child. A 20-month-old child in this group has an actual weight of 25 pounds. Which of the following is the residual weight, in pounds, for this child? a. -7.85 b. -4.60 c. 4.60 d. 5.00 e. 7.85
Copyright © 2010 Pearson Education, Inc. Slide 10 - 2 A least squares regression line was fitted to the weights (in pounds) versus age (in months) of a group of many young children. The equation of the line is Where is the predicted weight and t is the age of the child. A 20-month-old child in this group has an actual weight of 25 pounds. Which of the following is the residual weight, in pounds, for this child? a. -7.85 b. -4.60 c. 4.60 d. 5.00 e. 7.85
Copyright © 2010 Pearson Education, Inc. Chapter 10 Re-expressing Data: Get it Straight!
Copyright © 2010 Pearson Education, Inc. Slide 10 - 4 Straight to the Point We cannot use a linear model unless the relationship between the two variables is linear. If the data is not linear, you may need to re-expression the data, straightening bent relationships so that we can fit and use a simple linear model. Simple ways to re-express data are with logarithms, square roots and reciprocals. Re-expressions can be seen in everyday life: The Richter scale of earthquake strength is expressed in logs. Decibel scale for sound intensity is measured in logs. Gauges of shotguns are measured in square roots.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 5 Straight to the Point (cont.) The relationship between fuel efficiency (in miles per gallon) and weight (in pounds) for late model cars looks fairly linear at first:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 6 Straight to the Point (cont.) A look at the residuals plot shows a problem:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 7 Straight to the Point (cont.) We can re-express fuel efficiency as gallons per hundred miles (a reciprocal) and eliminate the bend in the original scatterplot:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 8 Straight to the Point (cont.) A look at the residuals plot for the new model seems more reasonable:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 9 The Goals of Re-expression are: Goal 1: Make the distribution of a variable (as seen in its histogram, for example) more symmetric.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 10 Goals of Re-expression (cont.) Goal 2: Make the spread of several groups (as seen in side-by-side boxplots) more alike, even if their centers differ.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 11 Goals of Re-expression (cont.) Goal 3: Make the form of a scatterplot more nearly linear.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 12 Goals of Re-expression (cont.) Goal 4: Make the scatter in a scatterplot spread out evenly rather than thickening at one end. This can be seen in the two scatterplots we just saw with Goal 3:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 13 The Ladder of Powers There is a family of simple re-expressions that move data toward our goals in a consistent way. This collection of re-expressions is called the Ladder of Powers. The Ladder of Powers orders the effects that the re-expressions have on data.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 14 The Ladder of Powers Ratios of two quantities (e.g., mph) often benefit from a reciprocal. The reciprocal of the data –1 An uncommon re-expression, but sometimes useful. Reciprocal square root –1/2 Measurements that cannot be negative often benefit from a log re-expression. Well use logarithms here 0 Counts often benefit from a square root re-expression. Square root of data values ½ Data with positive and negative values and no bounds are less likely to benefit from re-expression. Raw data1 Try with unimodal distributions that are skewed to the left. Square of data values 2 CommentNamePower
Copyright © 2010 Pearson Education, Inc. Slide 10 - 15 Plan B: Attack of the Logarithms When none of the data values is zero or negative, logarithms can be a helpful ally in the search for a useful model. Try taking the logs of both the x- and y-variable. Then re-express the data using some combination of x or log(x) vs. y or log(y).
Copyright © 2010 Pearson Education, Inc. Slide 10 - 16 Plan B: Attack of the Logarithms (cont.)
Copyright © 2010 Pearson Education, Inc. Slide 10 - 17 Multiple Benefits We often choose a re-expression for one reason and then discover that it has helped other aspects of an analysis. For example, a re-expression that makes a histogram more symmetric might also straighten a scatterplot or stabilize variance.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 18 Why Not Just Use a Curve? If theres a curve in the scatterplot, why not just fit a curve to the data?
Copyright © 2010 Pearson Education, Inc. Slide 10 - 19 Why Not Just Use a Curve? (cont.) The mathematics and calculations for curves of best fit are considerably more difficult than lines of best fit. Besides, straight lines are easy to understand. We know how to think about the slope and the y-intercept.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 20 Example: For each of the models listed below, predict y when x = 3: a. b. c. d. e.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 21 TI – Tips Re-expressing data Check the scatterplot of Yr and TUIT. Then check the residual plot. Dont forget to calculate the regression line so it fills the RESID list. Remember RESID is your y when plotting the residual plot. You decide because of the curve you need to do some re- expressing. But you dont know which one to do. Lets start by using the square root. Fill L1 with the square root of TUIT. Look at the scatterplot and residual plot (recalculate the regression line to fill the RESID list.) Still not good enough … change fill L1 with the log of TUIT. Scatterplot and residual plot again. Better?! Now, write the model of the re-expressed data. (it should have a log in it!)
Copyright © 2010 Pearson Education, Inc. Slide 10 - 22 What Can Go Wrong? Dont expect your model to be perfect. Dont stray too far from the ladder. Dont choose a model based on R 2 alone:
Copyright © 2010 Pearson Education, Inc. Slide 10 - 23 What Can Go Wrong? (cont.) Beware of multiple modes. Re-expression cannot pull separate modes together. Watch out for scatterplots that turn around. Re-expression can straighten many bent relationships, but not those that go up then down, or down then up.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 24 What Can Go Wrong? (cont.) Watch out for negative data values. Its impossible to re-express negative values by any power that is not a whole number on the Ladder of Powers or to re-express values that are zero for negative powers.
Copyright © 2010 Pearson Education, Inc. Slide 10 - 25 Homework: 1-17 odd (skip 3)
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# Lecture26 - Calorimetry Calorimetry an experimental method...
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Calorimetry Calorimetry : an experimental method for observing and measuring the flow of heat into and out of a system. The heat measured with constant volume ( q v ) is different from the heat measured with constant pressure ( q p ). The First Law tells us: Δ E = q + w If only PV -work is present, w = P Δ V . Then, Δ E = q P Δ V At constant V , Δ V = 0, thus Δ E = q v .
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Enthalpy ( H ) What is the heat flow at constant P ( q p )? Define a new state function enthalpy : H = E + P V Define a new state function Then, the change in enthalpy: Δ H = Δ E + Δ ( P V ) We already know: Δ E q P Δ V if only PV work is involved We already know: = , if only -work is involved. Also, Δ ( P V ) = P Δ V , at constant P . Therefore, Δ H = q P Δ V + P Δ V = q p Heat under constant V Heat under constant P Δ E = q v Δ H = q p h i d h i Exo thermic Endo thermic Δ H < 0, or Δ E < 0 Δ H > 0, or Δ E > 0 losing energy Ä gaining energy losing energy gaining energy
Heating/Cooling Curve of H 2 O at 1 atm R P
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Δ H
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# Thread: Tricky trigonometric word problem
1. ## Tricky trigonometric word problem
I've tried three different approaches to it, and it's taken me well over an hour and I CAN'T figure it out.
The pictures pretty much just for the diagram but in case you can't read the text.
19. The Great Pyramid of Khufu has a square base with a side length of about 230m. The four triangular faces of the pyramid are congruent and isosceles. The altitude of each triangular face makes an angle of 52 degrees. with the base. Find the measure of each base angle of the triangular faces to the nearest degree.
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2. Originally Posted by juliajank
I've tried three different approaches to it, and it's taken me well over an hour and I CAN'T figure it out.
The pictures pretty much just for the diagram but in case you can't read the text.
19. The Great Pyramid of Khufu has a square base with a side length of about 230m. The four triangular faces of the pyramid are congruent and isosceles. The altitude of each triangular face makes an angle of 52 degrees. with the base. Find the measure of each base angle of the triangular faces to the nearest degree.
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Those altitudes all meet directly above the center of the square base so dropping a perpendicular from the tip of the pyramid to the base, drawing a perpendicular from the center to the base to the center of one side of the base, and then from that point directly to the tip of the pyramid gives a right triangle in which one leg has length $\displaystyle 230/2= 115m$ and angle 52 degrees. The length of the altitude is given by sin(52)= 115/x so x= 115/sin(52). Now look at that triangular face. The altitude, one edge from the base to the tip, and half the base also form a right triangle. If the "base angle of the triangle is $\displaystyle \theta$, then we have $\displaystyle tan(\theta)= (115/sin(52))/115)= 1/sin(52)$ so $\displaystyle \theta= arctan(1/sin(52))= arctan(csc(52))$.
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# Congruent Triangle
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Congruent triangles are the triangles which have their respective corresponding sides and corresponding angles equal to each other. This implies that the corresponding sides of one triangle are equal to the corresponding sides of the other triangle. Similarly, corresponding angles of one triangle are equal to the corresponding angles of the other triangle, and then the two triangles are called as congruent triangles. Congruency between two triangles can be proved using congruency properties such as SAS, SSS, ASA, AAS and HL (only for right triangles).
Example 1: In triangle ABC, angle ABC is 35º, angle ACB is 75º and side BC = 5m. In triangle XYZ, angle XYZ is 35º, angle XZY is 75º and side YZ is 5m. Are ABC and XYZ congruent triangles?
According to ASA (Angle-included side-Angle congruency property), corresponding 2 angles and an included side are equal in both the triangles and hence they are congruent triangles.
Example 2: In triangle ABC, side AB = 3m, side AC = 4m and angle BAC is 62º. In triangle PQR, side PQ = 3m, PR= 4m and a
ngle QPR is 62º. Are ABC and PQR congruent triangles?
According to SAS (Side-included Angle-Side congruency property), corresponding 2 sides and an included angle are equal in both the triangles and hence they are congruent triangles.
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