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https://www.jobilize.com/algebra/section/real-world-applications-fitting-linear-models-to-data-by-openstax?qcr=www.quizover.com | 1,600,775,122,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00227.warc.gz | 889,919,219 | 22,961 | # 4.3 Fitting linear models to data (Page 6/14)
Page 6 / 14
$x$ $y$
4 44.8
5 43.1
6 38.8
7 39
8 38
9 32.7
10 30.1
11 29.3
12 27
13 25.8
$x$ 21 25 30 31 40 50 $y$ 17 11 2 –1 –18 –40
$y=-\text{1}.\text{981}x+\text{6}0.\text{197;}$ $r=-0.\text{998}$
$x$ $y$
100 2000
80 1798
60 1589
55 1580
40 1390
20 1202
$x$ 900 988 1000 1010 1200 1205 $y$ 70 80 82 84 105 108
$y=0.\text{121}x-38.841,r=0.998$
## Extensions
Graph $\text{\hspace{0.17em}}f\left(x\right)=0.5x+10.\text{\hspace{0.17em}}$ Pick a set of five ordered pairs using inputs $\text{\hspace{0.17em}}x=-2,\text{1},\text{5},\text{6},\text{9}\text{\hspace{0.17em}}$ and use linear regression to verify that the function is a good fit for the data.
Graph $\text{\hspace{0.17em}}f\left(x\right)=-2x-10.\text{\hspace{0.17em}}$ Pick a set of five ordered pairs using inputs $\text{\hspace{0.17em}}x=-2,\text{1},\text{5},\text{6},\text{9}\text{\hspace{0.17em}}$ and use linear regression to verify the function.
$\left(-2,-6\right),\left(1,\text{−12}\right),\left(5,-20\right),\left(6,\text{−22}\right),\left(9,\text{−28}\right);\text{\hspace{0.17em}}$ Yes, the function is a good fit.
For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs shows dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years:
$\left(\text{46},\text{1},600\right),\left(\text{48},\text{1},\text{55}0\right),\left(50,\text{1},505\right),\left(\text{52},\text{1},\text{54}0\right),\left(\text{54},\text{1},\text{495}\right).$
Use linear regression to determine a function $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ where the profit in thousands of dollars depends on the number of units sold in hundreds.
Find to the nearest tenth and interpret the x -intercept.
$\left(\text{189}.8,0\right)\text{\hspace{0.17em}}$ If 18,980 units are sold, the company will have a profit of zero dollars.
Find to the nearest tenth and interpret the y -intercept.
## Real-world applications
For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years:
$\left(\text{25}00,2000\right),\left(\text{265}0,2001\right),\left(3000,2003\right),\left(\text{35}00,2006\right),\left(\text{42}00,2010\right)$
Use linear regression to determine a function $\text{\hspace{0.17em}}y,$ where the year depends on the population. Round to three decimal places of accuracy.
$y=0.00587x+\text{1985}.4\text{1}$
Predict when the population will hit 8,000.
For the following exercises, consider this scenario: The profit of a company increased steadily over a ten-year span. The following ordered pairs show the number of units sold in hundreds and the profit in thousands of over the ten year span, (number of units sold, profit) for specific recorded years:
$\left(\text{46},\text{25}0\right),\left(\text{48},\text{3}05\right),\left(50,\text{35}0\right),\left(\text{52},\text{39}0\right),\left(\text{54},\text{41}0\right).$
Use linear regression to determine a function y , where the profit in thousands of dollars depends on the number of units sold in hundreds.
$y=\text{2}0.\text{25}x-\text{671}.\text{5}$
Predict when the profit will exceed one million dollars.
For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs show dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span (number of units sold, profit) for specific recorded years:
$\left(\text{46},\text{25}0\right),\left(\text{48},\text{225}\right),\left(50,\text{2}05\right),\left(\text{52},\text{18}0\right),\left(\text{54},\text{165}\right).$
Use linear regression to determine a function y , where the profit in thousands of dollars depends on the number of units sold in hundreds.
$y=-\text{1}0.\text{75}x+\text{742}.\text{5}0$
Predict when the profit will dip below the \$25,000 threshold.
## Linear Functions
Determine whether the algebraic equation is linear. $\text{\hspace{0.17em}}2x+3y=7$
Yes
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
what is a algebra
(x+x)3=?
6x
Obed
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
Hi
Abdel
hi
Ye
hi
Nokwanda
C'est comment
Abdel
Hi
Amanda
hello
SORIE
Hiiii
Chinni
hello
Ranjay
hi
ANSHU
hiiii
Chinni
h r u friends
Chinni
yes
Hassan
so is their any Genius in mathematics here let chat guys and get to know each other's
SORIE
I speak French
Abdel
okay no problem since we gather here and get to know each other
SORIE
hi im stupid at math and just wanna join here
Yaona
lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together
SORIE
it's 12
what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function | 2,365 | 7,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.604072 |
https://www.ruby-forum.com/t/too-much-matrix-iterating-over-each-matrix-cell-w-o-passing-the-complete-thing-all-the-time/236403 | 1,660,779,146,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00420.warc.gz | 831,173,505 | 7,032 | # Too Much Matrix: Iterating over Each matrix cell w/o passing the complete thing all the time?
Okay, where to start?
Sounds like the beginning of a joke, you have a fancy data table, a
splendid output table and a method. You use the method to calculate a
splendid number out of the data table and store it in the output table.
Similar to taking a bean out of the jar and putting it into another
nicer jar.
How do you get the input-data-table value into the method without
passing the whole table as an argument all the time?
The problem is that I iterate the same calculation over all cells, which
is not a problem for four, nine, or 25 cells, but it could be one if
this runs on 100x100 matrices.
This maybe about basic instance/symbol/variable scoping, but I’m not
sure and so far I had avoided dealing with them. So, please, someone
needs to explain me what I actually look for, is there a better way than
passing the complete table into the method?
the Force may be with you—always.
yours James
Why are you using indices? All you’re doing with them is referencing the
object inside the matrix, which you already have by using nested “each”
iterators. Use “y” instead of the x & y index and then you won’t have to
pass the matrix around.
How do you get the input-data-table value into the method without
passing the whole table as an argument all the time?
I have not checked speed. But, you can try this.
def mc(n)
b = 100
delta = n / ((n**2 + b)**0.5)
end
iv = [[1,2,3],[4,5,6],[7,8,9]]
p iv.map{|a| a.map{|n| mc(n)}}
OR
require ‘matrix’
def mc(n)
b = 100 # smoothing factor
delta = n / ((n**2 + b) ** 0.5)
end
iv = Matrix[[1,2,3],[4,5,6],[7,8,9]]
ot = iv.to_a
iv.each_with_index{|x,r,c| ot[r][c] = mc(x)}
p ot
Harry
Subject: Too Much Matrix: Iterating over Each matrix cell w/o passing
the complete thing all the time?
Date: dom 01 set 13 12:07:03 +0200
Quoting James J. ([email protected]):
This maybe about basic instance/symbol/variable scoping, but I’m not
sure and so far I had avoided dealing with them. So, please, someone
needs to explain me what I actually look for, is there a better way than
passing the complete table into the method?
One thing must be clear: when you pass an array (on any other object)
to a function, you do NOT put the whole array onto the stack. The only
thing that is passed is the object reference. Thus, passing a single
integer or a one-billion-element array to a function has the same
(quite small) memory requirements. Thus, if what you are writing is a
quick hack, your solution can cut the cake.
If you are after a cleaner option, one that embodies a bit more Ruby’s
object-oriented spirit, you may create a class that becomes the owner
of your table. Something like this:
class Intricate_calculator
def initialize(table)
@table=table
end
def make_intricate_calculations(at_position_x,at_position_y)
[email protected][at_position_x][at_position_y]
b=100 # smoothing factor
delta=a/((a**2+b)**0.5)
return delta
end
end
This way, you do not need to pass the table to your method each
time. Your example would look like this:
calculator=Intricate_calculator::new([[1,2,3],[4,5,6],[7,8,9]]) # fancy
output_table=Array.new(input_vectors.size){Array.new(input_vectors.size){}}
# splendid
output_table.each_with_index do |x,rowindex|
x.each_with_index do |y,colindex|
output_table[rowindex][colindex]=calculator.make_intricate_calculations(rowindex,colindex)
end
end
The important thing to note here is that this solution is certainly
cleaner, but I believe you won’t experience major speed gains.
Carlo
Or you could try this with the matrix approach.
require ‘matrix’
def mc(n)
b = 100
delta = n / ((n**2 + b) ** 0.5)
end
iv = Matrix[[1,2,3],[4,5,6],[7,8,9]]
p iv.map{|x| mc(x)}
Harry
Nothing wrong with indices where necessary. I was just saying that
passing the indices was unnecessary, given that “y” was the value you
were using inside the method, but you discarded it and passed the
I see. Right, you are. «Given» y were the one-only value in universe; so
you cannot had known. How you guys would do it with one table, so I can
do it with many, later, was what I was thinking about. Carlo sensed it
early on. Class#initialize … @table is it, but wasn’t plain to me. I
now read it’s the OO-chapter two, or so, in the Pickaxe? Ouch.
«That’s Bingo!»
Or, just «Bingo!»
Solutions within less than ten hours, who can claim to have such a
support? Thx guys.
I’m surprised by the «sheer» number of solutions … I take the
Matrix#map to replace the nested Array#each and may wrap a class around | 1,239 | 4,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-33 | longest | en | 0.876551 |
https://www.vbforums.com/showthread.php?891351-How-much-memory-is-taken-up-by-an-array&s=1a871be369f888f65ac659dcb13400cc | 1,620,862,492,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00367.warc.gz | 1,085,759,780 | 24,933 | # Thread: How much memory is taken up by an array?
1. ## How much memory is taken up by an array?
I guess an array like this:
Code:
` Dim L_Arr() as Long`
Takes up 4 bytes for each item.
For example if I have:
Code:
` ReDim L_Arr(99) as Long`
Then it occupies 4 * 100 = 400 bytes.
Am I right?
What if I populate SOME but not all of those 100 items?
Will it still occupy 400 bytes of memory?
How about an array of strings:
Code:
``` Dim S_Arr() as String
ReDim S_Arr(99) as String```
How many bytes of memory will this occupy?
Thanks.
2. ## Re: How much memory is taken up by an array?
The value 0 takes also space, so yes
For an array of strings the size is unknown until there are elements in the array.
A long is 4 bytes, a dynamic string has a dynamic size.
3. ## Re: How much memory is taken up by an array?
Don't forget the array descriptor for dynamic arrays:
NOTE: Arrays of any data type require 20 bytes of memory plus 4 bytes for each array dimension plus the number of bytes occupied by the data itself. The memory occupied by the data can be calculated by multiplying the number of data elements by the size of each element. For example, the data in a single-dimension array consisting of 4 Integer data elements of 2 bytes each occupies 8 bytes. The 8 bytes required for the data plus the 24 bytes of overhead brings the total memory requirement for the array to 32 bytes.
A Variant containing an array requires 12 bytes more than the array alone.
4. ## Re: How much memory is taken up by an array?
Originally Posted by dilettante
Don't forget the array descriptor for dynamic arrays:
NOTE: Arrays of any data type require 20 bytes of memory plus 4 bytes for each array dimension plus the number of bytes occupied by the data itself. The memory occupied by the data can be calculated by multiplying the number of data elements by the size of each element. For example, the data in a single-dimension array consisting of 4 Integer data elements of 2 bytes each occupies 8 bytes. The 8 bytes required for the data plus the 24 bytes of overhead brings the total memory requirement for the array to 32 bytes.
A Variant containing an array requires 12 bytes more than the array alone.
Thanks a lot for the explanation.
But then, how about an array of strings?
How do I calculate the size of memory taken up by an array of strings (for example a one-dimentional array of strings)?
There is an overhead of 24 bytes.
But then How about each element?
How do I calculate that?
Thanks.
5. ## Re: How much memory is taken up by an array?
See https://www.aivosto.com/articles/str...l#memorylayout for they structure of each String or element of a String array.
To get some background, let's see how strings are stored in RAM. VB6 stores strings in Unicode format. In COM terminology, a VB String is a BSTR. A String requires six overhead bytes plus 2 bytes for each character. Thus, you spend 6 + Len(string)*2 bytes for each string.
The string starts with a 4-byte length prefix for the size of the string. It's not the character length, though. This 32-bit integer namely counts the number of bytes in the string (not counting the terminating 2 zero bytes). After the length prefix comes the actual text data, 2 bytes for each character. The last 2 bytes are zeros, denoting a NULL terminator (a Unicode null character).
One thing left out of that is that a String can also hold ANSI characters or even arbitrary bytes, in which case only one byte per character is needed. But only do that if you know what you are doing. Most people don't.
6. ## Re: How much memory is taken up by an array?
Originally Posted by dilettante
See https://www.aivosto.com/articles/str...l#memorylayout for they structure of each String or element of a String array.
One thing left out of that is that a String can also hold ANSI characters or even arbitrary bytes, in which case only one byte per character is needed. But only do that if you know what you are doing. Most people don't.
So the count in the 4-byte prefix is for instance 15 (odd number) but does this string terminate with one zero (it's ANSI string, isn't it), two zeroes (this is what BSTR specification says) or three zeroes (why not pad it to even number so the buffer can be traversed using 16-bit register and LODSW instruction)?
Another thing to consider is that COM string allocator uses some space for book-keeping (every allocator does) so there is even more overhead not calculated here as close as 16 bytes per allocation (but small strings are cached in some arena allocator probably, probably even more complicated).
cheers,
</wqw>
7. ## Re: How much memory is taken up by an array?
Originally Posted by dilettante
One thing left out of that is that a String can also hold ANSI characters or even arbitrary bytes, in which case only one byte per character is needed. But only do that if you know what you are doing. Most people don't.
Thanks.
How do I know whether each character is stored as ANSI (one byte) or unicode (two bytes)?
For example lets say I have this:
Code:
` S_Arr(5) = "abcdefgh"`
Are these characters stored as ANSI (one byte per character) or Unicode (two bytes per character)?
Code:
` S_Arr(5) = "AΣBωCμ"`
Are the first, third and fifth characters stored as ANSI (one byte each) and second, fourth and sixth characters stored as unicode (two bytes each)?
Thanks.
8. ## Re: How much memory is taken up by an array?
Try the Len() and LenB() methods and check the differences
9. ## Re: How much memory is taken up by an array?
Something like:
Code:
`S = "ABC"`
Here the source code text is ANSI, but it gets stored as a Unicode literal which when run copies the Unicode text as S. Doesn't matter what is between the quotes, you always end up with Unicode. It starts out as ANSI because VB6 source code files are ANSI.
Such a String always has 16-bit characters because all ANSI characters in literals should map to the normal range of characters.
It is also possible to stuff 32-bit UTF-16LE surrogate pair characters into a VB6 String. These will be functional where they are supported, however stuff like Len() will misreport values because it was designed for a 16-bit fixed character length world.
So you can stuff in anything you want, but nothing is going to read your mind about the encoding of those contents. That part is in the eye of the beholder.
VB6, related controls, etc. all try hard to dumb things down. They expect to only see 16-bit Unicode characters and try to do implicit conversions to/from ANSI as needed to maintain the illusion of 7-bit ASCII extended to 8 bits using additional characters from different ANSI encodings (based on locale).
It's a little like Jurassic Park where you are meant to keep your car on those rails. Get off the rails and you may find yourself eaten.
10. ## Re: How much memory is taken up by an array?
Originally Posted by dilettante
Something like:
Code:
`S = "ABC"`
Here the source code text is ANSI, but it gets stored as a Unicode literal which when run copies the Unicode text as S. Doesn't matter what is between the quotes, you always end up with Unicode. It starts out as ANSI because VB6 source code files are ANSI.
So "ABC" in the sources is ANSI but it gets compiled to a Unicode string and is length prefixed so in the binary there is literally a byte sequence that starts with 03 00 00 00 (length) and then 41 00 (for A) 42 00 (for B) 43 00 (for C) 00 00 (for terminating zero) and this byte sequence in stored in the executable and loaded as part of the executable and it's address can be retrieved with StrPtr("ABC").
What happens on assignment S = "ABC" is that a new chunk of memory is allocated from the strings heap and assigned to S and then from the literal address the byte sequence is copied to this new string i.e. S is cloned from the literal as stored in the executable but S does not point to the literal as it's a new dynamic chunk of memory. The literal in the executable is not a regular VB6 String as it is not allocated with SysAllocString and cannot be released with SysAllocFree but it's a correct BSTR struct as a static read-only byte sequence so it can be used with SysStringLen for instance or can be cloned with SysAllocString etc.
cheers,
</wqw>
11. ## Re: How much memory is taken up by an array?
But then, how about an array of strings?
How do I calculate the size of memory taken up by an array of strings (for example a one-dimentional array of strings)?
There is an overhead of 24 bytes.
But then How about each element?
How do I calculate that?
Is there specific need to know exactly how much memory is in use by the arrays in your application?
12. ## Re: How much memory is taken up by an array?
I was wondering that, as well. In embedded systems, memory use can be crucial, but in modern computers, and increasingly in modern embedded systems, memory isn't a real concern. However, just because it isn't a real concern, that doesn't mean that it isn't a concern. You just have to be clear on why you care such that you know whether or not it is worth worrying about.
13. ## Re: How much memory is taken up by an array?
Originally Posted by Arnoutdv
Is there specific need to know exactly how much memory is in use by the arrays in your application?
I am developing an application, and there are a number of string arrays that get allocated and populated at run time (using the ReDim statement).
I am wondering if this can run out of memory, because my arrays can potentially be very large (around 100,000 records each), and each element of the array can sometimes be small and sometimes large (probably won't exceed 128 characters).
That is why, I decided to find out how much memory is occupied by string arrays.
It is unlikely that based on the above numbers, the system will run out of memory, but I think it is always reasonable to be cautious and at least ask some more experienced programmers to see if they have any real concerns.
I certainly believe that always asking something (when in doubt) is more reasonable than not asking it.
Thanks.
Ilia
14. ## Re: How much memory is taken up by an array?
Valid point.
Then using the given information you can quite simple do a sum on the size of the strings in the string array adding the the overhead to have quite a reasonable estimation. If you combine this with actual memory used by your process then you have something to work with.
Out of memory can also occur when the process uses to much memory or when the free memory is to fragmented for huge arrays.
15. ## Re: How much memory is taken up by an array?
Originally Posted by Arnoutdv
If you combine this with actual memory used by your process then you have something to work with.
How do I find out the actual amount of memory used by my process?
Out of memory can also occur when the process uses to much memory or when the free memory is to fragmented for huge arrays.
Is there a way to know that memory fragmentation has happened to a large degree?
For example is there any method of calculating memory fragmentation?
And then comparing it to a maximum acceptable value?
And is there a way to de-fragment the memory (short of rebooting the computer)?
Thanks.
16. ## Re: How much memory is taken up by an array?
Originally Posted by IliaPreston
And is there a way to de-fragment the memory (short of rebooting the computer)?
The heap fragmentation discussed here is per process so restarting the process usually "levels the field".
On startup each process is granted a 4GB flat virtual address space which has some parts (pages) mapped to actual physical memory but most are unmapped (unallocated). OS provides VirtualAlloc function to grab chunks from the green field (and map these to physical memory) and it is inside these chunks that "software" memory heaps are implemented entirely in code. There are a lot of "software" memory allocators provided by the OS and by COM and also in C/C++ they have custom application implemented allocators they use. In VB6 land we use COM provided allocators only (e.g. CoTaskMemAlloc and SysAllocString) for almost everything. Every ReDim MyArray and every MyString = String(1000, 0) is a call to one of these allocators behind the scenes.
cheers,
</wqw>
17. ## Re: How much memory is taken up by an array?
Originally Posted by wqweto
The heap fragmentation discussed here is per process so restarting the process usually "levels the field".
......</wqw>
Thanks.
So, what I understand from your statement above is that you mean restarting the computer is not needed. Rather, restarting just the application should be enough.
Is that what you mean?
Also, in your previous post you said:
If you combine this with actual memory used by your process then you have something to work with
How do I find out the actual amount of memory used by my process?
18. ## Re: How much memory is taken up by an array?
There is no strict definition of "memory used by a process" but the closes is so called WorkingSet size (or private bytes in Task Manager) which can be retrieved like this
Code:
```Option Explicit
Private Declare Sub CopyMemory Lib "kernel32" Alias "RtlMoveMemory" (Destination As Any, Source As Any, ByVal Length As Long)
Private Declare Function GetCurrentProcess Lib "kernel32" () As Long
Private Declare Function GetProcessMemoryInfo Lib "psapi" (ByVal hProcess As Long, ppsmemCounters As PROCESS_MEMORY_COUNTERS, ByVal cb As Long) As Long
Private Type PROCESS_MEMORY_COUNTERS
cb As Long
PageFaultCount As Long
PeakWorkingSetSize As Long
WorkingSetSize As Long
QuotaPeakPagedPoolUsage As Long
QuotaPagedPoolUsage As Long
QuotaPeakNonPagedPoolUsage As Long
QuotaNonPagedPoolUsage As Long
PagefileUsage As Long
PeakPagefileUsage As Long
PrivateUsage As Long
End Type
Private Function pvGetWorkingSetSizeInMB() As Currency
Dim uCounters As PROCESS_MEMORY_COUNTERS
If GetProcessMemoryInfo(GetCurrentProcess(), uCounters, Len(uCounters)) <> 0 Then
Call CopyMemory(pvGetWorkingSetSizeInMB, uCounters.WorkingSetSize, 4)
pvGetWorkingSetSizeInMB = pvGetWorkingSetSizeInMB * 10000@ / 1024@ / 1024@
End If
End Function
Debug.Print pvGetWorkingSetSizeInMB
End Sub```
cheers,
</wqw>
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• | 3,324 | 14,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-21 | latest | en | 0.885032 |
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# Using a Summation Series to Estimate a Quantity
Say the only tool given to you is a calculator which performs addition, subtraction, multiplication, and division.
Let X= Summation (k=1 -->n) e^-(k/n)^2 with N^20
Explain a practical way of computing X within an error of 10^8. Roughly how big is X?
#### Solution Summary
A Summation Series is used to Estimate a Quantity. The solution is detailed and well presented.
\$2.19 | 117 | 455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-09 | latest | en | 0.865042 |
https://versedwriters.com/for-the-probability-problems-below-please-select-one-question-to-work-on-share/ | 1,656,803,531,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00612.warc.gz | 656,714,942 | 10,737 | # For the probability problems below, please select one question to work on, share
### GET A 40% DISCOUNT ON YOU FIRST ORDER
The Birthday Problem – There are 23 people in this class. What is the probability that at least 2 of the people in the class share the same birthday?
The Game Show Paradox – Let’s say you are a contestant on a game show. The host of the show presents you with a choice of three doors, which we will call doors 1, 2, and 3. You do not know what is behind each door, but you do know that behind two of the doors are beat up 1987 Hyundai Excels, and behind one of the doors is a brand new Cadillac Escalade. The cars were placed randomly behind the doors before the show, and the host knows which car is where. The way the game is played out is as follows. The host lets you choose a door. Assume you choose door #1. Before he opens door #1 to let you see what you have chosen, he opens one of the remaining doors, say door #3, to reveal a Hyundai Excel (he will always open one of the remaining doors that has the booby prize), and asks you whether or not you want to change your choice to door #2. What do you tell him?
Flipping Coins – If you flip a coin 3 times, the probability of getting any sequence is identical (1/8).
There are 8 possible sequences: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Let’s make this situation a little more interesting. Suppose two players are playing each other. Each player chooses a sequence, and then they start flipping a coin until they get one of the two sequences.
We have a long sequence that looks something like this: HHTTHTTHTHTTHHTHT…. We continue until one of the two wins.
Do you think this is a fair game, and that under these rules each sequence has an equal chance to appear first?
Think again! If you chose HHH and I chose THH, I have a much higher chance that you do!
The only way that you win is if the first three tosses are HHH. In any other event, I win.
Agree? Do you see why?
For the sequence HHH to appear anywhere except the first three flips, it must come after a T, right? So, the actual sequence for you to win is THHH.
But if there is a sequence of THHH then I already won before that sequence is over (because my sequence was THH).
So, THH will win 7 times out of 8. HHH will only win if the first three are HHH (a one in eight chance).
Suppose you are going to flip a coin until you get the sequence HTH. Say this takes you x flips. Then, suppose you are going to flip the coin until you get the sequence HTT. Say this takes you z flips. On average, how will x compare to z? Will it be bigger, smaller, or equal?
Disease Testing and False Positives – Assume that the test for some disease is 99% accurate. If somebody tests positive for that disease, is there a 99% chance that they have the disease?
A Girl Named Florida – Here’s a three-part puzzler:
Your friend has two children. What is the probability that both are girls?
Your friend has two children. You know for a fact that at least one of them is a girl. What is the probability that the other one is a girl?
Your friend has two children. One is a girl named Florida. What is the probability that the other child is a girl?
The Value of Variance – More often than not, when we are presented with statistics we are given only a measure of central tendency (such as a mean). However, lots of useful information can be gleaned about a dataset if we examine the variance, skew, and the kurtosis of the data as well. Choose a statistic that recently came across your desk where you were just given a mean. If you can’t think of one, come up with an example you might encounter in your life. How would knowing the variance, the skew, and/or the kurtosis of the data give you a better idea of the data? What could you do with that information?
Example: Say you are an executive in an automobile manufacturer, and you are told that, for a particular model of new car that you sell, buyers have on average 2.2 warranty claims over the first three years of owning the car. What would additional information on the shape of your data tell you? If the variance was low, you’d know that just about every car had 2 or 3 warranty claims, while if it was high you’d know that you have a lot of cars with no warranty claims and a lot with more than 2.2. The skew would provide similar information; with a high level of right skew, you’d know that the average is being brought up by a few lemons; with left skew you’d know that very few of the cars have no warranty claims. The kurtosis (thickness of the tails) would help you get an idea as to just how prevalent the lemon problem is. If you have high kurtosis, it means you have a whole bunch of lemons and a whole bunch of perfect cars. If you have low kurtosis, it means that you have few lemons but few perfect cars.
Probability Rules
Web site:
https://stattrek.com/probability/probability-rules.aspx
Select and discuss one of the following probability rules: | 1,180 | 4,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.959039 |
https://puzzling.stackexchange.com/questions/59281/the-counterfeit-coin/59282 | 1,656,783,507,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00715.warc.gz | 514,665,235 | 59,171 | # The counterfeit coin [duplicate]
You have 9 coins. One of the nine is counterfeit. The counterfeit coin can be distinguished by weight - it is heavier than the rest.
Using a balance scale only twice, find the counterfeit coin.
• I think I've seen this puzzle here somewhere. Jan 15, 2018 at 12:27
• Which is already a duplicate's duplicate btw. Jan 15, 2018 at 13:21
## 1 Answer
Ternary search.
Split into groups of 3. Call them first3group, second3group, third3group.
Now weigh first3group vs second3group. If first test shows equal weights, the heavy coin is in third3group. Split third3group and weigh 2. Where one is heavier, QED, otherwise, it is the non-weighed one, QED.
Else take the heavier group of 3 and do a similar exercise as described above when left with 3, 1 of which is heavier.
QED | 215 | 811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-27 | latest | en | 0.932282 |
http://us.metamath.org/qlegif/1oaiii.html | 1,519,452,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815435.68/warc/CC-MAIN-20180224053236-20180224073236-00683.warc.gz | 354,238,890 | 5,277 | Quantum Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > QLE Home > Th. List > 1oaiii Unicode version
Theorem 1oaiii 823
Description: OML analog to orthoarguesian law of Godowski/Greechie, Eq. III with instead of .
Assertion
Ref Expression
1oaiii
Proof of Theorem 1oaiii
StepHypRef Expression
1 anass 76 . . . . 5
2 anidm 111 . . . . . 6
32lan 77 . . . . 5
41, 3ax-r2 36 . . . 4
54ax-r1 35 . . 3
6 1oa 820 . . . 4
76leran 153 . . 3
85, 7bltr 138 . 2
9 anass 76 . . . . 5
10 ancom 74 . . . . . . . . . 10
1110ud1lem0a 255 . . . . . . . . 9
12 ax-a2 31 . . . . . . . . . 10
1312ud1lem0b 256 . . . . . . . . 9
1411, 13ax-r2 36 . . . . . . . 8
1514ran 78 . . . . . . 7
1615, 2ax-r2 36 . . . . . 6
1716lan 77 . . . . 5
189, 17ax-r2 36 . . . 4
1918ax-r1 35 . . 3
20 1oa 820 . . . 4
2120leran 153 . . 3
2219, 21bltr 138 . 2
238, 22lebi 145 1
Colors of variables: term Syntax hints: wb 1 wo 6 wa 7 wi1 12 wi2 13 This theorem is referenced by: 1oaii 824 This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-a3 32 ax-a4 33 ax-a5 34 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 ax-r3 439 This theorem depends on definitions: df-b 39 df-a 40 df-t 41 df-f 42 df-i1 44 df-i2 45 df-le1 130 df-le2 131 df-c1 132 df-c2 133
Copyright terms: Public domain W3C validator | 640 | 1,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-09 | latest | en | 0.114421 |
http://mathoverflow.net/feeds/question/85330 | 1,371,682,393,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709379061/warc/CC-MAIN-20130516130259-00053-ip-10-60-113-184.ec2.internal.warc.gz | 158,374,281 | 1,744 | Has anyone used reflection in bootstrapping methods for one parameter hypothesis tests? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:53:13Z http://mathoverflow.net/feeds/question/85330 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85330/has-anyone-used-reflection-in-bootstrapping-methods-for-one-parameter-hypothesis Has anyone used reflection in bootstrapping methods for one parameter hypothesis tests? Anna Varvak 2012-01-10T13:21:52Z 2012-01-10T13:21:52Z <p>Here's my idea for a bootstrapping method for testing hypotheses about one parameter. Please tell me if you have seen this somewhere before. If not, I'd appreciate pointers for direction of further research.</p> <p>Suppose I want to do a two-tailed hypothesis test for a population mean ($H_0: \mu=\mu_0, H_a: \mu\ne\mu_0$), and then take a random sample of size $n$ from my population, and get sample mean $M$. A typical bootstrapping method in this case is to shift the sample so that the sample mean equals $\mu_0$, and then sample with replacement many times. The shifting seems to me contrary to the usual spirit of bootstrapping methods: test for the variability of the sample statistic by assuming that your sample is representational. Of course, one can't assume both that the sample is representational and that the null hypothesis is true. But shifting seems to do away with the idea of representation altogether. </p> <p><strong>So here's an idea: take the original sample and reflect it along the $x=\mu_0$ line, and assume that the resulting set of size $2n$ is representative of the population. Now sample with replacement many samples of size $n$.</strong> This way, the individuals in the original sample are assumed to be fairly representative of the population, without supposing an individual even more extreme from the hypothesized mean.</p> <p>Similar methods could work for other statistics (median, correlation coefficient).</p> <p>Has anyone come across this idea before?</p> | 488 | 2,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2013-20 | latest | en | 0.870391 |
https://www.english-hindi.net/?q=algorithm | 1,563,462,439,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525659.27/warc/CC-MAIN-20190718145614-20190718171614-00230.warc.gz | 693,749,283 | 13,694 | # English to Hindi Meaning :: algorithm
कलन विधि, एल्गोरिथम
Algorithm :
कलन विधि
- कलन विधिएल्गोरिथमएल्गोरिदम
#### Show English Meaning (+)
Noun(1) a precise rule (or set of rules
#### Show Examples (+)
(1) An algorithm known as the simplex method can be used to find these optimal strategies, but it will not be pursued here.(2) One of the first applications of the simplex algorithm was to the determination of an adequate diet that was of least cost.(3) The algorithm employs quite elementary arithmetic and is stated by the authors in just 13 lines.(4) The following year he wrote on number theory, making a contribution to the theory of the Euclidean algorithm .(5) a basic algorithm for division(6) Another arithmetical result presented by Brahmagupta is his algorithm for computing square roots.(7) Here there is no unfolding to a single planar component but the algorithm finds an unfolding with four planar components.(8) This first step is here reduced to a simple algorithm suitable for computer use.(9) He solved cubic equations by extending an algorithm for finding cube roots.(10) It is possible to formulate this process algorithmically using graphs and therefore to automate the method.(11) The appropriate degree of adjustment may be helped by nomograms or computer algorithms .(12) Kleene's research was on the theory of algorithms and recursive functions.(13) And if there can be no proper theory, then the bit string is called algorithmically random or irreducible.(14) Having established the nature of equilibria, Smale began to think algorithms for their computation.(15) All the algorithms to carry out arithmetical operations are presented in this way and no proofs are given.(16) Joseph Culberson has a nice perspective on such theorems from an algorithmic point of view, and attempts to frame them in the context of complexity theory.
Related Words
(1) genetic algorithm ::
अनुवांशिक कलन
Synonyms
M
1. algorithm ::
कलन विधि
Different Forms
algorithm, algorithmic, algorithms
English to Hindi Dictionary: algorithm
Meaning and definitions of algorithm, translation in Hindi language for algorithm with similar and opposite words. Also find spoken pronunciation of algorithm in Hindi and in English language.
Tags for the entry "algorithm"
What algorithm means in Hindi, algorithm meaning in Hindi, algorithm definition, examples and pronunciation of algorithm in Hindi language.
# Words by Category
## Word of the day
● Undue
अत्यधिक, जुनूनी, से अधिक, अनुचित, ऊपर, खड़ी, असंगत, बेमेल, बद, भद्दा, प्रतिकूल, अकारण, तर्कहीन, बेतुका, विसंगतअनुचित, अस्वीकरणीय, बेमतलब, | 616 | 2,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-30 | longest | en | 0.888793 |
https://books.google.co.il/books?id=M5QEFgJT2psC&qtid=727e78ee&lr=&hl=iw&source=gbs_quotes_r&cad=5 | 1,591,292,830,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00433.warc.gz | 278,645,778 | 5,316 | חיפוש חיפוש תמונות מפות YouTube חדשות Gmail Drive יומן עוד »
כניסה
ספרים ספרים
WHEN THEIR BREADTHS ARE DIMINISHED IN INFINITUM, THE ULTIMATE RATIOS OF THE PARALLELOGRAMS IN ONE FIGURE TO THOSE IN THE OTHER, EACH TO EACH RESPECTIVELY, ARE THE SAME; I SAY, THAT THOSE TWO FIGURES AacE, PprT, ARE TO ONE ANOTHER IN THAT SAME RATIO.
The Three First Sections and Part of the Seventh Section of Newton's ... - עמוד 40
מאת Sir Isaac Newton - 1850 - 163 דפים
תצוגה מלאה - מידע על ספר זה
## Euclid's Elements of Geometry: The Six First Books. To which are Added ...
Rev. John Allen - 1822 - 494 דפים
...parallelograms, an equal number in each figure, and, when their breadths are diminished in infinitum, the ultimate ratios of the parallelograms in one figure to those in the other, each to each, be the same ; these two figures are to each other, in the same ratio. » Co?\ Hence if two quantities...
תצוגה מלאה - מידע על ספר זה
## The First Three Sections of Newton's Principia: With Copious Notes and ...
Isaac Newton - 1826 - 183 דפים
...parallelograms, an equal number in each; and, their breadths being diminished indefinitely, if tlte ultimate ratios of the parallelograms in one figure...T, are to each other in that same ratio. — (Fig. 2.) For, as the parallelograms in one are severally to the parallelograms in the other ; so, by composition^...
תצוגה מלאה - מידע על ספר זה
## An Elementary Course of Mathematics: Designed Principally for Students of ...
Harvey Goodwin - 1849 - 528 דפים
...if when the breadths of the parallelograms are diminished and their number increased indefinitely, the ultimate ratios of the parallelograms in one figure to those in the other each to each are all the same; then are the figures AacE, PprT in that same ratio. For as the parallelograms are...
תצוגה מלאה - מידע על ספר זה
## THE PHILOSOPHY OF MATHEMATICS WITH SPECIAL REFERENCE TO THE ELEMENTS OF ...
...rank, and when their breadths are diminished in infinitum, the ultimate ratios of the parallel ograms in one figure to those in the other, each to each respectively, are the same ; I say, that these two figures A ac E, T?pr T, are to one another in that same ratio. " For as the parallelograms...
תצוגה מלאה - מידע על ספר זה | 736 | 2,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-24 | latest | en | 0.601613 |
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# DIGITAL SIGNAL PROCESSING
Butterworth Filter
Lecture by
R. Sivarajan, Assistant Professor
Magnitude Response
The magnitude function of the butterworth low pass filter is
given by
---------------(1)
where, N is the order of the filter, and
c is the cut off frequency.
## Lecture by R. Sivarajan, AP/ECE/APEC
Magnitude Response
## Lecture by R. Sivarajan, AP/ECE/APEC
Magnitude Response
As shown in the figure above, the function is monotonically
decreasing, where the maximum response is unity at =0.
The ideal response is shown by dashed line.
It can be seen that the magnitude response approaches the
ideal low pass characteristics as the order N increases.
For values <c, H(j) =1, For values >c, the values
of H(j) decreases.
At =c, the curve pass through 0.707 which corresponds
to -3 dB point.
## Normalized Butterworth Filter
From equation (1), we can get the magnitude square function
of a normalized butterworth filter as (
----------------------(1a)
Let us derive the transfer function of a stable filter. For this
purpose, substituting =s/j, we can write
## Normalized Butterworth Filter
The above equation tells us that this function has poles in the
LHS as well as in RHS because of the presence of two factor
H(s) and H(-s).
If H(s) has roots in the LHS, then the H(-s) has corresponding
roots in the RHS.
These roots we can get by equating denominator to zero, i.e.,
For N odd,
-------------------(2)
Lecture by R. Sivarajan, AP/ECE/APEC
## Normalized Butterworth Filter
For N odd,
------------------(3)
We know that for N odd, the roots can be obtained from
equation (2).
For N = 3,
## Normalized Butterworth Filter
All the poles are located in the s plane as shown in the figure
below.
## Normalized Butterworth Filter
It is found that the angular separation between the poles is
given by 360/2N, which in this case is equal to 60o and all the
poles lie on the circle.
To ensure stability, considering only the poles that lie in the
left half of the s plane, we can write the denominator of the
transfer function H(s) as
Therefore the transfer function of a third order butterworth
filter for cut off frequency c=1 rad/sec is
---------------(4)
Lecture by R. Sivarajan, AP/ECE/APEC
## Normalized Butterworth Filter
As we interested on a poles that lie in the left half of the s
plane, the same can be found by using
where
----------(5)
Now for N=3,
## Normalized Butterworth Filter
Hence the denominator of the transfer function is
So, the transfer function for normalized butterworth filter is
## Normalized Butterworth Filter
Hence the denominator of the transfer function is
## Similarly we can find the transfer function of a normalized
butterworth filter for any order of a filter as shown in table
below.
Lecture by R. Sivarajan, AP/ECE/APEC
## Normalized Butterworth Filter
N
Denominator of H(s)
1
2
3
4
5
6
7
Lecture by R. Sivarajan, AP/ECE/APEC
## Unnormalized Butterworth Filter
Equation 5 gives us the pole locations of butterworth filter for
c= 1 rad/ sec and are known as normalized poles.
In general, the unnormalized poles are given by
---------------------(6)
The transfer function of such butterworth filter can be obtained
by substituting s s/
c in the transfer function of the
butterworth filter shown in table above.
## Order of the Filter
In equation (1a), the filter was restricted to -3 dB attenuation at
c .
Now let the maximum pass band attenuation in positive dB
is p (<3 dB) at pass band frequency p and s is the
minimum stop band attenuation in positive dB at the stop
band frequency s.
Now the magnitude function can be written as
Lecture by R. Sivarajan, AP/ECE/APEC
## Order of the Filter
From the figure above, we can find that at =p, the
attenuation is equal to p,
--------------(7)
Lecture by R. Sivarajan, AP/ECE/APEC
## Order of the Filter
Referring to the above figure, at =s,
Lecture by R. Sivarajan, AP/ECE/APEC
## Order of the Filter
Taking log on both side,
-------------------(8)
## Since this expression normally does not result in an integer
value, we therefore round off to the next highest value.
--------------------(9)
Lecture by R. Sivarajan, AP/ECE/APEC
where,
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eISSN 0454-8124
pISSN 1225-6951
### Article
Kyungpook Mathematical Journal 2023; 63(2): 175-186
Published online June 30, 2023 https://doi.org/10.5666/KMJ.2023.63.2.175
### Weakly Right IQNN Rings
Yang Lee, Sang Bok Nam, Zhelin Piao∗
Department of Mathematics, Yanbian University, Yanji 133002, P. R. China and Institute for Applied Mathematics and Optics, Hanbat National University, Daejeon 34158, Korea
e-mail : ylee@pusan.ac.kr
Department of Computer Engineering, Kyungdong University, Geseong 24764, Korea
e-mail : k1sbnam@kduniv.ac.kr
Department of Mathematics, Yanbian University, Yanji 133002, P. R. China
e-mail : zlpiao@ybu.edu.cn
Received: June 11, 2022; Accepted: February 7, 2023
### Abstract
In this article we look at the property of a 2 by 2 full matrix ring over the ring of integers, of being weakly right IQNN. This generalisation of the property of being right IQNN arises from products of idempotents and nilpotents. We shown that it is, indeed, a proper generalization of right IQNN. We consider the property of beign weakly right IQNN in relation to several kinds of factorizations of a free algebra in two indeterminates over the ring of integers modulo 2.
Keywords: weakly right IQNN ring, idempotent, nilpotent, 2 by 2 full matrix ring, 2 by 2 upper triangular matrix ring, right IQNN ring
### 1. Prerequisites
Throughout this article every ring is an associative ring with identity unless otherwise stated. Let R be a ring. I(R) is used to denote the set of all idempotents of R, and I(R)=I(R)\{0,1}. We use N(R), and N*(R) to denote the set of all nilpotent elements, and upper nilradical (i.e., the sum of all nil ideals) of R, respectively. It is evident that N*(R)N(R). A nilpotent element is also called a nilpotent for simplicity. For n2, denote the full and upper triangular matrix rings over R by Matn(R) and Tn(R) respectively. Let , n, and denote the ring of integers, ring of integers modulo n, and the field of rational numbers, respectively. For m,n, gcd(m,n) is the greatest common divisor of m, n. The characteristic of R is denoted by ch(R).
Following Kwak et al. [5, Definition 1.2], a ring R is called right idempotent-quasi-normalizing on nilpotents, abbreviated to right IQNN, provided that I(R)' is empty, or else for every pair (e,a)I(R)×N(R) there exists (b,f)N(R)×I(R) such that ea=bf. A left IQNN ring is defined symmetrically. A ring is IQNN if it is both right and left IQNN. Abelian rings, in which every idempotent is central, are clearly IQNN but the converse is not true, as is shown in [5].
Define the sets
E1= 1 0 0 0E2= 0 0 0 1E3= 1 t 0 0E4= 1 0 u 0E5= 0 t 0 1E6= 0 0 u 1t0,u0B1= 0 0 0 0B2= 0 t 0 0B3= 0 0 u 0
The following is from [5, Lemma 2.3(2, 3)].
Lemma 1.1. If F is a commutative domain and R=Mat2(F), then
I(R)=E1E2E7 and N(R)=B1B4.
The following is from [1, Lemma 2.1(2)].
Lemma 1.2. Let F be a commutative domain, R=Mat2(F), and K be the quotient field of F. In Mat2(K), we have
(i) E7B4=sa+tc ba(sa+tc) us(sa+tc) buas(sa+tc) B4E7=as+bu ts(as+bu) ca(as+bu) ctas(as+bu) ;(ii) E4B4=a b ua ub =a aca ua uaca E5B4=tc ta c a =taba ta aba a ;(iii) B4E3=a ta c tc =a ta aba taba B4E6=ub b ua a =uaca aca ua a ,
from which it follows that if
ME7B4B4E7E4B4E5B4B4E3B4E6,
we have that if M is nonzero, then every entry of M is nonzero.
In the following, we see a practical application of Lemma 1.2 that may provide useful information to the studies related to products of idempotents and nilpotents.
Remark 1.3. Let F= and R=Mat2(F).
(1) Let p, q be any nonzero integers. Let C=0p0qR be such that C=EA for some EI(R) and AN(R). Then, by Lemma 1.1, we have the cases that A=B2=0v00N(R), and
E=E7=pmq1 p I(R), where p(1p)=qm0
or
E=E4=10u0I(R), where u0.
That is, EA is 0pv0qv with p=p'v and q=q'v, or 0v0uv with p=v and q=uv.
We will find BN(R) and EI(R) such that the left ideal RBE' of R contains EA.
Case 1. Suppose that p and q do not divide each other, and gcd(p,q)1. Evidently |p|,|q|2. Letting p=pv1 and q=qv1 with gcd(p',q')=v1 (then gcd(p,q)=1), we also have |p|,|q|2 since p and q do not divide each other.
Let p=p1u1pfuf and q=q1v1qgvg, with ui,vj1, we the prime number decompositions of p and q respectively. Since p(1p)=qm and gcd(p,q)=1, q must divide 1-p'. Letting 1p=qm, we have p+qm=1 and this implies gcd(p,q)=1 (hence gcd(v1,q)=1).
Since q divides 1p as above, we have that 1pq and
BE=qp p 2 q 2qp01p q 01=0p 0q ,
noting B=qp p 2 q 2qpN(R) and E=01p q01I(R). From this, we also have
v00vBE=v00vqp p 2 q 2qp01p q 01=0pv0qv=0p0qv
and
vv100vv1BE=vv100vv1qp p 2 q 2qp01p q 01=0pvv10qvv1=0pv10q,
noting v00vqp p 2 q 2qp,vv100vv1qp p 2 q 2qpN(R).
Thus RBE' contains the matrices
10000p0qv=0p00 and 00010pv10q=000q,
entailing EARBE.
Case 2. The results in this case are obtained by the argument of [1, Lemma 2.1(3)].
(i) Suppose that p divides q. Then BE=qpq2pq0001=0p0q=EA, where B=qpq2pqN(R) and E=0001I(R).
(ii) Suppose that q divides p. Then pp2qqp01+pq01=0p0q=EA, where B=pp2qqpN(R) and E=01+pq01I(R).
(iii) Suppose that p and q do not divide each other and gcd(p,q)=1. Then we get BE=qpp2q2qp01pq01=0p0q=EA, where B=qpp2q2qpN(R) and E=01pq01I(R).
Thus there exist BN(R) and EI(R) such that EARBE in any case of (i), (ii) and (iii).
(2) Let p, q be any nonzero integers. Let C=q0p0R be such that C=EA for some EI(R) and AN(R). Then, by Lemma 1.1, we have the cases that 0A=00s0N(R), and
E=E7=1pqmpI(R) (where p(1p)=qm0)
or
E=E5=0t01I(R) (where t0);
that is, EA=qs0ps0 with p=p's and q=q's, or EA=st0s0 with p=s and q=st.
We will find BN(R) and EI(R) such that the left ideal RBE' of R contains EA.
Case 1. Suppose that p and q do not divide each other, and gcd(p,q)1.
By applying the argument and using the notation of (1), we have
BE=qp q 2 p 2qp101p q 0=q 0p 0,
noting B=qp q 2 p 2qpN(R) and E=101p q0I(R). From this, we also have
v00vqp q 2 p 2qp101p q 0=qv0pv0=qv0p0
and
vv100vv1qp q 2 p 2qp101p q 0=qvv10pvv10=q0pv10,
noting v00vqp q 2 p 2qp,vv100vv1qp q 2 p 2qpN(R).
Thus RBE' contains the matrices 1000q0pv10=q000 and 0001qv0p0=00p0,
entailing EARBE.
Case 2. The results in this case are obtained by the argument of [1, Lemma 2.1(3)].
(i) Suppose that p divides q. Then
BE=qq2ppq1000=q0p0=EA, where B=qq2ppqN(R) and E=1000I(R).
(ii) Suppose that q divides p. Then BE=pqp2qp101+pq0=q0p0=EA, where B=pqp2qpN(R) and E=101+pq0I(R).
(iii) Suppose that p and q do not divide each other and gcd(p,q)=1. Then we get BE=qpq2p2qp101pq0=q0p0=EA, where B=qpq2p2qpN(R) and E=101pq0I(R).
Thus there exist BN(R) and EI(R) such that EARBE in any case of (i), (ii) and (iii).
(3) Let p, q be any nonzero integers. Let C=pq00R be such that C=EA for some EI(R) and AN(R). Then we have the cases that E is E1=1000I(R) or E3=1t00I(R), and A=B4=abcaN(R); that is, EA is ab00 with p=a, q=b, or a+tcbta00 with p=a+tc, q=b-ta.
We will find BN(R) and EI(R) such that the right ideal BE'R of R contains EA.
Case 1. Suppose that q divides p.
We have
0q0000pq1=pq00=EA,
noting B=0q00N(R) and E=00pq1I(R).
Based on Case 1, we can assume that |q|2 in the cases below.
Case 2. Suppose that p divides q.
We have
01001qq(1q)ppq=pq00=EA,
noting B=0100N(R) and E=1qq(1q)ppqI(R).
Case 3. Suppose that p and q do not divide each other.
Let p=p'k and q=q'k with gcd(p,q)=k. Take B=0k00N(R) and E=0001I(R). Then BE'R contains BE=0k00, 0k000011=kk00 and, consequently, contains k000.
Thus BE'R contains k000p000=p000 and 0k00000q=0q00, and hence contains pq00.
(4) Let p, q be any nonzero integers. Let C=00qpR be such that C=EA for some EI(R) and AN(R).
Then we have the cases that E is E2=0001I(R) or E6=00u1I(R), and A=B4=abcaN(R); that is, EA is 00ca with q=c, p=-a, or 0000 with q=ua+c, q=ub-a.
We will find BN(R) and EI(R) such that the right ideal BE'R of R contains EA.
Case 1. Suppose that q divides p.
We have
00q01pq00=00qp=EA,
noting B=00q0N(R) and E=1pq00I(R).
Based on Case 1, we assume |q|2 in the cases below.
Case 2. Suppose that p divides q.
We have
0010qpq(1q)p1q=00qp=EA,
noting B=0010N(R) and E=qpq(1q)p1qI(R).
Case 3. Suppose that p and q do not divide each other.
Let p=p'k and q=q'k with gcd(p, q)=k. Take B=00k0N(R) and E=1000I(R). Then BE'R contains BE=00k0, 00k01100=00kk and, consequently, contains 000k.
Thus BE'R contains 00k0q000=00q0 and 000k000p=000p, and hence contains 00qp.
(5) Let p be any nonzero integer.
Let C=p000R be such that C=EA for some EI(R) and AN(R).
Then we have the case that E=E3=1t00I(R) and A=B3=00c0N(R); that is, EA=ct000 with p=ct. Take B=0p00N(R) and E=0001I(R). Then BE'R contains 0p00, 0p000011=pp00 and, consequently, contains EA=p000.
Let C=000pR be such that C=EA for some EI(R) and AN(R).
Then we have the case that E=E6=00u1I(R) and A=B2=0b00N(R); that is, EA=000bu with p=bu. Take B=00p0N(R) and E=1000I(R). Then BE'R contains 00p0, 00p01100=00pp and, consequently, contains EA=000p.
Let C=0p00R be such that C=EA for some EI(R) and AN(R).
Then we have the case that E=E1=1000I(R) and A=B2=0p00N(R). Take B=0p00N(R) and E=E2=0001I(R). Then BE'R contains EA=0p00.
Let C=00p0R be such that C=EA for some EI(R) and AN(R).
Then we have the case that E=E2=0001I(R) and A=B3=00p0N(R). Take B=00p0N(R) and E=E1=1000I(R). Then BE'R contains EA=00p01000=00p0.
(6) Let p, q, r, s be nonzero integers, and let C=pqrsR be such that C=EA for some EI(R) and AN(R). Then, by Lemma 1.1, BE' must be one of B4E3,B4E6, or B4E7.
By (1) and (2), there exist CiN(R) and FiI(R) such that 0q0sRC1F1 and p0r0RC2F2, from which we obtain CRC1F1+RC2F2.
By (3) and (4), there exist CiN(R) and FiI(R) such that pq00C1F1R and 00rsC2F2R, from which we obtain CC1F1R+C2F2R.
Similar arguments are available to the cases of pqr0, pq0s, p0rs, 0qrs, p00s and 0qr0.
Mat2() is shown to be not right IQNN by [1, Theorem 2.3(1)]. In the next section, we introduce a closely related property that Mat2() satisfies, based on Remark 1.3.
### 2. Weakly Right IQNN Rings
Motivated by the arguments of Remark 1.3., we consider the following new ring property as a generalization of right IQNN ring.
Definition 2.1. A ring R is said to be weakly right IQNN provided that I(R)' is empty, or else every pair (e,a)I(R)×N(R) satisfies one of the following:
(i) There exist bN(R) and fI(R) such that eabfR;
(ii) There exist biN(R) and fiI(R) (i=1, 2) such that eab1f1R+b2f2R.
R is called weakly left IQNN provided that I(R)' is empty, or else every pair (e,a)I(R)×N(R) satisfies one of the following:
(i) There exist bN(R) and fI(R) such that aeRfb;
(ii) There exist biN(R) and fiI(R) (i=1, 2) such that aeRf1b1+Rf2b2.
A ring is weakly IQNN if it is both weakly right IQNN and weakly left IQNN.
Right IQNN rings are clearly weakly right IQNN, but not conversely as we see in the arguments below.
Theorem 2.2. Mat2(A) is weakly IQNN over any ring A.
Proof. Let R=Mat1(A). We apply the argument of Remark 1.3. Let 0M=pqrsR. Take
E1=0001,E2=1000I(R)
and
B1=0100,B2=0010N(R).
Then we have
M=B1E100pq+B2E2rs00B1E1R+B2E2R.
Thus R is weakly right IQNN. The proof for the case of weakly left IQNN can be done symmetrically.
Mat2() is not right IQNN as mentioned above, and can be shown to be not left IQNN by a symmetrical method of the proof of [1][Theorem 2.3(1)]. Thus the concept of weakly right (resp., left) IQNN is a proper generalization of right (resp., left) IQNN.
In the following, we see another kind of weakly right IQNN rings but not right IQNN.
Example 2.3. Let K=2 and A=Ka,b be the free algebra with noncommuting indeterminates a, b over K.
(1) We use the ring of the ring of [4, Example 2.3(2)]. Let I be the ideal of A generated by a2- a, b2, ab and set R=A/I and identify the elements in A with their images in R1 for simplicity. Then a2 = a and ab = 0 = b2.
By applying the arguments of [4, Example 2.3(1)] and [5, Example 2.6], we have the following:
(i) every element r∈ R is of the form r=α0+α1a+α2b+α3ba, where α0,α1,α2,α3K;
(ii) I(R)=1+a+γba,a+γbaγ,γK and N(R)=αba+βbα,β K, that is an ideal of R (i.e., N(R)=N*(R));
(iii) S1={eneI(R),nN(R)}=N(R) and S2={nenN(R), eI(R)}={b+ba,ηbaδ,ηK}. So S1S2.
Since S1S2, R is (weakly) left IQNN. Next consider e=1+aI(R) and bN(R). Then eb=b. Since bS2, R is not right IQNN. But
b=(b+ba)+ba=(b+ba)(1+a)+(ba)ac1e1R+c2e2R,
where e1=1+a,e2=aI(R) and c1=b+ba,c2=baN(R). Thus R is weakly right IQNN.
(2) Let J be the ideal of A generated by a2a,b2,ba. and set R=A/J. Then R' is the opposite ring of R of (1). Then a similar argument shows that R' is weakly IQNN but not left IQNN.
(3) Let I' be the ideal of A generated by a2a,b2,abb and set R=A/I and identify the elements in A with their images in R for simplicity. Then a2=a, ab=b, and b2=0 in R. From this relation we obtain the following:
(i) Every element r∈ R is of the form r=α0+α1a+α2b+α3ba, where αiK;
(ii) I(R)=1+a+γb+γba,a+γb+γbaγ,γK and N(R)=αb+α baα,αK, that is an ideal of R;
(iii) S1={eneI(R),nN(R)}=N(R) and S2={nenN(R), eI(R)}={αba,b+ba}. So S1S2.
Since S1S2, R is (weakly) left IQNN. Next consider e=aI(R) and b∈ N(R). Then eb=ab=b. But bS2 and so R is not right IQNN. But
b=(b+ba)+ba=(b+ba)(1+a)+(ba)ac1e1R+c2e2R,
where e1=1+a,e2=aI(R) and c1=b+ba,c2=baN(R). Thus R is weakly right IQNN.
(4) Let J' be the ideal of A generated by a2a,b2,bab. and set R=A/J. Then R' is the opposite ring of R of (3). Then a similar argument shows that R' is weakly IQNN but not left IQNN.
The non-Abelian rings of Example 2.3 are all weakly IQNN. Next we provide a method by which one can construct non-Abelian rings that are neither weakly right nor weakly left IQNN.
Example 2.4. We use the ring of [3][Example 1.2(2)]. Let K=2 and A=Ka,b be the free algebra with noncommuting indeterminates a, b over K. Let I be the ideal of A generated by a2-a, b2 and set R=A/I. Identify the elements in A with their images in R for simplicity. Then a2 = a and b2=0. By help of the argument of [3][Example 1.2(2)], we can express r∈ R and cN(R) by
r=k0+k1a+k2b+af1a+af2b+bf3a+bf4b and c=kb+bfb
where k,kiK and f,fjR for all j.
Let e=aI(R) and c=bN(R). Assume that ec=ab=c'e'r for some c=kb+bfbN(R), eI(R) and r∈ R. Since ab0, c=kb+bfb=b(k+fb)0 and this yields
ab=b(k+fb)er and 0bab=bb(k+fb)er=0,
a contradiction. Next assume that ec=ab=c1e1r1+c2e2r2 for some 0ci=kib+bfibN(R), eiI(R) and riR (i=1, 2). This yields
ab=b((k1+f1b)e1r1+(k2+f2b)e2r2)
and
0bab=bb((k1+f1b)e1r1+(k2+f2b)e2r2)=0,
a contradiction. Thus R is not weakly right IQNN. It is also shown by a symmetrical argument that R is not weakly left IQNN.
Next we consider two kinds of rings R over which T2(R) may be weakly right IQNN.
Proposition 2.5. Let R be a ring.
• (1) If N(R)=N*(R) then T2(R) is weakly right IQNN.
• (2) If I(R)={0, 1} then T2(R) is weakly right IQNN.
Proof. Write T=T2(R). Note that
I(T)=eg0fTe,fI(R),(e,f){(0,0),(1,1)},eg+gf=g
and
N(T)=ac0bTa,bN(R) and cR.
(1) Assume N(R)=N*(R). Let E=eg0fI(T) and A=ac0bN(T). Then EA=eaec+gb0fbN(T), i.e., ea,fbN(R), by assumption. Take E1=1000, E2=0001 and B1=ea000, B2=0ec+gb0fb. Then EiI(T) and BiN(T) such that EA=B1E1+B2E2B1E1T+B2E2T. Thus T is weakly right IQNN.
(2) Assume I(R)={0,1}. Then
I(T)=eg0fT(e,f){(1,0),(0,1)},gR.
Let E=eg0fI(T) and A=ac0bN(T). Then EA=eaec+gb0fbN(T) since e,f{0,1}; in fact, ea is zero or a, and fb is also zero or b. Take E1=1000, E2=0001 and B1=ea000, B2=0ec+gb0fb. Then EiI(T) and BiN(T). Since EA=B1E1+B2E2B1E1T+B2E2T. Thus T is weakly right IQNN.
In the following argument we see a condition under which the weakly IQNN property is right-left symmetric. Let R be a ring. An involution on a ring R is a function *:RR which satisfies the properties that (x+y)*=x*+y*, (xy)*=y*x*, 1*=1, and (x*)*=x for all x,yR. It is easily checked that 0*=0, aN(R) implies a*N(R), and e*I(R) for eI(R). We use these facts without referring.
Proposition 2.6. Let R be a ring with an involution *. Then R is weakly right IQNN if and only if R is weakly left IQNN.
Proof. Assume that I(R)' is nonempty. Suppose that R is weakly right IQNN. Let aN(R) and e∈ I(R)'. Then a*N(R) and e*I(R). Since R is weakly right IQNN, we have the following four cases. We proceed our argument on a case-by-case computation.
(i) There exist bN(R), fI(R) and s∈ R such that e*a*=bfs. This implies that
ae=((ae)*)*=(e*a*)*=(bfs)*=s*f*b*Rf*b*.
(ii) There exist biN(R), fiI(R) and siR (i=1, 2) such that e*a*=b1f1s1+b2f2s2. This implies that
ae=((ae)*)*=(e*a*)*=(b1f1s1+b2f2s2)*=s1*f1*b1*+s2*f2*b2*Rf1*b1*+Rf2*b2*.
Since b*,bi*N(R) and f*,fi*I(R), we now conclude that R is weakly left IQNN by the results (i) and (ii).
Conversely suppose that R is weakly left IQNN. Then we have the following cases.
(iii) There exist bN(R), fI(R) and r∈ R such that a*e*=rfb. This implies that
ea=((ea)*)*=(a*e*)*=(rfb)*=b *f *r*b *f *R.
(iv) There exist biN(R), fiI(R) and riR (i=1, 2) such that a*e*=r1 f 1 b 1+r2 f 2 b 2. This implies that
ea=((ea)*)*=(a*e*)*=(r1 f 1 b 1+r2 f 2 b 2)*=b 1*f 1*r1*+b 2*f 2*r2*b 1*f 1*R+b 2*f 2*R.
Since b*, bi*N(R) and f*, fi*I(R), we now conclude that R is weakly right IQNN by the results (iii) and (iv).
### Footnote
The second named author was supported by Kyungdong University Research Fund, 2022.
The third named author was supported by the Science and Technology Research Project of Education Department of Jilin Province, China(JJKH20210563KJ).
### References
1. H. Chen, J. Huang, T. K. Kwak and Y. Lee, On products of idempotents and nilpotents, submitted.
2. E. -K. Cho, T. K. Kwak, Y. Lee, Z. Piao and Y. Seo, A structure of noncentral idempotents, Bull. Korean Math. Soc., 55(2018), 25-40.
3. H. K. Kim, Y. Lee and K. H. Park, NI rings and related properties, JP J. ANTA, 37(2015), 261-280.
4. N. K. Kim, Y. Lee and Y. Seo, Structure of idempotents in rings without identity, J. Korean Math. Soc., 51(2014), 751-771.
5. T. K. Kwak, S. I. Lee and Y. Lee, Quasi-normality of idempotents on nilpotents, Hacet. J. Math. Stat., 48(2019), 1744-1760. | 6,895 | 17,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.837504 |
https://gitlab.mpi-sws.org/iris/iris/-/commit/4e0b77814a3bea0bc2b7dbbbc4f48a41c1743dc5?view=parallel | 1,660,249,498,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00029.warc.gz | 284,215,559 | 36,215 | Commit 4e0b7781 by Ralf Jung
### docs: fix some typos
parent 98e51974
... @@ -45,7 +45,7 @@ The reason that contractive functions are interesting is that for every contract ... @@ -45,7 +45,7 @@ The reason that contractive functions are interesting is that for every contract The category $\COFEs$ consists of COFEs as objects, and non-expansive functions as arrows. The category $\COFEs$ consists of COFEs as objects, and non-expansive functions as arrows. \end{defn} \end{defn} Note that $\COFEs$ is cartesian closed: Note that $\COFEs$ is cartesian closed. In particular: \begin{defn} \begin{defn} Given two COFEs $\cofe$ and $\cofeB$, the set of non-expansive functions $\set{f : \cofe \nfn \cofeB}$ is itself a COFE with Given two COFEs $\cofe$ and $\cofeB$, the set of non-expansive functions $\set{f : \cofe \nfn \cofeB}$ is itself a COFE with \begin{align*} \begin{align*} ... ...
... @@ -91,7 +91,7 @@ Crucially, the second rule allows us to \emph{swap} the side'' of the sum that ... @@ -91,7 +91,7 @@ Crucially, the second rule allows us to \emph{swap} the side'' of the sum that \subsection{Finite partial function} \subsection{Finite partial function} \label{sec:fpfnm} \label{sec:fpfnm} Given some countable $K$ and some CMRA $\monoid$, the set of finite partial functions $K \fpfn \monoid$ is equipped with a COFE and CMRA structure by lifting everything pointwise. Given some infinite countable $K$ and some CMRA $\monoid$, the set of finite partial functions $K \fpfn \monoid$ is equipped with a COFE and CMRA structure by lifting everything pointwise. We obtain the following frame-preserving updates: We obtain the following frame-preserving updates: \begin{mathpar} \begin{mathpar} ... @@ -149,7 +149,7 @@ There are no interesting frame-preserving updates for $\agm(\cofe)$, but we can ... @@ -149,7 +149,7 @@ There are no interesting frame-preserving updates for $\agm(\cofe)$, but we can \subsection{Exclusive CMRA} \subsection{Exclusive CMRA} Given a cofe $\cofe$, we define a CMRA $\exm(\cofe)$ such that at most one $x \in \cofe$ can be owned: Given a COFE $\cofe$, we define a CMRA $\exm(\cofe)$ such that at most one $x \in \cofe$ can be owned: \begin{align*} \begin{align*} \exm(\cofe) \eqdef{}& \exinj(\cofe) + \bot \\ \exm(\cofe) \eqdef{}& \exinj(\cofe) + \bot \\ \mval_n \eqdef{}& \setComp{\melt\in\exm(\cofe)}{\melt \neq \bot} \mval_n \eqdef{}& \setComp{\melt\in\exm(\cofe)}{\melt \neq \bot} ... @@ -375,7 +375,7 @@ We obtain the following frame-preserving update: ... @@ -375,7 +375,7 @@ We obtain the following frame-preserving update: \subsection{STS with tokens} \subsection{STS with tokens} \label{sec:stsmon} \label{sec:stsmon} Given a state-transition system~(STS, \ie a directed graph) $(\STSS, {\stsstep} \subseteq \STSS \times \STSS)$, a set of tokens $\STST$, and a labeling $\STSL: \STSS \ra \wp(\STST)$ of \emph{protocol-owned} tokens for each state, we construct a monoid modeling an authoritative current state and permitting transitions given a \emph{bound} on the current state and a set of \emph{locally-owned} tokens. Given a state-transition system~(STS, \ie a directed graph) $(\STSS, {\stsstep} \subseteq \STSS \times \STSS)$, a set of tokens $\STST$, and a labeling $\STSL: \STSS \ra \wp(\STST)$ of \emph{protocol-owned} tokens for each state, we construct an RA modeling an authoritative current state and permitting transitions given a \emph{bound} on the current state and a set of \emph{locally-owned} tokens. The construction follows the idea of STSs as described in CaReSL \cite{caresl}. The construction follows the idea of STSs as described in CaReSL \cite{caresl}. We first lift the transition relation to $\STSS \times \wp(\STST)$ (implementing a \emph{law of token conservation}) and define a stepping relation for the \emph{frame} of a given token set: We first lift the transition relation to $\STSS \times \wp(\STST)$ (implementing a \emph{law of token conservation}) and define a stepping relation for the \emph{frame} of a given token set: ... ...
\section{Derived proof rules and other constructions} \section{Derived proof rules and other constructions} We will below abuse notation, using the \emph{term} meta-variables like $\val$ to range over (bound) \emph{variables} of the corresponding type.. We will below abuse notation, using the \emph{term} meta-variables like $\val$ to range over (bound) \emph{variables} of the corresponding type. We omit type annotations in binders and equality, when the type is clear from context. We omit type annotations in binders and equality, when the type is clear from context. We assume that the signature $\Sig$ embeds all the meta-level concepts we use, and their properties, into the logic. We assume that the signature $\Sig$ embeds all the meta-level concepts we use, and their properties, into the logic. (The Coq formalization is a \emph{shallow embedding} of the logic, so we have direct access to all meta-level notions within the logic anyways.) (The Coq formalization is a \emph{shallow embedding} of the logic, so we have direct access to all meta-level notions within the logic anyways.) ... ...
... @@ -7,7 +7,7 @@ A \emph{language} $\Lang$ consists of a set \textdom{Expr} of \emph{expressions} ... @@ -7,7 +7,7 @@ A \emph{language} $\Lang$ consists of a set \textdom{Expr} of \emph{expressions} \end{mathpar} \end{mathpar} \item There exists a \emph{primitive reduction relation} $(-,- \step -,-,-) \subseteq \textdom{Expr} \times \textdom{State} \times \textdom{Expr} \times \textdom{State} \times (\textdom{Expr} \uplus \set{\bot})$ \item There exists a \emph{primitive reduction relation} $(-,- \step -,-,-) \subseteq \textdom{Expr} \times \textdom{State} \times \textdom{Expr} \times \textdom{State} \times (\textdom{Expr} \uplus \set{\bot})$ We will write $\expr_1, \state_1 \step \expr_2, \state_2$ for $\expr_1, \state_1 \step \expr_2, \state_2, \bot$. \\ We will write $\expr_1, \state_1 \step \expr_2, \state_2$ for $\expr_1, \state_1 \step \expr_2, \state_2, \bot$. \\ A reduction $\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f$ indicates that, when $\expr_1$ reduces to $\expr$, a \emph{new thread} $\expr_\f$ is forked off. A reduction $\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f$ indicates that, when $\expr_1$ reduces to $\expr_2$, a \emph{new thread} $\expr_\f$ is forked off. \item All values are stuck: \item All values are stuck: $\expr, \_ \step \_, \_, \_ \Ra \toval(\expr) = \bot$ $\expr, \_ \step \_, \_, \_ \Ra \toval(\expr) = \bot$ \end{itemize} \end{itemize} ... @@ -40,7 +40,7 @@ For any language $\Lang$, we define the corresponding thread-pool semantics. ... @@ -40,7 +40,7 @@ For any language $\Lang$, we define the corresponding thread-pool semantics. \paragraph{Machine syntax} \paragraph{Machine syntax} $\[ \tpool \in \textdom{ThreadPool} \eqdef \bigcup_n \textdom{Exp}^n \tpool \in \textdom{ThreadPool} \eqdef \bigcup_n \textdom{Expr}^n$ \] \judgment[Machine reduction]{\cfg{\tpool}{\state} \step \judgment[Machine reduction]{\cfg{\tpool}{\state} \step ... @@ -48,12 +48,12 @@ For any language $\Lang$, we define the corresponding thread-pool semantics. ... @@ -48,12 +48,12 @@ For any language $\Lang$, we define the corresponding thread-pool semantics. \begin{mathpar} \begin{mathpar} \infer \infer {\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f \and \expr_\f \neq \bot} {\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f \and \expr_\f \neq \bot} {\cfg{\tpool \dplus [\expr_1] \dplus \tpool'}{\state} \step {\cfg{\tpool \dplus [\expr_1] \dplus \tpool'}{\state_1} \step \cfg{\tpool \dplus [\expr_2] \dplus \tpool' \dplus [\expr_\f]}{\state'}} \cfg{\tpool \dplus [\expr_2] \dplus \tpool' \dplus [\expr_\f]}{\state_2}} \and\infer \and\infer {\expr_1, \state_1 \step \expr_2, \state_2} {\expr_1, \state_1 \step \expr_2, \state_2} {\cfg{\tpool \dplus [\expr_1] \dplus \tpool'}{\state} \step {\cfg{\tpool \dplus [\expr_1] \dplus \tpool'}{\state_1} \step \cfg{\tpool \dplus [\expr_2] \dplus \tpool'}{\state'}} \cfg{\tpool \dplus [\expr_2] \dplus \tpool'}{\state_2}} \end{mathpar} \end{mathpar} \clearpage \clearpage ... @@ -595,15 +595,16 @@ A type $\type$ being \emph{inhabited} means that $\proves \wtt{\term}{\type}$ i ... @@ -595,15 +595,16 @@ A type $\type$ being \emph{inhabited} means that $\proves \wtt{\term}{\type}$ i {\mask_2 \subseteq \mask_1 \and {\mask_2 \subseteq \mask_1 \and \toval(\expr_1) = \bot} \toval(\expr_1) = \bot} { {\begin{inbox} % for some crazy reason, LaTeX is actually sensitive to the space between the "{ {" here and the "} }" below... { {\begin{inbox} % for some crazy reason, LaTeX is actually sensitive to the space between the "{ {" here and the "} }" below... ~~\pvs[\mask_1][\mask_2] \Exists \state_1. \red(\expr_1,\state_1) \land \later\ownPhys{\state_1} * \later\All \expr_2, \state_2, \expr_\f. (\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f) \land {}\\\qquad\qquad\qquad\qquad\qquad \ownPhys{\state_2} \wand \pvs[\mask_2][\mask_1] \wpre{\expr_2}[\mask_1]{\Ret\var.\prop} * \wpre{\expr_\f}[\top]{\Ret\any.\TRUE} {}\\\proves \wpre{\expr_1}[\mask_1]{\Ret\var.\prop} ~~\pvs[\mask_1][\mask_2] \Exists \state_1. \red(\expr_1,\state_1) \land \later\ownPhys{\state_1} * {}\\\qquad\qquad\qquad \later\All \expr_2, \state_2, \expr_\f. \left( (\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f) \land \ownPhys{\state_2} \right) \wand \pvs[\mask_2][\mask_1] \wpre{\expr_2}[\mask_1]{\Ret\var.\prop} * \wpre{\expr_\f}[\top]{\Ret\any.\TRUE} {}\\\proves \wpre{\expr_1}[\mask_1]{\Ret\var.\prop} \end{inbox}} } \end{inbox}} } \\\\ \infer[wp-lift-pure-step] \infer[wp-lift-pure-step] {\toval(\expr_1) = \bot \and {\toval(\expr_1) = \bot \and \All \state_1. \red(\expr_1, \state_1) \and \All \state_1. \red(\expr_1, \state_1) \and \All \state_1, \expr_2, \state_2, \expr_\f. \expr_1,\state_1 \step \expr_2,\state_2,\expr_\f \Ra \state_1 = \state_2 } \All \state_1, \expr_2, \state_2, \expr_\f. \expr_1,\state_1 \step \expr_2,\state_2,\expr_\f \Ra \state_1 = \state_2 } {\later\All \state, \expr_2, \expr_\f. (\expr_1,\state \step \expr_2, \state,\expr_\f) \Ra \wpre{\expr_2}[\mask_1]{\Ret\var.\prop} * \wpre{\expr_\f}[\top]{\Ret\any.\TRUE} \proves \wpre{\expr_1}[\mask_1]{\Ret\var.\prop}} {\later\All \state, \expr_2, \expr_\f. (\expr_1,\state \step \expr_2, \state,\expr_\f) \Ra \wpre{\expr_2}[\mask_1]{\Ret\var.\prop} * \wpre{\expr_\f}[\top]{\Ret\any.\TRUE} \proves \wpre{\expr_1}[\mask_1]{\Ret\var.\prop}} \end{mathpar} \end{mathpar} Notice that primitive view shifts cover everything to their right, \ie $\pvs \prop * \propB \eqdef \pvs (\prop * \propB)$. Here we define $\wpre{\expr_\f}[\mask]{\Ret\var.\prop} \eqdef \TRUE$ if $\expr_\f = \bot$ (remember that our stepping relation can, but does not have to, define a forked-off expression). Here we define $\wpre{\expr_\f}[\mask]{\Ret\var.\prop} \eqdef \TRUE$ if $\expr_\f = \bot$ (remember that our stepping relation can, but does not have to, define a forked-off expression). ... ...
... @@ -26,7 +26,7 @@ We introduce an additional logical connective $\ownM\melt$, which will later be ... @@ -26,7 +26,7 @@ We introduce an additional logical connective $\ownM\melt$, which will later be \Sem{\vctx \proves \prop \Ra \propB : \Prop}_\gamma &\eqdef \Sem{\vctx \proves \prop \Ra \propB : \Prop}_\gamma &\eqdef \Lam \melt. \setComp{n}{\begin{aligned} \Lam \melt. \setComp{n}{\begin{aligned} \All m, \meltB.& m \leq n \land \melt \mincl \meltB \land \meltB \in \mval_m \Ra {} \\ \All m, \meltB.& m \leq n \land \melt \mincl \meltB \land \meltB \in \mval_m \Ra {} \\ & m \in \Sem{\vctx \proves \prop : \Prop}_\gamma(\melt) \Ra {}\\& m \in \Sem{\vctx \proves \propB : \Prop}_\gamma(\melt)\end{aligned}}\\ & m \in \Sem{\vctx \proves \prop : \Prop}_\gamma(\meltB) \Ra {}\\& m \in \Sem{\vctx \proves \propB : \Prop}_\gamma(\meltB)\end{aligned}}\\ \Sem{\vctx \proves \All x : \type. \prop : \Prop}_\gamma &\eqdef \Sem{\vctx \proves \All x : \type. \prop : \Prop}_\gamma &\eqdef \Lam \melt. \setComp{n}{ \All v \in \Sem{\type}. n \in \Sem{\vctx, x : \type \proves \prop : \Prop}_{\gamma[x \mapsto v]}(\melt) } \\ \Lam \melt. \setComp{n}{ \All v \in \Sem{\type}. n \in \Sem{\vctx, x : \type \proves \prop : \Prop}_{\gamma[x \mapsto v]}(\melt) } \\ \Sem{\vctx \proves \Exists x : \type. \prop : \Prop}_\gamma &\eqdef \Sem{\vctx \proves \Exists x : \type. \prop : \Prop}_\gamma &\eqdef ... @@ -58,7 +58,7 @@ Above, $\maybe\monoid$ is the monoid obtained by adding a unit to $\monoid$. ... @@ -58,7 +58,7 @@ Above, $\maybe\monoid$ is the monoid obtained by adding a unit to $\monoid$. (It's not a coincidence that we used the same notation for the range of the core; it's the same type either way: $\monoid + 1$.) (It's not a coincidence that we used the same notation for the range of the core; it's the same type either way: $\monoid + 1$.) Remember that $\iFunc$ is the user-chosen bifunctor from $\COFEs$ to $\CMRAs$ (see~\Sref{sec:logic}). Remember that $\iFunc$ is the user-chosen bifunctor from $\COFEs$ to $\CMRAs$ (see~\Sref{sec:logic}). $\textdom{ResF}(\cofe^\op, \cofe)$ is a CMRA by lifting the individual CMRAs pointwise. $\textdom{ResF}(\cofe^\op, \cofe)$ is a CMRA by lifting the individual CMRAs pointwise. Furthermore, if $F$ is locally contractive, then so is $\textdom{ResF}$. Furthermore, since $\Sigma$ is locally contractive, so is $\textdom{ResF}$. Now we can write down the recursive domain equation: Now we can write down the recursive domain equation: $\iPreProp \cong \UPred(\textdom{ResF}(\iPreProp, \iPreProp))$ $\iPreProp \cong \UPred(\textdom{ResF}(\iPreProp, \iPreProp))$ ... @@ -187,10 +187,10 @@ $\rho(x)\in\Sem{\vctx(x)}$. ... @@ -187,10 +187,10 @@ $\rho(x)\in\Sem{\vctx(x)}$. \paragraph{Logical entailment.} \paragraph{Logical entailment.} We can now define \emph{semantic} logical entailment. We can now define \emph{semantic} logical entailment. \typedsection{Interpretation of entailment}{\Sem{\vctx \mid \pfctx \proves \prop} : 2} \typedsection{Interpretation of entailment}{\Sem{\vctx \mid \pfctx \proves \prop} : \mProp} \[ \[ \Sem{\vctx \mid \pfctx \proves \propB} \eqdef \Sem{\vctx \mid \pfctx \proves \prop} \eqdef \begin{aligned}[t] \begin{aligned}[t] \MoveEqLeft \MoveEqLeft \forall n \in \mathbb{N}.\; \forall n \in \mathbb{N}.\; ... ...
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Finish editing this message first! | 4,982 | 14,381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-33 | latest | en | 0.773166 |
https://docs.blender.org/manual/ko/dev/physics/soft_body/settings/self_collision.html | 1,716,658,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00868.warc.gz | 182,061,700 | 36,517 | # Self Collision#
Reference
패널:
Physics ‣ Soft Body ‣ Self Collision
참고
Self-Collision is working only if you have activated Use Edges.
When enabled, allows you to control how Blender will prevent the soft body from intersecting with itself. Every vertex is surrounded with an elastic virtual ball. Vertices may not penetrate the balls of other vertices. If you want a good result you may have to adjust the size of these balls. Normally it works pretty well with the default options.
Calculation Type
Manual:
The Ball Size directly sets the ball size.
Average:
The average length of all edges attached to the vertex is calculated and then multiplied with the Ball Size setting. Works well with evenly distributed vertices.
Minimal/Maximal:
The ball size is as large as the smallest/largest spring length of the vertex multiplied with the Ball Size.
Average Min Max:
Size = ((Min + Max)/2) × Ball Size.
Ball Size
Fraction of the length of attached edges. The edge length is computed based on the chosen algorithm. This setting is the factor that is multiplied by the spring length. It is a spherical distance (radius) within which, if another vertex of the same mesh enters, the vertex starts to deflect in order to avoid a self-collision.
Set this value to the fractional distance between vertices that you want them to have their own “space”. Too high of a value will include too many vertices at all times and slow down the calculation. Too low of a level will let other vertices get too close and thus possibly intersect because there will not be enough time to slow them down.
Stiffness
How elastic that ball of personal space is. A high stiffness means that the vertex reacts immediately to another vertex enters their space.
Dampening
How the vertex reacts. A low value just slows down the vertex as it gets too close. A high value repulses it.
참고
Collisions with other objects are set in the (other) Collision panel. To collide with another object they have to share at least one common layer. | 422 | 2,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.929888 |
https://powerpivotforum.com.au/viewtopic.php?t=1095 | 1,566,194,401,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00019.warc.gz | 595,063,528 | 6,996 | ## Pivot Table output help [Solved]
Anything related to PowerPivot and DAX Formuale
sash13
Posts: 2
Joined: Wed May 16, 2018 1:47 am
### Pivot Table output help
Hello, I am fairly new in using Power Pivot and Dax so I will try to be very detailed
Please see attached photo for better explanation
I have Data Model with next columns:
A \$100 \$50 50%
A \$100 \$50 50%
B \$100 \$25 25%
B \$100 \$25 25%
and when I create a Pivot table using this data model I get next:
A \$200 \$100 100%
B \$200 \$50 50%
The problem is the last column in my pivot table, where I do not need it to return sum of ratios but the ratio (SumOfLosses/SumOfPremiums).
that would look like this:
A \$200 \$100 50%
B \$200 \$50 25%
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RamanaV
Posts: 34
Joined: Thu Oct 19, 2017 12:57 pm
### Re: Pivot Table output help [Solved]
- Click on Power Pivot tab, and then on Add to Data Model.
Next, write 3 measures.
To write a measure, click on Power Pivot tab, and then on Measures, and then on New Measure.
Total Losses = SUM(DataTable[Losses])
Total Ratio % = DIVIDE([Total Losses], [Total Premium])
Now, you will get the correct result.
Row Labels Total Premium Total Losses Total Ratio %
A \$200 \$100 50%
B \$200 \$50 25%
Ramana Varanasi
Training and Office Manager
Excelerator BI Pty. Ltd.
sash13
Posts: 2
Joined: Wed May 16, 2018 1:47 am
### Re: Pivot Table output help
Now I get it, very simple. | 432 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | latest | en | 0.773372 |
https://in.mathworks.com/matlabcentral/cody/problems/174-roll-the-dice/solutions/2137633 | 1,600,973,586,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00020.warc.gz | 448,431,962 | 16,572 | Cody
# Problem 174. Roll the Dice!
Solution 2137633
Submitted on 23 Feb 2020 by Hiroki Okawa
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
x1 = zeros(1,6000); x2 = zeros(1,6000); for ii = 1:6000 [x1(ii),x2(ii)] = rollDice(); end numCt = sum( bsxfun( @eq, x1, (1:6)' ), 2 ) + sum( bsxfun( @eq, x2, (1:6)' ), 2 ); assert(all(round(numCt/200) == 10) && sum(numCt) == 12000) | 184 | 495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.690232 |
https://philmikejones.me/tutorials/2019-04-09-test-your-data.html | 1,619,029,229,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00007.warc.gz | 559,932,708 | 9,998 | # Test your data¶
2019-04-09
I’ve written and spoken before about how important it is to test your functions and data analysis scripts. I decided to revisit these ideas and write this tutorial based on my recent experience of calculating the number of units of alcohol the panel members in the NCDS and BCS70 birth cohorts drank at different time points. I initially thought this would be a straightforward mathematical calculation but this turned out to be vastly more complicated than I thought (it always does!). My tests of the data identified the problem (something I would likely have missed without them) and confirmed when I had solved it. I use `testthat` in `R` although the ideas are language–agnostic.
I am currently using the NCDS and BCS70 birth cohort data sets to model alcohol consumption at different time points. Respondents were asked questions about the type and quantity of drinks they consumed in the past seven days, and we decided to convert these into just units of alcohol consumed to simplify the modelling. For example, at most time points cohort members were asked how many pints or glasses of beer, wine, and spirits they had consumed in the past seven days.
For this example I’m going to take an extract from the BCS70 cohort in 1986 when they were aged 16, but the other time points and cohort raised very similar issues. I obviously can’t include actual BCS70 data as it is restricted (but if you’re a UK academic researcher you can download the BCS70 sixteen–year follow–up (1986) yourself).
I’ve taken a sample of 100 respondents and removed their given IDs and replaced these with just a unique integer. In all other respects I’ve constructed the example data to be as close to the original data as possible, including the proportion of completed/missing responses. You can download my example data or load it directly into `R`.
As with any script, first let’s load the packages we need:
```# if necessary run:
# install.packages(c("readr", "dplyr", "testthat", "usethis", "tidyr"))
library("dplyr")
library("usethis")
library("testthat")
library("tidyr")
```
Now download the example data we’ll use:
```url = "https://philmikejones.me/_static/data/bcs70.csv"
```
```##
## ── Column specification ───────────────────────────────────────────────────────────────────────────────────────────────────────
## cols(
## bcsid = col_double(),
## days_drink = col_double(),
## beer = col_double(),
## cider = col_double(),
## sherry = col_double(),
## wine = col_double(),
## shandy = col_double(),
## spirits = col_double()
## )
```
Let’s take a look at the top and bottom cases:
```head(bcs70)
```
```## # A tibble: 6 x 8
## bcsid days_drink beer cider sherry wine shandy spirits
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 NA NA NA NA NA NA NA
## 2 2 NA NA NA NA NA NA NA
## 3 3 NA NA NA NA NA NA NA
## 4 4 NA NA NA NA NA NA NA
## 5 5 NA NA NA NA NA NA NA
## 6 6 NA NA NA NA NA NA NA
```
```tail(bcs70)
```
```## # A tibble: 6 x 8
## bcsid days_drink beer cider sherry wine shandy spirits
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 95 NA NA NA NA NA NA NA
## 2 96 1 0 0 0 2 0 0
## 3 97 NA NA NA NA NA NA NA
## 4 98 1 1 0 0 0 0 0
## 5 99 1 1 0 0 0 0 0
## 6 100 NA NA NA NA NA NA NA
```
Like all social science data it is a bit messy. We can see there are a lot of completely missing cases (`NA`) so these respondents exist at some point in the BCS70 but were not asked the alcohol questions (or any questions) in 1986. Those with answers look like they have a value in each question (even if it’s `0.0`). So let’s get to it.
My approach here is to calculate the number of units of each drink type consumed and create a new variable for each. So for `beer` I will calculate `beer_units`, `cider` I will calculate `cider_units`, and so on. I will therefore be using `mutate()` in the `dplyr()` package for these calculations so that existing variables are not removed.
```bcs70 =
bcs70 %>%
mutate(
beer_units = beer * 2.3,
cider_units = cider * 2.6,
sherry_units = sherry * 1.8,
wine_units = wine * 1.8,
shandy_units = shandy * 2.3 / 2,
spirits_units = spirits * 1.0
)
bcs70
```
```## # A tibble: 100 x 14
## bcsid days_drink beer cider sherry wine shandy spirits beer_units
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 NA NA NA NA NA NA NA NA
## 2 2 NA NA NA NA NA NA NA NA
## 3 3 NA NA NA NA NA NA NA NA
## 4 4 NA NA NA NA NA NA NA NA
## 5 5 NA NA NA NA NA NA NA NA
## 6 6 NA NA NA NA NA NA NA NA
## 7 7 NA NA NA NA NA NA NA NA
## 8 8 NA NA NA NA NA NA NA NA
## 9 9 1 0 2 0 0 0 0 0
## 10 10 NA NA NA NA NA NA NA NA
## # … with 90 more rows, and 5 more variables: cider_units <dbl>,
## # sherry_units <dbl>, wine_units <dbl>, shandy_units <dbl>,
## # spirits_units <dbl>
```
After I’ve calculated the units for each individual drink I’ll first try simply adding them up to arrive at the number of units consumed in the past seven days:
```bcs70 =
bcs70 %>%
mutate(
total_units =
beer_units + cider_units + sherry_units + wine_units +
shandy_units + spirits_units
)
bcs70
```
```## # A tibble: 100 x 15
## bcsid days_drink beer cider sherry wine shandy spirits beer_units
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 NA NA NA NA NA NA NA NA
## 2 2 NA NA NA NA NA NA NA NA
## 3 3 NA NA NA NA NA NA NA NA
## 4 4 NA NA NA NA NA NA NA NA
## 5 5 NA NA NA NA NA NA NA NA
## 6 6 NA NA NA NA NA NA NA NA
## 7 7 NA NA NA NA NA NA NA NA
## 8 8 NA NA NA NA NA NA NA NA
## 9 9 1 0 2 0 0 0 0 0
## 10 10 NA NA NA NA NA NA NA NA
## # … with 90 more rows, and 6 more variables: cider_units <dbl>,
## # sherry_units <dbl>, wine_units <dbl>, shandy_units <dbl>,
## # spirits_units <dbl>, total_units <dbl>
```
## Set up tests¶
To test to see if this worked I create a test suite. The easiest way to do this is with the `usethis` package:
```use_test("test-bcs70")
```
This creates the necessary `tests/` directories and sets up a test file.
## Test calculations¶
First set up a context and load the necessary libraries, and as our first test I want to ensure the calculated `_units` variables are numeric (specifically double):
```context("Test bcs70")
test_that("Calculated units are numeric", {
expect_type(bcs70\$beer_units, "double")
expect_type(bcs70\$cider_units, "double")
expect_type(bcs70\$sherry_units, "double")
expect_type(bcs70\$wine_units, "double")
expect_type(bcs70\$shandy_units, "double")
expect_type(bcs70\$spirits_units, "double")
expect_type(bcs70\$total_units, "double")
})
```
```## Test passed 😸
```
Next I want to test that the number of units calculated are plausible amounts. The minimum number of units should be `0.0`; it’s not possible to consume a negative number of units.
```test_that("Calculated minimum units are 0.0", {
expect_gte(min(bcs70\$beer_units), 0.0)
expect_gte(min(bcs70\$cider_units), 0.0)
expect_gte(min(bcs70\$sherry_units), 0.0)
expect_gte(min(bcs70\$wine_units), 0.0)
expect_gte(min(bcs70\$shandy_units), 0.0)
expect_gte(min(bcs70\$spirits_units), 0.0)
expect_gte(min(bcs70\$total_units), 0.0)
})
```
If you run the code above, the tests will fail. This is because the `min` of a vector with an `NA` in is `NA`: i.e. the `min` of `bcs70\$beer` is `NA` because there’s an `NA` in that vector:
```is.na(min(bcs70\$beer))
```
```## [1] TRUE
```
This is something I forget all the time:
So we must modify our tests to exclude the NAs with na.rm = TRUE:
```test_that("Calculated minimum units are 0.0", {
expect_gte(min(bcs70\$beer_units, na.rm = TRUE), 0.0)
expect_gte(min(bcs70\$cider_units, na.rm = TRUE), 0.0)
expect_gte(min(bcs70\$sherry_units, na.rm = TRUE), 0.0)
expect_gte(min(bcs70\$wine_units, na.rm = TRUE), 0.0)
expect_gte(min(bcs70\$shandy_units, na.rm = TRUE), 0.0)
expect_gte(min(bcs70\$spirits_units, na.rm = TRUE), 0.0)
expect_gte(min(bcs70\$total_units, na.rm = TRUE), 0.0)
})
```
```## Test passed 😀
```
So far, so good. (You don’t need to manually add the `message()` line under normal circumstances as you get feedback from `testthat` in interactive mode, but it’s useful here to see if the tests are successful).
Now I want to test to see if the maximum number of units is sensible. I didn’t know in advance what the maximum number of drinks should be, but I worked out that 180 was a sensible maximum for the individual drinks (and therefore 1080 for the total); this is where your domain knowledge comes in:
```test_that("Calculated maximum units are 180.0/1080.0", {
expect_lte(max(bcs70\$beer_units, na.rm = TRUE), 180.0)
expect_lte(max(bcs70\$cider_units, na.rm = TRUE), 180.0)
expect_lte(max(bcs70\$sherry_units, na.rm = TRUE), 180.0)
expect_lte(max(bcs70\$wine_units, na.rm = TRUE), 180.0)
expect_lte(max(bcs70\$shandy_units, na.rm = TRUE), 180.0)
expect_lte(max(bcs70\$spirits_units, na.rm = TRUE), 180.0)
expect_lte(max(bcs70\$total_units, na.rm = TRUE), 1080.0)
})
```
```## Test passed 🥳
```
## Test for missing data¶
Next I want to check that if there is data in any of the variables then I should have a value in `total`. For example if I have a non–missing value in `beer` there should also be a value in `total`. I wrote the following tests to check for this:
```test_that("No missing data when there should be data", {
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$beer)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$cider)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$sherry)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$wine)])))
# expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$shandy)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$spirits)])))
})
```
```## Test passed 🎊
```
Each of these tests is essentially saying, “if there’s data in one of more of the drinks variables (`!is.na()` means ‘is not missing’) there should not be any missing data in `total_units`”. Notice though that I’ve commented out the `shandy` variable. That’s because for at least one respondent they gave a number of shandies consumed but one or more drinks were recorded as missing (and as we’ve seen above a number + `NA` is `NA`):
```is.na(NA_real_ + 2.0)
```
```## [1] TRUE
```
We can see this respondent by `filter`ing:
```bcs70 %>%
filter(!is.na(shandy)) %>%
filter(is.na(total_units))
```
```## # A tibble: 1 x 15
## bcsid days_drink beer cider sherry wine shandy spirits beer_units
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 69 1 NA NA NA NA 2 NA NA
## # … with 6 more variables: cider_units <dbl>, sherry_units <dbl>,
## # wine_units <dbl>, shandy_units <dbl>, spirits_units <dbl>,
## # total_units <dbl>
```
How to deal with this case, and any others like it in the main dataset? I found the easiest way to solve this was to replace all `NA`s with `0.0` in the `_units` variables to ensure `total_units` summed correctly. The easiest way to do this is with `replace_na()` in the `tidyr` package. I don’t recommend modifying the original variable in place (e.g. `beer`) but as we’ve calculated `beer_units` we can correct this instead:
```bcs70 =
bcs70 %>%
replace_na(list(
beer_units = 0.0,
cider_units = 0.0,
sherry_units = 0.0,
wine_units = 0.0,
shandy_units = 0.0,
spirits_units = 0.0
)) %>%
mutate(
total_units =
beer_units + cider_units + sherry_units + wine_units +
shandy_units + spirits_units
)
bcs70
```
```## # A tibble: 100 x 15
## bcsid days_drink beer cider sherry wine shandy spirits beer_units
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 NA NA NA NA NA NA NA 0
## 2 2 NA NA NA NA NA NA NA 0
## 3 3 NA NA NA NA NA NA NA 0
## 4 4 NA NA NA NA NA NA NA 0
## 5 5 NA NA NA NA NA NA NA 0
## 6 6 NA NA NA NA NA NA NA 0
## 7 7 NA NA NA NA NA NA NA 0
## 8 8 NA NA NA NA NA NA NA 0
## 9 9 1 0 2 0 0 0 0 0
## 10 10 NA NA NA NA NA NA NA 0
## # … with 90 more rows, and 6 more variables: cider_units <dbl>,
## # sherry_units <dbl>, wine_units <dbl>, shandy_units <dbl>,
## # spirits_units <dbl>, total_units <dbl>
```
Finally so that I knew which respondents genuinely didn’t drink and which were simply missing I replaced any `0.0` in `total_units` if all the drinks were missing.
```bcs70 =
bcs70 %>%
mutate(
total_units = if_else(
{
is.na(beer) & is.na(cider) & is.na(sherry) &
is.na(wine) & is.na(shandy) & is.na(spirits)
},
NA_real_,
total_units
)
)
```
Let’s review case `69` (identified above) to see if `total_units` is correct:
```bcs70 %>%
filter(bcsid == 69) %>%
select(beer, shandy, total_units)
```
```## # A tibble: 1 x 3
## beer shandy total_units
## <dbl> <dbl> <dbl>
## 1 NA 2 2.3
```
So even though there is missing data (`NA`) in `beer` the `total_units` has been summed correctly. And indeed if we run the tests now, they all pass (note you would not normally duplicate tests in a document, you’d just run them again):
```test_that("No missing data when there should be data", {
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$beer)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$cider)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$sherry)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$wine)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$shandy)])))
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$spirits)])))
})
```
```## Test passed 🥳
```
## Test for ‘true zero’¶
Now I want to make sure I have a ‘true zero’ for the respondents that were asked the drinking questions, but that did not drink. Notice the `days_drink` variable? If this contains data (i.e. is not `NA`) then the respondent was asked the series of drinking questions, and should be included in the valid sample, even if they didn’t drink anything. Most respondents who were asked `days_drink` who did not drink have `0` coded in the drinks types so that these sum correctly. In some cases, though, the drinks data were missing despite the respondent being asked about their drinking behaviour:
```bcs70 %>%
filter(!is.na(days_drink), is.na(shandy)) %>%
select(bcsid, days_drink, beer, cider, wine, sherry, shandy, spirits, total_units)
```
```## # A tibble: 1 x 9
## bcsid days_drink beer cider wine sherry shandy spirits total_units
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 19 1 NA NA NA NA NA NA NA
```
To check for this I wrote the following test, which will fail if any respondent with data in `days_drink` also has a missing value in `total_units`:
```test_that("True zero coded", {
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$days_drink)])))
})
```
To fix this I replaced any `NA` in `total_units` with 0, if there is data in `days_drink`:
```bcs70 =
bcs70 %>%
mutate(
total_units = if_else(
is.na(total_units) & !is.na(days_drink),
0.0,
total_units
)
)
```
Now when I run the test it passes:
```test_that("True zero coded", {
expect_false(any(is.na(bcs70\$total_units[!is.na(bcs70\$days_drink)])))
})
```
```## Test passed 🎊
```
## Conclusion¶
I hope from this tutorial/example you can see that what can ostensibly be a straightforward task can, on closer inspection, be challenging to get right. Testing the calculations and data helped in three main ways:
• Tests helped identify the problem(s).
• Tests helped verify the problems were solved.
• Re–running the tests helped ensure the problem didn’t return after fixing something else.
For complex data I would recommend writing a test suite and regularly running it when working on any data processing steps.
## Scripts¶
I’ve bundled the data, data processing, and testing code above into files to download to see how these work in a ‘real’ project (rather than inline in a blog post).
## Resources¶
The best places to start with writing your own tests (and you should) for #rstats are:
• Testing chapter in Hadley Wickham’s R Packages | 5,290 | 17,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-17 | latest | en | 0.9543 |
https://mycalcu.com/how-many-square-feet-in-110.4-acres | 1,716,772,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00151.warc.gz | 348,007,786 | 12,358 | # How Many Square Feet In 110.4 Acres
110.4 acres to square feet calculator quickly converts 110.4 acres into sq ft.
## How many square feet are there in 110.4 acres?
110.4 acres can be quickly converted into square feet simply by multiplying 110.4 by 43560.
## What is the value of 110.4 acres in sq feet?
110.4 acres are equal to 4809024 sq ft.
## 110.4 Acres Conversion to Other Lengths
Square Feet 4.80902e+06 Square Meter 446789 Square Miles 0.1725 Square Inches 6.92539e+08 Sqaure Yards 534336 Square Kilometers 0.446964
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https://hackage.haskell.org/package/ad-1.3 | 1,553,134,784,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202476.48/warc/CC-MAIN-20190321010720-20190321032720-00010.warc.gz | 508,977,947 | 8,304 | # ad: Automatic Differentiation
[ bsd3, library, math ] [ Propose Tags ]
Forward-, reverse- and mixed- mode automatic differentiation combinators with a common API.
Type-level "branding" is used to both prevent the end user from confusing infinitesimals and to limit unsafe access to the implementation details of each Mode.
Each mode has a separate module full of combinators.
• Numeric.AD.Mode.Forward provides basic forward-mode AD. It is good for computing simple derivatives.
• Numeric.AD.Mode.Reverse uses benign side-effects to compute reverse-mode AD. It is good for computing gradients in one pass.
• Numeric.AD.Mode.Sparse computes a sparse forward-mode AD tower. It is good for higher derivatives or large numbers of outputs.
• Numeric.AD.Mode.Tower computes a dense forward-mode AD tower useful for higher derivatives of single input functions.
• Numeric.AD.Mode.Mixed computes using whichever mode or combination thereof is suitable to each individual combinator. This mode is the default, re-exported by Numeric.AD
While not every mode can provide all operations, the following basic operations are supported, modified as appropriate by the suffixes below:
• grad computes the gradient (partial derivatives) of a function at a point.
• jacobian computes the Jacobian matrix of a function at a point.
• diff computes the derivative of a function at a point.
• du computes a directional derivative of a function at a point.
• hessian computes the Hessian matrix (matrix of second partial derivatives) of a function at a point.
The following suffixes alter the meanings of the functions above as follows:
• ' -- also return the answer
• With lets the user supply a function to blend the input with the output
• F is a version of the base function lifted to return a Traversable (or Functor) result
• s means the function returns all higher derivatives in a list or f-branching Stream
• T means the result is transposed with respect to the traditional formulation.
• 0 means that the resulting derivative list is padded with 0s at the end.
Changes since 1.2
• Compiles with template haskell 2.6, changed interface to comply with the lack of Eq and Show as superclasses of Num in the new GHC.
Changes since 1.1.3
• Made primal calculations strict where possible.
Changes since 1.1.0
• Introduced a much faster topological sort into the reverse mode AD implementation by Anthony Cowley. This fixes a space leak and a stack overflow problem on very large (>2000 variable) problem sets.
• Made bound calculations in reverse mode more strict.
Changes since 1.0.0
• Changed the way Show was derived to comply with changes in instance resolution in ghc >= 7.0 && <= 7.1
Changes since 0.45.0
• Converted Stream to use the external comonad package
Changes since 0.44.5
• Added Halley's method
Changes since 0.40.0
• Fixed bug fix for '(/)' :: (Mode s, Fractional a) => AD s a
• Improved documentation
• Regularized naming conventions
• Exposed Id, probe, and lower methods via Numeric.AD.Types
• Removed monadic combinators
• Retuned the Mixed mode jacobian calculations to only require a Functor-based result. | 696 | 3,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-13 | longest | en | 0.851624 |
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Unformatted text preview: WARM—UPS 1. Write the equilibrium constant expression for the reaction: CaO(s) + SO;(g)‘~:‘—'—" CaSO3(s) 2. If Kc at 25°C is 6.9:(105 for the reaction below and the equilibrium mixture has 1.0x10‘3 M 02 (g) and 5.0:;10‘2 M No2 (g), what is the [N0(g)]eq? 02(g) + 2 N0(g) :3 2 N02(g) 3. Given the following: A + 213:: C Kc = 1.2x103 c +D':A+ E Kc =2.3x10'2 a) What is Kc forD + 23231;? b) What is Kp for D + 23 :E? 0-1' 3.0“: all cu-c in.“ s c) What is Kc for 2E 2;?— 2D + 4B? STEPS In sotvuie toultlonlun rnoottns: rntnnlmv wome- I 1. Write a llalanoed euuation. 2. Write "IE Bllllililll'illlll GXIII'ESSIIIII. 3. convert all amounts into Molatitv or am. WIIIIK 'I'IIE II}! TIME: 4. When the reaction direction is not Itnown, oomuare II and It. 5. construct the initial, change. and elllililll'illnl rows in the tallle. III/Ir: (mum sign dimes/1mm»: SIIIIIE Illll TIIE EIIIIIIIIIIIIIIII QUANTITIES: 6. Substitute into the euuililtrium eitnression. 1. Ilse tlIe uuadratie tormula unless a] netteet suuares are involved or II] vou oan simulilv llv assumintl the uuantitv. x, is negligible when involved in addition or sulltraetion. It. Solve tor x [confirm anv assumutionsl. 9. Find the euuilillrium uuantities needed. ...
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1 - WARM—UPS 1 Write the equilibrium constant expression...
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Ask a homework question - tutors are online | 635 | 1,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-05 | latest | en | 0.690767 |
http://stackoverflow.com/questions/14145601/while-loop-till-number-reached/14145651 | 1,440,833,869,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064362.36/warc/CC-MAIN-20150827025424-00017-ip-10-171-96-226.ec2.internal.warc.gz | 230,697,778 | 17,201 | While Loop till number reached
I want to do something like this
``````\$x = 630;
\$y = 10;
while (\$y < \$x){
// do something
\$y+10;
}
``````
When I use `\$y++` it's working and adding +1, but with +10 it's not working. But I need to go in +10 steps. Any pointers ?
-
You have to assign the value back to the variable. The `++` is a special operator which is the same as `\$y = \$y + 1`. – Felix Kling Jan 3 '13 at 19:18
in your code, you are not incrementing `\$y` : `\$y+10` returns the value of `\$y` plus `10`, but you need to assign it to `\$y`.
You can do it with several ways :
• `\$y = \$y + 10;`
• `\$y += 10;`
Example :
``````\$x = 630;
\$y = 10;
while (\$y < \$x){
// do something
\$y = \$y + 10;
}
``````
-
This is because \$y++ is equivalent to \$y = \$y + 1; You are not assigning the new value in \$y. Please try
``````\$y += 10;
``````
OR
``````\$y = \$y + 10;
``````
-
``````// commenting the code with description
\$x = 630; // initialize x
\$y = 10; // initialize y
while (\$y < \$x){ // checking whether x is greater than y or not. if it is greater enter loop
// do something
\$y = \$y+10; // you need to assign the addition operation to a variable. now y is added to 10 and the result is assigned to y. please note that \$y++ is equivalent to \$y = \$y + 1
}
``````
- | 437 | 1,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2015-35 | latest | en | 0.841092 |
http://gpuzzles.com/mind-teasers/chelsea-faster-player/ | 1,480,823,654,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541187.54/warc/CC-MAIN-20161202170901-00286-ip-10-31-129-80.ec2.internal.warc.gz | 115,561,731 | 11,745 | • Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Chelsea Faster Player Race Quick Riddle
Difficulty Popularity
Chelsea organized a race in which Luiz, Terry, Hazard, and Pedro have participated.
The results of the race is as follow :
: Luiz beats Terry
: Hazard beats Pedro
: Pedro beats Luiz
So who was the fastest in the race ?
Discussion
Suggestions
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Fake Coin Brain Teaser
Difficulty Popularity
In front of you, there are 9 coins. They all look absolutely identical, but one of the coins is fake. However, you know that the fake coin is lighter than the rest, and in front of you is a balance scale. What is the least number of weightings you can use to find the counterfeit coin?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Find KIller From Calendar Puzzle
Difficulty Popularity
In a friend's get together party on New Year's Eve, a power cut takes place unexpectedly. When the lights go live again, it is found that Joseph, one of the friends has been stabbed to death. Since no one else had the access to the party, the killer is one of the friends only.
Police are called and they search the surroundings for any evidence, but can't find any. Upon checking the pockets of Joseph's pants, they find a pocket calendar and they find out that it has been marked and can also find some numbers written on it.
You can see the calendar attached with this question.
You don't know the name of the friends who were there at the party. Can you still decipher the name of the killer through the picture?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Unique Brain Teaser
Difficulty Popularity
It takes ten minutes to fry a steak (five minutes for each side). You are frying the steaks in a pan that can accommodate only two steaks at one time. What is the least amount of time by which you can fry all the three steaks you have?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Quick Fire Riddle
Difficulty Popularity
There is uniqueness between these five names
Nell Edna swati itaws and Ellen.
what uniqueness am i talking about ?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Playing Cards Puzzle
Difficulty Popularity
I purchased five decks of card from the market and mixed them.
Now i have ( 5 * 52 = 260 cards).
What is minimum number of cards, i need to take out from the above 260 cards to guarantee at least one 'four of a kind'.
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : May The Fastest Car Win
Difficulty Popularity
John is out with his class of 25 boys to a local park. Each guy has a remote controlled car with them. The park has a racetrack that allows 5 cars to be raced at once. Their teacher, Mr. Ted, declares that the top three fastest cars get ice cream.
How many races are required to determine the 3 fastest cars?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Tricky Math Race Riddle
Difficulty Popularity
Rooney, Hernandez, and Robin race each other in a 100 meters race. All of them run at a constant speed throughout the race.
Rooney beats Hernandez by 20 meters.
Hernandez beats Robin by 20 meters.
How many meters does Rooney beat Robin by ?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Cool Equation Riddle
Difficulty Popularity
If 1 + 9 + 11 = 1, Then what is the value of
12 + 11 + 9 = ?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Smart Tic Tac Toe Brain Teaser
Difficulty Popularity
How will you place six crosses on a Tic Tac Toe board without forming any three-in-a-row combination?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Trick Answer Riddle
Difficulty Popularity
In Era of Abraham Lincoln, there is a world famous surgeon whose more than 90% surgeries are successful and yet almost all of his patients die.
Explain ?
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How many words can a pen with half refil... | 1,119 | 4,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-50 | longest | en | 0.955898 |
http://www.physicsforums.com/showthread.php?s=7f383aa1e5b81c0fb166cba42350b0cd&p=4142000 | 1,369,135,898,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699924051/warc/CC-MAIN-20130516102524-00011-ip-10-60-113-184.ec2.internal.warc.gz | 667,355,761 | 7,117 | ## positive definite function,semi-definite functions
Hello all!
*please explain the terms 'positive definite function' and 'semi-definite function'.*
CONTEXT:
I am reading a book on the stability analysis of non-linear models.
In the chapter for 'neighborhood stability analysis',I came across the "Lyapunov function V".
V has the following properties:
1.V is positive definite.
2.dV/dt is negative semi-definite(stable valley)
3.dV/dt is positive semi-definite(unstable valley)
I understand the usual hilltop valley visualization,but please explain the terms 'positive definite function' and 'semi-definite function'. Any level of math is understandable.
**Is there a connect between 'positive definite function' and 'positive definite matrix'?**
Tags stability analysis | 163 | 777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.769852 |
https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/BettiCharacters/html/_character_lp__Character__Decomposition_cm__Character__Table_rp.html | 1,716,712,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00561.warc.gz | 313,614,272 | 3,279 | next | previous | forward | backward | up | index | toc
# character(CharacterDecomposition,CharacterTable) -- recover character from decomposition
## Description
Use this function to recover a character from its decomposition into a linear combination of the irreducible characters in a character table. The shortcut d*T is equivalent to the command character(d,T).
As an example, we construct the character table of the symmetric group on 3 elements, then use it to decompose the character of the action of the same symmetric group permuting the variables of a standard graded polynomial ring.
i1 : s = {2,3,1} o1 = {2, 3, 1} o1 : List i2 : M = matrix{{1,1,1},{-1,0,2},{1,-1,1}} o2 = | 1 1 1 | | -1 0 2 | | 1 -1 1 | 3 3 o2 : Matrix ZZ <-- ZZ i3 : R = QQ[x_1..x_3] o3 = R o3 : PolynomialRing i4 : P = {1,2,3} o4 = {1, 2, 3} o4 : List i5 : T = characterTable(s,M,R,P) o5 = Character table over R | 2 3 1 ----+----------- X0 | 1 1 1 X1 | -1 0 2 X2 | 1 -1 1 o5 : CharacterTable i6 : acts = {matrix{{x_2,x_3,x_1}},matrix{{x_2,x_1,x_3}},matrix{{x_1,x_2,x_3}}} o6 = {| x_2 x_3 x_1 |, | x_2 x_1 x_3 |, | x_1 x_2 x_3 |} o6 : List i7 : A = action(R,acts) o7 = PolynomialRing with 3 actors o7 : ActionOnGradedModule i8 : c = character(A,0,10) o8 = Character over R (0, {0}) => | 1 1 1 | (0, {1}) => | 0 1 3 | (0, {2}) => | 0 2 6 | (0, {3}) => | 1 2 10 | (0, {4}) => | 0 3 15 | (0, {5}) => | 0 3 21 | (0, {6}) => | 1 4 28 | (0, {7}) => | 0 4 36 | (0, {8}) => | 0 5 45 | (0, {9}) => | 1 5 55 | (0, {10}) => | 0 6 66 | o8 : Character i9 : d = c/T o9 = Decomposition table | X0 X1 X2 -----------+------------ (0, {0}) | 1 0 0 (0, {1}) | 1 1 0 (0, {2}) | 2 2 0 (0, {3}) | 3 3 1 (0, {4}) | 4 5 1 (0, {5}) | 5 7 2 (0, {6}) | 7 9 3 (0, {7}) | 8 12 4 (0, {8}) | 10 15 5 (0, {9}) | 12 18 7 (0, {10}) | 14 22 8 o9 : CharacterDecomposition i10 : c === d*T o10 = true | 804 | 1,850 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.449215 |
https://adainteraktif.com/qa/quick-answer-how-many-lights-can-be-on-a-15-amp-circuit.html | 1,611,524,399,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00483.warc.gz | 211,334,700 | 8,687 | # Quick Answer: How Many Lights Can Be On A 15 Amp Circuit?
## How many 60 watt bulbs can be on a 15 amp circuit?
The main limitation to the number of recessedfixtures you can run at the same time is the ratingof the circuit breaker that controls the circuit.
Each fixture witha 60-watt incandescent or halogen bulb draws about 1/2 amp, so a15-amp breaker for a standard lighting circuit would be ableto handle 30 of them..
## How many 100 watt bulbs can be on a 15 amp circuit?
Number of Lights For 100-watt fixtures, you can have a maximum of 12 divided by 0.9A, or 13 fixtures. If the lights will be used less than three hours at a time, you could install 16 fixtures on the circuit.
## How many outlets can be on a 15 amp breaker?
10 outletsGenerally, homes have eight to 10 outlets on a 15 amp breaker but not all are used at the same time. Lights and outlets are normally wired on separate branch circuits, which means that the lights will be protected by a separate circuit breaker to the one which protects the outlets.
## Is it OK to mix 12 and 14 gauge wire?
The amp capacity for 12-gauge wires is 20 amps, and while you can technically use 14-gauge wires on 20-amp circuits, it is not recommended.
## How many outlets can you put on a 14 2 wire?
10 receptaclesfor both 15A and 20A circuits the rule of thumb is a maximum of 10 receptacles per circuit. 15 amp breaker circuit 14/2 wire. Some say to limited total number of outlets to 10, meaning the total number of consistent in-use outlets, correct?
## Can I put lights and outlets on the same circuit?
Basic answer to your question of can a mixture of lights and receptacles be installed on a single circuit is yes. Check with local authorities on those limitations but yes it can be done. … Also be sure to install a GFCI receptacle, not a conventional duplex receptacle.
## What size wire is used for lights?
Wire Gauges and UsesWire UseRated AmpacityWire GaugeLow-voltage Lighting and Lamp Cords10 Amps18 GaugeExtension Cords13 Amps16 GaugeLight Fixtures, Lamps, Lighting Runs15 Amps14 GaugeReceptacles, 110-volt Air Conditioners, Sump Pumps, Kitchen Appliances20 Amps12 Gauge7 more rows
## How many watts can a 1mm cable take?
Choosing the correct size cableConductor SizeCurrentMaximum power (Watts)1.0 mm210 ampsUp to 2400 Watts1.25 mm213 ampsUp to 3120 Watts1.5 mm215 ampsUp to 3600 Watts2.5 mm220 ampsUp to 4800 Watts1 more row
## How many watts can a 15 amp switch handle?
1400 wattsSome 15-amp wall switches are only rated for 120 volts. A 15-amp switch can support 1400 watts, while a 20-amp switch can support up to 2400 watts.
## How many amps does a 100 watt bulb use?
0.90 AmpsA 100 Watts bulb draws 0.90 Amps.
## How many amps does a 100 watt LED light bulb use?
0.90 AmpsA 100 Watts bulb draws 0.90 Amps.
## How many outlets and lights can be on one circuit?
One rule of thumb is to assign a maximum draw of 1.5 amps to each receptacle, which allows for 10 receptacles on a 20-amp circuit.
## How many LED lights can be on a 15 amp circuit?
Breaker Limitations Each CFL or LED bulb typically gives the same amount of light as a 60-watt incandescent bulb while drawing 10 watts or less, which is equivalent to a current draw of 1/12 amp. Thus a 15-amp circuit can safely control 180 or more fixtures that use CFL or LED bulbs.
## Can I use 12 gauge wire on a 15 amp breaker?
Because it’s thinner and lighter, 14-gauge wire is easier to run than 12-gauge wire. However, 12-gauge wire is acceptable on both 15- and 20-amp circuits, so some electricians use it exclusively when wiring a house.
## What appliances need a dedicated circuit?
Appliances that need a dedicated circuit include:Electric ranges.Wall ovens.Refrigerators.Large Microwaves.Freezers.Dishwashers.Garbage disposals.Toaster ovens.More items…
## What size cable do I need for lighting circuit?
Lighting circuits are generally run in 1mm2 two-core-and-earth cable, but particularly long circuits can use 1.5mm2 cable to compensate for the drop in voltage experienced on long cable runs.
## How many 75 watt lights can be on a 15 amp circuit?
30 lights30 lights at 75W is 2250 total, 15 amps on a 230V breaker 3450 available. generally we use 80% for lighting loads 2750 is left so this would pass code in the U.S. (Your local regulations may be different).
## Can I run a refrigerator on a 15 amp circuit?
Most appliance manufacturers thus recommend their refrigerators to be installed on a dedicated circuit, meaning an independent one that only serves the refrigerator, with a 15 or 20 amp circuit breaker or time-delay fuse.
## Is 1mm cable OK for lighting?
1mm unless exceptionly long runs or other unusuall circumstances need larger. 1mm will be run at less than half its rating in most circumstances, if we consider 1.5 to be needed for lighting, then by the same logic for a 32 amp radial we should select cable suitable for about 60 amps !
## What can a 15 amp breaker handle?
A 15 Amp breaker can handle up to 1,800 Watts. A 20 Amp breaker can handle up to 2,400 Watts. Many of the breakers you see in the panel run entire rooms. This includes things like standard outlets and built-in lighting systems.
## How many appliances can be on a 15 amp circuit?
The total load exceeds 1,800 watts for a 15-amp circuit. (120 volts x 15 amps = 1,800 watts.) Look for the amp rating of the circuit in tiny numbers on the circuit breaker switch or fuse to determine how many outlets you can have on a 15-amp circuit. For a 20-amp circuit, the load limit is 2,400 watts. | 1,407 | 5,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.869532 |
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A044072 Numbers k such that string 2,3 occurs in the base 4 representation of k but not of k-1. 3
%I #16 Nov 27 2021 04:52:20
%S 11,27,43,59,75,91,107,123,139,155,171,203,219,235,251,267,283,299,
%T 315,331,347,363,379,395,411,427,459,475,491,507,523,539,555,571,587,
%U 603,619,635,651,667,683,779,795,811,827,843,859,875,891,907,923,939,971
%N Numbers k such that string 2,3 occurs in the base 4 representation of k but not of k-1.
%C Members are of form 16k+11, but not all such numbers are in the sequence. The first missing number is 187.
%H Michael S. Branicky, <a href="/A044072/b044072.txt">Table of n, a(n) for n = 1..10000</a>
%t Flatten[Position[Partition[Table[If[MemberQ[Partition[IntegerDigits[n, 4], 2, 1], {2, 3}], 1, 0], {n, 1000}], 2, 1], {0, 1}]] + 1 (* _Vincenzo Librandi_, Aug 19 2015 *)
%o (Python)
%o from sympy.ntheory.factor_ import digits
%o def has23(n): return "23" in "".join(map(str, digits(n, 4)[1:]))
%o def ok(n): return has23(n) and not has23(n-1)
%o print([k for k in range(972) if ok(k)]) # _Michael S. Branicky_, Nov 27 2021
%Y Differs from A106839.
%Y Cf. A044453.
%K nonn,base
%O 1,1
%A _Clark Kimberling_
%E a(48) and beyond from _Michael S. Branicky_, Nov 27 2021
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SQL Question
# sql lowest running balance in a group
I've been trying for days to solve this problem to no solution.
I want to get the lowest running balance in a group.
Here is a sample data
The running balance is imaginary and is not part of the table.
the running balance is also computed dynamically.
the problem is I want to get the lowest running balance in a Specific month (January)
so the output should be 150 for memberid 10001 and 175 for memberid 10002 as highlighted in the image.
my desired out put should be
memberid | balance
10001 | 150
10002 | 175
Is that possible using sql query only?
PS. Using c# to compute lowest running balance is very slow since I have more than 100,000 records in my table.
I've updated the question.
The answer provided by Mihir Shah gave me the idea how solve my problem. His answer takes to much time to process making it much slower than my computation on my c# program. here is my answer to get the minimum lowest value in a speific group (specific month) with a running value or running total without sacrificing a lot of performance.
with IniMonth1 as
(
select a.memberid, a.iniDeposit, a.iniWithdrawal,
(cast(a.iniDeposit as decimal(10,2)) - cast(a.iniWithdrawal as decimal(10,2))) as RunningTotal
from
(
select b.memberid, sum(b.depositamt) as iniDeposit, sum(b.withdrawalamt) as iniWithdrawal
from savings b
where trdate < '01/01/2016'
group by b.memberid
) a /*gets all the memberid, sum of deposit amount and withdrawal amt from the beginning of the savings before the specific month */
where cast(a.iniDeposit as decimal(10,2)) - cast(a.iniWithdrawal as decimal(10,2)) > 0 /*filters zero savings */
)
,DetailMonth1 as
(
select a.memberid, a.depositamt,a.withdrawalamt,
(cast(a.depositamt as decimal(10,2)) - cast(a.withdrawalamt as decimal(10,2))) as totalBal,
Row_Number() Over(Partition By a.memberid Order By a.trdate Asc) RowID
from savings a
where
a.trdate >= '01/01/2016'
and
a.trdate <= '01/31/2016'
and (a.depositamt<>0 or a.withdrawalamt<>0)
) /* gets all record within the specific month and gives a no of row as an id for the running value in the next procedure*/
,ComputedDetailMonth1 as
(
select a.memberid, min(a.runningbalance) as MinRunningBal
from
(
select a.rowid, a.memberid, a.totalbal,
(
sum(b.totalbal) +
(case
when c.runningtotal is null then 0
else c.runningtotal
end)
)as runningbalance , c.runningtotal as oldbalance
from DetailMonth1 a
inner join DetailMonth1 b
on b.rowid<=a.rowid
and a.memberid=b.memberid
left join IniMonth1 c
on a.memberid=c.memberid
group by a.rowid,a.memberid,a.totalbal,c.runningtotal
) a
group by a.memberid
) /* this gets the running balance of specific month ONLY and ADD the sum total of IniMonth1 using join to get the running balance from the beginning of savings to the specific month */
/* I then get the minimum of the output using the min function*/
, OldBalanceWithNoNewSavingsMonth1 as
(
select a.memberid,a.RunningTotal
from
IniMonth1 a
left join
DetailMonth1 b
on a.memberid = b.memberid
where b.totalbal is null
)/*this gets all the savings that is not zero and has no transaction in the specific month making and this will become the default value as the lowest if the member has no transaction in the specific month. It also gets the minimum transaction in the month */
,finalComputedMonth1 as
(
select a.memberid,a.runningTotal as MinRunTotal from OldBalanceWithNoNewSavingsMonth1 a
union
select b.memberid,b.MinRunningBal from ComputedDetailMonth1 b
)/*unions the minimum running total with the clients wiht no current transaction */
select * from finalComputedMonth1 order by memberid /* display the final output */
I have more than 600k savings record on my savings table
Surprisingly the performance of this code is very efficient.
It takes almost 2hrs to compute using my c# program.
This code makes only 2 secs and at most 9 secs just to compute everything. i Just display to c# for another 2secs.
The output was tested and compared with my computation using my c#
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The box and whiskers plot shows the weights, in kilograms, of randomly selected packages in a mail truck. Which of the following conjectures does the plot support? A. The median weight of the packages in the truck is about 30 kilograms. B. About three fourths of the packages in the truck weight between 20 and 50 kilograms. C. One half of the packages in the truck weigh more than 50 kilograms. D. The truck can hold no more than 80 packages.
Mathematics
• Stacey Warren - Expert brainly.com
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Lecture9
Lecture9 - Advanced Mathematical Programming IE417 Lecture...
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Unformatted text preview: Advanced Mathematical Programming IE417 Lecture 9 Dr. Ted Ralphs IE417 Lecture 9 1 Reading for This Lecture Chapter 6, Section 1-2 IE417 Lecture 9 2 Lagrangian Duality IE417 Lecture 9 3 The Primal Problem Given functions f : R n R , g : R n R m , and h : R n R l , consider the constrained optimization problem P , which we will now call the primal problem : min f ( x ) s.t. g ( x ) h ( x ) = 0 x X Here, X is a set that implicitly enforces additional constraints. There is usually more than one way to define X and this choice can be important, as we will see. IE417 Lecture 9 4 The Dual Problem We can now formulate the following dual problem D : max ( ,v ) s.t. where ( ,v ) = inf { ( x,,v ) : x X } . How do we interpret this? IE417 Lecture 9 5 Weak Duality Theorem 1. Let x be a feasible solution to the primal problem P and let ( ,v ) be a solution to the dual problem D . Then f ( x ) ( ,v ) ....
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Ask a homework question - tutors are online | 442 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-09 | latest | en | 0.767701 |
https://www.techwhiff.com/issue/choose-the-absolute-value-inequality-that-corresponds--5603 | 1,686,235,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00281.warc.gz | 1,095,749,036 | 12,210 | # Choose the absolute value inequality that corresponds to the given compound inequality.–5 < x + 2 < 9
###### Question:
Choose the absolute value inequality that corresponds to the given compound inequality.
–5 < x + 2 < 9
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### Double Points! (i have no clue how to do this) Find the value of the variables in the figure.
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### Jon wais In takes and is paid $3 an hour. In a 40 hour period he earned$235 in tips. What were his total earnings
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### All federal laws related to criminal offenses as well as how these crimes are prosecuted in court are contained in the _____. A. U.S. Code B. Declaration of Independence C. Bill of Rights D. Constitution
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### Which item below best exemplies the overall mood of the story? a "There were no attendants at home; they had absconded to make merry in honor of the time." b "I scarcely laid the first tier of the masonry when I discovered that the intoxication of Fortunato had in a great measure worn off." c "It was now midnight, and my task was drawing to a close. I had completed the eight, the ninth, and the tenth tier." d "At the most remote end of the crypt there appeared another less spacious. Its walls ha
Which item below best exemplies the overall mood of the story? a "There were no attendants at home; they had absconded to make merry in honor of the time." b "I scarcely laid the first tier of the masonry when I discovered that the intoxication of Fortunato had in a great measure worn off." c "It wa... | 1,017 | 4,322 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-23 | longest | en | 0.880453 |
https://www.gmatprepnow.com/module/gmat-word-problems/video/923 | 1,660,318,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00522.warc.gz | 698,374,368 | 13,599 | # Lesson: Introduction to Sequences
## Comment on Introduction to Sequences
### Each term in sequence S is
Sir i am having doubt in Solving this problem
Each term in sequence S is determined by multiplying the prior term by 2 and dividing that product by 3. What is the 100th term of the sequence S?
(1) The sum of the first 2 terms is 15
(2) The first term of the sequence is 9
### Hi Rajkumar,
Hi Rajkumar,
I just posted a solution to that question here: https://gmatclub.com/forum/each-term-in-sequence-s-is-determined-by-mult...
Cheers,
Brent
### Sir, but in statement 2,
Sir, but in statement 2,
If we multiply the first term by 2/3 we are getting the second term as 6, and next terms 4,8/3......etc.Here the common difference varies.Hence how it is possible to find 100th term?
### Here's how:
Here's how:
term1 = 9
term2 = (9)(2/3) = 6
term3 = (9)(2/3)(2/3) = 4
term4 = (9)(2/3)(2/3)(2/3) = 8/3
term5 = (9)(2/3)(2/3)(2/3)(2/3) = 16/9
.
.
.
If you recognize that we can keep doing this indefinitely, then you can see that we COULD determine the value of term100
We might also see the pattern and recognize that term100 = (9)(2/3)^99
### Hi Brent,
Hi Brent,
A little stuck on this problem: I don’t know how to finish it after the third step below
A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?
a) -24(5^12)
b) -23(5^12)
c) -24(5^6)
d) 23(5^12)
5) 24(5^12)
The nth term of an a.p. is given by: a+(n−1)d
a = first term (happen to be looking for A)
n = nth term we are looking for (A is the eighth term)
d = the common difference (4 x 5^12 is the common difference)
a + (8-1)4 x 5^12
a + 7(4 x 5^12)
a + 28 x 5^12
### You're right to say the
You're right to say the difference between any two successive values is 4(5^12). However, I think there's a problem with how you set up your equation.
Here's one approach that doesn't rely on our knowledge of arithmetic progressions:
Given: G and H are equal to 5^12 and 5^13 respectively
5^13 - 5^12 = 5^12(5 - 1) = (5^12)(4) = 4(5^12)
So, the DIFFERENCE between any two successive values is 4(5^12).
In other words, to find the NEXT term in the sequence, we just ADD 4(5^12) to the term before for it.
Conversely, to find the PREVIOUS term in the sequence, we just SUBTRACT 4(5^12).
We have:
H = 5^13
G = 5^12
F = 5^12 - 4(5^12)
E = 5^12 - 4(5^12) - 4(5^12) = 5^12 - 8(5^12)
D = 5^12 - 8(5^12) - 4(5^12) = 5^12 - 12(5^12)
Following this PATTERN, we get:
C = 5^12 - 16(5^12)
B = 5^12 - 20(5^12)
A = 5^12 - 24(5^12)
5^12 - 24(5^12) = 1(5^12) - 24(5^12)
= -23(5^12)
Does that help?
Cheers,
Brent
### A.....(4)(512)..... B.....(4)
A.....(4)(5^12)..... B.....(4)(5^12)..... C....(4)(5^12)... D......(4)(5^12)..... E.....(4)(5^12)..... F.....(4)(5^12).... G.....(4)(5^12)....H
5^12 – (4)( 5^12) x 6 (times 6 because it’s 6 steps to A from F
5^12 – (24)( 5^12)
= -23(512)
Got it! But I Don’t understand why the equation “a+(n−1)d” won’t work? It’s always worked. Is it because it is going backwards (negative)?
I find this on the Forum:
The user is obviously using the equation I have always used but his setup is a little different as he seems to be equating it to 5^12. The part where he is equating it to 5^12 is where I am a little stuck/confused.
G = A + (n-1) d
5^12 = A + (7-1) 4 (5^12)
As n=7, because its the 7th term in the sequence
Therefore,
A = (5^12) - 24 (5^12)
= (5^12) (1-24) Taking 5^12 common
= - 23 (5^12)
### Quote: "I Don’t understand
Quote: "I Don’t understand why the equation “a+(n−1)d” won’t work? It’s always worked. Is it because it is going backwards (negative)?"
The formula will work; it just takes some fiddling at the end.
We know that the difference (d) = 4(5^12)
At this point, we can apply the formula term_n = a + (n−1)d, where a = the first term
If we recognize that G (aka 5^12) is the same as term_7, then we can write:
So, term_7 = a + (7−1)[4(5^12)]
Since term_7 = 5^12, we can write: 5^12 = a + (7−1)[4(5^12)]
Simplify: 5^12 = a + (6)[4(5^12)]
Simplify: 5^12 = a + 24(5^12)
Solve for a to get: a = -23(5^12)
----THAT'S THE EASIER VERSION------------
Notice that it's more work if we use term_8 (aka H)
term_8 = 5^13
So, we can write: 5^13 = a + (8−1)[4(5^12)]
Simplify: 5^13 = a + (7)[4(5^12)]
Simplify: 5^13 = a + 28(5^12)
Now what?
From here, we need to recognize that 5^13 = 5(5^12)
So, we can write: 5(5^12) = a + 28(5^12)
So, a = 5(5^12) - 28(5^12) = -23(5^12)
Cheers,
Brent
### Never mind that last comment.
Never mind that last comment. I was way overthinking it somehow for far too long without realizing/seeing the obvious.The nth term of an a.p. is given by: a+(n−1)d or n = a+(n−1)d. N=N.
### Hey Brent,
Hey Brent,
considering this Q:
https://gmatclub.com/forum/if-sequences-s-has-240-terms-what-is-the-239th-term-of-s-160762.html
If the second stmt gave a formula that wasn´t the same, would it be possible to solve it?
Thanks,
Philipp
Yes. There are some ways to create a statement 2 that would result in an answer of C
For example...
Statement 2) term1 = 5
In this case, the combined statements would be sufficient.
Cheers,
Brent
### Hi Brent,
Hi Brent,
Isn't the formula for the question in the video is:
t(n)=3+7*(n-1)
EXAMPLES
t(1)=3+7*(1-1)=0
t(2)=3+7*(2-1)=10
t(3)=3+7*(3-1)=17
...
t(41)=3+7*(41-1)=283
### Yes, that's the logic we used
Yes, that's the logic we used to find term41. We multiply 7 by 1 less than the term number.
I try not to throw out formulas without explaining the rationale behind them. Instead, I prefer that students recognize the pattern and then make the same conclusion that you made.
Cheers,
Brent
### So, is it always true for
So, is it always true for this type of question to use: last term minus 1st term = n (in this example, 41-1 = 40), multiplied by the increment (in this case, 7) plus the value of the 1st term? Thanks!
### In general, if the first term
In general, if the first term in a sequence is "a" and each subsequent term is equal to the previous term PLUS some constant value (d), we can say that:
term_n = a + d(n-1)
Cheers,
Brent
### This is the same *formula*
This is the same *formula* referred to at 03:45 - correct? Thanks.
That's correct.
### https://gmatclub.com/forum
https://gmatclub.com/forum/the-price-of-darjeeling-tea-in-rupees-per-kilogram-is-100-0-10n-317562.html
### Hi Brent,
Hi Brent,
Can you help in this question please ?
https://gmatclub.com/forum/in-the-sequence-shown-a-n-a-n-1-k-where-2-n-15-and-k-126119.html#p1029520
Thanks,
Karaan
### Hey Brent,
Hey Brent,
I'm stuck on the below. I see the pattern in your explanation but something isn't clicking in regards to the "Each term in the sequence is equal to the SQUARE of term before it". Can you help with this?
https://gmatclub.com/forum/a-sequence-of-numbers-a1-a2-a3-is-defined-as-follows-a1-3-a2-220319.html
We have:
term3 = 15
term4 = 15^2
term5 = 15^4 = (15^2)^2
term6 = 15^8 = (15^4)^2
term7 = 15^16 = (15^8)^2
term8 = 15^32 = (15^16)^2
etc
Notice that term6 = (term5)^2
That is, 15^8 = (15^4)^2
Likewise, term7 = (term6)^2
That is, 15^16 = (15^8)^2
And so on.
Does that help?
### Ohhh! Yes, I see this. After
Ohhh! Yes, I see this. After letting my brain rest I went back and it all made sense. Thank you!
### https://gmatclub.com/forum
https://gmatclub.com/forum/the-table-above-shows-the-results-of-a-survey-of-100-voters-who-each-89187.html
can I get the solution for this question
### What happened... I started at
What happened... I started at t0 instead of t1 for this question and got E.
https://gmatclub.com/forum/in-the-sequence-above-each-term-is-9-more-than-the-previous-term-wha-209261.html
Unless stated otherwise, the first term must be t1.
### https://gmatclub.com/forum
https://gmatclub.com/forum/what-is-the-thousandth-term-of-s-a-certain-sequence-of-numbers-324942.html
How, I don't see how, why is n 1000 in your solution? I can say n = 10 and n^2 = 100, why is it sufficient?
The question asked us to find the "thousandth term of S"
In other words, we want to find the value of term_1000
Statement 1: For every n, the nth term of S is n².
In other words, term_n = n²
For example: term_1 = 1² = 1
term_2 = 2² = 4
term_3 = 3² = 9
term_4 = 4²
Continuing we get term_1000 = 1000² = 1,000,000
Since we can find the precise value of term_1000, statement 1 is sufficient.
Does that help? | 2,939 | 8,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-33 | latest | en | 0.933938 |
https://web2.0calc.com/questions/help_27989 | 1,610,756,334,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00176.warc.gz | 636,455,516 | 5,614 | +0
# help
0
339
1
+496
Use the graph that shows the solution to f(x) = g(x).
f(x)= - 3/4x^2 + 3x+1
g(x) = 2^x
what is the solution to f(x) = g(x)?
x=0
x=1
x=2.
x=4
Apr 9, 2019
#1
+114040
+1
The problem here is to try to pick the correct graph
The key is to recognize that g(x) = 2^x is never below the x axis
So....the second graph is incorrect
So....it appears that the intersection points on the first graph are x = 0 and x = 2
Does this make sense ???
Apr 9, 2019 | 189 | 492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-04 | latest | en | 0.89449 |
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https://litfile.me/3-phase-transformer-wiring-diagrams-for-electrical/ | 1,566,126,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00453.warc.gz | 530,202,461 | 34,250 | # 3 Phase Transformer Wiring Diagrams For Electrical
In Wiring Diagram225 views
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Top | 1,547 | 7,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-35 | latest | en | 0.895956 |
https://justaaa.com/finance/126183-what-is-the-value-of-a-european-swap-option-that | 1,695,864,293,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00553.warc.gz | 376,830,498 | 9,745 | Question
# What is the value of a European swap option that gives the holder the right to...
What is the value of a European swap option that gives the holder the right to enter into a 3-year
annual-pay swap in 4 years where a fixed rate of 5% is paid and LIBOR is received? The swap
principal is \$10 million. Assume that the yield curve is flat at 5% per annum with annual
compounding and the volatility of the swap rate is 20%.
A yield curve flattens at a place where both the options (LIBOR as well as fixed rate pay) are equal to one another.
A volatility of a swap ratio is the rate at which the LIBOR is valued. Hence, the value of the swap option will be the amount saved due to opting for LIBOR over the fixed interest part.
Hence, LIBOR % = fixed rate - (volatility of swap rate * fixed interest rate) = 5% - 5*20% = 5-1 = 4%.
therefore, value of European swap option for 3years = 10million * (5-4)% * 3 = \$ 0.30million.
#### Earn Coins
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https://metanumbers.com/1521669000000 | 1,600,765,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00303.warc.gz | 492,224,180 | 8,198 | ## 1521669000000
1,521,669,000,000 (one trillion five hundred twenty-one billion six hundred sixty-nine million) is an even thirteen-digits composite number following 1521668999999 and preceding 1521669000001. In scientific notation, it is written as 1.521669 × 1012. The sum of its digits is 30. It has a total of 15 prime factors and 392 positive divisors. There are 398,854,400,000 positive integers (up to 1521669000000) that are relatively prime to 1521669000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 30
• Digital Root 3
## Name
Short name 1 trillion 521 billion 669 million one trillion five hundred twenty-one billion six hundred sixty-nine million
## Notation
Scientific notation 1.521669 × 1012 1.521669 × 1012
## Prime Factorization of 1521669000000
Prime Factorization 26 × 3 × 56 × 59 × 8597
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 15 Total number of prime factors rad(n) 15216690 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,521,669,000,000 is 26 × 3 × 56 × 59 × 8597. Since it has a total of 15 prime factors, 1,521,669,000,000 is a composite number.
## Divisors of 1521669000000
392 divisors
Even divisors 336 56 28 28
Total Divisors Sum of Divisors Aliquot Sum τ(n) 392 Total number of the positive divisors of n σ(n) 5.11843e+12 Sum of all the positive divisors of n s(n) 3.59676e+12 Sum of the proper positive divisors of n A(n) 1.30572e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23356e+06 Returns the nth root of the product of n divisors H(n) 116.538 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,521,669,000,000 can be divided by 392 positive divisors (out of which 336 are even, and 56 are odd). The sum of these divisors (counting 1,521,669,000,000) is 5,118,431,358,240, the average is 130,572,228,52.,653.
## Other Arithmetic Functions (n = 1521669000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 398854400000 Total number of positive integers not greater than n that are coprime to n λ(n) 6232100000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56312135380 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 398,854,400,000 positive integers (less than 1,521,669,000,000) that are coprime with 1,521,669,000,000. And there are approximately 56,312,135,380 prime numbers less than or equal to 1,521,669,000,000.
## Divisibility of 1521669000000
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 2 0 3
The number 1,521,669,000,000 is divisible by 2, 3, 4, 5, 6 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1521669000000)
Base System Value
2 Binary 10110001001001010100010100100101101000000
3 Ternary 12101110200201200101120010
4 Quaternary 112021022202210231000
5 Quinary 144412334231000000
6 Senary 3123013423344520
8 Octal 26111242445500
10 Decimal 1521669000000
12 Duodecimal 206aaba52140
20 Vigesimal 2j8g1b5000
36 Base36 jf1mfmtc
## Basic calculations (n = 1521669000000)
### Multiplication
n×i
n×2 3043338000000 4565007000000 6086676000000 7608345000000
### Division
ni
n⁄2 7.60834e+11 5.07223e+11 3.80417e+11 3.04334e+11
### Exponentiation
ni
n2 2315476545561000000000000 3523388879607261309000000000000000000 5361431633043101708804721000000000000000000000000 8158324311621063534135170999349000000000000000000000000000000
### Nth Root
i√n
2√n 1.23356e+06 11502 1110.66 273.19
## 1521669000000 as geometric shapes
### Circle
Diameter 3.04334e+12 9.56093e+12 7.27428e+24
### Sphere
Volume 1.47587e+37 2.90971e+25 9.56093e+12
### Square
Length = n
Perimeter 6.08668e+12 2.31548e+24 2.15196e+12
### Cube
Length = n
Surface area 1.38929e+25 3.52339e+36 2.63561e+12
### Equilateral Triangle
Length = n
Perimeter 4.56501e+12 1.00263e+24 1.3178e+12
### Triangular Pyramid
Length = n
Surface area 4.01052e+24 4.15235e+35 1.24244e+12
## Cryptographic Hash Functions
md5 41a2ea05308e844051a597fbf727ea10 0908559317dcdacec293e02aa9cc56273eac4b2b e9742abaaa352c915e6b00d78e2e4371288e31d6acf5fa2070230cf0bac12876 b203e88862a6f31d6eb4b50491004f718adb84de12911d15f51a5c7122f501f81f3a83b854eed1c43e4a425f42429b1c84ca841f3b67e1d70de9bcde4d739e9e 61b7da13f8b40938babf7b2072ed07a221efd260 | 1,745 | 4,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-40 | latest | en | 0.758502 |
https://socratic.org/questions/how-do-you-solve-the-following-system-8x-6y-3-2x-3y-7 | 1,571,792,688,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00008.warc.gz | 709,768,214 | 7,073 | # How do you solve the following system?: -8x -6y =3, -2x +3y = -7
Nov 13, 2015
Isolate one variable via adding/subtracting the equations and determine its value, then substitute that back into the equations to determine the other variable.
#### Explanation:
When confronted with a system of equations that consists of an equal number of variables and equations, it behooves us to attempt to manipulate the equations such that we can isolate and determine the value of one of the variables, which normally allows us to substitute that value for the variable in the equation.
In this case, our system of equations consists of:
A) $- 8 x - 6 y = 3$
B) $- 2 x + 3 y = - 7$
Now our manipulation begins. How can we add or subtract multiples of $A$ and $B$ from/to each other, such that we isolate a variable? In order to do this we must cancel out one variable; thus, we must add/subtract the equations such that we either have 0$x$ or 0$y$.
From examining the equation, we determine there are two (well, technically 3) ways to do this:
1) $A - 4 B$: yields $- 8 x - 6 y + 8 x - 12 y = 3 + 28 \implies - 18 y = 31 \implies y = - \frac{31}{18}$
2)$4 B - A$: yields $- 8 x + 12 y + 8 x + 6 y = - 28 - 3 \implies 18 y = - 31 \implies y = - \frac{31}{18}$
3)$A + 2 B$: yields $- 8 x - 6 y - 4 x + 6 y = 3 - 14 \implies - 12 x = - 11 \implies x = \frac{11}{12}$
Note that 1 and 2 are essentially the same operation, simply reordered.
Although above we have already determined the solution (e.g. $x = \frac{11}{12} , y = - \frac{31}{18}$), it should still be illustrated how one would substitute the determined value for a variable back into the original system in order to obtain the other variable(s).
We might prefer operation $3$ here, as it requires the least amount of multiplying. Thus, from operation 3 we determine $x = \frac{11}{12}$. Now we will substitute $x = \frac{11}{12}$ back into equation B to determine $y$. Of note is that we could also insert the value for $x$ into equation A, but with equation B we completely cancel the $x$ coefficient, which seems marginally easier.
B) $- 2 x + 3 y = - 7 \implies - 2 \left(\frac{11}{12}\right) + 3 y = - 7 \implies - \frac{11}{6} + 3 y = - 7 \implies 3 y = - \frac{31}{6} \implies y = - \frac{31}{18}$
The solution we have obtained is $x = \frac{11}{12} , y = - \frac{31}{18}$
If we wish, then at this point we may substitute the values for $x$ and $y$ into equation A to ensure they solve both (since we have determined they solve B)
A) $- 8 x - 6 y = 3 \implies - 8 \left(\frac{11}{12}\right) - 6 \left(- \frac{31}{18}\right) = 3 \implies - \frac{22}{3} + \frac{31}{3} = 3 \implies \frac{9}{3} = 3 \implies 3 = 3$
The solution checks out. | 858 | 2,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 24, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2019-43 | longest | en | 0.858361 |
https://hackernoon.com/quantum-computing-explained-94125280fabb | 1,722,737,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00436.warc.gz | 233,094,586 | 45,905 | Quantum Computing Explainedby@shaanray
# Quantum Computing Explained
April 15th, 2019
Advances in computing require appropriate hardware. Though computers have become smaller and more powerful over time, the power of regular computers (also known as classical computers) is limited.
### Introduction
Advances in computing require appropriate hardware. Though computers have become smaller and more powerful over time, the power of regular computers (also known as classical computers) is limited.
Quantum computers are a new generation of computers built to solve the problem of exponential scaling (for example, finding the optimal solution to a problem in which there are too many possibilities for a classical computer to analyze).
### Background on Classical Computers
Classical computers have many components (including main memory, arithmetic unit, control unit, and others). They represent, process, and control data through these components. Computer chips contain modules, which include logic gates, which in turn include transistors.
A computer module is a collection of electronic circuits on a circuit board. The logic gates are tiny computers within the computer itself. They look at two bits and push one of them out as an output. Their job is to read any input, in order to produce an output. A transistor is a switch that either allows or denies information to pass through it. The combinations of the logic gates form modules that allow for the basic functions of a computer. If we think of the transistor as an electric switch, the electricity is moving from one place to another when the switch is on. If the switch is off, then electrons are blocked.
Computer components are getting smaller. A typical scale for transistors today is 14 nanometers, which is 500 times smaller than a red blood cell. As these transistors decrease in size, electrons can move to the other side of a blocked passage, resulting in it not being blocked at all. This process is called ‘quantum tunneling’, and it is slowing down our technological progress.
Our computers are based on a binary system (also known as a base 2 numeral system), which uses 0 and 1 as bits. A bit, derived from ‘binary unit’, is a unit of information for a computer that holds the values of 0 or 1. For example, a 64-bit computer can work with 64 binary numbers at a time.
Combinations of these bits are used to represent more complex data and operations. The logic gate performs a Boolean function, producing a single binary output. Boolean logic is a division of algebra that is used to create true and false statements. Since classical computers operate in binary, their logic is expressed in Boolean terms. The computer uses operators such as AND, OR and NOT to express the value and return a true or false output. For example, if you have values for x and y and the logical expression states x AND y, the computer would return true if both are true, while if it said x OR y, it would return true if at least one were true.
An easy programming example, minus the programming language would be:
x = 2, y = 4
x AND y are greater than 1: this would return true.
x OR y are greater than 2: this would also return true, as y is greater than 2.
In a classical computer, a true statement could (for example) return a value of 1, while a false statement could return a value of 0. This forms the basis of classical computing: though most calculations require more than one simple true/false statement, classical computers are large combinations of these binary statements. This is what everything from clicking a mouse to opening a browser rests on.
### What Are Quantum Computers
For certain problems that a classical computer solves, you may need a very small number of logic gates. However, if you want to find the factors of an extremely large number, it is going to take a large magnitude of logic gates (which a classical computer will not have).
A quantum bit, known as a qubit, is a computer bit that is able to hold two different states at once, meaning that it can hold a position of 0 and 1 at the same time. A regular bit can only hold one of the two at a specific time.
A qubit can be any two-level quantum system — a spin and a magnetic field, or a single photon (particle representing a quantum of light). This system’s possible states are 0 and 1. Within the quantum realm, the qubit can be in any proportion of both states at once; this phenomenon is called superposition.Superposition allows quantum computers to analyze far more possibilities than a classical computer. As soon as you test the value of the photon by sending it through a filter, it needs to decide on vertical or horizontal polarization. You cannot predict if it will decide on position 0 or 1. When in superposition, the photon is in some combination of both 0 and 1 simultaneously. However, as soon as you measure its value, it collapses into one of the defined states.
An exhibit of an IBM quantum computer.
### Numerical Example
Think of 4 bits that can be on or off. Such a system would have 2⁴ (so, 16) possible combinations. In a traditional setting, you can use only one of these. However, qubits could hold all of these 16 combinations at once. This number grows exponentially with every added qubit.
### Entanglement
Qubits can hold the property of ‘entanglement’, a close connection between qubits that allows them to react to a change in the other’s state instantly no matter how far apart they are. Therefore, when measuring one entangled qubit, you can directly conclude the properties of its partner without having to test it.
### Qubit Manipulation
As a traditional logic gate receives a simple set of inputs, it produces one definite output. The quantum gate manipulates an input of superpositions, rotates probabilities, and finally produces a determined state as its output. To break down the steps, a quantum computer:
1. Sets up qubits
2. Applies qubit gates to entangle them and manipulate probabilities
3. Measures the outcome, collapsing the qubits into one defined state: a sequence of 0s and 1s. This means no more superposition.
This way, all calculations in the setup are done at the exact same time.
### Simplistic Overview
The essential power of a quantum computer is that you can consider many states simultaneously. In order to make it work, its algorithm must be able to produce an end state that is readable (so, the information that you read out at the end cannot have superpositions). This means that quantum computers require a more complex algorithm design to be useful.
### Where Quantum Computers are Effective
Quantum computers will most likely not replace our home computers. However, they can be superior in areas such as data searching. To find something in a database, a classical computer must test every one of its entries. A quantum computer will take the square root of that time to come up with the same answer.Quantum computers can challenge existing IT security measures. Currently, data is protected by various levels of encryption. In this case, you can give everyone a public key to encode messages only you can decode. While technically this public key can be used to calculate your secret private key by the use of trial and error on a classical computer, it would take far too long to be worth anyone’s time. A quantum computer with exponentially higher speed will be able to do it much faster.
### Two conceptual explanations of how quantum computers work
Example 1
You have a 10 person dinner party. You need to figure out how to seat everyone. There are 10! = 3,628,800 ways of doing so (where ‘10!’ is pronounced ‘ten factorial’ and represents 10 x 9 x 8 x…x 1. A classical computer will have to go through each of the 3.6 million ways individually and then compare them to figure out the best optimization.
A quantum computer would:
1. Take the qubits and go into a superposition of all possible states and configurations
2. When this problem is encoded into a quantum computer, the encoder applies a phase on each of the states. When the ways are in phase, the amplitudes add. When they are out of phase, the amplitudes cancel. This is a similar idea to noise cancelling headphones, which create noise to phase out outer noise.
3. You use interference to amplify some answers, while cancelling the others, finally approaching one answer.
Example 2
Here we have a maze. Imagine you are in the center of it and want to get out at either of the two exits. You can start exploring each path, one by one. After a lot of tries, you will finally get out of the maze.
Now, imagine you have with you several clones of yourself. Everyone can start exploring the different ways, and one clone will directly find the correct way out. You and your clones were exploring all the different paths at once, meaning that you were all in different places at the same time. You were in a superposition of states, like a qubit. This allows you to find the best solution possible, quickly.
These simplified, conceptual examples help explain why quantum computers (when created) will be immensely helpful in solving large, complex problems. | 1,965 | 9,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.931081 |
https://in.mathworks.com/matlabcentral/answers/371292-deleting-an-irregular-row-column-from-2-d-array?s_tid=prof_contriblnk | 1,679,569,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00555.warc.gz | 362,944,847 | 25,756 | # Deleting an irregular "row"/"column" from 2-d array
1 view (last 30 days)
Knut on 5 Dec 2017
Edited: Matt J on 5 Dec 2017
I have a 2d array, and I want to eliminate the same number of elements in each row (or column) (e.g. 1 element), but the index of that element is itself av vector. This operation would be relevant for doing seam-carving (content-aware image resize).
Is there any neat way of doing this without my array collapsing into a 1d vector?
A = (1:3)'*(1:3);
A =
1 2 3
2 4 6
3 6 9
b = [1; 2; 2];
for r=(1:3)
tmp = A(r,:);
tmp(b(r)) = [];
Y(r,:) = tmp;
end
Y =
2 3
2 6
3 9
I seem to have a working method, but it is kind of cumbersome as I have to (?) make sure that the linear indexing into Y is strictly monotonic:
Y = A';
Y(sub2ind(size(A'), b, (1:3)')) = [];
Y = reshape(Y, size(A,2)-1, size(A,1))'
Anything neater?
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### Answers (1)
Matt J on 5 Dec 2017
Edited: Matt J on 5 Dec 2017
I think your solution is probably the best, except that I would avoid the extra calls to size() and ctranspose(). The latter are particularly expensive.
Y = A.';
[m,n]=size(Y);
Y( sub2ind([m,n], b, (1:3)) )=[];
Y = reshape(Y, m-1, n).'
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## How do you wire a light with two white and two black wires?
Take your always hot black and put it on your light switch, than take the other black in that box and put it on the light switch. Take your white that is in the same cable as your always hot black and wire nut it with your other white. Now go up to your light box.
## Why would both black and white wires be hot?
Here’s a rundown of electrical wires: The black wire is the “hot” wire, it carries the electricity from the breaker panel into the switch or light source. The white wire is the “neutral” wire, it takes any unused electricity and current and sends it back to the breaker panel.
## How do you wire a ceiling fan with two white wires?
Quote from Youtube video: The white is neutral. The green wire is for the ground. And a red blue or striped wire can act as a conductor to carry power to the fans light kit.
## Why are there 2 sets of wires in one outlet?
When an outlet receptacle falls in the middle of a circuit run rather than at the end, there are generally five wires in the outlet box. Two cables are hot wires—one bringing power in, the other carrying it onward to the next receptacle. Two cables are neutral and serve the same function.
## Why does my light fixture have double wires?
to allow you to control the bulbs separately, but from the same circuit (each black to a different switch, whites together).
## How do you wire a light with two sets of wires?
Connect the lead wires from the fixture to the black hot wire in the electrical box by twisting a wire connector to the ends of the lead and hot wires. Repeat these connections with the two white neutral wires from the light fixture and the white neutral wire in the light fixture box.
## What is a wire that is both black and white?
Answer: Well, the black one is the wire that’s actually termed as the hot wire. It carries electricity to your light source or switches from your breaker panel. Then the white is one is known as the common or neutral wire. This wire sends back unused electricity to that same breaker panel.
## What happens if you mix up hot and neutral wires?
This happens when the hot and neutral wires get flipped around at an outlet, or upstream from an outlet. Reversed polarity creates a potential shock hazard, but it’s usually an easy repair. Any \$5 electrical tester will alert you to this condition, assuming you have a properly grounded three-prong outlet.
## What happens if you connect the wrong wires on a light fixture?
But here’s the catch: If you connect the circuit wires to the wrong terminals on an outlet, the outlet will still work, but the polarity will be backward. When this happens, a lamp, for example, will have its bulb socket sleeve energized rather than the little tab inside the socket.
## Which wire is hot when both are same color?
In most modern fixtures the neutral wire will be white and the hot wire is red or black. In some types of fixtures, both wires will be the same color. In this case, the neutral wire is always identified by some means. In some cases, there will be small writing on the wiring case.
## What happens when two live wires touch?
You will receive a shock if you touch two wires at different voltages at the same time. You will receive a shock if you touch a live wire and are grounded at the same time. When a circuit, electrical component, or equipment is energized, a potential shock hazard is present.
## What happens when two hot wires touch?
A short circuit happens when a “hot” wire (black) touches another hot wire or touches a “neutral” wire (white) in one of your outlets. When these two wires touch, a large amount of current flows, creating more heat than the circuit can handle, so it shuts off.
## Why are there two black wires?
Black means hot, white signifies neutral, and green indicates ground. However, if you need to rewire a light switch or a plug socket, you may occasionally come across two black wires. It’s essential that you determine which black wire is hot before proceeding.
## Can you put two black wires together?
This is the typical way to connect a switch. The 2 blacks connected are “power in, power out” to the next switch in the circuit. The pig-tail to the dimmer connected to the two blacks is also typical.
## Why are the black and white wires connected together?
Because the neutral is re-tasked to be a hot, it must be marked with a few wraps of tape. White is used for always-hot because another rule requires this. That’s so when you’re at the other end, it’s easier to detect that the white Is hot, because it’s always hot.
## How do you wire a light fixture with two white wires?
Quote from the video:
Quote from Youtube video: The two ground wires from the two cables are connected together but the ground wire from the light was not connected together with those two ground wires.
## Why are there two hot wires?
The reason for multiple hot/neutral wires for one outlet is that the outlets are daisy-chained together. This means hot/neutral is only coming from one of the wires and it is being sent to the other wire.
## Why are all neutral wires tied together?
The neutrals were all tied together to make one group so it doesnt matter which one makes it to the switch.
## Can neutral wires touch each other?
If you touch the neutral wire in a live circuit, whether it be a lamp, an appliance or something else, it is the same as touching the active wire. It is only “safe” to touch the neutral wire when there is no current flowing, just as it is “safe” to touch the earth wire (when one exists).
## What happens if black and white wires are reversed?
So, what happens with reversed polarity? In reversed polarity, both the hot wire and neutral wire get switched, causing the electric current to flow backward, entering the appliance through the neutral terminal instead of the hot terminal, which energizes the appliance even when off. | 1,334 | 6,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-14 | longest | en | 0.915033 |
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how to determine the capacity of a stone mining mill. Ball mill capacity caculation capacityth capacity calculation of ball mill capacity calculation of ball mill calculate ball mill grinding capacity metallurgist mar the sizing of ball mills and ball milling circuits from laboratory grinding tests is largely a question of,ball mill capacity caculation email protected
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### how to determine the capacity of a crusher
how to determine the capacity of a crusher. Popular Searches . cip flow rate calculation Know More. nbsp 0183 32 specifying the clean-in-place cip mode for uf dupontthis flowrate can be specified in the 168 design 168 window as shown in figure 4 from the recycle flow rate and duration the cip system design considerations for cleaning pharmaceutical 8 aug 2017 determining What determines
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### how to determine the capacity of a crusher
The width of the jaw determines capacity. Jaw crusher output gradation is controlled by the closed side setting. This is the adjustable opening at Read more. How to Calculate the Production Capacity of Double Roll Crusher Production capacity is an important index measuring the performance of double roll crusher as it has a direct relationship with the selection of all the parameters
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### How To Determine The Crushing Capacity Of Jaw Crusher
How To Determine The Capacity Of A Crusher. How to determine the capacity of a crusher Calculating jaw crusher capacity melbourne a jaw crusher with a closed setting of 3 in produces 50 tons per hour of crushed stone determine the amount of stone produced in tons crusher the capacity of a roll crusher will vary with 9the kind of stone 9the size of feed 9the size of the finished product 9the
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### how to determine the capacity of a crusher
detailed calculation of capacity and power of a roll crusher calculate belt conveyor capacity used in stone crusher, buyers for rock phosphate malaysia stone crusher machine in turkey how to calculate carrying capacity How Can Determine the Chemical Component of Impact Crusher 28 Nov 2013 Before determining the chemical component of the hammers of impact life of the hammers
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### how to determine the capacity of a crusher Joniatuinen
Crusher Capacity Calculation jaw crusher capacity calculation calculating circulating load around screen and crusher kenya crushers the formula for this how to determine the capacity of a crusher detailed calculation of capacity and power of a roll crusher calculate belt conveyor capacity used in stone crusher, crusher capacity.
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### Stone Crusher Plant Crusher Plant Capacity
Stone Crushing Plant Capacity. CRUSHER PLANT crushes such as limestone, granite, basalt and so on to product aggregates for ready mix concrete, building material, construction site and others. Various final product sizes are available with customizing crushing plant system solution. You can choose Stationary crusher plant type or Portable crushing plant (semi mobile crushing plant) with
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### how to calculate impact crusher capacity
Stone crusher is my company in absorbing foreign advanced technology on the mill capacity » how to mill a cylinder head on a vertical mill » how to calculate . Read more. Crushing Machine. The output is decided by the processing capacity of stone crusher. But under the conditions of materials' volume and density, we can calculate the output. Read more. Chapter 5. Gyratory and Cone
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### How To Determine Impact Crusher Capacity
How To Determine The Capacity Of An Impact Crushers. Calculation Capacity Crusher rusher How To Calculate Stone Crusher Capacity wffofoundationHow To Determine Impact Crusher Capacity 13 Feb 2016 Impact crusher or impactor crusher is the ideal stone not only improves the production capacity and crushingcalculation model of a cone Read more.
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### how to determine the capacity of a crusher
how to determine the capacity of a crusher. Popular Searches . cip flow rate calculation Know More. nbsp 0183 32 specifying the clean-in-place cip mode for uf dupontthis flowrate can be specified in the 168 design 168 window as shown in figure 4 from the recycle flow rate and duration the cip system design considerations for cleaning pharmaceutical 8 aug 2017 determining What determines
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### With Capacity Of Crushing T How To Use A Rock Crusher
The solutions of set up quarrying stone crusher unitAccording to the capacity and layout, we design Tesmec Model TRS 1475 Rock Hawg. In fact, the flooding was so rapid that the large rock crushing machine was left the property, purchased the site and set up the lake as a scuba training park. . HSE Quarries Safe. Read More. Mobile Crusher Jxsc Rock Crusher. The mobile stone crusher plant is
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### determine the crushing efficiency roll crusher
How To Determine The Capacity Of A Stone Crusher. how to determine the crushing capacity of jaw crusher what reason cause the abrasion of jaw crusher spare parts 2019 10 18ensp0183enspjaw crusher is the main machine in stone crushing process its
More | 1,525 | 7,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.846479 |
https://kateathome.com/bake/what-temp-does-sugar-water-boil.html | 1,642,926,446,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00475.warc.gz | 398,451,720 | 19,889 | # What temp does sugar water boil?
Contents
The higher the concentration of sugar in your solution, the higher the boiling point of this solution. As an example, water boils at 100°C (212°F), however, a solution with 80% sugar and only 20% water boils at 112°C (233°F).
## What temperature does sugar syrup boil at?
We use syrup cooked to soft ball stage to make Italian meringue, Pâte à bombe, fudge and fondant icing. Firm ball occurs when sugar syrup is cooked to 118°C (245°F) – 121°C (250°F). It is used for making caramel candy, fruit jams, and Italian meringue.
## Can you boil sugar water?
Heating the sugar and water together won’t ruin the syrup — it just takes longer to heat. It’s not necessary to bring the water to a boil. Once the sugar is dissolved, let the syrup cool.
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## How long does it take for sugar water to boil?
In a high-sided saucepan over medium-high heat, bring cold water and sugar to a boil. Turn the heat to low and stir constantly until the sugar dissolves completely and the mixture is clear, approximately 3 to 5 minutes. Remember – the longer you boil it, the thicker the syrup will be when cooled.
## What does putting sugar in boiling water do?
The addition of sugar to boiling water forms a paste, which sticks to skin and intensifies burns. It is a tactic commonly used in prisons, where it is described as “napalm” due to the way it attaches to skin and burns.
## What temp does water boil?
A liquid at high pressure has a higher boiling point than when that liquid is at atmospheric pressure. For example, water boils at 100 °C (212 °F) at sea level, but at 93.4 °C (200.1 °F) at 1,905 metres (6,250 ft) altitude. For a given pressure, different liquids will boil at different temperatures.
## Why does sugar water boil at a higher temperature?
Boiling point of sugar solutions
The higher the concentration of sugar in your solution, the higher the boiling point of this solution.
## What is the easiest way to determine temperature of boiling sugar syrup?
This CANDY – SYRUP TEMPERATURE CHART indicates two different methods of determining when the solution has been cooked to its proper sugar concentration, depending on the recipe being made: One is by TEMPERATURE using a candy thermometer to record the syrup’s boiling point in degrees F; the other is by using the COLD …
## Can you drink sugar water?
IT IS A BETTER SPORTS DRINK: Most sports drinks pack glucose sources to energise you. Only drinking sugar and water serves the purpose of replenishing your body better. This is also healthier because sometimes unknown substances are used to enhance your post activity high.
## What does sugar water do for bees?
Granulated sugar dissolved into water, mimics natural plant nectar. While it is not exactly the same nutritionally, it is very similar in sweetness. And, honey bees are accustom to collecting liquid food. When making sugar water for your bees, it is important to only use white sugar.
## How long does it take for sugar to reach 300 degrees?
Just put your candy thermometer in the pan and watch for it to rise to 300 degrees. It takes about 10 minutes from the time it starts to really boil. After it’s reached 300 degrees, remove it from the heat and add whatever flavor you want, stirring constantly then add food coloring.
## At what temperature does sugar caramelise?
Caramelization is what happens to pure sugar when it reaches 338° F. A few tablespoons of sugar put in a pan and heated will eventually melt and, at 338° F, start to turn brown. At this temperature, the sugar compounds begin to break down and new compounds form.
## How long does it take for sugar to get to the hard crack stage?
Continue boiling, uncovered, not stirring the mixture but shaking the saucepan occasionally to distribute the heat as the mixture turns amber in about 9 minutes, then darker amber as it registers 305 degrees on the thermometer (hard-crack stage ), about 3 minutes longer.
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## Why do prisoners throw boiling water and sugar?
Combining boiling water and sugar is known in prison circles as “napalm”. The mixture sticks to the skin and intensifies burns, one of the principal effects of jelly-like napalm bombs.
## How long will sugar water last in the refrigerator?
Sugar water does not stay fresh very long. It lasts just 4-5 days in the refrigerator and can spoil quickly outside depending on how hot it is.
## Is sugar water good for skin?
Sugar scrubs are touted as creating soft, smooth skin, but these are much too harsh for facial skin. Stick with using sugar scrubs only on the body, and consider alternatives that are safer for your face. The goal of a facial scrub is to gently exfoliate your skin — not irritate it. | 1,079 | 4,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.920146 |
https://www.mgvcursos.com.br/2020/01/09/what-you-need-to-do-about-battery-definition-physics-starting-in-the-next-10-minutes/ | 1,618,429,395,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00624.warc.gz | 1,017,371,919 | 10,213 | # What You Need to Do About Battery Definition Physics Starting in the Next 10 Minutes
## Top Battery Definition Physics Choices
Axis the volume of a segment, so the number of absolutely free charges in it’s nAx. The wire functions as a type of charge pipe whereby charge can flow. Drift Velocity Good conductors have large quantities of completely free charges inside them.
## Why Almost Everything You’ve Learned About Battery Definition Physics Is Wrong
At the present time, job opportunities in many areas of the Earth sciences are a lot superior than average. There are not any emissions created from using renewable resources also. Distance of Closest Approach inside this problem you’ll be requested to come across the closest approach between two charges.
## Where to Find Battery Definition Physics
Closely review your requirements, study the challenges ahead and do what’s right for you. Quite simply, it requires a whole loop for something to acquire the job finished. essay online An instance of the usage of the head-to-tail procedure is illustrated below.
It’s possible to purchase a new membership any chance to restore access to the whole breadth of Scribd’s catalog. You need to be sure to show all your work as partial solutions may get partial credit. Earlier in Lesson 1, the usage of an electric possible diagram was discussed.
This circuit is composed of D-cell and a light bulb. The movement of charge during the internal circuit necessitates energy because it is an uphill movement in a direction that’s against the electric field. Because electric potential distinction is expressed in units of volts, it can be known as the voltage.
There are a lot of different topics that are also important. Credit for those answers will be contingent on the grade of your solutions and the explanations you give. The specifics of the past two stages are usually considered within the subject of fracture mechanics.
## Choosing Battery Definition Physics Is Simple
The formula must correct any issues and it might be an equation, logical relation which might be either number or symbols. cheap research papers In this way, you’ll not just build understanding of the appropriate physics topics, but in addition the mathematical principles involved with the problems you’ll be requested to solve during the test. Still, theories can never be fully proven.
Before you’re able to commence studying, you will want to find an excellent comprehension of your present understanding. As stated above, all students in schools that are near a fast food restaurant or that have limited access to exercise opportunities share an elevated prospect of obesity. Each one of the topics covered in the course is going to be given to students in the form of typewritten text which could possibly be seen in the Lecture Notes on the class website.
Many schools teach these as the very same class, or only teach one particular topic. In an attempt to help students and instructors in preparing for the true exam, work is ongoing to construct an AP-style problem set that might be used freely in classrooms for this function. Last, don’t neglect to get used to the vital resources which are going to be available to your during the exam.
## What Battery Definition Physics Is – and What it Is Not
After it’s over, you’ve got summer break to anticipate. The simple fact they are offered on the exact same day is a sign of the way that they are normally paired. Should you need accurate figure (accurate to the last decimal place) as a consequence, then we advise you to use an actual calculator.
## The Appeal of Battery Definition Physics
As mentioned above, the very first step is to simplify the circuit by replacing both parallel resistors with a single resistor which has an equivalent resistance. Thus, the electric current flow doesn’t occur upon the P-N junction. It is the total amount of resistance a single resistor would have to have in order to equal the total effect of the selection of resistors that exist in the circuit.
## The Battery Definition Physics Trap
To predict fatigue, the total stress history has to be transformed into a stress spectrum that may be associated with fatigue within the next step of the analysis. When evaluating test results, it’s important to think about statistical effects too. Since the stress is based on the excitation frequency, the fatigue evaluation can be created in the frequency domain using, as an example, power-spectral density procedures.
If you own a teacher like that at your school, and they teach an AP subject you’re interested in, you will probably have a rather high prospect of passing. Despite such high pass prices, AP Physics 2 is among the absolute most difficult AP exams. You wish to take AP classes you believe you’ll succeed in.
Do a little research before registering for an AP class and make certain it’s the correct fit for you. Much of its content is comparable to the old AP Physics B program. If you’re thinking about college engineering but aren’t certain if it’s appropriate for you, taking AP Physics C might be a great way to find out whether you are up to the challenge and take pleasure in the topics you would continue to explore in college.
## Lies You’ve Been Told About Battery Definition Physics
The process is validated against simulated along with data generated by way of a physics code related to the quark masses of protons. For starters, for quantum computation, the experiment would want to get scaled up from a single device to likely a whole lattice of those. Linear molecules are normally non-polar.
The book has a few minor shortcomings. In fact, it’s in fact incredibly straightforward to trust in both God and evolution. Lucky you if you in the event you get a remarkable teacher.
## The Foolproof Battery Definition Physics Strategy
We are then requested to learn the current by way of this circuit. This is a good example of a combination circuit. Look carefully at the circuit above and you’ll understand that the current has to experience the path it’s in.
The illustration indicates a very easy parallel circuit. A parallel circuit differs. An electrical circuit is a sort of network with a closed loop, which gives a return path for the current.
The use of batteries isn’t restricted to the households, it’s also utilized in several other applications. Batteries exist in nearly all households. Some rechargeable batteries will endure for months and a few will endure for ages. | 1,258 | 6,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.920926 |
https://whomadewhat.org/what-do-quartiles-tell-us/ | 1,718,226,076,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00589.warc.gz | 553,657,738 | 12,430 | Quartiles tell us about the spread of a data set by breaking the data set into quarters, just like the median breaks it in half. For example, consider the marks of the 100 students below, which have been ordered from the lowest to the highest scores, and the quartiles highlighted in red.
Example: 5, 7, 4, 4, 6, 2, 8 And the result is: Quartile 1 (Q1) = 4. Quartile 2 (Q2), which is also the Median, = 5.
Subsequently, What does the first quartile tell us about data?
The fact that 72.5 is the 1st quartile tells us that a quarter of the data values are less than 25, and the rest of them are higher than 25. … and the 3rd quartile will be the median of this half of the list. So the 3rd quartile is 75.5.
Also, How do you calculate Q1 and Q3?
Q1 is the median (the middle) of the lower half of the data, and Q3 is the median (the middle) of the upper half of the data. (3, 5, 7, 8, 9), | (11, 15, 16, 20, 21). Q1 = 7 and Q3 = 16.
What does the first quartile represent?
The first quartile, denoted by Q1 , is the median of the lower half of the data set. This means that about 25% of the numbers in the data set lie below Q1 and about 75% lie above Q1 . The third quartile, denoted by Q3 , is the median of the upper half of the data set.
Last Review : 14 days ago.
## How are quartiles calculated?
The quartile measures the spread of values above and below the mean by dividing the distribution into four groups. A quartile divides data into three points—a lower quartile, median, and upper quartile—to form four groups of the dataset.
## How do you find Q1 Q2 Q3 in statistics?
Q1 is the “middle” value in the first half of the rank-ordered data set. Q2 is the median value in the set. Q3 is the “middle” value in the second half of the rank-ordered data set.
## What is the Q1 of a data set?
Q1 is the middle value in the first half of the data set. Since there are an even number of data points in the first half of the data set, the middle value is the average of the two middle values; that is, Q1 = (3 + 4)/2 or Q1 = 3.5. Q3 is the middle value in the second half of the data set.
## What does Q1 and Q3 represent?
The values that divide each part are called the first, second, and third quartiles; and they are denoted by Q1, Q2, and Q3, respectively. Q1 is the “middle” value in the first half of the rank-ordered data set. Q2 is the median value in the set. Q3 is the “middle” value in the second half of the rank-ordered data set.
## How do you find Q1 and Q2 in statistics?
Q1 is the “middle” value in the first half of the rank-ordered data set. Q2 is the median value in the set. Q3 is the “middle” value in the second half of the rank-ordered data set.
## What does the 3rd quartile represent?
The third quartile, denoted by Q3 , is the median of the upper half of the data set. This means that about 75% of the numbers in the data set lie below Q3 and about 25% lie above Q3 .
## What is third quartile example?
An Example In other words, the median is: (7 + 8)/2 = 7.5. Here the median is (15 + 15)/2 = 15. Thus the third quartile Q3 = 15.
## How are quartiles used in real life?
Some companies use the quartiles to benchmark other companies. For example, the median company pay for a given position is set at the first quartile of the top 20 companies in that region. The quartiles and IQR information is typically used when you create a box-plot of your data set.
## How do you find Q2 in statistics?
– Quartile 1 (Q1) = (4+4)/2 = 4.
– Quartile 2 (Q2) = (10+11)/2 = 10.5.
– Quartile 3 (Q3) = (14+16)/2 = 15.
## How do you find the quartiles of a data set?
– Quartile 1 (Q1) = (4+4)/2 = 4.
– Quartile 2 (Q2) = (10+11)/2 = 10.5.
– Quartile 3 (Q3) = (14+16)/2 = 15.
## What does the first quartile tell you?
The fact that 72.5 is the 1st quartile tells us that a quarter of the data values are less than 25, and the rest of them are higher than 25. … and the 3rd quartile will be the median of this half of the list. So the 3rd quartile is 75.5.
## What is a 3rd quartile?
The third quartile, denoted by Q3 , is the median of the upper half of the data set. This means that about 75% of the numbers in the data set lie below Q3 and about 25% lie above Q3 .
## How do you find the Q1 of a data set?
Q1 is the middle value in the first half of the data set. Since there are an even number of data points in the first half of the data set, the middle value is the average of the two middle values; that is, Q1 = (3 + 4)/2 or Q1 = 3.5.
## How much data is between Q1 and Q3?
3. Providing insight into interesting properties of the data. 34Since Q1 and Q3 capture the middle 50% of the data and the median splits the data in the middle, 25% of the data fall between Q1 and the median, and another 25% falls between the median and Q3.
## How do you find quartiles in statistics?
Find Quartiles: Examples Step 1: Put the numbers in order: 2, 5, 6, 7, 10, 12 13, 14, 16, 22, 45, 65. Step 2: Count how many numbers there are in your set and then divide by 4 to cut the list of numbers into quarters. There are 12 numbers in this set, so you would have 3 numbers in each quartile. | 1,509 | 5,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.926839 |
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# Credit Card Interest Rates
## What is Credit Card Interest Rate
The rate at which a credit card company charges interest on your borrowings is known as the credit card interest rate. Also referred to as finance charges, the rate of interest on credit cards is usually expressed in the form of an Annual Percentage Rate (APR).
Credit card companies usually charge different rates of interest depending on your usage. For instance, the interest charge on a credit card will be different for purchases, loans, cash advances, and EMIs.
Although the rate of credit card interest can vary from one type of card to another, it is not dependent on the user’s credit score or history. This effectively means that two different individuals with varying credit scores holding the same credit card would be charged the same amount of interest.
## How to Calculate Credit Card Interest Rate
You can use an online credit card interest calculator to arrive at the interest charged on a credit card. Alternatively, you can also do it manually through the use of a specific formula to arrive at the credit card monthly interest. Here’s a closer look.
### Formula to Calculate Credit Card Interest Rate
The formula that’s typically used to calculate credit card interest is as follows -
Credit card interest = [(Total number of days x Transaction amount x Credit card interest rate per month x 12 months)] ÷ 365 days
For instance, let’s say that you’ve spent Rs. 20,000 to purchase some products through your credit card. And that the total number of days from the first transaction date to the credit card payment due date is 28 days. The interest rate charged by the credit card provider is 3% per month. So, by applying the formula, we get the credit card interest as -
Credit card interest = [(28 x Rs. 20,000 x 36% p.a)] ÷ 365 days = Rs. 552.33.
## When is the Credit Card Interest Rate Charged
The credit card monthly interest rate is charged only under the following circumstances -
• When you repay only the minimum amount due on the card
• When you pay an amount that is lower than the outstanding due
• When you do not make your credit card payment
• When you use your credit card to withdraw cash from an ATM
• When you purchase items on EMI
## Credit Card Interest Rate Charged by Top Banks
Credit card interest rate, also known as a ‘finance charge’, is the rate that the card issuers charge on the borrowed amount and is depicted as an annual percentage rate or an APR. However, these interest rates apply to those cardholders whose outstanding amount has not been paid in full each month and vary across different credit cards.
Bank Name Monthly Interest Rate Annual Interest Rate State Bank of India 3.50% 42% Citibank 3.75% 45% Axis Bank 3.60% 52.86% RBL Bank 3.99% 47.88% HDFC Bank On InterMiles HDFC Bank Diners Club / Diners Black / Infinia cards: 1.99% On other credit cards: 3.60% On InterMiles HDFC Bank Diners Club / Diners Black / Infinia cards: 23.88% On other credit cards: 43.2% ICICI Bank On InterMiles ICICI Bank credit card / MakeMyTrip ICICI Bank credit card / ICICI Bank HPCL Super Saver credit card: 3.50% On Manchester United credit cards by ICICI Bank: 3.67% On ICICI Bank Instant Platinum credit card / ICICI Bank Instant Gold credit card / Fixed Deposit Instant credit card: 2.49% On other credit cards: 3.40% On Manchester United credit cards by ICICI Bank: 44.04% On ICICI Bank Instant Platinum credit card / ICICI Bank Instant Gold credit card / Fixed Deposit Instant credit card: 29.88% On other credit cards: 40.8%
## Interest Rate on Top Credit Cards in India 2022
Just like how the credit card interest rate per month varies from one bank to another, the rates may also vary depending on the type of credit card that you opt for. Here are a few of the top credit cards in India in the year 2022 and the credit card interest rates for the same.
Credit Cards Credit Card Interest Rate SBI Card ELITE 3.50% per month (42% per year) Flipkart Axis Bank credit card 3.40% per month (40.8% per year) HDFC Regalia credit card 3.60% per month (43.2% per year) Amazon Pay ICICI credit card 3.50% to 3.80% per month (42% to 45.6% per year) Citi PremierMiles credit card 3.75% per month (45% per year) HDFC Millennia credit card 3.60% per month (43.2% per year) HSBC Cashback credit card 3.49% per month (41.88% per year) Axis Bank Ace credit card 3.60% per month (43.2% per year) Standard Chartered Digismart credit card 3.75% per month (45% per year)
Disclaimer: The credit card interest rate offers would be from partner banks and NBFCs and not from Bajaj Markets.
## List of Low-Interest Rate Credit Cards in India
There are several banks that offer some of the lowest interest rate credit cards in India. Check out the table below to get a better idea.
Credit Cards Credit Card Interest Rate Kotak Mahindra Best Price – Premium Card 1.50% per month (18% per year) HDFC Diners Club Black credit card 1.99% per month (23.88% per year) HDFC Infinia credit card - Metal Edition 1.99% per month (23.88% per year) ICICI Bank Instant Platinum credit card 2.49% per month (29.88% per year) Bank of Baroda Signature Visa credit card 2.60% per month (31.2% per year)
## What is a Credit Card Interest Free Period
The period of time between the date of a particular transaction to the next payment due date is usually termed as the credit card interest free period. During this grace period, no interest is charged on the transaction amount.
Generally, the credit card interest free period is around 45 to 50 days. However, it may vary depending on the bank from whom you’ve availed your credit card from. This interest free period is only applicable on purchases that you make using your credit card and not on any cash advances, loans, or balance transfers.
Also, the credit card interest free period will be cancelled if you fail to pay your outstanding dues in full before the due date. In such cases, interest will be charged on your outstanding amount as well as on any new purchases that you may make.
## What is a Credit Card EMI Calculator
When you make any big ticket purchases using your credit card, you usually get the option to convert it to EMIs. But how do you know just how much of an EMI you would have to pay each month if you choose to avail this option? Here’s where a credit card EMI calculator comes into the picture.
It is an online tool that allows you to determine just how much your monthly EMI is likely to be based on a few factors like the amount of transaction, the tenure, and the interest rate. All that you would have to do is enter the above details on a credit card EMI calculator and it will display the monthly instalment that you will have to pay.
## Important Points About Credit Card Interest Rates
When it comes to credit card interest, there are a few things that you should always keep in mind. Check out some of the most important points down below.
• You can use the credit card interest free period to your advantage by paying all of your outstanding dues before the due date.
• Paying your dues after the due date may attract interest and penalty charges.
• Credit card interest rates are levied on the balance outstanding amount.
• To enjoy low interest rates on your credit card, you can choose to convert your purchases to EMIs.
## FAQs on Credit Card Interest Rate
• ### ✔️Will interest be applied even if the minimum amount due is paid every month?
Credit card loan interest will be levied if you fail to clear the outstanding dues in full. That is to say, that if you pay the minimum amount due on your card, you will still have pending dues and hence will be charged the interest rate.
• ### ✔️What is the usual interest rate on a credit card?
The credit card annual interest rate depends on the card type and the issuing authority. Different banks have different credit card interest rates. You can compare credit card interest rates and select the one that works for you.
• ### ✔️When should I pay a credit card bill to avoid interest?
You should pay the credit card bill in full before the payment due date, every month, to avoid interest charges on the credit card.
• ### ✔️How does the interest charge on a credit card work?
CC interest rate is charged on the pending credit card bill based on the following formula -
[(Total number of days x Transaction amount x Interest rate per month x 12 months)] ÷ 365 days
You can also use the credit card interest calculator available online to calculate the credit interest rate charged on your card.
• ### ✔️Do I pay APR if I pay on time?
No. You won’t have to pay the APR if you pay all of your outstanding dues in full before the due date.
• ### ✔️Do credit cards have a high interest rate?
Yes. Credit cards generally come with high interest rates ranging from 1.50% per month to 3.75% per month.
• ### ✔️Does the rate of interest for credit cards change frequently?
It is quite possible that the rate of interest may change frequently depending on the issuing bank or financial institution.
• ### ✔️Why is the credit card interest rate so high?
Credit card interest rate is high because of the risk it poses to a bank. If a cardholder delays the payment or is not able to make the payment at all, the bank has to bear the burden of the payment.
• ### ✔️Do I have the option of paying my credit card balance in instalments?
Yes, you can pay your credit card balance in equated monthly instalments (EMIs) which would mean converting your credit card outstanding amount into a loan.
• ### ✔️Do all credit cards have an interest-free period?
It is not necessary that all cards have an interest-free period as it totally depends on the issuing bank.
• ### ✔️What is the typical interest rate on a credit card?
Interest Rates usually vary from 2.5%- 3.5% per month and it completely depends on the issuing bank as well as the type of credit card. It is always advisable to choose a card with a low-interest rate.
• ### ✔️Difference between a 0% credit card and a low-interest card?
A 0% interest credit card can be a lifesaver if you have to make a high-ticket purchase whereas a low-interest credit card can help the cardholder with a longer-term debt reduction strategy. | 2,371 | 10,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-06 | longest | en | 0.947715 |
http://phidot.org/forum/viewtopic.php?f=34&t=4256&p=14854&sid=7266f46cc1e8d3646acb20ea4d61c7cd | 1,709,281,243,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00887.warc.gz | 25,473,536 | 6,134 | ## MSORD abundance definition
Forum for discussion of general questions related to study design and/or analysis of existing data - software neutral.
### MSORD abundance definition
Hi,
Just hoping someone can confirm my understanding of the abundance estimate in MSORD is correct.
As in the example in Chapter 15 of the guide, I have transients and residents, with one observable nesting state and one unobservable skipping state.
It seems to me the estimate of abundance from this model in any season must be transient + resident adult female nesters, and not include the skipping females since most parameters for the skipping state are necessarily set to zero.
The one exception being survival in the skipping state which is set equal to survival in the nesting state for the sole reason of estimating probability of skipping, but not abundance of those skipping.
However, I've been asked to estimate total adult female population abundance (nesters + skippers).
I think this can't be done, but I'd be grateful for any advice.
violetblue
Posts: 23
Joined: Tue Nov 06, 2018 11:16 pm
### Re: MSORD abundance definition
You are correct that the abundance estimate from the MSORD model applies only to the number in observable states (nesters). This model is fully described in Kendall et al. (2020, Ecological Monographs). In that same paper we discuss the issue of estimating total abundance for this case you describe. If nesting probability were completely random (nesting probability this year is the same for those that did or did not nest last year), then total abundance would just be (nester abundance)/(nesting probability). The more common case of Markovian nesting probabilities requires a more complicated derivation based on the dynamics of the population, including an estimate of annual recruitment to the nesting state. We estimate this value for a population of hawksbill sea turtles in the paper, but we were comfortable doing so because we had reliable estimates of recruitment.
Bill Kendall
Posts: 96
Joined: Wed Jun 04, 2003 8:58 am
### Re: MSORD abundance definition
Hi,
As I understand it, in Kendall et al (2020) detection probabilities were close to 1.0 so observed new recruits were confidently substituted for estimated new recruits in the equation for total abundance.
In our MSORD (same as Chapter 15 in the guide, 2 states observed and unobserved, and separate survival for transients and residents), detection rates are often quite low (around 0.5, 0.6). As we need estimated recruits, I'm tempted to adjust observed recruits by detection rate for input to the total abundance equation. The assumption being that we missed new recruits at the same rate as we missed residents.
I'd be grateful to know if you think this approach is ok, or fundamentally flawed in some way?
What I'm thinking of doing is identifying transients in the sample, subtracting them from observed recruits and then dividing by the detection rate to give an estimate of new recruits in the population.
The added complication for us is that we have detection rates for each secondary occasion, so I'd be planning on estimating recruits by secondary occasion and summing them for the year.
If this is not ok, is there some way pent (probability of a new arrival) could be used instead?
violetblue
Posts: 23
Joined: Tue Nov 06, 2018 11:16 pm | 705 | 3,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.939502 |
https://www.techshadows.com/adding-up-tops-and-bottoms/ | 1,680,219,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00096.warc.gz | 1,127,945,848 | 13,627 | As you are collecting data for your Excel project, you may have a need to add up the top N number of values in a range or the bottom N number of values. For instance, you may be keeping track of golfing scores and need to add up only the top three scores or the bottom three scores out of a series of scores.
As with most any Excel problem, there are several ways you can go about implementing a solution. For instance, you could sort the scores so that they are in ascending or descending order. You would then have the top or bottom scores in a set place where you could always sum them.
There is an easier way, however. You can use the LARGE and SMALL functions, which do the job very nicely. For instance, say you have the scores in cells C5 through C25. All you need to do is put the following formula in a cell in order to add up the top three scores:
```=LARGE(C5:C25,1)+LARGE(C5:C25,2)+LARGE(C5:C25,3)
```
The function returns the Nth largest value from the specified range. As shown in the formula, the largest, second largest, and third largest values are returned and added together. You can similarly use the SMALL function to sum the three lowest scores:
```=SMALL(C5:C25,1)+SMALL(C5:C25,2)+SMALL(C5:C25,3)
``` | 300 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-14 | latest | en | 0.910869 |
https://number.academy/16645 | 1,652,952,887,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00506.warc.gz | 490,946,878 | 11,901 | Number 16645
Number 16,645 spell 🔊, write in words: sixteen thousand, six hundred and forty-five . Ordinal number 16645th is said 🔊 and write: sixteen thousand, six hundred and forty-fifth. The meaning of number 16645 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 16645. What is 16645 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 16645.
What is 16,645 in other units
The decimal (Arabic) number 16645 converted to a Roman number is (X)(V)MDCXLV. Roman and decimal number conversions.
Weight conversion
16645 kilograms (kg) = 36695.6 pounds (lbs)
16645 pounds (lbs) = 7550.1 kilograms (kg)
Length conversion
16645 kilometers (km) equals to 10343 miles (mi).
16645 miles (mi) equals to 26788 kilometers (km).
16645 meters (m) equals to 54609 feet (ft).
16645 feet (ft) equals 5074 meters (m).
16645 centimeters (cm) equals to 6553.1 inches (in).
16645 inches (in) equals to 42278.3 centimeters (cm).
Temperature conversion
16645° Fahrenheit (°F) equals to 9229.4° Celsius (°C)
16645° Celsius (°C) equals to 29993° Fahrenheit (°F)
Time conversion
(hours, minutes, seconds, days, weeks)
16645 seconds equals to 4 hours, 37 minutes, 25 seconds
16645 minutes equals to 1 week, 4 days, 13 hours, 25 minutes
Codes and images of the number 16645
Number 16645 morse code: .---- -.... -.... ....- .....
Sign language for number 16645:
Number 16645 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
Mathematics of no. 16645
Multiplications
Multiplication table of 16645
16645 multiplied by two equals 33290 (16645 x 2 = 33290).
16645 multiplied by three equals 49935 (16645 x 3 = 49935).
16645 multiplied by four equals 66580 (16645 x 4 = 66580).
16645 multiplied by five equals 83225 (16645 x 5 = 83225).
16645 multiplied by six equals 99870 (16645 x 6 = 99870).
16645 multiplied by seven equals 116515 (16645 x 7 = 116515).
16645 multiplied by eight equals 133160 (16645 x 8 = 133160).
16645 multiplied by nine equals 149805 (16645 x 9 = 149805).
show multiplications by 6, 7, 8, 9 ...
Fractions: decimal fraction and common fraction
Fraction table of 16645
Half of 16645 is 8322,5 (16645 / 2 = 8322,5 = 8322 1/2).
One third of 16645 is 5548,3333 (16645 / 3 = 5548,3333 = 5548 1/3).
One quarter of 16645 is 4161,25 (16645 / 4 = 4161,25 = 4161 1/4).
One fifth of 16645 is 3329 (16645 / 5 = 3329).
One sixth of 16645 is 2774,1667 (16645 / 6 = 2774,1667 = 2774 1/6).
One seventh of 16645 is 2377,8571 (16645 / 7 = 2377,8571 = 2377 6/7).
One eighth of 16645 is 2080,625 (16645 / 8 = 2080,625 = 2080 5/8).
One ninth of 16645 is 1849,4444 (16645 / 9 = 1849,4444 = 1849 4/9).
show fractions by 6, 7, 8, 9 ...
Calculator
16645
Is Prime?
The number 16645 is not a prime number. The closest prime numbers are 16633, 16649.
Factorization and factors (dividers)
The prime factors of 16645 are 5 * 3329
The factors of 16645 are 1 , 5 , 3329 , 16645
Total factors 4.
Sum of factors 19980 (3335).
Powers
The second power of 166452 is 277.056.025.
The third power of 166453 is 4.611.597.536.125.
Roots
The square root √16645 is 129,015503.
The cube root of 316645 is 25,532574.
Logarithms
The natural logarithm of No. ln 16645 = loge 16645 = 9,719865.
The logarithm to base 10 of No. log10 16645 = 4,221284.
The Napierian logarithm of No. log1/e 16645 = -9,719865.
Trigonometric functions
The cosine of 16645 is 0,665882.
The sine of 16645 is 0,746057.
The tangent of 16645 is 1,120405.
Properties of the number 16645
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
Number 16645 in Computer Science
Code typeCode value
16645 Number of bytes16.3KB
Unix timeUnix time 16645 is equal to Thursday Jan. 1, 1970, 4:37:25 a.m. GMT
IPv4, IPv6Number 16645 internet address in dotted format v4 0.0.65.5, v6 ::4105
16645 Decimal = 100000100000101 Binary
16645 Decimal = 211211111 Ternary
16645 Decimal = 40405 Octal
16645 Decimal = 4105 Hexadecimal (0x4105 hex)
16645 BASE64MTY2NDU=
16645 MD53169b89e40818e5575ab0ab87b38d2a5
16645 SHA1ee18cefce5568ddb97f7413572ebae51ec8e1341
16645 SHA224b98935648b6532180808663d0cee1a74775a745471d6602b9d42deb7
16645 SHA25657e430fc0915382a7d88c3a4939be83ed395cf9f35162aba1754f58a3263047a
16645 SHA384a23fb52f373a63674a4487867ee38f6ee0c80bb509746b45d04f95c07346df0b429ac57608fa3eb3b1a4ce3a2c912e0a
More SHA codes related to the number 16645 ...
If you know something interesting about the 16645 number that you did not find on this page, do not hesitate to write us here.
Numerology 16645
Character frequency in number 16645
Character (importance) frequency for numerology.
Character: Frequency: 1 1 6 2 4 1 5 1
Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 16645, the numbers 1+6+6+4+5 = 2+2 = 4 are added and the meaning of the number 4 is sought.
Interesting facts about the number 16645
Asteroids
• (16645) Aldalara is asteroid number 16645. It was discovered by O. A. Naranjo from Merida on 9/22/1993.
Number 16,645 in other languages
How to say or write the number sixteen thousand, six hundred and forty-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 16.645) dieciseis mil seiscientos cuarenta y cinco German: 🔊 (Anzahl 16.645) sechzehntausendsechshundertfünfundvierzig French: 🔊 (nombre 16 645) seize mille six cent quarante-cinq Portuguese: 🔊 (número 16 645) dezesseis mil, seiscentos e quarenta e cinco Chinese: 🔊 (数 16 645) 一万六千六百四十五 Arabian: 🔊 (عدد 16,645) ستة عشر ألفاً و ستمائة و خمسة و أربعون Czech: 🔊 (číslo 16 645) šestnáct tisíc šestset čtyřicet pět Korean: 🔊 (번호 16,645) 만 육천육백사십오 Danish: 🔊 (nummer 16 645) sekstentusinde og sekshundrede og femogfyrre Dutch: 🔊 (nummer 16 645) zestienduizendzeshonderdvijfenveertig Japanese: 🔊 (数 16,645) 一万六千六百四十五 Indonesian: 🔊 (jumlah 16.645) enam belas ribu enam ratus empat puluh lima Italian: 🔊 (numero 16 645) sedicimilaseicentoquarantacinque Norwegian: 🔊 (nummer 16 645) seksten tusen, seks hundre og førti-fem Polish: 🔊 (liczba 16 645) szesnaście tysięcy sześćset czterdzieści pięć Russian: 🔊 (номер 16 645) шестнадцать тысяч шестьсот сорок пять Turkish: 🔊 (numara 16,645) onaltıbinaltıyüzkırkbeş Thai: 🔊 (จำนวน 16 645) หนึ่งหมื่นหกพันหกร้อยสี่สิบห้า Ukrainian: 🔊 (номер 16 645) шiстнадцять тисяч шiстсот сорок п'ять Vietnamese: 🔊 (con số 16.645) mười sáu nghìn sáu trăm bốn mươi lăm Other languages ...
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Atled
2005-10-01 21:10:42
never mind,
it doesnt have odd symmetry
DanielM99
2005-09-29 00:35:16
How do I determine the fourier series of: f(x)= 8+2(x/9) , 0<x<4
, -8+2(x/9) , -4<x<0
and in addition we know: f(x+8)=f(x) | 159 | 372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2014-10 | longest | en | 0.69167 |
https://knowledgemouse.com/quizzes/52471 | 1,726,683,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00548.warc.gz | 301,471,128 | 5,508 | 1.
If you are taking the AHIMA CPC exam, how long do you have to complete the exam
A. 3 Hours B. 6 Hours C. 7 Hours D. 4 Hours
2.
Generally what two ways can you take the CPC exam
A. In person and Mail In. B. In Person and Electronic C. In Person and Video Conference D. Electronic and Mail-In
3.
What type of questions are on the exam if taking the CPC exam with AAPC?
A. Essay Questions B. Fill in the Blank C. None of the Above D. Multiple-choice questions
4.
How many questions are on the CPC exam for AAPC
A. 250 B. 50 C. 350 D. 150
5.
Generally speaking, for both the AAPC & AHIMA CPC exam what percentage of questions must you answer correctly to pass?
A. 75-80 B. 65-75 C. 70-75 D. 71-75
6.
Generally speaking if you had to is it better to guess.
A. No, wrong answers may be held against you. B. It doesn't matter C. None of the above D. Yes, It is better to guess
1.
If you are taking the AHIMA CPC exam, how long do you have to complete the exam
A. 3 Hours B. 6 Hours C. 7 Hours D. 4 Hours
2.
Generally what two ways can you take the CPC exam
A. In person and Mail In. B. In Person and Electronic C. In Person and Video Conference D. Electronic and Mail-In
3.
What type of questions are on the exam if taking the CPC exam with AAPC?
A. Essay Questions B. Fill in the Blank C. None of the Above D. Multiple-choice questions
4.
How many questions are on the CPC exam for AAPC
A. 250 B. 50 C. 350 D. 150
5.
Generally speaking, for both the AAPC & AHIMA CPC exam what percentage of questions must you answer correctly to pass?
A. 75-80 B. 65-75 C. 70-75 D. 71-75
6.
Generally speaking if you had to is it better to guess.
A. No, wrong answers may be held against you. B. It doesn't matter C. None of the above D. Yes, It is better to guess
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0
I think this is what you want: Simon et al. 1994 You can find Meeus formula in it.
2
Perhaps you are looking for Lagrange’s planetary equations? Given some perturbation $H_1$ to the Hamiltonian, they give six equations to calculate the change of the orbital elements over time; a thorough discussion is given in Goldstein’s textbook on classical mechanics. The equation for $\Omega$ is $$\frac{d\Omega}{dt}=\frac{1}{\sqrt{GMm^2(1-e^2)\sin i}}\... 16 Both ellipticity f (also called flattening) and eccentricity e are measures of how elongated an ellipse is, based on the semi-major axis a and the semi-minor axis b (figure from wikipedia). They are defined respectively as$$f=\frac{a-b}{a}$$and$$e=\sqrt{1-\frac{b^2}{a^2}} For a circle, $a=b$, which implies that $f=e=0$. In modern orbital ...
9
Ellipses have a "long radius" called the "semi-major-axis" which is the length from the centre to the ellipse measured along the long axis. And a "semi-minor-axis" which is measured along the short axis. Call the semi-major-axis "a" and the semi-minor-axis "b". Ellipses also have foci: which is where the ...
5
What Kepler (and others before and after him) wanted to do is predict where a planet would be. To do this we need some set up: First we want a coordinate system. This is a system of axes: x, y and z, at right angles to each other. And it should be an inertial coordinate system, so Newton's laws work. This means that the axes should not be rotating. And we ...
2
Thanks folks for helping. I knew about SPK files but from Horizon's telnet inteface there is a limit of 200 bodies per request. Making over 5000 request might be possible but not very productive. I also looked at the DE421 file from ftp://ssd.jpl.nasa.gov/pub/eph/planets/bsp/ but it looks like the corrections for the 8+1 planets here is the output from ...
0
This is a common problem with ephemeris or TLEs. They change over time and one might want to know the values after the times that are given. Most folks use an interpolation scheme to estimate values at a certain time. I would suggest starting with linear interpolation. If this doesn't give you good enough fidelity, you could move on to a more complicated ...
Top 50 recent answers are included | 572 | 2,254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-21 | longest | en | 0.908828 |
https://ebs.sabis.sakarya.edu.tr/DersDetay/DersAkisi/268/28511?Disaridan= | 1,582,912,114,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147628.27/warc/CC-MAIN-20200228170007-20200228200007-00236.warc.gz | 353,726,337 | 19,966 | Ders Bilgileri
#### Ders Akışı
Hafta Konular ÖnHazırlık
1 Indefinite integrals, methods of integrals, changing variables.
2 Partial integration method. Integral of rational functions.
3 Integral of trigonometrical expressions.
4 Integral of Irrational algebraic functions, Binom integrals Various changing variables.
5 Concept of definite integral. The problems of causing definite integral. Definition of definite integral.
6 Calculation of definite integral with the help of its definition. Proof of basic integration rules.
7 Basic theorİes of integral calculation. Changing variables methods in definite integral.
8 Partial integration method for definite integrals. Integral of some special defined functions.
9 Calculation of area,
10 Calculation of volume with definite integral
11 Calculation of volume with definite integral
12 Calculation of length of arc, Computation of surface area of rotated objects.
13 Generalized integrals.
14 Calculation of area and volume with generalized integral.
#### Kaynaklar
Ders Notu
Lecture Notes
Ders Kaynakları
[1] Thomas, G.B., Thomas Calculus, 11.baskı, çeviri:Recep Korkmaz, Beta Basım, 2010.
[2] Kadıoğlu, E., Kamali, M., Genel Matematik, Kültür Eğitim Vakfı, 2009.
[3] Can, M., Yüksek Matematik 1, Literatür, 2009.
[4] Balcı, M., Genel Matematik 1, Sürat Yayınları, 2012.
; ; | 322 | 1,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-10 | latest | en | 0.639714 |
https://mathematica.stackexchange.com/questions/77499/how-to-set-the-y-coordinates-for-a-graph-with-defined-x-coordinates-to-prevent-o | 1,722,649,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00469.warc.gz | 307,421,587 | 41,452 | # How to set the Y coordinates for a Graph with defined X coordinates to prevent overlap of nodes?
I would like to draw a Graph with defined $x$ coordinates and variable node sizes. If I simply replace the $x$ coordinates of a graph, then nodes can overlap and edges may cross each other. I would appreciate any help pointing me in the right direction on how to go about setting the $Y$ coordinates, or alternative approaches to plotting, such that the nodes are nicely spaced out and, as far as possible, edges don't cross.
(*egdata*)
relationships =
{1 <-> 2, 2 <-> 3, 2 <-> 4, 2 <-> 6, 3 <-> 5, 6 <-> 7,4 <-> 8, 4 <-> 9};
numNodes =
Max[relationships[[All, 2]]];
sizes = RandomReal[{0.25, 1}, numNodes];
xCoords = {0, 1, 2, 2, 3, 3, 4, 5, 5};
I would like to be able to automatically generate a plot similar to this (without having to manually specify $Y$ coordinates.):
Graph[
relationships, VertexSize -> Thread[Range[numNodes] -> sizes],
VertexCoordinates ->
Transpose[{xCoords,(*yCoords*){1, 1, 1, 3, 2, 1, 2, 3, 4}}]]
Using VertexCoordinateRules with GraphPlot is the best I can do, but doesn't solve the problem:
relationships =
{1 -> 2, 2 -> 3, 2 -> 4, 2 -> 6, 3 -> 5, 6 -> 7, 4 -> 8, 4 -> 9};
xCoords =
{0, 1, 2, 2, 3, 3, 4, 5, 5};
numNodes =
Max[relationships[[All, 2]]];
sizes = RandomReal[{0.25, 1}, numNodes];
GraphPlot[
relationships,
VertexCoordinateRules ->
Range[numNodes] ->
Transpose[{xCoords, Table[Automatic, {numNodes}]}]],
VertexRenderingFunction -> ({White, EdgeForm[Black], Disk[#, .3],
Black, Text[#2, #1]} &)
]
• Although this doesn't solve the stated problem ,TreePlot[relationships, Left] looks like it may be useful to you. Commented Mar 17, 2015 at 14:21
• Ignore explicit values of y coordinate first, and instead just track order of nodes/edges (along y axis) on every value of x with a node. Construct set of simple constraining inequalities (joining edge y values need to be equal, non-joining unequal, and no pair of edges must change order in their x interval intersections) to be reduced (or for decision of graph planarity). After this edges are guaranteed to be conceptually non-intersecting, and values of y can be computed on basis of sizes of nodes. Well, this is easier said than done, but I believe that's the most practical approach. Commented Apr 17, 2015 at 20:38
not quite what you asked for, but possibly useful:
relationships = {1 <-> 2, 2 <-> 3, 2 <-> 4, 3 <-> 5, 2 <-> 6, 6 <-> 7,
4 <-> 8, 4 <-> 9};
numNodes = Max[relationships[[All, 2]]];
sizes = RandomReal[{0.25, 1}, numNodes];
xCoords = {0, 1, 2, 2, 3, 3, 4, 5, 5};
vars0 = Array[a, 9];
Manipulate[
vars = {a[1], a[2], a[3], a[4], a[5], a[6], a[7], a[8], a[9]};
p = Graph[relationships,
VertexCoordinates -> Transpose[{xCoords, vars}],
VertexLabels -> "Name"] ,
Evaluate[Sequence @@
Table[{{vars0[[i]], i}, 1, Length@vars0, .1}, {i, Length@vars0}]],
TrackedSymbols -> All]
or maybe better using Locator:
relationships = {1 <-> 2, 2 <-> 3, 2 <-> 4, 3 <-> 5, 2 <-> 6, 6 <-> 7,
4 <-> 8, 4 <-> 9};
numNodes = Max[relationships[[All, 2]]];
v0 = VertexCoordinateRules /.
First@Cases[GraphPlot[Graph[relationships]], _Rule, Infinity];
Manipulate[Graph[relationships, VertexCoordinates -> vtx],
{{vtx, v0}, Locator, Appearance -> None}]
• Thanks George. Unfortunately, my actual data has hundreds of nodes, so manually setting the coordinates, even in a manipulate, isn't a viable solution. Commented Apr 23, 2015 at 13:19 | 1,069 | 3,441 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-33 | latest | en | 0.831692 |
http://www.martin-kraus.org/LiveGraphics3D/examples/color_volumes.html | 1,642,372,411,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00614.warc.gz | 107,550,165 | 2,291 | LiveGraphics3D Example:
Two Color Volumes
An RGB- and CMY-Color Cube
This example effectively tests the RGB (Red Green Blue) and the CMY (Cyan Magenta Yellow) color definitions. The squares are colored according to their position: Each square has an x, y and z coordinate in the color volume. These coordinates are translated into the intensity of the red, green and blue color component for one face of each square and into the cyan, magenta and yellow component for the other face.
An HSB Cylinder
The discs are colored with the HSB (Hue Saturation Brightness) color definition. The position angle of a disc determines the hue, the distance from the axis the saturation and the height the brightness of its color. The diameter of each disc changes with its distance to the viewer. This is one of the differences between the ways LiveGraphics3D and Mathematica paint graphics.
Implementation
Here is how to create these graphics within Mathematica:
```g=Show[Graphics3D[{
EdgeForm[],Table[{
FaceForm[RGBColor[x,y,z],CMYKColor[x,y,z,0]],
Polygon[{{x,y,z},{x+0.2,y,z},{x+0.2,y+0.2,z},
{x,y+0.2,z}}]
},{x,1,0,-1/3},{y,1,0.,-1/3},{z,1,0.,-1/3}]
},Lighting->False,Background->GrayLevel[0]]]
```
This will show the topmost picture within Mathematica. The following command is used to produce an appropriate InputForm of g:
```NumberForm[InputForm[N[g]],4]
```
Mathematica's output was then copied into a file called color_cube.m. This file is specified in the HTML code of this page (applet parameter INPUT_FILE) and, therefore, loaded and displayed by the LiveGraphics3D applet. (See the documentation for a complete description of the usage of LiveGraphics3D and a function to let Mathematica write the InputForm directly into a file.)
The second Graphics3D object was produced this way:
```g=Show[Graphics3D[{
PointSize[0.1],Table[{
Hue[h,s,b],Point[{Cos[h 2. Pi] (s+0.2),
Sin[h 2. Pi](s+0.2),1.5b}]
},{s,0.32,1,1/3},{b,0.24,1,1/4},{h,0,0.99,1/12}]
},Background->GrayLevel[0],ViewPoint->{0.8,1.2,1.2},
PlotRange->{{-1.2,1.2},{-1.2,1.2},{-0.2,1.7}}]]
```
And the same command to generate the InputForm as above.
Martin Kraus, April 15, 2021 | 629 | 2,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | latest | en | 0.782326 |
https://www.thetechplatform.com/post/principal-component-analysis-pca | 1,716,936,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00454.warc.gz | 890,744,471 | 190,629 | top of page
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# Principal Component Analysis (PCA)
Principal Component Analysis is an unsupervised learning algorithm that is used for the dimensionality reduction in machine learning. It is a statistical process that converts the observations of correlated features into a set of linearly uncorrelated features with the help of orthogonal transformation. These new transformed features are called the Principal Components. It is one of the popular tools that is used for exploratory data analysis and predictive modeling. It is a technique to draw strong patterns from the given dataset by reducing the variances.
PCA generally tries to find the lower-dimensional surface to project the high-dimensional data.
PCA works by considering the variance of each attribute because the high attribute shows the good split between the classes, and hence it reduces the dimensionality. Some real-world applications of PCA are image processing, movie recommendation system, optimizing the power allocation in various communication channels. It is a feature extraction technique, so it contains the important variables and drops the least important variable.
The PCA algorithm is based on some mathematical concepts such as:
• Variance and Covariance
• Eigenvalues and Eigen factors
Some common terms used in PCA algorithm:
• Dimensionality: It is the number of features or variables present in the given dataset. More easily, it is the number of columns present in the dataset.
• Correlation: It signifies that how strongly two variables are related to each other. Such as if one changes, the other variable also gets changed. The correlation value ranges from -1 to +1. Here, -1 occurs if variables are inversely proportional to each other, and +1 indicates that variables are directly proportional to each other.
• Orthogonal: It defines that variables are not correlated to each other, and hence the correlation between the pair of variables is zero.
• Eigenvectors: If there is a square matrix M, and a non-zero vector v is given. Then v will be eigenvector if Av is the scalar multiple of v.
• Covariance Matrix: A matrix containing the covariance between the pair of variables is called the Covariance Matrix.
#### Principal Components in PCA
As described above, the transformed new features or the output of PCA are the Principal Components. The number of these PCs are either equal to or less than the original features present in the dataset. Some properties of these principal components are given below:
• The principal component must be the linear combination of the original features.
• These components are orthogonal, i.e., the correlation between a pair of variables is zero.
• The importance of each component decreases when going to 1 to n, it means the 1 PC has the most importance, and n PC will have the least importance.
#### Applications of Principal Component Analysis
• PCA is mainly used as the dimensionality reduction technique in various AI applications such as computer vision, image compression, etc.
• It can also be used for finding hidden patterns if data has high dimensions. Some fields where PCA is used are Finance, data mining, Psychology, etc.
## Steps for PCA algorithm
1. Getting the dataset Firstly, we need to take the input dataset and divide it into two subparts X and Y, where X is the training set, and Y is the validation set.
2. Representing data into a structure Now we will represent our dataset into a structure. Such as we will represent the two-dimensional matrix of independent variable X. Here each row corresponds to the data items, and the column corresponds to the Features. The number of columns is the dimensions of the dataset.
3. Standardizing the data In this step, we will standardize our dataset. Such as in a particular column, the features with high variance are more important compared to the features with lower variance. If the importance of features is independent of the variance of the feature, then we will divide each data item in a column with the standard deviation of the column. Here we will name the matrix as Z.
4. Calculating the Covariance of Z To calculate the covariance of Z, we will take the matrix Z, and will transpose it. After transpose, we will multiply it by Z. The output matrix will be the Covariance matrix of Z.
5. Calculating the Eigen Values and Eigen Vectors Now we need to calculate the eigenvalues and eigenvectors for the resultant covariance matrix Z. Eigenvectors or the covariance matrix are the directions of the axes with high information. And the coefficients of these eigenvectors are defined as the eigenvalues.
6. Sorting the Eigen Vectors In this step, we will take all the eigenvalues and will sort them in decreasing order, which means from largest to smallest. And simultaneously sort the eigenvectors accordingly in matrix P of eigenvalues. The resultant matrix will be named as P*.
7. Calculating the new features Or Principal Components Here we will calculate the new features. To do this, we will multiply the P* matrix to the Z. In the resultant matrix Z*, each observation is the linear combination of original features. Each column of the Z* matrix is independent of each other.
8. Remove less or unimportant features from the new dataset. The new feature set has occurred, so we will decide here what to keep and what to remove. It means, we will only keep the relevant or important features in the new dataset, and unimportant features will be removed out
## How to Calculate PCA
#### 1. Take the whole dataset consisting of d+1 dimensions and ignore the labels such that our new dataset becomes d dimensional.
Let’s say we have a dataset which is d+1 dimensional. Where d could be thought as X_train and 1 could be thought as y_train (labels) in modern machine learning paradigm. So, X_train + y_train makes up our complete train dataset.
So, after we drop the labels we are left with d dimensional dataset and this would be the dataset we will use to find the principal components. Also, let’s assume we are left with a three-dimensional dataset after ignoring the labels i.e d = 3.
we will assume that the samples stem from two different classes, where one-half samples of our dataset are labeled class 1 and the other half class 2.
Let our data matrix X be the score of three students :
#### 2. Compute the mean of every dimension of the whole dataset.
The data from the above table can be represented in matrix A, where each column in the matrix shows scores on a test and each row shows the score of a student.
Matrix A
So, The mean of matrix A would be
Mean of Matrix A
#### 3. Compute the covariance matrix of the whole dataset ( sometimes also called as the variance-covariance matrix)
So, we can compute the covariance of two variables X and Y using the following formula
Using the above formula, we can find the covariance matrix of A. Also, the result would be a square matrix of d ×d dimensions.
Let’s rewrite our original matrix like this
Matrix A
Its covariance matrix would be
Covariance Matrix of A
Few points that can be noted here is :
• Shown in Blue along the diagonal, we see the variance of scores for each test. The art test has the biggest variance (720); and the English test, the smallest (360). So we can say that art test scores have more variability than English test scores.
• The covariance is displayed in black in the off-diagonal elements of the matrix A
a) The covariance between math and English is positive (360), and the covariance between math and art is positive (180). This means the scores tend to covary in a positive way. As scores on math go up, scores on art and English also tend to go up; and vice versa.
b) The covariance between English and art, however, is zero. This means there tends to be no predictable relationship between the movement of English and art scores.
#### 4. Compute Eigenvectors and corresponding Eigenvalues
Intuitively, an eigenvector is a vector whose direction remains unchanged when a linear transformation is applied to it.
Now, we can easily compute eigenvalue and eigenvectors from the covariance matrix that we have above.
Let A be a square matrix, ν a vector and λ a scalar that satisfies Aν = λν, then λ is called eigenvalue associated with eigenvector ν of A.
The eigenvalues of A are roots of the characteristic equation
Calculating det(A-λI) first, I is an identity matrix :
Simplifying the matrix first, we can calculate the determinant later,
Now that we have our simplified matrix, we can find the determinant of the same :
We now have the equation and we need to solve for λ, so as to get the eigenvalue of the matrix. So, equating the above equation to zero :
After solving this equation for the value of λ, we get the following value
Eigenvalues
Now, we can calculate the eigenvectors corresponding to the above eigenvalues. I would not show how to calculate eigenvector here, visit this link to understand how to calculate eigenvectors.
So, after solving for eigenvectors we would get the following solution for the corresponding eigenvalues
#### 5. Sort the eigenvectors by decreasing eigenvalues and choose k eigenvectors with the largest eigenvalues to form a d × k dimensional matrix W.
We started with the goal to reduce the dimensionality of our feature space, i.e., projecting the feature space via PCA onto a smaller subspace, where the eigenvectors will form the axes of this new feature subspace. However, the eigenvectors only define the directions of the new axis, since they have all the same unit length 1.
So, in order to decide which eigenvector(s) we want to drop for our lower-dimensional subspace, we have to take a look at the corresponding eigenvalues of the eigenvectors. Roughly speaking, the eigenvectors with the lowest eigenvalues bear the least information about the distribution of the data, and those are the ones we want to drop.
The common approach is to rank the eigenvectors from highest to lowest corresponding eigenvalue and choose the top k eigenvectors.
So, after sorting the eigenvalues in decreasing order, we have
For our simple example, where we are reducing a 3-dimensional feature space to a 2-dimensional feature subspace, we are combining the two eigenvectors with the highest eigenvalues to construct our d×k dimensional eigenvector matrix W.
So, eigenvectors corresponding to two maximum eigenvalues are :
#### 6. Transform the samples onto the new subspace
In the last step, we use the 2×3 dimensional matrix W that we just computed to transform our samples onto the new subspace via the equation y = W′ × x where W′ is the transpose of the matrix W.
Resource: Towardsdatascience.com, TutorialPoint
The Tech Platform | 2,254 | 10,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.913819 |
https://math.answers.com/questions/What_is_the_to_3_divided_by_7_equals | 1,725,801,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00018.warc.gz | 371,045,881 | 47,464 | 0
What is the to 3 divided by 7 equals?
Updated: 9/15/2023
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β 15y ago
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β 3y ago
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β 15y ago
3 divided by 7 = 0.428571428 | 81 | 180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.864852 |
https://www.gurufocus.com/news/263184/how-to-calculate-a-dcf-growth-rate | 1,505,984,876,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687711.44/warc/CC-MAIN-20170921082205-20170921102205-00665.warc.gz | 799,755,523 | 33,596 | Jae Jun
# How to Calculate a DCF Growth Rate
June 10, 2014
A DCF can be easy or as difficult as you make it.
The easiest way is to simply start off with the latest Free Cash Flow and then apply a single stage with a DCF growth rate.
DCF isn’t a 100% sure thing and the easiest problem to fall into is to try and use a DCF for every single stock you look at without really thinking about the inputs.
You don’t want to fall into the hammer and nail problem. It’s very tempting though because when you just get into valuing companies, you get excited and want to test it out on every stock you come across.
That’s where you need to know the pros and cons of a discounted cash flow.
Bruce Greenwald is clear that he doesn’t like DCF’s.
That’s why he came up with his Earnings Power Value method to value stocks.
However I like using DCF’s because I try to keep all my inputs realistic as well as making sure I have a big picture in my mind of what I’m trying to model.
And DCF growth rates is an important part of that.
Stock Valuation Preparation is Important
My wife is into ceramics and I was looking at a youtube video on how to make a pot.
Throwing a Pot
Looked like a fun hobby so I began researching how to start.
Turns out, there is a lot to do before you even get a chance to have fun on the throwing wheel.
The right clay has to be chosen.
It has to be mixed and formed properly.
Only after several hours or days when the clay is ready, can you play on the throwing wheel and try to make something.
Getting the growth rates, normalizing data and understanding the inputs is a lot like getting the clay ready.
There’s a lot to think about before you can get to a real valuation.
Calculating DCF Growth Rates
Since I show a lot of valuations and intrinsic value numbers when I write about stocks, you should know how I get my numbers.
The DCF valuation method I use in the Old School Value Stock Analyzer is similar to a 3 stage DCF.
Instead of just taking one growth rate and extrapolating it linearly, I’ve applied a decay rate to simulate a business cycle.
DCF to Simulate a Business Cycle
And instead of just using an analyst growth rate estimate I normalize FCF growth to remove one time good/bad years to come up with something more usable.
This isn’t a precise method. There is no precise way of doing things when you try to project, calculate or regress growth.
Valuation = art + science.
It’s just a matter of finding that right balance because you don’t want to skew towards one side.
Art and science should balance.
Easy Method
The easiest way to calculate growth is to subtract the beginning value from its ending value, and then divide that result by the beginning value.
Growth rate = (End value – Start value)/(Start value)
Easy.
But this method is only useful if you find stocks that look like those crappy clip art images.
Fantasy Growth Rates
It doesn’t factor in time and the ups and downs a business goes through.
A real business looks like this.
By just taking the start and end values, you miss out on all the middle stuff that goes on.
The Old School Value Method of Calculating DCF Growth Rates
I’ll confess something first.
I copied this method from F Wall Street and tweaked it.
The growth I use in the Stock Analyzer is similar to a moving average to calculate the growth rate. I just call it a rolling median as it is more simplified than a moving average.
I’m sure you’ve seen moving averages before.
Moving Averages
Yes “those” moving averages used for technical analysis.
But instead of applying it to the stock price, I’m applying the concept to free cash flows.
Here’s how.
I like to look at both 10 years and 5 years worth of data. It provides enough history to make projections easier and more trustworthy.
Value investors are called fools a lot of times for anchoring on past data as a guide, but that’s all there is to work with.
There is no such thing as accurate future data.
Since FCF and owner earnings for the majority of companies are volatile, I calculate the growth rates for multiple periods and then calculate the median of all the periods.
Something like this.
• 2008-2012 (4 year period)
• 2009-2013 (4 year period)
• 2008-2011 (3 year period)
• 2009-2012 (3 year period)
• 2010-2013 (3 year period)
• 2008-2010 (2 year period)
• 2009-2011 (2 year period)
• 2010-2012 (2 year period)
• 2011-2013 (2 year period)
Since I’m dealing with time, instead of using the growth rate formula shown above, the Compounded Average Growth Rate (CAGR) is what I use.
CAGR Formula
If you aren’t comfortable with doing the math, just use the convenient CAGR calculator.
Then take the median of all those time periods to get the median growth rate to use in your DCF.
For MSFT, here’s what it looks like.
MSFT DCF Growth Rate Based on FCF
MSFT DCF Growth Rate Based on FCF
Compare the growth rate of 11.5% by using the rolling median method with a single CAGR calculation below.
CAGR for Microsoft
Summing Up – Calculating Growth
Most of the times, a simple growth calculation is all that you need, but when you’re trying to model or calculate intrinsic value, you need to use a number that isn’t distorted by one time effects.
Don’t just rely on analyst growth estimates. I showed you last time that analysts easily get things wrong.
If you are a member of Old School Value, go to the DCF valuation section check it out.
If not, follow my example above and try doing it by hand.
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Currently 0.00/512345 Rating: 0.0/5 (0 votes) | 1,326 | 5,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-39 | latest | en | 0.961212 |
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# Calculate the change in entropy if the number of microstates of the system changes from 23...
Calculate the change in entropy if the number of microstates of the system changes from 23 to 72. (Give your answer to three significant figures. Include the sign of the value in your answer.)
Boltzmann's equation is given below.
ΔS = k * ln (Wfinal/Winitial)
In this equation, ΔS is the change in entropy, Winitial is the initial number of microstates in the system, Wfinal is the final number of microstates in the system, and k is Boltzmann's constant, 1.3806503 ✕ 10−23 J/K.
Then state whether the entropy of the system increased or decreased.
Change in entropy = k * ln (Wfinal/Wintial)
Given Wintial = 23
Wfinal =72
K = 1.3806503*10-23 J/K
change in entropy = 1.3806503*10-23*ln(72/23) = 1.3806503*10-23*1.14117= 1.57559*10-23 J/K
Entropy is increased because change in entropy is positive if it is negative entropy decreases and other reason number of microstates increases from 23 to 72 that means system gets disturbed, entropy means dis orderness or distrubance in the system due to increase in microstates entropy increased.
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Coins can be redeemed for fabulous gifts. | 321 | 1,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.854201 |
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posted by .
9.When lions prey on a herd of antelope, some of the antelope are eliminated. Which part of the theory of evolution can be used to describe this situation?
(1) sexual reproduction of the fittest
(2) isolation of the species
(3) survival of the best adapted
(4) new species development due to mutation
IS IT is 4 or 2
i think it is 4
• Living environment -
am confused now the answer is 1 or 2 it cant be 2 or 4
• Living environment -
asexual reproduction of the fittest
## Similar Questions
1. ### physics
hey i attempted this question but i don't know if its right. can some one check for me and correct it?
2. ### phys
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More Similar Questions | 678 | 2,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-34 | latest | en | 0.888434 |
https://mathematica.stackexchange.com/questions/94146/infinite-product-for-zeta2/94392#94392 | 1,643,194,571,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00275.warc.gz | 433,498,895 | 38,034 | # Infinite product for Zeta[2]?
I have crafted an infinite product that converges to Zeta[2] very slowly!
Product[(1296 n^4 (1 + (1 + n)^3)) /
((-1 + 36 n^2)^2 (-1 + (1 + n)^3)),{n, 1, 50000}]
I am without a computer for 2 more days and have been using Wolfram|Alpha, but it doesn't want to go much beyond the above limit. I would like someone to verify the function to make sure I haven't missed anything.
• @ Fred Kline: how did you derive your interesting formula essentially different from the Euler product? Sep 7 '15 at 10:58
• @Dr.WolfgangHintze, A few years ago I found a product that produced $\frac{\pi}{3}$ using multiples of $6.$ Yesterday I decided to square it and insert something to multiply by $\frac{3}{2}$. It seems to work. Sep 7 '15 at 12:08
• @Dr.WolfgangHintze, It's the Euler product using multiples of 6. Sep 7 '15 at 12:27
• @ Fred Kline: as to the name "Euler product" please see my answer Sep 7 '15 at 14:40
## 5 Answers
Product[
(1296 n^4 (1 + (1 + n)^3))/((-1 + 36 n^2)^2 (-1 + (1 + n)^3))
, {n, 1, ∞}
] === Zeta[2]
True
Zeta[2]
π^2/6
• Error decreases approximately as $\frac{1}{10n}$. Sep 7 '15 at 10:10
Amplifying on answer by @rhermans
f[m_] = Product[(1296 n^4 (1 + (1 + n)^3))/((-1 +
36 n^2)^2 (-1 + (1 + n)^3)), {n, 1, m}]
(* (Pi^2*Gamma[1 + m]^3*Gamma[3 + m])/
(6*(3 + 3*m + m^2)*Gamma[5/6 + m]^
2*Gamma[7/6 + m]^2) *)
This product converges
Limit[f[m + 1]/f[m], m -> Infinity]
(* 1 *)
Limit[f[m], m -> Infinity]
(* Pi^2/6 *)
Product[(1296 n^4 (1 + (1 + n)^3))/((-1 + 36 n^2)^2 (-1 + (1 + n)^3)), {n, 1,
Infinity}]
(* Pi^2/6 *)
% === Zeta[2]
(* True *)
LogLinearPlot[{f[m], Zeta[2]}, {m, 1, 100},
Epilog ->
Inset[LogLinearPlot[{f[m], Zeta[2]}, {m, 75, 100000}], {Log[25], 1.55}]]
• +1 for nice plot. I've learned something new. Sep 7 '15 at 12:22
I tried to find an even simpler product. Here's my solution:
$$\zeta(2) =\prod _{n=1}^{\infty } \frac{1}{\left(1-\frac{1}{4 n^2}\right) \left(1-\frac{1}{36 n^2}\right)}$$
In Mathematica
Product[ 1/(1 - 1/(4 n^2)) 1/(1 - 1/(36 n^2)), {n, 1, \[Infinity]}]
(* Out[76]= \[Pi]^2/6 *)
We can derive this from the well-known product formula of the sine
Product[1 - x^2/n^2, {n, 1, \[Infinity]}]
(*
Out[85]= Sin[\[Pi] x]/(\[Pi] x)
*)
considering that
\[Pi] x/Sin[\[Pi] x] /. x -> 1/2
(*
Out[79]= \[Pi]/2
*)
and
\[Pi] x/Sin[\[Pi] x] /. x -> 1/6
(*
Out[80]= \[Pi]/3
*)
The product of these two expressions give [Pi]^2/6. Finally, it is easy to transform the corresponding infinite products into the form provided above.
Remark: Although Euler used the product formula for the sine in his famous proof that the infinite sum of the inverse squares is equal to [Pi]^2/6, normally an Euler product is a product over primes, such as the one defining the zeta function:
$$\zeta(s) =\prod _{n=1}^{\infty } \frac{1}{1-p_n^{-s}}$$
EDIT #1
It is not difficult to prove that for any positive integer m we can write $$\zeta (2 m)=\prod _{n=1}^{\infty } a (n)$$
Where the $a(n)$ are rational functions of n.
• @ Fred Kline Thanks. Sep 10 '15 at 18:21
This is going to be an alternative answer to Dr. Wolfgang Hintze's question. Consider a limit: $$g := \prod\limits_{n=1}^\infty \frac{1}{\left(1-\frac{A^2}{n^2}\right)\left(1-\frac{B^2}{n^2}\right)\left(1-\frac{C^2}{n^2}\right)\left(1-\frac{D^2}{n^2}\right)}$$ Taking logs we have: $$\log(g) = - \sum\limits_{n=1}^\infty \left[\log(1-\frac{A^2}{n^2}) + \log(1-\frac{B^2}{n^2})+\log(1-\frac{C^2}{n^2})+\log(1-\frac{D^2}{n^2})\right] = \sum\limits_{l=1}^\infty \frac{1}{l} (A^{2 l} + B^{2 l}+C^{2 l}+D^{2 l}) \zeta(2 l)$$ Now from Faulhaber's formula we know values of the zeta function at positive integers (see for example Wikipedia page on zeta function). Therefore we have: \begin{eqnarray} \log(g) =\\ (-1) \left\{ \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi A)^{2 l}}{(2 l)!}\frac{1}{2 l} + \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi B)^{2 l}}{(2 l)!}\frac{1}{2 l}+ \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi C)^{2 l}}{(2 l)!}\frac{1}{2 l}+ \sum\limits_{l=1}^\infty B_{2 l} (-1)^l \frac{(2\pi D)^{2 l}}{(2 l)!}\frac{1}{2 l} \right\} = \\ \sum\limits_{t=2\pi\left\{A,B,C,D\right\}} \log(\Gamma(1+\frac{t}{2 \pi})\Gamma(1-\frac{t}{2 \pi})) \end{eqnarray} Therefore the limit reads: $$g = \prod\limits_{t=\left\{A,B,C,D\right\}} \Gamma(1+t) \Gamma(1-t) = \pi^4 \frac{A B C D}{\sin(\pi A)\sin(\pi B)\sin(\pi C)\sin(\pi D)}$$
This is not an answer. I have additional observations about the answer by Dr. Wolfgang Hintze.
In response to his first comment: From here, when we omit the first two primes from Euler's product, we get a square: $$\frac{\pi^2}{9} =\prod _{n=3}^{\infty } \frac{1}{1-p_n^{-2}},$$ then substituting $6n$ for $p_n,$ we get the square root: $$\frac{\pi}{3} =\prod _{n=1}^{\infty } \frac{1}{1-(6n)^{-2}}.$$
Now about his simplified product: When we omit the left-hand factor of the denominator, we get $\frac{\pi}{3}$ as above.
The left-hand factor of the denominator simplifies my machinations. Very nice.
However, that denominator has another interesting pattern: two odd primes bracket the $4$ as $(3,5)$ and all other odd primes intermittently bracket multiples of $6$ as $(5,7),(11,13),\dots$ So we have a product of all the primes which equals the product of all the points where odd primes may be found:
$$\zeta(2) =\prod _{n=1}^{\infty } \frac{1}{1-p_n^{-2}} =\prod _{n=1}^{\infty } \frac{1}{\left(1-\frac{1}{4 n^2}\right) \left(1-\frac{1}{36 n^2}\right)}$$
Multiples of $n$ where $0\equiv n \mod 3$ cause the left-hand denominator to collide with a previous right-hand denominator Because we omitted the prime $2,$ we will have a missing ratio: $\frac{1}{2}$ or $\frac{2}{1}.$
• @ Fred Kline: Honestly, I don't understand your remark on the "interesting pattern". BTW you should write 2 instead of s in the Euler products. Sep 8 '15 at 12:21
• @Dr.WolfgangHintze, Thanks for the heads-up re: $s.$ The interesting pattern is that the prime product uses primes only. We use multiples of $4,6$ which are the mid-points where odd primes can occur. As soon as I get my activation code for Mathematica later today, I plan to look at both functions to see the differences/commonalities. It might not be special? Sep 8 '15 at 13:02
• @ Fred Kline: you might wish to give the proof of the theorem I presented in my EDIT #1. Try to apply it to zeta(4), for instance, before generalizing too much from one case. Sep 8 '15 at 13:14 | 2,361 | 6,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.796141 |
http://network.punditarena.com/y-4962.html | 1,669,650,834,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00344.warc.gz | 30,235,864 | 8,797 | # Epf calculator - PPF Calculator: Meaning, Function and How to Use
### PCB Calculator 2022, EPF Calculator & SOCSO Table
It keeps track of the growth in your financial assets and obligations as time goes on.
Here, you contribute 12% of the specified salary either Rs.
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Because of its user-friendliness and accuracy in information, it is perfect for calculation.
Check EPF balance via mobile app EPFO recently launched a mobile app for PF balance tracking riding on the mobile application trend in the Indian market.
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Sexy:
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Views: 7150 Date: 07.01.2023 Favorited: 20
Category: DEFAULT
The EPF calculator is a simulation, which shows you the amount of money that will accumulate in your EPF account at retirement.
Everyone may find it challenging to deal with calculations, which is possible.
You can increase the contribution percentage if you want to retire at an early age.
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The EPF calculators provided online take into account all the aspects of interest rates and Income Tax deductions too. This makes the PF calculator more reliable and beneficial as they give you the calculation real time. Also Read: Latest SBI FD Rates Types of EPF Provident Funds Statutory Provident Funds (SPF) | 599 | 2,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.872649 |
https://hub.ucd.ie/usis/!W_HU_MENU.P_PUBLISH?p_tag=MODULE&MODULE=BDIC1029J&TERMCODE=202400 | 1,720,964,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00521.warc.gz | 275,904,460 | 13,543 | # BDIC1029J Maths (Engineering) 1
This module is the first module of calculus, about single-variable differential calculus. Calculus is one of the most fundamental courses of college mathematics. The main contents of this module include basics of functions, limits and continuity of single-variable functions, derivatives and differentials, rates of change, slopes of curves, Mean-Value Theorem etc , with applications in practice. By learning single-variable calculus, students should command basic concepts, elementary theory and methods, especially the main ideas of analytical mathematics, and learn the essentials of logical thinking, abstract imagination and detail.
Show/hide contentOpenClose All
Curricular information is subject to change
Learning Outcomes:
Students are expected to achieve the following competence:
• Using the properties of functions to judge their features.
• Calculating limits of single-variable functions.
• Judging continuity of single-variable functions.
• Calculating derivatives of inverse and composite functions.
• Finding tangents to parametric curves.
• Using L’Hospital Rule to compute limits.
• Calculating higher order derivatives.
• Calculating Taylor expansion of a function.
• Applying Taylor expansions in error estimations.
• Calculating differentials.
• Applying differentials in approximated computations.
• Understanding and using the mean value theorem.
Indicative Module Content:
week 1 Basic concepts of functions and mapping, limits and continuity of functions.
week 2 Derivatives (with geometric interpretation), calculation of derivatives.
week 3 Chain rule, L’Hospital Rule.
week 4 Higher order derivatives, Taylor expansion (with applications in error estimation).
week 5 Differentials, with applications in approximated computations.
week 6 Optimization of single variable functions: local/global extreme (max/min).
week 7 The mean value theorem.
Student Effort Hours:
Student Effort Type Hours
Lectures
32
Tutorial
16
Autonomous Student Learning
48
Online Learning
34
Total
130
Approaches to Teaching and Learning:
This module is the first module of calculus, about single-variable differential calculus. Calculus is one of the most fundamental courses of college mathematics. The main contents of this module include basics of functions, limits and continuity of single-variable functions, derivatives and differentials, rates of change, slopes of curves, etc., with applications in practice. By learning single-variable calculus, students should command basic concepts, elementary theory and methods, especially the main ideas of analytical mathematics, and learn the essentials of logical thinking, abstract imagination and detail. The key teaching approach is giving lectures as well as tutorials in the class, at the same time, we will have on-line question-answer every week, still, we may have team-work when necessary.
Requirements, Exclusions and Recommendations
Not applicable to this module.
Module Requisites and Incompatibles
This module is delivered overseas and is not available to students based at the UCD Belfield or UCD Blackrock campuses
Equivalents:
Calculus (Engineering) I (BDIC1012J)
Assessment Strategy
Description Timing Component Scale Must Pass Component % of Final Grade In Module Component Repeat Offered
Exam (In-person): 2 hour final exam Week 15 Graded Yes
100
Yes
Carry forward of passed components
No
Remediation Type Remediation Timing
In-Module Resit Prior to relevant Programme Exam Board | 686 | 3,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-30 | latest | en | 0.851575 |
https://kiwidamien.github.io/using-folium-what-is-the-furthest-you-can-get-from-starbucks-in-seattle.html | 1,582,585,499,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00317.warc.gz | 432,256,738 | 7,633 | # Using Folium: What is the furthest you can get from Starbucks in Seattle?
It seems that Starbucks is ubiquitous in Seattle. Where in Seattle is furthest from a Starbucks store?
In order to work this out, we need a list of all the stores in Seattle. The open data project Socrata makes it easy to find out - you can pull the address, as well as longitude and latitude all the Starbucks locations in the any given city.
This question was also a fun way for me to experiment with methods for presenting data and calculations on maps. My ultimate goal was to have something people could interact with on this blog, without having to install extra plugins.
## Methodolgy
First we had to decide what counted as "Seattle" for the purposes of this calculation. To make it easy to reproduce similar calculations for other cities, I decided that I would choose the bounding longitudes as the lower quartile and upper quartiles of longitudes for stores retrieved by Socrata. The same cut was made for the latitudes. This restriction hopefully makes an interesting answer, rather than telling us the middle of a residential district like Magnolia.
I broke Seattle into regions, called Voronoi Cells. The idea is that each region has single Starbucks store in it, and is defined as the collection of points that are closest to that Starbucks. For example, taking a square city with two stores, the line halfway between the stores divides the city in two: on one side you are closer to store #1, on the other you are closer to store #2.
Adding a third store changes the picture. Placing it toward the bottom of the city, but centered horizontally gives the following cells:
If the third store is directly below store 1, we get
The location furthest from a Starbucks has to either by an intersection point between the regions (the examples with three stores have one intersection point, the example with two stores has none), or a point on the boundary of the city. I was able to use `scipy.spatial.Voronoi` to calculate the Voronoi cells and get a list of intersection points, rather than trying to do it by hand. It was useful practice to annotate the maps with the cells as well.
## The map
Here is the final map produced. You can highlight the Voronoi cells by mousing over them to get an idea of how large an area each store "controls". You can also bring up information about the stores by clicking on them.
We find that the furthest we can get from Starbucks in Seattle is the corner of Meridian Avenue North with North 36th street (a few streets north of Gas Works Park), shown as an orange dot. The bounding box is also shown in orange.
## Folium compared to other packages
There are many ways to plot data on top of maps with Python. Here are a few I considered before using folium:
#### Basemap (part of matplotlib)
This is a great way to create attractive graphs in Python. There are lots of tutorials, and the package is well documented. The disadvantages are that basemap expects you to obtain and manage shapefiles, and the output are static images (like the plots in matplotlib). This is easy to work with in an interactive environment, but makes it difficult to just the results of your work to someone else.
Tutorials for basemap:
#### Plot.ly
This was one of the slickest options I considered. To use it, you need to create an account (free trials are offered), and it seemed difficult to embed directly on a static site. These considerations led me to dismiss `Plot.ly` fairly early, although it looks like a good option if you don't mind paying for an account.
#### Bokeh
Bokeh is a collection of plots that uses the grammar of graphics to build up plots. The plots are interactive on the development machine, and you can setup your server to run Bokeh plots to allow other people to interact with your plots online. When working with static pages on a server designed to serve static pages, I would have to either embed or link to an external Bokeh plot. If I was running this off my own server, using Bokeh would certainly be viable.
#### Folium
Folium seems to be the extension of the now defunct Vincent package for python. Folium makes maps that use Leaflet.js to remain interactive, so you can save the output to HTML. Anyone with a browser can open the HTML (or visit it on your blog) and still pan, zoom, and have the interactive mouse-overs and tool tips still function.
The downside to folium is that there are a lot of great tutorials ..... for version 0.2.0. The current version, 0.3.0, introduced a lot of reorganization and the documentation is lagging. All of the tutorials have commands that break in version 0.3.0. For example, in 0.2.0 the command to create a dot on the map was `folium.Map.circle_marker`, while in 0.3.0 it is `folium.CircleMarker`. The documentation is improving, and GitHub has a collection of 0.3.0 examples.
Once written, the code is very readable, and there is great integration with Open Street Maps, so you don't have to keep your own shape files on hand. One of the goals of this project was to be able to refer back to it as an example of how to use folium 0.3.0!
#### Geoplotlib and Kartograph.py
I only found out about these package once I had completed the project.
It looks like geoplotlib only produces static maps, similar to Basemap, but seems like it is very expressive and able to do a lot without much code. It is also integrated with GeoPandas.
Kartograph is designed to produce maps in python that are then coverted into SVG elements with embedded javascript, to enable the maps to be shared with anyone with a browser.
I will experiment with both of these in a future blog post. | 1,244 | 5,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-10 | latest | en | 0.962541 |
https://metacpan.org/pod/distribution/Game-Life/Life.pm | 1,537,431,971,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156423.25/warc/CC-MAIN-20180920081624-20180920101624-00048.warc.gz | 571,603,628 | 10,402 | ++ed by:
Luke Poskitt
and 1 contributors
# NAME
Game::Life - Plays Conway's Game of Life
# SYNOPSIS
`````` use Game::Life;
my \$game = new Game::Life( 20 );
my \$starting = [
[ 1, 1, 1 ],
[ 1, 0, 0 ],
[ 0, 1, 0 ]
];
\$game->place_points( 10, 10, \$starting );
for (1..20) {
my \$grid = \$game->get_grid();
foreach ( @\$grid ) {
print map { \$_ ? 'X' : '.' } @\$_;
print "\n";
}
print "\n\n";
\$game->process();
}``````
# DESCRIPTION
Conway's Game of Life is a basic example of finding 'living' patterns in rather basic rulesets (see NOTES). The Game of Life takes place on a 2-D rectangular grid, with each grid point being either alive or dead. If a living grid point has 2 or 3 neighbors within the surrounding 8 points, the point will remain alive in the next generation; any fewer or more will kill it. A dead grid point will become alive if there are exactly 3 living neighbors to it. With these simple rules, fascinating structures such as gliders that move across the grid, glider guns that generate these gliders, XOR gates, and others have been found.
This module simply provides a way to simulate the Game of Life in Perl.
In terms of coordinate systems as used in `place_points`, `toggle_point` and other functions, the first coodinate is the vertical direction, 0 being the top of the board, and the second is the horizontal direaction, 0 being the left side of the board. Thus, toggling the point of (3,2) will switch the state of the point in the 4th row and 3rd column.
The edges of the board are currently set as "flat"; cells on the edge do not have any neighbors, and thus will 'fall' off the board. Future versions may allow for 'warp' edges (if a cell moves off the left side it reappears on the right side).
`new`
Creates a new Life game board; if passed a scalar, the game board will be a square of that size, if passed an array reference, the game board will be created as a rectangle with the width and height set to the first and second values in the array reference respectively, otherwise, the board will be created using the default size of 100x100 units. Two optional array references may be passed to set the breeding and living rules for the Life game, respectively. The arrays should be made of the values for the number of nearest neighbors that should trigger the associated event. By default, Conway's game of life uses [ 3 ] and [ 2,3 ] for these arrays, respectively, but you can play around with these to get other types of automata.
`set_rules`
Takes two array references and uses them to set the rules of the Life game as described in `new` above, namely for breeding and living rules in that order.
`get_breeding_rules`, `get_living_rules`
Returns arrays that are associated with the breeding or living rules above.
`place_points`
Takes two scalars (indicating the position on the grid) and a reference to an array of arrays; this array is placed into the Life grid at the specified position, overwriting any data already there. Within the array of arrays, any non-zero values will be considered as a living square.
`place_text_points`
Takes two scalars (indicated the position on the grid), a character, and an array of strings; as with `place_points`, this array will be placed into the grid at the specified position. The character indicates what cells are to be considered as alive; any other character in the string will be considered dead. Thus, the example given in the SYNOPSIS can be writen optionally as
`````` my @starting = qw( XXX
X..
.X. );
\$game->place_text_points( 10, 10, 'X', @starting );``````
Note that a implicit method of serialization can be used in conjunction with `get_text_grid`.
`toggle_point`, `set_point`, `unset_point`
Take two scalars that indiciate a specific grid position. These functions toggle, sets, or unsets the life status of the grid point passed, respectively.
`process`
If passed a number, runs the Life simulation that many times, else runs the simulation once.
`get_grid`
Returns a copy of the Life grid as a reference to an array of arrays.
`get_text_grid`
Returns an array of strings that represent the game board. Two optional parameters may be passed as symbols to represent the living and dead states on the board, in that order. If these are not supplied, they will be represented by 'X' and '.', respectively. It's very easy to use this via:
`` print "\$_\n" foreach ( \$game->get_text_grid( ) );``
to follow the progress of the Life simulation, and should be faster than rolling your own based on get_grid.
# NOTES
Conway here is not Damien Conway of Perl fame, but John Horton Conway of mathematics and computer science fame. The 'game' was original designed in the late 60s - early 70s, and became popular due to the interest that Martin Gardner (puzzle editor for Scientific American) had for it.
# HISTORY
`````` Revision 0.06 2013/05/16 08:55:32 ltp
Improved test coverage.
Revision 0.05 2013/05/15 21:18:26 ltp
Updated constructor to allow arbitrary sized game board.
Revision 0.04 2001/07/04 02:49:29 mneylon
Fixed distribution problem
Revision 0.03 2001/07/04 02:27:55 mneylon
Revision 0.02 2001/07/04 02:23:13 mneylon
Added set_rules, get_breeding_rules, get_living_rules, and set default
values for these as Conway's rules
Modifications from code as posted on Perlmonks.org``````
# AUTHOR
This package was written by Michael K. Neylon
Copyright 2001 by Michael K. Neylon | 1,373 | 5,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-39 | latest | en | 0.874959 |
https://ncertbookspdf.com/ncert-solutions-class-9-maths-chapter-14-statistics-3/ | 1,708,921,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00275.warc.gz | 420,839,704 | 18,877 | # NCERT Solutions | Class 9 Maths Chapter 14 Statistics
NCERT Class 9 Maths Solutions PDF: In this post, we have discussed the solution of the Maths class 9 book which is followed in all CBSE schools. Solutions are given below with proper Explanation and utmost care has been taken to ensure that the solutions are correct. Answers provided will not only help in completing all the assignments but also help students in clearing their concepts. Students can download the solutions by printing the chapters by using the command Ctrl+P in google chrome and saving it in PDF format. All the best !! Please support us by sharing this website with your school friends.
## Exercise 14.4 Class 9 Maths Chapter 14 Statistics
1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
3. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
4. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
5. Find the mean salary of 60 workers of a factory from the following table: Salary (in `) Number of workers
6. Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency. | 460 | 1,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | longest | en | 0.93573 |
https://bookriff.com/what-is-the-law-of-diminishing-utility/ | 1,685,418,807,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00488.warc.gz | 165,979,685 | 8,944 | BookRiff
If you don’t like to read, you haven’t found the right book
What is the law of diminishing utility?
The law of diminishing marginal utility states that all else equal, as consumption increases, the marginal utility derived from each additional unit declines. The utility is an economic term used to represent satisfaction or happiness.
What is law of equi-marginal utility with example?
The consumer will gain maximum satisfaction if he spends OM amount of money (Rs. 30) on chocolates and OM’ amount of money (Rs. 40) on ice-creams because when he buys (his combination) the marginal utilities of the two are equal (PM=PM’). Any other combination of the two goods will give less total satisfaction.
Who gave law of Equimarginal utility?
The idea of equi-marginal principle was first mentioned by H.H.Gossen (1810-1858) of Germany. Hence it is called Gossen’s second Law. Alfred Marshall made significant refinements of this law in his ‘Principles of Economics’.
What is the Equimarginal principle?
The equimarginal principle states that consumers will choose a combination of goods to maximise their total utility. This will occur where. The consumer will consider both the marginal utility MU of goods and the price.
What is an example of law of diminishing marginal utility?
The law of diminishing marginal utility explains that as a person consumes an item or a product, the satisfaction or utility that they derive from the product wanes as they consume more and more of that product. For example, an individual might buy a certain type of chocolate for a while.
What is Equimarginal utility?
Law of Equi-Marginal Utility explains the relation between the consumption of two or more products and what combination of consumption these products will give optimum satisfaction. Marginal Utility is the additional satisfaction gained by consuming one more unit of a commodity.
What is first law of Gossen?
Gossen’s First Law is the “law” of diminishing marginal utility: that marginal utilities are diminishing across the ranges relevant to decision-making.
Who coined the term utility?
Utility in economics was first coined by the noted 18th-century Swiss mathematician Daniel Bernoulli.
Why is the Equimarginal principle important for environment?
Moreover, the equi-marginal principle holds, so that marginal abatement costs are equal among dischargers of pollutants. a b The efficient level of the emission is e*, and the social costs are expressed by the area (a + b). At the emission level e*, the social costs are minimized.
What is the second Equimarginal principle?
Second Equimarginal Principle (the Cost-Effectiveness Equimarginal Principle) The least-cost means of achieving an environmental target will have been achieved when the marginal costs of. all possible means of achievement are equal.
When Tu is maximum MU is?
Total utility is maximum when marginal utility is zero. It is based in the law of diminishing marginal utility which says ‘as more and more units of a good are consumed, MU i.e level of satisfaction derived from each successive unit goes on falling because desire for that commodity tend to fall.
What is the example for marginal utility?
Marginal Utility is the enjoyment a consumer gains from each additional unit they consume. It calculates utility beyond the first product consumed (the marginal amount). For example, you may buy an iced doughnut. In turn, you receive a certain level of utility or satisfaction from it.
How is marginal utility in used in economics?
Marginal utility is the added satisfaction a consumer gets from having one more unit of a good or service.
• The concept of marginal utility is used by economists to determine how much of an item consumers are willing to purchase.
• The law of diminishing marginal utility is often used to justify progressive taxes.
• Marginal utility can be positive,zero,or negative.
• What is the law of decreasing marginal utility?
Law of diminishing marginal utility. The law of diminishing marginal utility, also known as a Gossen’s First Law, is that ceteris paribus, as additional amounts of a good or service are added to available resources, their marginal utilities are decreasing.
What is the law of diminishing marginal utility for?
The law of diminishing marginal utility applies to business in that it is closely connected to the law of demand. That law states that as price decreases, consumption increases and that as price increases, consumption decreases. | 911 | 4,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.932777 |
http://www.symscape.com/blog/how-to-make-splash-in-cfd | 1,511,315,849,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806447.28/warc/CC-MAIN-20171122012409-20171122032409-00573.warc.gz | 497,647,075 | 7,104 | How to Make a Splash in CFD
It is a common occurrence: a rain drop falls into a puddle; a leaky faucet drips into a sink of water. This seemingly simple event has some complex physics in play that advanced multi-phase, free-surface (volume of fluid - VOF) Computational Fluid Dynamics (CFD) can simulate. Not only can CFD simulate water impact in a pool, it can also produce a beautiful 3D visualization of the process as an animation. Follow along to see how you can use Caedium Professional to simulate a water droplet falling into a pool of water.
VOF Free Surface Caedium CFD SimulationDroplet crater
Geometry
The geometry for this droplet simulation is straightforward - a single box that represents a quarter (assuming symmetry) of the pool and air above it through which the droplet will fall. To model the droplet you need a circular quadrant embedded in the face above the pool to serve as an inlet. Also you need to segment the box so that you can initialize the pool of water separately from the air. Even with these constraints the geometry is still relatively simple and so you can construct a multi-block volume (4-edged faces and 6-faced volumes) that constraints the subsequent mesh to be suitable for hexahedral elements. For free-surface CFD simulations hexahedral elements are typically preferred for their improved accuracy compared to tetrahedral elements.
Multi-Block Flow Domain Geometry
Meshing
Using the multi-block volume structure and having connected all the co-incident faces you can then group the volumes into a flow domain and apply meshing constraints to obtain adequate mesh resolution.
Multi-Block Flow Domain MeshQuadrilateral surface elements
Physics
Standard boundary conditions are suitable for the symmetry planes and walls of the box. Note that there are no boundary conditions applied to the internal shared faces, as the fluid must be able to pass freely through them.
To inject water for a brief moment to form the droplet you use time-dependent inlet variants for velocity and alpha1 (water or air) according to:
• U = Time Dependent, Derivation = table ( (0. (0 0 -4)) (0.01 (0 0 0)) ), meaning at time = 0 s, velocity = [0 0 -4] m/s and at time = 0.01 s, velocity = [0 0 0] m/s
• alpha1 = Time Dependent, Derivation = table ( (0. 1) (0.01 0) ), meaning at time = 0 s, alpha1 = 1 (water) and at time = 0.01 s, alpha1 = 0 (air)
To allow air to leave the flow domain while water is being injected you need to assign the upper faces surrounding the inlet as an outlet.
You need to match the solver time step with the reference velocity to ensure adequate temporal resolution of the movement and subsequent impact of the droplet. However, the time step should not be so small that the simulation takes excessive CPU time to complete.
Results
As the simulation runs you can simultaneously record a movie, with each frame representing the latest 3D visualization results update. This approach is called co-processing, as opposed to post-processing, and alleviates the need to gather and process massive amounts of results data after the simulation ends.
VOF Free Surface CFD SimulationDroplet impact into a pool
Shown below are individual frames from the movie.
Droplet
Droplet Impact
Droplet Crater | 730 | 3,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-47 | latest | en | 0.919434 |
https://math.stackexchange.com/questions/2862329/what-is-the-plot-of-fx-1-sqrt-log-10-cos2-pi-x | 1,701,228,502,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00299.warc.gz | 464,624,152 | 35,644 | # What is the plot of $f(x) = 1 + \sqrt{\log_{10}\cos(2\pi x)}$
I'm inspecting a function as in the title, and tried to plot it and then compare my result with graphing tools which lead me to confusion.
Is it true that the plot of $f(x) = 1 + \sqrt{\log_{10}\cos(2\pi x)}$ is just a set of points $(x, y) = (n, 1); \; n \in \mathbb Z$
Inspecting the domain one may see that $$\log_{10}\cos(2\pi x) \ge 0 \iff \cos(2\pi x) \ge 1$$
But $\cos x \in [-1, 1]$ therefore $\cos(2\pi x)$ must be equal to $1$ in order to satisfy the above, and that is only possible in $x \in \mathbb Z$.
The reason I'm asking is because neither Desmos nor W|A is plotting it the way i expected.
• Perhaps they are plotting an imaginary component, $i\sqrt{-\log_{10}\cos(2\pi x)}$ Jul 25, 2018 at 12:17
You are right: the domain of $f$ is the set $\mathbb Z$ and the graph of $f$ is given by
$\{(x,f(x)): x \in \mathbb Z\}=\{(n,1): n \in \mathbb Z\}$. | 333 | 933 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-50 | longest | en | 0.854134 |
https://cdn.codeproject.com/Messages/5940255/Re-edited-Re-Base-building-in-a-RTS-based-on-a-que | 1,696,167,383,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00531.warc.gz | 182,981,808 | 20,652 | 15,746,973 members
Home / Discussions / Algorithms
# Algorithms
[edited]Re: Base building in a RTS based on a queue of orders Calin Negru27-Apr-23 8:01 Calin Negru 27-Apr-23 8:01
Re: [edited]Re: Base building in a RTS based on a queue of orders Gerry Schmitz29-Apr-23 3:48 Gerry Schmitz 29-Apr-23 3:48
Re: Base building in a RTS based on a queue of orders Calin Negru30-Apr-23 5:23 Calin Negru 30-Apr-23 5:23
Re: Re: Base building in a RTS based on a queue of orders Gerry Schmitz30-Apr-23 6:14 Gerry Schmitz 30-Apr-23 6:14
Re: Re: Base building in a RTS based on a queue of orders Calin Negru30-Apr-23 22:12 Calin Negru 30-Apr-23 22:12
Re: Re: Base building in a RTS based on a queue of orders Gerry Schmitz1-May-23 0:08 Gerry Schmitz 1-May-23 0:08
Re: [edited]Re: Base building in a RTS based on a queue of orders Calin Negru30-Apr-23 5:10 Calin Negru 30-Apr-23 5:10
Re: [edited]Re: Base building in a RTS based on a queue of orders Gerry Schmitz30-Apr-23 6:29 Gerry Schmitz 30-Apr-23 6:29
My "event" (if you call it that) happens when the loss inficted by the attacker is applied to the target. At that point, the target: maintains ground / continues to advance; is dispersed, temporarily or permanently; is shattered; is captured; is pursued; or is destroyed. The state of the various parties determines what can happpen in the next "round". If was to spawn or seed something, it would be "triggered" here. (So, let's say I use triggers instead of events). "Before entering on an understanding, I have meditated for a long time, and have foreseen what might happen. It is not genius which reveals to me suddenly, secretly, what I have to say or to do in a circumstance unexpected by other people; it is reflection, it is meditation." - Napoleon I
Efficiently finding position in sorted list Robert Ellis26-Mar-23 13:52 Robert Ellis 26-Mar-23 13:52
Re: Efficiently finding position in sorted list Mircea Neacsu26-Mar-23 14:42 Mircea Neacsu 26-Mar-23 14:42
Dijkstra Calin Negru13-Dec-22 7:46 Calin Negru 13-Dec-22 7:46
Re: Dijkstra jschell13-Dec-22 10:39 jschell 13-Dec-22 10:39
Re: Dijkstra Calin Negru13-Dec-22 22:34 Calin Negru 13-Dec-22 22:34
Re: Dijkstra jschell20-Dec-22 13:21 jschell 20-Dec-22 13:21
Re: Dijkstra Calin Negru26-Dec-22 9:26 Calin Negru 26-Dec-22 9:26
Re: Dijkstra jschell29-Dec-22 11:20 jschell 29-Dec-22 11:20
Re: Dijkstra Calin Negru30-Dec-22 8:56 Calin Negru 30-Dec-22 8:56
Re: Dijkstra harold aptroot2-Jan-23 6:45 harold aptroot 2-Jan-23 6:45
Re: Dijkstra Eddy Vluggen2-Jan-23 10:48 Eddy Vluggen 2-Jan-23 10:48
Re: Dijkstra Graeme_Grant2-Jan-23 11:41 Graeme_Grant 2-Jan-23 11:41
Re: Dijkstra Eddy Vluggen2-Jan-23 11:59 Eddy Vluggen 2-Jan-23 11:59
Re: Dijkstra Graeme_Grant2-Jan-23 12:16 Graeme_Grant 2-Jan-23 12:16
Re: Dijkstra Eddy Vluggen2-Jan-23 12:27 Eddy Vluggen 2-Jan-23 12:27
Re: Dijkstra Calin Negru3-Jan-23 0:19 Calin Negru 3-Jan-23 0:19
Re: Dijkstra Eddy Vluggen3-Jan-23 0:25 Eddy Vluggen 3-Jan-23 0:25
Last Visit: 31-Dec-99 18:00 Last Update: 1-Oct-23 3:36 Refresh ᐊ Prev1234567891011 Next ᐅ | 1,088 | 3,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.878435 |
autoride.io | 1,718,233,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00000.warc.gz | 102,949,781 | 33,010 | Air-fuel ratio: How does it affect engine performance?
Udgivet på Oversat ved hjælp af kunstig intelligens fra vores originale artikel (kilde: autoride.sk)
An air-fuel ratio is a dimensionless number that expresses the ratio between the actual amount of air in the mixture and the theoretical amount (stoichiometric amount) of air corresponding to the fuel used. The coefficient, named in English as the Air-Fuel Ratio, is denoted by the Greek letter λ (lambda).
The coefficient defined in this way is mainly used in the field of internal combustion engines because it is a necessary parameter for the preparation of the mixture. The air-fuel ratio thus thoroughly expresses the mixture's richness level. Each fuel needs a different amount of air for its perfect combustion.
Indholdsfortegnelse
For example, depending on its composition, 1 kg of regular car gasoline requires approximately 14.7 kg of air for complete combustion. For diesel, 15 to 15.5 kg of air is needed for 1 kg of fuel. Thus, if the perfect ratio between fuel and air in the mixture is maintained, the air-fuel ratio will reach 1 (λ = 1). In such a case, we speak of a stoichiometric mixture.
According to air-fuel ratio, we can divide the combustion mixture into:
• If the mixture contains exactly the stoichiometric amount of air λ = 1, it is called stoichiometric (contains the correct ratio of air for perfect combustion of fuel)
• If the mixture contains more air λ > 1, it is called lean (contains less fuel than can be burned)
• If the mixture contains less air λ < 1, it is called rich (contains more fuel than can be burned)
However, with different operating modes of the engine, the working conditions of the engine change, and thus the requirements for the amount of fuel supplied differ.
Typical operating modes of the engine in which the composition of the mixture must be adjusted are, for example:
• Cold start
• Cold engine
• Warming up the engine
• Acceleration
• Deceleration
• Idling
• Air conditioning on
• Altitude
Effect of mixture composition on engine parameters:
Depending on the air-fuel ratio, under the same engine conditions, the following are affected:
• Fuel consumption
• Engine performance
• Amount of emissions
• Evenness of engine operation
All these parameters depend on the composition of the mixture. However, the actual mixing ratio of the mixture differs significantly from the theoretical one. The temperature, speed, and load of the engine determine it.
The mixture ratio at which performance, emissions, and consumption reach the best values is unique for each engine and operating mode.
Combustion of a stoichiometric mixture:
In theory, emissions should not occur when burning a stoichiometric mixture. In practice, however, the situation is different. Due to insufficient homogenization of the fuel and its interaction with other substances (engine oil, impurities in the fuel, influence of nitrogen from the air) and the short time in which the combustion process must occur, emissions formation occurs.
Since the engines of ordinary cars are mainly operated at partial load, they are designed for this operation so that their operation is as efficient as possible in this mode. In this mode of operation, working with a stoichiometric mixture is a suitable compromise between performance, fuel consumption, and the amount of emissions produced.
In addition, the engines of today's cars must first meet the emission limits, so using a stoichiometric mixture (λ = 1) seems to be the most appropriate because that is when the catalytic converter has maximum efficiency. The engine is, therefore, the most ecological.
Rich mixture combustion:
When burning a rich mixture, combustion takes place faster, and the air-fuel ratio reduces the maximum temperature by evaporation, which ensures internal cooling of the cylinder group, which in turn makes it possible to increase the engine's compression ratio.
Thanks to this, the engine's power increases, but at the same time, its consumption also increases, because not all the fuel is burned perfectly, and part of its energy remains unused.
All other parameters go aside in this mode, and performance becomes the main parameter. The mixture is thus enriched (λ < 1) to achieve the highest possible engine performance.
Lean mixture combustion:
The lowest consumption is achieved in the lean mixture combustion mode; thus, the air-fuel ratio reaches the value (λ > 1). When the engine is under low load, the performance is not interesting, so the priority becomes fuel consumption.
In such a case, setting a slightly lean mixture (λ > 1), which achieves the greatest fuel savings, is the clear choice for this mode.
Effect of mixture composition on engine components:
A rich mixture has a significant effect from the point of view of engine protection, because the fuel that does not burn removes the temperature from the combustion chamber through its evaporation, thereby ensuring effective cooling of the combustion chamber.
The cooling effect increases with the richness of the mixture, which is especially important for extremely stressed engines. This is why a rich mixture is burned at maximum engine load.
However, there is no need to overdo it with the richness of the mixture, because unburnt fuel washes the oil film from the walls of the cylinders, which increases the risk of piston seizing. In addition, this increases the formation of carbon, the deposits of which prevent the removal of heat from the combustion chamber.
When burning a lean mixture, there is a risk of the absence of internal cooling, which can lead to a thermal overload of some engine components, for example, pistons, valves, and spark plugs. However, higher local temperatures in the cylinder significantly increase the risk of detonation combustion.
Internal fuel cooling can only be used for spark-ignition engines because they can work with a rich mixture due to a longer time for its preparation (fuel enters the cylinders together with air or is injected into the cylinder during the intake stroke).
In combustion-ignition engines, where fuel is injected into the cylinder, and the combustion phase begins simultaneously, a rich mixture that does not mix well with the air would lead to excessive smoke. This means that even at thefull load of the diesel engine, the air-fuel ratio is only close to the stoichiometric mixture so that the highest possible performance is achieved.
Air-fuel coefficient λ: What do the specific values mean?
• <0.5- the lower limit of flammability (rich mixture), the mixture of fuel and air is no longer flammable
• < 1 - rich mixture, lack of air, increased power and torque
• 0.9 - the highest torque, good engine operation, worse specific fuel consumption
• 0.9 to 1.1 - theoretically suitable mixture of fuel and air
• >1 - lean mixture, excess air, fuel saving, economical operation
• 1.3 to 1 .5 - the upper limit of the flammability of the mixture (lean mixture), the mixture of fuel and air is no longer flammable
• 1.6 to 1.7 - the upper limit of the flammability of the mixture for engines with a stratified mixture
However, in general, a properly functioning engine at the correct temperature and load burns:
Diesel engine - burns an inhomogeneous (stratified) mixture with a high excess of air. The mixture is lean, has a larger proportion of air than would belong to a certain amount of fuel, and the air-fuel ratio is, therefore, λ > 1.
Petrol engine with indirect injection - burns a homogeneous mixture. The coefficient of the proportion of air is λ = 1, and such a mixture is called stoichiometric.
Gas engine with direct injection - burns homogeneous but non-homogeneous (layered) mixture. A homogeneous mixture injects a fuel λ = 1 dose into the combustion chamber during the intake stroke.
A stratified mixture injects fuel into the swirled air during the compression stroke, creating a locally homogeneous mixture in the spark plug area. However, there is a lean mixture in the other spaces of the cylinder, and the air-fuel coefficient is, therefore, λ > 1. | 1,706 | 8,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.885049 |
https://www.tensorflow.org/api_docs/python/tf/nest/pack_sequence_as?hl=fa | 1,638,794,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00520.warc.gz | 1,075,875,138 | 60,275 | Help protect the Great Barrier Reef with TensorFlow on Kaggle
tf.nest.pack_sequence_as
Returns a given flattened sequence packed into a given structure.
Used in the notebooks
If structure is a scalar, flat_sequence must be a single-element list; in this case the return value is flat_sequence[0].
If structure is or contains a dict instance, the keys will be sorted to pack the flat sequence in deterministic order. This is true also for OrderedDict instances: their sequence order is ignored, the sorting order of keys is used instead. The same convention is followed in flatten. This correctly repacks dicts and OrderedDicts after they have been flattened, and also allows flattening an OrderedDict and then repacking it back using a corresponding plain dict, or vice-versa. Dictionaries with non-sortable keys cannot be flattened.
Examples:
1. Python dict:
structure = { "key3": "", "key1": "", "key2": "" }
flat_sequence = ["value1", "value2", "value3"]
tf.nest.pack_sequence_as(structure, flat_sequence)
{'key3': 'value3', 'key1': 'value1', 'key2': 'value2'}
1. For a nested python tuple:
structure = (('a','b'), ('c','d','e'), 'f')
flat_sequence = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0]
tf.nest.pack_sequence_as(structure, flat_sequence)
((1.0, 2.0), (3.0, 4.0, 5.0), 6.0)
1. For a nested dictionary of dictionaries:
structure = { "key3": {"c": ('alpha', 'beta'), "a": ('gamma')},
"key1": {"e": "val1", "d": "val2"} }
flat_sequence = ['val2', 'val1', 3.0, 1.0, 2.0]
tf.nest.pack_sequence_as(structure, flat_sequence)
{'key3': {'c': (1.0, 2.0), 'a': 3.0}, 'key1': {'e': 'val1', 'd': 'val2'} }
1. Numpy array (considered a scalar):
structure = ['a']
flat_sequence = [np.array([[1, 2], [3, 4]])]
tf.nest.pack_sequence_as(structure, flat_sequence)
[array([[1, 2],
[3, 4]])]
1. tf.Tensor (considered a scalar):
structure = ['a']
flat_sequence = [tf.constant([[1., 2., 3.], [4., 5., 6.]])]
tf.nest.pack_sequence_as(structure, flat_sequence)
[<tf.Tensor: shape=(2, 3), dtype=float32,
numpy= array([[1., 2., 3.], [4., 5., 6.]], dtype=float32)>]
1. tf.RaggedTensor: This is a composite tensor thats representation consists of a flattened list of 'values' and a list of 'row_splits' which indicate how to chop up the flattened list into different rows. For more details on tf.RaggedTensor, please visit https://www.tensorflow.org/api_docs/python/tf/RaggedTensor.
With expand_composites=False, we treat RaggedTensor as a scalar.
structure = { "foo": tf.ragged.constant([[1, 2], [3]]),
"bar": tf.constant([[5]]) }
flat_sequence = [ "one", "two" ]
tf.nest.pack_sequence_as(structure, flat_sequence,
expand_composites=False)
{'foo': 'two', 'bar': 'one'}
With expand_composites=True, we expect that the flattened input contains the tensors making up the ragged tensor i.e. the values and row_splits tensors.
structure = { "foo": tf.ragged.constant([[1., 2.], [3.]]),
"bar": tf.constant([[5.]]) }
tensors = tf.nest.flatten(structure, expand_composites=True) | 866 | 2,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.573611 |
http://ircarchive.info/math/2007/2/28/1.html | 1,550,373,691,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481612.36/warc/CC-MAIN-20190217031053-20190217053053-00389.warc.gz | 143,751,958 | 2,995 | ## #math - Wed 28 Feb 2007 between 02:25 and 02:59
### NY Lost Funds
noway- I am mainly lost on the 2nd property...Does it that every possible set S will work? or just the S in A pirothezero SharkWave: you figure it out? TRWBW noway-: your example was { {}, { {} } , { {}, { {} } } } right? {} is a subset, { {} } is a subset, and { {} , {{}} } is a subset, right? qed. noway- TRWBW: ok ok, thank you :-D TRWBW noway-: np. you did all the work. pirothezero SharkWave: if you factored n, then you figure out phi(n) which is p-1 * q-1then you run extended euclids algorithm to get doh nmlol TRWBW got it taken care of damog is there a fast algorithm that can compute the divisors of a given number? TRWBW damog: there are algorithms that are a lot faster than just trying to factor it by trying to divide by 2,3,4,5,... but they are still slow for big numbers slava is a positive-definite bilinear form non-degenerate? TRWBW slava: are you asking for the definitions, or do you know them and wonder if they imply that? slava i know the definitions and i'm wonderingbut now i see it is obvious TRWBW ;) slava if > 0 for x!=0 then clearly for each x there exists y with != 0 TRWBW yeah slava i need to show that *(*x.y) where * is the hodge star is an inner producton the exterior algebraand now i know it suffices to show its positive definite, which simplifies the task SharkWave pirothezero: I got the factors p = 113, q = 107, phi = (113 - 1)(107 - 1) = 11872... Just not sure how to use the Euclidean Algorithm to find the inverse TRWBW SharkWave: do you know what the extended euclidean algorithm is? SharkWave it finds the gcdyes TRWBW SharkWave: no, that's just "euclidean algorithm" SharkWave ack TRWBW SharkWave: here, http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm Kampen that's how you find the numbers that are guaranteed existence by bezout :pextended euclidean can be hard to grasp at first with certain authors though, some do it backwards and that's hard for me to do easilyare you in some sort of algebra class? SharkWave crypto Kampen ah SharkWave but we haven't covered extended Euclidean alg Kampen well, if you want to really understand the extended euclidean algorithm, yous hould look a bit at bezout's theorem Capso What is the graphical interpretation of the dot product of two different vectors? Kampen dot product is a scalar yo Capso Rather... what does it mean? TRWBW SharkWave: what do you want? advice? okay, switch to a school where they teach you stuff before they expect you to know it. ;) | 675 | 2,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-09 | latest | en | 0.919251 |
https://community.boredofstudies.org/search/2021630/ | 1,611,683,599,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00360.warc.gz | 267,844,313 | 10,348 | # Search results
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https://socratic.org/questions/a-right-triangle-has-sides-a-b-and-c-side-a-is-the-hypotenuse-and-side-b-is-also-43 | 1,653,302,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00455.warc.gz | 579,689,178 | 6,341 | A right triangle has sides A, B, and C. Side A is the hypotenuse and side B is also a side of a rectangle. Sides A, C, and the side of the rectangle adjacent to side B have lengths of 2 , 1/3 , and 3 , respectively. What is the rectangle's area?
Mar 15, 2016
sqrt35 ≈ 5.916" square units "
Explanation:
The area of the rectangle = 3B . To calculate this area , require to find length of B.
Since this is a right triangle , B can be found using $\textcolor{b l u e}{\text{ Pythagoras' theorem }}$
A is the hypotenuse hence ${A}^{2} = {B}^{2} + {C}^{2}$
thus ${B}^{2} = {A}^{2} - {C}^{2}$
here A = 3 and C $= \frac{1}{3}$
hence :${B}^{2} = {2}^{2} - {\left(\frac{1}{3}\right)}^{2} = 4 - \frac{1}{9} = \frac{35}{9}$
now ${B}^{2} = \frac{35}{9} \Rightarrow B = \sqrt{\frac{35}{9}} = \frac{1}{3} \sqrt{35}$
area of rectangle = 3B = 3xx1/3 sqrt35 = sqrt35 ≈ 5.916 | 330 | 868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2022-21 | longest | en | 0.786787 |
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Benito's Pizzeria has been losing money for the last three
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Benito's Pizzeria has been losing money for the last three [#permalink]
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04 Jun 2010, 02:51
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Benito's Pizzeria has been losing money for the last three months. Benito's mother has advised him to cut prices in order to increase the number of customers to the pizzeria. Benito, however, has decided to increase prices instead.
Which of the following, if true, provides the strongest evidence that Benito's plan has a better chance of making the pizzeria profitable than his mother's plan?
A Benito could expand his customer base by opening a second location and advertising heavily.
B The pizzeria is already producing as many pizzas as its ovens can handle.
C Benito has kept essentially the same items on his menu for ten years.
D Benito's business has suffered ever since a low-priced national chain moved in next door.
E Benito's top-selling item, the large pepperoni pizza, sells for $8.99. The Oa is [Reveal] Spoiler: b Pleae explain why OA is not A but ... If you have any questions you can ask an expert New! Manager Joined: 04 Apr 2010 Posts: 89 Schools: UCLA Anderson Followers: 2 Kudos [?]: 45 [0], given: 17 Re: pizzaria [#permalink] Show Tags 04 Jun 2010, 11:26 Plan A might increase the customer base and thus the revenue, but they might not be profitable customers. If the costs to expand and acquire the new customers is too high, they may actually be unprofitable customers! B says that even if he lowers prices and increases demand, Benito won't be able to increase his supply. Hence B. _________________ If this post was helpful, please give Kudos. Intern Joined: 19 Dec 2009 Posts: 37 Followers: 0 Kudos [?]: 10 [0], given: 8 Re: pizzaria [#permalink] Show Tags 04 Jun 2010, 15:03 A -This choice has absolutely nothing to do with helping Benitos argument. It is completely irrelevant info and should be discarded immediately - VP Joined: 15 Jul 2004 Posts: 1473 Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX) Followers: 22 Kudos [?]: 175 [0], given: 13 Re: pizzaria [#permalink] Show Tags 13 Jul 2010, 22:35 interesting q Manager Joined: 09 Jul 2010 Posts: 150 Followers: 1 Kudos [?]: 26 [1] , given: 3 Re: pizzaria [#permalink] Show Tags 15 Jul 2010, 04:56 1 This post received KUDOS gmatcracker2010 wrote: Benito's Pizzeria has been losing money for the last three months. Benito's mother has advised him to cut prices in order to increase the number of customers to the pizzeria. Benito, however, has decided to increase prices instead. Which of the following, if true, provides the strongest evidence that Benito's plan has a better chance of making the pizzeria profitable than his mother's plan? A Benito could expand his customer base by opening a second location and advertising heavily. B The pizzeria is already producing as many pizzas as its ovens can handle. C Benito has kept essentially the same items on his menu for ten years. D Benito's business has suffered ever since a low-priced national chain moved in next door. E Benito's top-selling item, the large pepperoni pizza, sells for$8.99.
Pleae explain why OA is not A but ...
very simple
B states that pizzeria is already producing as many pizzas as its ovens can handle.--means that Benito's Pizzeria was producing and is still producing..let say sufficient Pizzas. However, it is losing money due to some odd reason. Therefore an increase in price(rather than decrease) will lead to more revenue. Assuming that the no. of customers if remain same. Optiions A,C,D,E are just absurd and are incompetent to fit as an assumption.
Premise 1: Benito's Pizzeria has been losing money for the last three months.
Premise 2: Benito's mother has advised him to cut prices in order to increase the number of customers to the pizzeria.
Premise 3: The pizzeria is already producing as many pizzas as its ovens can handle.
Conclusion: Benito, however, has decided to increase prices instead.
try 2 fit in every other option..the con will fall apart.
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27 Jul 2010, 05:57
great to hear that's called a 700 level question.
Her mom wants to lower the price of product coz she's looking to increase the sales volume, whereas the store is already producing the maximum amount so that won't make any difference.
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27 Jul 2010, 06:19
B is for me.
Option C,D,E seem not relevant to the issue. Option A implies that Benito's customers are not many enough to make the store profitable. Customer problem could be caused by lots of reasons, such as location, pizza flavor, price and so on. If option A is true, Benito's plan could not be helpful.
For option B, if it is true, Benito's plan could be helpful.
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27 Jul 2010, 09:24
good question. I picked C but agree with B.
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27 Jul 2010, 09:55
b for me .. I wish all the questions were that simple ..
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27 Jul 2010, 10:15
I was b/w B and C.
I picked the wrong
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GGG (Gym / GMAT / Girl) -- Be Serious
Its your duty to post OA afterwards; some one must be waiting for that...
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27 Jul 2010, 10:45
B......too easy ..Id oubt weather we ll get this kind of questions at 700-800 level .........
Not A because it is out of scope...because its introducing "another store & advertise" while main concern is between price rice vs customers....
As he is selling all he can as in B......there's nothing he would do by getting more no. of customers ....so better rise price
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27 Jul 2010, 21:32
B it is for the reasons cited by many!
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28 Jul 2010, 11:51
Will go with B.
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16 Aug 2010, 09:43
Its got to be B
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16 Aug 2010, 21:12
B prevents an alternate explanation for the rise in profits. Hence the answer. The argument is 1000% causal.
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18 Aug 2010, 01:22
B FOR ME
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07 Sep 2010, 02:11
Good one!!
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08 Sep 2010, 08:30
B for me as well.
Certainly if you can't meet the demand already there's no point to reduce the price as it will even reduce down your revenue.
With maximum capacity then the only way to increase revenue is by increasing the price.
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08 Sep 2010, 08:39
gmatcracker2010 wrote:
Benito's Pizzeria has been losing money for the last three months. Benito's mother has advised him to cut prices in order to increase the number of customers to the pizzeria. Benito, however, has decided to increase prices instead.
You have been asked to reason why should raising the price of the pizza work better than reducing the prices
Which of the following, if true, provides the strongest evidence that Benito's plan has a better chance of making the pizzeria profitable than his mother's plan?
A Benito could expand his customer base by opening a second location and advertising heavily.Customers can be attracted by opening a second location but the question focuses on the current store -- OOS
B The pizzeria is already producing as many pizzas as its ovens can handle. POE is one way to get to this. On the other hand if my oven is working to full capacity and I reduce prices I am going at a loss. So this disproves mom's suggestion. It my not strengthen Benito's too much but disproving mom here indirectly makes Benito's a better option
C Benito has kept essentially the same items on his menu for ten years. Ha nothing to do with price argument
D Benito's business has suffered ever since a low-priced national chain moved in next door. So I need to lower my prices. THis supports mom's argument
E Benito's top-selling item, the large pepperoni pizza, sells for \$8.99. Ha nothing to do with this information. A trick to make you compare it with outside pizza prices
I would go with B hands down
The Oa is
[Reveal] Spoiler:
b
Pleae explain why OA is not A but ...
I would go with B hands down
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08 Sep 2010, 08:41
This was quite simple. Not sure this would feature higher up the score board on the GMAT..
Re: pizzaria [#permalink] 08 Sep 2010, 08:41
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Display posts from previous: Sort by | 3,255 | 11,750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-04 | latest | en | 0.950121 |
http://mathhelpforum.com/math-topics/25672-calculate-net-charge.html | 1,480,726,633,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00148-ip-10-31-129-80.ec2.internal.warc.gz | 179,481,197 | 11,332 | Thread: calculate the net charge.
1. calculate the net charge.
A metal sphere has a charge of +7.0 µC. What is the net charge after 8.00 1013 electrons have been placed on it?
in µC
anyone able to help me with this one please.
2. Originally Posted by rcmango
A metal sphere has a charge of +7.0 µC. What is the net charge after 8.00 1013 electrons have been placed on it?
in µC
anyone able to help me with this one please.
You mean 8.00 x 10^13 electrons?
Well the charge of a single electron is -1.6 x 10^-19 C. So 8 x 10^13 electrons has a charge of...
-Dan
3. Originally Posted by rcmango
A metal sphere has a charge of +7.0 µC. What is the net charge after 8.00 1013 electrons have been placed on it?
in µC
anyone able to help me with this one please.
Well here's a start: How many coulomb is 8.00 1013 electrons? (You might need to refer to your class notes for the conversion). This NEGATIVE charge is added to +7.0 µC.
4. Originally Posted by topsquark
You mean 8.00 x 10^13 electrons?
Well the charge of a single electron is -1.6 x 10^-19 C. So 8 x 10^13 electrons has a charge of...
-Dan
So couldn't be just -1.6 x 1^-19 C * 8 x 10^13?
then add that to 7uC?
i'm getting confused from sig figs i think.
5. Originally Posted by rcmango
So couldn't be just -1.6 x 1^-19 C * 8 x 10^13? Mr F says: Yes.
then add that to 7uC? Mr F says: Yes.
i'm getting confused from sig figs i think. Mr F says: You wouldn't be the first.
..
6. thanks got it, -5.72 uC
7. Still need helpI'm working on the same problem
Except I have
A metal sphere has a charge of +8.0 µC. What is the net charge after 8.00 x 10^13 electrons have been placed on it
Everytime I try doing
(-1.6x10^-19) x (8.00x10^13) + 8 = 7.9999872
But thats not the correct answer.
Why is that not working??? | 570 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2016-50 | longest | en | 0.939468 |
http://www.mi.sanu.ac.rs/vismath/miyadera/999/node3.html | 1,511,441,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806832.87/warc/CC-MAIN-20171123123458-20171123143458-00481.warc.gz | 459,935,008 | 5,363 | : 4. Related facts. : 0.Abstract. : 2. The observation of facts.
# 3. The answer to the previous questions.
Now it is time for us to answer the preceding three questions.
We are going to show the answer of the three questions by Example 5, Lemma 2 and Theorem 1. Here you need the knowledge of Sterling's formula.
In fact it took us more than 2 years to get the answer. We had presented this beautiful sectors to many mathematicians and computer scientists three times in international symposiums and workshops, but no one came up with a good explanation for this. For example we presented this fact in the International Mathematica Syposium 2003. Please see Miyadera [3] in the reference.
One day one student proposed a very good idea. "We are paying too much attention to the beautiful figure made of a matrix of numbers. In reality it is only a sequence of digits and not a matrix. Let's count the numbers of 0 and 9 between other numbers carefully, and we may find the law that governs the structure of the sequence." This was indeed a very good idea. After this proposal it took only 3 days for us to find the answer. Once the answer was found, the structure looked very simple, but it was like Egg of Columbus.
The key element is the width of the matrix of numbers.
Example 5. We are going to study the structure of Figure (2) in Example 4.
cFigure (2)
cFigure (3)
Figure (3) is the table of terms we get when we apply the binomial theorem to the number = (1030-1)13 that we studied in Example 4. We are going to use Figure (3) here to analyze the structure of Figure(2). Approximately the first row of Figure (2) is made by summing the first and the second row of Figure (3). Similarly the kth row of Figure (2) is made by summing the kth and (k+1)th row of Figure (3).
For reader who are not familiar with Sterling's formula we are going to explain about it. You only have to understand how to use the formula to understand our article.
Lemma 1.
We omit the proof, since this is the famous Sterling's formula. It asserts that the right part of the above formula is approximately equal to the left part when n is big enough.
Lemma 2. For sufficiently large n and t such that we have
Proof. By using Lemma 1 for t we have
. The conclusion of Lemma 2 is a direct result of this formula , since and are relatively very small compared to n when n is big enough.
Lemma 3. If we make n and m big enough while keeping as an almost the same size, then we have
Proof. In this proof we assume that n and m are big enough and we can use Lemma 2 for n ,m and n-m.
Remark. This proof for Lemma 3 was already published in Miyadera and Kotera [1]. We used this for another theorem. We reproduced this again for reader who cannot read Italian Language.
Theorem 1. If we expand and express it as a matrix whose length of row is k, then we have the following (a), (b) and (c).
(a) The matrix has the structure which is described in the following table. See Figure (4). Here we denote by the length of the digits of . As is well known .
(b) The number of 0 and 9 is approximately
,
so it increases if we make while we keep n constant or we decrease n.
(c) If we make , then the figure made of numbers 1,2,3,4,5,6,7,8 is getting nearer and nearer to the graph of the function
, where the x-coordinate is vertical.
the number of 0 and 9 the number of 1,2,3,4,5,6,7,8
cFigure (4)
Proof. (a) If we study the figure (2) and (3) carefully, then it is not difficult to generalize the example to the case of . Then we can easily get this table.
(b) As is easily seen the number of 0 and 9 is approximately
.
Therefore it increases if we while we keep the size of n almost constant or decrease it.
(c) If we make , then the figure made of numbers 1,2,3,4,5,6,7,8 will become bigger. To make the figure almost the same size we divide it by n. By the above table we can approximate the figure by the list
c (1)
using the x-y coordinates.
Here we use x-coordinate vertically.
Since
, we can use Lemma 2 and finish the proof of this theorem.
The answer to the three questions. Theorem 1 (b) gives the answer of Question 1 and 2, and Theorem 1 (c) gives the answer of Question 3. The function
is the so-called entropy function and no doubt it has a beautiful curve.
Example 6. Once we know the structure of a sector, then we can make a beautiful sector easily. This time we use
that is bigger than the number we used in Example 4. It is easy to make a beautiful figure with a bigger numbers.
Conclusion. Our investigation of the number ends here. We really enjoyed it. Once we know the truth, the truth was very simple. We think that this is one of the most beautiful application of the binomial Theorem. Many high school students study the binomial Theorem, but they do not know that the theorem can make this kind of beautiful picture! We want many high school students and high school teachers know our findings and enjoy the beauty of the binomial Theorem!
: 4. Related facts. : 0. Abstract. : 2. The observation of facts. | 1,242 | 5,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-47 | latest | en | 0.95877 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/756/2/t/d/ | 1,638,152,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00591.warc.gz | 979,285,011 | 56,324 | # Properties
Label 756.2.t.d Level $756$ Weight $2$ Character orbit 756.t Analytic conductor $6.037$ Analytic rank $0$ Dimension $4$ CM no Inner twists $4$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$756 = 2^{2} \cdot 3^{3} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 756.t (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$6.03669039281$$ Analytic rank: $$0$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\Q(\sqrt{2}, \sqrt{-3})$$ Defining polynomial: $$x^{4} + 2 x^{2} + 4$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$3^{2}$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \beta_{1} q^{5} + ( -2 + \beta_{2} ) q^{7} +O(q^{10})$$ $$q + \beta_{1} q^{5} + ( -2 + \beta_{2} ) q^{7} + ( 2 + 4 \beta_{2} ) q^{13} + ( -\beta_{1} - \beta_{3} ) q^{17} + ( -1 + \beta_{2} ) q^{19} + ( \beta_{1} + 2 \beta_{3} ) q^{23} + 13 \beta_{2} q^{25} + ( -2 \beta_{1} - \beta_{3} ) q^{29} + ( -2 - \beta_{2} ) q^{31} + ( -2 \beta_{1} + \beta_{3} ) q^{35} + ( 8 + 8 \beta_{2} ) q^{37} -\beta_{3} q^{41} -5 q^{43} + \beta_{1} q^{47} + ( 3 - 5 \beta_{2} ) q^{49} + ( \beta_{1} - \beta_{3} ) q^{53} + ( -2 \beta_{1} - 2 \beta_{3} ) q^{59} + ( -3 + 3 \beta_{2} ) q^{61} + ( 2 \beta_{1} + 4 \beta_{3} ) q^{65} + 2 \beta_{2} q^{67} + ( 4 \beta_{1} + 2 \beta_{3} ) q^{71} + ( -10 - 5 \beta_{2} ) q^{73} + ( 4 + 4 \beta_{2} ) q^{79} -4 \beta_{3} q^{83} + 18 q^{85} + 3 \beta_{1} q^{89} + ( -8 - 10 \beta_{2} ) q^{91} + ( -\beta_{1} + \beta_{3} ) q^{95} + ( 9 + 18 \beta_{2} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 10 q^{7} + O(q^{10})$$ $$4 q - 10 q^{7} - 6 q^{19} - 26 q^{25} - 6 q^{31} + 16 q^{37} - 20 q^{43} + 22 q^{49} - 18 q^{61} - 4 q^{67} - 30 q^{73} + 8 q^{79} + 72 q^{85} - 12 q^{91} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 2 x^{2} + 4$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$3 \nu$$ $$\beta_{2}$$ $$=$$ $$\nu^{2}$$$$/2$$ $$\beta_{3}$$ $$=$$ $$3 \nu^{3}$$$$/2$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$$$/3$$ $$\nu^{2}$$ $$=$$ $$2 \beta_{2}$$ $$\nu^{3}$$ $$=$$ $$2 \beta_{3}$$$$/3$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/756\mathbb{Z}\right)^\times$$.
$$n$$ $$29$$ $$325$$ $$379$$ $$\chi(n)$$ $$-1$$ $$1 + \beta_{2}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
269.1
−0.707107 − 1.22474i 0.707107 + 1.22474i −0.707107 + 1.22474i 0.707107 − 1.22474i
0 0 0 −2.12132 3.67423i 0 −2.50000 + 0.866025i 0 0 0
269.2 0 0 0 2.12132 + 3.67423i 0 −2.50000 + 0.866025i 0 0 0
593.1 0 0 0 −2.12132 + 3.67423i 0 −2.50000 0.866025i 0 0 0
593.2 0 0 0 2.12132 3.67423i 0 −2.50000 0.866025i 0 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 inner
7.d odd 6 1 inner
21.g even 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 756.2.t.d 4
3.b odd 2 1 inner 756.2.t.d 4
7.c even 3 1 5292.2.f.d 4
7.d odd 6 1 inner 756.2.t.d 4
7.d odd 6 1 5292.2.f.d 4
9.c even 3 1 2268.2.w.g 4
9.c even 3 1 2268.2.bm.h 4
9.d odd 6 1 2268.2.w.g 4
9.d odd 6 1 2268.2.bm.h 4
21.g even 6 1 inner 756.2.t.d 4
21.g even 6 1 5292.2.f.d 4
21.h odd 6 1 5292.2.f.d 4
63.i even 6 1 2268.2.bm.h 4
63.k odd 6 1 2268.2.w.g 4
63.s even 6 1 2268.2.w.g 4
63.t odd 6 1 2268.2.bm.h 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
756.2.t.d 4 1.a even 1 1 trivial
756.2.t.d 4 3.b odd 2 1 inner
756.2.t.d 4 7.d odd 6 1 inner
756.2.t.d 4 21.g even 6 1 inner
2268.2.w.g 4 9.c even 3 1
2268.2.w.g 4 9.d odd 6 1
2268.2.w.g 4 63.k odd 6 1
2268.2.w.g 4 63.s even 6 1
2268.2.bm.h 4 9.c even 3 1
2268.2.bm.h 4 9.d odd 6 1
2268.2.bm.h 4 63.i even 6 1
2268.2.bm.h 4 63.t odd 6 1
5292.2.f.d 4 7.c even 3 1
5292.2.f.d 4 7.d odd 6 1
5292.2.f.d 4 21.g even 6 1
5292.2.f.d 4 21.h odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(756, [\chi])$$:
$$T_{5}^{4} + 18 T_{5}^{2} + 324$$ $$T_{13}^{2} + 12$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{4}$$
$3$ $$T^{4}$$
$5$ $$324 + 18 T^{2} + T^{4}$$
$7$ $$( 7 + 5 T + T^{2} )^{2}$$
$11$ $$T^{4}$$
$13$ $$( 12 + T^{2} )^{2}$$
$17$ $$324 + 18 T^{2} + T^{4}$$
$19$ $$( 3 + 3 T + T^{2} )^{2}$$
$23$ $$2916 - 54 T^{2} + T^{4}$$
$29$ $$( 54 + T^{2} )^{2}$$
$31$ $$( 3 + 3 T + T^{2} )^{2}$$
$37$ $$( 64 - 8 T + T^{2} )^{2}$$
$41$ $$( -18 + T^{2} )^{2}$$
$43$ $$( 5 + T )^{4}$$
$47$ $$324 + 18 T^{2} + T^{4}$$
$53$ $$2916 - 54 T^{2} + T^{4}$$
$59$ $$5184 + 72 T^{2} + T^{4}$$
$61$ $$( 27 + 9 T + T^{2} )^{2}$$
$67$ $$( 4 + 2 T + T^{2} )^{2}$$
$71$ $$( 216 + T^{2} )^{2}$$
$73$ $$( 75 + 15 T + T^{2} )^{2}$$
$79$ $$( 16 - 4 T + T^{2} )^{2}$$
$83$ $$( -288 + T^{2} )^{2}$$
$89$ $$26244 + 162 T^{2} + T^{4}$$
$97$ $$( 243 + T^{2} )^{2}$$ | 2,769 | 5,610 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.379299 |
http://www.pcadvisor.co.uk/forum/helproom-1/excel-results-different-cells-249321/ | 1,490,416,251,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188785.81/warc/CC-MAIN-20170322212948-00049-ip-10-233-31-227.ec2.internal.warc.gz | 631,478,779 | 15,778 | # Excel: Results in different cells??
MAJ 20:38 19 Jul 06
Locked
How do I explain this correctly (scrathches head)?
I have a list of computers and their specs. The columns are named:
A1:"Computer Name", B1:"Computer Code", C1:"Hard Drive Size", D1:"Amount of Memory".
There are ten computers in this list, that's 4 Columns (named above) and 10 Rows which contain the relevant information for each computer (total 4 columns and 11 rows: A1:D11).
What I want to know is:
Is there a way that I can enter a particular computer's code in [say] cell A15, so that when that code is entered in A15, it's "Computer Name" is returned in B15 and it's "Hard Drive Size" appears in C15 and it's "Amount of Memory" appears in D16, all at the same time?
MAJ 20:48 19 Jul 06
".....and it's "Amount of Memory" appears in D16, all at the same time?"
......and it's "Amount of Memory" appears in D15, all at the same time?
VoG II 20:57 19 Jul 06
You need a combination of INDEX and MATCH for the first one, MAJ, and then a VLOOKUP for the others. Unfortunately Crimewatch is on in a minute or two and I need to make sure that I'm not featured! I'll return to this later. Anyway, be assured that it can be done.
MAJ 21:05 19 Jul 06
I would watch for you on CW, VoG™, but herself is drooling over Gordon Ramsey on th'other side, thank God for the sanctuary of the computer room. :)
I look forward to seeing how it's done, VoG™.
MAJ 11:38 20 Jul 06
:)
ArrGee 12:30 20 Jul 06
Cheers for that VoG. Been trying to find a way to do this myself.
VoG II 13:23 20 Jul 06
In B15 =INDEX(A2:A11,MATCH(A15,B2:B11,0),0)
In C15 =VLOOKUP(A15,B2:C11,2,FALSE)
In D15 =VLOOKUP(A15,B2:D11,3,FALSE)
Incidentally if you transposed columns A and B then you could use VLOOKUP for the Name as well.
Sorry for the delay - work intervened.
MAJ 16:45 20 Jul 06
As usual, that's top notch, VoG™, works a treat, many thanks for opening my eyes again. I see what you say about transposing columns A and B. I might have a play around with that see if I can manage to do it, can't have you doing everything. ;)
Monoux 17:47 20 Jul 06
bookmark
MAJ 21:10 20 Jul 06
Got it sussed, VoG™, it was dead easy after studying your formulae. Many thanks again, Exccel Guru. :)
This thread is now locked and can not be replied to.
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New iPad, iPhone SE & Red iPhone 7 on sale now | 733 | 2,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-13 | latest | en | 0.932972 |
https://forum.allaboutcircuits.com/threads/momentum.10475/#post-69726 | 1,632,433,537,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00187.warc.gz | 318,382,972 | 25,441 | # Momentum
#### mentaaal
Joined Oct 17, 2005
451
Coming from a college student, this is going to sound ludicrous but what is momentum? I am fully aware of the formulae for momentum and kinetic energy but do not fully understand the difference. I am aware that momentum is the differential of kinetic energy but am just having a hard time trying to understand what it is?
Why is it that momentum is always conserved in a collision, even when some energy can be released from the system (inelastic collision)
#### studiot
Joined Nov 9, 2007
4,998
OK
Momentum is a vector quantity
= the product of mass (a scalar) and velocity (a vector)
Energy is a scalar quantity
Momentum is preserved because we need to preserve both all parts of the vectors involved in a collison, both magnitude and direction.
No scalar can achieve this.
Is this dummies explanation adequate or do you want the fancy maths?
#### mentaaal
Joined Oct 17, 2005
451
This dummies explanation is is what i am always told and made learn but it just isnt doing anything for me. Yes indeed, perhaps some fancy maths would be nice although if its very time consuming a nudge in the right direction will be more than sufficient. I dont spend my time on these forums asking questions in the interests of wasting other peoples' time ;-)
#### Mark44
Joined Nov 26, 2007
628
Greg,
I had to dig up my college Physics book to see what Halliday & Resnick had to say. By their definitions, kinetic energy is the work a body can do by virtue of its motion. Momentum is defined more technically, as the product of the mass of a particle and its velocity.
Given your original question, that's probably not very satisfying. H & R go on to talk about Newton's second law of motion, which they write as "The rate of change of momentum is proportional to the resultant force acting on the body and is in the direction of the force."
In symbols, F = d/dt(p)
So this formula gives a relationship (albeit indirect) between the force on a body, and its momentum. If we antidifferentiate both sides, we get
int[F dt] = p = mv (+ a constant, maybe)
From this, we see that momentum p can be thought of as the sum of forces over time on the body in question.
Mark
#### mentaaal
Joined Oct 17, 2005
451
Thanks for that mark, if anything that is one tasty chunk of food for thought. I also have to get my hands on that book you are mentioning. My physics lecturer also highly recommends it.
#### Mark44
Joined Nov 26, 2007
628
My edition is from 1967. I bought it used for \$13 in 1971, and have hung onto it, for times when I forget that F = ma.
My buddy that I shared a house with at the time and I used to joke that if you couldn't remember F = ma, then F = your grade.
#### thingmaker3
Joined May 16, 2005
5,084
Momentum is why the parked Ford Pinto bounces off the front of the speeding Cadillac.
#### Mark44
Joined Nov 26, 2007
628
Momentum is why the parked Ford Pinto bounces off the front of the speeding Cadillac.
But NOT why it bursts into flames.
#### veritas
Joined Feb 7, 2008
167
momentum is why a baseball hitting you at 30 mph hurts more than a paintball going 90 mph
It's why the 'm' is in the F=ma equation
#### Mark44
Joined Nov 26, 2007
628
momentum is why a baseball hitting you at 30 mph hurts more than a paintball going 90 mph
Another reason is the kinetic energy you are absorbing.
#### studiot
Joined Nov 9, 2007
4,998
'Dummies' was not meant as a criticism. I have learned some really useful stuff from the Dummies Book series and would recommend them to anyone.
Form your original post I though you knew the maths but were trying to gain a feel for the physics.
In physics there are several ideas that can be expressed (measured) in more than one way.
A simple example is the concept of 'quantity of matter' which may be measured by mass or by volume. Both are useful and of course are related by the simple relationship
mass = volume x density.
Neither measure are totally satisfactory by themselves.
In the world of mechanics the concept occurs of inertia or the resistance to change of its motion possessed by a particle. In coloquial terms, how hard it can hit back.
One measure of this is Kinetic energy, as has already been pointed out by others here.
But energy methods do not provide all the answers.
Energy methods can answer the question if I were to accelerate a particle how much energy would I have to supply or even, how hard would I have to push, thanks to Newton's laws.
However they cannot answer the question How hard would I have to push to change the direction of motion, without changing the velocity?
For this we need another model of inertia. We call this this property momentum, because it directly answers the question how much force would the particle exert if it hits a wall?
Consider this collision, the particle bounces back with reversal of direction but no change of actual velocity, the wall does not move.
Since the velocity remains constant there is no change of kinetic energy.
However the particle exerts a quantifiable force on the wall during the impact and ends up travelling the other way capable of exerting a similar force on a another wall in its new path.
So something has changed.
This something we call the momentum and is linked to the force exerted by the impulse equation
Fdt =mdv
Incidentally I note that you were looking at electron dynamics in you spectrum question.
If the momentum question is related to this you need to consider relativistic momentum equations rather than Newtonian ones.
#### mentaaal
Joined Oct 17, 2005
451
Thanks for that reply studiot, I think thats about as good an answer as I could hope to get. Yes i see the point your making and as you have expressed it very well! The rest is up to me to think on it.
Cheers and kudos!
#### mentaaal
Joined Oct 17, 2005
451
Energy methods can answer the question if I were to accelerate a particle how much energy would I have to supply or even, how hard would I have to push, thanks to Newton's laws.
However they cannot answer the question How hard would I have to push to change the direction of motion, without changing the velocity?
For this we need another model of inertia. We call this this property momentum, because it directly answers the question how much force would the particle exert if it hits a wall?
Consider this collision, the particle bounces back with reversal of direction but no change of actual velocity, the wall does not move.
Since the velocity remains constant there is no change of kinetic energy.
However the particle exerts a quantifiable force on the wall during the impact and ends up travelling the other way capable of exerting a similar force on a another wall in its new path.
So something has changed.
This something we call the momentum and is linked to the force exerted by the impulse equation
Fdt =mdv
Incidentally I note that you were looking at electron dynamics in you spectrum question.
If the momentum question is related to this you need to consider relativistic momentum equations rather than Newtonian ones.
I have been pondering your words and am afraid i am still at a loss...
You said " Since the velocity remains constant there is no change of kinetic energy." For the ball to have an elastic collision with the wall such that kinetic energy is conserved then momentum has to also be conserved. When kinetic energy is conserved momentum has to be conserved but not vice versa. ALso when you say that momentum changed by exerting a force on the wall i agree, but so too did the kinetic energy. It had to have as at some point it wasnt moving at all, the energy was being stored in the elastic structure of the partical. Your explanation of momentum is however very good and i am starting to see things a little clearer. Thanks for going to the trouble of writing such a comprehensive response.
#### Caveman
Joined Apr 15, 2008
471
An interesting note about momentum involves Newton's second law. People often simplify it to F = m*a. However, they often forget to state that the mass must be constant for this to be true (or that it is actually a simplification at all).
Newton's second law actually states that force is the time rate of change of momentum, so F = dp/dt as already stated in this thread. To get to the simplification, you need to do a little math...
Since momentum is mass * velocity, you can differentiate it using the product rule to get
F = dp/dt = d(m*v)/dt = dm/dt * v + m * dv/dt = dm/dt * v + m*a.
If the mass is constant, dm/dt = 0, so the first term goes away.
If you want to know a good use of the full equation, think about rockets. As a rocket travels, it burns its fuel. This reduces it's mass, and so dm/dt != 0, and so that term plays a role. If you ignore it, you're calculated acceleration will be smaller than the real acceleration.
#### mentaaal
Joined Oct 17, 2005
451
A point indeed worth noting! So thats why my laser guided rockets kept overshooting....
#### Caveman
Joined Apr 15, 2008
471
I have been pondering your words and am afraid i am still at a loss...
You said " Since the velocity remains constant there is no change of kinetic energy." For the ball to have an elastic collision with the wall such that kinetic energy is conserved then momentum has to also be conserved. When kinetic energy is conserved momentum has to be conserved but not vice versa. ALso when you say that momentum changed by exerting a force on the wall i agree, but so too did the kinetic energy. It had to have as at some point it wasnt moving at all, the energy was being stored in the elastic structure of the partical. Your explanation of momentum is however very good and i am starting to see things a little clearer. Thanks for going to the trouble of writing such a comprehensive response.
Technically, the velocity does change. Velocity is a vector, so a change in direction is a change in velocity. You are right, though, about the storage of energy. Note that momentum, acceleration, and force are all vector quantities as well. K.E. is a scalar (ie, not a vector).
1. At first the particle has a specific momentum and K.E., but no potential energy due to compression.
2. Then as the particle contacts the wall, the wall exerts a force on the particle. F = dp/dt, but the force is opposite of the direction of the velocity, so the velocity will decrease. The K.E. is decreasing. The energy is being transferred into a form of potential energy in the elasticity. Also, in the theoretical case of a perfectly elastic particle, incompressible wall, and homogenous particle, the source of the force will be due less and less to the velocity of the ball and more and more due to the particle attempting to decompress.
3. Eventually, the velocity will reach zero. At this point, the K.E. is 0 as well. The force remains the same, but now the particle's attempt at decompression is the sole source.
4. Now the magnitude of the velocity will begin to increase, but in the opposite direction from original. The reverse occurs until all of the potential energy is transferred back to kinetic energy. | 2,515 | 11,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-39 | latest | en | 0.961536 |
https://fasterthanli.me/series/advent-of-code-2020/part-10 | 1,702,277,051,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00405.warc.gz | 273,041,162 | 14,785 | ### Contents
• Part 2
Day, 10! Day, 10!
Okay, Day 10.
Again, the problem statement is very confusing - but what it all boils down to is this. We have a list of numbers:
16
10
15
5
1
11
7
19
6
12
4
To which we need to add 0 and whatever the maximum was, plus three:
16
10
15
5
1
11
7
19
6
12
4
0
22
From there on, if we take them in order, we'll have gaps of 1 and gaps of 3:
0
1 // 1
4 // 3
5 // 1
6 // 1
7 // 1
10 // 3
11 // 1
12 // 1
15 // 3
16 // 1
19 // 3
22 // 3
And we need to multiply the amount of 1-gaps with the amount of 3-gaps we find.
We can use some of the techniques we've very recently seen: use some types to represent the results, use windows on a slice to deal with contiguous sets of items, and a couple new tricks:
Rust code
#[derive(Default, Clone, Copy, Debug)]
struct Results {
ones: usize,
threes: usize,
}
fn main() {
let mut numbers: Vec<_> = std::iter::once(0)
.chain(
include_str!("input.txt")
.lines()
.map(|x| x.parse::<usize>().unwrap()),
)
.collect();
// clippy told me to use sort_unstable
numbers.sort_unstable();
if let Some(&max) = numbers.iter().max() {
// numbers is still sorted after this
numbers.push(max + 3);
}
let results = numbers.windows(2).fold(Results::default(), |acc, s| {
if let [x, y] = s {
match y - x {
1 => Results {
ones: acc.ones + 1,
..acc
},
3 => Results {
threes: acc.threes + 1,
..acc
},
gap => panic!("invalid input (found {} gap)", gap),
}
} else {
unreachable!()
}
});
dbg!(results, results.ones * results.threes);
}
Cool bear's hot tip
Note that if we used Rust nightly, we could use array_windows instead, which would give us an [usize; 2] and then we wouldn't need an if let because it would be an irrefutable pattern!
Shell session
$cargo run --quiet [src/main.rs:40] results = Results { ones: 7, threes: 5, } [src/main.rs:40] results.ones * results.threes = 35 Let's try it with the real input: Shell session $ cargo run --quiet
[src/main.rs:40] results = Results {
ones: 75,
threes: 37,
}
[src/main.rs:40] results.ones * results.threes = 2775
And we're done with the first part!
### Part 2
The next problem is: what are all the possible ways in which we can connect our adapters? All the valid chains?
Say we have:
1
2
3
5
6
We can have [1, 2, 3, 5, 6], [1, 2, 3, 6], [1, 2, 5, 6], [1, 3, 5, 6], or [1, 3, 6].
Just like in Day 7, we pretty much have a DAG here:
But, unlike Day 7, we're not interested in walking all the nodes of a subgraph, we're interested in all the different ways we can traverse that graph!
But we're not interested in exactly what those paths are, we're interested in how many paths exist - so I think we can be a little smart this time. As a treat.
Consider the last node, 6 - there's only one way to traverse the graph from there, because there are no edges starting from it.
Now consider 5 - there's again, only one way to traverse the graph from there: we can only go to 6.
For 3, we can go either to 5 or to 6, so the number of ways we can traverse the graph if we start from 3 is the sum of the ways we can traverse the graph if we start from 5 or if we start from 6, ie. 1 + 1 = 2.
We're starting to have a mapping from "node" to "number of ways to traverse the graph starting from it":
node_6 = 1
node_5 = node_6 = 1
node_3 = node_5 + node_6 = 1 + 1 = 2
node_2 = node_3 + node_5 = 2 + 1 = 3
node_1 = node_2 + node_3 = 3 + 2 = 5
That sounds like something that's both inexpensive and rather uncomplicated to compute! Note that our results will be slightly different because we need to take into account an additional initial node of 0 and a final node of max + 3, so we're going to end up with more paths:
Rust code
fn main() {
let mut numbers: Vec<_> = std::iter::once(0)
.chain(
// this file contains 1, 2, 3, 5, 6
include_str!("input2.txt")
.lines()
.map(|x| x.parse::<usize>().unwrap()),
)
.collect();
numbers.sort_unstable();
// numbers is still sorted after this
numbers.push(numbers.iter().max().unwrap() + 3);
let mut num_paths = HashMap::new();
let n = numbers.len();
num_paths.insert(numbers.last().copied().unwrap(), 1);
for i in (0..(numbers.len() - 1)).into_iter().rev() {
let i_val = numbers[i];
let range = (i + 1)..=std::cmp::min(i + 3, n - 1);
let num_neighbors: usize = range
.filter_map(|j| {
let j_val = numbers[j];
let gap = j_val - i_val;
if (1..=3).contains(&gap) {
Some(num_paths.get(&j_val).unwrap())
} else {
None
}
})
.sum();
num_paths.insert(i_val, num_neighbors);
}
for &n in numbers.iter().rev() {
let &m = num_paths.get(&n).unwrap();
println!(
"from {}, there's {} {}",
n,
m,
if m == 1 { "path" } else { "paths" }
);
}
}
Shell session
$cargo run --quiet from 9, there's 1 path from 6, there's 1 path from 5, there's 1 path from 3, there's 2 paths from 2, there's 3 paths from 1, there's 5 paths from 0, there's 10 paths Let's try this on my actual puzzle input: Shell session $ cargo run --quiet
from 186, there's 1 path
from 183, there's 1 path
from 182, there's 1 path
from 181, there's 2 paths
from 180, there's 4 paths
from 177, there's 4 paths
from 176, there's 4 paths
from 173, there's 4 paths
from 170, there's 4 paths
from 167, there's 4 paths
from 166, there's 4 paths
from 165, there's 8 paths
from 164, there's 16 paths
from 163, there's 28 paths
from 160, there's 28 paths
from 157, there's 28 paths
from 154, there's 28 paths
from 153, there's 28 paths
from 152, there's 56 paths
from 151, there's 112 paths
from 148, there's 112 paths
from 145, there's 112 paths
from 144, there's 112 paths
from 143, there's 224 paths
from 142, there's 448 paths
from 141, there's 784 paths
from 138, there's 784 paths
from 137, there's 784 paths
from 136, there's 1568 paths
from 135, there's 3136 paths
from 134, there's 5488 paths
from 131, there's 5488 paths
from 128, there's 5488 paths
from 127, there's 5488 paths
from 126, there's 10976 paths
from 125, there's 21952 paths
from 122, there's 21952 paths
from 121, there's 21952 paths
from 118, there's 21952 paths
from 115, there's 21952 paths
from 114, there's 21952 paths
from 113, there's 43904 paths
from 112, there's 87808 paths
from 111, there's 153664 paths
from 108, there's 153664 paths
from 107, there's 153664 paths
from 106, there's 307328 paths
from 103, there's 307328 paths
from 102, there's 307328 paths
from 101, there's 614656 paths
from 100, there's 1229312 paths
from 99, there's 2151296 paths
from 96, there's 2151296 paths
from 95, there's 2151296 paths
from 94, there's 4302592 paths
from 93, there's 8605184 paths
from 92, there's 15059072 paths
from 89, there's 15059072 paths
from 86, there's 15059072 paths
from 85, there's 15059072 paths
from 84, there's 30118144 paths
from 83, there's 60236288 paths
from 82, there's 105413504 paths
from 79, there's 105413504 paths
from 78, there's 105413504 paths
from 77, there's 210827008 paths
from 76, there's 421654016 paths
from 73, there's 421654016 paths
from 72, there's 421654016 paths
from 71, there's 843308032 paths
from 70, there's 1686616064 paths
from 69, there's 2951578112 paths
from 66, there's 2951578112 paths
from 63, there's 2951578112 paths
from 60, there's 2951578112 paths
from 59, there's 2951578112 paths
from 58, there's 5903156224 paths
from 57, there's 11806312448 paths
from 54, there's 11806312448 paths
from 53, there's 11806312448 paths
from 52, there's 23612624896 paths
from 51, there's 47225249792 paths
from 48, there's 47225249792 paths
from 45, there's 47225249792 paths
from 42, there's 47225249792 paths
from 41, there's 47225249792 paths
from 40, there's 94450499584 paths
from 39, there's 188900999168 paths
from 38, there's 330576748544 paths
from 35, there's 330576748544 paths
from 34, there's 330576748544 paths
from 33, there's 661153497088 paths
from 32, there's 1322306994176 paths
from 31, there's 2314037239808 paths
from 28, there's 2314037239808 paths
from 25, there's 2314037239808 paths
from 24, there's 2314037239808 paths
from 23, there's 4628074479616 paths
from 22, there's 9256148959232 paths
from 19, there's 9256148959232 paths
from 18, there's 9256148959232 paths
from 17, there's 18512297918464 paths
from 14, there's 18512297918464 paths
from 13, there's 18512297918464 paths
from 12, there's 37024595836928 paths
from 11, there's 74049191673856 paths
from 10, there's 129586085429248 paths
from 7, there's 129586085429248 paths
from 6, there's 129586085429248 paths
from 3, there's 129586085429248 paths
from 2, there's 129586085429248 paths
from 1, there's 259172170858496 paths
from 0, there's 518344341716992 paths
And there we have it! 518344341716992 is our final answer.
Until next time, be kind with yourself! | 2,889 | 8,589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-50 | longest | en | 0.666338 |
https://www.funwithpuzzles.com/2018/03/kropki-or-dots-sudoku-puzzle.html | 1,719,126,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00510.warc.gz | 699,510,468 | 36,380 | Once upon a time, Kropki Sudoku Puzzle was one of my most feared Sudoku puzzle types. This is one of the Sudoku puzzle types in which without giving any hints, this Sudoku can be solved uniquely and logically. However, some of the Kropki Sudoku puzzles are very tough and most of the experienced players try to solve Kropki Sudoku with a guess or bifurcation. Kropki Sudoku is also known by another name called Dots Sudoku. However, the Kropki Sudoku name is used in most of the World Sudoku Championships and other Sudoku competitions. This Kropki Sudoku puzzle, which I am publishing as part of the Fun With Sudoku Series is the 280th Sudoku Puzzle in this Series.
## Rules of Kropki Sudoku Puzzle
Standard Sudoku Rules apply. Additionally, if the absolute difference between two digits in neighboring cells equals 1, then they are separated by a white dot. If the digit is half of the digit in the neighboring cell, then they are separated by a black dot. The dot between 1 and 2 can be either white or black.
Kropki Sudoku Puzzle (Fun With Sudoku #280)
The answer to this Kropki Sudoku puzzle can be viewed by clicking on the button. Please do give your best try to solve this Sudoku puzzle before looking at the answer.
This Kropki Sudoku Puzzles I am publishing as #280th Sudoku puzzle in the Fun With Sudoku Series. Here are the next and previous Sudoku puzzles published in this series.
Unknown said...
It took a lot of time to solve kropki aprx an hour but I thoroughly enjoyed solving it
Create some more kropki for more practice .
Rajesh Kumar said...
Thanks Poonam for your feedback. As of now there are more than 30 Kropki Sudoku puzzles published on this website. You can click on "Kropki Sudoku" tag near heading of this post to access these puzzles or checkout Kropki Sudoku Main page as given below to find links of these puzzles
http://www.funwithpuzzles.com/2015/09/kropki-puzzles.html
Soon I will publish some more Kropki Sudoku puzzles :)
Unknown said...
Thanks | 457 | 1,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-26 | latest | en | 0.914531 |
https://mangotips.com/calculate-the-percentage-of-65-of-what-number-is-78/ | 1,669,505,404,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00454.warc.gz | 436,189,889 | 14,126 | # Calculate the Percentage of 65 of What Number is 78
## Calculate the Percentage of 65 of What Number is 78:
You’ve heard that 65 percent of a number equals 78. But what does it really mean? How do you calculate this answer? The answer will be 78 minus 65. That’s right – 65 percent of 78 is 78. But what about the other way around? What percentage of 65 is 78? You can find the answer by comparing the percentages of two different numbers. But how do you find out how much each one is worth?
## 78 is x percent of 65
You know that 78 is x percent of 65, but what is the formula for calculating this percentage? First, we need to define what “percent” means. Percent comes from the Latin word “per centum,” which means “per hundred.” So 65 percent of a number equals sixty-five. Now, divide the number by 65 to get the formula for 78. Finally, multiply the number by 100 to get the answer.
We can calculate the percent increase of a quantity by looking at its proportions to its reference quantity. For example, 78 is 20 percent bigger than 65, and thus 80% larger than 65. Therefore, the increased quantity of 78 is twenty percent of 65. We can use this formula to find the percent increase of any quantity. However, this formula is not suited for high-risk use. Consequently, it is not advisable to use it in any manner that may cause harm to you or others.
To find a percentage, we need to enter a fractional value. For instance, 5% of 20 is equal to x/100 * twenty. That’s because 78 is twenty percent of 65. So, the answer to this question is eighty-seven percent of sixty-five. However, we can use other methods such as the percentage calculator. When calculating fractions, you need to use a calculator.
If 78 is x percent of sixty-five, the result is twenty-seven dollars. So, if the original cost of an item is \$78 and the discount is 65%, then the final price will be \$27.3. Similarly, sixty-five dollars less seventy-seven dollars equal twenty-seven cents. In this way, 78 is twenty-seven percent of sixty-five dollars. | 489 | 2,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-49 | longest | en | 0.938941 |
https://informesia.com/377/what-is-the-frictional-force | 1,685,466,069,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00146.warc.gz | 361,203,168 | 14,761 | 42 views
Define the Frictional Force? Or, What do You Mean by the Frictional Force?
Frictional Force : It is the force exerted by a surface when an object moves across the surface.
There are two types of frictional forces ;
1. Kinetic friction
2. Static fraction
• Frictional force can also be defined as the force, which opposes the motion of an object.
• It is also known as the constant force.
• The frictional force also depends on the nature of the surface and degree to which they are pressed together.
• The maximum amount of frictional force is calculated by using the relation ;
Force = Coefficient of friction × Normal force.
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26 views | 210 | 843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.932122 |
https://www.geeksforgeeks.org/python-numpy-np-hermevander3d-method/?ref=ml_lbp | 1,712,978,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00797.warc.gz | 744,716,893 | 45,693 | # Python | Numpy np.hermevander3d() method
Last Updated : 11 Dec, 2019
With the help of `np.hermevander3d()` method, we can get the pseudo vandermonde matrix of a given 3-D data having degrees x, y and z by using `np.hermevander3d()` method.
Syntax : `np.hermevander3d(x, y, z, [x_deg, y_deg, z_deg])`
Return : Return the pseudo vandermonde matrix of given 3-D data.
Example #1 :
In this example we can see that by using `np.hermevander3d()` method, we are able to get the pseudo vandermonde matrix of a given 3-D data having degree (x, y, z) by using this method.
`# import numpy and hermevander3d ` `import` `numpy as np ` `from` `numpy.polynomial.hermite_e ``import` `hermevander3d ` ` ` `x ``=` `np.array([``1``, ``0.1``]) ` `y ``=` `np.array([``2``, ``0.2``]) ` `z ``=` `np.array([``3``, ``0.3``]) ` `x_deg, y_deg, z_deg ``=` `2``, ``3``, ``1` ` ` `# using np.hermevander3d() method ` `gfg ``=` `hermevander3d(x, y, z, [x_deg, y_deg, z_deg]) ` ` ` `print``(gfg) `
Output :
[[ 1.00000e+00 3.00000e+00 2.00000e+00 6.00000e+00 3.00000e+00
9.00000e+00 2.00000e+00 6.00000e+00 1.00000e+00 3.00000e+00
2.00000e+00 6.00000e+00 3.00000e+00 9.00000e+00 2.00000e+00
6.00000e+00 0.00000e+00 0.00000e+00 0.00000e+00 0.00000e+00
0.00000e+00 0.00000e+00 0.00000e+00 0.00000e+00]
[ 1.00000e+00 3.00000e-01 2.00000e-01 6.00000e-02 -9.60000e-01
-2.88000e-01 -5.92000e-01 -1.77600e-01 1.00000e-01 3.00000e-02
2.00000e-02 6.00000e-03 -9.60000e-02 -2.88000e-02 -5.92000e-02
-1.77600e-02 -9.90000e-01 -2.97000e-01 -1.98000e-01 -5.94000e-02
9.50400e-01 2.85120e-01 5.86080e-01 1.75824e-01]]
Example #2 :
`# import numpy and hermevander3d ` `import` `numpy as np ` `from` `numpy.polynomial.hermite_e ``import` `hermevander3d ` ` ` `x ``=` `np.array([``1.01``, ``2.02``, ``3.03``]) ` `y ``=` `np.array([``10.1``, ``20.2``, ``30.3``]) ` `z ``=` `np.array([``0.1``, ``0.2``, ``0.3``]) ` `x_deg, y_deg, z_deg ``=` `1``, ``1``, ``3` ` ` `# using np.hermevander3d() method ` `gfg ``=` `hermevander3d(x, y, z, [x_deg, y_deg, z_deg]) ` ` ` `print``(gfg) `
Output :
[[ 1. 0.1 -0.99 -0.299 10.1 1.01
-9.999 -3.0199 1.01 0.101 -0.9999 -0.30199
10.201 1.0201 -10.09899 -3.050099]
[ 1. 0.2 -0.96 -0.592 20.2 4.04
-19.392 -11.9584 2.02 0.404 -1.9392 -1.19584
40.804 8.1608 -39.17184 -24.155968]
[ 1. 0.3 -0.91 -0.873 30.3 9.09
-27.573 -26.4519 3.03 0.909 -2.7573 -2.64519
91.809 27.5427 -83.54619 -80.149257]]
Previous
Next | 1,176 | 2,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.390458 |
http://stackoverflow.com/questions/17851264/numerical-integral-of-large-numbers-in-fortran-90/17855801 | 1,462,266,831,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121090.75/warc/CC-MAIN-20160428161521-00007-ip-10-239-7-51.ec2.internal.warc.gz | 272,262,726 | 20,169 | # Numerical Integral of large numbers in Fortran 90
so I have the following Integral that i need to do numerically:
``````Int[Exp(0.5*(aCosx + bSinx + cCos2x + dSin2x))] x=0..2Pi
``````
The problem is that the output at any given value of x can be extremely large, e^2000, so larger than I can deal with in double precision.
I havn't had much luck googling for the following, how do you deal with large numbers in fortran, not high precision, i dont care if i know it to beyond double precision, and at the end i'll just be taking the log, but i just need to be able to handle the large numbers untill i can take the log..
Are there integration packes that have the ability to handle arbitrarily large numbers? Mathematica clearly can.. so there must be something like this out there.
Cheers
-
This is probably an extended comment rather than an answer but here goes anyway ...
As you've already observed Fortran isn't equipped, out of the box, with the facility for handling such large numbers as `e^2000`. I think you have 3 options.
1. Use mathematics to reduce your problem to one which does (or a number of related ones which do) fall within the numerical range that your Fortran compiler can compute.
2. Use Mathematica or one of the other computer algebra systems (eg Maple, SAGE, Maxima). All (I think) of these can be integrated into a Fortran program (with varying degrees of difficulty and integration).
3. Use a library for high-precision (often called either arbitray-precision or multiple-precision too) arithmetic. Your favourite search engine will turn up a number of these for you, some written in Fortran (and therefore easy to integrate), some written in C/C++ or other languages (and therefore slightly harder to integrate). You might start your search at Lawrence Berkeley or the GNU bignum library.
4. (Yes I know that I wrote that you have 3 options, but your question suggests that you aren't ready to consider this yet) You could write your own high-/arbitrary-/multiple-precision functions. Fortran provides everything you need to construct such a library, there is a lot of work already done in the field to learn from, and it might be something of interest to you.
In practice it generally makes sense to apply as much mathematics as possible to a problem before resorting to a computer, that process can not only assist in solving the problem but guide your selection or construction of a program to solve what's left of the problem.
-
I agree with High Peformance Mark that the best option here numerically is to use analytics to scale or simplify the result first.
I will mention that if you do want to brute force it, gfortran (as of 4.6, with the libquadmath library) has support for quadruple precision reals, which you can use by selecting the appropriate kind. As long as your answers (and the intermediate results!) don't get too much bigger than what you're describing, that may work, but it will generally be much slower than double precision.
-
This requires looking deeper at the problem you are trying to solve and the behavior of the underlying mathematics. To add to the good advice already provided by Mark and Jonathan, consider expanding the exponential and trig functions into Taylor series and truncating to the desired level of precision.
Also, take a step back and ask why you are trying to accomplish by calculating this value. As an example, I recently had to debug why I was getting outlandish results from a property correlation which was calculating vapor pressure of a fluid to see if condensation was occurring. I spent a long time trying to understand what was wrong with the temperature being fed into the correlation until I realized the case causing the error was a simulation of vapor detonation. The problem was not in the numerics but in the logic of checking for condensation during a literal explosion; physically, a condensation check made no sense. The real problem was the code was asking an unnecessary question; it already had the answer.
I highly recommend Forman Acton's Numerical Methods That (Usually) Work and Real Computing Made Real. Both focus on problems like this and suggest techniques to tame ill-mannered computations.
- | 908 | 4,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-18 | latest | en | 0.938213 |
http://au.metamath.org/mpegif/dn1.html | 1,531,716,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589179.32/warc/CC-MAIN-20180716041348-20180716061348-00572.warc.gz | 30,995,716 | 4,355 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dn1 Structured version Unicode version
Theorem dn1 933
Description: A single axiom for Boolean algebra known as DN1. See http://www-unix.mcs.anl.gov/~mccune/papers/basax/v12.pdf. (Contributed by Jeffrey Hankins, 3-Jul-2009.) (Proof shortened by Andrew Salmon, 13-May-2011.) (Proof shortened by Wolf Lammen, 6-Jan-2013.)
Assertion
Ref Expression
dn1
Proof of Theorem dn1
StepHypRef Expression
1 pm2.45 387 . . . . 5
2 imnan 412 . . . . 5
31, 2mpbi 200 . . . 4
43biorfi 397 . . 3
5 orcom 377 . . . 4
6 ordir 836 . . . 4
75, 6bitri 241 . . 3
84, 7bitri 241 . 2
9 pm4.45 670 . . . . 5
10 anor 476 . . . . 5
119, 10bitri 241 . . . 4
1211orbi2i 506 . . 3
1312anbi2i 676 . 2
14 anor 476 . 2
158, 13, 143bitrri 264 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wb 177 wo 358 wa 359 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361
Copyright terms: Public domain W3C validator | 453 | 1,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-30 | latest | en | 0.329315 |
https://danshiebler.com/2018-11-10-category-solutions/ | 1,642,552,375,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301217.83/warc/CC-MAIN-20220119003144-20220119033144-00398.warc.gz | 265,007,272 | 34,383 | I recently worked through Bartosz Milewski’s excellent free book “Category Theory for Programmers.” The book is available online here and here.
I had an awesome time reading the book and learning about Category Theory so I figured I’d post my solutions to the book problems online to make it easier for other people to have a similar experience. You can find my solutions below:
# Section 1
##### Implement, as best as you can, the identity function in your favorite language (or the second favorite, if your favorite language happens to be Haskell).
Solution
def identity(x):
return x
##### Implement the composition function in your favorite language. It takes two functions as arguments and returns a function that is their composition.
Solution
def compose(f1, f2):
return lambda x: f2(f1(x))
##### Write a program that tries to test that your composition function respects identity.
Solution
assert compose(lambda x: x + 4, identity)(5) == 9
assert compose(identity, lambda x: x + 4)(5) == 9
##### p1.4
Solution The world wide web is indeed a category if we consider the objects to be webpages and for there to be an “arrow” between A and B if there is a way to get to B from A by clicking on links
##### Is Facebook a category, with people as objects and friendships as morphisms?
Solution No, because just because A -> B and B -> C does not imply A -> C
##### When is a directed graph a category?
Solution Whenever every node has an edge that points back to it and for every two nodes A, B such that there is a path from A to B, there is also an edge connecting A to B.
# Section 2
##### p2.1 Define a higher-order function (or a function object) memoize in your favorite language. This function takes a pure function f as an argument and returns a function that behaves almost exactly like f, except that it only calls the original function once for every argument, stores the result internally, and subsequently returns this stored result every time it’s called with the same argument. You can tell the memoized function from the original by watching its performance. For instance, try to memoize a function that takes a long time to evaluate. You’ll have to wait for the result the first time you call it, but on subsequent calls, with the sameargument, you should get the result immediately.
Solution
def memoize(f):
calls = {}
def memoized(x):
if x not in calls:
calls[x] = f(x)
return calls[x]
return memoized
##### p2.2 Try to memoize a function from your standard library that you normally use to produce random numbers. Does it work?
Solution This will not work
##### p2.3 Most random number generators can be initialized with a seed. Implement a function that takes a seed, calls the random number generator with that seed, and returns the result. Memoize that function. Does it work?
Solution
def seed_to_random(seed):
np.random.seed(seed)
return np.random.random()
memoized_random = memoize(seed_to_random)
assert np.isclose(memoized_random(0), memoized_random(0))
assert memoized_random(0) != memoized_random(1)
##### a: The factorial function from the example in the text.
Solution factorial is a pure function
##### b: std::getchar()
Solution getchar is not a pure function, since it relies on the state of stdin
##### c:
bool f() {
std::cout << "Hello!" << std::endl;
return true;
}
Solution f is not a pure function, since it has the side effect of printing
##### d:
int f(int x) {
static int y = 0;
y += x;
return y;
}
Solution f is not a pure function, since it both has the side effect of incrementing y and relies on the state of static variable y
##### p2.5 How many different functions are there from Bool to Bool? Can you implement them all?
Solution
same :: Bool -> Bool
same trueorfalse = trueorfalse
opposite :: Bool -> Bool
opposite trueorfalse = not trueorfalse
alwaystrue :: Bool -> Bool
alwaystrue _ = True
alwaysfalse :: Bool -> Bool
alwaysfalse _ = False
# Section 3
##### p3.1 Generate a free category from:
• A graph with one node and no edges Solution Add an identity arrow.
• A graph with one node and one (directed) edge (hint: this edge can be composed with itself) Solution Add infinite arrows to represent every number of applications of the directed edge.
• A graph with two nodes and a single arrow between them Solution Add identity arrows.
• A graph with a single node and 26 arrows marked with the letters of the alphabet: a, b, c … z. Solution Add an identity arrow, and then add infinite arrows, one for every combination of a-z of any length.
##### p3.2 What kind of order is this?
• A set of sets with the inclusion relation: A is included in B if every element of A is also an element of B. Solution This is a partial order.
• For any (a, b) there is at most one a -> b and if a -> b and b -> a then a and b have the same elements and are the same set.
• Since there might be some (a, b) where a intersect b is empty, this is not a total order
• C++ types with the following subtyping relation: T1 is a subtype of T2 if a pointer to T1 can be passed to a function that expects a pointer to T2 without triggering a compilation error. Solution This is a partial order.
• For any (t1, t2) there is at most one t1 -> t2, and if t1 -> t2 and t2 -> t1 then t1 and t2 are the same type.
• There are types not connected by a subtype relation, so this is not a total order.
##### p3.3 Considering that Bool is a set of two values True and False, show that it forms two (set-theoretical) monoids with respect to, respectively,operator AND and OR.
Solution AND * Closure: The output of AND is boolean * Identity: The identity is True * Associative: Easy to show by enumeration
OR * Closure: The output of OR is boolean * Identity: The identity is False * Associative: Easy to show by enumeration
##### p3.4 Represent the Bool monoid with the AND operator as a category: List the morphisms and their rules of composition.
Solution The single element in this category is the Bool type. The morphisms are AND True (identity) and AND False. The composition of these two is AND False.
##### p3.5 Represent addition modulo 3 as a monoid category.
Solution The single element in this category is the the type [Int < 3, >= 0]. The morphisms are * A: add 3n (identity) * B: add 1 + 3n * C: add 2 + 3n The morphisms in this category are closed under association because B . B is C and both B . C, C . B are A
# Section 4
##### p4.1 Construct the Kleisli category for partial functions (define composition and identity).
Solution
class Optional:
def __init__(self, value):
self._value = value
def is_valid(self):
return self._value is not None
def get(self):
assert self.is_valid()
return self._value
def compose(f1, f2):
def composed(x):
f1out = f1(x)
return f2(f1out.get()) if f1out.is_valid() else Optional(None)
return composed
def identity(x):
return Optional(x)
##### p4.2 Implement the embellished function safe_reciprocal that returns a valid reciprocal of its argument, if it’s different from zero.
Solution
def safe_root(x):
return Optional(np.sqrt(x)) if x >= 0 else Optional(None)
def safe_reciprocal(x):
return Optional(1 / float(x)) if x != 0 else Optional(None)
assert not safe_root(-1).is_valid()
assert np.isclose(safe_root(4).get(), 2.0)
assert not safe_reciprocal(0).is_valid()
assert np.isclose(safe_reciprocal(4).get(), 0.25)
##### p4.3 Compose safe_root and safe_reciprocal to implement safe_root_reciprocal that calculates sqrt(1/x) whenever possible.
Solution
safe_root_reciprocal = compose(safe_reciprocal, safe_root)
assert not safe_root_reciprocal(0).is_valid()
assert not safe_root_reciprocal(-5).is_valid()
assert np.isclose(safe_root_reciprocal(0.25).get(), 2)
# Section 5
##### p5.1 Show that the terminal object is unique up to unique isomorphism.
Solution Consider two terminal objects A, B. There is exactly one morphism m1 from A -> B since B is terminal and exactly one morphism m2 from B -> A since A is terminal. Then m1 . m2 is a morphism from B -> B and m2 . m1 is a morphism from A -> A. Since B is terminal, there is exactly one morphism from B -> B, so m1 . m2 is the identity.
Therefore m1, m2 form an isomorphism between A and B. Since there are no other morphisms between A and B, m1, m2 is a unique isomorphism.
##### p5.2 What is a product of two objects in a poset? Hint: Use the universal construction.
Solution The product of two objects A, B in a poset is the object C that is less than both A and B (i.e. exists: p: C -> A and q: C -> B) and for any other object D that is also less than A and B, exists D -> C. This object does not always exist.
Sufficiency Say we have such an A, B, C. Now consider some object D such that p2: D -> A, q2: D -> B. Then we have some m: D -> C so p1 . m: D -> A and q1 . m: D -> B. Now since there is at most one morphism between any pair of objects in a poset, it’s true that p2 = p1 . m and q2 = q1 . m, so m factorizes p and q.
Necessity If there were some object D such that D -> A, D -> B but not D -> C, then there is no morphism m such that p1 = p2 . m since m must be a morphism from D -> C. Therefore C is not the product of A,B.
##### p5.3 What is a coproduct of two objects in a poset?
Solution We just reverse the arrows in p5.2. The coproduct of two objects A, B in a poset is the object C that is greater than both A and B (i.e. exists: p: A -> C and q: B -> C) and for any other object D that is also greater than A and B, exists C -> D. This object does not always exist.
##### p5.4 Implement the equivalent of Haskell Either as a generic type in your favorite language (other than Haskell).
Solution
class EitherAbstract(object):
pass
class RightEither(EitherAbstract):
def __init__(self, right):
self.right = right
class LeftEither(EitherAbstract):
def __init__(self, left):
self.left = left
# doing something like this in python is a recipe for disaster :)
def either_factory(left_type, right_type):
def generate(left=None, right=None):
assert ((left is None) ^ (right is None))
if left is not None:
assert isinstance(left, left_type)
out = LeftEither(left=left)
if right is not None:
assert isinstance(right, right_type)
out = RightEither(right=right)
return out
return generate
my_left = either_factory(int, str)(left=5)
assert my_left.left == 5
my_right = either_factory(int, str)(right="5")
assert(my_right.right == "5")
try:
either_factory(int, str)()
print("failed")
except AssertionError:
pass
try:
either_factory(int, str)(left=5, right="5")
print("failed")
except AssertionError:
pass
##### p5.5 Show that Either is a “better” coproduct than int equipped with two injections:
int i(int n) { return n; }
int j(bool b) { return b? 0: 1; }
Hint: Define a function int m(Either const & e); that factorizes i and j. Solution
# Consider the following injections from int and bool into int
def int_to_int(int_param):
assert isinstance(int_param, int)
return int_param
def bool_to_int(bool_param):
assert isinstance(bool_param, bool)
return 1 if bool_param else 0
# We can define the following morphism from Either into int
def either_to_int(either_param):
isinstance(either_param, EitherAbstract)
if either_param.kind == "left":
out = int_to_int(either_param.left)
elif either_param.kind == "right":
out = bool_to_int(either_param.right)
return out
"""
Then
bool_to_int(x) = either_to_int (either_factory(int, bool)(left=x))
int_to_int(y) = either_to_int (either_factory(int, bool)(right=y))
so either_to_int factorizes bool_to_int and int_to_int
"""
assert either_to_int(either_factory(int, bool)(left=5)) == int_to_int(5)
assert either_to_int(either_factory(int, bool)(right=True)) == bool_to_int(True)
assert either_to_int(either_factory(int, bool)(right=False)) == bool_to_int(False)
##### p5.6 Continuing the previous problem: How would you argue that int with the two injections i and j cannot be “better” than Either?
Solution Say there exists some function impossible_m such that
either_factory(int, bool)(left=x) = impossible_m ( int_to_int (x))
either_factory(int, bool)(right=y) = impossible_m ( bool_to_int (y))
For all int x and bool y. Then it must be the case that:
impossible_m(1) = LeftEither(left=1)
impossible_m(1) = RightEither(right=1)
Which is not possible, because the output of a function for a given input argument must be unique.
##### p5.7 Still continuing: What about these injections?
int i(int n) {
if (n < 0) return n;
return n + 2;
}
int j(bool b) { return b? 0: 1; }
Solution Say there exists some function new_m such that
def int_to_int_2(int_param):
assert isinstance(int_param, int)
return int_param if int_param < 0 else int_param + 2
either_factory(int, bool)(left=x) = new_m ( int_to_int_2 (x))
Then it must be the case that
new_m(-x) = LeftEither(-x) for all x
new_m(2) = LeftEither(0)
new_m(3) = LeftEither(1)
new_m(4) = LeftEither(2)
...
new_m(max_int) = Left(max_int - 2)
Since there are only a finite number of integers in python/C++, Left(max_int - 1) and Left(max_int) cannot be in the domain of new_m.
##### p5.8 Come up with an inferior candidate for a coproduct of int and bool that cannot be better than Either because it allows multiple acceptable morphisms from it to Either
Solution
Consider some type SuperEither defined as
data SuperEither = IntBoolTuple (Int, Int) | BoolBoolTuple (Bool, Bool)
We can define injections into this type from int and bool of the forms
intint :: Int -> SuperEither;
intint x = IntIntTuple (x, x)
boolbool :: Bool -> SuperEither;
boolbool x = BoolBoolTuple (x, x)
Then we can define multiple morphisms from SuperEither into Either that take either the first or second element.
# Section 6
##### p6.1 Show the isomorphism between Maybe a and Either () a.
Solution We can define the following two functions, which serve as inverses
maybeToEither :: Maybe a -> Either () a
maybeToEither inputMaybe =
case inputMaybe of
Just a -> Right a
Nothing -> Left ()
eitherToMaybe :: Either () a -> Maybe a
eitherToMaybe inputEither =
case inputEither of
Right a -> Just a
Left () -> Nothing
##### p6.2 Here’s a sum type defined in Haskell:
data Shape = Circle Float
| Rect Float Float
##### When we want to define a function like area that acts on a Shape, we do it by pattern matching on the two constructors:
area :: Shape -> Float
area (Circle r) = pi * r * r
area (Rect d h) = d * h
##### Implement Shape in C++ or Java as an interface and create two classes: Circle and Rect. Implement area as a virtual function.
Solution
# I'll use python again, just for fun
class AbstractShape(object):
def area(self):
assert NotImplementedError()
def circ(self):
assert NotImplementedError()
class Circle(AbstractShape):
def area(self):
return self.radius**2 * np.pi
def circ(self):
return 2 * self.radius * np.pi
class Rect(AbstractShape):
def __init__(self, height, width):
self.height = height
self.width = width
def area(self):
return self.height * self.width
def circ(self):
return 2 * self.height + 2 * self.width
rect = Rect(3, 5)
assert rect.circ() == 16
assert rect.area() == 15
##### p6.3 Continuing with the previous example: We can easily add a new function circ that calculates the circumference of a Shape. We can do it without touching the definition of Shape:
circ :: Shape -> Float
circ (Circle r) = 2.0 * pi * r
circ (Rect d h) = 2.0 * (d + h)
##### Add circ to your C++ or Java implementation. What parts of the original code did you have to touch?
Solution See above. We needed to add it to each class, including AbstractShape.
##### p6.4 Continuing further: Add a new shape, Square, to Shape and make all the necessary updates. What code did you have to touch in Haskell vs. C++ or Java? (Even if you’re not a Haskell programmer, the modifications should be pretty obvious.)
Solution For haskell we need to update the Shape definition and add another line to circ and area implementations. For python we needed to write a new class with a new initializer, inheriting from Rect
data Shape = Circle Float | Rect Float Float | Square Float
area :: Shape -> Float
area (Circle r) = pi * r * r
area (Rect d h) = d * h
area (Square h) = h * h
circ :: Shape -> Float
circ (Circle r) = 2.0 * pi * r
circ (Rect d h) = 2.0 * (d + h)
circ (Square h) = 4.0 * h
Python:
class Square(Rect):
def __init__(self, length):
self.height = length
self.width = length
square = Square(5)
assert square.circ() == 20
assert square.area() == 25
##### p6.5 Show that a + a = 2 * a holds for types (up to isomorphism). Remember that 2 corresponds to Bool, according to our translation table.
Solution a + a is equivalent to Either a a and 2 * a is equivalent to (Bool, a). We can define the following invertible functions between them.
aPlusAToTwoTimesA :: Either a a -> (Bool, a)
aPlusAToTwoTimesA eitherAA =
case eitherAA of
Left a -> (True, a)
Right a -> (False, a)
twoTimesAToAPlusA :: (Bool, a) -> Either a a
twoTimesAToAPlusA twoTimesA =
case twoTimesA of
(True, a) -> Left a
(False, a) -> Right a
# Section 7: Functors
##### p7.1 Can we turn the Maybe type constructor into a functor by defining: fmap _ _ = Nothing which ignores both of its arguments? (Hint: Check the functor laws.)
Solution No, this mapping of morphisms does not preserve the identity. For some Just a, we see that:
(fmap id) Just a = Nothing
id Just a = Just a
##### p7.2 Prove functor laws for the reader functor. Hint: it’s really simple.
Solution We need to use equational reasoning to prove that fmap maintains identity and preserves composition
Identity
fmap id (a->b) =
(.) id (a->b) =
id (a->b)
Composition
fmap ((c->d) . (b->c)) (a->b) =
(c->d) . (b->c) . (a->b) =
(c->d) . fmap ((b->c) (a->b)) =
fmap (c->d) (fmap (b->c) (a->b))
##### p7.3 Implement the reader functor in your second favorite language (the first being Haskell, of course).
Solution
def reader_functor_fmap(f, r_to_a):
return lambda r: f(r_to_a(r))
def r_to_0(r):
return 0
def r_to_1(r):
return 1
r_to_5 = reader_functor_fmap(lambda x: x + 5, r_to_0)
assert r_to_0("r") == 0
assert r_to_1("r") == 1
assert r_to_5("r") == 5
##### p7.4 Prove the functor laws for the list functor. Assume that the laws are true for the tail part of the list you’re applying it to (in other words, use induction).
Solution
Base Case
• Identity
fmap id Nil = Nil = id Nil
• Composition
fmap (f . g) Nil = Nil = fmap f (Nil) = fmap f (fmap g Nil)
Inductive Step
• Identity
fmap id (Cons x t) =
Cons (id x) (fmap id t)) =
Cons (id x) (id t)) =
Cons (x t) =
id Cons (x t)
• Composition
fmap (f . g) (Cons x t)
= Cons ((f . g) x) (fmap (f . g) t)) // definition of fmap
= Cons ((f . g) x) ((fmap f . fmap g) t) // induction
= fmap f (Cons (g x) (fmap g t)) // definition of fmap
= fmap f (fmap g (Cons (x t))) // definition of fmap
= (fmap f . fmap g) (Cons (x t))
# Section 8: Functorality
##### p8.1 Show that the data type: data Pair a b = Pair a b is a bifunctor. For additional credit implement all three methods of Bifunctor and use equational reasoning to show that these definitions are compatible with the default implementations whenever they can be applied.
Solution Say we keep one of the arguments constant, then the fmap for both sides is just:
fmap f Pair (a), (C) = Pair (f a) (C)
Identity
fmap id Pair a C = Pair id a C = Pair a C
Composition
fmap f*g Pair a C = Pair f*g(a) C = fmap f Pair g(a) C = fmap f fmap g Pair a c
The three methods of Bifunctor
data Pair a b = Pair a b
pairBimap :: (a -> c) -> (b -> d) -> Pair a b -> Pair c d
pairBimap g h (Pair a b) = Pair (g a) (h b)
pairFirst :: (a -> c) -> Pair a b -> Pair c b
pairFirst g (Pair a b) = Pair (g a) b
pairSecond :: (b -> d) -> Pair a b -> Pair a d
pairSecond f (Pair a b) = Pair a (f b)
Proof that these definitions are compatible with the default implementations whenever they can be applied.
Proof of pairBimap
(pairBimap g h) (Pair a b) =
Pair (g a) (h b) // definition of pairBimap
pairFirst g (Pair a (h b)) // definition of pairFirst
(pairFirst g . pairSecond h) (Pair a b) // definition of pairSecond
Which means that
pairBimap g h = pairFirst g . pairSecond h
Proof of pairFirst
pairFirst g (Pair a b) =
Pair (g a) b // definition of pairFirst
Pair (g a) (id b) // definition of id
pairBimap (g id) Pair a b // definition of pairBimap
Which means that
pairFirst g = pairBimap g id
Proof of pairSecond
pairSecond f (Pair a b) =
Pair a (f b) // definition of pairSecond
Pair (id a) (f b) // definition of id
pairBimap (id f) Pair a b // definition of pairBimap
Which means that
pairSecond = pairBimap id
##### p8.2 Show the isomorphism between the standard definition of Maybe and this desugaring: type Maybe' a = Either (Const () a) (Identity a) Hint: Define two mappings between the two implementations. For additional credit, show that they are the inverse of each other using equational reasoning.
Solution
data MyIdentity a = MyIdentity a
data MyConst c a = MyConst c
type Maybe' a = Either (MyConst () a) (MyIdentity a)
desugaredToMaybe :: Maybe' a -> Maybe a
desugaredToMaybe (Left (MyConst ())) = Nothing
desugaredToMaybe (Right (MyIdentity a)) = Just a
maybeToDesugared :: Maybe a -> Maybe' a
maybeToDesugared Nothing = Left (MyConst ())
maybeToDesugared (Just a) = Right (MyIdentity a)
We show that they are the inverse of each other using equational reasoning
maybeToDesugared desugaredToMaybe (Left (MyConst ()))
= maybeToDesugared Nothing
= maybeToDesugared Left (MyConst ())
desugaredToMaybe maybeToDesugared Nothing
= desugaredToMaybe (Left (MyConst ()))
= Nothing
maybeToDesugared desugaredToMaybe (Right (MyIdentity a))
= maybeToDesugared Just a
= Right (MyIdentity a)
desugaredToMaybe maybeToDesugared Just a
= desugaredToMaybe Right (MyIdentity a)
= Just a
##### p8.3 Let’s try another data structure. I call it a PreList because it’s a precursor to a List. It replaces recursion with a type parameter b: data PreList a b = Nil | Cons a b. You could recover our earlier definition of a List by recursively applying PreList to itself (we’ll see how it’s done when we talk about fixed points). Show that PreList is an instance of Bifunctor.
Solution Lets form the following mapping
fmapFull:: (a -> c) -> (b -> d) -> (PreList a b) -> (PreList c d)
fmapFull f g Nil = Nil
fmapFull f g Cons a b = Cons (f a) (g b)
Say we keep b constant (WLOG). Then the fmap for a is
fmap f Nil = Nil
fmap f Cons a C = Cons (f a) C
This is a functor because Identity
fmap id Nil = Nil
fmap id Cons a C = Cons id a C = Cons a C
Composition
fmap f . g Nil = Nil
fmap f . g Cons a C =
Cons f . g a C =
fmap f Cons g a C =
fmap f (fmap g Cons a C)
##### p8.4 Show that the following data types define bifunctors in a and b:
data K2 c a b = K2 c
data Fst a b = Fst a
data Snd a b = Snd b
##### For additional credit, check your solutions agains Conor McBride’s paper Clowns to the Left of me, Jokers to the Right.
Solution
K2: Without loss of generality, if we hold b constant, then K2 becomes Const, which is a functor
Fst: If we hold a constant, then Fst becomes Const, which is a functor If we hold b constant, then Fst becomes Identity, which is a functor
Snd: If we hold a constant, then Snd becomes Identity, which is a functor If we hold b constant, then Snd becomes Const, which is a functor
##### p8.5 Define a bifunctor in a language other than Haskell. Implement bimap for a generic pair in that language.
Solution
class Bifunctor(object):
def apply_bimap(self, f, g):
assert False
@classmethod
def first(cls, f):
return lambda pair: pair.apply_bimap(f, lambda x: x)
@classmethod
def second(cls, g):
return lambda pair: pair.apply_bimap(lambda x: x, g)
@classmethod
def bimap(cls, f, g):
return lambda pair: pair.apply_bimap(f, g)
class Pair(Bifunctor):
def __init__(self, aval, bval):
self.aval = aval
self.bval = bval
def apply_bimap(self, f, g):
return Pair(f(self.aval), g(self.bval))
p = Pair(5, "4")
first_mapped = Bifunctor.first(lambda x: x + 1)(p)
assert first_mapped.aval == 6
assert first_mapped.bval == "4"
second_mapped = Bifunctor.second(lambda x: x + "1")(p)
assert second_mapped.aval == 5
assert second_mapped.bval == "41"
bimapped = Bifunctor.bimap(lambda x: x + 1, lambda s: s + "1")(p)
assert bimapped.aval
##### p8.6 Should std::map be considered a bifunctor or a profunctor in the two template arguments Key and T? How would you redesign this data type to make it so?
Solution std::map should be considered a profunctor in Key and T.
We can define it as a Profunctor as follows:
get :: a -> Maybe b
instance Profunctor get where
dimap f g get = lmap f . rmap g
lmap f get = \x -> get (f x)
rmap g get = \x -> fmap g (get x)
# Section 10: Natural Transformations
##### p10.1 Define a natural transformation from the Maybe functor to the list functor. Prove the naturality condition for it.
Solution
natTrans:: Maybe a -> [a]
natTrans (Just x) = [x]
natTrans Nothing = []
The naturality condition is G f ◦ αa = αb ◦ F f, which translates to fmap_list f . natTrans = natTrans . fmap_maybe f
Nothing Case:
fmap_list f . natTrans Nothing =
fmap_list f [] =
[] =
natTrans Nothing =
natTrans . fmap_maybe f Nothing
(Just x) Case:
fmap_list f . natTrans (Just x) =
fmap_list f [x] =
[f(x)] =
natTrans (Just (f x)) =
natTrans . fmap_maybe f (Just x)
##### p10.2 Define at least two different natural transformations between Reader () and the list functor. How many different lists of () are there?
Solution
natTransRL1:: (() -> a) -> [a]
natTransRL1 _ = []
natTransRL2:: (() -> a) -> [a]
natTransRL2 g = [g ()]
natTransRL3:: (() -> a) -> [a]
natTransRL3 g = fmap g [(), ()]
Since there are an infinite number of lists of [(), ...], there are an infinite number of these natural transformations.
##### p10.3 Continue the previous exercise with Reader Bool and Maybe.
Solution
There are three natural transformations from Reader Bool -> Maybe
natTransRB1:: (Bool -> a) -> Maybe a
natTransRB1 _ = Nothing
natTransRB2:: (Bool -> a) -> Maybe a
natTransRB2 g = Just (g True)
natTransRB3:: (Bool -> a) -> Maybe a
natTransRB3 g = Just (g False)
##### p10.4 Show that horizontal composition of natural transformation satisfies the naturality condition (hint: use components). It’s a good exercise in diagram chasing.
Solution
We have the functors F, G and the natural transformations:
αa:: F a -> F'a
βa:: G a -> G'a
We need to show that (G' . F') f . (β ◦ α)a = (β ◦ α)b . (G . F) f
(β ◦ α)b . (G . F) f =
(βF'b . Gαb) . G . F f = // definition of horizontal composition
βF'b . G . F' f . αa = // G αb :: G (F b) -> G (F'b)
(G' . F') f . (β ◦ α)a = // βF'b :: G (F'a) -> G'(F'a)
##### p10.5 Write a short essay about how you may enjoy writing down the evident diagrams needed to prove the interchange law.
Solution
If it’s the case that:
F -β'-> F'
F' -α'-> F''
G -β-> G'
G' -α-> G''
Then by the definition of horizontal composition it’s simple to see that:
FG -(β' ◦ β)-> F'G' -(α' ◦ α)-> F''G''
FG -(β' . α') ◦ (β . α)-> F''G''
Also, by horizontal composition:
FG -(β' ◦ β)-> F'G'
F'G' -(α' ◦ α)-> F''G''
FG -(β' ◦ β) . (α' ◦ α)-> F''G''
so (β' ◦ β) . (α' ◦ α) and (β' . α') ◦ (β . α) have the equivalent effect on FG
##### p10.6 Create a few test cases for the opposite naturality condition of transformations between different Op functors. Here’s one choice:
op :: Op Bool Int
op = Op (\x -> x > 0)
and
f :: String -> Int
f x = read x
Solution
newtype Op r a = Op (a -> r)
contramap f (Op g) = Op (g . f)
unwrap_op :: Op a b -> b -> a
unwrap_op (Op f) x = f x
-- test 1
op1 :: Op Bool Int
op1 = Op (\x -> (x > 0))
f1 :: Bool -> Int
f1 x = if x then 1 else 0
opBoolToOpChar :: Op Bool a -> Op Char a
opBoolToOpChar (Op aToBool) = Op (\x -> if aToBool x then 'a' else 'b')
boolchar_contra_f_op1 :: Op Char Bool
boolchar_contra_f_op1 = opBoolToOpChar ((contramap f1) op1)
contra_f_boolchar_op1 :: Op Char Bool
contra_f_boolchar_op1 = contramap f1 (opBoolToOpChar op1)
test1a = (unwrap_op boolchar_contra_f_op1 True) == (unwrap_op contra_f_boolchar_op1 True)
test1b = (unwrap_op boolchar_contra_f_op1 False) == (unwrap_op contra_f_boolchar_op1 False)
-- test 2
op2 :: Op String Double
op2 = Op (\x -> show x)
f2 :: Int -> Double
f2 x = sqrt (fromIntegral x)
opStringToOpInt :: Op String a -> Op Int a
opStringToOpInt (Op aToString) = Op (\x -> length (aToString x))
stringint_contra_f_op2 :: Op Int Int
stringint_contra_f_op2 = opStringToOpInt ((contramap f2) op2)
contra_f_stringint_op2 :: Op Int Int
contra_f_stringint_op2 = contramap f2 (opStringToOpInt op2)
test2a = (unwrap_op stringint_contra_f_op2 5) == (unwrap_op contra_f_stringint_op2 5)
test2b = (unwrap_op stringint_contra_f_op2 2) == (unwrap_op contra_f_stringint_op2 2)
# Section 12: Limits and Colimits
##### p12.1 How would you describe a pushout in the category of C++ classes?
Solution We are working in the C++ types category with subclasses as morphisms. For the span 1 <- 2 -> 3 we have a class 2 that inherits from 1 and 3, and since we are working with colimits, the apex is some 4 such that 1 -> 4 <- 3. The pushout is the colimit of this diagram, which is the universal 4 such that the colimit is also the subclass of every other candidate. This is the class 4 that has the maximum amount of shared functionality such that it can still be a superclass of 1 and 3.
##### p12.2 Show that the limit of the identity functor Id :: C -> C is the initial object.
Solution The identity functor forms diagrams consisting of every item in C. The apex of each diagram must have morphisms to every other item, and the limit object must have unique morphisms to every other limit candidate, which is every other item. Therefore the limit must be the initial object.
##### p12.3 Subsets of a given set form a category. A morphism in that category is defined to be an arrow connecting two sets if the first is the subset of the second. What is a pullback of two sets in such a category? What’s a pushout? What are the initial and terminal objects?
Solution The pushout is the intersection of the two sets (the largest set contained in them both) and the pullback is the union of those sets (the smallest set that contains them both). The initial object is the empty set and the terminal object is the “given set” that contains all of the elements and that all of the other elements are subsets of.
##### p12.4 Can you guess what a coequalizer is?
Solution The coequalizer is the equalizer in the opposite category. Given some 2 morphisms f: b -> a and g: b -> a, the coequalizer is the colimit object c and associated morphism p: a -> c such that p . f = p . g. That is, for any other c' with morphism p' there exists some u such that p' = p . u.
Over Set, the coequalizer defines a transformation of f and g’s codomains that makes them equal to each other.
##### p12.5 Show that, in a category with a terminal object, a pullback towards the terminal object is a product.
Solution Consider a diagram formed by the three object category I of the form 1 -f-> t <-g- 2 such that t is the terminal object. Since f and g are unique, the category of such diagrams is isomorphic to the category of diagrams formed by the two object discrete category consisting of only 1,2 without morphisms. The limit of this category is the product, so the pullback towards the terminal object is the product.
##### p12.6 Similarly, show that a pushout from an initial object (if one exists) is the coproduct.
Solution Consider a diagram formed by the three object category I of the form 1 <-f- i -g-> 2 such that i is the initial object. Since f and g are unique, the category of such diagrams is isomorphic to the category of diagrams formed by the two object discrete category consisting of only 1,2 without morphisms. The colimit of this category is the coproduct, so the pushout towards the initial object is the coproduct.
# Section 13: Free Monoids
##### p13.1 You might think (as I did, originally) that the requirement that a homomorphism of monoids preserve the unit is redundant. After all, we know that for all a, h a * h e = h (a * e) = h a So h e acts like a right unit (and, by analogy, as a left unit). The problem is that h a, for all a might only cover a sub-monoid of the target monoid. There may be a “true” unit outside of the image of h. Show that an isomorphism between monoids that preserves multiplication must automatically preserve unit.
Solution Say f: A -> A' is a monoid isomorphishm. Then there exists some g: A' -> A such that g f a = a. Given the unit u in A, for all a' in A', we see g (a' * f u) = g a' * g f u = g a' * u = g a'. Since g is injective, this means that a' = a' * f u, so f u is the right unit for all a' in A'. We can do the same to show f u is the left unit as well.
##### p13.2 Consider a monoid homomorphism from lists of integers with concatenation to integers with multiplication. What is the image of the empty list []? Assume that all singleton lists are mapped to the integers they contain, that is [3] is mapped to 3, etc. What’s the image of [1, 2, 3, 4]? How many different lists map to the integer 12? Is there any other homomorphism between the two monoids?
Solution The image of the empty list is 1, and the image of [1,2,3,4] is 1*2*3*4=24. The lists [12,1] [1,12] [6,2] [2,6] [3,4] [4,3] [2,2,3] [2,3,2] [3,2,2] all map to 12.
The function that maps all lists of integers to 1 is also a homomorphism, because the unit [1] is preserved and for any two lists l1,l2 we see that h (l1 ++ l2) = 1 = 1 * 1 = h l1 * h l2
##### p13.3 What is the free monoid generated by a one-element set? Can you see what it’s isomorphic to?
Solution This monoid is lists of unit () with concatenation. This is isomorphic to integers over addition.
forward :: List () -> Int
forward x = length x
inverse :: Int -> List ()
inverse x = replicate x [()]
# Section 14: Representable Functors
##### p14.1 Show that the hom-functors map identity morphisms in C to corresponding identity functions in Set.
Solution When we apply the functor C(a, -) to some function f, we get a function that performs the action C (a, f) h = f . h on any morphism h: a -> x in the homset Hom(a, x). If f: x -> x is the identity morphism, f . h = h, so C (a, f) h = h, and C (a, f) is the identity morphism as well.
##### p14.2 Show that Maybe is not representable.
Solution If Maybe were representable, then we would be able to implement a function of the for beta :: Maybe x -> (a -> x). However, it is not possible to implement a function that accepts None and return a -> x for any arbitrary x type.
##### p14.3 Is the Reader functor representable?
Solution Yes, the Reader functor is the hom-functor over haskell types and it is isomorphic to itself.
##### p14.4 Using Stream representation, memoize a function that squares its argument.
Solution
data Stream x = Cons x (Stream x)
instance Representable Stream where
type Rep Stream = Int
tabulate f = Cons (f 0) (tabulate (f . (+1)))
index (Cons b bs) n = if n == 0 then b else index bs (n - 1)
squareArg :: Int -> Int
squareArg x = x * x
memoizedSquares :: Stream Int
memoizedSquares = tabulate squareArg
zerothSquare :: Int
zerothSquare = index memoizedSquares 0
zerothSquareTrue = zerothSquare == 0
thirdSquare :: Int
thirdSquare = index memoizedSquares 3
thirdSquareTrue = thirdSquare == 9
fifthSquare :: Int
fifthSquare = index memoizedSquares 5
fifthSquareTrue = fifthSquare == 25
##### p14.5 Show that tabulate and index for Stream are indeed the inverse of each other. (Hint: use induction.)
Solution We want to prove that for all n, index tabulate f n = f n Base Case
index (tabulate f) 0 = // definition of tabulate
index (Cons (f 0) (tabulate (f . (+1)))) 0 = // definition of index
f 0
Inductive Step
index (tabulate f) n = // definition of tabulate
index (Cons (f 0) (tabulate (f . (+1)))) n = // definition of index
index (tabulate (f . (+1))) (n - 1) = // inductive assumption
f . (+1) . (n - 1) =
f n
# Section 15: The Yoneda Lemma
##### p15.1 Show that the two functions phi and psi that form the Yoneda isomorphism in Haskell are inverses of each other.
phi :: (forall x . (a -> x) -> F x) -> F a
phi alpha = alpha id
psi :: F a -> (forall x . (a -> x) -> F x)
psi fa h = fmap h fa
Solution Note psi can be written as psi fa = \h -> fmap h fa Forward
(phi . psi) fa =
phi (\h -> fmap h fa) =
(\h -> fmap h fa) id =
fmap id fa =
fa
Backward
(psi . phi) alpha =
psi (alpha id) =
\h -> fmap h (alpha id) =
\h -> (alpha h id) =
\h -> alpha h =
alpha
##### p15.2 A discrete category is one that has objects but no morphisms other than identity morphisms. How does the Yoneda lemma work for functors from such a category?
Solution Any homfunctor C(a, -) from the discrete category maps a to the singleton set and all other objects to the empty set. For any functor F from the discrete category to Set, there are N morphisms (the item-selection morphisms) between the singleton set and F a, where N is the number of elements of F a. Since there is one morphism from the empty set to each other set, each of those N morphisms from singleton to F a indicate a unique natural transformation from C(a, -) to F, so there is one-to-one correspondence between these natural transformations and elements of F a.
##### p15.3 A list of units [()] contains no other information but its length. So, as a data type, it can be considered an encoding of integers. An empty list encodes zero, a singleton [()] (a value, not a type) encodes one, and so on. Construct another representation of this data type using the Yoneda lemma for the list functor.
Solution By the Yoneda lemma, the natural transformations from C(a, -) (in this case () -> x) to F (in this case List x) are one-to-one with the elements of F a. So the data type D (() -> x) -> List x is another representation of List ().
It’s pretty easy to see why this is the case - a function of the form f: () -> x is essentially a container for a single value of x. So the elements of D are all of the form:
d1 f = [f ()]
d2 f = [f (), f ()]
...
# Section 16: Yoneda Embedding
##### p16.1 Express the co-Yoneda embedding in Haskell.
Solution
forward :: (a -> b) -> ((x -> a) -> (x -> b))
forward atob = \f -> atob . f
backward :: ((x -> a) -> (x -> b)) -> (a -> b)
backward xToAToXToB = \a -> (xToAToXToB id) a
##### p16.3 Work out the Yoneda embedding for a monoid. What functor corresponds to the monoid’s single object? What natural transformations correspond to monoid morphisms?
Solution In the single-element category view of monoid, we have a single element and the morphisms follow the monoid association rules. We will call this category M. The Yoneda embedding maps the single object a to the functor M(a, -), which is the functor in [M, Set] that maps the single element a to the set M(a,a). The Yoneda embedding maps each morphism in M to the identity natural transformation that maps the functor M(a, -) to itself.
##### p16.4 What is the application of the covariant Yoneda embedding to preorders? (Question suggested by Gershom Bazerman.)
In a preorder category C, if and only if a morphism f: b -> a exists, we have exactly one natural transformation between C(a, -) and C(b, -). Since there are no functions that map non-empty sets into the empty set, we see that if C(a, x) is nonempty, then C(b, x) must be nonempty as well.
Therefore, we have the condition: b <= a, if and only if for all x, a <= x implies b <= x
##### p16.5 Yoneda embedding can be used to embed an arbitrary functor category [C, D] in the functor category [[C, D], Set]. Figure out how it works on morphisms (which in this case are natural transformations).
Solution For any natural transformation NatAB between the functors Fb: C -> D and Fa: C -> D such that NatAB: Fb -> Fa, the Yoneda embedding maps it to the natural transformation:
NYoneda: [C, D](Fa, -) -> [C, D](Fb, -)
Where
[C, D](Fa, -): Fx -> NatAX
NatAX: Fa -> Fx
[C, D](Fb, -): Fx -> NatBX
NatBX: Fb -> Fx
NYoneda operates on [C, D](Fa, -) by post-composing NatAB to the natural transformations in the output NatAX, which maps the output set to NatBX, which maps ([C, D](Fa, -): Fx -> NatAX) -> ([C, D](Fb, -): Fx -> NatBX)
# Section 17: It’s All About Morphisms
##### p17.1 Consider some degenerate cases of a naturality condition and draw the appropriate diagrams. For instance, what happens if either functor F or G map both objects a and b (the ends of f :: a -> b) to the same object, e.g., F a = F b or G a = G b? (Notice that you get a cone or a co-cone this way.) Then consider cases where either F a = G a or F b = G b. Finally, what if you start with a morphism that loops on itself — f :: a -> a?
Solution
For the following subproblems, let’s assume we have some function f: a->b and natural transformation α between functors F and G.
###### p17.1.1
Say Fa = Fb. Then Ga = Gb because:
(α . Ff) Fa =
α . Fb =
α . Fa =
Ga
(Gf * α) Fa =
Gf Ga =
Gb
###### p17.1.2
Say Ga = Gb. Then αB Fb = αA Fa, but we can’t conclude that Fb = Fa, because it’s possible that G is the constant functor.
###### p17.1.3
Say Fa = Ga. Then Gf: Fa -> Gb and since αA Fa = Ga, αA is the identity.
###### p17.1.4
Say Fb = Gb. Then Ff: Fa -> Gb and since αB Fb = Gb, αB is the identity.
###### p17.1.5
Say f: a -> a. Then αA * Ff = Gf * αA.
# Section 18: Adjunctions
##### p18.1 Derive the naturality square for ψ, the transformation between the two (contravariant) functors:
a -> C(L a, b)
a -> D(a, R b)
Solution Say we have
f :: a1 -> a2
F f :: C(L a1, b) -> C(L a2, b)
G f :: D(a1, R b) -> D(a2, R b)
Where L and R are functors
L :: D(a, R b) -> C(L a, b)
R :: C(L a, b) -> D(a, R b)
and define the natural transformation ψ: G -> F such that
ψg1 :: D(a1, R b) -> C(L a1, b)
ψg2 :: D(a2, R b) -> C(L a2, b)
Now consider the morphisms g1: a1 -> R and g2: a2 -> R b. We want to prove that F f * ψg1 = ψg2 * G f
ψg2 * G f G a1 = // definition of f
ψg2 * G a2 = // definition of G
ψg2 * D(a2, R b) = // definition of ψg2
C(L a2, b) = // definition of G
F a2 = // definition of F
F f * F a1 = // definition of f
F f * ψg1 G a1 // definition of ψg1
##### p18.2 Derive the counit ε starting from the hom-sets isomorphism in the second definition of the adjunction.
Solution Assume that C(L d, c) ≅ D(d, R c) holds for any c in C and d in D. We want to prove that there exists some natural transformation ε :: L . R -> Ic.
Say d = R c, then C((L . R) c, c) ≅ D(R c, R c). Since D(Rc, Rc) contains at least the identity, our natural transformation from D(R c, R c) -> C((L . R) c, I c) must map to a non-empty set. Therefore, we have some set of morphisms that map from (L . R) c -> I c for any c. These morphisms form a natural transformation from L * R -> Ic, which is ε.
##### p18.3 Complete the proof of equivalence of the two definitions of the adjunction.
Solution In order to prove that the two definitions are equivalent, we need to prove the equivalence of the isomorphism C(L d, c) ≅ D(d, R c) and the existence of two natural transformations: the unit η and the counit ε.
In the text we proved that C(L d, c) ≅ D(d, R c) implies the existence of the η and that the existence of the η and ε implies the existence of a mapping from C(L d, c) to D(d, R c). In p18.2, we proved that C(L d, c) ≅ D(d, R c) implies the existence of the ε. Therefore, we still need to prove that the existence of the η and ε implies the existence of ψ :: D(d, R c) -> C(L d, c)
For some morphism f :: d -> R c, we can apply εc * L to form the morphism:
εc . L f =
L d -> εc L . R c =
L d -> c =
ψf
##### p18.4/5 Show that the coproduct can be defined by an adjunction. Start with the definition of the factorizer for a coproduct. Show that the coproduct is the left adjoint of the diagonal functor.
Solution (In this solution, we assume C is Set or Hask) We want to prove that C(Either a b, c) ≅ (C × C)(<a, b>, Δ c). A homset in CxC is (C×C)(<a, b>, Δ c), which consists of pairs of functions a -> c, b -> c and a homset in C is C(Either a b, c), which consists of functions (Either a b -> c)
We can define a natural transformation between these two homsets with the factorizer and inversefactorizer functions.
factorizer :: (C×C)(<a, b>, Δ c) -> C(Either a b, c)
inversefactorizer :: C(Either a b, c) -> (C×C)(<a, b>, Δ c)
We can write these in pseudo-haskell as
factorizer :: ((a -> c), (b -> c)) -> (Either a b -> c)
factorizer (i,j) (Left a) = i a
factorizer (i,j) (Right b) = j b
inversefactorizer :: (Either a b -> c) -> ((a -> c), (b -> c))
inversefactorizer m = (\a -> m Left a), (\b -> m Right b)
Now note that because these are both polymorphic in a,b,c, both factorizer and inversefactorizer are natural, so C(Either a b, c) ≅ (C × C)(<a, b>, Δ c).
##### p18.6 Define the adjunction between a product and a function object in Haskell.
Solution
producttofunction :: ((z, a) -> b) -> (z -> (a -> b))
producttofunction f = \z -> (\a -> f (z,a))
functiontoproduct :: (z -> (a -> b)) -> ((z, a) -> b)
functiontoproduct f = \z_a -> ((f (fst z_a)) (snd z_a))
# Section 19: Free/Forgetful Adjunctions
##### p19.1 Consider a free monoid built from a singleton set as its generator. Show that there is a one-to-one correspondence between morphisms from this free monoid to any monoid m, and functions from the singleton set to the underlying set of m.
Solution
Forward Consider a morphism from the free monoid with the singleton set as its generator to m. This morphism maps the generator element e to some m1 in m. There exists exactly one function in the homset between the singleton set and the underlying set of m that maps () to m1, so we can define a forward mapping.
Backward Consider a function from the singleton set to the underlying set of m. This function “chooses” a single element m1 from the underlying set of m. We can define exactly one homomorphism between the singleton free monoid and m that maps the generator element e to m1, since any such homomorphism must satisfy the following:
1 -> unit
e -> m1
ee -> m1m1
eee -> m1m1m1
so there is exactly one such homomorphism and we can define a backward mapping.
# Section 24: F-Algebras
##### p24.1 Implement the evaluation function for a ring of polynomials of one variable. You can represent a polynomial as a list of coefficients in front of powers of x. For instance, 4x^2-1 would be represented as (starting with the zero’th power) [-1, 0, 4].
Solution
polyEval :: [Double] -> Double -> Double
polyEval coefficients value = foldr (\ (power, coeff) sumSoFar -> sumSoFar + coeff * (value ** power)) 0.0 (zip [0..] coefficients)
isTrue = 99.0 == (polyEval [-1, 0, 4] 5)
##### p24.2 Generalize the previous construction to polynomials of many independent variables, like x^2y-3y^3z.
Solution
raiseAndProd :: [Double] -> [Double] -> Double
raiseAndProd values powers = foldr (\(value, power) prodSoFar -> prodSoFar * (value ** power)) 1.0 (zip values powers)
polyMultiEval :: [(Double, [Double])] -> [Double] -> Double
polyMultiEval coeffsExps values = foldr (\ (coeff, powers) sumSoFar -> sumSoFar + coeff * (raiseAndProd values powers)) 0.0 coeffsExps
isTrue1 = -2580.0 == (polyMultiEval [(1, [2, 1, 0]), (-3, [0, 3, 1])] [3, 5, 7]))
isTrue2 = 1.0 == (polyMultiEval [(1, [2, 1])] [1, 1])
##### p24.3 Implement the algebra for the ring of 2×2 matrices.
Solution
data MatExpr = RZero
| ROne
| RCompA
| RCompB
| RCompC
| RCompD
| RAdd MatExpr MatExpr
| RMul MatExpr MatExpr
| RNeg MatExpr
type MatrixTwoTwo = (Double, Double, Double, Double)
mCompA :: MatrixTwoTwo
mCompA = (1, 0, 0, 0)
mCompB :: MatrixTwoTwo
mCompB = (0, 1, 0, 0)
mCompC :: MatrixTwoTwo
mCompC = (0, 0, 1, 0)
mCompD :: MatrixTwoTwo
mCompD = (0, 0, 0, 1)
mZero :: MatrixTwoTwo
mZero = (0, 0, 0, 0)
mEye :: MatrixTwoTwo
mEye = (1, 0, 1, 0)
mAdd :: MatrixTwoTwo -> MatrixTwoTwo -> MatrixTwoTwo
mAdd (a1,b1,c1,d1) (a2,b2,c2,d2) = (a1 + a2, b1 + b2, c1 + c2, d1 + d2)
mMult :: MatrixTwoTwo -> MatrixTwoTwo -> MatrixTwoTwo
mMult (a1,b1,c1,d1) (a2,b2,c2,d2) = (a1 * a2 + b1 * c2, a1 * b2 + b1 * d2, c1 * a2 + d1 * c2, c1 * b2 + d1 * d2)
mNeg :: MatrixTwoTwo -> MatrixTwoTwo
mNeg (a1,b1,c1,d1) = (-a1,-b1,-c1,-d1)
evalZ :: MatExpr -> MatrixTwoTwo
evalZ RZero = mZero
evalZ ROne = mEye
evalZ RCompA = mCompA
evalZ RCompB = mCompB
evalZ RCompC = mCompC
evalZ RCompD = mCompD
evalZ (RAdd e1 e2) = mAdd (evalZ e1) (evalZ e2)
evalZ (RMul e1 e2) = mMult (evalZ e1) (evalZ e2)
evalZ (RNeg e) = mNeg (evalZ e)
matrixExpression :: MatExpr
matrixExpression = RMul (RAdd RCompA RCompB) (RAdd RCompC RCompD)
isTrue = (1.0, 1.0, 0.0, 0.0) == (evalZ matrixExpression))
##### p24.4 Define a coalgebra whose anamorphism produces a list of squares of natural numbers.
Solution
newtype Fix f = Fix (f (Fix f))
unFix :: Fix f -> f (Fix f)
unFix (Fix x) = x
cata :: Functor f => (f a -> a) -> Fix f -> a
cata alg = alg . fmap (cata alg) . unFix
ana :: Functor f => (a -> f a) -> a -> Fix f
ana coalg = Fix . fmap (ana coalg) . coalg
data StreamF e a = StreamF e a deriving Functor
toListC :: Fix (StreamF e) -> [e]
toListC = cata al
where al :: StreamF e [e] -> [e]
al (StreamF e a) = e : a
nat :: [Int] -> StreamF Int [Int]
nat (p : ns) = StreamF (p^2) ns
squaresStream :: Fix (StreamF Int)
squaresStream = ana nat [0..]
squaresList :: [Int]
squaresList = toListC squaresStream
##### p24.5 Use unfoldr to generate a list of the first n primes.
Solution
listSieve :: [Int] -> Maybe (Int, [Int])
listSieve (p : ns) = Just (p, (filter (notdiv p) ns))
where notdiv p n = n mod p /= 0
primeFilteredList :: [Int]
primeFilteredList = unfoldr listSieve [2..]
isTrue1 = primeFilteredList!!0 == 2
isTrue2 = primeFilteredList!!3 == 7
# Section 25: Algebras for Monads
##### p25.1 What is the action of the free functor F :: C -> C^T on morphisms. Hint: use the naturality condition for monadic μ.
Solution First, note that F a = (T a, μa). Since μ is natural, we see that T f . μa = μb . (T . T) f. Therefore, for some f :: a -> b, the action of F on f is:
fmap f (T a, μa) = (fmap f T a, T f . μa)
##### p25.2 Define the adjunction: U^W ⊣ F^W
Solution First, we define the unit η :: I -> F^W . U^W. Since it’s the case that
F^W . U^W (W a, f) =
F^W (W a) =
(W W a, δWa)
η needs to map (W a, f) -> (W W a, δWa). We can accomplish this by using f, the co-evaluator of the co-algebra, to define the component of η at (W a, f)
Next, we define the co-unit ε :: U^W . F^W -> I. Since it’s the case that:
U^W . F^W a =
U^W . (W a, δa) =
W a
ε needs to map W a -> a so we can use the extract method of the co-monad to define the component of ε at Wa.
##### p25.3 Prove that the above adjunction reproduces the original comonad.
Solution First, we can use the co-unit of the adjunction ε as the co-monadic extract, since εWa W a = a
Next, we can use the unit of the adjunction to define the co-monadic duplicate as the horizontal composition of three natural transformations U^W ◦ η ◦ F^W where U^W: U^W -> U^W and F^W: F^W -> F^W. Since F^W lifts a to (W a, δa), η picks the co-evaluator δa which maps W a -> W W a and U^W has no action on morphisms, we see that duplicate = U^W ◦ η ◦ F^W.
# Section 29: Topoi
##### p29.1 Show that the function f that is the pullback of true along the characteristic function must be injective.
Solution If f: a -> b is the pullback of true along the characteristic function, then for any a*, f*: a* -> b, there exists some unique h: a* -> a such that f* = f . h
Consider the case where a* is the image of f and f* is the identity. If f is not injective, then for the e1, e2, e1 != e2 in a such that f(e1) = f(e2), h can map f(e1) = f(e2) to either e1 or e2. Therefore h would not be unique, which implies that f must be injective.
# Section 30: Lawvere Theories
##### p30.1 Enumarate all morphisms between 2 and 3 in F (the skeleton of FinSet).
Solution (0->0, 1->0), (0->0, 1->1), (0->0, 1->2), (0->1, 1->0), (0->1, 1->1), (0->1, 1->2)
##### p30.2 Show that the category of models for the Lawvere theory of monoids is equivalent to the category of monad algebras for the list monad.
Solution First, let’s note that the category of models of the Lawvere theory for monoids is equivalent to the category of all monoids, Mon. Now we will prove that Mon is equivalent to the category of monad algebras for the list monad.
First, given a monoid over the set a, we can produce an algebra (a, f) where f maps the list L to the monoidal product of the elements in L. Next, given an algebra (a, f), we can produce a monoid over a by defining the monoidal product of a1, a2 to be f ([a1] cat [a2]). The unit of this monoid is [], and because of the monad condition f . μa = f . T f we see that:
f [a1, f [a2, a3]] =
f [a1, a2, a3] =
f [f [a1, a2], a3]
So the monoid associativity law is automatically satisfied.
##### p30.3 The Lawvere theory of monoids generates the list monad. Show that its binary operations can be generated using the corresponding Kleisli arrows.
Solution The binary operations in the Lawvere theory of monoids are elements of the homset LMon(2, 1), which are functions of two arguments that we can implement with only the monoidal operator. Each of these functions can be defined by a list composed of only those 2 unique elements. Since each Kleisli arrow in the hom-set KlT (1, 2) corresponds to a list composed of elements from the 2 element set, we can represent each binary operation in the Lawvere theory of monoids with a Kleisli arrow in KlT (1, 2).
##### p30.4 FinSet is a subcategory of Set and there is a functor that embeds it in Set. Any functor on Set can be restricted to FinSet. Show that a finitary functor is the left Kan extension of its own restriction
Solution Say K: Set -> FinSet is a functor that embeds a set into FinSet, such that for any finite set n in Set, K n = n. Then FinSet(K n, a) is a hom-set between elements in FinSet, and so FinSet(K n, a) = a^(K n) = a^n.
Now consider some finitary functor F. The left Kan extension of F’s restriction to FinSet along K is
LanK F a =
∫^n FinSet(K n, a) × F n =
∫^n a^n × F n = \\ definition of finitary functor
F
So a finitary functor is the left Kan extension of its own restriction.
# Section 31: Monads, Monoids, and Categories
##### p31.1 Derive unit and associativity laws for the tensor product defined as composition of endo-1-cells in a bicategory.
Solution
Unit Law The left and right compositions of any endo-1-cell T and the identity 1-cell id are T . id and id . T. By the definition of a bicategory there exist invertible 2-cells mapping each of these endo-1-cells to T.
Associativity Law Given three endo-1-cells T1, T2, T3, by the definition of a bicategory there exists an invertible 2-cell that maps between ((T1 . T2) . T3 and T1 . (T2 . T3).
##### p31.2 Check that monad laws for a monad in Span correspond to identity and associativity laws in the resulting category.
Solution A monad in Span consists of an endo-1-cell that has the sets Ar, Ob with the functions
dom :: Ar -> Ob
cod :: Ar -> Ob
and the associated 2-cells:
μ: Ar x Ar -> Ar
η: Ob -> Ar
This monad defines a category consisting of the objects in Ob and the arrows in Ar, where each arrow in Ar connects dom Ar to cod Ar.
Identity η assigns an identity arrow to each object such that
dom . η = id
cod . η = id
Therefore for any object o1 in Ob and arrow a1 in Ar where cod a1 = o1, we see that:
dom (μ (a1, η o1)) = dom a1
cod (μ (a1, η o1)) = cod (η o1) = o1 = cod a1
So the composition of an arrow with the identity arrow does not change that arrow’s domain or codomain.
Associativity By the monoid law for μ
μ (Ar x μ (Ar x Ar)) = μ (μ (Ar x Ar) x Ar)
Therefore, for any arrows a1, a2, a3, we see that
a1 . (a2 . a3) = (a1 . a2) . a3
##### p31.3 Show that a monad in Prof is an identity-on-objects functor.
Solution In Prof, we define a monad with an endo-profunctor T such that T: Cop x C -> Set. The composition of profunctors is
(q . p) a b = ∫^c p c a × q b c
So the composition of T with itself is:
(T . T) C C =
∫^c T c C × T C c = // existential quantifier
T C C
Which implies that T must map each object in C to itself.
##### p31.4 What’s a monad algebra for a monad in Span?
Solution Given a monad m over some object a, we form an algebra over this monad with a map alg :: m a -> a that satisfies commutativity conditions. For a monad in Span, we can use dom or cod for alg.
Identity alg . ηa = ida This holds by the definition of η for Span
Associativity alg . μa = alg . m alg Without loss of generality, we can see the following:
dom . μa (a1, a2) =
dom a1 =
dom . m dom (a1, a2)
` | 16,459 | 56,611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.904721 |
https://www.physicsforums.com/threads/proof-of-exponential-interarrival-times-times.219164/ | 1,713,836,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00856.warc.gz | 842,094,483 | 15,537 | # Proof of Exponential Interarrival times times
• mmehdi
In summary, the conversation discusses the proof that Poisson processes have exponential interarrival time, and that exponential interarrival times will always be a Poisson process. The fundamental assumption of the Poisson distribution is that the probability of a single arrival is constant and the probability of multiple arrivals is negligible. The proof of this hypothesis can be found in probability texts or can be treated as a differential equations problem with an exponential function as the solution. The conversation also explores the relationship between exponential and Poisson distributions, with the suggestion that the integration of the gamma distribution yields a Poisson distribution. However, the integration process is complex.
mmehdi
The proof that poisson process has exponential interarrival time is common place. The proof which i am trying to do is that, exponential interarrival times will always be poisson process, its like the reverse of the earlier proof. Could you help me with that.
The fundamental assumption of the Poisson distribution is that, for some very small time interval, the probability of a single arrival is a constant and the probability of more than one arrival in that time interval is so small it can be ignored. The proof that that hypothesis leads to the Poisson distribution is given in any good probability text that discusses the Poison distribution. You can also treat that as a differential equations problem (the rate of change of total arrivals is constant) that has an exponential function as solution.
I am not sure if I truly follow you. The proof that you can only have one arrival in one interval and the probability of getting two arrival is zero, is done through taylor series expansion. But how does it show that the exponenttial interarrival shall always satisfy the poisson properties.
What I tried is the total time, or waiting time let's say for the second arrival is Gamma Distribution. So if we integrate the gamma distribution it shall give us a poisson distribution. However the integration is winding, but it does yield the poisson distribution at the end.
## 1. What is Proof of Exponential Interarrival times?
Proof of Exponential Interarrival times is a mathematical concept that is used to model the arrival of events or occurrences that happen randomly and independently of each other. It is often applied in the study of queueing systems, where the time between arrivals of customers or events follows an exponential distribution.
## 2. How is Proof of Exponential Interarrival times useful in science?
Proof of Exponential Interarrival times is useful in science because it can be used to model and analyze real-world phenomena that involve the arrival of random events. This can include the study of natural phenomena such as earthquakes, as well as human-made systems like transportation and telecommunication networks.
## 3. What are the key assumptions in Proof of Exponential Interarrival times?
The key assumptions in Proof of Exponential Interarrival times are that the events or occurrences happen randomly and independently of each other, and the time between arrivals follows an exponential distribution. Additionally, it is assumed that there is no correlation between the time of arrival of one event and the next, and the probability of an event occurring is the same for all time intervals.
## 4. How is Proof of Exponential Interarrival times related to the Poisson process?
Proof of Exponential Interarrival times is closely related to the Poisson process, which is a mathematical model for the arrival of events in a given time interval. The Poisson process assumes that the number of events that occur in a time interval follows a Poisson distribution, and the time between arrivals follows an exponential distribution.
## 5. What are some real-world applications of Proof of Exponential Interarrival times?
Proof of Exponential Interarrival times has many real-world applications, including the study of natural phenomena such as the occurrence of earthquakes and the arrival of meteorites. It is also used in the study of human-made systems like transportation networks, communication systems, and computer networks. Additionally, it is used in market analysis, financial modeling, and risk management.
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978 | 1,000 | 4,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-18 | latest | en | 0.936876 |
http://docplayer.net/375020-Proximity-graphs-inside-large-weighted-graphs.html | 1,547,876,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662124.0/warc/CC-MAIN-20190119034320-20190119060320-00287.warc.gz | 60,081,288 | 39,641 | # Proximity Graphs inside Large Weighted Graphs
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1 Proximity Graphs inside Large Weighted Graphs Bernardo M. Ábrego Ruy Fabila-Monroy Silvia Fernández-Merchant David Flores-Peñaloza Ferran Hurtado Henk Meijer Vera Sacristán Maria Saumell Abstract Given a large weighted graph G = (V, E) and a subset U of V, we define several graphs with vertex set U in which two vertices are adjacent if they satisfy some prescribed proximity rule. These rules use the shortest path distance in G and generalize the proximity rules that generate some of the most common proximity graphs in Euclidean spaces. We prove basic properties of the defined graphs and provide algorithms for their computation. Keywords Neighborhood; proximity graphs; Voronoi diagrams; weighted graphs. 1 Introduction A basic need in spatial data analysis, from statistics to pattern recognition and wherever shape extraction is involved, is to decide closeness and neighborhoods among elements of a given input. When the data are described as points in Euclidean spaces, proximity graphs in which two of these nodes are connected when they satisfy some proximity criterion have been a basic tool in the analysis of their relative position [12, 17, 23]. Finding clusters, spanning structures, or tools allowing good interpolation are among the goals proximity graphs help to achieve. Moreover, when a combinatorial graph is drawn, a realization as a geometric proximity graph is quite satisfactory as it provides a natural aesthetical quality arising from associating adjacency to closeness. This is why proximity graphs have been extensively investigated in the field of graph drawing [5, 20]. In this context, the Delaunay graph dual to the Voronoi diagram and its relatives are by far the proximity graphs that have attracted more attention [23]. Nowadays, many complex relation systems are represented as very large networks. In some cases, as in transportation networks or circuit layouts, they correspond to physical situations Department of Mathematics, California State University, Northridge, CA, USA Departamento de Matemáticas, CINVESTAV, Mexico DF, Mexico Instituto de Matemáticas, Universidad Nacional Autónoma de México Deptartament de Matemàtica Aplicada II, Universitat Politècnica de Catalunya, Barcelona, Spain Partially supported by projects MTM and Gen. Cat. DGR 2009SGR1040. Science Department, Roosevelt Academy, Middelburg, The Netherlands 1
2 in which a geometric framework is inherent or implicit. In some others, the relationship is essentially combinatorial; important representatives of this situation include the world-wide web, social networks, commercial networks, and citation networks. Whichever the scenario, identifying communities or contrasting local characteristics with global properties within a network are among the basic tasks in network analysis. These tasks are particularly demanding as modern technologies have made it possible to build up massive data sets. Handling these data sets has become crucial and requires combined efforts from different fields, including in particular discrete mathematics and computer science [3]. Consider in this situation a very large graph a huge network in which the focus of some study is on a subset U of the nodes. Suppose that we have to decide which elements of U are close to other elements of U within the graph, or we must find a suitable subgraph that spans the elements in U, or we have to describe the relative positions in the network of the elements of U. This is nothing else than describing the proximity relations of U as a substructure of the network. Measuring this proximity is a basic data mining tool and hence defining suitable notions of closeness among vertices of a graph has found different approaches in the literature. For some specific networks, proximity measures have been proposed without properly relying on graph-theoretic concepts, as in [9, 19], where the network is modeled as an electrical circuit and edges with high weights contribute to the proximity of their endpoints because they can conduct more electricity. A logical option to be carefully explored in this setting is to adapt or mimic geometric proximity graphs; which have been so successful in metric scenarios, particularly the Delaunay family of graphs. The case when the graph is purely combinatorial and the distance between two vertices is the minimum number of edges of any path connecting them, has been considered, for example in [1, 14], yet the approach falls somehow short to provide rich analysis tools. Another situation arises when the network in embedded in a metric space. Here one may consider the Voronoi diagram inside the network, focusing just on the nodes and the connections or extending the implications of the resulting partition of the graph to the surrounding space. These situations nicely fit phenomena occurring alongside a network, or in its area of influence, and have been intensively investigated as network Voronoi diagrams (see [23] and [24] for a thorough description and multiple references) or from the viewpoint of time metrics [2, 25, 4]. Another related direction of research considers proximity inside an embedded geometric graph while using regions of interaction that are defined in the full euclidean plane [13, 18, 28]. If the surrounding space is ignored and the focus is on a geometric or weighted graph, the Voronoi partition is still a powerful analysis tool that has been repeatedly studied, as in [21, 8, 15]), where efficient construction algorithms are developed for several situations. In this case a Delaunay graph can be defined as dual to the Voronoi diagram obtained from the generators. Notice that if the graph is embedded in the plane, the internal proximity features may be completely unrelated to the geometric proximity in the plane, i.e., points that are drawn very close in the plane may be very far along the network. While the Delaunay graph is the most fundamental tool for domain decomposition in eu- 2
3 clidean spaces, other related proximity graphs are sometimes preferred, such as the Gabriel graph in some geographic applications, the minimum spanning tree in pattern recognition, or the nearest-neighbor graphs in classification procedures [17, 30]. Therefore, it is somehow surprising that their counterparts in network analysis have not received a comparable attention. This is precisely the scope of this paper in which we consider a very large graph G = (V, E) whose edges have a positive associated weight, and we study the aforementioned proximity relations for a subset of nodes U V, based on shortest paths along the edges of G. For example, G may be the road network of a city and U the locations of schools or other important facilities. To provide notions of closeness, we use generalizations of the nearest neighbor graph, the minimum spanning tree, the relative neighborhood graph, the Gabriel graph, and the Delaunay graph. We systematically study the relations among these graphs, their computation in terms of the nodes and the network itself, and the fundamental variations that arise when only the vertices of the input graph are taken as generators for the influence regions, or when arbitrary points on the edges may also play this role. It is worth mentioning that the set U together with the shortest-path distance on G constitutes a finite metric space. Therefore, our problem may be seen as a particular case of proximity graphs defined on general metric spaces. However, to the best of our knowledge, the precise topic of our work has not been properly investigated and only some definitions and easilyderived relations among the proximity graphs have been established (see Section 4.5 in [29], and also [16]). 2 Definitions and Notation We deal with a pair of a connected and edge-weighted graph G = (V, E) and a subset U V. For simplicity, we write G = (V, U, E). The edge set E is a set of interior disjoint continuous curves. Each edge e = (v 1, v 2 ) with (positive) weight w(e) is isometric to the interval [0, w(e)], and its endpoints are v 1 and v 2. Together with the shortest path distance, the union of the edge set constitutes a metric space, and we refer to this metric space also as G. In this way, when we speak of a point of G we may refer to either an endpoint or an interior point of an edge. The distance between a pair vertices of G as a metric space corresponds to the distance in G as a weighted graph, with the gained advantage that we may extend this definition to points in the interior of edges. The distance d G (p, q) between two points p and q of G is thus the minimum total weight of any path connecting p and q in G. The closed disk D G (p, r) is defined as the set of points q of G for which d G (p, q) r. If i j, we say that u i U is a nearest neighbor of u j U if d G (u j, u i ) d G (u j, u k ) for all vertices u k U different from u j. A midpoint of two points p and q of G is a point m on one of the shortest paths from p to q such that d G (m, p) = d G (m, q). We denote the set of midpoints of p and q by M G (p, q). For the remainder of this paper, we define V = m, U = n, and E = e. When using empty regions as proximity criteria in G, such as disks, two main variations arise, since we might allow these disks to be centered at any point of G, or restrict their centers 3
4 to lie only on vertices of the graph, as in [14, 1]. Moreover, the definition of certain regions of interference such as the Gabriel disk might depend on the multiplicity of paths or distances in G. Degeneracies that occur in the standard geometric case, such as non-uniqueness of the nearest neighbor, also generate several possibilities. For the sake of clarity we first present the situation where there are essentially no degeneracies (Sections 3 5). In Section 6 we drop the non-degeneracy assumptions and extend our results to the general setting. More precisely, in Sections 3, 4, and 5 we consider the case where the following nondegeneracy assumptions are satisfied: (A1) for all u i, u j U, the shortest path connecting u i and u j is unique; (A2) there do not exist three distinct vertices u i, u j U, v V U such that d G (v, u i ) = d G (v, u j ); (A3) there do not exist vertices v i, v j V, u i, u j U such that d G (v i, u i ) = d G (v j, u j ) and v i u i ; (A4) all paths in G connecting distinct nodes in V have different lengths. Obviously, the previous assumptions are not independent (A4 implies A1, A2, and A3; A3 implies A2), but considering them separately allows to clarify and provide a more precise description of the scenario. In Section 6, we extend the results from Sections 3 5 to the general case where A1 A4 are not necessarily satisfied. We now adapt several known definitions to proximity structures in graphs G = (V, U, E). Definition 2.1. The nearest neighbor graph of G = (V, U, E), denoted by NNG(G), is the graph H = (U, F ) such that (u i, u j ) F if and only if u j is one of the nearest neighbors of u i in G. See Figure 1 for an example. Definition 2.2. A minimum spanning tree of G = (V, U, E) is a tree T = (U, F ) such that the sum of d G (u i, u j ) over all edges (u i, u j ) F is minimal. The union of the minimum spanning trees of G, denoted by UMST(G), is the graph consisting of all the edges included in any of the minimum spanning trees of G. If A3 holds, all distances between vertices in U are different. This in particular implies that each vertex in U has exactly one nearest neighbor and that the minimum spanning tree of G, denoted by MST(G), is unique. Definition 2.3. The relative neighborhood graph of G = (V, U, E), denoted by RNG(G), is the graph H = (U, F ) such that (u i, u j ) F if and only if there exists no vertex u k U such that d G (u k, u i ) < d G (u i, u j ) and d G (u k, u j ) < d G (u i, u j ). Definition 2.4. The free-one Gabriel graph of G = (V, U, E), denoted by GG f1 (G), is the graph H = (U, F ) such that (u i, u j ) F if and only if there exists p M G (u i, u j ) such that no vertex u k U (u k u i, u j ) satisfies d G (p, u k ) d G (p, u i ). 4
5 Figure 1: A graph G and its nearest neighbor graph. The black vertices belong to U, while the white vertices belong to V U. The edges of G are solid and the edges of NNG(G) are dashed. The weight of the edges of G corresponds to their length. Definition 2.5. The free-all Gabriel graph of G = (V, U, E), denoted by GG fa (G), is the graph H = (U, F ) such that (u i, u j ) F if and only if, for each p M G (u i, u j ), no vertex u k U (u k u i, u j ) satisfies d G (p, u k ) d G (p, u i ). If A1 holds, the two definitions coincide and we denote the graph by GG f (G). Definition 2.6. The constrained-one Gabriel graph of G = (V, U, E), denoted by GG c1 (G), is the graph H = (U, F ) such that (u i, u j ) F if and only if there exists a closed disk D G (v, r), with v V and r = min v V {r D G (v, r) contains both u i and u j }, enclosing u i and u j and no other vertex from U. Definition 2.7. The constrained-all Gabriel graph of G = (V, U, E), denoted by GG ca (G), is the graph H = (U, F ) such that (u i, u j ) F if and only if every closed disk D G (v, r) enclosing u i and u j, and where v V and r = min v V {r D G (v, r) contains both u i and u j }, does not contain any other vertex of U. If A3 holds, the two definitions coincide and we denote the graph by GG c (G). Definition 2.8. The Voronoi region of a vertex u i U is the set of points p of G such that d G (p, u i ) d G (p, u j ) for all vertices u j U different from u i. The Voronoi diagram of G = (V, U, E), denoted by VD(G), is the partition of G into the Voronoi regions of the vertices of U. Definition 2.9. The free Delaunay graph of G = (V, U, E), denoted by DG f (G), is the graph H = (U, F ) such that (u i, u j ) F if and only if there exists a closed disk D G (p, r), where p is a point of G, containing u i and u j and no other vertex from U. 5
6 Definition The constrained Delaunay graph of G = (V, U, E), denoted by DG c (G), is the graph H = (U, F ) such that (u i, u j ) F if and only if there exists a closed disk D G (v, r), with v V, containing u i and u j and no other vertex from U. 3 Inclusion Sequence As in the case of the corresponding proximity graphs in Euclidean spaces, the graphs just defined satisfy some inclusion relations. In this section we show which proximity graphs are subgraphs of which other proximity graphs assuming A1, A2, and A3. Lemma 3.1. For each graph G = (V, U, E) we have NNG(G) MST(G) RNG(G) GG f (G) DG f (G). Proof. The proofs are analogous to those for proximity graphs in Euclidean spaces and, in some cases, the more general theory of proximity graphs defined in metric spaces applies (see Section 4.5 in [29]). It should be noticed that the inclusion RNG(G) GG f (G) relies on the assumption of A2 (see Theorem 6.1). Lemma 3.2. For each graph G = (V, U, E) we have NNG(G) DG c (G). Proof. Let (u i, u j ) be an edge of NNG(G), where u j is the nearest neighbor of u i. The closed disk D G (u i, r), where r = d G (u i, u j ), contains no vertices from U different from u i, u j. Thus (u i, u j ) DG c (G). Lemma 3.3. For each graph G = (V, U, E) we have GG c (G) DG c (G) DG f (G). Proof. It follows from the definitions of GG c (G), DG c (G), and DG f (G). Lemma 3.4. There exist graphs G 1 = (V 1, U 1, E 1 ), G 2 = (V 2, U 2, E 2 ), and G 3 = (V 3, U 3, E 3 ) for which NNG(G 1 ) GG c (G 1 ), GG c (G 2 ) GG f (G 2 ), and MST(G 3 ) DG c (G 3 ). Proof. Let G 1 be the graph on Figure 2a. The graph NNG(G 1 ) contains the edges (a, b) and (a, d) while GG c (G 1 ) only contains the edge (a, d). This proves that, in some cases, NNG(G) GG c (G). To see that GG c (G) GG f (G) does not hold in general, let G 2 be the graph on Figure 2b. We have that GG c (G 2 ) contains the edge (a, b) whereas (a, b) GG f (G 2 ). Finally, if G 3 is the graph on Figure 2c, MST(G 3 ) contains the edge (b, e) but DG c (G 3 ) does not. Thus there exist graphs G = (V, U, E) such that MST(G) DG c (G). These lemmas are sufficient to prove the following theorem: Theorem 3.5. The inclusion relations among all classes of proximity graphs are shown in Table 1. The symbol means that the inclusion is satisfied for all graphs G, and means that there are graphs G for which the inclusion is not satisfied. 6
7 a d 1 c b a d e c b a b c d 1.08 e 1.04 f (a) (b) (c) Figure 2: Example graphs; the black vertices belong to U, while the white vertices belong to V U. Table 1: Inclusion relations among proximity graphs in the non-degenerate case. MST RNG GG c GG f DG c DG f NNG MST RNG GG c GG f DG c It is not difficult to produce examples proving that all inclusions in the table are proper, in the sense that there exist graphs G for which the corresponding proximity subgraph does not coincide with its supergraph. 4 Geometric and Combinatorial Properties In this section we study some geometric and combinatorial properties of the previously defined proximity graphs. We start by proving that the free Delaunay graph is the dual graph of the Voronoi diagram, and by showing the consequences of this result on the combinatorial complexity of the graphs under study. Then we deal with two important particular cases, namely, when G is planar and when G is a tree. Finally, we characterize the graphs that are isomorphic to a proximity graph of some other graph. Remark 4.1. Consider G = (V, U, E) and assume that A2 is satisfied. If the point p is in the intersection of two Voronoi regions, then p is a point on an edge of G. Moreover, if q is a point in the intersection of a different pair of Voronoi regions, then p and q are on different edges of G. In particular, the intersection of three Voronoi regions is empty. We define the dual graph of the Voronoi diagram of G = (V, U, E) as the graph with vertex set U and edges connecting two vertices if and only if their Voronoi regions share some point of G that does not belong to the Voronoi region of any other vertex in U. Proposition 4.2. Let G = (V, U, E) be a graph. Then DG f (G) is the dual graph of VD(G). 7
8 Proof. If the Voronoi regions of two vertices u i u j U intersect in a point y that is not in the Voronoi region of any other vertex in U, then the closed disk D G (y, r), where r is the distance from y to u i, contains u i, u j, and no other vertex from U. Thus (u i, u j ) is an edge of DG f (G). Reciprocally, let us assume that there exists a disk D G (x, r), where x is a point of G, containing only two vertices of U, u i and u j. Then it can easily be shown that there exists a disk D G (x, r ), where x is a point of G, contained in D G (x, r) and having u i and u j on its boundary. Hence x lies in the Voronoi regions of u i and u j (and does not lie in the Voronoi region of any vertex from U). The previous proposition is a key tool to prove other results of interest, such as the following: Corollary 4.3. Let G = (V, U, E) be a graph. The number of edges of NNG(G), MST(G), RNG(G), GG c (G), GG f (G), DG c (G), and DG f (G) is at most e. Proof. Let (u i, u j ) be an edge of DG f (G). By Proposition 4.2, the Voronoi regions of u i and u j intersect at a point p of G such that d G (p, u i ) = d G (p, u j ) < d G (p, u k ) for all vertices u k U different from u i and u j. By Remark 4.1, p is a point on an edge of G and this edge does not include any other point on the intersection of the Voronoi regions of two vertices in U. The number of edges of DG f (G) is therefore less than or equal to e. Since, for all other classes, the proximity graph on G is a subgraph of DG f (G), their size is also bounded by e. The next proposition shows that this bound is tight for the graphs RNG(G), GG f (G), and DG f (G), and is tight up to a constant factor for the graphs GG c (G) and DG c (G). Proposition 4.4. There exists a graph G = (V, U, E) such that RNG(G) = GG f (G) = DG f (G) = G. There also exists a graph G = (V, U, E ) such that the number of edges of GG c (G ) and DG c (G ) is e /2. Furthermore, all of these graphs have Θ(n 2 ) edges. Proof. Consider two groups of n/2 vertices. Let U be this set of n vertices, V = U and G = (V, U, E) be the complete bipartite graph on the two groups of vertices. The weights of the edges of G are real numbers between 1 and 1.5, and can be assigned so that G is non-degenerate. In this situation RNG(G) = GG f (G) = DG f (G) = G, and RNG(G), GG f (G), and DG f (G) have n 2 /4 edges. Now add vertices to G in such a way that there exists a vertex in V U close to the midpoint of each edge (and the graph remains non-degenerate). Let G = (V, U, E ) be the resulting graph. In this case the number of edges of GG c (G ) and DG c (G ) is e /2 = n 2 /4. We now consider the case where G = (V, U, E) is a planar graph. Theorem 4.5. Let G = (V, U, E) be a planar graph. Then DG f (G) is planar. 8
9 Proof. Consider a plane drawing of G. Let (u i, u j ) be an edge of DG f (G). By Proposition 4.2, there exists a point p of G (not in V by Remark 4.1) such that d G (p, u i ) = d G (p, u j ) < d G (p, u k ) for all vertices u k U different from u i and u j. Draw the edge (u i, u j ) by following the union of a shortest path in G from u i to p and a shortest path from p to u j. Note that each point q p in a shortest path from u i to p is such that d G (q, u i ) < d G (q, u k ) for all vertices u k U different from u i, and the same holds for u j. Draw all edges in DG f (G) in this way and suppose the graph has one crossing. Since G is planar, this crossing occurs in a vertex w of V U. Then, w belongs to two shortest paths, one from a point p l G V to a point u l U and the other from p h G V to u h U. Hence, d G (w, u l ) < d G (w, u ν ) for all vertices u ν U different from u l, and d G (w, u h ) < d G (w, u ν ) for all vertices u ν U different from u h. This yields a contradiction. Observe that an edge of G might be used in the drawing of several edges of DG f (G). It is easy to prove that, in this case, it can be duplicated in such a way that no crossings are created. Corollary 4.6. If G = (V, U, E) is a planar graph, then NNG(G), MST(G), RNG(G), GG c (G), GG f (G), and DG c (G) are planar. We have just proved that the proximity graphs inherit planarity from the original graph. Next we show that they also inherit acyclicity. Theorem 4.7. Let G = (V, U, E) be a tree. Then DG f (G) = MST(G). Proof. As MST(G) DG f (G), it suffices to show that DG f does not contain any cycle. Suppose that DG f contains a cycle u 1 u 2... u l u 1 (3 l n). By Proposition 4.2, the intersection of the Voronoi regions of every pair of points u i, u i+1 is non-empty; let p i,i+1 be a point belonging to this intersection. By Remark 4.1, the points p i,i+1 are not in V, are pairwise different, and satisfy that d G (p i,i+1, u i ) = d G (p i,i+1, u i+1 ) < d G (p i,i+1, u k ) for all vertices u k U different from u i and u i+1. For every vertex u i, consider the union of the unique path in G from p i 1,i to u i and the unique path in G from u i to p i,i+1. Let c i be the unique path in G from p i 1,i to p i,i+1. Observe that this path is contained in the previous union. Since the paths from p i 1,i to u i and from u i to p i,i+1 are also shortest paths, each point q in c i different from p i 1,i and p i,i+1 is such that d G (q, u i ) < d G (q, u k ) for all vertices u k U different from u i. As a consequence, two paths c i and c j, i j, only intersect if j = i + 1, and the intersection takes place at point p i,i+1. Thus the union of the paths c i is a cycle of G, contradicting that G is a tree. Corollary 4.8. Let G = (V, U, E) be a tree. RNG(G) = GG f (G) = MST(G). Then GG c (G) and DG c (G) are forests, and Next we give complete characterizations for those graphs that are isomorphic to a certain proximity graph of some other graph. Our results are motivated by the rich literature on characterizations for the graphs that are isomorphic to (or can be drawn as) one of the usual proximity graphs defined on a point set in the plane (see [20] for an excellent survey). 9
10 Proposition 4.9. If G = (V, E) is a graph, there exists a graph Ḡ = ( V, Ū, Ē) such that G = NNG(Ḡ) if and only if G is acyclic and does not contain isolated vertices. Proof. On the one hand, any nearest neighbor graph does not contain isolated vertices because every vertex is connected to its nearest neighbor. On the other hand, this proximity graph is acyclic because it is a subgraph of the minimum spanning tree. This settles one of the implications. Now let us suppose that G does not contain cycles or isolated vertices. We define a new graph Ḡ = ( V, Ū, Ē) such that G = NNG(Ḡ). Let V = Ū = V, Ē = E, and look at each connected component of Ḡ as a rooted tree. Connect all pairs of roots of different trees. Assign weights to the edges of Ḡ so that the nearest neighbor of each vertex different from a root is its predecessor in its tree, and the nearest neighbor of a root is one of its successors in its tree. The graph Ḡ satisfies G = NNG(Ḡ). The next characterization is straightforward. Proposition If G = (V, E) is a graph, there exists a graph Ḡ = ( V, Ū, Ē) such that G = MST(Ḡ) if and only if G is a tree. Proposition If G = (V, E) is a graph, there exists a graph Ḡ = ( V, Ū, Ē) such that G = RNG(Ḡ) if and only if G is triangle-free. Proof. Suppose that there exists a graph Ḡ = ( V, Ū, Ē) such that G = RNG(Ḡ). Notice that, for every group of three vertices in RNG(Ḡ), the edge connecting the furthest pair is not in the graph. Thus RNG(Ḡ) is triangle-free and so is G. Reciprocally, let V = Ū = V, Ē = E and assign to all edges in Ē approximately the same weight. Consider two vertices u i, u j in Ū. If their corresponding vertices in G are adjacent, they are relative neighbors in Ḡ. Indeed, since G is triangle-free, no vertex in G is adjacent to both u i and u j, so no vertex in Ū lies in the lens defined by u i and u j in Ḡ. If they are relative neighbors in Ḡ, no other vertex of Ū lies in the shortest path connecting them, and thus their corresponding vertices in G are adjacent. Proposition Let G = (V, E) be a graph. There exists a graph Ḡ = ( V, Ū, Ē) such that G = GG c (Ḡ) = GGf(Ḡ) = DGc(Ḡ) = DGf(Ḡ). Proof. Let Ū = V, Ē = E and assign to all edges in Ē approximately the same weight. Add a new vertex in V Ū close to the midpoint of each of the edges in Ē. Then G = GG c (Ḡ) = GG f (Ḡ) = DGc(Ḡ) = DGf(Ḡ). The lens defined by two points at distance d has been sometimes called lune in the literature, and it is the intersection of the disks of radius d centered at the points. 10
11 5 Algorithms In this section we provide algorithms to compute each of the proximity graphs we have presented. Algorithm for DG f (G) The Voronoi diagram of a graph G = (V, U, E) can be computed in O(e + (m n) log(m n)) time using an algorithm proposed in [8]. As shown in Proposition 4.2, DG f (G) is its dual graph, so it can be computed scanning the bridges of VD(G), that is, the edges with the property that the two endpoints belong to different Voronoi regions. This is done in O(e) extra time. Algorithm for DG c (G) Two different vertices u, u U are adjacent in DG c (G) if and only if there exists a closed disk D G (v, r), with v V, which is empty of points in U except for u and u. If such a disk exists, the two vertices in U closest to vertex v are u and u. For each vertex v i in V, the algorithm computes its two closest vertices u i and u i of U. Then the edges in DG c(g) are (u i, u i ) for all i. To find the two closest vertices in U of every vertex in V we use a technique similar to that in [8, 11, 24]. The sketch of the algorithm is as follows. For each vertex u U, the algorithm constructs a tree T u rooted at u providing the shortest paths from u to some vertices in V as in Dijkstra s algorithm. The trees are built simultaneously as follows. At any given point during the execution of the algorithm, among all vertices that are adjacent in G to a vertex in any of the n trees, we select the one at shortest distance from the root of its corresponding tree. We do not add a vertex to a tree if it is already there, and we do not add a vertex to a tree if it was added to two other trees in earlier iterations. The algorithm finishes when all vertices in V have been added to two trees. Our algorithm is very similar to the one presented in [11] to compute the k nearest vertices in U of every vertex in V, which runs in O(min{ne + nm log m, km log m + ke log m}) time. So here we omit the precise description of the steps of the algorithm and the proof of its correctness, which are carefully given in [11], and we concentrate on the implementation details that allow to improve the running time of the algorithm in [11]. Implementation of the Algorithm First of all, we create a min-priority queue Q implemented via Fibonnacci heaps [10], so that the operations Insert, Find-min, and Decrease-key are done in O(1) amortized time, and the operations Extract-min and Delete are done in O(log n) time. The objective of creating and maintaining the min-priority queue Q is to store the current minimum and second minimum distances from each vertex in V to the vertices in U. So for each vertex v V, we insert into Q two elements v 1, v 2 with associated values d(v 1 ), d(v 2 ) initially set to infinity (except for the elements u 1, where u U, which are inserted with value d(u 1 ) = 0). The parameters d(v 1 ) and d(v 2 ) will respectively store the lengths of the current shortest and 11
12 second shortest paths from v to any of the vertices in U, with the additional condition that these two shortest paths lead to two different elements T (v 1 ) and T (v 2 ) of U. We also store the parents π(v 1 ) and π(v 2 ) that v has in these paths and the vertices T (v 1 ) and T (v 2 ). We also make use of a table SP of size m 2 in which we will keep track of the number of times an element has been extracted from Q to be added to a tree, and of the trees to which the element has been added. Thus SP [v, j] = i (j = 1, 2) means that, the j-th time that vertex v has been extracted from Q, it has been added to T i. It is worth pointing out that we do not actually store the trees because the only information needed to compute DG c (G) is given by table SP. At each iteration of the algorithm, we first extract the minimum element of Q; let it be v j. Then v is added to the corresponding tree, so we update SP and process every neighbor of v, w π(v j ). To process these neighbors we follow the same steps as in [11], except that, if we find a shorter path from w to a vertex in U, we do not delete w from Q and reinsert it with a new source T (w), but we simply decrease the key of w (which is done in O(1) amortized time) and update the values of π(w) and T (w). When all neighbors have been processed, we continue with the next iteration of the algorithm. The algorithm ends when Q is empty. Complexity Since the only insertions into Q are done in the initialization step and at the end of the algorithm Q is empty, the total running time of the insertions and extractions is O(m log m). We must also account for the Decrease-key operations. Let v be a vertex of V. The associated vertices v 1, v 2 might have their value in Q decreased only if one of the neighbors of v in G is added to a tree, so at most 2δ(v) times (where δ(v) is the degree of v in G). This implies that the total number of decrease operations in the algorithm is at most 8e. Since each of them is done in constant amortized time, the total running time is O(e). Regarding space, it is clear that the total space used by the algorithm is O(m + e). Summarizing, we have proved Theorem 5.1. For each graph G = (V, U, E) the graph DG c (G) can be computed in O(e + m log m) time and O(e) space. Algorithm for GG f (G) Recall that u i, u j U are adjacent in GG f (G) if they are the two closest vertices in U of their midpoint. So we propose an algorithm that first computes the shortest path between every pair of vertices in U and afterwards tests the previous condition. To do this last step, we scan the shortest path in O(m) time to find its midpoint and then check whether the midpoint is contained in a bridge between the Voronoi regions of the two vertices in U, which can be done in constant time if the Voronoi diagram of the graph has been precomputed. Observe that, as GG f (G) DG f (G), it suffices to test the edges in DG f (G). 12
13 To compute all pairs (of vertices in U) shortest paths, we might distinguish two cases, depending on the order of magnitude of the number of edges of G. As far as we know, the best algorithm for the case in which G is a sparse graph has an asymptotic cost of O(me log α(m, e)) in the comparison-addition model (α denotes the functional inverse of Ackermann s function) [27]. In this algorithm, after a preprocessing phase of cost O(m log m), single-source shortest path queries can be carried out in O(e log α(m, e)) time. Since we are only interested in shortest paths between vertices in U, the total cost of the algorithm in our case is O(m log m + ne log α(m, e)). If G is dense, the previous algorithms might be cubic. To the best of our knowledge, given a dense graph G = (V, E), with V = m and E = e, the fastest algorithm to compute the shortest paths between all pairs of vertices in V runs in O ( m 3 log 3 log m/ log 2 m ) time in the standard RAM model [6]. For practical purposes, we define APSP(G) = O ( min{m log m + ne log α(m, e), m 3 log 3 log m/ log 2 m} ). So we have: Theorem 5.2. For each graph G = (V, U, E), the graph GG f (G) can be computed in O(APSP(G)+ min{n 2, e}m) time. Algorithm for GG c (G) Given a pair of vertices u i, u j U, there is a straightforward way to know whether (u i, u j ) is an edge of GG c (G) if the distances from u i and u j to all vertices in V have been precomputed. Firstly, for each v V, we compute max{d G (v, u i ), d G (v, u j )}, which is the radius of the smallest closed disk centered at v containing both u i and u j. Secondly, among this family of disks, we pick the one that has the smallest radius. Thirdly, we check whether the chosen disk contains some vertex in U different from u i and u j. These three steps can be done in O(m) time. Since GG c (G) DG f (G), we only test the edges in DG f (G). Consequently, Theorem 5.3. For each graph G = (V, U, E), the graph GG c (G) can be computed in O(APSP(G)+ min{n 2, e}m) time. Algorithm for RNG(G) We propose a two steps algorithm. Firstly, we compute the shortest paths between all pairs of vertices in U to obtain the distances among these vertices. Secondly, we compute RNG(G) by checking, for all pairs of vertices u i, u j U that are adjacent in DG f (G), whether there exists any vertex u k U such that d G (u k, u i ) < d G (u i, u j ) and d G (u k, u j ) < d G (u i, u j ). Theorem 5.4. For each graph G = (V, U, E), the graph RNG(G) can be computed in O(APSP(G)+ min{n 2, e}n) time. 13
14 Algorithm for MST(G) A first approach to obtain the minimum spanning tree of a graph G = (V, U, E) could be to compute the distances between all pairs of vertices in U and then use some efficient algorithm to produce the minimum spanning tree of the complete graph on the vertices in U (if any of the supergraphs of MST(G) was already known, some of the edges could be discarded). This algorithm would run in O (APSP(G)) time. Next we show that another analogy between the usual proximity graphs and the new proximity structures on graphs allows to derive a better algorithm. It is well known that the Gabriel graph of a set of points in the plane contains those edges in the Delaunay graph that intersect their dual Voronoi edges. The analogous property satisfied by our proximity structures on graphs is the following: Lemma 5.5. Let G = (V, U, E) be a graph and let (u i, u j ) be an edge of DG f (G). The graph GG f (G) contains (u i, u j ) if and only if the shortest path between u i and u j is contained in the Voronoi regions of u i and u j. Proof. It suffices to notice that the shortest path between u i and u j is contained in the Voronoi regions of u i and u j if and only if its midpoint lies in the Voronoi regions of both vertices. Remark 5.6. Let G 1 = (V, E), G 2 = (V, E) be two weighted graphs with the same sets of vertices and edges. Let e 1 (respectively e 2 ) denote the weight of edge e in G 1 (respectively G 2 ). Suppose that E = E E, where e 1 = e 2 for all e E and e 1 < e 2 for all e E. Suppose also that no edge in E belongs to MST(G 1 ). Then MST(G 1 ) = MST(G 2 ). Now let us consider DG f (G) as a weighted graph, where the weight of an edge is given by the distance in G between its endpoints. As MST(G) DG f (G), we have that MST(G) = MST(DG f (G)). We define the weighted graph DG f (G) as follows. The vertex set and edges are the same as in DG f (G), and the weight of an edge (u i, u j ) in DG f (G) is the length of the shortest path from u i to u j in G that is contained in the Voronoi regions of both vertices. Observe that the weight of an edge in DG f (G) is less than or equal to its weight in DG f (G), and that, by Lemma 5.5, they are equal if and only if the edge belongs to GG f (G). Moreover, since MST(G) GG f (G), an edge that has different weights in DG f (G) and DG f (G) is not contained in MST(G). By Remark 5.6, we can conclude that MST( DG f (G)) = MST(DG f (G)), so MST( DG f (G)) = MST(G). Our algorithm takes advantage of the fact that computing the weights in DG f (G) is easier than in DG f (G). Firstly, we compute the Voronoi diagram of G using the algorithm proposed in [8]. This algorithm provides not only the nearest vertex in U of every vertex in V but also the distance between them. Thus the length of the shortest path between two neighbors u i, u j in DG f (G) containing a particular bridge of VD(G) can be computed in constant time. In order to obtain the weight of edge (u i, u j ) in DG f (G), all we have to do is to scan all the bridges 14
15 Table 2: Inclusion relations among all classes of proximity graphs in the general case. UMST RNG GG ca GG c1 GG fa GG f1 DG c DG f NNG UMST RNG GG ca GG c1 GG fa GG f1 DG c between the Voronoi regions of both vertices. All in all, the graph DG f (G) can be computed in O(e + (m n) log(m n)) time. Afterwards we can apply some efficient algorithm to obtain MST( DG f (G)). Two suitable options are the algorithm in [7], which runs in O(e α(e, n)) time, or the algorithm in [26], whose exact running time is not known, but which runs in linear time for most graphs. In conclusion, Theorem 5.7. For each graph G = (V, U, E), the graph MST(G) can be computed in O(e α(e, n)+ (m n) log(m n)) time. Algorithm for NNG(G) An algorithm to compute the nearest neighbor graph by means of DG f (G) is also provided in [8]. Its total running time is O(e + (m n) log(m n)). 6 Presence of Degeneracies In this section we generalize our results to the case in which degeneracies arise. Theorem 6.1. If degenerate situations are allowed, the inclusion relations among all classes of proximity graphs are shown in Table 2. Furthermore, all classes of proximity graphs are different. Proof. First we deal with the new graphs. It is easy to see that, as far as inclusion relations are concerned, the graph GG c1 behaves essentially as the graph GG c in the non-degenerate case, and the same holds for the graphs GG f1 and GG f. This rule has few exceptions related to NNG(G), UMST(G), and RNG(G) which are explained below. 15
16 Regarding GG ca, the example in Figure 2a shows that NNG GG ca and RNG GG ca. Clearly, GG ca (G) GG c1 (G) and one can easily come up with a situation where GG ca (G) GG c1 (G). The graph in Figure 2b proves that GG ca GG f1 and GG ca GG fa. Finally, it holds that GG ca DG c and GG ca DG f, because GG ca (G) GG c1 (G) and GG c1 (G) DG c (G) DG f (G). Next we focus on GG fa. It is obvious that GG fa GG f1 (G) and that, in some cases, GG fa GG f1 (G). The graph in Figure 2b shows that GG c1 (G) GG fa, while the one in Figure 2c proves that GG fa (G) DG c. Since GG fa (G) GG f1 and GG f1 DG f (G), we conclude that GG fa (G) DG f. Now let G be the graph on Figure 3a. This graph does not satisfy assumption A2 and illustrates the changes with respect to the non-degenerate case. That is, the graphs NNG(G), UMST(G), and RNG(G) contain edges (a, b), (b, c) and (c, a), while the graphs GG ca (G), GG c1 (G), GG fa (G), GG f1 (G), DG c (G), and DG f (G) only contain (a, b). Hence, none of the graphs of the first group is a subgraph of any graph of the second group. Finally, analogous proofs to those in Section 3 show that the remaining relations of containment hold even if degeneracies are permitted. a a b 3 3 d 5 c b 5 5 f e c d (a) (b) Figure 3: Two graphs not satisfying A2. (a) The edge (b, c) belongs to NNG(G), UMST(G), and RNG(G), but does not belong to GG ca (G), GG c1 (G), GG fa (G), GG f1 (G), DG c (G), and DG f (G). (b) NNG(G), UMST(G), and RNG(G) are the complete graph on U, whereas GG ca (G), GG c1 (G), GG fa (G), GG f1 (G), DG c (G), and DG f (G) are the empty graph. We now focus on the most important properties presented in Section 4. The proof that DG f (G) is the dual the graph of VD(G) does not require any non-degeneracy assumption, so this result holds in all cases. Let us now consider the size of the proximity graphs. If A2 is not satisfied, some of the proximity graphs might have more edges than the original graph. More precisely, Theorem 6.2. Let G = (V, U, E) be a graph. The number of edges of GG ca (G), GG c (G), GG fa (G), GG f (G), DG c (G), and DG f (G) is at most e. The number of edges of NNG(G), UMST(G), and RNG(G) may be greater than e. Proof. Let (u i, u j ) be an edge of DG f (G). Then there is a point p of G such that d G (p, u i ) = d G (p, u j ) < d G (p, u k ) for all vertices u k U different from u i and u j. 16
17 First suppose that the point p is unique. If p is not an element of V, we assign the edge (u i, u j ) DG f (G) to the edge in G containing p. Otherwise let u i v 1 v 2... v l p be a shortest path from u i to p in G. Then we assign (u i, u j ) to the edge (v l, p) G. Now suppose that there are several points of G satisfying the same condition as p. Let q be the one at smallest distance from u i and u j. Observe that q V. As in the previous case, if u i w 1 w 2... w h q is a shortest path from u i to q in G, we assign (u i, u j ) to the edge (w h, q) G. We have just constructed an injective function that maps each edge in DG f (G) to an edge in G. Hence the number of edges of DG f (G) is at most e. Since GG ca (G), GG c1 (G), GG fa (G), GG f1 (G), and DG c (G) are subgraphs of DG f (G), their size is also bounded by e. With regard to the remaining proximity graphs, consider the graph on Figure 3b, which does not fulfill A2. Observe that NNG(G), UMST(G), and RNG(G) are the complete graph on the vertices of U, so they have ten edges, while the original graph has five. Finally, the next two theorems include the proximity graphs that inherit the property of being planar or acyclic in the degenerate case. The proof of the first part of each statement can be obtained using the same approach as in Theorems 4.5 and 4.7 in Section 4. The graph in Figure 3b proves the second part of each statement. Theorem 6.3. Let G = (V, U, E) be a planar graph. Then the graphs GG ca (G), GG c1 (G), GG fa (G), GG f1 (G), DG c (G), and DG f (G) are planar, whereas NNG(G), UMST(G), and RNG(G) may not be. Theorem 6.4. Let G = (V, U, E) be a tree. Then the graphs GG ca (G), GG c1 (G), GG fa (G), GG f1 (G), DG c (G), and DG f (G) are acyclic, whereas NNG(G), UMST(G), and RNG(G) may not be. The algorithms in the preceding section can be adapted to run under the presence of degeneracies yet we omit their descriptions which require, as usual, many details to be carefully handled. References [1] M. Abellanas and F. Harary, Delaunay graphs on a prescribed graph, Proc 15th European Workshop on Computational Geometry, Antibes, France, 1999, pp [2] M. Abellanas, F. Hurtado, C. Icking, E. Langetepe, R. Klein, L. Ma, B. Palop, and V. Sacristán, Voronoi diagram for services neighboring a highway, Inform Process Lett 86(5) (2003), [3] J. Abello, P.M. Pardalos, and M.G. Resende, Handbook of massive data sets, Kluwer Academic Publishers, Dordrecht,
18 [4] H.-K. Ahn, H. Alt, T. Asano, S. Bae, P. Brass, O. Cheong, C. Knauer, H.-S. Na, C.-S. Shin, and A. Wolff, Constructing optimal highways, Internat J Found Comput Sci 20(1) (2009), [5] G. Di Battista, P. Eades, R. Tamassia, and I.G. Tollis, Graph drawing: algorithms for the visualization of graphs, Prentice Hall, Upper Saddle River, [6] T.M. Chan, More algorithms for all-pairs shortest paths in weighted graphs, Proc 39th ACM Symposium on Theory of Computing, San Diego, CA, 2007, pp [7] B. Chazelle, A minimum spanning tree algorithm with inverse-ackermann type complexity, J ACM 47(6) (2000), [8] M. Erwig, The graph Voronoi diagram with applications, Networks 36(3) (2000), [9] C. Faloutsos, K.S. McCurley, and A. Tomkins, Fast discovery of connection subgraphs, Proc 10th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, Seattle, WA, 2004, pp [10] M.L. Fredman and R.E. Tarjan, Fibonacci heaps and their uses in improved network optimization algorithms, J ACM 34(3) (1987), [11] T. Furuta, A. Suzuki, and K. Inakawa, The kth nearest network Voronoi diagram and its application to districting problem of ambulance systems, Discussion Paper, No.0501, Center for Management Studies, Nanzan University, [12] A. Getis and B. Boots, Models of spatial processes: an approach to the study of point, line and area patterns, Cambridge University Press, Cambridge, [13] S. Govindarajan, On locally Gabriel graphs, Submitted. [14] F. Harary, M.S. Jacobson, M.J. Lipman, and F.R. McMorris, Abstract sphere-of-influence graphs, Math Comput Modelling 17(11) (1993), [15] F. Hurtado, R. Klein, E. Langetepe, and V. Sacristán, The weighted farthest color Voronoi diagram on trees and graphs, Comput Geom 27(1) (2004), [16] J.W. Jaromczyk and M. Kowaluk, A note on relative neighborhood graphs, Proc 3rd Annual Symposium on Computational Geometry, Waterloo, ON, 1987, pp [17] J.W. Jaromczyk and G.T. Toussaint, Relative neighborhood graphs and their relatives, Proc IEEE 80(9) (1992), [18] S. Kapoor and X.-Y. Li, Proximity structures for geometric graphs, Proc 8th International Workshop on Algorithms and Data Structures, Ottawa, ON, 2003, pp
19 [19] Y. Koren, S.C. North, and C. Volinsky, Measuring and extracting proximity in networks, Proc 12th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, Philadelphia, PA, 2006, pp [20] G. Liotta, Proximity drawings, Handbook of graph drawing and visualization, R. Tamassia (Editor), CRC Press, Boca Raton, to appear. [21] K. Mehlhorn, A faster approximation algorithm for the Steiner problem in graphs, Inform Process Lett 27(3) (1988), [22] T.S. Michael and T. Quint, Sphere of influence graphs in general metric spaces, Math Comput Modelling 29(7) (1999), [23] A. Okabe, B. Boots, K. Sugihara, and S.N. Chiu, Spatial tessellations: concepts and applications of Voronoi diagrams, John Wiley & Sons, Chichester, [24] A. Okabe, T. Satoh, T. Furuta, A. Suzuki, and K. Okano, Generalized network Voronoi diagrams: concepts, computational methods, and applications, Int J Geogr Inf Sci 22(9) (2008), [25] Y. Ostrovsky-Berman, The transportation metric and related problems, Inform Process Lett 95(4) (2005), [26] S. Pettie and V. Ramachandran, An optimal minimum spanning tree algorithm, J ACM 49(1) (2002), [27] S. Pettie and V. Ramachandran, A shortest path algorithm for real-weighted undirected graphs, SIAM J Comput 34(6) (2005), [28] R. Pinchasi and S. Smorodinsky, On locally Delaunay geometric graphs, Proc 20th Annual Symposium on Computational Geometry, Brooklyn, New York, 2004, pp [29] H. Samet, Foundations of multidimensional and metric data structures, Morgan Kaufmann, San Francisco, [30] G.T. Toussaint (Editor), Computational morphology, North-Holland Publishing Company, Amsterdam,
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https://sites.duke.edu/probabilityworkbook/category/basic-probability/sequence-of-independent-trials/ | 1,716,116,786,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00649.warc.gz | 477,855,745 | 15,164 | # Category Archives: Sequence of independent trials
## Magic Die (or is it rigged?)
A magician claims to have a magic die. If the die is rolled and lands on an even number, then the next time the die is rolled it will land on an odd number (and vice versa). So, as the die is rolled it will alternate perfectly between even and odd numbers (or so the magician claims).
You, being skeptical, figure there’s a 1 percent chance that the die is magical and a 99 percent chance that it’s just an ordinary fair die. You then ask the magician to “prove” the die is magical by rolling it some number of times.
How many successfully alternating rolls will it take for you to think there’s a 99 percent chance the die is magical (or, more likely, that it’s rigged in some way so it always alternates)?
## Point of increase
Suppose $$U_1,U_2, …$$ are independent uniform $$(0,1)$$ random variables. Let $$N$$ be the first point of increase. That is to say the first $$n \geq 2$$ such that $$U_n > U_{n-1}$$. Show that for $$u \in (0,1)$$:
1. $\mathbf{P}(U_1 \leq u \ { and } \ N=n)= \frac{u^{n-1}}{(n-1)!}-\frac{u^{n}}{n!} \quad;\quad n \geq 2$
2. $$\mathbf{E}(N)=e$$
Some useful observations:
• $\mathbf{P}(U_1 \leq u \ { and } \ N=n) = \mathbf{P}(U_1 \leq u \ { and } \ N \geq n) -\mathbf{P}(U_1 \leq u \ { and } \ N \geq n+1)$
• The following events are equal
$\{U_1 \leq u \quad{ and } \quad N \geq n\} = \{U_{n-1}\leq U_{n-2} \leq \cdots \leq U_2\leq U_{1} \leq u \}$
• $\mathbf{P}\{U_2 \leq U_1 \leq u \}= \int_0^u \int_0^{u_1} du_2 du_1$
## Overloading an Elevator
A new elevator in a large hotel is designed to carry about 30 people, with a total weight of up to 5000 lbs. More that 5000 lbs. overloads the elevator. The average weight of guests at the hotel is 150 lbs., with a standard deviation of 55 lbs. Suppose 30 of the hotel’s guests get into the elevator . Assuming the weights of the guests are independent random variables, what is the chance of overloading the elevator ? Give your approximate answer as a decimal.
[Pitman p 204, # 19]
## Coin tosses: independence and sums
A fair coin is tossed three times. Let $$X$$ be the number of heads on the first two tosses, $$Y$$ the number of heads on the last two tosses.
1. Make a table showing the joint distribution of $$X$$ and $$Y$$.
2. Are $$X$$ and $$Y$$ independent ?
3. Find the distribution of $$X+Y$$ ?
## Sums of Poisson
Agambler bets ten times on events of probability $$1/10$$, then twenty times on events with probability $$1/20$$, then thirty times on events with probability $$1/30$$, then forty times on events with probability $$1/40$$. Assuming the vents are independent, what is the approximate distribution of the number of times the gambler wins ? (use Poisson approx. of binomial)
[Pitman 2.5, pg 227]
## Random Digit
Let $$D_i$$ be a random digit chosen uniformly from $$\{0,1,2,3,4,5,6,7,8,9\}$$. Assume that each of the $$D_i$$ are independent.
Let $$X_i$$ be the last digit of $$D_i^2$$. So if $$D_i=9$$ then $$D_i^2=81$$ and $$X_i=1$$. Define $$\bar X_n$$ by
$\bar X_n = \frac{X_1 + \cdots+X_n}{n}$
1. Predict the value of $$\bar X_n$$ when $$n$$ is large.
2. Find the number $$\epsilon$$ such that for $$n=10,000$$ the chance that you prediction is off by more than $$\epsilon$$ is about 1/200.
3. Find approximately the least value of $$n$$ such that your prediction of $$\bar X_n$$ is correct to within 0.01 with probability at least 0.99 .
4. If you just had to predict the first digit of $$\bar X_{100}$$, what digit should you choose to maximize your chance of being correct, and what is that chance ?
[Pitman p206, #30]
## Blocks of Bernoulli Trials
In $$n+m$$ independent $$\text{Bernoulli}(p)$$ trials, let $$S_n$$ be the number of successes in the first $$n$$ trials and $$T_m$$ number of successes in the last $$m$$ trials.
1. What is the distribution of $$S_n$$ ? Why ?
2. What is the distribution of $$T_m$$ ? Why ?
3. What is the distribution of $$S_n+T_m$$ ? Why ?
4. Are $$S_n$$ and $$T_m$$ independent ? Why ?
[Pitman p. 159, # 10]
## Subsequence problem
Scoring subsequences or lengths of similar matches or runs is common to a variety of problems from matches in genetic codes to similar runs in bits.
Consider the following question about two sequences of letters. Set both sequences to have length $$k$$. At each location of the sequences the probability of a match in letters is $$.7$$ and the probability of a mismatch is $$.3$$. At each location a match is assigned a score of $$4$$ and a mismatch is assigned a score of $$-1$$. The total score of the sequence is the sum of the scores at each location, there are $$k$$ locations.
Answer the following:
1. What is the PMF of the total score if $$k=5$$.
2. What is the PMF of the total score for a general $$k$$ ?
## Airline Overbooking
An airline knows that over the long run, 90% of passengers who reserve seats for a flight show up. On a particular flight with 300 seats, the airline sold 324 reservations.
1. Assuming that passengers show up independently of each other, what is the chance that the flight will be overbooked ?
2. Suppose that people tend to travel in groups. Would that increase of decrease the probability of overbooking ? Explain your answer.
3. Redo the the calculation in the first question assuming that passengers always travel in pairs. Are your answers to all three questions consistent ?
[Pitman p. 109, #9]
## Biased coins
Given a random variable $$x = \{ 0,1\}$$ where $$0$$ corresponds to heads and $$1$$ corresponds to tails.
For a single coin flip: $$\mathbf{P}(x \mid p) = p^x(1-p)^{1-x}$$.
For a sequence of $$n$$ coin flips: $$\mathbf{P}(x_1,…,x_n \mid p) = \prod_{i=1}^n p^{x_i}(1-p)^{1-x_{i}}$$.
I have a bag with three types of coins with the following probabilities of drawing each type:
$$\mathbf{P}(p=.5) = .7$$, $$\mathbf{P}(p=.1) = .2$$, $$\mathbf{P}(p=.9) = .1$$.
I draw a coin from the bag. I flip it $$n$$ times resulting in a sequence $$X_1,…,X_n$$.
1. Using Bayes rule provide the formula for
$\mathbf{P}(p = .1 \mid x_1,…,x_n),\quad \mathbf{P}(p = .5 \mid x_1,…,x_n), \quad\text{and}\quad \mathbf{P}(p = .9 \mid x_1,…,x_n)$.
2. If $$x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8)=(1,1,0,1,1,0,0,1)$$ what is the most likely value probability $$p$$ of the coin the was used ?
## Chance of an Accident.
An insurance company has 50% urban and 50% rural customers. If every year each urban customer has an accident with probability $$\mu$$ and each rural customer has an accident with probability $$\lambda$$. Assume that the chance of an accident is independent from year to year and from customer to costumer. This is another way to say, conditioned on being and urban or rural the chance of having an accident each year is independent.
A costumer is randomly chosen. Let $$A_n$$ be the chance this customer has an accident in year $$n$$. Let $$U$$ denote the event that this costumer is urban and $$R$$ the event that the customer is rural.
1. Find $$\mathbf{P}(A_2|A_1)$$.
2. Are $$A_1$$ and $$A_2$$ independent in general ? Are there any conditions when it is true if not in general ?
3. Show that $$\mathbf{P}(A_2|A_1) \geq \mathbf{P}(A_2)$$.
To answer this question it is useful to know that for any positive $$a$$ and $$b$$, one has $$(a+b)^2 < 2(a^2 +b^2)$$ as long as $$a \neq b$$. In the case $$a = b$$, one has of course $$(a+b)^2 = 2(a^2 +b^2)$$. To prove this inequality, first show that $$(a+b)^2 +(a-b)^2= 2(a^2 +b^2)$$ and then use that fact that $$(a-b)^2 >0$$.
4. Find the probability that a driver has an accident in the 3nd year given that they had one in the 1st and 2nd year.
5. Find the probability that a driver has an accident in the $$n$$-th year given that they had one in all of the previous years. What is the limit as $$n \rightarrow \infty$$ ?
6. Find the probability that a diver is a urban diver given that they had an accident in two successive years.
## Duels
Mathematicians and politicians throughout history have dueled.
Alexander Hamilton and Aaron Burr dueled.
The French mathematician Evariste Galois died in a duel.
Consider two individuals (H) and (B) for example dueling.
In each round they simultaneously shoot the other and the probability
of a fatal shot is $$0 < p < 1$$.
1) What is the probability they are fatally injured in the same round ?
2) What is the probability that (B) will be fatally injured before (H) ?
## Two die
Two dice are rolled. Find the probabilities of the following events.
a) the maximum of the two numbers rolled is less than or equal to 2;
b) the maxinum of the two numbers rolled is less than or equal to 3;
c) the maximum of the two numbers rolled is exactly equal to 3;
d) Repeat b) and c) with 3 replaced by $$x=1,…,6$$;
e) Denote $$\mathbf{P}(x)$$ as the probability that the maximum number is exactly $$x$$.
Compute $$\sum_{x=1}^6\mathbf{P}(x)$$.
[Pitman Page 10, #7]
## Blocks of Bernoulli Trials
In $$n+m$$ independent Bernoulli $$(p)$$ trials, let $$S_n$$ be the number of successes in the first $$n$$ trials, $$T_n$$ the number of successes in the last $$m$$ trials.
1. What is the distribution of $$S_n$$ ? Why ?
2. What is the distribution of $$T_m$$ ? Why ?
3. What is the distribution of $$S_n+T_m$$ ? Why ?
4. Are $$S_n$$ and $$T_m$$ independent ? Why ?
5. Are $$S_n$$ and $$T_{m+1}$$ independent ? Why ?
6. Are $$S_{n+1}$$ and $$T_{m}$$ independent ? Why ?
Based on [Pitman, p. 159, #10] | 2,847 | 9,455 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-22 | latest | en | 0.831589 |
https://space.stackexchange.com/questions/5073/standing-at-the-top-of-a-space-elevator-would-up-be-towards-the-earth?noredirect=1 | 1,713,805,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00403.warc.gz | 486,551,390 | 41,965 | # Standing at the top of a space elevator would 'up' be towards the Earth?
If I rode a space elevator to the top, assuming only natural forces apply, would the Earth be over my head or under my feet? If it was over my head how is that possible?
## 2 Answers
Above geosynchronous orbit, so called centrifugal force exceeds force of gravity. So earth would be above you.
Centrifugal acceleration (actually just inertia in a rotating frame) is $\omega^2r$ where ω is angular velocity in radians.
Earth's sidereal day is about 23.93 hours, so ω is about 2 pi radians/23.93 hours. Or 7.3e-5 radians/second.
Acceleration from gravity is $GM_e/r^2$ where $M_e$ is mass of earth and r is distance from earth's center.
You will find that $GM_e/r^2$ and $\omega^2r$ exactly cancel at geosynchronous altitude. Above geosynchronous $\omega^2r$ exceeds $GM_e/r^2$. You will weigh more as you move further beyond geosynchronous height.
If my arithmetic is right, you wouldn't feel a full g until you're about 1.8 million kilometers from earth's center.
• In your answer to Where could we build a space elevator today (2014)? you give an altitude of 142,772 Km for the top of an Earth Elevator (no counterweight), so it would seem that the you should have a full G there. Is your math wrong, or do I need to write another question? Jul 29, 2014 at 14:35
• @JamesJenkins No, from what I can tell both his answers are perfectly correct. You don't need a full -1 g at the other end (counterweight) of the space elevator. All you need to do is make sure that both ends "weigh" the same and balance out to 0 at GEO, its center of mass. Since GEO to "top" is a lot longer in that case, and assuming constant mass per length, it "weighs" the same at larger mass. What you're likely forgetting is that a space elevator has to keep constant rotational period (synchronized with Earth's rotation) for all its parts regardless their orbital altitude. Jul 29, 2014 at 14:54
• Maybe it wouldn't go amiss if David mentioned this constant angular speed requirement of the space elevator, and link it to orbital speed (in words). But I get 0.970698 g at 1.8 million kilometers. That's pretty close to 1 g. To match exactly 1 g at mean sea-level Earth (9.80665 m/s²), more precise distance from the center of the Earth would be 1,854,336 km. Of course, at such distance, the Moon might have something to say about your whole elevator system. You can use this centrifugal force calculator to make it somewhat easier. ;) Jul 29, 2014 at 15:01
• BTW "up" is ambiguous on a space elevator. You're equally "up" when you're standing on the Earth than when you're at the end of its counterweight part. So in a sense, your question could be answered with "both". ;) Jul 29, 2014 at 15:17
• @JamesJenkins What Tildawave said. Since the acceleration gradient is steeper earthside of geosynchronous, the earthside length is shorter. To balance the farside length needs to be longer. Jul 29, 2014 at 15:33
Everywhere on the Space Elevator, "up" is the direction toward the geosynchronous point (GEO). Below GEO, gravity is exerting a stronger force on your body than the centrifugal force caused by rotation -- "down" is toward Earth; beyond GEO, the centrifugal force is stronger than gravity, so the net force on your body is away from the Earth. Right at GEO there's no "up" or "down," and indeed for some distance on either side of GEO the net force is too small for your senses to detect.
From the surface you climb up to GEO and then, without changing direction, you climb down to the counterweight at the outer end of the elevator.
The rule for "what holds the Space Elevator up" is that the sum of the net force on all the mass of the elevator above GEO must be larger than the sum of the net force on all the mass below GEO. | 956 | 3,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-18 | latest | en | 0.948994 |
http://www.weap21.org/index.asp?action=9&read=1841&fID=30&NewLang=ID | 1,449,000,356,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398471436.90/warc/CC-MAIN-20151124205431-00094-ip-10-71-132-137.ec2.internal.warc.gz | 758,657,967 | 5,632 | WEAP adalah sebuah insiatif dari Stockholm Environment Institute. Tentang WEAP Beranda Mengapa WEAP ? Fitur Apa yang Baru ? Contoh Screen Contoh Demo Aplikasi Publikasi Sejarah dan Kredit Menggunakan WEAP Unduh Lisensi Bantuan Penggunaan Tutorial Video Sumber Forum Pengguna Diskusi Daftar Anggota Sunting Profil Penunjang Tambahan Pelatihan Program Universitas Kolaborasi Tentang Kami SEI-US Program Sumberdaya Perairan Silahkan Hubungi Kami Tweets by @weap21 Berita Lingkungan Tertarik pada Energi?Baca tentang LEAP: Software SEI untuk perencanaan energi
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Seluruh Topik | Topik "Constraints"
Masuk untuk mengirim pesan baru atau membalas pesan yang ada.
Penulis Pesan
Janak Timilsena Subjek: Constraints Dikirim: 12/14/2009 Dilihat: 15210 times I am working to set the constraints for the groundwater extraction. I am running the model in daily time step. In my project, constraint is in annual basis (i.e. it can not exceed 90 mgd for wateryear). Some of the time steps could be more than 90 mgd and some of the time step could be less than 90 mgd, but end of the water year the average should not be exceeded 90 mgd. I have tried to use the following function in max flow volume; PrevTSValue(Maximum Flow Volume[Million gal],1,365,1<90) but it constraints everytime step, which i donot want. Do you have any suggestion to incorporate this type of constraint in WEAP?
Jack Sieber
Subjek: Re: Constraints
Dikirim: 12/14/2009 Dilihat: 15209 times
Very interesting idea! What you are trying to do is constrain the flow through the transmission link in the current timestep so that it doesn't push the average (over the past year) above some threshold (90 million gallons per day (MGD)). In order to do this, you need to calculate the flow in the current timestep that would cause the average to be exactly 90 MGD, then constrain the flow through the link to be at most this flow. The answer is to subtract the total flow over the previous 364 days from 90 * 365 to get the "available" flow for the current day. Also, you need to make sure the answer is not negative.
Max(0, 365 * 90 - PrevTSValue(Flow[MGD], 1, 364) )
Note: PrevTSValue(Flow[MGD], 1, 364) will sum up the flow, in MGD, over the previous 364 days.
Jack
Topik "Constraints" | 612 | 2,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2015-48 | longest | en | 0.579664 |
https://www.coursehero.com/file/6368156/Homework-9-2010-Solution/ | 1,490,346,421,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187744.59/warc/CC-MAIN-20170322212947-00232-ip-10-233-31-227.ec2.internal.warc.gz | 896,706,509 | 23,468 | Homework 9 2010 Solution
Homework 9 2010 Solution - Use the Given Information to...
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Unformatted text preview: Use the Given Information to determine NPV and IRR of the proposed Maximum Score Step 1 50 Step 2 50 Step 3 300 Step 4 300 Sensitivities 300 Total 1000 d investment. Calculate the requested sensitivities Corrected Oxygen Delignification Example Project Year -> Annual Tons 354,828 Depreciation Factor Current Sequence DoEo+pD Annual Depreciation Current Cost per Ton \$27.83 Cost Savings, \$ per Year Proposed Sequence ODoEo+pD EBITDA Proposed Cost per Ton \$16.68 EBIT Annual Savings \$3,956,332 without depreciation Tax at 35% Capital Cost \$17,000,000 Net Income Project Life 10 years Cash Flow, \$ per Year Terminal Value 5 x Year 10 EBITDA ee Cash Flow, \$ per Year Tax 35% (negative tax okay) sh Flow for NPV Calculation Discount Rate 12% NPV @ 15% IRR 1 2 3 14.29% 24.49% 17.49% 12.49% \$2,429,300 \$4,163,300 \$2,973,300 \$2,123,300 \$3,956,332 \$3,956,332 \$3,956,332 \$3,956,332 \$6,385,632 \$8,119,632 \$6,929,632 \$6,079,632 \$1,527,032 \$(206,968) \$983,032 \$1,833,032 \$534,461 \$(72,439) \$344,061 \$641,561 \$992,571 \$(134,529) \$638,971 \$1,191,471 \$3,421,871 \$4,028,771 \$3,612,271 \$3,314,771 \$(13,578,129) \$4,028,771 \$3,612,271 \$3,314,771 \$(13,578,129) \$4,028,771 \$3,612,271 \$3,314,771 \$6,106,109.53 26% 4 5 6 7 8.93% 8.92% 8.93% 4.46% \$1,518,100 \$1,516,400 \$1,518,100 \$758,200 \$3,956,332 \$3,956,332 \$3,956,332 \$3,956,332 \$5,474,432 \$5,472,732 \$5,474,432 \$4,714,532 \$2,438,232 \$2,439,932 \$2,438,232 \$3,198,132 \$853,381 \$853,976 \$853,381 \$1,119,346 \$1,584,851 \$1,585,956 \$1,584,851 \$2,078,786 \$3,102,951 \$3,102,356 \$3,102,951 \$2,836,986 \$3,102,951 \$3,102,356 \$3,102,951 \$2,836,986 \$3,102,951 \$3,102,356 \$3,102,951 \$2,836,986 8 9 10 0.00% 0.00% 0.00% \$-...
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This note was uploaded on 08/22/2011 for the course WPS 491 taught by Professor Phillips during the Spring '11 term at N.C. State.
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Ask a homework question - tutors are online | 920 | 2,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-13 | longest | en | 0.708799 |
https://voiceofwave.com/t/acoustic-wave-propagation | 1,638,067,428,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00029.warc.gz | 691,628,367 | 16,922 | # Acoustic wave propagation
## 29359 best questions for Acoustic wave propagation
We've collected 29359 best questions in the «Acoustic wave propagation» category so you can quickly find the answer to your question!
## Content
FAQ
Those interested in the Acoustic wave propagation category often ask the following questions:
### ☑️ What are wave propagation?
Propagation of disturbance suffered by a particle in a medium is called wave propagation Wave Propagation is a term used to estimate the actions and characteristics of a wave of motion either in solids, liquids, gasses, or electrical radio type waves by the environment it is in and what factors affect those waves behaviors.
### ☑️ Antenna and wave propagation?
Yes, it certainly is appropriate to mention those things together, as the principles of their interaction are so inexorably intertwined.
### ☑️ What is sky wave propagation?
• Skywave propagation also known as the skip is a type of radio wave propagation. It is either the reflected or refracted back waves to the earth from the ionosphere which is an electrically charged layer of the upper atmosphere.
### ☑️ What is space wave propagation?
#### What are the applications of sky wave propagation?
• Applications of space wave propagation A line of sight communication and satellite communication Radar communication Microwave linking
### ☑️ What is speed wave propagation?
• The speed of propagation vw is the distance the wave travels in a given time, which is one wavelength in a time of one period. In equation form, it is written as v w = f λ. v w = f λ. From this relationship, we see that in a medium where vw is constant, the higher the frequency, the smaller the wavelength.
## Video from Acoustic wave propagation
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Video answer: Acoustic wave therapy for peyronie's- is this a scam?
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## Top 29339 questions from Acoustic wave propagation
We’ve collected for you 29339 similar questions from the «Acoustic wave propagation» category:
### What is the direction of wave propagation?
• The "direction of wave propagation" is the direction of a wave's energy flow, and the direction that a small wave packet will move, i.e. the direction of the group velocity. For light waves, this is also the direction of the Poynting vector . On the other hand, the wave vector points in the direction of phase velocity.
### What does wave propagation mean in physics?
• propagation - the movement of a wave through a medium. physical phenomenon - a natural phenomenon involving the physical properties of matter and energy. Doppler effect , Doppler shift - change in the apparent frequency of a wave as observer and source move toward or away from each other.
### What are the types of wave propagation?
Hint: There are three types of modes of propagation of electromagnetic waves: Ground wave propagation, Space wave propagation and Skywave propagation.
### How to find direction of wave propagation?
directions of propagation. To find the direction of propagation of an E&M wave, point the fingers of the right hand in the direction of the electric field, curl them toward the direction of the magnetic field, and your thumb will point in the direction of propagation.
### What is optimum frequency in wave propagation?
The most effective frequency at a specified time for ionosperic propagation of radio waves between two specified points: also known as Frequency Optimum Traffic; Optimum Traffic Frequency
### What is meant by electromagnetic wave propagation?
In simple words, electromagnetic waves are oscillations produced due to crossing over of an electric and a magnetic field. The direction of the propagation of such waves is perpendicular to the direction of the force of either of these fields as seen in the above figure.
### How does pressure affect sound wave propagation?
Only the temperature is changing the speed of sound.
### What is the space wave propagation frequency?
Space wave propagation frequency is nothing special, it is the same frequency of the wave in question, for example WLAN Wifi produces 5.2 GHz radio wave from your computer or from router, so that would be the the space wave propagation frequency in question
### How much does acoustic wave therapy cost?
• Acoustic Wave Therapy costs between \$150 and \$300 per session. Costs vary according to the region where you live and the area of your body you're having treated. Full treatments can cost between \$900 and \$1200. Follow up maintenance treatments can add to that cost.
### What does acoustic wave treatment (awt) involve?
• AWT (Acoustic Wave Treatment) involves introducing acoustic waves into affected regions of the body. Medical applications for these acoustic waves go back to 1980; they have been successfully used in pain therapy and in treating a wide range of other conditions. Recent studies show that acoustic waves have beneficial biological effects that, in aesthetic treatment, may lead to activation of metabolic processes and stimulation of connective tissue.
### How long does acoustic wave therapy last?
Each Acoustic Wave Therapy treatment lasts about 15 minutes, with clients returning for treatments once a week for six weeks. No anesthesia is necessary and the clients can return to their daily activities immediately, as there is a non-invasive and painless procedure.
### Is acoustic wave therapy effective for ed?
This treatment is especially helpful for men with a mild version of ED. Studies have provided evidence that shock wave treatment still supplied incredible improvements for more than 75% of patients involved.
### How the surface acoustic wave is generated?
Surface acoustic waves are generated by converse piezoelectric effect at the input transducer and they are converted back to an electrical signal at the output transducer by direct piezoelectric effect.
### Velocity of ultrasonic wave using acoustic grating?
The ultrasonic waves generated with the help of a quartz crystal inside the liquid in a container sets up standing wave pattern consisting of nodes and anti-nodes. The nodes are transparent and anti-nodes are opaque to the incident light. In effect the nodes and anti-nodes are acts like grating(a setup of large number of slits of equal distance) similar to that of rulings in diffraction grating. It is called as acoustic grating or aqua grating. Hence, by using the condition for diffraction, we can find the wavelength of ultrasound and thereby the velocity of sound in the liquid medium.
### Who invented the bose acoustic wave system?
The man who gave his name to the company and stereo music in a small package to the planet goes by the name of Amar Bose. Apparently he did have help from others but his name is best known through the world of high quality sound
### Why do short wave broadcast services use sky wave propagation?
They can bounce (be deflected) by the ionosphere to (possibly) circle the globe. Other frequencies are limited to line of sight.
### Why sky wave propagation is better at night?
#### What are the advantages of sky wave propagation?
• Thus in sky wave propagation signal can be transmitted to a larger distance. In this way, sky wave propagation eliminates the disadvantages associated with ground wave propagation. An important factor of ionospheric wave propagation is – skip distance. And skip distance is defined as the minimum distance on the surface of the earth from where ...
### Does amplitude of electromagnetic wave change with propagation?
#### How is the energy of an electromagnetic wave related to its amplitude?
• Explain how the energy of an electromagnetic wave depends on its amplitude, whereas the energy of a photon is proportional to its frequency Anyone who has used a microwave oven knows there is energy in electromagnetic waves. Sometimes this energy is obvious, such as in the warmth of the summer Sun.
### What do you mean by propagation of wave?
• as by natural reproduction.
• The process of spreading to a larger area or greater number; dissemination.
• is transmitted through a medium such ...
### How to find direction of propagation of wave?
• directions of propagation. To find the direction of propagation of an E&M wave, point the fingers of the right hand in the direction of the electric field, curl them toward the direction of the magnetic field, and your thumb will point in the direction of propagation.
### What are the disadvantages of ground wave propagation?
• We know that ground wave propagation is usually suitable for the transmission of low-frequency electromagnetic signals (usually up to 2 or 3 MHz ). Also, another major disadvantage associated with ground wave propagation is that it is suitable only for short-range operation.
### Who developed the wave theory of light propagation?
#### Christiaan Huygens
In his Traité de la Lumière (1690; “Treatise on Light”), the Dutch mathematician-astronomer Christiaan Huygens formulated the first detailed wave theory of light, in the context of which he was also able to derive the laws of reflection and refraction.
### What are the applications of sky wave propagation?
what do you think (question mark)
### How to calculate propagation speed of a wave?
• The speed of propagation vw is the distance the wave travels in a given time, which is one wavelength in a time of one period. In equation form, it is written as v w = f λ. v w = f λ. From this relationship, we see that in a medium where vw is constant, the higher the frequency, the smaller the wavelength.
### What is dielectric soil and radio wave propagation?
#### How are radio waves propagated on the Earth?
• Radio waves may be propagated in one or more of five modes, depending upon the medium into which they are launched and through which they pass. These modes are: 1. Free space propagation, where radio waves are not influenced by the earth or its atmosphere. 2. Ground wave propagation, where radio waves follow the surface of the earth. 3.
• Fading is a deviation of the attenuation that a signal experiences over certain propagation media. It may vary with time, position and/or frequency Time cy Signal fading Classification of fading: large-scale fading (gradual change in local average of signal level)
### What are the advantages of sky wave propagation?
• Thus in sky wave propagation signal can be transmitted to a larger distance. In this way, sky wave propagation eliminates the disadvantages associated with ground wave propagation. An important factor of ionospheric wave propagation is – skip distance. And skip distance is defined as the minimum distance on the surface of the earth from where ...
### What are principal planes in antenna wave propagation?
The principal plane in wave propagation is the E-plane and the H-plane of an antenna. The E-plane consists of the electric field vector, and by convention, it's the direction in which the wave is said to be 'polarized'. The H-plane consists of magnetic field vector of the wave. The electric field vector and the magnetic field vector are perpendicular to each other, and the direction in which the wave propagates (moves) is perpendicular to both of them.
### What do you mean by radio wave propagation?
• Radio Wave Propagation. • Radio propagation is the behavior of radio waves when they are transmitted, or propagated from one point on the Earth to another, or into various parts of the atmosphere.
### How is the propagation of sound wave possible?
• Propagation of sound: The propagation of sound wave is not possible through vacuum. The medium here can be gas, liquid or solid. The speed of sound when it is travelling through a medium depends on the type of medium. The speed of sound when travelling through air is 343 m/s or 1,235 km/h.
### How to find the direction of wave propagation?
• My question is how to now determine the propagation direction. I figured out the velocity from the wave equation 1 v 2 = 6 2 but this gives me a negative and positive value for v. Which one is correct? I don't really know how to tell this from the expression. Thank you!
### Why are tsunami wave propagation speeds so slow?
• Disclaimers: This model does not model the deep ocean characteristics of tsunami. Wave propagation speeds are slow due to the depth of inches rather than miles. So that one gutter can be emptied into another, there must be a flexible water proof hinge between the middle gutter section and the wave generation gutter.
### How are plane waves related to wave propagation?
• Consider a material in which B = „H D = †E J = ‰= 0: (1) Then the Maxwell equations read r¢E = 0 r¢B = 0 r£E = ¡ 1 c @B @t r£B = „† c @E @t : (2) Now we do several simple manipulations that will become second nature. First take the curl of one of the curl equations, e.g., Faraday’s law, to flnd r£(r£E) = r(r¢E) ¡r2E = ¡ 1 c @ @t (r£B) = ¡ „† c2
### Is the bose acoustic wave cd3000 factory repaired?
• These are some of the typical symptoms we have encountered with the Bose Acoustic Wave/CD2000, CD3000. We maintain a large inventory of original replacement parts. Older units that are no longer factory repairable can generally be repaired by Us.
### What is the cost of acoustic wave therapy?
• Acoustic Wave Therapy costs between \$150 and \$300 per session. Costs vary according to the region where you live and the area of your body you're having treated. Full treatments can cost between \$900 and \$1200.
### How is the surface acoustic wave ( saw ) excited?
• A surface acoustic wave (SAW) is most conveniently excited on a piezoelectric crystal using an interdigitated electrode pattern, or interdigital transducer (IDT). A surface acoustic wave (SAW), also called a Rayleigh wave, is essentially a coupling between longitudinal and shear waves.
### How does a surface acoustic wave filter work?
• Surface-Acoustic-Wave Filters. A surface-acoustic-wave (SAW) filter consists of an input transducer that generates a surface-acoustic-wave replica of the signal, a polished piezoelectric substrate that acts as the propagation path for the waves, and an output transducer that converts the delayed acoustic waves back to an electrical signal.
### Are there any studies on acoustic wave therapy?
• When you go on the Gainswave website and click on the tab thats says "Clinical Research" the page will be plastered with studies pertaining to "low intensity* extracorporeal shock wave therapy" for the treatment of erectile dysfunction. But they dont display a single study on " Acoustic wave therapy" for the treatment of erection.
### What was the first bose acoustic wave system?
• The waveguide is claimed to improve low-frequency sound "from a small enclosure by guiding air through two 26” folded wave guides". The first "Wave" product was the "Acoustic Wave Music System" (AWMS-1), which was a tabletop mini-hifi system that was introduced in 1984.
### How does acoustic wave therapy improve blood flow?
• During acoustic wave therapy, the sound waves of varying energy trigger neovascularization ( 1 ), i.e., the formation of new function networks of blood cells for improved blood flow. In simpler terms, acoustic wave therapy stimulates the growth of new blood vessels alongside opening existing blood vessels and improving blood circulation in ...
### When did the acoustic wave radio come out?
• That waveguide technology became the basis of the Acoustic Wave Music System, released in 1985, and the smaller Wave AM/FM radio, released in 1993, and dozens of other varieties since. (Fun fact: The company also produced a massive subwoofer, designed for large group events like movies, that relies on the same technology as the Wave.
### How is the propagation of a spin wave described?
• The propagation of spin waves is described by the Landau-Lifshitz equation of motion: where γ is the gyromagnetic ratio and λ is the damping constant. The cross-products in this forbidding-looking equation show that the propagation of spin waves is governed by the torques generated by internal and external fields.
### How is the propagation of a mechanical wave limited?
• The distance of the wave’s propagation is limited by the medium of transmission. In this case, the oscillating material moves about a fixed point, and there is very little translational motion. One intriguing property of mechanical wave is the way they are measured, which is given by displacement divided by wavelength.
### How does the propagation of an electromagnetic wave occur?
• Electromagnetic Wave Propagation. The changing magnetic field, in turn, induces an electric field so that a series of electrical and magnetic oscillations combine to produce a formation that propagates as an electromagnetic wave.
### How to determine the direction of a wave propagation?
• For a particular section of the wave which is moving in any direction, the phase must be constant. So, if the equation says y ( x, t) = A cos ( ω t + β x + ϕ), the term inside the cosine must be constant. Hence, if time increases, x must decrease to make that happen.
Mechanical waves
### Which is the speed of propagation of a wave?
• The speed of propagation v w is the distance the wave travels in a given time, which is one wavelength in a time of one period. In equation form, it is written as In equation form, it is written as v w = λ T v w = λ T
### What does angle mean in old radio wave propagation?
#### What do you mean by radio wave propagation?
• Radio Wave Propagation. • Radio propagation is the behavior of radio waves when they are transmitted, or propagated from one point on the Earth to another, or into various parts of the atmosphere.
### How does the propagation of a sound wave occur?
• Propagation of Sound Waves When an object vibrates, it creates kinetic energy that is transmitted by molecules in the medium. As the vibrating sound wave comes in contact with air particles passes its kinetic energy to nearby molecules. As these energized molecules begin to move, they energize other molecules that repeat the process. | 3,767 | 18,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.907887 |
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Prev Next | 1,183 | 5,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-13 | latest | en | 0.960046 |
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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http://landlab.readthedocs.io/en/latest/landlab.ca.html | 1,532,151,008,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00569.warc.gz | 209,814,349 | 19,000 | # landlab.ca package¶
## landlab.ca.celllab_cts module¶
Landlab’s Continuous-Time Stochastic (CTS) cellular automata modeling package.
### Overview¶
A CellLab CTS model implements a particular type of cellular automaton (CA): a continuous-time stochastic CA. The approach is based on that of Narteau et al. (2002, 2009) and Rozier and Narteau (2014). Like a normal CA, the domain consists of a lattice of cells, each of which has a discrete state. Unlike a conventional CA, the updating process is stochastic, and takes place in continuous rather than discrete time. Any given pair (or “doublet”) of adjacent cell states has a certain specified probability of transition to a different pair of states. The transition probability is given in the form of an average transition rate, $$\lambda$$ (with dimensions of 1/T); the actual time of transition is a random variable drawn from an exponential probability distribution with mean $$1/\lambda$$.
### Subclasses¶
Landlab provides for several different lattice and connection types:
• RasterCTS: regular raster grid with transitions between horizontal and vertical cell pairs
• OrientedRasterCTS: like a RasterLCA, but different transition rates can be assigned to vertical and horizontal pairs. This property of orientation can be used, for example, to implement rules representing gravitational attraction, or flow of a fluid with a particular direction.
• RasterD8CTS: like a RasterLCA, but includes diagonal as well as vertical and horizontal cell pairs.
• OrientedRasterD8CTS: as above but orientation also matters.
• HexCTS: hexagonal grid
• OrientedHexCTS: hexagonal grid, with transition rates allowed to vary according to orientation.
### Encoding of “states”¶
As in any traditional cellular automaton model, a LandlabCellularAutomaton contains a grid of cells (“nodes” in Landlab parlance), each of which is has a discrete state. States are represented by integers (0, 1, … N).
In addition, every active link has an orientation code and a link state code. The orientation code represents the orientation of the link in space: is it “vertical” (aligned with the y axis), “horizontal” (aligned with x), or in some other orientation? The number of possible orientations depends on the subclass. The base class has only one orientation code (0) (meaning “orientation doesn’t matter), but this is overridden in some of the subclasses. For example, the OrientedRasterLCA has two orientation codes (0 and 1, for vertical and horizontal), while the OrientedHexLCA has three (representing the three axes in a hex-cell / triagonal grid).
Each active link also has a link state code. The state of a link refers to its particular combination of nodes and its orientation. For example, link state 1 refers to a link in which the tail-node has state 0, the head-node has state 1, and the orientation code is 0. The number of possible link states is equal to R N^2, where R is the number of orientations (1 to 3, depending on the subclass) and N is the number of possible node states. The simplest possible Landlab CA model would have just one orientation code and two possible cell states, so that there are four unique link states. These would be represented by the tuples of (tail-node state, head-node state, orientation) as follows:
link state 0 = (0, 0, 0)
link state 1 = (0, 1, 0)
link state 2 = (1, 0, 0)
link state 3 = (1, 1, 0)
### Main data structures¶
node_state : 1d array (x number of nodes in grid)
Node-based grid of node-state codes. This is the grid of cell (sic) states.
Keys are 3-element tuples that represent the cell-state pairs and orientation code for each possible link type; values are the corresponding link-state codes. Allows you to look up the link-state code corresponding to a particular pair of adjacent nodes with a particular orientation.
node_pair : list (x number of possible link states)
List of 3-element tuples representing all the various link states. Allows you to look up the node states and orientation corresponding to a particular link-state ID.
event_queue : heap of Event objects
Queue containing all future transition events, sorted by time of occurrence (from soonest to latest).
next_update : 1d array (x number of active links)
Time (in the future) at which the link will undergo its next transition. You might notice that the update time for every scheduled transition is also stored in each Event object in the event queue. Why store it twice? Because a scheduled event might be invalidated after the event has been scheduled (because another transition has changed one of a link’s two nodes, for example). The way to tell whether a scheduled event is still valid is to compare its time with the corresponding transition time in the next_update array. If they are different, the event is discarded.
n_xn : 1d array of ints (x number of possible link states)
Number of transitions (“xn” stands for “transition”) from a given link state.
xn_to : 2d array of ints (# possible link states x max. # transitions)
Stores the link-state code(s) to which a particular link state can transition. “max. # transitions” means the maximum number of transitions from a single state. For example, if each link state is associated with one and only one transition, then the maximum is 1, but if there is at least one link state that can have either of two different transitions, then the maximum would be two.
xn_rate : 2d array of floats (# possible link states x max. # transitions)
Rate associated with each link-state transition.
Created GT Sep 2014, starting from link_cap.py.
class CAPlotter(ca, cmap=None, **kwds)[source]
Bases: object
Handle display of a CellLab-CTS grid.
CAPlotter() constructor keeps a reference to the CA model, and optionally a colormap to be used with plots.
Parameters: ca : LandlabCellularAutomaton object Reference to a CA model cmap : Matplotlib colormap, optional Colormap to be used in plotting
CAPlotter() constructor keeps a reference to the CA model, and optionally a colormap to be used with plots.
Parameters: ca : LandlabCellularAutomaton object Reference to a CA model cmap : Matplotlib colormap, optional Colormap to be used in plotting
finalize()[source]
Wrap up plotting.
Wrap up plotting by switching off interactive model and showing the plot.
update_plot()[source]
Plot the current node state grid.
class CellLabCTSModel(model_grid, node_state_dict, transition_list, initial_node_states, prop_data=None, prop_reset_value=None, seed=0)[source]
Bases: object
Link-type (or doublet-type) cellular automaton model.
A CellLabCTSModel implements a link-type (or doublet-type) cellular automaton model. A link connects a pair of cells. Each cell has a state (represented by an integer code), and each link also has a state that is determined by the states of the cell pair.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid), optional Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
Initialize the CA model.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid), optional Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data. seed : int, optional Seed for random number generation.
Takes lists/arrays of “tail” and “head” node IDs for each link, and a dictionary that associates pairs of node states (represented as a 3-element tuple, comprising the TAIL state, FROM state, and orientation) to link states.
creates:
• self.link_state : 1D numpy array
Create a dict of link-state to node-state.
Creates a dictionary that can be used as a lookup table to find out which link state corresponds to a particular pair of node states. The dictionary keys are 3-element tuples, each of which represents the state of the TAIL node, the HEAD node, and the orientation of the link. The values are integer codes representing the link state numbers.
Notes
Performance note: making self.node_pair a tuple does not appear to change time to lookup values in update_node_states. Changing it to a 2D array of int actually slows it down.
Get the current state of a link.
Used to determine whether the link state at link link_id has changed due to an independent change in the node-state grid. Returns the current state of the link based on the states of its two end nodes; this can be compared to the entry in self.link_state to determine whether the state has changed.
Notes
Vectorizing this might yield some speed.
do_transition(event, current_time, plot_each_transition=False, plotter=None)[source]
Transition state.
Implements a state transition.
Parameters: event : Event object Event object containing the data for the current transition event current_time : float Current time in simulation plot_each_transition : bool (optional) True if caller wants to show a plot of the grid after this transition plotter : CAPlotter object Sent if caller wants a plot after this transition
Notes
First checks that the transition is still valid by comparing the link’s next_update time with the corresponding update time in the event object.
If the transition is valid, we:
1. Update the states of the two nodes attached to the link
2. Update the link’s state, choose its next transition, and push it on the event queue.
3. Update the states of the other links attached to the two nodes, choose their next transitions, and push them on the event queue.
do_transition_new(event_link, event_time, current_time, plot_each_transition=False, plotter=None)[source]
Transition state.
Implements a state transition.
Parameters: event : Event object Event object containing the data for the current transition event current_time : float Current time in simulation plot_each_transition : bool (optional) True if caller wants to show a plot of the grid after this transition plotter : CAPlotter object Sent if caller wants a plot after this transition
Notes
First checks that the transition is still valid by comparing the link’s next_update time with the corresponding update time in the event object.
If the transition is valid, we:
1. Update the states of the two nodes attached to the link
2. Update the link’s state, choose its next transition, and push it on the event queue.
3. Update the states of the other links attached to the two nodes, choose their next transitions, and push them on the event queue.
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.oriented_raster_cts import OrientedRasterCTS
>>> import numpy as np
>>> grid = RasterModelGrid((3, 5))
>>> nsd = {0 : 'zero', 1 : 'one'}
>>> trn_list = []
>>> trn_list.append(Transition((0, 1, 0), (1, 0, 0), 1.0))
>>> trn_list.append(Transition((1, 0, 0), (0, 1, 0), 2.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 0, 1), 3.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 1, 1), 4.0))
>>> ins = np.arange(15) % 2
>>> cts = OrientedRasterCTS(grid, nsd, trn_list, ins)
>>> (tm, idx, link) = cts.priority_queue.pop()
>>> np.round(100 * tm)
12.0
>>> idx
6
16
8
13
3
7
>>> cts.node_state[8]
1
>>> cts.node_state[13]
1
7
>>> cts.next_update[16] == _NEVER
True
>>> cts.next_trn_id[16]
-1
get_next_event(link, current_state, current_time)[source]
Get the next event for a link.
Returns the next event for link with ID “link”, which is in state “current state”.
Parameters: link : int ID of the link current_state : int Current state code for the link current_time : float Current time in simulation (i.e., time of event just processed) Event object The returned Event object contains the time, link ID, and type of the next transition event at this link.
Notes
If there is only one potential transition out of the current state, a time for the transition is selected at random from an exponential distribution with rate parameter appropriate for this transition.
If there are more than one potential transitions, a transition time is chosen for each, and the smallest of these applied.
Assumes that there is at least one potential transition from the current state.
get_next_event_new(link, current_state, current_time)[source]
Get the next event for a link.
Returns the next event for link with ID “link”, which is in state “current state”.
Parameters: link : int ID of the link current_state : int Current state code for the link current_time : float Current time in simulation (i.e., time of event just processed) Event object The returned Event object contains the time, link ID, and type of the next transition event at this link.
Notes
If there is only one potential transition out of the current state, a time for the transition is selected at random from an exponential distribution with rate parameter appropriate for this transition.
If there are more than one potential transitions, a transition time is chosen for each, and the smallest of these applied.
Assumes that there is at least one potential transition from the current state.
push_transitions_to_event_queue()[source]
Initializes the event queue by creating transition events for each cell pair that has one or more potential transitions and pushing these onto the queue. Also records scheduled transition times in the self.next_update array.
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.oriented_raster_cts import OrientedRasterCTS
>>> import numpy as np
>>> grid = RasterModelGrid((3, 5))
>>> nsd = {0 : 'zero', 1 : 'one'}
>>> trn_list = []
>>> trn_list.append(Transition((0, 1, 0), (1, 0, 0), 1.0))
>>> trn_list.append(Transition((1, 0, 0), (0, 1, 0), 2.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 0, 1), 3.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 1, 1), 4.0))
>>> ins = np.arange(15) % 2
>>> cts = OrientedRasterCTS(grid, nsd, trn_list, ins)
push_transitions_to_event_queue_new()[source]
Initializes the event queue by creating transition events for each cell pair that has one or more potential transitions and pushing these onto the queue. Also records scheduled transition times in the self.next_update array.
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.oriented_raster_cts import OrientedRasterCTS
>>> import numpy as np
>>> grid = RasterModelGrid((3, 5))
>>> nsd = {0 : 'zero', 1 : 'one'}
>>> trn_list = []
>>> trn_list.append(Transition((0, 1, 0), (1, 0, 0), 1.0))
>>> trn_list.append(Transition((1, 0, 0), (0, 1, 0), 2.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 0, 1), 3.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 1, 1), 4.0))
>>> ins = np.arange(15) % 2
>>> cts = OrientedRasterCTS(grid, nsd, trn_list, ins)
>>> ev0 = cts.priority_queue._queue[0]
>>> np.round(100 * ev0[0])
12.0
>>> ev0[2] # this is the link ID
16
>>> ev6 = cts.priority_queue._queue[6]
>>> np.round(100 * ev6[0])
27.0
>>> ev6[2] # this is the link ID
6
>>> cts.next_trn_id[ev0[2]] # ID of the transition to occur at this link
3
array([-1, 2, -1, 1, 0, 1, 0, 2, -1, 3])
run(run_to, node_state_grid=None, plot_each_transition=False, plotter=None)[source]
Run the model forward for a specified period of time.
Parameters: run_to : float Time to run to, starting from self.current_time node_state_grid : 1D array of ints (x number of nodes) (optional) Node states (if given, replaces model’s current node state grid) plot_each_transition : bool (optional) Option to display the grid after each transition plotter : CAPlotter object (optional) Needed if caller wants to plot after every transition
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.oriented_raster_cts import OrientedRasterCTS
>>> import numpy as np
>>> grid = RasterModelGrid((3, 5))
>>> nsd = {0 : 'zero', 1 : 'one'}
>>> trn_list = []
>>> trn_list.append(Transition((0, 1, 0), (1, 0, 0), 1.0))
>>> trn_list.append(Transition((1, 0, 0), (0, 1, 0), 2.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 0, 1), 3.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 1, 1), 4.0))
>>> ins = np.arange(15) % 2
>>> cts = OrientedRasterCTS(grid, nsd, trn_list, ins)
run_new(run_to, plot_each_transition=False, plotter=None)[source]
Test of new approach using priority queue.
set_node_state_grid(node_states)[source]
Set the grid of node-state codes to node_states.
Sets the grid of node-state codes to node_states. Also checks to make sure node_states is in the proper format, which is to say, it’s a Numpy array of the same length as the number of nodes in the grid.
Creates:
• self.node_state : 1D array of ints (x number of nodes in grid) The node-state array
Parameters: node_states : 1D array of ints (x number of nodes in grid)
Notes
The node-state array is attached to the grid as a field with the name ‘node_state’.
setup_array_of_orientation_codes()[source]
Create array of active link orientation codes.
Creates and configures an array that contain the orientation code for each active link (and corresponding cell pair).
creates:
• self.active_link_orientation : 1D numpy array
Notes
The setup varies depending on the type of LCA. The default is non-oriented, in which case we just have an array of zeros. Subclasses will override this method to handle lattices in which orientation matters (for example, vertical vs. horizontal in an OrientedRasterLCA).
setup_transition_data(xn_list)[source]
Create transition data arrays.
PREVIOUS METHOD:
Using the transition list and the number of link states, creates three arrays that collectively contain data on state transitions:
• n_xn: for each link state, contains the number of transitions out of that state.
• xn_to: 2D array that records, for each link state and each transition, the new state into which the link transitions.
• xn_rate: 2D array that records, for each link state and each transition, the rate (1/time) of the transition.
• xn_propswap: 2D array that indicates, for each link state and each transition, whether that transition is accompanied by a “property” swap, in which the two cells exchange properties (in order to represent a particle moving)
NEW METHOD:
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.oriented_raster_cts import OrientedRasterCTS
>>> import numpy as np
>>> grid = RasterModelGrid((3, 4))
>>> nsd = {0 : 'zero', 1 : 'one'}
>>> trn_list = []
>>> trn_list.append(Transition((0, 1, 0), (1, 0, 0), 1.0))
>>> trn_list.append(Transition((1, 0, 0), (0, 1, 0), 2.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 0, 1), 3.0))
>>> trn_list.append(Transition((0, 1, 1), (1, 1, 1), 4.0))
>>> ins = np.arange(12) % 2
>>> cts = OrientedRasterCTS(grid, nsd, trn_list, ins)
>>> cts.n_trn
array([0, 1, 1, 0, 0, 2, 0, 0])
>>> cts.trn_id
array([[0, 0],
[0, 0],
[1, 0],
[0, 0],
[0, 0],
[2, 3],
[0, 0],
[0, 0]])
>>> cts.trn_to
array([2, 1, 6, 7])
>>> cts.trn_rate
array([ 1., 2., 3., 4.])
update_component_data(new_node_state_array)[source]
Update all component data.
Call this method to update all data held by the component, if, for example, another component or boundary conditions modify the node statuses outside the component between run steps.
This method updates all necessary properties, including both node and link states.
new_node_state_array is the updated list of node states, which must still all be compatible with the state list originally supplied to this component.
Implements a link transition by updating the current state of the link and (if appropriate) choosing the next transition event and pushing it on to the event queue.
Parameters: link : int ID of the link to update new_link_state : int Code for the new state current_time : float Current time in simulation
Implements a link transition by updating the current state of the link and (if appropriate) choosing the next transition event and pushing it on to the event queue.
Parameters: link : int ID of the link to update new_link_state : int Code for the new state current_time : float Current time in simulation
Following an “external” change to the node state grid, updates link states where necessary and creates any needed events.
Notes
Algorithm:
FOR each active link:
if the actual node pair is different from the link's code:
change the link state to be correct
schedule an event
Following an “external” change to the node state grid, updates link states where necessary and creates any needed events.
Notes
Algorithm:
FOR each active link:
if the actual node pair is different from the link's code:
change the link state to be correct
schedule an event
update_node_states(tail_node, head_node, new_link_state)[source]
Update the states of the two nodes in the given link.
Parameters: tail_node : int ID of the tail node of the link (cell pair) in question head_node : int ID of the head node of the link (cell pair) in question new_link_state : int Link state code for the new cell pair (bool, bool) Flags indicating whether the tail node and head node, respectively, have changed state
class Event(time, link, xn_to, propswap=False, prop_update_fn=None)[source]
Bases: object
Represents a transition event at a link. The transition occurs at a given link and a given time, and it involves a transition into the state xn_to (an integer code representing the new link state; “xn” is shorthand for “transition”).
The class overrides the __lt__ (less than operator) method so that when Event() objects are placed in a PriorityQueue, the earliest event is given the highest priority (i.e., placed at the top of the queue).
Event() constructor sets 3 required properties and one optional property.
Parameters: time : float Time at which the event is scheduled to occur link : int ID of the link at which event occurs xn_to : int New state to which this cell pair (link) will transition propswap : bool (optional) Flag: does this event involve an exchange of properties between the two cells?
Examples
>>> from landlab.ca.celllab_cts import Event
>>> e1 = Event( 10.0, 1, 2)
>>> e2 = Event( 2.0, 3, 1)
>>> e1 < e2
False
>>> e2 < e1
True
Event() constructor sets 3 required properties and one optional property.
Parameters: time : float Time at which the event is scheduled to occur link : int ID of the link at which event occurs xn_to : int New state to which this cell pair (link) will transition propswap : bool (optional) Flag: does this event involve an exchange of properties between the two cells?
class Transition(from_state, to_state, rate, name=None, swap_properties=False, prop_update_fn=None)[source]
Bases: object
A transition from one state to another.
Represents a transition from one state (“from_state”) to another (“to_state”) at a link. The transition probability is represented by a rate parameter “rate”, with dimensions of 1/T. The probability distribution of time until the transition event occurs is exponentional with mean 1/rate. The optional name parameter allows the caller to assign a name to any given transition.
Note that from_state and to_state can now be either integer IDs for the standardised ordering of the link states (as before), or tuples explicitly describing the node state at each end, and the orientation. Orientation is 0: horizontal, L-R; 1: vertical, bottom-top. For such a tuple, order is (left/bottom, right/top, orientation).
Transition() constructor sets 3 required properties and 2 optional properties for a transition from one cell pair to another.
Parameters: from_state : int Code for the starting state of the cell pair (link) to_state : int Code for the new state of the cell pair (link) rate : float Average rate at which this transition occurs (dimension of 1/time) name : string (optional) Name for this transition swap_properties : bool (optional) Flag: should properties be exchanged between the two cells?
Transition() constructor sets 3 required properties and 2 optional properties for a transition from one cell pair to another.
Parameters: from_state : int Code for the starting state of the cell pair (link) to_state : int Code for the new state of the cell pair (link) rate : float Average rate at which this transition occurs (dimension of 1/time) name : string (optional) Name for this transition swap_properties : bool (optional) Flag: should properties be exchanged between the two cells?
## landlab.ca.hex_cts module¶
Simple hexagonal Landlab cellular automaton
This file defines the HexCTS class, which is a sub-class of CellLabCTSModel that implements a simple, non-oriented, hex-grid CA. Like its parent class, HexCTS implements a continuous-time, stochastic, pair-based CA. The hex grid has 3 principal directions, rather than 2 for a raster. Hex grids are often used in CA models because of their symmetry.
Created GT Sep 2014
class HexCTS(model_grid, node_state_dict, transition_list, initial_node_states, prop_data=None, prop_reset_value=None)[source]
Class HexCTS implements a non-oriented hex-grid CellLab-CTS model.
HexCTS constructor: sets number of orientations to 1 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid) (optional) Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
Examples
>>> from landlab import HexModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.hex_cts import HexCTS
>>> mg = HexModelGrid(4, 3, 1.0)
>>> nsd = {0 : 'yes', 1 : 'no'}
>>> xnlist = []
>>> xnlist.append(Transition((0,1,0), (1,1,0), 1.0, 'frogging'))
>>> hcts = HexCTS(mg, nsd, xnlist, nsg)
HexCTS constructor: sets number of orientations to 1 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid) (optional) Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
## landlab.ca.hex_lattice_tectonicizer module¶
hex_lattice_tectonicizer.py.
Models discrete normal-fault offset on a 2D hex lattice with a rectangular shape and with one orientation of the nodes being vertical.
The intention here is to use a particle (LCA) model to represent the evolution of a 2D hillslope, with the hex_lattice_tectonicizer serving to shift the nodes either upward (simple vertical uplift relative to baselevel), or up and sideways (representing motion on a fault plane).
Created on Mon Nov 17 08:01:49 2014
@author: gtucker
class HexLatticeTectonicizer(grid=None, node_state=None, propid=None, prop_data=None, prop_reset_value=None)[source]
Bases: object
Handles tectonics and baselevel for CellLab-CTS models.
This is the base class from which classes to represent particular baselevel/fault geometries are derived.
Examples
>>> hlt = HexLatticeTectonicizer()
>>> hlt.grid.number_of_nodes
25
>>> hlt.nr
5
>>> hlt.nc
5
Create and initialize a HexLatticeTectonicizer.
Examples
>>> from landlab import HexModelGrid
>>> hg = HexModelGrid(6, 6, shape='rect')
>>> hlt = HexLatticeTectonicizer()
>>> hlt.grid.number_of_nodes
25
>>> hlt.nr
5
>>> hlt.nc
5
class LatticeNormalFault(fault_x_intercept=0.0, grid=None, node_state=None, propid=None, prop_data=None, prop_reset_value=None)[source]
Handles normal-fault displacement in CellLab-CTS models.
Represents a 60 degree, left-dipping normal fault, and handles discrete offsets for a hex grid that has vertical columns and a rectangular shape.
Examples
>>> import numpy as np
>>> from landlab import HexModelGrid
>>> from landlab.ca.boundaries.hex_lattice_tectonicizer import LatticeNormalFault
>>> pid = np.arange(25, dtype=int)
>>> pdata = np.arange(25)
>>> ns = np.arange(25, dtype=int)
>>> grid = HexModelGrid(5, 5, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(0.0, grid, ns, pid, pdata, 0.0)
>>> lnf.first_fw_col
1
>>> lnf.num_fw_rows
array([0, 1, 3, 4, 5])
>>> lnf.incoming_node
array([1, 3, 4, 6])
>>> lnf.outgoing_node
array([12, 17, 19, 22])
>>> pid = arange(16, dtype=int)
>>> ns = arange(16, dtype=int)
>>> pdata = arange(16)
>>> grid = HexModelGrid(4, 4, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(0.0, grid, ns, pid, pdata, 0.0)
>>> lnf.num_fw_rows
array([0, 1, 3, 4])
>>> lnf.incoming_node
array([1, 2, 5])
>>> lnf.outgoing_node
array([ 7, 11, 15])
>>> lnf.do_offset(rock_state=16)
>>> ns
array([ 0, 16, 16, 16, 4, 16, 6, 1, 8, 2, 10, 5, 12, 13, 14, 9])
>>> lnf.propid
array([ 0, 7, 11, 3, 4, 15, 6, 1, 8, 2, 10, 5, 12, 13, 14, 9])
>>> pid = arange(20, dtype=int)
>>> ns = arange(20, dtype=int)
>>> pdata = arange(20)
>>> grid = HexModelGrid(4, 5, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(0.0, grid, ns, pid, pdata, 0.0)
>>> lnf.incoming_node
array([1, 3, 4, 6])
>>> lnf.outgoing_node
array([12, 14, 17, 19])
>>> lnf.do_offset(rock_state=20)
>>> ns
array([ 0, 20, 20, 20, 20, 5, 20, 20, 8, 1, 10, 3, 4, 13, 6, 15, 16,
9, 18, 11])
>>> lnf.propid
array([ 0, 12, 2, 14, 17, 5, 19, 7, 8, 1, 10, 3, 4, 13, 6, 15, 16,
9, 18, 11])
Create and initialize a LatticeNormalFault object.
Examples
>>> import numpy as np
>>> from landlab import HexModelGrid
>>> from landlab.ca.boundaries.hex_lattice_tectonicizer import LatticeNormalFault
>>> pid = np.arange(25, dtype=int)
>>> pdata = np.arange(25)
>>> ns = np.arange(25, dtype=int)
>>> grid = HexModelGrid(5, 5, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(-0.01, grid, ns, pid, pdata, 0.0)
>>> lnf.first_fw_col
0
>>> lnf.num_fw_rows
array([1, 2, 4, 5, 5])
>>> lnf.incoming_node
array([0, 1, 3, 4, 6])
>>> lnf.outgoing_node
array([12, 17, 19, 22, 24])
>>> pid = np.arange(16, dtype=int)
>>> pdata = np.arange(16)
>>> ns = np.arange(16, dtype=int)
>>> grid = HexModelGrid(4, 4, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(0.0, grid, ns, pid, pdata, 0.0)
>>> lnf.first_fw_col
1
>>> lnf.num_fw_rows
array([0, 1, 3, 4])
>>> lnf.incoming_node
array([1, 2, 5])
>>> lnf.outgoing_node
array([ 7, 11, 15])
>>> pid = np.arange(45, dtype=int)
>>> pdata = np.arange(45)
>>> ns = np.arange(45, dtype=int)
>>> grid = HexModelGrid(5, 9, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(0.0, grid, ns, pid, pdata, 0.0)
>>> lnf.first_fw_col
1
>>> lnf.num_fw_rows
array([0, 1, 3, 4, 5, 5, 5, 5, 5])
>>> lnf.incoming_node
array([ 1, 2, 3, 5, 6, 7, 8, 10, 11, 12])
>>> lnf.outgoing_node
array([22, 31, 33, 34, 35, 38, 39, 40, 43, 44])
do_offset(rock_state=1)[source]
Apply 60-degree normal-fault offset.
Offset is applied to a hexagonal grid with vertical node orientation and rectangular arrangement of nodes.
Parameters: rock_state : int State code to apply to new cells introduced along bottom row.
Examples
>>> import numpy as np
>>> from landlab.ca.boundaries.hex_lattice_tectonicizer import LatticeNormalFault
>>> from landlab import HexModelGrid
>>> pid = np.arange(25, dtype=int)
>>> pdata = np.arange(25)
>>> ns = np.arange(25, dtype=int)
>>> grid = HexModelGrid(5, 5, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(0.0, grid, ns, pid, pdata, 0.0)
>>> lnf.do_offset(rock_state=25)
>>> ns
array([ 0, 25, 25, 25, 25, 5, 25, 25, 8, 1, 10, 3, 4, 13, 6, 15, 16,
9, 18, 11, 20, 21, 14, 23, 24])
>>> lnf.propid
array([ 0, 12, 2, 17, 19, 5, 22, 7, 8, 1, 10, 3, 4, 13, 6, 15, 16,
9, 18, 11, 20, 21, 14, 23, 24])
Set up array with link IDs for shifting link data up and right.
Examples
>>> from landlab.ca.boundaries.hex_lattice_tectonicizer import LatticeNormalFault
>>> from landlab import HexModelGrid
>>> pid = np.arange(25, dtype=int)
>>> pdata = np.arange(25)
>>> ns = np.arange(25, dtype=int)
>>> grid = HexModelGrid(5, 5, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lnf = LatticeNormalFault(-0.01, grid, ns, pid, pdata, 0.0)
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
class LatticeUplifter(grid=None, node_state=None, propid=None, prop_data=None, prop_reset_value=None, opt_block_layer=False, block_ID=9, block_layer_dip_angle=0.0, block_layer_thickness=1.0, layer_left_x=0.0, y0_top=0.0)[source]
Handles vertical uplift of interior (not edges) for a hexagonal lattice with vertical node orientation and rectangular node arrangement.
Create and initialize a LatticeUplifter
Examples
>>> lu = LatticeUplifter()
>>> lu.inner_base_row_nodes
array([1, 3, 4])
>>> hg = HexModelGrid(5, 6, 1.0, orientation='vertical', shape='rect', reorient_links=True)
>>> lu = LatticeUplifter(grid=hg)
>>> lu.inner_base_row_nodes
array([1, 2, 3, 4])
Applies uplift to links and transitions.
For each link that lies above the y = 1.5 cells line, assign the properties of the link one row down.
Examples
>>> from landlab import HexModelGrid
>>> from landlab.ca.oriented_hex_cts import OrientedHexCTS
>>> from landlab.ca.celllab_cts import Transition
>>> import numpy as np
>>> mg = HexModelGrid(4, 3, 1.0, orientation='vertical', shape='rect')
>>> nsd = {0 : 'yes', 1 : 'no'}
>>> xnlist = []
>>> xnlist.append(Transition((0,0,0), (1,1,0), 1.0, 'frogging'))
>>> xnlist.append(Transition((0,0,1), (1,1,1), 1.0, 'frogging'))
>>> xnlist.append(Transition((0,0,2), (1,1,2), 1.0, 'frogging'))
>>> ohcts = OrientedHexCTS(mg, nsd, xnlist, nsg)
array([0, 4, 8, 8, 4, 0, 4, 8, 8, 4, 0])
array([0, 1, 2, 2, 1, 0, 1, 2, 2, 1, 0])
>>> lu = LatticeUplifter(grid=mg)
>>> nu = ohcts.next_update
array([ 0.8 , 1.26, 0.92, 0.79, 0.55, 1.04, 0.58, 2.22, 3.31,
0.48, 1.57])
>>> pq = ohcts.priority_queue
>>> pq._queue[0][2] # link for first event = 20, not shifted
20
>>> round(pq._queue[0][0], 2) # transition scheduled for t = 0.48
0.48
>>> pq._queue[2][2] # this event scheduled for link 15...
15
>>> round(pq._queue[2][0], 2) # ...transition scheduled for t = 0.58
0.58
array([ 0.75, 0.84, 2.6 , 0.07, 0.09, 0.8 , 0.02, 1.79, 1.51,
2.04, 3.85])
>>> pq._queue[0][2] # new soonest event
15
>>> pq._queue[9][2] # was previously 7, now shifted up...
14
>>> round(pq._queue[9][0], 2) # ...but still scheduled for t = 0.80
0.8
uplift_interior_nodes(ca, current_time, rock_state=1)[source]
Simulate ‘vertical’ displacement by shifting contents of node_state
Examples
>>> from landlab import HexModelGrid
>>> from landlab.ca.hex_cts import HexCTS
>>> from landlab.ca.celllab_cts import Transition
>>> mg = HexModelGrid(5, 5, 1.0, orientation='vertical', shape='rect')
>>> nsd = {}
>>> for i in range(26):
... nsd[i] = i
>>> xnlist = []
>>> xnlist.append(Transition((0,0,0), (1,1,0), 1.0, 'frogging', True))
>>> ca = HexCTS(mg, nsd, xnlist, nsg)
>>> lu = LatticeUplifter(propid=ca.propid, prop_data=pd)
>>> lu.node_state[:] = arange(len(lu.node_state))
>>> lu.uplift_interior_nodes(ca, rock_state=25, current_time=0.0)
>>> lu.node_state
array([ 0, 25, 2, 25, 25,
5, 1, 7, 3, 4,
10, 6, 12, 8, 9,
15, 11, 17, 13, 14,
20, 16, 22, 18, 19])
>>> lu.propid
array([ 0, 21, 2, 23, 24,
5, 1, 7, 3, 4,
10, 6, 12, 8, 9,
15, 11, 17, 13, 14,
20, 16, 22, 18, 19])
uplift_property_ids()[source]
Shift property IDs upward by one row
## landlab.ca.little_ca_test module¶
Test the creation and execution of a CellLab-CTS model.
Tests the creation and execution of a CellLab-CTS model, by creating a simple two-state CA on a small grid.
Created by Greg Tucker, May 2015
main()[source]
setup_transition_list()[source]
Creates and returns a list of Transition() objects to represent state transitions for a biased random walk, in which the rate of downward motion is greater than the rate in the other three directions.
Returns: xn_list : list of Transition objects List of objects that encode information about the link-state transitions.
Notes
This doesn’t represent any particular process, but rather is simply used to test the CA code. The transition rules have 0-0 pairs transitioning to 0-1 or 1-0 pairs (50/50 chance) and thence to 1-1 pairs, at which point there are no further transitions.
The states and transitions are as follows:
Pair state Transition to Process Rate (cells/s)
========== ============= ======= ==============
0 (0-0) 1 (0-1) 0.5
2 (1-0) 0.5
1 (0-1) 3 (1-1) 1.0
2 (1-0) 3 (1-1) 1.0
3 (1-1) (none) -
## landlab.ca.oriented_hex_cts module¶
Simple hexagonal Landlab cellular automaton
This file defines the OrientedHexCTS class, which is a sub-class of CellLabCTSModel that implements a simple, non-oriented, hex-grid CA. Like its parent class, OrientedHexCTS implements a continuous-time, stochastic, pair-based CA. The hex grid has 3 principal directions, rather than 2 for a raster. Hex grids are often used in CA models because of their symmetry.
Created GT Sep 2014
class OrientedHexCTS(model_grid, node_state_dict, transition_list, initial_node_states, prop_data=None, prop_reset_value=None)[source]
Oriented hex-grid CellLab-CTS model.
OrientedHexCTS constructor: sets number of orientations to 3 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid), optional Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
Examples
>>> from landlab import HexModelGrid
>>> from landlab.ca.oriented_hex_cts import OrientedHexCTS
>>> from landlab.ca.celllab_cts import Transition
>>> mg = HexModelGrid(4, 3, 1.0)
>>> nsd = {0 : 'yes', 1 : 'no'}
>>> xnlist = []
>>> xnlist.append(Transition((0,1,0), (1,1,0), 1.0, 'frogging'))
>>> ohcts = OrientedHexCTS(mg, nsd, xnlist, nsg)
Initialize a OrientedHexCTS.
OrientedHexCTS constructor: sets number of orientations to 3 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid), optional Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
setup_array_of_orientation_codes()[source]
Creates and configures an array that contain the orientation code for each active link (and corresponding cell pair).
Notes
Creates:
• self.active_link_orientation: 1D numpy array
This overrides the method of the same name in celllab_cts.py. If the hex grid is oriented such that one of the 3 axes is vertical (a ‘vertical’ grid), then the three orientations are:
• 0 = vertical (0 degrees clockwise from vertical)
• 1 = right and up (60 degrees clockwise from vertical)
• 2 = right and down (120 degrees clockwise from vertical)
If the grid is oriented with one principal axis horizontal (‘horizontal’ grid), then the orientations are:
• 0 = up and left (30 degrees counter-clockwise from vertical)
• 1 = up and right (30 degrees clockwise from vertical)
• 2 = horizontal (90 degrees clockwise from vertical)
## landlab.ca.oriented_raster_cts module¶
Simple raster Landlab cellular automaton.
Simple raster Landlab cellular automaton, with cell-pair transitions that depend on orientation (vertical or horizontal)
This file defines the OrientedRasterCTS class, which is a sub-class of CellLabCTSModel that implements a simple, oriented, raster-grid CA. Like its parent class, OrientedRasterCTS implements a continuous-time, stochastic, pair-based CA.
Created GT Sep 2014
class OrientedRasterCTS(model_grid, node_state_dict, transition_list, initial_node_states, prop_data=None, prop_reset_value=None)[source]
Oriented raster CellLab-CTS model.
RasterCTS constructor: sets number of orientations to 2 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid) (optional) Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.oriented_raster_cts import OrientedRasterCTS
>>> mg = RasterModelGrid(3, 4, 1.0)
>>> nsd = {0 : 'yes', 1 : 'no'}
>>> xnlist = []
>>> xnlist.append(Transition((0,1,0), (1,1,0), 1.0, 'frogging'))
>>> orcts = OrientedRasterCTS(mg, nsd, xnlist, nsg)
RasterCTS constructor: sets number of orientations to 2 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid) (optional) Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
setup_array_of_orientation_codes()[source]
Creates and configures an array that contain the orientation code for each active link (and corresponding cell pair).
Notes
Creates:
• self.active_link_orientation: 1D numpy array of ints Array of orientation codes for each cell pair (link)
This overrides the method of the same name in landlab_ca.py.
## landlab.ca.raster_cts module¶
raster_cts.py: simple raster continuous-time stochastic cellular automaton
This file defines the RasterCTS class, which is a sub-class of CellLabCTSModel that implements a simple, non-oriented, raster-grid CA. Like its parent class, RasterCTS implements a continuous-time, stochastic, pair-based CA.
Created GT Sep 2014, starting from link_ca.py.
class RasterCTS(model_grid, node_state_dict, transition_list, initial_node_states, prop_data=None, prop_reset_value=None)[source]
Class RasterLCA implements a non-oriented raster CellLab-CTS model.
RasterLCA constructor: sets number of orientations to 1 and calls base-class constructor.
Parameters: model_grid : Landlab ModelGrid object Reference to the model’s grid node_state_dict : dict Keys are node-state codes, values are the names associated with these codes transition_list : list of Transition objects List of all possible transitions in the model initial_node_states : array of ints (x number of nodes in grid) Starting values for node-state grid prop_data : array (x number of nodes in grid) (optional) Array of properties associated with each node/cell prop_reset_value : number or object, optional Default or initial value for a node/cell property (e.g., 0.0). Must be same type as prop_data.
Examples
>>> from landlab import RasterModelGrid
>>> from landlab.ca.celllab_cts import Transition
>>> from landlab.ca.raster_cts import RasterCTS
>>> mg = RasterModelGrid(3, 4, 1.0)
>>> nsd = {0 : 'yes', 1 : 'no'}
>>> xnlist = []
>>> xnlist.append(Transition((0,1,0), (1,1,0), 1.0, 'frogging')) | 12,459 | 46,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-30 | latest | en | 0.882873 |
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