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https://codegolf.stackexchange.com/questions/166623/generate-a-deck-of-cards/166829 | 1,718,287,811,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00569.warc.gz | 158,448,599 | 76,110 | # Generate a deck of cards
Here's an array representing a standard deck of cards, including two Jokers.
[
"AS", "2S", "3S", "4S", "5S", "6S", "7S", "8S", "9S", "10S", "JS", "QS", "KS",
"AD", "2D", "3D", "4D", "5D", "6D", "7D", "8D", "9D", "10D", "JD", "QD", "KD",
"AH", "2H", "3H", "4H", "5H", "6H", "7H", "8H", "9H", "10H", "JH", "QH", "KH",
"AC", "2C", "3C", "4C", "5C", "6C", "7C", "8C", "9C", "10C", "JC", "QC", "KC",
"J", "J"
]
It's composed in this manner:
• There are four suits; hearts, spades, diamonds, and clubs (H, S, D, C).
• Each suit has one card for the numbers 2 to 10, plus 4 'picture' cards, Ace, Jack, Queen, and King (A, J, Q, K).
• For each combination of suit and value, there should be one item in the array, which is a string, and is made up of the value followed by the suit (Whitespace between these is permitted).
• On top of that, there are two Joker cards ('J').
• Write in any language you please.
• Golf it up! Try to produce this output in the smallest number of bytes.
• It does not matter what order the output is in.
• I was just wondering if you could get 23456789 by multiplying 2 smaller numbers... only to discover it's prime! Commented Jun 11, 2018 at 11:57
• @match Since the order of the output doesn't matter, maybe you can still create it with smaller numbers by ending it with 2 or 4 or anything else so it isn't a prime anymore. Commented Jun 11, 2018 at 13:19
• The playing field isn't very level when you've excluded a majority of languages
– Jo King
Commented Jun 12, 2018 at 9:55
• Rules in comments don't count. If we cannot print the result to STDOUT (and that's a big if, as it overrides our defaults for I/O and prevents languages without functions and string arrays from participating), that rule has to be stated explicitly in the challenge spec. Commented Jun 12, 2018 at 15:05
• That doesn't rule out a string representation of such an array, which could be printed to STDOUT. If not for the comment section, I never would have guessed this wasn't allowed. Commented Jun 12, 2018 at 23:50
{{i}ddd}iiiic{d}iiic{ddd}dddddc{iiii}iicddddddc{ddd}ddddddc{iiii}iicddc{dddd}c{iiii}iic{i}dc{ddddd}dc{iiiii}dc{d}ddddc{ddd}dddddc{iiiii}dc{d}dddc{ddd}ddddddc{iiiii}dc{d}ic{dddd}c{iiiii}dciic{ddddd}dc{iiii}iiic{d}iic{ddd}dddddc{iiii}iiic{d}iiic{ddd}ddddddc{iiii}iiicdddc{dddd}c{iiii}iiic{i}ddc{ddddd}dc{iii}iiiciic{ddd}dddddc{iii}iiiciiic{ddd}ddddddc{iii}iiic{i}dddc{dddd}c{iii}iiic{ii}ddc{ddddd}dc{ii}ddc{ii}dddc{ddd}dddddc{ii}ddc{ii}ddc{ddd}ddddddc{ii}ddc{ii}iic{dddd}c{ii}ddc{iii}iiic{ddddd}dc{ii}dc{i}iiiiiic{ddd}dddddc{ii}dc{ii}dddc{ddd}ddddddc{ii}dc{ii}ic{dddd}c{ii}dc{iii}iic{ddddd}dc{ii}c{i}iiiiic{ddd}dddddc{ii}c{i}iiiiiic{ddd}ddddddc{ii}c{ii}c{dddd}c{ii}c{iii}ic{ddddd}dc{ii}ic{i}iiiic{ddd}dddddc{ii}ic{i}iiiiic{ddd}ddddddc{ii}ic{ii}dc{dddd}c{ii}ic{iii}c{ddddd}dc{ii}iic{i}iiic{ddd}dddddc{ii}iic{i}iiiic{ddd}ddddddc{ii}iic{ii}ddc{dddd}c{ii}iic{iii}dc{ddddd}dc{ii}iiic{i}iic{ddd}dddddc{ii}iiic{i}iiic{ddd}ddddddc{ii}iiic{ii}dddc{dddd}c{ii}iiic{iii}ddc{ddddd}dc{ii}iiiic{i}ic{ddd}dddddc{ii}iiiic{i}iic{ddd}ddddddc{ii}iiiic{i}iiiiiic{dddd}c{ii}iiiic{iii}dddc{ddddd}dc{ii}iiiiic{i}c{ddd}dddddc{ii}iiiiic{i}ic{ddd}ddddddc{ii}iiiiic{i}iiiiic{dddd}c{ii}iiiiic{ii}iiiiiic{ddddd}dc{ii}dddcdc{ii}dc{ddd}dddddc{ii}dddcdc{ii}c{ddd}ddddddc{ii}dddcdc{ii}iiiic{dddd}c{ii}dddcdc{iii}iiiiic{ddddd}dc{iiii}iic{dddd}ddc{iiii}iic
Try it online!
• This looks quite a lot longer than the generated text. Am I missing something? Commented Mar 22, 2021 at 10:34
• @AJFaraday Deadfish~ is a very inconvenient lanugage. See the wiki Commented Mar 23, 2021 at 4:27
• WHYYYYYYYYYYYYY! Commented Mar 23, 2021 at 13:01
• Why not? I didn't write this by hand. Commented Mar 23, 2021 at 19:43
# Vyxal, 24 bytes
₀ɾḢ«∧9₂«fJ«¬⟩C«Ẋvṅ‛jjfJɾ
Try it Online!
## How?
₀ɾḢ«∧9₂«fJ«¬⟩C«Ẋvṅ‛jjfJɾ
₀ɾ # Push a list in the range [1, 10]
Ḣ # Remove the first: [2, 3, 4, ..., 10]
«∧9₂« # Push compressed string "jqka"
fJ # Convert to list of characters and add to the other list: [2, 3, 4, ..., 10, "j", "q", "k", "a"]
«¬⟩C« # Push compressed string "hsdc"
Ẋ # Cartesian product of list and string: [[2, "h"], [2, "s"], ..., ["a", "c"]]
vṅ # Join each together: ["2h", "2s", ..., "ac"]
‛jjf # Push string "jj" and convert to list of characters: ["j", "j"]
J # Append: ["2h", "2s", ..., "ac", "j", "j"]
ɾ # Uppercase each: ["2H", "2S", ..., "AC", "J", "J"]
"J":"J":[c++[s]|c<-"10":map pure"AJQK23456789",s<-"SDHC"]
Try it online!
# Retina, 49 42 bytes
AJQK2345678910
1?.
$0H,$0S,$0D,$0C,
$J,J Credit to @mazzy's comment on my Java answer. -7 bytes thanks to @ovs. Try it online. Explanation: Start with "AJQK2345678910": AJQK2345678910 For each character (with "10" as single character): Replace it with "cH,cS,cD,cC" where c is this character: 1?.$0H,$0S,$0D,$0C, And append "J,J" $
J,J
• You can shorten this by 4 bytes by removing capturing groups and replacing $1 with $&.
– ovs
Commented Jun 11, 2018 at 12:43
• 42 bytes
– ovs
Commented Jun 11, 2018 at 12:46
• @ovs Thanks. Was about to remove the parenthesis at (1?.) and use $0, but the appending of J,J is certainly shorter. Figured it would be possible, but I don't golf that often in Retina, so didn't knew how to append and hadn't had to time to search in the docs more thoroughly. PS: What's the difference between $& and $0 in this case, if there is any? Commented Jun 11, 2018 at 12:58 • It seems like there is no difference, but $& is used more often in the Retina docs, so I got used to it.
– ovs
Commented Jun 11, 2018 at 13:57
• $& comes from Perl, where it is the string matched by the last successful pattern match. (In Perl 6, you would use $/ with the prefix stringification op ~ instead) Commented Jun 11, 2018 at 17:41
# VBA + RegExp, 135 bytes
A declared function that takes no input and outputs to the out var, o.
This subroutine depends upon the Microsoft VBScript Regular Expression 5.5 reference.
Sub a(o)
Set r=New RegExp
r.Global=1
r.Pattern="(\w+)"
o=Split(r.Replace("A 2 3 4 5 6 7 8 9 10 J Q K","$1S$1H $1D$1C")&" J J")
End Sub
Note: The terminal " on line 4 is not counted in the bytecount as it is not necessary and is only included for SO's syntax highlighting
# 8th, 156152 149 bytes
Code
[] ( >s a:push ) 2 10 loop ["J","Q","K","A"] a:+ ( nip dup ["C","D","S","H"] ( nip s:+ >r dup ) a:each drop ) a:each ["J","J"] ( r> a:push ) 52 times
Array result is available on TOS
Output
["J","J","AH","AS","AD","AC","KH","KS","KD","KC","QH","QS","QD","QC","JH","JS","JD","JC","10H","10S","10D","10C","9H","9S","9D","9C","8H","8S","8D","8C","7H","7S","7D","7C","6H","6S","6D","6C","5H","5S","5D","5C","4H","4S","4D","4C","3H","3S","3D","3C","2H","2S","2D","2C"]
Ungolfed code
\ make array ["2","3","4","5","6","7","8","9","10"]
[] ( >s a:push ) 2 10 loop
\ make array ["2","3","4","5","6","7","8","9","10","A","J","Q","K"]
["J","Q","K","A"] a:+
\ make 13 cards for each suit and store each card on r-stack
( nip dup ["C","D","S","H"] ( nip s:+ >r dup ) a:each drop ) a:each
\ make array result reading previous cards from r-stack
["J","J"] ( r> a:push ) 52 times
\ array result is available on TOS. Display it and get rid of intermediate stuff
. cr
reset
# MATLAB R2018a, 82 57 bytes
Thanks to @David:
s=["A" 2:10 "J" "Q" "K"]+["D";"S";"H";"C"];[s(:);"J";"J"]
My old code:
X=["J","J"];for n=["A",2:10,"J","Q","K"]for s=["S","H","C","D"]X=[X,n+s];end;end;X
I figure it doesn't need an explanation but in case someone's interested:
X=["J","J"] // Set up first two Jokers
for n=["A",2:10,"J","Q","K"] // Set up array for A, 2, 3... to K
for s=["S","H","C","D"] // Set up suits
X=[X,n+s]; // "Glue" value and suit together and add to array
end
end
X // Print out the result
Output:
X =
1×54 string array
Columns 1 through 11
"J" "J" "AS" "AH" "AC" "AD" "2S" "2H" "2C" "2D" "3S"
Columns 12 through 22
"3H" "3C" "3D" "4S" "4H" "4C" "4D" "5S" "5H" "5C" "5D"
Columns 23 through 33
"6S" "6H" "6C" "6D" "7S" "7H" "7C" "7D" "8S" "8H" "8C"
Columns 34 through 43
"8D" "9S" "9H" "9C" "9D" "10S" "10H" "10C" "10D" "JS"
Columns 44 through 54
"JH" "JC" "JD" "QS" "QH" "QC" "QD" "KS" "KH" "KC" "KD"
• +1, I didn't know about using " for strings! This can be shortened in newer version of Matlab (=>2016b) s=["A" 2:10 "J" "Q" "K"]+["D";"S";"H";"C"];[s(:);"J";"J"] Commented Jun 13, 2018 at 21:20
# Bash (if Unicode is allowed), 45 bytes
echo -e \\U1F0{{A..D}{{1..9},{A..E}},{CF,DF}}
...produces:
🂡 🂢 🂣 🂤 🂥 🂦 🂧 🂨 🂩 🂪 🂫 🂬 🂭 🂮 🂱 🂲 🂳 🂴 🂵 🂶 🂷 🂸 🂹 🂺 🂻 🂼 🂽 🂾 🃁 🃂 🃃 🃄 🃅 🃆 🃇 🃈 🃉 🃊 🃋 🃌 🃍 🃎 🃑 🃒 🃓 🃔 🃕 🃖 🃗 🃘 🃙 🃚 🃛 🃜 🃝 🃞 🃏 🃟
Try it online!
This should work wherever Unicode is supported (ex: LANG=en_US.UTF8, not LANG=C). I'm not sure, however, if this meets the conditions of the rules. In particular:
For each combination of suit and value, there should be one item in the array, which is a string, and is made up of the value followed by the suit (Whitespace between these is permitted).
The Unicode answer has one string per element, and the images indeed show a value followed by a suit. The example output, however, shows two separate chars per element, so there's room for doubt.
If the Unicode approach is disallowed, the code can be easily trimmed down to 34 bytes:
d=({A,K,Q,J,{10..2}}{S,H,D,C} J J)
Try it online!
...but that is no shorter than the existing answer by glenn jackman (albeit sorted).
Regardless, Unicode is a technique worth mentioning. If you're dealing out a deck of cards, you might as well use the nice little images which are probably already on your computer, even in a text-mode terminal.
As for the question about the rules, I think the best answer is for someone to develop a question format which intrinsically includes an automated unit test. Maybe someday...
• Welcome to the site. Unlike other forums all answers must include a complete solution to the problem, and I think you are correct that this does not technically fit the specifications of the problem. If you would like to add a complete solution to your answer you can do that (and keep your partial solution as a bonus) but without a complete solution answers are subject to deletion. Commented Jul 9, 2018 at 17:05
• Anyways, the unicode solution you have it prints out the knight card. Commented Jul 23, 2018 at 22:57
# Gol><>, 46 44 bytes
"CDSH"4F:j9FL2+noao|j"AKQJ"4F$ooao||S"JnJ"; Try it online! Edit: Saved 2 bytes thanks to Jo King Explanation: "CDSH"4F:j9FL2+noao|j"AKQJ"4F$ooao||S"JnJ";
"CDSH" //Init the 4 suites
4F //Loop to iterate over the 4 suites
:j //Pop a copy of the current suite into the next loop
9FL2+noao| //Loop for the output of cards 2-10
j //Pass current suite into next loop & pop it from the stack
"AKQJ" //Init the 4 pictures
4F$ooao| //Loop 4 times to output the picture cards | //Jump back to the for loop to iterate over all suites S"JnJ"; //Output the two J and terminate programm # C (gcc), 86 bytes f(i){for(i=54;i--;)printf(i<52?"1%c%c "+(i>3):"J ","023456789JQKA"[i/4],"HSDC"[i%4]);} Try it online! # MASM 5 MACRO (x86 machine code), 109 100 bytes D LABEL BYTE IRPC r,234567890JQKA IRPC s,DSHC IF'&r'EQ'0' DB'1' ENDIF DB'&r','&s ' ENDM ENDM DB'J J' Since machine code has no formal definition for "an array of variable length strings", this returns the elements as characters in contiguous memory, delimited by a separator byte (20H) with the label D as a pointer to the location in memory. Obviously, this would all be a lot simpler and smaller (76 bytes) if we could use 'T' for 10, which would make all cards exactly 2 bytes in size and simply be an indexable array of DW (double-words). That's your Kolmogorov rub I suppose! Here's a short test program that shows one possible way this could be utilized: MOV DX, OFFSET D MOV AH, 9 INT 21H MOV AH, 4CH INT 21H Output 2D 2S 2H 2C 3D 3S 3H 3C 4D 4S 4H 4C 5D 5S 5H 5C 6D 6S 6H 6C 7D 7S 7H 7C 8D 8S 8H 8C 9D 9S 9H 9C 10D 10S 10H 10C JD JS JH JC QD QS QH QC KD KS KH KC AD AS AH AC J J # Factor + poker, 64 bytes <deck> hand>card-names [ "T""10"replace ] map { "J""J"} append . Try it online! Factor has a poker vocabulary that gets most of the way there; just have to change T to 10 and add the jokers. # Rockstar, 129 99 bytes say "J J" cut "CHSD" in S while S roll S in X say 10+X cut "A23456789JQK" in V while V say roll V+X Try it here (Code will need to be pasted in) say "J\nJ" :Output the jokers cut "CHSD" in S :Split "CHSD" into array S while S :While S is not empty roll S in X : Pop the first element into variable X say 10+X : Output 10 appended with X cut "..." in V : Split "A23456789JQK" into array V while V : While V is not empty say : Output roll V : Pop the first element from V +X : Append X # CJam, 45 bytes "A"]8,2f+'0f+"10"a+"JQK"++"SDHC"m*::+["J"_]+p Try it online! # Jelly, 24 bytes ⁵ḊṾ€;“AJQK”p“HSDC”;⁾JJẎ€ Try it online! # JavaScript (Node.js), 126 bytes [..."SDHC"].map(a=>Array.apply(0,Array(13)).map((b,i)=>(i==0?'A':i>9?'JQK'[i-10]:i+1)+a)).reduce((a,b)=>a.concat(b),['J','J']) Try it online! # C# (.NET Core), 140 bytes x=>{var a=new string[54];a[0]=a[1]="J";int i=2;foreach(var s in"SDCH")foreach(var n in"023456789AJQK")a[i++]=(n=='0'?"10":n+"")+s;return a;} Try it online! -10 bytes thanks to Kevin Cruijssen -1 byte by taking unused input -18 bytes thanks to Corak • You can remove the {} of the foreach, and .ToString() can be +"". Commented Jun 11, 2018 at 12:10 • 116 bytes by creating a port of my Java answer: o=>(new System.Text.RegularExpressions.Regex("(1?.)").Replace("AJQK2345678910","$1H,$1S,$1D,$1C,")+"J,J").Split(','). Credit to @mazzy's comment. Commented Jun 11, 2018 at 12:17 • @KevinCruijssen huh, I thought I needed the {} because of the nested foreach. Thanks! I'm not going to use your java port given it's such a drastically different approach, I'll let you post that yourself. Also just realised I can take unused input to shave a byte Commented Jun 11, 2018 at 12:33 • I wonder if something could be done with new[]{"J","J"}.Concat(from s in "SDCH" from n in "023456789AJQK" select (n=='0'?"10":n+"")+s);, but the using alone would probably make it unpractical. Commented Jun 11, 2018 at 14:31 • @Corak Ah, you're right the .ToArray() should be added for this particular challenge, in which case it's 119 bytes. But the System.Collections.Generic isn't necessary in this case. As you can see in the TIO I linked, with just the using System.Linq; it works, so only that one should be counted towards to byte-count. Commented Jun 12, 2018 at 11:49 # Red, 88 bytes foreach s{HSDC}[foreach c['A 2 3 4 5 6 7 8 9 10 'J 'Q 'K][prin c print s]]prin['J"^/"'J] Try it online! ## As a list: # Red, 100 bytes d:[]foreach s{HSDC}[foreach c['A 2 3 4 5 6 7 8 9 10 'J 'Q 'K][append d rejoin[c s]]]append d["J""J"] Try it online! # Stax, 21 bytes Å|R╦¥Ä└╥¿‼≈┴2½#d☺Iwa▐ Run and debug it Explanation: 9{^$m"AJQK"+"HSDC"|*'JQPm Full program, unpacked
Stack ([] is array, "" string '' char): empty
9{ m Map each element in range [1, 9]
^ Increment
$Stringify Stack: ["2", "3", "4", "5", "6", "7", "8", "9", "10"] "AJQK"+ Add "AJQK" Stack: ["2", "3", "4", "5", "6", "7", "8", "9", "10", 'A', 'J', 'Q', 'K'] "HSDC" Push suits Stack: "HSDC" ["2", "3", "4", "5", "6", "7", "8", "9", "10", 'A', 'J', 'Q', 'K'] |* Cartesian join Stack: [["2", 'H'], ["2", 'S'], ..., "AH", "AS", ...] 'J Push "J" Stack: "J" [...] Q Peek and print with newline Stack: "J" [...] P Pop and print with newline Stack: [...] m Map, implicitly print each with a newline # SimpleTemplate, 69 bytes This is a template-like language I've been developing, all written in PHP. The instructions on how to use it are available in the Github page. J J{@setx"SDHC","23456789JQKA"}{@eachx.1 asN}{@eachx.0}{@echo" ",N,_} ## Ungolfed: J J {@set chars "SDHC", "23456789JQKA"} {@each chars.1 as number} {@each chars.0 suite} {@echo " ", number, suite} {@/} {@/} Due to a series of bugs, this is not the most optimal solution. A shorter (non-working but should work) version below (51 bytes): J J{@each"23456789JQKA"asN}{@each"SDHC"} {@echoN,_} This, however, doesn't work. # vim, 80 bytes aoA J Q K<C-V><ESC>:r !seq 2 10<C-V><CR>o<C-V><ESC>V10kJ:s/ /H /g<ESC> yy3P34lrSjrDjrC:g/./:norm dd@" oJ J<ESC> <C-V> is 0x16. <CR> is 0x0d. <ESC> is 0x1B ### Annotated a " insert vim code into the buffer that: oA J Q K<C-V><ESC> " generates face cards :r !seq 2 10<C-V><CR>o<C-V><ESC> " generates 2-10 V10kJ " joins into one line :s/ /H /g<ESC> " adds suit (hearts) yy3P " make 3 copies of code 34lrSjrDjrC " change suit for each copy :g/./:norm dd@" " replace each line with the results from running it o J J<ESC> " Jokers Try it online! # Google Sheets, 86 Bytes A worksheet function that returns an array of "cards" to the the calling cell and the 53 cells to its right. =Split(RegExReplace("A 2 3 4 5 6 7 8 9 10 J Q K","(\w+)","$1S $1H$1D $1C ")&" J J"," Note: This function ends with a trailing space # Twig, 83 82 bytes J J{%for I in(2..9)|merge('AKQJ'|split)%} {{I~'SDHC'|split|join(' '~I)}}{%endfor%} This outputs the suits grouped by the "number". You can test it on: https://twigfiddle.com/lz9cuf Old version: # Julia, 63 bytes [["1"^(k<'1')*"$k$z"for k="A234567890JQK"for z="HSDC"];"J";"J"] ## Explanation [ [ "1"^(k<'1') # "1" if k is lexicographically before '1' # which here only happens if k is '0' *"$k$z" # concatenated to k and z for k="A234567890JQK" # for k in ranks for z="HSDC" # and z in suits ] ;"J" # then concatenate 2 jokers ;"J" ] # Yabasic, 123 bytes Dim r$(54)
r$(1)="J" r$(2)="J"
For i=0To 51
r$(i+3)=Mid$("JQKA2345678910",1+i/4,1+(i/4>=12))+Mid$("CDHS",Mod(i,4)+1,1) Next Try it online! # Groovy, 6866 60 bytes This is my first stab at a golf challenge, hope I get everything right. 68 byte version [*2..10,*'AKQJ'.grep()].collectMany{n->'HSDC'.collect{n+it}}+['J']*2 66 byte version [*2..10,*'AKQJ'.grep()].inject([]){a,e->a+'HSDC'*.plus(e)}+['J']*2 Try it here 60 byte version 'HSDC'.grep().sum{[*2..10,*'AKQJ'.grep()]*.plus(it)}+['J']*2 Try it here • Welcome to Code Golf! Nice work! One thing, tho, it's usually preferable to add a link to an online interpreter so folks can verify your solution works. Here's one for your solution: groovyconsole.appspot.com/script/5194941521199104 (You might like to add that to your answer). Commented Jun 14, 2018 at 15:29 # Cjam, 29 26 bytes B,2>"AJQK"+"SDHC"m*"JJ"+N* Try it online! Explanation: B,2> e# Generate the range [2, 10] "AJQK"+ e# add the list with "AJQK" e# which generates all the 13 cards "SDHC" e# Push the variants m* e# Cartesian product "JJ"+ e# Add 2 J's to the list N* e# And join the list items with newlines Note: 1. The previous count was incorrect; it should be 29 instead of 28. 2. The challenge said the order doesn't matter, so 3 bytes off! # Plain TeX, 348 bytes It's just text printed as if it were an array, but that is close enough I think. It's not like TeX has native arrays anyway (although I do use some form of arrays in the code). Short answer: \newcount\i\let\c\csname\let\e\endcsname\let\t\the\def~#1,#2,#3{\expandafter\def\c#2\t#1\e{#3}}\i3 ~\i,a,K~\i,s,D\i4 ~\i,a,Q~\i,s,C\i5 ~\i,a,J~\i,s,S\i2 ~\i,a,A~\i,s,H\newcount\v[\loop\v2{\loop"\t\v\c s\t\i\e", \ifnum\v<10\advance\v1\repeat}{\loop"\c a\t\v\e\c s\t\i\e", \ifnum\v<5\advance\v1\repeat}\break\ifnum\i<5\advance\i1\repeat "J", "J"]\bye Long answer, with comments as explanation: %First shorten some commands \let\c\csname \let\e\endcsname \let\t\the %then define the function todefine and fill the arrays \def ~#1,#2,#3{\expandafter\def\c #2\t#1\e{#3}} %main counter in this program \newcount\i \i3 %array a contains A, K, Q, J, the 'picture cards' %array s contains D, C, S, H, the suits %indexing is 2 to 5 so the same counter as for the 2...10 can be used ~\i,a,K ~\i,s,D \i4 ~\i,a,Q ~\i,s,C \i5 ~\i,a,J ~\i,s,S %2 is the last one as we want i to be 2 after this \i2 ~\i,a,A ~\i,s,H %the other counter \newcount\v %outer loop loops over the suits [\loop %2 is the starting point for the arrays and the numbers \v2 %brackets to ensure local use of \v %first loop iterates over the numbers { \loop %this prints v and the current suit "\t\v \c s\t\i\e", \ifnum\v<10 %\advance \v by 1 \advance\v1 \repeat } %...second loop over the picture cards { \loop "\c a\t\v\e \c s\t\i\e", \ifnum\v<5 \advance\v1 \repeat } %to prevent the output escaping the file on the right border \break \ifnum\i<5 \advance\i1 \repeat %then print the jokers "J", "J" ] \bye # Pyth - 32 bytes K+r2hTc"JQKA"1\J\JJ"SDHC"VJVK+HN Explanation: K+r2hTc"JQKA"1\J\JJ"SDHC"VJVK+HN K Assign variable K to r2hT Range from 2 to 10 plus 1 + Plus c"JQKA"1 Length 1 substrings of "JQKA" \J\J Print 1 character string "J" twice J"SDHC" Assign J to "SDHC" VJ For N in J VK For H in K +HN Implicitly print H+N # PicoLisp, 181 171 bytes (replaced ifn with if) (setq c '(A 2 3 4 5 6 7 8 9 10 J Q K)) (de z (s l)(if (<> l NIL) (cons (pack (car l) s) (z s (cdr l))))) (setq d (conc (z 'C c) (z 'D c) (z 'H c) (z 'S c) (list "J" "J"))) Sample output: bash$ pil +
` | 8,018 | 21,410 | {"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} | 2.859375 | 3 | CC-MAIN-2024-26 | latest | en | 0.936755 |
https://www.keyword-suggest-tool.com/search/motor+ground+wire+sizing/ | 1,596,996,581,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00458.warc.gz | 733,378,503 | 8,687 | # Motor ground wire sizing
Motor ground wire sizing keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
## Keyword Suggestions
Sizing motor ground wire
Motor ground wire sizing per nec
( Please select at least 2 keywords )
#### Most Searched Keywords
## Websites Listing
We found at least 10 Websites Listing below when search with motor ground wire sizing on Search Engine
### Ground Wire Size Calculator
The Ground Conductor Size Calculator will calculate the proper ground conductor size for grounding raceways and equipment based on ampere rating or setting of automatic overcurrent protection device in circuit ahead of equipment. This is based on NEC (NFPA 70®: National Electrical Code® (NEC®), 2014 Edition) Table 250.122.
http://wiresizecalculator.net/calculators/groundconductor.htm
DA: 22 PA: 32 MOZ Rank: 54
### Motor Wire Size Calculator
Motor wire size and circuit protection calculator. A good field reference for maintenance and troubleshooting is Ugly's Electric Motor and Controls (Ugly's Electric Motors And Controls, 2014 Edition).This calculator is based on information found in this reference guide, and can help electricians troubleshoot maintenance and control circuits.
http://wiresizecalculator.net/calculators/motorwiresizecalculator.htm
DA: 22 PA: 40 MOZ Rank: 62
### Ground Wire Size Chart - Wire Size Calculator
Ground Wire Size Table Source: NFPA 70, National Electrical Code, Table 250-95 Rating or Setting of Automatic Overcurrent Device in Circuit Ahead of Equipment, Conduit, etc.
http://wiresizecalculator.net/tables/groundconductortable.htm
DA: 22 PA: 32 MOZ Rank: 54
### Motor frame grounding conductor sizing - Electric motors ...
The NEC states that equipment grounding conductor should be sized according to the rating of the overcurrent device feeding the equipment. It seems like the logic is that in event of a fault within the motor causing ground current to flow in equipment ground conductor, we want the cable to be protected over a range of time and currents that is bounded by the protection.
DA: 16 PA: 15 MOZ Rank: 31
### NEC Code for Size of Cable for Motor - EEP
NEC Code 430.6 (A) (Size of Cable for Group of Motors or Elect.Load) Cables or Feeder which is supplying more than one motors other load(s), shall have an ampacity not less than 125 % of the full-load current rating of the highest rated motor plus the sum of the full-load current ratings of all the other motors in the group, as determined by 430.6(A).
https://electrical-engineering-portal.com/nec-code-for-size-of-cable-for-motor
DA: 33 PA: 37 MOZ Rank: 70
### Motor Data - Three-Phase AC Motors - Wire Size Calculator
Note: The wire size in this table is based on 75°C terminations and insulation. See Maximum Allowable Ampacities for Conductors in Raceway, Cable or Earth (30°C) for other insulation ratings.. Motor Wire Sizing. The NEC requires circuits supplying single motors to have an ampacity rating of greater than or equal to 125% of the full-load current rating of the motor.
http://wiresizecalculator.net/tables/threephasemotordata.htm
DA: 22 PA: 31 MOZ Rank: 53
### Wire Size Calculator
– The calculator allows entry of a voltage drop, but caution should be used when doing so to make sure you are calculating the wire sizes in accordance with NEC article 210-19 (FPN No. 4.) The NEC allows a maximum of a 3% voltage drop on the main branch of a circuit at the farthest outlet of power and 5% total to both feeders and branch ...
DA: 17 PA: 18 MOZ Rank: 35
### BASIC NEC CODE RULES AND DESIGN PRACTICE
Load: Motor 3 HP, 1 phase, 230V run ground wire with all motors I = 17A NEC 430=148 Wire Size = 17A = 21.5A .8 NEC 310-16 #10 is next size wire good for 30 Amp CB = (I)(2 to 2.5) NEC 430-152 CB Size = (17 A) (2) = 34A USE 35A circuit breaker. Conduit Size 3 #10, (2 power,1 ground), Table 3A, THW, 1/2" conduit will take 4 #10 Example
https://www.keyword-suggest-tool.com
DA: 28 PA: 28 MOZ Rank: 28
### Online Wire Size Calculators & Tables
Voltage - Enter the voltage at the source of the circuit. Single-phase voltages are usually 115V or 120V, while three-phase voltages are typically 208V, 230V or 480V. Amperes - Enter the maximum current in amps that will flow through the circuit. For motors, it is recommended to multiply the nameplate FLA by 1.25 for wire sizing.
http://wiresizecalculator.net/
DA: 22 PA: 22 MOZ Rank: 23 | 1,130 | 4,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-34 | latest | en | 0.819408 |
https://programmer.help/blogs/bzoj2400-spoj-839-optimal-marks-minimum-secant.html | 1,685,753,467,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00226.warc.gz | 502,260,084 | 4,587 | # BZOJ2400:[Spoj 839]Optimal Marks (Minimum secant)
Portal
Topic:
Given an undirected graph, the point values of some points have already been given, and the point values of some points can be freely chosen.Defines an edge with a weight of two points XOR.These points are required to be assigned a value so that the weights of the last edge are the smallest, and the point values at this time are output. If there are multiple schemes that make the weights of the edge the smallest, then the output makes the point values and the smallest schemes.
Questions:
That's a good question.
First, XOR bits are independent and can be calculated separately.
It is found that for each consideration, the equivalent is to assign a value of 0,1 to an undetermined point, so that the two adjacent points are different or minimized.
The original problem is essentially equivalent to dividing points into S,T sets, where the contribution of edges between points of different sets is 1, and then becomes the template of the smallest cut.
Then there's how to handle point values and minimizations.Similarly, for the residual network shrink point, it is found that the strong connection of S set at this time is the minimum point set, so it is good to dfs the origin directly.
```#include<bits/stdc++.h>
using namespace std;
inline int rd(){
char ch=getchar();int i=0,f=1;
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){i=(i<<1)+(i<<3)+ch-'0';ch=getchar();}
return i*f;
}
const int Maxn=5e2+50,Maxm=6e3+50;
int n,m,lev[Maxn],cur[Maxm],vis[Maxn],vt,a[Maxn],a2[Maxn],pos;
queue<int>q;
long long ans;
const int src=0,des=501,INF=0x3f3f3f3f;
struct G{
int g[Maxn],v[Maxm],c[Maxm],nt[Maxm],ec;
G(){ec=1;}
inline void add(int x,int y,int o){++ec;nt[ec]=g[x];g[x]=ec;v[ec]=y;c[ec]=o;}
inline void init(){
}
inline bool bfs(){
while(!q.empty())q.pop();
for(int i=0;i<=n;i++)lev[i]=0;
lev[des]=0;lev[src]=1;q.push(src);
while(!q.empty()){
int u=q.front();q.pop();
for(int j=g[u];j;j=nt[j]){
if(!c[j]||lev[v[j]])continue;
lev[v[j]]=lev[u]+1;q.push(v[j]);
if(v[j]==des)return true;
}
}
return false;
}
inline int dinic(const int &x,int f){
if(x==des)return f;
int rs=0;
for(int &j=cur[x];j;j=nt[j]){
if(!c[j]||(lev[v[j]]!=lev[x]+1))continue;
int o=dinic(v[j],min(f-rs,c[j]));
rs+=o;c[j]-=o;c[j^1]+=o;
if(rs==f)return rs;
}
return (lev[x]=0,rs);
}
inline void dfs(int now){
vis[now]=vt;a2[now]|=(1<<pos);
for(int j=g[now];j;j=nt[j]){
if(!c[j])continue;
if(vis[v[j]]!=vt)dfs(v[j]);
}
}
inline int maxflow(){
int rs=0;
while(bfs()){
int t;memcpy(cur,g,sizeof(cur));
while((t=dinic(src,INF)))rs+=t,memcpy(cur,g,sizeof(cur));
}
++vt;dfs(src);return rs;
}
}g,g2;
int main(){
n=rd(),m=rd();
for(int i=1;i<=n;i++)a[i]=rd();
for(int i=1;i<=m;i++){
int x=rd(),y=rd(); | 864 | 2,759 | {"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-2023-23 | latest | en | 0.676686 |
https://www.airmilescalculator.com/distance/crl-to-krk/ | 1,621,068,430,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00498.warc.gz | 672,714,338 | 41,992 | # Distance between Brussels (CRL) and Kraków (KRK)
Flight distance from Brussels to Kraków (Brussels South Charleroi Airport – Kraków John Paul II International Airport) is 678 miles / 1092 kilometers / 590 nautical miles. Estimated flight time is 1 hour 47 minutes.
Driving distance from Brussels (CRL) to Kraków (KRK) is 803 miles / 1292 kilometers and travel time by car is about 12 hours 2 minutes.
## Map of flight path and driving directions from Brussels to Kraków.
Shortest flight path between Brussels South Charleroi Airport (CRL) and Kraków John Paul II International Airport (KRK).
## How far is Kraków from Brussels?
There are several ways to calculate distances between Brussels and Kraków. Here are two common methods:
Vincenty's formula (applied above)
• 678.484 miles
• 1091.914 kilometers
• 589.587 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 676.386 miles
• 1088.537 kilometers
• 587.763 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Brussels South Charleroi Airport
City: Brussels
Country: Belgium
IATA Code: CRL
ICAO Code: EBCI
Coordinates: 50°27′33″N, 4°27′13″E
B Kraków John Paul II International Airport
City: Kraków
Country: Poland
IATA Code: KRK
ICAO Code: EPKK
Coordinates: 50°4′39″N, 19°47′5″E
## Time difference and current local times
There is no time difference between Brussels and Kraków.
CEST
CEST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 122 kg (269 pounds).
## Frequent Flyer Miles Calculator
Brussels (CRL) → Kraków (KRK).
Distance:
678
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
678
Round trip? | 493 | 1,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-21 | latest | en | 0.779061 |
http://perplexus.info/show.php?pid=9423&cid=54577 | 1,539,818,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511314.51/warc/CC-MAIN-20181017220358-20181018001858-00443.warc.gz | 280,992,335 | 4,351 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Date Deduction II (Posted on 2015-02-12)
Not counting Saturdays and Sundays, determine the date occurring precisely 1000 days after January 1, 2015. How about precisely 2015 days after January 1, 2015?
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
solution Comment 1 of 1
As we're counting only weekdays, 1000 days is 200 weeks. Once we restore the weekends, that's 1400 days or about 3.83 years. Three years which include one leap year (2016) contain 1096 days, leaving 304 days beyond Jan. 1, 2018. That's about 10 months.
31+28+31+30+31+30+31+31+30+31 = 304 exactly
If we had to add zero days, it would be Jan. 1. But 304 more brings it to Nov. 1, 2018.
The second part, 2015 weekdays after Jan. 1, 2015:
That's 403 weeks or 2821 calendar days or about 7.72 years, including two leap years (2016 and 2020). Seven such years add to 7*365 + 2 = 2557 days, leaving 264 remaining of the 2821.
31+28+31+30+31+30+31+31 = 243
264 - 243 = 21
The date is Sept 21+1, 2022, i.e., Sept 22.
Posted by Charlie on 2015-02-12 15:59:49
Search: Search body:
Forums (0) | 401 | 1,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-43 | latest | en | 0.872243 |
https://ask.csdn.net/questions/661642 | 1,571,720,780,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987798619.84/warc/CC-MAIN-20191022030805-20191022054305-00422.warc.gz | 381,412,687 | 26,118 | Key Insertion
Description
As an employee of the Macrohard Company, you have been asked to implement the new data structure that would be used to store some integer keys.
The keys must be stored in a special ordered collection that can be considered as an array A, which has an infinite number of locations, numbered starting from 1. Initially all locations are empty. The following operation must be supported by the collection: Insert(L, K), where L is the location in the array and K is some positive integer value.
The operation must be processed as follows:
If A[L] is empty, set A[L] ← K.
If A[L] is not empty, perform Insert(L + 1, A[L]) and after that set A[L] ← K.
Given N integer numbers L1 , L2 , . . . , LN you have to output the contents of the array after a sequence of the following operations:
Insert(L1 , 1)
Insert(L2 , 2)
. . .
Insert(LN , N)
Input
The first line of the input file contains N --- the number of Insert operations and M --- the maximal position that can be used in the Insert operation (1 <= N <= 131 072, 1 <= M <= 131 072).
Next line contains N integer numbers L i that describe Insert operations to be performed (1 <= Li <= M ).
Output
Output the contents of the array after a given sequence of Insert operations. On the first line print W --- the number of the greatest location that is not empty. After that output W integer numbers --- A[1], A[2], . . . , A[W ]. Output zeroes for empty locations.
Sample Input
5 4
3 3 4 1 3
Sample Output
6
4 0 5 2 3 1
1个回答
Key Insertion
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K-th Number 怎么求解
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K-th Number
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GitHub开源的10个超棒后台管理面板
100 个网络基础知识普及，看完成半个网络高手
C语言实现推箱子游戏
Java 的每个基本类型都对应了一个包装类型，比如说 int 的包装类型为 Integer，double 的包装类型为 Double。基本类型和包装类型的区别主要有以下 4 点。
C语言这么厉害，它自身又是用什么语言写的？
SpringBoot注解梳理
2019年10月全国程序员工资统计，一半以上的职位5个月没招到人。
Java 网络爬虫，就是这么的简单
58道JavaScript题，看看你能全对不？
“我想学习人工智能与机器学习，该从何做起？”
Java 爬虫遇到需要登录的网站，该怎么办？ | 597 | 1,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-43 | longest | en | 0.695596 |
http://aamt.edu.au/Student-activities/National-Maths-Day/National-Maths-Day-2014/A-twist-on-Gardner-s-two-children-problem | 1,527,307,881,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867309.73/warc/CC-MAIN-20180526033945-20180526053945-00441.warc.gz | 5,766,249 | 7,975 | ## Featured resource
### Check the Clues A: Pre-Primary
Cooperative Group Problem Solving Cards for Mathematics
Members: $22.40 inc.GST Others:$ 28.00 inc.GST
Home > Student activities > National Maths Day > National Maths Day 2014 > A twist on Gardner's two children problem
# A twist on Gardner's two children problem
Gardner's two children problem:
Mr Jones has two children. The older child is a girl. What is the probability that both children are girls?
Mr Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
The twist: Foshee’s two children problem:
Mr Ng has two children. One is a boy born on a Tuesday. What is the probability that both children are boys?
Surely knowing that one child was born on a Tuesday has nothing to do with it! Surprisingly enough, the additional information about the day of the week of the birthday does make a difference to the probability.
Read a detailed explanation in the Australian Mathematics Teacher journal Volume 70 Number 2 2014 entitled 'Gardner's two children problems and variations: Puzzles with conditional probability and sample spaces'. | 258 | 1,160 | {"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} | 3.046875 | 3 | CC-MAIN-2018-22 | latest | en | 0.956222 |
http://r.prevos.net/euler-problem-24/ | 1,508,390,197,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00759.warc.gz | 301,428,065 | 15,748 | # Lexicographic Permutations: Euler Problem 24
Euler Problem 24 asks to develop lexicographic permutations which are ordered arrangements of objects in lexicographic order. Tushar Roy of Coding Made Simple has shared a great introduction on how to generate lexicographic permutations.
## Euler Problem 24 Definition
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
## Brute Force Solution
The digits 0 to 9 have $10! = 3628800$ permutations (including combinations that start with 0). Most of these permutations are, however, not in lexicographic order. A brute-force way to solve the problem is to determine the next lexicographic permutation of a number string and repeat this one million times.
nextPerm <- function(a) {
# Find longest non-increasing suffix
i <- length(a) while (i > 1 && a[i - 1] >= a[i])
i <- i - 1
# i is the head index of the suffix
# Are we at the last permutation?
if (i <= 1) return (NA)
# a[i - 1] is the pivot
# Find rightmost element that exceeds the pivot
j <- length(a)
while (a[j] <= a[i - 1])
j <- j - 1
# Swap pivot with j
temp <- a[i - 1]
a[i - 1] <- a[j]
a[j] <- temp
# Reverse the suffix
a[i:length(a)] <- rev(a[i:length(a)])
return(a)
}
numbers <- 0:9
for (i in 1:(1E6 - 1)) numbers <- nextPerm(numbers)
This code takes the following steps:
1. Find largest index $i$ such that $a_{i-1} < a_i$.
1. If no such index exists, then this is already the last permutation.
2. Find largest index $j$ such that $j \geq i$ and $a_j > a_{i-1}$.
3. Swap $a_j$ and $a_{i-1}$.
4. Reverse the suffix starting at $a_i$.
## Combinatorics
A more efficient solution is to use combinatorics, thanks to MathBlog. The last nine digits can be ordered in $9! = 362880$ ways. So the first $9!$ permutations start with a 0. By extending this thought, it follows that the millionth permutation must start with a 2.
$\lfloor (1000000 - 1) / 9! \rfloor = 2$
From this rule, it follows that the 725761st permutation is 2013456789. We now need 274239 more lexicographic permutations:
$(1000000 - 1) - (2 \times 9!) = 274239$
We can repeat this logic to find the next digit. The last 8 digits can be ordered in 40320 ways. The second digit is the 6th digit in the remaining numbers, which is 7 (2013456789).
$\lfloor 274239 / 8! \rfloor = 6$
$274239 - (6 \times 7!) = 32319$
This process is repeated until all digits have been used.
numbers <- 0:9
n <- length(numbers)
remain <- 1E6 - 1
for (i in 1:n) {
j <- floor(remain / factorial(n - i))
remain <- remain %% factorial(n - i)
numbers <- numbers[-(j + 1)]
} | 873 | 2,885 | {"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": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.720795 |
https://projects.coin-or.org/Coopr/browser/coopr.pyomo/trunk/coopr/pyomo/base/numvalue.py?rev=3071 | 1,620,709,693,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991641.5/warc/CC-MAIN-20210511025739-20210511055739-00311.warc.gz | 490,353,007 | 10,913 | # source:coopr.pyomo/trunk/coopr/pyomo/base/numvalue.py@3071
Last change on this file since 3071 was 3071, checked in by wehart, 11 years ago
Bug fixes to param management.
File size: 10.5 KB
Line
1# _________________________________________________________________________
2#
3# Coopr: A COmmon Optimization Python Repository
4# Copyright (c) 2008 Sandia Corporation.
6# Under the terms of Contract DE-AC04-94AL85000 with Sandia Corporation,
7# the U.S. Government retains certain rights in this software.
9# _________________________________________________________________________
10
11__all__ = ['value', 'NumericValue', 'as_numeric', 'NumericConstant', 'ZeroConstant', 'OneConstant', 'is_constant']
12
13import plugin
14#import pyutilib.component.core
15import pyutilib.math
16import sys
17from set_types import Reals
18
19
20##------------------------------------------------------------------------
21##
22## Standard types of expressions
23##
24##------------------------------------------------------------------------
25
26def create_name(name, ndx):
27 if ndx is None:
28 return name
29 if type(ndx) is tuple:
30 tmp = str(ndx).replace(', ',',')
31 return name+"["+tmp[1:-1]+"]"
32 return name+"["+str(ndx)+"]"
33
34def value(obj):
35 """
36 A utility function that returns the value of an expression
37 # NOTE: This is a terminal class, so no need to pop the keyword/argument
38 # I would actually argue that we just need to make the inputs named
39 # keywords, to by-pass all of this overhead. """
40 if obj is None:
41 return None
42 elif type(obj) in [bool,int,long,float,str]:
43 return obj
44 if not isinstance(obj, NumericValue):
45 raise ValueError, "Object "+str(obj)+" is not a NumericValue object"
46 tmp = obj()
47 if tmp is None:
48 raise ValueError, "No value for uninitialized NumericValue object "+obj.name
49 return tmp
50
51def is_constant(obj):
52 """
53 A utility function that returns a boolean that indicates whether the object is a constant
54 """
55 if obj is None:
56 return False
57 elif type(obj) in [bool,int,long,float]:
58 return True
59 if not isinstance(obj, NumericValue):
60 return False
61 return obj.is_constant()
62
63def as_numeric(obj):
64 """Verify that this obj is a NumericValue or intrinsic value"""
65 if isinstance(obj,NumericValue):
66 return obj.as_numeric()
67 if type(obj) in [bool,int,long,float]:
68 return NumericConstant(value=obj)
69 raise ValueError, "This object is not a numeric value: "+str(obj)
70
71
72class NumericValue(object):
73 """An object that contains a numeric value"""
74
75 __hash__ = None
76
77 def __init__(self, **kwargs):
78
79 self.name = 'unknown'
80 self.domain = None
81
82 # a value of None indicates unassigned; None is also
83 # (treated as) a member of all domains.
84 value = None # validated prior to assigning to self.value
85
86 # because this is a terminal class (not passing keyword arguments
87 # further to the 'object' base class), there is a good reason in terms
88 # of performance - given how often NumericValue objects are created -
89 # to avoid the usual technique of invoking kwargs.pop(name,default).
90 # manipulating the keyword dictionary takes too much time.
91 for keyword, keywordvalue in kwargs.items():
92 if keyword is 'value':
93 value = keywordvalue
94 elif keyword is 'within':
95 self.domain = keywordvalue
96 elif keyword is 'domain':
97 self.domain = keywordvalue
98 elif keyword is 'name':
99 self.name = keywordvalue
100 # this class can't deal with any other keywords - in principle,
101 # we could throw an exception for debug/diagnostic purposes.
102
103 if (value and pyutilib.math.is_nan(value)):
104 value = pyutilib.math.nan
105
106 # a numeric value is only constrained by the domain,
107 # if specified. avoid invoking set_value() here to
108 # accelerate object construction, which takes
109 # significant time due to the typical number of
110 # occurrences. similarly, we are avoiding the
111 # invocation of _valid_value.
112 if (value is not None) and (self.domain is not None) and (value not in self.domain):
113 raise ValueError, "Numeric value `"+str(value)+"` is not in domain "+str(self.domain)
114
115 self.value = value
116
117 self.model = None
118
119 def is_constant(self):
120 return True
121
122 def fixed_value(self):
123 return False
124
125 def as_numeric(self):
126 return self
127
128 def pprint(self, ostream=None):
129 raise IOError, "NumericValue:pprint is not defined" #pragma:nocover
130
131 def display(self, ostream=None):
132 self.pprint(ostream=ostream)
133
134 def reset(self): #pragma:nocover
135 pass #pragma:nocover
136
137 def set_value(self, val):
138 if self._valid_value(val):
139 self.value=val
140
141 def __call__(self, exception=True):
142 return self.value
143
144 def __float__(self):
145 tmp = self.__call__()
146 if tmp is None:
147 raise ValueError, "Cannot coerce variable `"+self.name+"' to float because it is uninitialized."
148 return float(tmp)
149
150 def __int__(self):
151 tmp = self.__call__()
152 if tmp is None:
153 raise ValueError, "Cannot coerce variable `"+self.name+"' to integer because it is uninitialized."
154 return int(tmp)
155
156 def _valid_value(self,value,use_exception=True):
157 #print "HERE X", self.domain is None
158 ans = value is None or self.domain is None or value in self.domain
159 if not ans and use_exception:
160 raise ValueError, "Numeric value `"+str(value)+"` is not in domain "+str(self.domain)
161 return ans
162
163 def X__getattr__(self,name):
164 #print "GETATTR",name
165 #d = self.__dict__
166 try:
167 return self.__dict__[name]
168 except:
169 pass
170 try:
171 return self.__dict__["_attr_"+name]
172 except:
173 pass
174 raise AttributeError, "Unknown attribute `"+str(name)+"' for object with type "+str(type(self))
175
176 def X__setattr__(self,name,val):
177 ##print "SETATTR",name,val
178 if name == "__class__":
179 self.__class__ = val
180 return
181 if name[0] == "_":
182 self.__dict__[name] = val
183 return
184 if name in self._standard_attr:
185 self.__dict__["_attr_"+name] = val
186 return
187 raise AttributeError, "Unallowable attribute `"+name+"`"
188 #self._standard_attr[name] = val
189
190 def __lt__(self,other):
191 """Less than operator
192
193 (Called in response to 'self < other' or 'other > self'.)
194 """
195 return plugin.ExpressionFactory('<', [self,as_numeric(other)])
196
197 def __gt__(self,other):
198 """Greater than operator
199
200 (Called in response to 'self > other' or 'other < self'.)
201 """
202 return plugin.ExpressionFactory('>', [self,as_numeric(other)])
203
204 def __le__(self,other):
205 """Less than or equal operator
206
207 (Called in response to 'self <= other' or 'other >= self'.)
208 """
209 return plugin.ExpressionFactory('<=', [self,as_numeric(other)])
210
211 def __ge__(self,other):
212 """Greater than or equal operator
213
214 (Called in response to 'self >= other' or 'other <= self'.)
215 """
216 return plugin.ExpressionFactory('>=', [self,as_numeric(other)])
217
218 def __eq__(self,other):
219 """Equal to operator
220
221 (Called in response to 'self = other'.)
222 """
223 return plugin.ExpressionFactory('=', [self,as_numeric(other)])
224
227
228 (Called in response to 'self + other'.)
229 """
231
232 def __sub__(self,other):
233 """ Binary subtraction
234
235 (Called in response to 'self - other'.)
236 """
237 return expr.generate_expression('sub',self,other)
238
239 def __mul__(self,other):
240 """ Binary multiplication
241
242 (Called in response to 'self * other'.)
243 """
244 return expr.generate_expression('mul',self,other)
245
246 def __div__(self,other):
247 """ Binary division
248
249 (Called in response to 'self / other'.)
250 """
251 return expr.generate_expression('div',self,other)
252
253 def __pow__(self,other):
254 """ Binary power
255
256 (Called in response to 'self ** other'.)
257 """
258 return expr.generate_expression('pow',self,other)
259
262
263 (Called in response to 'other + self'.)
264 """
266
267 def __rsub__(self,other):
268 """ Binary subtraction
269
270 (Called in response to 'other - self'.)
271 """
272 return expr.generate_expression('rsub',self,other)
273
274 def __rmul__(self,other):
275 """ Binary multiplication
276
277 (Called in response to 'other * self'.)
278 """
279 return expr.generate_expression('rmul',self,other)
280
281 def __rdiv__(self,other):
282 """ Binary division
283
284 (Called in response to 'other / self'.)
285 """
286 return expr.generate_expression('rdiv',self,other)
287
288 def __rpow__(self,other):
289 """ Binary power
290
291 (Called in response to 'other ** self'.)
292 """
293 return expr.generate_expression('rpow',self,other)
294
295 def __neg__(self):
296 """ Negation
297
298 (Called in response to '- self'.)
299 """
300 return expr.generate_expression('neg',self)
301
302 def __pos__(self):
303 """ Positive expression
304
305 (Called in response to '+ self'.)
306 """
307 return self
308
309 def __abs__(self):
310 """ Absolute value
311
312 (Called in response to 'abs(self)'.)
313 """
314 return expr.generate_expression('abs',self)
315
316
317class NumericConstant(NumericValue):
318 """An object that contains a constant numeric value"""
319
320 def __init__(self, **kwds):
321 if not 'domain' in kwds:
322 kwds['domain'] = Reals
323 NumericValue.__init__(self,**kwds)
324
325 def fixed_value(self):
326 return True
327
328 def __call__(self, exception=True):
329 return self.value
330
331 def __str__(self):
332 return str(self.value)
333
334 def __eq__(self,other):
335 """Equal to operator
336
337 (Called in response to 'self == other'.)
338 """
339 return self.value == other
340
341 def pprint(self, ostream=None):
342 if ostream is None:
343 ostream = sys.stdout
344 print >>ostream, str(self),
345
346 def simplify(self, model):
347 return self #pragma:nocover
348
349
350ZeroConstant = NumericConstant(value=0)
351OneConstant = NumericConstant(value=1)
352
353import expr
Note: See TracBrowser for help on using the repository browser. | 3,260 | 11,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-21 | latest | en | 0.20752 |
https://www.syrianspock.com/functional%20programming/software/2018/02/03/functional-cpp-adventures-function-composition-on-steroids/ | 1,566,136,121,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313889.29/warc/CC-MAIN-20190818124516-20190818150516-00152.warc.gz | 988,122,034 | 4,745 | # Functional Programming in C++: Composition
In the previous post, we discussed the optional idiom introduced to C++ starting from C++17. Although optional is not implemented as a monad in C++, we can treat it as such and implement the missing parts to achieve greater function composition.
• define types,
• compose them,
• apply functions to them.
This pattern is powerful as it enables better function composition, you’ll see why later, bear with me.
Many functions we write in our code can fail. We saw previously how three patterns can be used to express failure. Here we will focus on the solution involving optional because optional already allows us to define types, so that fulfills the first element in the description of the monad pattern I gave above. Let’s see how we can build up the other two missing elements of the monad pattern.
# Toy problem
Let’s say I want to implement the following mathematical function:
You might ask yourself why on Earth would do we need to compute such an evil function that is only valid for non-zero numbers between -1 and 1? Well.. life is unfair, get over it.
One approach to implementing this would be using a single function:
std::optional<double> my_single_function(double x)
{
if (x >= -1 && x <= 1 && x != 0) { return 1 / asin(x); }
else { return {}; }
}
and that looks fine.
# Breaking it down
For the sake of discussion, let’s assume I’m writing a math library and would like to reuse each building block (arcsine and inverse) with safe input domain checks. The function can be defined as the composition of two functions, the inverse and the arcsine:
The implementation can be broken down to reflect this:
std::optional<double> inverse(double x)
{
if (x != 0) { return 1 / x; }
else { return {}; }
}
std::optional<double> arcsine(double x)
{
if (x >= -1 && x <= 1) { return asin(x); }
else { return {}; }
}
std::optional<double> my_function(double x)
{
if (const auto y = arcsine(x)) { return inverse(y.value()); }
else { return {}; }
}
Well that looks uglier and less efficient. We introduced smaller building blocks which makes the code more modular, but function composition sucks. The return value of the first call to inverse needs to be checked, and only if it contains some value do we feed it to the following call to arcsine. There is too much noise introduced by the use of optional, and noise builds up as we chain functions, what can we do about it?
In Haskell, this problem could be solved by writing the following:
inverse :: Float -> Maybe Float
inverse 0 = Nothing
inverse x = Just(1 / x)
arcsine :: Float -> Maybe Float
arcsine x
| -1 <= x && x <= 1 = Just(asin x)
| otherwise = Nothing
f x = x >>= arcsine >>= inverse
Haskell provides the >>= operator for monads (including the Maybe monad) which is called the bind operator. This allows us to chain functions of type signature a -> M a: functions that take a type as input and return a type wrapped in a monad. Functions can be chained easily and if some step fails, the failure propagates through.
# Bind operator in C++
We can implement the bind operator for optional like this:
template <typename T, typename funcType>
auto operator>>=(const std::optional<T>& x, funcType &&f)
{
if (x) { return f(x.value()); }
else { return decltype(f(std::declval<T>())){}; }
}
Then we can redefine my_function more efficiently as:
std::optional<double> my_function(double x)
{
return (std::optional<double>(x) >>= arcsine) >>= inverse;
}
which looks a lot like Haskell. With the bind operator we can achieve more efficient function composition using the optional idiom. This fulfills the 2 last elements of the monad pattern as defined above.
# Summary
Monads are a very powerful pattern that brings function composition to the next level. Implementing the bind operator is crucial in using the monad pattern in C++. This is clearly not perfect, and mostly an attempt to understand how the monad pattern can make C++ code more readable and maintainable. A few interesting things to try next are:
• implementing more syntactic sugar (an equivalent to the do notation in Haskell),
• defining the bind operator for other templated types in C++ such as std::variant. | 967 | 4,303 | {"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} | 2.90625 | 3 | CC-MAIN-2019-35 | latest | en | 0.793996 |
https://www.physicsforums.com/threads/hubble-equation-function-of-time.870044/ | 1,508,257,437,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00494.warc.gz | 988,208,302 | 17,631 | # B Hubble equation function of time
1. May 2, 2016
### Einstein's Cat
Is there an equation which explains how the recessional velcoity of a galaxy changes with time? Furthermore, does an equation exist that describes the universe's accelerating expansion; is it the Friedman equation?
2. May 2, 2016
3. May 2, 2016
### Chronos
The concept of 'absolute' time has plagued cosmologists and philosophers since the 'beginning' of time. Einstein helped clear up the mystery with GR, but, the notion of absolute time continues to befuddle both mathematicians and philosophers.
4. May 3, 2016
### Einstein's Cat
How does proper distance change with time? Is there an equation to describe that? I looked on Wikipedia and found no such thing
5. May 3, 2016
### Chronos
From the perspective of physicists, the concepts of distance and time have no absolute meaning. Neither can exist independendent of the other. Hermann Minkowski is credited as the first to understand this odd state of affairs between space and time and coined the term spacetime to describe the interchangeability of these four fundamental dimensions of the universe back in 1908. Needless to say this greatly influenced the thinking of Einstein. See http://www.physicsoftheuniverse.com/topics_relativity_spacetime.html for a brief discussion.
6. May 3, 2016
### Einstein's Cat
I see, so proper distance changes with the expansion of the universe and thus with time because theoretical observers along it would appear to have relative velocity to a relative stationary observers? Therefore, meaning that the relatively moving observers would have a different interpretations of cosmological time and thus the proper distance would appear to be longer? Is this correct please?
Last edited by a moderator: May 4, 2016
7. May 3, 2016
### George Jones
Staff Emeritus
If $D$ is the (now) proper distance between us and a distant galaxy, then
$$D \left( t \right) = R \left( t \right) \chi,$$
where $R \left( t \right)$ is the scale factor for the universe (a solution to Friedmann's equation), $\chi$ is the constant comoving coordinate difference between us and the galaxy, and $t$ is cosmological time.
Einstein's Cat, are you familiar with calculus? Assuming you are, the rate of change of proper distance is given by
$$\frac{dD}{dt} = \frac{dR}{dt} \chi.$$
If the scale factor is known, then so is its rate of change $dR/dt$ , thus giving the rate change of proper distance, $dD/dt$.
Multiplying the left side of the above equation by one in the form $1=R/R$, and using $D = R \chi$ gives the Hubble relation
$$\frac{dD}{dt} = \frac{dR}{dt} \frac{R}{R} \chi = H D,$$
where the Hubble parameter (a function of time) is give by
$$H = R \frac{dR}{dt}.$$ | 687 | 2,737 | {"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-2017-43 | longest | en | 0.913926 |
https://www.theoptionsguide.com/price-to-earnings-ratio.aspx | 1,656,273,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271864.14/warc/CC-MAIN-20220626192142-20220626222142-00487.warc.gz | 1,099,426,160 | 10,282 | # Price-to-Earnings Ratio (P/E Ratio)
The price-to-earnings ratio, or simply P/E ratio, is a often used metric in stock valuation. Also known as earnings multiple, multiple, or simply p/e (or pe).
The P/E ratio is obtained by dividing the price per share by the earnings per share.
Earnings per share in this case refers to the last twelve months' earnings. The P/E ratio derived this way is also known as trailing P/E.
If we use next year's estimated earnings instead, we will be calculating the projected or forward P/E.
#### Interpreting the P/E Ratio
One simple way to understand P/E is that it gives the number of years the company will need to generate enough value to cover the cost the stock at the current market price (assuming no growth in earnings).
Like any business, the value of a stock is directly related to the company's ability to generate cash. Thus, in a sense, a lower price-earnings ratio often suggests value.
The P/E ratio also reflects the market's expectation regarding the future performance of the stock. Higher price-earnings ratio indicates higher expectations for the company.
Using the P/E ratio, we can compare the relative earning power of the companies regardless of their size or stock price.
On the surface, a \$50 stock may seem more expensive than a \$20 stock but if the \$50 stock earns \$5 a share while the \$20 stock earns only \$1, using the P/E ratio, you will be able to see that the \$20 stock is twice as expensive as the \$50 stock.
#### What is a good P/E ratio?
There maybe no such thing as a good price-to-earnings ratio. When P/E is high, one can either say its too expensive or argue that growth prospects are good. On the other hand, when P/E is low, one can say that it is a value play or that the company's future is not too bright.
Moreover, even the average P/E varies across companies in different industries.
#### Conclusion
The P/E ratio is just one of the many financial ratios that are used to evaluate stocks. As it is an often quoted metric, it seems like an all important ratio to many investors but in reality, it is definitely inadequate to base your investing decision on just this metric alone.
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### Investing in Growth Stocks using LEAPS® options
If you are investing the Peter Lynch style, trying to predict the next multi-bagger, then you would want to find out more about LEAPS® and why I consider them to be a great option for investing in the next Microsoft®.... [Read on...]
### Effect of Dividends on Option Pricing
Cash dividends issued by stocks have big impact on their option prices. This is because the underlying stock price is expected to drop by the dividend amount on the ex-dividend date....[Read on...]
### Bull Call Spread: An Alternative to the Covered Call
As an alternative to writing covered calls, one can enter a bull call spread for a similar profit potential but with significantly less capital requirement. In place of holding the underlying stock in the covered call strategy, the alternative....[Read on...]
### Dividend Capture using Covered Calls
Some stocks pay generous dividends every quarter. You qualify for the dividend if you are holding on the shares before the ex-dividend date....[Read on...]
### Leverage using Calls, Not Margin Calls
To achieve higher returns in the stock market, besides doing more homework on the companies you wish to buy, it is often necessary to take on higher risk. A most common way to do that is to buy stocks on margin....[Read on...]
Day trading options can be a successful, profitable strategy but there are a couple of things you need to know before you use start using options for day trading.... [Read on...]
### What is the Put Call Ratio and How to Use It
Learn about the put call ratio, the way it is derived and how it can be used as a contrarian indicator.... [Read on...]
### Understanding Put-Call Parity
Put-call parity is an important principle in options pricing first identified by Hans Stoll in his paper, The Relation Between Put and Call Prices, in 1969. It states that the premium of a call option implies a certain fair price for the corresponding put option having the same strike price and expiration date, and vice versa.... [Read on...]
### Understanding the Greeks
In options trading, you may notice the use of certain greek alphabets like delta or gamma when describing risks associated with various positions. They are known as "the greeks".... [Read on...]
### Valuing Common Stock using Discounted Cash Flow Analysis
Since the value of stock options depends on the price of the underlying stock, it is useful to calculate the fair value of the stock by using a technique known as discounted cash flow.... [Read on...]
#### Options Strategy Finder
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Risk Warning: Stocks, futures and binary options trading discussed on this website can be considered High-Risk Trading Operations and their execution can be very risky and may result in significant losses or even in a total loss of all funds on your account. You should not risk more than you afford to lose. Before deciding to trade, you need to ensure that you understand the risks involved taking into account your investment objectives and level of experience. Information on this website is provided strictly for informational and educational purposes only and is not intended as a trading recommendation service. TheOptionsGuide.com shall not be liable for any errors, omissions, or delays in the content, or for any actions taken in reliance thereon.
General Risk Warning: The financial products offered by the company carry a high level of risk and can result in the loss of all your funds. You should never invest money that you cannot afford to lose. | 1,402 | 6,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-27 | latest | en | 0.953941 |
https://www.andymatthewsphotography.com/what-is-difference-between-qualitative-and-quantitative-analysis/ | 1,701,203,118,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00043.warc.gz | 736,993,055 | 13,560 | # What is difference between qualitative and quantitative analysis?
## What is difference between qualitative and quantitative analysis?
Qualitative data analysis is based on classification of objects (participants) according to properties and attributes whereas quantitative analysis is based on classification of data based on computable values. Qualitative analysis is subjective whereas quantitative is objective.
## What are data analysis techniques?
The systematic application of statistical and logical techniques to describe the data scope, modularize the data structure, condense the data representation, illustrate via images, tables, and graphs, and evaluate statistical inclinations, probability data, to derive meaningful conclusions, is known as Data Analysis.
## What is the goal of qualitative analysis chemistry?
Qualitative analysis can tell you whether an atom, ion, functional group, or compound is present or absent in a sample, but it doesn’t provide information about its quantity. Quantification of a sample, in contrast, is called quantitative analysis.
## What is qualitative analysis sport?
Qualitative analysis is a method used to evaluate technique in the performance of sports (or exercise) skills. It relies on a knowledge of the relevant sport and sports skill, as well as an in-depth knowledge of ‘principles of movement’. Such analysis should be conducted using a chosen analysis model.
## What is qualitative data?
Qualitative data describes qualities or characteristics. It is collected using questionnaires, interviews, or observation, and frequently appears in narrative form. For example, it could be notes taken during a focus group on the quality of the food at Cafe Mac, or responses from an open-ended questionnaire.
## What is qualitative analysis?
Qualitative Analysis is the determination of non-numerical information about a chemical species, a reaction, etc. Examples would be observing that a reaction is creating gas that is bubbling out of solution or observing that a reaction results in a color change.
## How is qualitative analysis used in everyday life?
Forensic scientists may use qualitative chemical analysis to identify substances left at crime scenes, which can be vital in convicting criminals. Qualitative chemical analysis also may be used in the home. For instance, a person might tell the difference between margarine and butter in how quickly they melt.
## Is temperature qualitative or quantitative?
Quantitative Information – Involves a measurable quantity—numbers are used. Some examples are length, mass, temperature, and time. Quantitative information is often called data, but can also be things other than numbers. Qualitative Information – Involves a descriptive judgment using concept words instead of numbers.
## How qualitative research can be used in sports?
The results of qualitative research can inform stakeholders about facilitators and obstacles to exercise, motivation and adherence, the influence of experiences, beliefs, disability and capability on physical activity, exercise engagement and performance, and to test strategies that maximise physical performance.
## What is the importance of qualitative research in education?
Qualitative research is very important in educational research as it addresses the “how” and “why” research questions and enables deeper understanding of experiences, phenomena and context. Qualitative research allows you to ask questions that cannot be easily put into numbers to understand human experience.
## Why is qualitative analysis important?
Qualitative Analysis allows you to interpret the information in non-mathematical ways and often focuses on the how and why part of the survey.
## What is the process of qualitative data analysis?
Qualitative data analysis involves the identification, examination, and interpretation of patterns and themes in textual data and determines how these patterns and themes help answer the research questions at hand. Likely to change and adapt as the study evolves and the data emerges.
## What is quantitative analysis in sport?
Quantitative research is experimental and objective whereas qualitative research is explorative and is not in numerical form. Quantitative research is used to identify evidence of cause and effect relationships and is used to collect data from a larger population than qualitative research (Downey & Ireland, 1979).
## What are the steps of qualitative data analysis?
Qualitative data analysis requires a 5-step process:
• Prepare and organize your data. Print out your transcripts, gather your notes, documents, or other materials.
• Review and explore the data.
• Create initial codes.
• Review those codes and revise or combine into themes.
• Present themes in a cohesive manner. | 861 | 4,804 | {"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-2023-50 | longest | en | 0.926341 |
https://www.conversion-metric.org/length/hectometer-to-millimeter | 1,560,865,733,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998724.57/warc/CC-MAIN-20190618123355-20190618145355-00057.warc.gz | 716,008,568 | 9,284 | # Hectometer to Millimeter Conversion (hm to mm)
Please enter hectometer (hm) value of length unit to convert hectometer to millimeter.
1 hm = 100000 mm
Swap» Millimeter to Hectometer
hm: Hectometer, mm: Millimeter
## How Many Millimeter in a Hectometer?
There are 100000 millimeter in a hectometer.
1 Hectometer is equal to 100000 Millimeter.
1 hm = 100000 mm
### Hectometer Definition
A hectometer is one of the least commonly used in modern metric system unit of length or distance. One hectometer is equal to 100 meters or 1/10th of kilometer. Hectometer can be considered a practical unit for measuring small distances or the dimensions of relatively large objects like very large premises, large water reservoirs, small pool length, etc. Along with linear measurement, hectometer is sometimes used as a unit of volumetric measurement.
Convert Hectometer
### Millimeter Definition
Millimeter is one the most commonly used units of length which equals to 1/1000th of a meter. It is a unit of SI applicable for measuring small distances and small length. According to SI rules, the unit symbol is mm.
Convert Millimeter
### About hm to mm Converter
This is a very easy to use hectometer to millimeter converter. First of all just type the hectometer (hm) value in the text field of the conversion form to start converting hm to mm, then select the decimals value and finally hit convert button if auto calculation didn't work. Millimeter value will be converted automatically as you type.
The decimals value is the number of digits to be calculated or rounded of the result of hectometer to millimeter conversion.
You can also check the hectometer to millimeter conversion chart below, or go back to hectometer to millimeter converter to top.
### Hectometer to Millimeter Conversion Examples
```1 hm = 100000 Millimeter
Example for 3 Hectometer:
3 Hectometer = 3 (Hectometer)
3 Hectometer = 3 x (100000 Millimeter)
3 Hectometer = 300000 Millimeter
Example for 0.8 Hectometer:
0.8 Hectometer = 0.8 (Hectometer)
0.8 Hectometer = 0.8 x (100000 Millimeter)
0.8 Hectometer = 80000 Millimeter
Example for 0.0098 Hectometer:
0.0098 Hectometer = 0.0098 (Hectometer)
0.0098 Hectometer = 0.0098 x (100000 Millimeter)
0.0098 Hectometer = 980 Millimeter
```
### Hectometer to Millimeter Conversion Chart
HectometerMillimeter
1 hm100000 mm
2 hm200000 mm
3 hm300000 mm
4 hm400000 mm
5 hm500000 mm
6 hm600000 mm
7 hm700000 mm
8 hm800000 mm
9 hm900000 mm
10 hm1000000 mm
11 hm1100000 mm
12 hm1200000 mm
13 hm1300000 mm
14 hm1400000 mm
15 hm1500000 mm
16 hm1600000 mm
17 hm1700000 mm
18 hm1800000 mm
19 hm1900000 mm
20 hm2000000 mm
21 hm2100000 mm
22 hm2200000 mm
23 hm2300000 mm
24 hm2400000 mm
25 hm2500000 mm
26 hm2600000 mm
27 hm2700000 mm
28 hm2800000 mm
29 hm2900000 mm
30 hm3000000 mm
31 hm3100000 mm
32 hm3200000 mm
33 hm3300000 mm
34 hm3400000 mm
35 hm3500000 mm
36 hm3600000 mm
37 hm3700000 mm
38 hm3800000 mm
39 hm3900000 mm
40 hm4000000 mm
41 hm4100000 mm
42 hm4200000 mm
43 hm4300000 mm
44 hm4400000 mm
45 hm4500000 mm
46 hm4600000 mm
47 hm4700000 mm
48 hm4800000 mm
49 hm4900000 mm
50 hm5000000 mm
HectometerMillimeter
50 hm5000000 mm
55 hm5500000 mm
60 hm6000000 mm
65 hm6500000 mm
70 hm7000000 mm
75 hm7500000 mm
80 hm8000000 mm
85 hm8500000 mm
90 hm9000000 mm
95 hm9500000 mm
100 hm10000000 mm
105 hm10500000 mm
110 hm11000000 mm
115 hm11500000 mm
120 hm12000000 mm
125 hm12500000 mm
130 hm13000000 mm
135 hm13500000 mm
140 hm14000000 mm
145 hm14500000 mm
150 hm15000000 mm
155 hm15500000 mm
160 hm16000000 mm
165 hm16500000 mm
170 hm17000000 mm
175 hm17500000 mm
180 hm18000000 mm
185 hm18500000 mm
190 hm19000000 mm
195 hm19500000 mm
200 hm20000000 mm
205 hm20500000 mm
210 hm21000000 mm
215 hm21500000 mm
220 hm22000000 mm
225 hm22500000 mm
230 hm23000000 mm
235 hm23500000 mm
240 hm24000000 mm
245 hm24500000 mm
250 hm25000000 mm
255 hm25500000 mm
260 hm26000000 mm
265 hm26500000 mm
270 hm27000000 mm
275 hm27500000 mm
280 hm28000000 mm
285 hm28500000 mm
290 hm29000000 mm
295 hm29500000 mm
#### Hectometer to Millimeter Common Values
• 1 hm = 100000 mm
• 3 hm = 300000 mm
• 0.8 hm = 80000 mm
• 7 hm = 700000 mm
• 25 hm = 2500000 mm
• 6 hm = 600000 mm
• 5 hm = 500000 mm
• 41 hm = 4100000 mm
• 2 hm = 200000 mm
• 8 hm = 800000 mm
• 1000 hm = 100000000 mm
• 0.006 hm = 600 mm
• 4 hm = 400000 mm
• 3.2 hm = 320000 mm
• 10 hm = 1000000 mm
• 14 hm = 1400000 mm
• 14.56 hm = 1456000 mm
• 0.21 hm = 21000 mm
• 0.0098 hm = 980 mm | 1,511 | 4,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-26 | longest | en | 0.807412 |
https://mathequalslove.blogspot.com/2018/04/monday-must-reads-volume-39.html | 1,550,513,966,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487624.32/warc/CC-MAIN-20190218175932-20190218201932-00299.warc.gz | 610,039,559 | 33,515 | Math = Love: Monday Must Reads: Volume 39
## Monday, April 23, 2018
### Monday Must Reads: Volume 39
Happy Monday! The school year is winding down (at least at my school), and I am definitely excited about the upcoming summer. One of my favorite things to do during the summer is dream about what my classroom will be like the next year. It is my hope that you will find an idea to dream about doing either this year or next in this week's collection of Monday Must Reads!
Jocelle Skov shares an awesome multiplying monomials pyramid puzzle.
If you're teaching transformations, be sure to check out Lisa Richardson's awesome activity involving both dice and spinners! This looks so fun! Be sure to check out her blog post for more details.
Looking for some creative practice for order of operations? Check out the calendars that Beth Self has her students create!
Nickolas Corley inspires with an entire year's worth of Desmos projects in Algebra 1. These are so lovely!
Texas Math Teacher recommends a fun lesson involving skew dice (affiliate link). I didn't even know these were a thing, so I'm super excited by the idea!
Julia Anker shares an awesome test question that pits Desmos against TI.
Jonathan Lind makes the 3D distance formula come alive with a 3D coordinate plane made of straws.
Angela Manzo inspires with a fun, hands-on data collection activity involving origami frogs.
Tori Cox shares an awesomely beautiful card sort.
Solenne Abaziou created an awesome station in her classroom for early finishers. Each week, she puts up a new math game and a new tech activity for students to complete. How awesome is this?!?
I absolutely love this connecting representations task shared by Kaitlin. I feel like this would really et students thinking about what the different parts of an equation mean.
Mark Kaercher shares another use for that giant stack of paper plates that has been sitting in my cabinet ever since we made paper plate unit circles in trig a few years ago.
Hugo Sierra recommends introducing piecewise functions by having students graph their names.
Robert Kaplinsky shares a photo of some pizza prices that would spark some awesome conversations in math class.
I absolutely love that Mr. Derstein created a Surface Area Auction to give his students some fun, purposeful practice.
When I taught percent composition this year in chemistry, it was a little boring. If I ever teach it again, I will take a note from Michelle Vanhala and teach it using brown lunch bags full of candy!
Becci shares some photos of some awesome candy dilation projects.
Attention Calculus Teachers: How can you not love Dorothy Rich's idea of pretzel slope fields?
Amber Longhi shares some photos of some awesome geometry projects. | 579 | 2,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-09 | latest | en | 0.94375 |
https://osiris.utwente.nl/student/OnderwijsCatalogusSelect.do?selectie=cursus&cursus=202001362&collegejaar=2021&taal=nl | 1,701,448,229,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00090.warc.gz | 512,221,073 | 6,069 | Sluiten Help Print
Cursus: 202001362
202001362Algebra
Cursus informatie
Cursus202001362
Studiepunten (ECTS)3,5
CursustypeOnderwijseenheid
VoertaalEngels
Contactpersoondr. J.W. Polderman
E-mailj.w.polderman@utwente.nl
Docenten
Vorige 1-5 van 96-9 van 9 Volgende 4
Docent dr.ir. P. van 't Hof Docent dr. J. de Jong Docent dr.ir. R. Langerak Examinator dr. G. Loho Docent dr. L. Pehlivan
Collegejaar2021
Aanvangsblok
2A
OpmerkingMinor students: register for the minor!
AanmeldingsprocedureZelf aanmelden via OSIRIS Student
Inschrijven via OSIRISJa
Cursusdoelen
body { font-size: 9pt; font-family: Arial } table { font-size: 9pt; font-family: Arial } After completion of the course, the student is able to Apply the (extended) Euclidean algorithm to integers and polynomials. recognise and apply properties of groups to derive the symmetry group of a given set and to calculate the number of orbits with the aid of Burnside’s Theorem. work with ring and integral domains recognise under what conditions an integral domain is a field to work with the concept of vector space over an arbitrary field work with the classification of finite fields work with simple examples of linear code investigate and understand when two algebraic structures are isomorphic use and explain all components of RSA public key encryption, including all algorithmic underpinnings
Inhoud
body { font-size: 9pt; font-family: Arial } table { font-size: 9pt; font-family: Arial } This course provides an introduction to abstract algebra and some of its applications. Three families of algebraic structures are introduced and studied: groups, rings and fields. Roughly speaking groups are generalisations of the set of integers equipped with addition. Rings are generalisations of the set of integers equipped with both addition and multiplication. Fields are generalisations of the set of rational numbers equipped with addition, multiplication and division by nonzero numbers. The generalisations go as far as finite structures and structures in which the operation(s) are non-commutative. The backbones of the course are permutations, finite fields. Permutations and symmetries are used in certain counting problems, such as in how many essentially different ways, taking into account the symmetries, can we paint the faces of a cube using three colours. The theory of groups and rings is used to analyse a specific public key crypto system known as RSA. Also the algorithmic principles behind RSA are treated, e.g. the Euclidean algorithm and modular exponentiation. As an application of the theory of finite fields, a brief introduction to finite codes is provided. Topics covered in this course are: Integers and integers module n Group Theory, Orbits, Stabilizers Cosets & Lagrange Theorem Counting problems: Burnside’s theorem Rings and Fields Polynomial Rings Finite Fields Linear codes RSA Public Key Encryption
Participating study
Bachelor Applied Mathematics
Participating study
Module
Module 7
Verplicht materiaal
Book
Contemporary Abstract Algebra, Joseph A. Galian, 8th ed., Cengage Learning, 2012. ISBN: 978-1133599708
Aanbevolen materiaal
-
Werkvormen
Assessment Hoorcollege Opdracht Werkcollege
Toetsen
Algebra
Sluiten Help Print | 778 | 3,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.751519 |
https://igeeksmagz.com/questions-72 | 1,669,724,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00810.warc.gz | 349,718,041 | 7,476 | # App that does your homework
App that does your homework can be a helpful tool for these students. So let's get started!
## The Best App that does your homework
In this blog post, we will show you how to work with App that does your homework. definite integrals are used for finding the value of a function at a specific point. There are two types: definite integrals of first and second order. The definite integral of the first order is sometimes called the definite integral from the left to evaluate an area under a curve, whereas the definite integral of the second order is used to find an area under a curve between two values. Definite integrals can be solved by using integration by parts. This equation says that you can break your integral into two parts, one on each side of the equals sign, which will cancel out giving you just the value of your integral. You can also use complex numbers in the denominator to simplify things even more! If you want to solve definite integrals by hand, following these steps should get you going: Step 1: Find your area under the graph by drawing small rectangles where you want to find your answer. Step 2: Evaluate your integral by plugging in numbers into each rectangle. Step 3: Add up all your rectangles' areas and divide by n (where n is the number of rectangles). This will tell you how much area you evaluated for this particular function.
A word problem is a mathematical expression with one or more unknowns that must be solved for the value of the expression to be determined. The term "word problem" comes from the fact that such equations are written as words on a piece of paper. Word problems are common in everyday life, and they can often be solved using basic arithmetic. However, some word problems may require more advanced skills, such as an understanding of variables and operations. Word problems are also an important part of standardized tests. In order to successfully solve word problems, it is important to first understand their structure. Next, it is necessary to break down the problem into smaller parts. This is so that one can identify the values of the unknowns and make assumptions about them. Once this has been done, one can proceed to solve the problem by manipulating the values of the unknowns to arrive at a solution.
When you are dealing with a specific equation (one that has been written down in a specific way), it is often possible to solve it by eliminating one of the variables. For example, if you are given the equation: This can be simplified to: By multiplying both sides by '3', it becomes clear that the variable 'x' must be eliminated. This means that you can now simply put all the numbers on either side of the 'x' in place of their letters, and then solve for 'y'. This will give you: So, if you know what 'y' is and what all the other numbers are, you can solve for 'y'. This process is called elimination. You should always try to eliminate any variables from an equation first before trying to solve it, because sometimes doing so will simplify the equation enough to make it easier to work with.
A division solver is a program that solves division problems. There are many different types of division solvers, including ones that use a calculator, those that use an algorithm, and those that work with the number line. The simplest type of division solver is one that uses a calculator. This type of solver is typically used for problems where the number being divided by is simple, such as 2 divided by 3 or 12 divided by 5. This type of solver cannot handle more complicated problems, so it must be combined with an algorithm to solve them. An algorithm can take complex numbers and divide them into smaller parts. It can then retrieve the partial product from these parts and add all the partial products together to produce the final answer. Different divisions solvers use different algorithms for solving division problems. Some use the algorithm called long division (also known as decimal division) while others use the algorithm called long multiplication (also known as integer division). Regardless of how they work, all of them can be used to solve simple division problems.
The rule of equilateral triangles is that the three sides must be equal. Two-to-one correspondence: When you take two shapes like a circle and a square and connect them with lines to form a rectangle. The rule of rectangle is that the four sides must be equal. Three-to-one correspondence: When you take three shapes like an ellipse and a triangle and connect them with lines to form a parallelogram. The rule of parallelogram is that the five sides must be equal. Two-way correspondence: When you have two shapes like a circle and a square or two rectangles or two triangles and then connect them with lines to form another shape such as an equilateral triangle or rectangle or ellipse or parallelogram. Addition math problems: Adding numbers from one set to another set is the most common arithmetic math problem you will encounter in school. You can do it by adding sets
App is great since it not only solves a plethora of problems by just taking a picture, it even explains step by how to solve the problems down to the nitty gritty (if you played the 10 dollars for plus that is). Only down side is that plus costs 10 dollars but that’s probably only a problem since I’m broke so 5 stars
Cora Hayes
absolutely fantastic for those who are teachers and students. And the best part of this app is that it is completely free and offline. It does all the math correctly. it has helped me out a lot of times. KEEP UP THE GOOD WORK AND TRY ADDING MORE LANGUAGES NEXT TIMES. PEACE
Beatriz Mitchell
Take pic of math problem Axis of symmetry solver calculator Algebra long division solver College math problems with answers How to solve logarithmic functions Free online math problem solver | 1,197 | 5,893 | {"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} | 4.59375 | 5 | CC-MAIN-2022-49 | latest | en | 0.957026 |
https://www.jw.org/run/ivyasohowe/bibiliya/nwt/ibitabu/imigani/25/ | 1,713,658,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00880.warc.gz | 767,482,280 | 55,303 | # Imigani 25:1-28
• Kugumya ibanga (9)
• Amajambo acaguwe neza (11)
• Kutinjirira abandi mu buzima (17)
• Kurunda amakara ku mutwe w’umwansi (21, 22)
• Agakuru keza kameze nk’amazi apfutse (25)
25 N’iyi nyene ni imigani ya Salomo+, yimuwe* n’abantu ba Hezekiya+ umwami w’Ubuyuda: Ubuninahazwa bw’Imana ni ukugumiza ikintu ari ibanga+,ubuninahazwa bw’abami na bwo ni ugusuzuma ikintu bitonze. Kurya ijuru rigera hejuru cane, n’ubwimbike bw’isi bugashika kure cane,ni ko n’ivyiyumviro* vy’umwami ata wobitahura. Kura imicafu mw’ifeza,na yo izoheza ibe iyityoroye neza+. Kura umubisha imbere y’umwamimaze intebe yiwe y’ubwami izohangama mu bugororotsi+. Ntiwitere iteka imbere y’umwami+,kandi ntiwishire hamwe n’abahambaye+, kuko yopfuma akubwira ati: «Ingo hano», hako akumaramaza imbere y’abanyacubahiro+. Ntushingishe urubanza gihutihuti.None wokora iki hamwe mugenzawe yokumaramaza+? Genda mubiyage na mugenzawe+,ariko ntuhishure ivyo wabwiwe mw’ibanga*+, 10 kugira ngo uwukwumviriza ntakumaramazecanke ngo usange wakwiragije amajambo mabi* udashobora kugarura. 11 Ijambo rivuzwe ku gihe kibereye+rimeze nk’amapome y’inzahabu ari mu dusharizo tw’ifeza. 12 Ku muntu yakira neza impanuro+, umuhanuzi mwizaameze nk’ihereni ry’inzahabu n’agasharizo k’inzahabu nzizanziza. 13 Kurya kw’akabeho k’urubura* mu gihe c’iyimbura,ni ko n’intumwa yizigirwa imeze ku wayitumye,kuko iruhurira shebuja+. 14 Umuntu yiratira ingabire atigeze atanga*+ameze nk’ibicu n’umuyaga bitarimwo imvura. 15 Ukwihangana kurashobora kugondoza umuntu afise ububasha*,kandi amajambo y’ubwitonzi arashobora kumena igufa+. 16 Niwaronka ubuki, urarya ubwo ukeneye gusa,kuko niwarya bwinshi uzobudahwa+. 17 Ntiwamize ikirenge cawe mu nzu ya mugenzawe,kugira ntuhave umurambira maze akakwanka. 18 Uwagiriza ibinyoma mugenziwe+ameze nk’ubuhiri* n’inkota be n’umwampi usongoye. 19 Kwizigira umuntu atari umwizigirwa* mu gihe c’amagorwabimeze nko guhekenya n’iryinyo rivunitse canke gutambuka n’ikirenge kimugaye. 20 Kuririmbira umuntu atuntuye ku mutima+,ni nko kwambura impuzu mu gihe c’imbehocanke gusuka inkarihe ku munyu witwa alikali*. 21 Nimba umwansi wawe ashonje, numuhe ico arya.Nimba anyotewe, numuhe amazi yo kunywa+, 22 kuko uzoba umurunze amakara yaka ku mutwe*+,kandi Yehova azogushimira. 23 Umuyaga wo mu buraruko uzana imvura,ururimi ruvuga urusaku na rwo rukazana ishavu+. 24 Wopfuma wibera mu mfuruka y’igisenge c’inzu,hako ubana mu nzu imwe n’umugore w’amahane*+. 25 Agakuru keza kavuye mu gihugu ca kurekameze nk’amazi apfutse ku muntu arushe+. 26 Umugororotsi yemera kwoshwa* n’umubishaameze nk’iriba ryuzuye urwondo be n’isoko ricafuye. 27 Si vyiza kurya ubuki bwinshi cane+,kandi si vyiza kwironderera ubuninahazwa+. 28 Umuntu adashobora gucungera ishavu ryiwe+ameze nk’igisagara babomoye kigasigara kitagira uruhome.
### Utujambo tw'epfo
Canke «yegeranijwe.»
Canke «umutima.»
Canke «ntuhishure amabanga y’abandi.»
Canke «urusaku rubabaza.»
Canke «shelegi.» Shelegi ni utuvunjivunji twera cane dukoragurika nk’imvura. Mu gifaransa batwita neige.
Mu giheburayo havuga ngo «ingabire y’ikinyoma.»
Canke «kugondoza umukomanda.»
Canke «ubuhiri bw’intambara.»
Hashobora kandi kuvugwa ngo «umuntu w’umuhemu.»
Ni ibintu bimeze nk’umunyu bakoresha mu kumesa no mu kwoza ibintu.
Ni ukuvuga gutuma uwo muntu yoroha nk’aho woba ushongesheje ishavu ryiwe.
Canke «aguma atorana amahinyu.»
Canke «yemera kwumvikana.» Mu giheburayo havuga ngo «agenda aradandabagirana imbere.» | 1,490 | 3,488 | {"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-2024-18 | latest | en | 0.070015 |
https://studylib.net/doc/5870896/thompson-construction--subset-construction | 1,713,770,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818081.81/warc/CC-MAIN-20240422051258-20240422081258-00563.warc.gz | 517,238,636 | 12,204 | Thompson Construction, Subset Construction
```SUBSET CONSTRUCTION(CONVERSION OF NFA TO DFA)
It is used for converting an NFA having epsilon transition into
a DFA. The Algorithm for this is as follows:Let the DFA be D & its transition table be DTran.Dstates
represents the states of the DFA.Let NFA be N.
Initially e-closure(s){start state of NFA} is the only state in
Dstate and it is unmarked.
While there is an unmarked state T in Dstate do begin
Mark T;
For each input symbol a do begin
U=e-closure(move(T,a));
If U is not in Dstates then
Add U as unmarked state to Dstate.
Dtran[T,a]=U;
End
End
A state D is an accepting state if it is a set of NFA
containing atleast one accepting state of the N.
states
The following example was shown:NFA given was:INPUT(TERMINAL OR E(EPSILON))
STATES
->0(s)
1
2
3
4
5
6
7
8
9
10(f)
a
b
{3}
{8}
-
{5}
{9}
{10}
-
e (Epsilon)
{1,7}
{4,2}
{6}
{6}
{1,7}
-
s denotes the start state
f denotes it is a final state
The e-closures are calculated as follows:
eclosure(0)={0,1,2,4,7}=A=start state in corresponding DFA
eclosure(move(A,a))={1,2,3,4,6,7,8}=B
eclosure(move(A,b))={1,2,4,5,6,7}=C
eclosure(move(B,a))=B
eclosure(move(B,b))={1,2,5,6,7,9}=D
eclosure(move(C,a))=B
eclosure(move(C,b))=C
eclosure(move(D,a))=B
eclosure(move(D,b))={1,2,4,5,6,7,10}=E=final state
eclosure(move(E,a))=B
eclosure(move(E,b))=C
The Corresponding DFA is:INPUT(TERMINAL)
a
b
B
B
B
B
B
C
D
C
E(final)
C
STATE
->A
B
C
D
E
``` | 537 | 1,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.810718 |
http://sosmath.com/CBB/viewtopic.php?p=233937 | 1,371,614,525,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707440693/warc/CC-MAIN-20130516123040-00033-ip-10-60-113-184.ec2.internal.warc.gz | 230,903,737 | 8,558 | S.O.S. Mathematics CyberBoard
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Post subject: Probability HelpPosted: Sun, 10 Jun 2012 02:29:05 UTC
S.O.S. Newbie
Joined: Sun, 10 Jun 2012 02:27:27 UTC
Posts: 3
In a betting game a player gets to 3 chances to play after betting 6 dollars. The player must roll a 4 on a die. If he doesn't he loses. If he does then he can advance to the next round where he can then draw 3 cards from a deck of 8 consisting of 2 kings, two queens, two jacks and 2 jokers. If he draws the 2 jokers then he wins the game (does't matter what the third card is). The prize for winning the game is 30 dollars plus the 6 he originally used to play the game. So in total 36 dollars.
a) what's the probability of rolling a 4?
is it 1/6?
b) what's the probability of drawing 2 jokers?
is it (2C2)(6C1)/(8C2)
which is, 6/28
c) what is the probability of winning?
is it
(1/6)(6/28)
which is 1/28 ?
d) What is the expected value of one bet?
is it
E(x)= ( ∑ xi)(P(x))=((-6.00)(27/28))((30.00)(1/28))
= -1215/196 dollars?
e) Calculate his expected number of wins in those 3 trails
is it (3)(1/28) ?
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Post subject: Re: Probability HelpPosted: Sun, 10 Jun 2012 06:39:14 UTC
Member of the 'S.O.S. Math' Hall of Fame
Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7961
Location: Lexington, MA
Hello, Chee!
A few corrections . . .
Quote:
In a betting game a player gets 3 chances to play after betting 6 dollars.
The player must roll a 4 on a die. If he doesn't he loses.
If he does, then he can advance to the next round where he can then draw 3 cards
from a deck of 8 consisting of 2 kings, two queens, two jacks and 2 jokers.
If he draws the 2 jokers, then he wins the game (does't matter what the third card is).
The prize for winning the game is 30 dollars plus the 6 he originally used to play the game.
So in total 36 dollars.
a) What's the probability of rolling a 4?
Is it 1/6? . Yes!
b) What's the probability of drawing 2 jokers?
Is it (2C2)(6C1)/(8C2)? . no
There are: outcomes.
Quote:
c) What is the probability of winning?
is it (1/6)(6/28)? . no, for two reasons.
First, the probability of winning a game is: .
Second, for his $6 bet, he gets to play three times. I assume that if he wins, he stops playing. There are three ways he can. Win on the 1st try: . Win on the 2nd try: . Win on the 3rd try: . Therefore: . Also: . Quote: d) What is the expected value of one bet? He can expect to win$30 with probability
He can expect to lose $6 with probability . . . . . . . . . . . . . He can expect to lose an average of about$4.11 per game.
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Post subject: Re: Probability HelpPosted: Sun, 10 Jun 2012 13:25:38 UTC
S.O.S. Newbie
Joined: Sun, 10 Jun 2012 02:27:27 UTC
Posts: 3
Soroban wrote:
First, the probability of winning a game is: .
Second, for his $6 bet, he gets to play three times. I assume that if he wins, he stops playing. There are three ways he can. Win on the 1st try: . Win on the 2nd try: . Win on the 3rd try: . Therefore: . Also: .[/size] isn't the probability on the second and third still (1/56)? why on the win on the second try you multiplied (1/56) by (55/56) and then on the third you squared (55/56) then multiplied (1/56)? is there a formula or something? Top Post subject: Re: Probability HelpPosted: Sun, 10 Jun 2012 18:00:33 UTC Member of the 'S.O.S. Math' Hall of Fame Joined: Sun, 24 Jul 2005 20:12:39 UTC Posts: 3724 Location: Ottawa Ontario Soroban wrote: a) What's the probability of rolling a 4? Is it 1/6? . Yes! Not if "for 3 rolls"...problem's wording seems to allow 3 rolls. _________________ Walked over to a beggar...he gave me a quarter... Top Post subject: Re: Probability HelpPosted: Mon, 11 Jun 2012 00:57:40 UTC S.O.S. Newbie Joined: Sun, 10 Jun 2012 02:27:27 UTC Posts: 3 Soroban wrote: First, the probability of winning a game is: . Second, for his$6 bet, he gets to play three times.
I assume that if he wins, he stops playing.
There are three ways he can.
Win on the 1st try: .
Win on the 2nd try: .
Win on the 3rd try: .
Therefore: .
Also: .[/size]
but wait, aren't there many possibilities on how to win on the second try? One possibility, roll the die & don't get 4 (you lost so second try) then roll the die and get 4, then draw 2 jokers. Second possibility roll die and get 4, then don't draw 2 jokers (you lost so second try), roll 4, then draw 2 jokers? which results in two different probabilities of winning on the second attempt. can you apply the geometric distribution equation? :s
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Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA | 1,756 | 6,031 | {"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} | 3.921875 | 4 | CC-MAIN-2013-20 | latest | en | 0.93153 |
https://en.wikipedia.org/wiki/Sample_matrix_inversion | 1,481,439,219,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544140.93/warc/CC-MAIN-20161202170904-00450-ip-10-31-129-80.ec2.internal.warc.gz | 836,130,765 | 9,064 | # Sample matrix inversion
Sample matrix inversion (or direct matrix inversion) is an algorithm that estimates weights of an array (adaptive filter) by replacing the correlation matrix Ru with its estimate. Using K samples ${\displaystyle x(k),k=1,2,\dots ,K-1}$, an unbiased estimate of Ru, the correlation matrix of the array signals, may be obtained by means of a simple averaging scheme:
${\displaystyle {\hat {R}}_{u}(k)=(1/k)\sum (x(k)x^{H}(k)).}$ | 114 | 453 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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-2016-50 | latest | en | 0.863886 |
https://www.pathstoliteracy.org/strategies/ueb-lesson-9-grade-1-symbol-indicator-used-ueb | 1,582,861,160,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146940.95/warc/CC-MAIN-20200228012313-20200228042313-00415.warc.gz | 850,911,122 | 18,427 | # UEB Lesson 9: Grade 1 Symbol Indicator used in UEB
This post is the ninth in a series of lessons on teaching students who are braille readers to make the transition from EBAE (English Braille American Edition) to UEB (Unified English Braille). All lessons are by Catherine Summ and Suzanne Cappiello, who both work as Education Consultants for the Department of Rehabilitative Services (DORS-BESB) in CT.
See the full UEB Curriculum with all lessons.
### Goal:
The student will be able to identify the grade 1 symbol indicator that is now used in the UEB code.
The grade 1 symbol indicator will always be shown as dots 5,6.
Any character following the grade 1 symbol indicator is not a contraction. Dots 5,6 should be used when a letter stands by itself. Dots 5,6 are not required before a single a, I or o, which do not have contraction forms.
A double dot 5,6 means the next word is in grade 1, and a triple dot 5,6 signals the beginning of a passage that will not include any contractions. The grade 1 passage is terminated with dots 5,6 and 3.
For further explanation of the changes refer to the handout “Overview of Changes from Current Literary Braille to Unified English Braille” authored by Braille Authority of North America, March 2013, www.brailleauthority.org. Refer to the UEB manual for a full explanation of the braille code.
## Procedure:
Teacher prepares the examples below in braille. Students practice reading the examples below, and mark the grade 1 indicator.
1. She labelled all of the math books with letter m stickers.
2. John F. Kennedy was the 35th president of the United States.
3. The audience chanted: “Action, action, we want action! A-c-t-i-o-n!”
4. Lucy M. Montgomery wrote the “Anne of Green Gables” series of books.
5. The correct spelling of our first president’s name is G-e-o-r-g-e W-a-s-h-i-n-g-t-o-n and it will be on the spelling test.
6. The word knowledge starts with a k, not an n.
7. Vowels are a, e, i, o and u.
8. The map has an x where the treasure should be.
### Additional Activities: Writing Practice
Dictate the following words, phrases, and sentences to students. Feel free to add your own.
1. The word hippopotamus is spelled h-i-p-p-o-p-o-t-a-m-u-s.
2. He needed a c in order to spell crabapple in the Scrabble game.
3. Making a U turn on Main Street is illegal.
4. The h on the map indicates a house.
5. Write y for yes and n for no.
6. I’m wearing a v neck sweater.
7. We are approaching an S curve on the road.
8. Everyone in the class earned a B on the exam.
9. The baseball team chanted: “r-u-n, r-u-n, r-u-n!” when the batter hit the ball.
### Assessment:
Dictate the following words, phrases, and sentences to students. Add your own. Teacher compiles a spreadsheet for each student to document progress and compile data.
1. The correct way to spell Jersey in New Jersey is J-e-r-s-e-y.
2. Every cake box is marked with a c and every pie box is marked with a p.
3. George W. Bush was the 43rd president of the United States.
4. The class sang: “B-i-n-g-o, b-i-n-g-o, b-i-n-g-o, and Bingo was his name-O!”
5. The s indicates where to write your signature.
6. The Bay City Rollers released the song “S-A-T-U-R-D-A-Y Night” in 1974.
7. Do not forget the p in raspberry.
8. The mayor’s full name is Pat R. Russell.
9. All the moving boxes going to the kitchen should have a k on them.
10. X marks the spot.
### Educational Activity and Game:
#### Spelling Bee Road Race
Supplies: game tokens, dice, embossed graph paper, poster board, index cards each containing a contracted word
Make a game board using embossed graph paper. Cut the graph paper into strips and glue it to the poster board, or a similar piece of paper. Feel free to make a road pattern of your choice. Using raised stickers, puffy paint, or another tactile marker to randomly mark squares with a symbol for selecting a card. Mark where the road begins and where the road ends.
Students take turns rolling the dice and moving their game piece on the road. If a student lands on a “card” spot they should select a card from the pile. The student reads the word on their card and then brailles the word in its uncontracted format in a sentence using the appropriate grade 1 indicator. If they do this correctly they get to move ahead one square and then wait for their next turn. If they do this incorrectly then they have to move back a square. The person who reaches the finish line first wins.
Download the Duxbury file(Thank you to Karen Carl for creating and sharing this!)
Posted on October 1, 2015
Updated on: February 7, 2018 | 1,144 | 4,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-10 | longest | en | 0.886776 |
https://www.usna.edu/Users/cs/roche/courses/s13si486d/probs/059.php.html | 1,540,150,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514314.87/warc/CC-MAIN-20181021181851-20181021203351-00106.warc.gz | 1,136,610,185 | 2,567 | Problem 59
This is the archived version of this course from the Spring 2013 semester. You might find something more recent by visitning my teaching page.
# Subgraph selection with different probability
Due: March 29
Points: 3
In the randomized MST algorithm we saw in class, each edge is selected to go into H with probability $$\frac{1}{2}$$.
What if we made this probabilitiy slightly lower or slightly higher, for example selecting edges for the subgraph with probability $$\frac{1}{3}$$ or $$\frac{4}{5}$$ or something else. I don't need you to do the full analysis, but tell me how lowering or raising this probability would affect:
• The size (nodes, edges) of the random subgraph H
• The size (nodes, edges) of the recursively-computed minimum spanning forest F or H
• The size (nodes, edges) of the subgraph $$G''$$ that results from finding the F-light edges in H
• The expected running time of a single step in the algorithm.
• The expected running time of the entire randomized algorithm.
So for each of the things above, say how it would change if the probability were increased or if it were decreased, and why. (Note that many of the above factors are related to each other!) | 280 | 1,196 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-43 | latest | en | 0.932591 |
https://www.chegg.com/homework-help/algebra-and-trigonometry-9th-edition-chapter-12.6-solutions-9780321716569 | 1,529,616,880,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864300.98/warc/CC-MAIN-20180621211603-20180621231603-00287.warc.gz | 759,489,904 | 16,581 | Skip Navigation
Algebra and Trigonometry (9th Edition) View more editions Solutions for Chapter 12.6
• 9148 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Graph the equation: y =3x + 2
Sample Solution
Chapter: Problem:
• Step 1 of 3
The equation of the given line y = 3x + 2 is in the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
So, the slope of the line is 3 and the y-intercept is 2.
• Step 2 of 3
For getting a point on the line, substitute the value –1 for x and find the corresponding y-coordinate.
Thus, we have one point on the graph of the line as (–1, –1).
Start from the point (–1, –1), go 1 unit to the right and then up 3 units. We get to the point (0, 2).
• Step 3 of 3
The line y = 3x + 2 can be graphed by plotting the points and drawing a line passing through them.
Corresponding Textbook
Algebra and Trigonometry | 9th Edition
9780321716569ISBN-13: 0321716566ISBN: Michael SullivanAuthors:
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# News from Mathnasium of Meadowvale
### Mental Math Checkup: Is Your Child on Pace for Math Success?
Nov 2, 2017
Can your child answer these mental math questions? The results may surprise you!
If they can solve problems at and above grade level, they may be ready for a challenge. If they are unable to answer questions at grade level or below, they're likely in need of extra help.
Grade 1: 11 + 12 = _______
Grade 2: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = ______
Grade 3: How much is 99 plus 99 plus 99?
Grade 4: Count by 13/4 from 0 to 7.
Grade 6: Halfway through the second quarter, how much of the game is left?
Grade 7: How much is 6 1/2% of 250?
Grade 8: On a certain map, 6 inches represent 25 miles. How many miles do 15 inches represent?
Algebra: When you take 3 away from twice a number, the answer is 8. What is the number?
Functions: What is the absolute value of the point (3, 4)? | 313 | 1,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-50 | latest | en | 0.86802 |
https://www.sugarprocesstech.com/phosphoric-acid-h3-po4-calculator/ | 1,685,650,687,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00077.warc.gz | 1,094,183,864 | 22,433 | # Phosphoric Acid (H3 PO4 ) Dosing in Sugar Industry | Online Calculator
Contents
## Role of Phosphoric Acid (H3 PO4 ) Dosing in Sugar Process with Online Calculator
Phosphoric acid occurs in the cane juice as both soluble phosphates and in combination as protein in the cell material. But both phosphates are not participate in reaction, only the soluble phosphates take part in the reaction of defecation.
If the phosphate content is less than 200 to 250 mg/P2O in juice, it is beneficial to make additions of inorganic phosphate to raise the juice phosphate level to above 250-300 mg/kg.
The phosphoric acid added to the juice precipitates part of the colloids and coloring matter that it contains. The precipitate formed with lime is mainly Tricalcium Phosphate.
### The benefits of maintaining an optimum level of phosphate in juice are as:
a) Greater colloidal elimination
b) Less calcium in clear juice;
c) Faster mud floc formation, with more rapid settling;
d) Better clarification with lighter colored clarified juice.
Disadvantages of too much phosphate addition can be higher chemical costs, higher lime consumption and higher mud volume (and hence higher filter mud sugar loss).
#### Action Due to phosphoric Acid Addition:
The major chemical reaction is that of the Ca++ cation with the phosphate ion to form phosphate Intermediates and to precipitate tricalcium phosphate Ca 3(PO4)2.
We have to know the activity of calcium and phosphate ions and how the phosphate components behave in the adsorption and absorption of organic colloids and other non-sugars.
It is after reacting with the available Ca++ and then further with lime is one of the major reaction responsible for juice clarification. In the beginning the Ca++ ion reacts with HPO4 to form CaHPO4. This is second order reaction & hence it is fast reaction,
Ca2++ + HPO– –4→ CaHPO4
Second stage phosphate ion reacts with free Ca ion to form the precipitate of tricalcium phosphate. This a higher order reaction, hence takes time to complete
3Ca++ + 2PO– —4 → Ca3 (PO4)2 TriCalcium Phosphate
3 Ca (OH) 2 + 2 H3 PO4 ———- Ca3 (PO4) 2 + 6 H2O
### Steps in Calculation for required quantity of Phosphoric acid to add in Raw juice:
#### Required Data:
a) Crushing Rate ( M ) in T/hr
b) Percentage of mixed juice ( P ) = Take 100%
c) Density of juice ( D )(Generally take 1.06 gm/ml for raw juice)
d) Required phosphoric acid (in P2O5 Content) dosing in PPM ( R )to raw juice ( To know this analyze P2O5 content in mixed juice before addition than we can concluded to how much dosing required to raw juice).
e) Percentage of Ortho phosphoric acid chemical ( H ) ( i.e Percentage of H3PO4 purity in used chemical)
#### Calculation Part:
Calculate mixed juice quantity (Q ) in Ltrs/hr.
Q = ( M x 1000)/ D
P2O5 requirement in ( T ) Kg/hr (1ppm=1 mg/lit)
T = ( Q x R) / 1000000
Calculate P2 O5 % ( Y ) in given chemical H3PO4
2 H3PO4 → P2O5 + 3H2O
2(3+31+64) (2*31+5*16) + (3*2+3*16)
196 142 54
i.e 196 parts of H3PO4 gives 142 parts P205)
Y = ( H x 142)/196
H3PO4 required quantity in kg/Hr (T x 100)/Y
### Phosphoric Acid (H3 PO4 ) Online Calculator :
Some Related Articles:
Flocculants Used in Sugar Processing | Flocculant Dosing Calculation.
Juice Sulphitation Process | Methods of Sugar Cane Juice Sulphitation.
Concepts of Juice Defecation Process and its Mechanization.
Sugar Industry Related Important Websites | sugar Technology.
Sugar Tech | Sugar Technology related articles .
Sugar Industry Equipment Design and Drawing Online Calculations.
Thumb Rules for Sugar factory Equipment Design Sizing.
Hi friends Thanks for reading. I Hope you liked it. Give feed back, comments and please share it
## 15 thoughts on “Phosphoric Acid (H3 PO4 ) Dosing in Sugar Industry | Online Calculator”
#### U.Rajesh
(April 4, 2017 - 1:26 pm)
Good keep it up
Website is useful for Sugar technologists
(April 4, 2017 - 2:00 pm)
thank you sir
#### S S R Ayyangar
(January 16, 2021 - 4:38 pm)
When Mixed juice % cane above 100 , calculation not showing .Set right the formula above 100 % Mixed juice accordingly
#### siva alluri
(January 17, 2021 - 8:24 am)
Actually, we calculate Phosphoric acid dosing on the basis of cane only. It not depends on the MJ%cane. So calculation purpose we take 100% only
(April 5, 2017 - 5:18 am)
THANKS LOT OF USE SUGAR TECHNOLOGISTS
#### Brijendra singh
(May 28, 2017 - 4:32 am)
Very interested topic good
#### Sageer Ahmed
(April 12, 2018 - 9:27 pm)
Very nice and much useful calculation for sugar mills.
Thanks
#### siva alluri
(April 16, 2018 - 1:22 pm)
Thank you Mr.Sageer Ahmed
#### Sageer Ahmed
(July 12, 2018 - 5:49 pm)
Very useful calculation for sugar mills engineering & operations professionals.
Thanks
#### siva alluri
(July 15, 2018 - 1:41 pm)
Thank you Mr.Sageer Ahmed
#### Deep Malik
(September 16, 2019 - 10:36 am)
Very useful calculations and also other contents of the website. Good for sugar technologists.
#### siva alluri
(September 16, 2019 - 2:43 pm)
welcome Deep Malik
(November 13, 2020 - 7:19 pm)
Very useful articals,good work
Thanks
#### siva alluri
(November 14, 2020 - 5:15 am)
Welcome
#### Darlene Managuit
(March 3, 2022 - 8:36 am)
I want to ask where should phosphoric acid be applied to have a better clarification. In mixed juice before or after liming? In syrup clarifier? | 1,494 | 5,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-23 | longest | en | 0.908824 |
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Topic: Answer to Dik T. Winter
Replies: 441 Last Post: Feb 5, 2013 6:25 AM
Messages: [ Previous | Next ]
mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Answer to Dik T. Winter
Posted: Jul 2, 2009 5:03 AM
On 2 Jul., 09:06, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:
> On Jul 2, 12:01 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 2 Jul., 04:03, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
>
> > > My answer to your question above is similar. The question whether a countable
> > > union of countably many sets is countable is implied by the axiom of countable
> > > choice, which is not provable from ZF. So negating the axiom of countable
> > > choice (which implies negating the axiom of choice) we can have situations
> > > where the conclusion is false.
>
> > Feferman and Levy showed that the statement that the set of all real
> > numbers is the union of a denumerable set of denumerable sets cannot
> > be refuted.
>
> > So does it depend on the chosen axioms, whether R is countable?
>
> No, countable choice is a theorem of ZF. It is attributed to
> Hausdorff "countable unions of countable sets are not necessarily
> dissimilar from uncountable sets". However, to him is attributed a
> more direct statement.
>
> R can always be considered countable in for example some constructible
> universe containing the real numbers via Skolemization. That's where
> it's a countable model, even if R is still uncountable. The model is
> supposed to be some big ordinal, bigger than any it models, but there
> is no such thing in regular set theories.
There is no model of ZFC. My question is: Is the set of real numbers
uncountable in ZF, is it uncountale in ZFC. Is a countable union of
contable sets uncountable in ZF, is it uncountable in ZFC?
That boils down to the question: Does Cantor's proof show the same as
Feferman and Levy showed? Or does the result depend on the chosen
axioms?
Regards, WM
Date Subject Author
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6/20/09 Guest | 5,472 | 12,139 | {"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-2018-17 | longest | en | 0.896411 |
http://www.jiskha.com/display.cgi?id=1335724801 | 1,496,150,741,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615105.83/warc/CC-MAIN-20170530124450-20170530144450-00028.warc.gz | 688,375,973 | 3,917 | # statistics
posted by on .
do you want to own your won candy store? start up costs (units in 1,000) for a random sample of candy stores are: (95, 173, 129, 95, 75, 94, 116, 100, 85)
Use an 80% confidence interval to test the null hypothesis that the mean start up cost is 118. Justify answer.
• statistics - ,
80% = mean ± 1.28 SEm
SEm = SD/√n
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
I'll let you do the calculations.
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
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## 廣義線性模型/範例3 : SGD: Maximum margin separating hyperplane
• 透過用SGD方法進行訓練的線性分類器：SGDClassifier，在兩種類別的數據集中，繪出每個類別中的最大邊界與超平面使兩種分類被區隔開來
## 一、SGD Classifier
SGD Classifier為一個利用梯度下降法SGD (Stochastic gradient descent)進行訓練的線性分類器(默認為SVM)，模型每次迭代過後會計算樣本loss function梯度，並依據梯度值進行learning rate的更新，learning rate亦會經過迭代次數越多，更新的幅度會越少
## 二、引入函式與模型
• make_blobs為聚類數據生成器
```import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDClassifier
from sklearn.datasets import make_blobs```
## 三、數據產生與模型訓練
• 利用make_blobs產生50個點且分為兩種類別，此語法回傳X為點座標，Y為相對應點的所屬類別
```X, Y = make_blobs(n_samples=50, centers=2, random_state=0, cluster_std=0.60)
tempx = []
tempy = []
for i in range(50):
tempx.append(X[i,0])
tempy.append(X[i,1])
plt.scatter(tempx,tempy)```
• SGDClassifier中的loss function採用hinge使其為線性SVM分類器，alpha為penalty(預設為L2)的參數
```clf = SGDClassifier(loss="hinge", alpha=0.01, max_iter=200)
clf.fit(X, Y)```
## 四、Hyperplane
• clf.decision_function為樣本點到hyperplane的函數距離
• 利用matplotlib中繪製等高線圖的語法contour來繪製hyperplane與最大邊界
```# plot the line, the points, and the nearest vectors to the plane
xx = np.linspace(-1, 5, 10)
yy = np.linspace(-1, 5, 10)
X1, X2 = np.meshgrid(xx, yy)
Z = np.empty(X1.shape)
for (i, j), val in np.ndenumerate(X1):
x1 = val
x2 = X2[i, j]
p = clf.decision_function([[x1, x2]])
Z[i, j] = p[0]
print(Z)
levels = [-1.0, 0.0, 1.0]
linestyles = ['dashed', 'solid', 'dashed']
colors = 'k'
plt.contour(X1, X2, Z, levels, colors=colors, linestyles=linestyles)
plt.scatter(X[:, 0], X[:, 1], c=Y, cmap=plt.cm.Paired,
edgecolor='black', s=20)
plt.axis('tight')
plt.show()```
## 五、原始碼列表
Python source code: plot_sgd_separating_hyperplane.py
https://scikit-learn.org/stable/auto_examples/linear_model/plot_sgd_separating_hyperplane.html
```print(__doc__)
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDClassifier
from sklearn.datasets import make_blobs
# we create 50 separable points
X, Y = make_blobs(n_samples=50, centers=2, random_state=0, cluster_std=0.60)
# fit the model
clf = SGDClassifier(loss="hinge", alpha=0.01, max_iter=200)
clf.fit(X, Y)
# plot the line, the points, and the nearest vectors to the plane
xx = np.linspace(-1, 5, 10)
yy = np.linspace(-1, 5, 10)
X1, X2 = np.meshgrid(xx, yy)
Z = np.empty(X1.shape)
for (i, j), val in np.ndenumerate(X1):
x1 = val
x2 = X2[i, j]
p = clf.decision_function([[x1, x2]])
Z[i, j] = p[0]
levels = [-1.0, 0.0, 1.0]
linestyles = ['dashed', 'solid', 'dashed']
colors = 'k'
plt.contour(X1, X2, Z, levels, colors=colors, linestyles=linestyles)
plt.scatter(X[:, 0], X[:, 1], c=Y, cmap=plt.cm.Paired,
edgecolor='black', s=20)
plt.axis('tight')
plt.show()```
You can’t perform that action at this time. | 1,076 | 2,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-10 | latest | en | 0.195278 |
https://philosophy-question.com/library/lecture/read/43846-what-are-the-angles-in-a-3-4-5-triangle | 1,679,601,977,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00412.warc.gz | 532,053,407 | 7,965 | # What are the angles in a 3 4 5 triangle?
## What are the angles in a 3 4 5 triangle?
Interior Angles Because it is a right triangle one angle is obviously 90°. The other two are approximately 36.
## What are the angles of a Pythagorean triple?
Obviously, the most interesting pythagorean triple (a,b,c) would be one for which the corresponding triangle (with integer side lengths a,b,c) has angles 90°, 60° and 30° (π2,π3 and π6).
## How do you find the angle of Pythagoras?
Examples
1. Step 1 The two sides we know are Opposite (300) and Adjacent (400).
2. Step 2 SOHCAHTOA tells us we must use Tangent.
3. Step 3 Calculate Opposite/Adjacent = 300/400 = 0.
## What angles are in a triangle?
The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°.
## How do you find the degree of an angle without a protractor?
How to Calculate Angles Without a Protractor
1. Mark Two Points on the Line Opposite the Angle.
2. Measure the Line.
3. Use the Sine Formula.
4. Calculate the Angle.
## Are right angles 90 degrees?
Acute angles are less than 90 degrees, obtuse angles are more than 90 degrees, and right angles are exactly 90 degrees.
## What are different types of angles?
Different Types of Angles
• Zero Angles.
• Acute Angles.
• Right Angles.
• Obtuse Angles.
• Straight Angles.
• Reflex Angles.
• Complete Angle.
## What can I use if I don't have a protractor?
However, if you don't have a protractor handy, you can determine the size of an angle using the basic geometric principles of triangles. You'll need a scientific calculator to solve the equations. Most smartphones are equipped with them, but you can also download a free app or access a free calculator online.
## Can you make your own protractor?
Protractors are a tool used in mathematics to measure angles in degrees. You might need a protractor to work on homework or make construction plans, so learning how to make your own can come in handy. You can make a protractor by printing from a template or folding a square of paper to always have one on hand.
## Which angle is 60 degree?
A 60 degrees angle is an acute angle because it is less than 90 degrees.
## What do you call a 90 degree angle?
Angles that are 90 degrees (θ = 90°) are right angles. • Angles that are 180 degrees (θ = 180°) are known as straight angles. • Angles between 180 and 360 degrees (180°< θ < 360°) are called reflex angles.
## What are the 6 different types of angles?
Types of Angles
• Acute Angle: An angle whose measure is less than 90° is called an acute angle. ...
• Right Angle: An angle whose measure is 90° is called right angle. ...
• Obtuse Angle: An angle whose measure is greater than 90° but less than 180° is called an obtuse angle. ...
• Straight Angle: ...
• Reflex Angle: ...
• Zero Angle:
## Why should you never argue with a 90 degree angle?
It's always right. Because it is an inanimate object and cannot present proper support for its arguments. You should've said "acute".
## What are 4 ways to name an angle?
An angle can be named in different ways:
• by a number or letter written inside the angle.
• by the name of the vertex.
• by the vertex and a point on each ray.
## How do we name angles?
(1) We can name angles by using THREE capital letters like: ABC or DEF. The middle letter is called the VERTEX of the angle. The above angles are read "angle ABC" and "angle DEF." This leads us to the second way we can name angles.
## How do you find the vertex angle?
Properties. A vertex angle in a polygon is often measured on the interior side of the vertex. For any simple n-gon, the sum of the interior angles is π(n − 2) radians or 180(n − 2) degrees.
## Does an angle have a vertex?
Vertex (of an angle) The vertex of an angle is the common endpoint of two rays that form the angle.
## Which is called the vertex of an angle?
The point about which an angle is measured is called the angle's vertex, and the angle. associated with a given vertex is called the vertex angle. In a polygon, the (interior, i.e., measured on the interior side of the vertex) are generally denoted or .
## Which is the vertex of angle XYZ?
A Vertex is the point that 2 or more lines connect. When an angle is written given the letters of the lines/ points, the point where the lines connect are always the middle letter. For ∠XYZ, the vertex would be point Y, because it is the middle letter./span>
## How many vertex are there in an angle?
Similarly every angle has a vertex. So, the number of vertices to an angle is 1.
## What is the vertex of an angle example?
Vertex: The common end point at which the two rays meet to form an angle is called the vertex. Here, the point O is the vertex of ∠AOB. We can find angles in various things around us, such as in a pair of scissors, a hockey stick, a chair. | 1,239 | 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} | 4.6875 | 5 | CC-MAIN-2023-14 | latest | en | 0.872921 |
https://chrisalbon.com/python/basics/logical_operations/ | 1,620,951,077,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989616.38/warc/CC-MAIN-20210513234920-20210514024920-00041.warc.gz | 196,778,437 | 4,318 | # Logical Operations
### Create some simulated variables
x = 6
y = 9
z = 12
### x or y
x or y
6
### x and y
x and y
9
### not x
not x
False
### x is equal to y
x == y
False
### x is not equal to y
x != y
True
### One is less than two
1 < 2
True
### Two is less than or equal to four
2 <= 4
True
### Three is equal to five
3 == 5
False
### Three is not equal to four
3 != 4
True
### x is less than y which is less than z
x < y < z
True
### x is less than y and y is less than z
x < y and y < z
True | 186 | 545 | {"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} | 3.015625 | 3 | CC-MAIN-2021-21 | latest | en | 0.870911 |
https://studysoup.com/tsg/910046/elementary-statistics-12-edition-chapter-11-problem-8 | 1,611,285,065,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529080.43/warc/CC-MAIN-20210122020254-20210122050254-00109.warc.gz | 551,262,774 | 12,339 | ×
Get Full Access to Elementary Statistics - 12 Edition - Chapter 11 - Problem 8
Get Full Access to Elementary Statistics - 12 Edition - Chapter 11 - Problem 8
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# Get answer: Questions 610 refer to the sample data in the following table (based on data
ISBN: 9780321836960 18
## Solution for problem 8 Chapter 11
Elementary Statistics | 12th Edition
• Textbook Solutions
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Elementary Statistics | 12th Edition
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3
Problem 8
Questions 610 refer to the sample data in the following table (based on data from the Pew Research Center). Randomly selected subjects were asked about the use of marijuana for medical purposes. Assume that we want to use a 0.05 significance level to test the claim that response to the question is independent of gender. In Favor Oppose Don't Know Men 538 167 29 Women 557 186 33 Is the hypothesis test lefttailed, righttailed, or twotailed?
Step-by-Step Solution:
Step 1 of 3
Week 8: Normal Distribution 25 October 2016 Basic Statistics for Research Professor HK Dong Wendy Liu Probability density function f(x) – describes the distribution of probability for the continuous random variable. It has the properties The total area under the probability density curve is 1 area under the probability density curve between a and b for all x Normal distribution – bell-shaped density curve Mean = µ Standard deviation = σ Normal distribution w/ mean of µ and st. dev of σ is denoted by N(µ, σ) Standard normal distribution: N(0,1) Groundwork for Inferential Statistics Sampling distribution of a statistic o distribution of all possible observations of the statistic for samples of
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321836960
Elementary Statistics was written by and is associated to the ISBN: 9780321836960. This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. The answer to “Questions 610 refer to the sample data in the following table (based on data from the Pew Research Center). Randomly selected subjects were asked about the use of marijuana for medical purposes. Assume that we want to use a 0.05 significance level to test the claim that response to the question is independent of gender. In Favor Oppose Don't Know Men 538 167 29 Women 557 186 33 Is the hypothesis test lefttailed, righttailed, or twotailed?” is broken down into a number of easy to follow steps, and 76 words. The full step-by-step solution to problem: 8 from chapter: 11 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. Since the solution to 8 from 11 chapter was answered, more than 229 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 121 chapters, and 3629 solutions.
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Unlock Textbook Solution | 705 | 2,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-04 | latest | en | 0.879433 |
https://www.chegg.com/homework-help/evaluate-solid-ellipsoid-hint-let-x-au-y-bv-z-cw-integrate-a-chapter-14.8-problem-21e-solution-9780321388506-exc | 1,553,439,856,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00359.warc.gz | 724,316,500 | 26,797 | # Student Solutions Manual Part 2 for University Calculus (1st Edition) Edit edition Problem 21E from Chapter 14.8: Evaluate over the solid ellipsoid (Hint:...
We have solutions for your book!
Chapter: Problem:
Evaluate
over the solid ellipsoid
(Hint: Let x = au,y = bv, and z = cw. Then integrate over an appropriate region in uvw-space.)
Step-by-step solution:
Chapter: Problem:
• Step 1 of 4
Consider the triple integral:
Where, E is the solid ellipsoid:
The objective is to evaluate this triple integral on the specified region.
Let Then, the solid ellipsoid is,
And, the triple integral will become:
• Chapter , Problem is solved.
Corresponding Textbook
Student Solutions Manual Part 2 for University Calculus | 1st Edition
9780321388506ISBN-13: 032138850XISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: University Calculus: Alternate 1st Edition Textbook Solutions | 218 | 903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-13 | latest | en | 0.70899 |
http://mathhelpforum.com/advanced-algebra/40026-orthonormal-basis-linear-combinations.html | 1,481,444,342,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544358.59/warc/CC-MAIN-20161202170904-00508-ip-10-31-129-80.ec2.internal.warc.gz | 172,182,720 | 10,752 | # Thread: Orthonormal basis and linear combinations
1. ## Orthonormal basis and linear combinations
Let $\theta$ be a fixed real number and let \textbf{x}_1=\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right] and \textbf{x}_2=\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right].
Given a vector $\textbf{y}$ in $R^2$, write it as a linear combination $c_1\textbf{x}_1+c_2\textbf{x}_2$.
It also asked me to prove that $(\textbf{x}_1,\textbf{x}_2)$ was an orthonormal basis for $R^2$ so I did that.
However,
$\textbf{y}=c_1\textbf{x}_1+c_2\textbf{x}_2$. <\textbf{y},\textbf{x}_1>=\textbf{y}\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right] and <\textbf{y},\textbf{x}_2>=\textbf{y}\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right]
Can someone assist me on where to go from here? Thank you!
2. Hello,
Originally Posted by pakman
Let $\theta$ be a fixed real number and let \textbf{x}_1=\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right] and \textbf{x}_2=\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right].
Given a vector $\textbf{y}$ in $R^2$, write it as a linear combination $c_1\textbf{x}_1+c_2\textbf{x}_2$.
It also asked me to prove that $(\textbf{x}_1,\textbf{x}_2)$ was an orthonormal basis for $R^2$ so I did that.
However,
$\textbf{y}=c_1\textbf{x}_1+c_2\textbf{x}_2$. <\textbf{y},\textbf{x}_1>=\textbf{y}\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right] and <\textbf{y},\textbf{x}_2>=\textbf{y}\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right]
Can someone assist me on where to go from here? Thank you!
I'm not sure what you're trying to get ?
Did you prove that it was an orthonormal basis ?
If you're looking for $$ and $$, then... :
$=$
Because the scalar product is bilinear, this is equal to :
$+=c_1+c_2$
But $=||x||^2$. Here, $||x_1||=1$ because it's an orthonormal basis.
And $=0$ because it's an orthonormal basis.
Therefore, $=c_1 \cdot 1^2+c_2 \cdot 0=c_1$
If you prefer, the scalar product of y with the n-th vector of an orthonormal basis represents the n-th coordinate of y in this basis. (this is extra info and I'm not sure I explained it well ) | 818 | 2,356 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 30, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-50 | longest | en | 0.682273 |
https://solvedlib.com/a-discharging-rc-circuit-has-its-charge-reduced,312532 | 1,695,368,470,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00373.warc.gz | 574,836,859 | 16,885 | # A discharging RC circuit has its charge reduced from 20C to 15C in 20 seconds. If...
###### Question:
a discharging RC circuit has its charge reduced from 20C to 15C in 20 seconds. If the resistance of the circuit is 10 ohms. What is the capacitance.
pls show work.
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https://ccrl.chessdom.com/ccrl/4040/cgi/engine_details.cgi?match_length=30&print=Details&each_game=1&eng=Houdini%203%2032-bit | 1,600,974,106,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00134.warc.gz | 315,614,813 | 5,066 | Contents: CCRL 40/15 Downloads and Statistics September 19, 2020 Testing summary: Total: 1'178'235 games played by 2'711 programs White wins: 406'134 (34.5%) Black wins: 300'016 (25.5%) Draws: 472'085 (40.1%) White score: 54.5%
## Engine Details
Options Show each game results
Houdini 3 32-bit (3135+20
−20
)Quote
Author: Robert Houdart (Belgium) Link: Homepage
This is one of the 19 Houdini versions we tested: Compare them!
Opponent Elo Diff Results Score LOS Perf – Strelka 5.5 32-bit 3087 +15−15 (−48) 19.5 − 12.5(+13−6=13) 60.9%19.5 / 32 100.0% +21 = = = = = = 1 0 = 1 1 1 = 1 1 1 0 0 = = 0 = 1 1 = = 0 1 0 1 1 1 – Strelka 5.1 32-bit 3081 +15−15 (−54) 19.5 − 12.5(+10−3=19) 60.9%19.5 / 32 100.0% +7 = = 1 0 = = = = 1 = 1 = 1 1 = = = = = 1 = 0 1 = 1 0 = 1 = = 1 = – Critter 1.4 32-bit 3072 +14−14 (−63) 20 − 12(+9−1=22) 62.5%20.0 / 32 100.0% +3 = = = = = 1 1 = = = = = = 1 1 = = 1 0 1 = = = = = = 1 1 = 1 = = – Critter 1.6a 32-bit 3067 +16−16 (−68) 20 − 12(+11−3=18) 62.5%20.0 / 32 100.0% +4 0 = = = 1 = 1 = 1 = 1 = = = = = 1 1 = = = 1 = 1 = 1 = 0 0 = 1 1 – Stockfish 2.2.2 32-bit 3050 +14−14 (−85) 21.5 − 10.5(+13−2=17) 67.2%21.5 / 32 100.0% +15 = = 1 1 0 1 1 = = 1 = 1 = = 1 0 1 = 1 = = = 1 1 = = = = 1 1 = = – Komodo 5 32-bit 3046 +16−16 (−89) 19 − 13(+12−6=14) 59.4%19.0 / 32 100.0% −30 1 = 1 = = = 1 = 0 1 = 1 0 1 = = 1 = = = 1 1 0 = 1 1 0 0 = 0 = 1 – Stockfish 2.3.1 32-bit 3039 +17−17 (−96) 21 − 11(+11−1=20) 65.6%21.0 / 32 100.0% −7 1 1 1 = 1 = = 1 = = = = 1 = = = = = 1 = 1 = = = 0 = 1 = = 1 1 = – Rybka 4.1 32-bit 3030 +12−12 (−105) 21 − 11(+16−6=10) 65.6%21.0 / 32 100.0% +2 = = = = 1 = 1 = 1 1 0 1 1 1 0 1 1 0 = 1 1 0 1 0 = = 1 = 0 1 1 1 – Vitruvius 1.11C 32-bit 3030 +15−14 (−105) 21 − 11(+12−2=18) 65.6%21.0 / 32 100.0% −17 = = = 1 = 0 = = = = 1 0 = = = 1 1 = 1 = 1 1 = 1 = = = 1 1 = 1 1 – IvanHoe 9.46h 32-bit 3023 +14−13 (−112) 20 − 12(+10−2=20) 62.5%20.0 / 32 100.0% −42 = = = = = 0 = = = = 1 = = 1 1 = = 0 1 = 1 = = = 1 1 1 1 = = 1 = – Komodo 3 32-bit 3022 +14−14 (−113) 21 − 11(+15−5=12) 65.6%21.0 / 32 100.0% −10 = 1 0 1 = 0 1 1 = 1 1 1 0 = = 1 1 = = 1 = = 0 1 = = 1 0 1 1 1 = – DeepSaros 2.3j 32-bit 3020 +15−15 (−115) 21.5 − 10.5(+15−4=13) 67.2%21.5 / 32 100.0% −6 = 1 = 1 1 = = = 1 0 1 = 0 1 = 1 1 1 = = 1 1 1 = = 0 1 = = 1 1 0 – Bouquet 1.6 32-bit 3016 +18−18 (−119) 21.5 − 8.5(+16−3=11) 71.7%21.5 / 30 100.0% +24 1 1 0 1 = 1 = = 0 1 1 1 0 1 1 1 = 1 = = = 1 1 1 = 1 1 = = = – RobboLito 0.21Q 32-bit 3013 +18−18 (−122) 21 − 9(+15−3=12) 70.0%21.0 / 30 100.0% +7 1 1 = 1 0 1 = = = = = 1 = 1 1 1 = 1 0 1 = 1 1 1 0 = = 1 1 = – Bouquet 1.5 32-bit 3010 +18−17 (−125) 23.5 − 8.5(+18−3=11) 73.4%23.5 / 32 100.0% +32 = = 1 0 1 0 1 = = 1 1 1 1 0 1 1 1 = = 1 = 1 1 = 1 = 1 = 1 = 1 1 – Sting SF 2 32-bit 3002 +17−17 (−133) 23.5 − 8.5(+17−2=13) 73.4%23.5 / 32 100.0% +17 1 1 = 1 = 1 0 = 1 1 0 = = = 1 = 1 1 1 1 = 1 1 1 = 1 = 1 = = 1 = – DeepSaros 3.0 32-bit 2998 +18−18 (−137) 24 − 8(+19−3=10) 75.0%24.0 / 32 100.0% +36 1 0 1 0 1 0 = = = = 1 1 1 1 1 1 1 = 1 = 1 1 1 1 = = 1 = 1 1 1 = – Gull II b2 32-bit 2982 +17−18 (−153) 25.5 − 6.5(+21−2=9) 79.7%25.5 / 32 100.0% +61 1 = = = = = 1 = 1 1 1 1 1 = 1 1 1 = 1 0 1 1 1 1 1 1 1 1 1 0 1 = – Gull II 32-bit 2978 +18−18 (−157) 21.5 − 8.5(+15−2=13) 71.7%21.5 / 30 100.0% −20 = = 1 1 1 1 0 1 1 = 1 1 = = = 1 1 = 1 1 1 = = 1 = = = 0 = 1 – Hiarcs 14 2961 +10−10 (−174) 25 − 7(+19−1=12) 78.1%25.0 / 32 100.0% +14 1 = 1 = 1 = 1 1 1 1 1 1 = 1 1 = 1 = 1 = 1 1 = = 1 = = 0 1 1 1 = – Fritz 13 2960 +11−11 (−175) 25.5 − 6.5(+19−0=13) 79.7%25.5 / 32 100.0% +19 1 = 1 = 1 = = 1 1 1 1 1 = 1 1 1 1 = 1 = = = = 1 1 = 1 = 1 1 = 1 – Naum 4.2 32-bit 2958 +10−10 (−177) 24.5 − 7.5(+17−0=15) 76.6%24.5 / 32 100.0% −10 1 = 1 = = = 1 1 = = 1 = = = 1 1 1 = 1 1 1 1 = = 1 = 1 1 1 = 1 = – Chiron 1.5 32-bit 2940 +16−15 (−195) 23 − 9(+16−2=14) 71.9%23.0 / 32 100.0% −58 = = 1 1 1 = = 1 1 = 1 1 = = 1 1 = = = 1 = 1 = = 1 = 0 1 1 1 0 1 – Hannibal 1.3 32-bit 2929 +17−17 (−206) 34 − 10(+25−1=18) 77.3%34.0 / 44 100.0% −29 = 1 1 1 = 1 1 = 1 1 = = 1 = = = 1 1 1 1 = = = 1 = 1 1 = = 1 1 1 1 1 = 0 = = 1 1 1 1 1 = – Rybka 2.3.2a 32-bit 2912 +9−9 (−223) 27 − 5(+22−0=10) 84.4%27.0 / 32 100.0% +23 = 1 1 1 1 1 1 = = 1 = 1 1 1 1 1 = 1 1 = 1 1 1 1 = 1 1 1 = = = 1 – Deep Junior 13.3 32-bit 2909 +14−15 (−226) 25 − 7(+19−1=12) 78.1%25.0 / 32 100.0% −39 1 1 = = 1 0 = 1 1 1 1 = 1 = 1 1 = = = = = 1 = = 1 1 1 1 1 1 1 1
### Rating changes with played games
Created in 2005-2013 by CCRL team Last games added on September 19, 2020 | 2,942 | 4,503 | {"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-2020-40 | latest | en | 0.662676 |
https://www.fmaths.com/square-root/often-asked-how-to-integrate-square-root.html | 1,621,362,825,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991288.0/warc/CC-MAIN-20210518160705-20210518190705-00367.warc.gz | 810,331,539 | 10,205 | # Often asked: How To Integrate Square Root?
## What is the integration of a square?
Example 2. We complete the square in the quadratic expression: x2−x+2=x2−2⋅x⋅12+(12)2−(12)2+2=(x−12)2−14+2=(x−12)2+74=(x−12)2+(√72)2. we find the integral: ∫dxx2−x+2=∫dx(x−12)2+(√72)2=∫duu2+(√72)2=1√72arctanu√72+C=2√7arctanx−12√72+C=2√7arctan2x−1√7+C.
## How do you integrate?
So the integral of 2 is 2x + c, where c is a constant. A “S” shaped symbol is used to mean the integral of, and dx is written at the end of the terms to be integrated, meaning “with respect to x”. This is the same “dx” that appears in dy/dx. To integrate a term, increase its power by 1 and divide by this figure.
## What is a good sentence for integrate?
1: to form into a whole: unite Her music integrates jazz and rock. 2: to make a part of a larger unit They help integrate immigrants into the community. 3: desegregate The schools are being integrated.
## How do you integrate a shadow self?
How Do We Integrate Our Shadow?
1. Your shadow self is part of you. But it doesn’t define you.
2. Journal your revelations. As you discover hidden parts of yourself, it is crucial to write it out.
3. Pay attention to your reactions.
4. Don’t second guess or intellectualize your intuition.
5. Shadow Work Is a Lifetime Process.
## What is the square root 0f?
The square root of a number is a number that, when multiplied by itself, equals the desired value. So, for example, the square root of 49 is 7 (7×7=49). The process of multiplying a number times itself is called squaring. List of Perfect Squares.
You might be interested: Question: How To Remove Square Root?
NUMBER SQUARE SQUARE ROOT
9 81 3.000
10 100 3.162
11 121 3.317
12 144 3.464 | 525 | 1,715 | {"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} | 4.21875 | 4 | CC-MAIN-2021-21 | latest | en | 0.866589 |
https://www.funwithpuzzles.com/2018/05/thermometer-sudoku-puzzle.html?showComment=1525675697196 | 1,555,627,690,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00398.warc.gz | 701,526,850 | 37,134 | ## Monday, May 7, 2018
Thermometer Sudoku or Thermo Sudoku is one of very interesting Sudoku Variation. Thermo Sudoku puzzles are very much fun to solve and these puzzles requires some deep logical calculations. Moreover Thermo Sudoku can be designed to look visually appealing. I have been creating many Thermo Sudoku and Arrow Sudoku puzzles in last one year. Most of these puzzles are submitted for some of the forthcoming Sudoku books. However there is great number of leftovers for these Sudoku puzzles which I will be slowly publishing on this website. Currently this Thermo Sudoku I am publishing as part of Fun With Sudoku Series as 331st Sudoku Puzzle in this Series.
Theme of this Thermo Sudoku puzzle is Squares as there are four central squares in the Sudoku grid. This Thermo Sudoku, I will rate as tough which will requires some good logical deductions along with lots of pencil marks to solve this Thermo Sudoku.
## Rules of Thermometer Sudoku Puzzle
Standard Sudoku Rules apply. Additionally some thermometer shapes are in the grid, digits must be strictly increasing from the round bulb to the flat end.
Thermometer Sudoku Puzzle (Fun With Sudoku #331)
Answer of this Themro Sudoku #331 puzzle can be viewed by clicking on button. Please do give your best try to solve this Sudoku puzzle before looking at the answer.
This Thermometer Sudoku Puzzles I am publishing as #331st Sudoku puzzle in Fun With Sudoku Series. Here are the next and previous Sudoku puzzles published in this series.
1. 32:24 (logical solve). Tough but enjoyable sudoku
2. Finished in third attempt
Time taken was aprx 45 minutes
That is why I didn't comment
This is tough Thermo Sudoku and requires some of very narrow calculations. However this Thermo Sudoku completely logically solvable.
3. 21:35.Had to use uniqueness twice to complete the sudoku .Very hard one.
1. Yes, this was hard one. However I solved this Thermo Sudoku logically by fixing positions of 1 and 2 in central box.
4. Real hard one .. broke once and did it again with total 39 mins
It was challenging.
A comment doesn't cost a thing. Please drop a comment below to boost the author's morale. | 463 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-18 | longest | en | 0.941059 |
http://cambodia-financial-market.blogspot.com/2011/12/rate-of-return-basics.html | 1,532,180,900,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00579.warc.gz | 62,106,221 | 18,793 | Friday, December 09, 2011
Rate of Return Basics
CFA Level 1 - Portfolio Management
The Required Rate of Return
The required rate of return is the nominal rate of return that an investor needs in order to make an investment worthwhile.
This return varies over time and is comprised of the following:
• Real risk-free rate
Real risk-free rate of return
The real risk-free rate of return (Rf) is the minimum return an investor requires. This rate does not take into account expected inflation and the capital market environment.
Formula 17.1 Real risk free rate (Rf) = (1 + nominal risk-free rate) - 1 (1 + inflation rate)
Example: Real risk-free rate of return
Determine the real risk-free rate if the nominal risk-free rate is 8% and the inflation rate is 3%.
Rf = (1 + 0.08) - 1 = 4.85%
(1 + 0.03)
Nominal risk-free rate of return (Rnominal)
This is simply the real risk-free rate of return adjusted for inflation.
Formula 17.2 Nominal risk-free rate = (1 + risk-free rate) x (1 + rate of inflation) - 1
Example: Nominal risk-free rate of return
Determine the nominal risk-free rate of return if the risk-free rate is 3% and the rate of inflation is 3%. | 311 | 1,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-30 | latest | en | 0.800405 |
https://www.biblewheel.com/content.php?46-On-Integrity-as-the-Highest-Value&s=107706bb4d06a162d890a225135b28e9 | 1,596,763,434,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00585.warc.gz | 575,559,380 | 18,550 | • ### Recent Forum Posts
#### 9 1 1
Count the number of the beast = 1 6 8 0
The first occurrence of 1 6 8 0 completes at digit # 1 3 8 0 1,
3 .(1 4 . . . .
Scott Today, 03:05 PM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
and not to be overlooked:
"nineveh" = 121
"tarshish" = 1210
by fleeing to Tarshish he
sylvius Yesterday, 11:37 PM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
Interesting
https://www.rug.nl/research/portal/f...6738/Jonah.pdf
Wouldn't this be
sylvius Yesterday, 10:41 PM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
322, number of words in tjhe chapters 3 and 4 of Jonah, is gematria of "keves" = lamb of God that carries the sin of the world
sylvius Yesterday, 12:35 PM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
The Atbash of "neder" is 112, the number of words in the second chapter of Jonah.
(Atbash: Tav = 1, Shin =2 , Resh = 3 , ......
sylvius Yesterday, 08:37 AM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
Also Jacob uttereed a vow,
Genesis 28:20-22,
And Jacob uttered a vow, saying, "If God will be with me, and He will guard me
sylvius 08-04-2020, 10:39 PM
#### 9 1 1
Count the number = 1 0 7 4
2 5 8 + 1 0 7 4 + 2 5 8 = 1 5 9 0
6 6 6 + 2 5 8 + 6 6 6 = 1 5 9 0
2 5
Scott 08-04-2020, 11:17 AM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
the book of Jonah is written with 254 + 434 = 688 words.
254 = "neder" = vow
434 = 'delet"= door , coinciding
sylvius 08-04-2020, 02:07 AM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
Numbers 13:15 is lower in gematria, viz. 254
For the tribe of Gad, Geuel the son of Machi.
https://www.biblewheel.com/GR/GR_Database.php
sylvius 08-03-2020, 11:00 PM
#### Announcement
Good day everyone.
The Chinese have a saying/curse. "May you live in interesting times." Well, these are interesting times, and
thebluetriangle 08-03-2020, 10:34 AM
#### 9 1 1
I've known for months now that 2 7 2 4{Nov.7/20), and 8 6 5 8,
which is the sum of the words in Panin's N.T.
translation
Scott 08-03-2020, 10:32 AM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
this is bullshit, at least IMHO
https://www.biblewheel.com/GR/GR_499.php
sylvius 08-03-2020, 04:10 AM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
Aha!
After the plague:
Isaiah 60:19,
You shall no longer have the sun for light by day, and for brightness, the
sylvius 08-03-2020, 02:24 AM
#### Anyone knows the highest and lowest gematria of a single verse in the Tanakh
Numbers 25:19, וַיְהִ֖י אַֽחֲרֵ֣י הַמַּגֵּפָ֑ה, "vay'hi acharei hamagefah", might be the lowest in gematra, viz. 383
sylvius 08-02-2020, 10:52 PM
#### 9 1 1
The number 2 1 6 occurs first in Pi from digit # 9 9 1 to # 9 9 3
Digit # 7 7 7 + 2 1 6 digits = digit # 9 9 3
The 2 1 6
Scott 08-02-2020, 04:11 PM
• # On Integrity as the Highest Value
INTEGRITY:
1. The state of being whole, undivided, perfect in composition; unity, wholeness, completeness.
2. The quality of being honest and having strong moral principles; moral uprightness.
Integrity is the grand unifying concept of all knowledge and philosophy. It is the way, the means, and the end all in one. It is the ultimate value that subsumes all other values and explains why they are valuable. It is what defines a self or any entity that exists. Its absence literally entails disintegration, corruption, and confusion. Integrity is the root of all that is good and true. It is an end in itself.
There has been much philosophical, psychological, and ethical discussion about the meaning of integrity. The word carries many rich and diverse overtones but there is one idea that underlies them all, unity. The concept that unifies all philosophy and knowledge is itself that which it describes, unity. It is self-coherent and self-reflective. It is the most stable of all conceptions. It is fundamental to the very concept of being itself.
I have been fascinated with the concept of unity and self-reflective self-coherence for many years. This article welcomes the New Year 2013 as the Year of Love and Unity and gives a brief overview of the lines of research I will be pursuing. There is no limit to what could be written on this theme. This article is an outline of my basic insights and intuitions that move me to think that integrity is the key that opens every door. It unifies everything from the most fundamental meaning of existence itself (Ontology) to the highest value any sentient being can know (Love) which is itself the ultimate unity, the mystical vision that All is One.
Ontology: The Integrity of Being
Ontology is the branch of metaphysics that deals with the nature of being, what it means for a thing to be, to exist. There are two kinds of entities - elementary and compound. An elementary (or fundamental) entity is not made up of parts. Most, if not all fundamental entities are theoretical constructs; all "things" we actually encounter in this world via our senses are compound entities. A fundamental object with no parts is a simple unity, like a zero dimensional point or elementary particles like electrons and quarks (though the latter may be found to be composite like other particles that were once thought to be elementary like protons and neutrons). The relation between the parts and the whole raises a lot of questions. Aristotle wrote about degrees of integrity, thinking for example that a rigid body had more integrity than a jointed one. The concept of integrity carries the connotation of a harmonious integrated relation between the parts that constitute a thing. A thing has, by definition, ontological integrity which is really a redundancy because without integrity a thing does not exist as such.
Logic: The Integrity of Thought
Logic and ontology are closely related. The three classical Laws of Logic are formal expressions of the integrity implied by the idea of existence, what it means for a thing to be what it is:
• The Law of Identity: It is always true that A is A.
• The Law of Non-Contradiction: It is never true that A is not A.
• The Law of Excluded Middle: It is always true that a thing is either A or not A.
MRI of the anterior cingulate cortex which signals conflict.
All philosophy, science, and rational discourse are based on these three laws of logic which in turn are based on the idea of integrity. This logic is built into our brains - it has a neurological basis. Conflicting thoughts, beliefs, or desires evoke a painful state of mind called cognitive dissonance which is to the mind as physical pain is to the body. There is a compelling evolutionary explanation for this function. Without it, our minds would quickly fall into delusion and disintegrate like the bodies of children born with congenital insensitivity to pain who repeatedly injure themselves. The biological basis of cognitive dissonance is found in the anterior cingulate cortex which is involved in error control, conflict management, and motivation. There is evidence "suggesting that the more it signals conflict, the more dissonance a person experiences and the more their attitudes may change." Cognitive dissonance is a critical evolutionary adaptation that makes us what we are; Homo Sapiens, that is, the rational species of the great ape genus Homo. The healthy response to cognitive dissonance is to change our beliefs and behavior to better accord with reality. Unfortunately, not all responses are healthy. Perhaps the most notorious example of what happens when you break the relation between integrity and logic is seen in the Monica Lewinski case when President Clinton questioned the meaning of "is" to hide his broken integrity. I documented the disintegrating effect of habitual suppression of cognitive dissonance in my article The Art of Rationalization: A Case Study of Christian Apologist Rich Deem.
Epistemology: The Integrity of Knowledge
Epistemology explores the question "How do we know anything?". Like all branches of philosophy, it has a long history with a thousand rabbit trails one could follow. But for this overview, I focus on the essential answer: our confidence in any conclusion is based directly on how many independent lines of evidence support it. It is the unity of multiplied independent and diverse witnesses that convince. For example, by sight I may think there is a pool of water in the desert, but when I try to touch it I realize it was a mirage. There was a lack of consilience between my different senses. When all my senses cohere, I have confidence that the thing sensed is real and this confidence is greatly amplified if I find my observations cohere with those of others. This is very practical, ancient wisdom. It governs our courts of law, our daily lives, and is oft repeated in biblical verses such as "in the mouth of two or three witnesses shall ever word be established" (2 Corinthians 13:1). Epistemology, ontology, and logic are fundamentally based on the concept of integrity which is why they are mutually interdependent, as explained by Hugh G. Gauch, Jr. his book Scientific Method in Practice (quote available online):
In ordinary discourse, ontology, epistemology, and logic are reasonably distinct and recognizable topics within philosophy. But at the point where discourse begins, those topics fuse together. The reason is that epistemology presumes ontology, because what we know depends on what exists. But also ontology presumes epistemology, because what we can become aware of depends on our sensory and cognitive faculties. And logic is operating in any rational discourse.
Careful distinctions are required for discourse, but all reality is ultimately one. It is this integral nature of reality that guides science, as we shall presently see.
Consilience: The Integrity of Science
Scientific theories are designed to give a comprehensive explanation of a body of facts in terms of a few elementary principles. Successful theories reveal the underlying unity of apparently diverse phenomena. A Theory of Everything, often called the Holy Grail of physics, would give a fully unified explanation of all physical phenomena within a single theoretical framework based on a small set of axioms. Newton made the first great step in this direction with his three laws of motion. James Clerk Maxwell unified Optics with Electromagnetism. Einstein unified Mechanics with Electromagnetism in his theory of relativity and simultaneously exposed the fundamental unity of space and time. The historical trajectory of science is clear - it is continuously approaching a full integration and unification of all knowledge. I expand on this theme in my article The Logic of Love: A Natural Theory of Morality where also I show that the very essence of science, objectivity, is based on the concept of invariance which reveals the unity that defines the thing that really is. E. O. Wilson explored what this ultimate unity of all knowledge - from astronomy to zoology - might look like in his excellent book Consilience: The Unity of Knowledge.
Morality: The Integrity of Action
Conscience is simply a function of, or one aspect of, the integrity of being. It arises from this integrity and therefore is it. Since conscience is dependent on the source of being (Being), it never leaves the integrity of being. This function is an inclusive one - even though conscience appears to deal mostly with our "social" orientation, the being's integrity deals with our entire existence.
Reflections of Being by Peter Ralston (pg. 67)
There is good reason the word "integrity" means both "wholeness" and "moral soundness." Morality flows from the native integrity of the self. Integrity is the ontology of morality. Nietzsche caught a glimpse of this in his chapter called The Despisers of the Body in Thus Spake Zarathustra:
Behind your thoughts and feelings, my brother, there stands a mighty ruler, an unknown sage - whose name is Self. In your body he dwells; he is your body. There is more reason in your body than in your best wisdom.
We live in an age of great moral confusion as we transition from theism to atheism. Two thousand years of Christian theism has twisted our innate moral sense away from its natural foundation in our own selves and replaced it with arbitrary rules dictated by an inscrutable god. Atheists react by denying there are any objective moral values at all, not knowing there is a third way. Both the theist and the atheist are wrong. This error is so deeply engrained that some Christian apologists, most notably William Lane Craig, are confident that merely asserting there would be no morals without God to "ground" them is sufficient to prove that God exists. There is a great irony here. He exposes the fundamental logical incoherence of his argument in his article Keeping Moral Epistemology and Moral Ontology Distinct. He simultaneously insists that "keeping the distinction between moral epistemology and moral ontology clear is the most important task in formulating and defending a moral argument for God’s existence of the of the type I defend" even as he admits that "we do not need to know or even believe that God exists in order to discern objective moral values." Simply stated, Craig destroys the natural unity of being and knowing which we explored above. Just as Bill Clinton had to obfuscate the meaning of "is" to hide his damaged integrity, so Bill Craig must attack the inextricable unity of ontology and epistemology to foist his proof of God upon his philosophically unsophisticated audience. His error is as obvious as it is egregious. Objective moral values are real and are based on the objective ontology of what we are - the integrity that defines us as a self and as an integral member of society. The moral value of any action depends upon the nature of that action and how it affects others. We know if something is moral because of what it is and how it affects the unity of self and others. Ontology and epistemology, though separable in principle for careful philosophical discourse, are inseparable in practice. I explain this in more detail in my article The Golden Rule and the Foundation of Objective Morality.
Love: The Integrity of Consciousness
[M]ature love is union under the condition of preserving one’s integrity, one’s individuality. Love is an active power in man; a power which breaks through the walls which separate man from his fellow men, which unites him with others; love makes him overcome the sense of isolation and separateness, yet it permits him to be himself, to retain his integrity. In love the paradox occurs that two beings become one and yet remain two.
~ The Art of Loving by Erich Fromm (1956)
The essential nature of love is unity. Self love and love for others are one love. Love is the root and reason of morality. It explains all things. It is the only reason to live and the only answer to the question "to be or not to be." It flows naturally from our ontology as conscious beings - indeed, love is the consciousness of unity of self and others, as explained in my article The Logic of Love: A Natural Theory of Morality where I develop an objective scientific theory of morality. Philosophers and theologians have long perceived a close relation between the highest value, oft referred to as the Good, and the concept of Being in and of itself. Evil is often defined as a privation of good, a loss of being. Anything that threatens the integrity of a self, such as disease, age, or violence, is seen as a kind of evil. In Biblical terms, "Love does no harm to a neighbor; therefore love is the fulfillment of the law" (Romans 13:10). Love is the highest value, and integrity is the sure guide that will lead every person home. Trust your heart. Be very sensitive to cognitive dissonance. Avoid rationalizations. Seek the truth that springs from the integrity of your own body.
Let 2013 be the Year of Love and Unity. Pursue it with all the integrity of your heart and mind.
Love is an end in itself.
.
Leave Comment | 3,831 | 16,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-34 | latest | en | 0.845999 |
https://pdglive.lbl.gov/Particle.action?init=0&node=M300&home=MXXX050 | 1,723,123,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00105.warc.gz | 359,107,252 | 14,897 | #### Further States
This section contains states observed by a single group or states poorly established that thus need confirmation. QUANTUM NUMBERS, MASSES, WIDTHS, AND BRANCHING RATIOS
$\mathbf {{{\mathit X}{(360)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{?+})$
$\mathbf {{{\mathit X}{(1070)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(0{}^{++})$
$\mathbf {{{\mathit X}{(1110)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(even{}^{++})$
$\mathbf {{{\mathit f}_{{{0}}}{(1200-1600)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{++})$
$\mathbf {{{\mathit X}{(1420)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 2{}^{+}(0{}^{++})$
$\mathbf {{{\mathit X}{(1545)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{++})$
$\mathbf {{{\mathit X}{(1575)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(1{}^{--})$
$\mathbf {{{\mathit X}{(1600)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 2{}^{+}(2{}^{++})$
$\mathbf {{{\mathit X}{(1650)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(?{}^{?-})$
$\mathbf {{{\mathit X}{(1730)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{?+})$
$\mathbf {{{\mathit f}_{{{2}}}{(1750)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{++})$
$\mathbf { \Gamma\mathrm {({{\mathit K}} {{\overline{\mathit K}}})} }$
$\mathbf { \Gamma\mathrm {({{\mathit \gamma}} {{\mathit \gamma}})} }$
$\mathbf { \Gamma\mathrm {({{\mathit \pi}} {{\mathit \pi}})} }$
$\mathbf { \Gamma\mathrm {({{\mathit \eta}} {{\mathit \eta}})} }$
$\mathbf {{{\mathit X}{(1775)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(?{}^{-+})$
$\mathbf {{{\mathit X}}(1850 - 3100)}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(1{}^{--})$
$\mathbf {{{\mathit X}{(1855)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit X}{(1870)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(2{}^{??})$
$\mathbf {{{\mathit a}_{{{3}}}{(1875)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(3{}^{++})$
$\mathbf {B( {{\mathit a}_{{{3}}}{(1875)}} \rightarrow {{\mathit f}_{{{2}}}{(1270)}} {{\mathit \pi}})/B( {{\mathit a}_{{{3}}}{(1875)}} \rightarrow {{\mathit \rho}} {{\mathit \pi}})}$
$\mathbf {B( {{\mathit a}_{{{3}}}{(1875)}} \rightarrow {{\mathit \rho}_{{{3}}}{(1690)}} {{\mathit \pi}})/B( {{\mathit a}_{{{3}}}{(1875)}} \rightarrow {{\mathit \rho}} {{\mathit \pi}})}$
$\mathbf {{{\mathit a}_{{{1}}}{(1930)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(1{}^{++})$
$\mathbf {{{\mathit X}{(1935)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(1{}^{-?})$
$\mathbf {{{\mathit \rho}_{{{2}}}{(1940)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(2{}^{--})$
$\mathbf {{{\mathit \omega}_{{{3}}}{(1945)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(3{}^{--})$
$\mathbf {{{\mathit a}_{{{2}}}{(1950)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(2{}^{++})$
$\mathbf {{{\mathit \omega}{(1960)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(1{}^{--})$
$\mathbf {{{\mathit b}_{{{1}}}{(1960)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(1{}^{+-})$
$\mathbf {{{\mathit h}_{{{1}}}{(1965)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(1{}^{+-})$
$\mathbf {{{\mathit f}_{{{1}}}{(1970)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(1{}^{++})$
$\mathbf {{{\mathit X}{(1970)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit X}{(1975)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit \omega}_{{{2}}}{(1975)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(2{}^{--})$
$\mathbf {{{\mathit a}_{{{2}}}{(1990)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(2{}^{++})$
$\mathbf {\Gamma\mathrm {({{\mathit \gamma}} {{\mathit \gamma}})} \Gamma\mathrm {({{\mathit \pi}^{+}} {{\mathit \pi}^{-}} {{\mathit \pi}^{0}})} / \Gamma\mathrm {(total)}}$
$\mathbf {{{\mathit \rho}{(2000)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(1{}^{--})$
$\mathbf {{{\mathit f}_{{{2}}}{(2000)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{++})$
$\mathbf {{{\mathit X}{(2000)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(?{}^{?+})$
$\mathbf {{{\mathit X}{(2000)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(4{}^{++})$
$\mathbf {{{\mathit \eta}{(2010)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{-+})$
$\mathbf {{{\mathit \pi}_{{{1}}}{(2015)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(1{}^{-+})$
$\mathbf {{{\mathit a}_{{{0}}}{(2020)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(0{}^{++})$
$\mathbf {{{\mathit X}{(2020)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit h}_{{{3}}}{(2025)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(3{}^{+-})$
$\mathbf {{{\mathit b}_{{{3}}}{(2030)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(3{}^{+-})$
$\mathbf {{{\mathit a}_{{{2}}}{(2030)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(2{}^{++})$
$\mathbf {{{\mathit a}_{{{3}}}{(2030)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(3{}^{++})$
$\mathbf {{{\mathit \eta}_{{{2}}}{(2030)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{-+})$
$\mathbf {B({{\mathit a}_{{{2}}}} {{\mathit \pi}})_{\mathit L=0}/B({{\mathit a}_{{{2}}}} {{\mathit \pi}})_{\mathit L=2}}$
$\mathbf {B({{\mathit a}_{{{0}}}} {{\mathit \pi}})/B({{\mathit a}_{{{2}}}} {{\mathit \pi}})_{\mathit L=2}}$
$\mathbf {B({{\mathit f}_{{{2}}}} {{\mathit \eta}})/B({{\mathit a}_{{{2}}}} {{\mathit \pi}})_{\mathit L=2}}$
$\mathbf {{{\mathit f}_{{{3}}}{(2050)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(3{}^{++})$
$\mathbf {{{\mathit f}_{{{0}}}{(2060)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{++})$
$\mathbf {{{\mathit \pi}{(2070)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(0{}^{-+})$
$\mathbf {{{\mathit X}{(2075)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(1{}^{+?})$
$\mathbf {{{\mathit X}{(2080)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit X}{(2080)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(3{}^{-?})$
$\mathbf {{{\mathit a}_{{{1}}}{(2095)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(1{}^{++})$
$\mathbf {B( {{\mathit a}_{{{1}}}{(2095)}} \rightarrow {{\mathit f}_{{{1}}}{(1285)}} {{\mathit \pi}}) / B( {{\mathit a}_{{{1}}}{(2095)}} \rightarrow {{\mathit a}_{{{1}}}{(1260)}})}$
$\mathbf {{{\mathit \eta}{(2100)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{-+})$
$\mathbf {{{\mathit X}{(2100)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(0{}^{??})$
$\mathbf {{{\mathit X}{(2110)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(3{}^{-?})$
$\mathbf {{{\mathit X}{(2120)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(0{}^{??})$
$\mathbf {{{\mathit f}_{{{2}}}{(2140)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{++})$
$\mathbf {{{\mathit X}{(2150)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(2{}^{+?})$
$\mathbf {{{\mathit a}_{{{2}}}{(2175)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(2{}^{++})$
$\mathbf {{{\mathit \eta}{(2190)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{-+})$
$\mathbf {{{\mathit \omega}_{{{2}}}{(2195)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(2{}^{--})$
$\mathbf {{{\mathit X}{(2210)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit X}_{{{2}}}{(2210)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{++})$
$\mathbf {{{\mathit h}_{{{1}}}{(2215)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(1{}^{+-})$
$\mathbf {{{\mathit \rho}_{{{2}}}{(2225)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(2{}^{--})$
$\mathbf {{{\mathit \rho}_{{{4}}}{(2230)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(4{}^{--})$
$\mathbf {{{\mathit b}_{{{1}}}{(2240)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(1{}^{+-})$
$\mathbf {{{\mathit f}_{{{2}}}{(2240)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{++})$
$\mathbf {{{\mathit b}_{{{3}}}{(2245)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(3{}^{+-})$
$\mathbf {{{\mathit \eta}_{{{2}}}{(2250)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{-+})$
$\mathbf {{{\mathit \pi}_{{{4}}}{(2250)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(4{}^{-+})$
$\mathbf {{{\mathit \omega}_{{{4}}}{(2250)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(4{}^{--})$
$\mathbf {{{\mathit \omega}_{{{5}}}{(2250)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(5{}^{--})$
$\mathbf {{{\mathit \omega}_{{{3}}}{(2255)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(3{}^{--})$
$\mathbf {{{\mathit a}_{{{4}}}{(2255)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(4{}^{++})$
$\mathbf {{{\mathit a}_{{{2}}}{(2255)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(2{}^{++})$
$\mathbf {{{\mathit X}{(2260)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(4{}^{+?})$
$\mathbf {{{\mathit \rho}{(2270)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{+}(1{}^{--})$
$\mathbf {{{\mathit a}_{{{1}}}{(2270)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(1{}^{++})$
$\mathbf {{{\mathit h}_{{{3}}}{(2275)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(3{}^{+-})$
$\mathbf {{{\mathit a}_{{{3}}}{(2275)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(3{}^{++})$
$\mathbf {{{\mathit \pi}_{{{2}}}{(2285)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(2{}^{-+})$
$\mathbf {{{\mathit \omega}_{{{3}}}{(2285)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(3{}^{--})$
$\mathbf {{{\mathit \omega}{(2290)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(1{}^{--})$
$\mathbf {{{\mathit f}_{{{2}}}{(2295)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(2{}^{++})$
$\mathbf {{{\mathit f}_{{{3}}}{(2300)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(3{}^{++})$
$\mathbf {{{\mathit f}_{{{1}}}{(2310)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(1{}^{++})$
$\mathbf {{{\mathit \eta}{(2320)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{-+})$
$\mathbf {{{\mathit \eta}_{{{4}}}{(2330)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(4{}^{-+})$
$\mathbf {{{\mathit \omega}{(2330)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(1{}^{--})$
$\mathbf {{{\mathit X}{(2340)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit \pi}{(2360)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(0{}^{-+})$
$\mathbf {{{\mathit X}{(2360)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(4{}^{+?})$
$\mathbf {{{\mathit X}{(2440)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(5{}^{-?})$
$\mathbf {{{\mathit a}_{{{6}}}{(2450)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 1{}^{-}(6{}^{++})$
$\mathbf {{{\mathit X}{(2540)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(0{}^{++})$
$\mathbf { \Gamma\mathrm {({{\mathit \gamma}} {{\mathit \gamma}})} {\times } B({{\mathit K}} {{\overline{\mathit K}}}) }$
$\mathbf {{{\mathit X}{(2600)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {B( {{\mathit J / \psi}} \rightarrow {{\mathit \gamma}} {{\mathit X}{(2600)}}) {\times } B( {{\mathit X}{(2600)}} \rightarrow {{\mathit f}_{{{0}}}{(1500)}} {{\mathit \eta}^{\,'}}) {\times } B( {{\mathit f}_{{{0}}}{(1500)}} \rightarrow {{\mathit \pi}^{+}} {{\mathit \pi}^{-}})}$
$\mathbf {B( {{\mathit J / \psi}} \rightarrow {{\mathit \gamma}} {{\mathit X}{(2600)}}) {\times } B( {{\mathit X}{(2600)}} \rightarrow {{\mathit X}{(1540)}} {{\mathit \eta}^{\,'}}) {\times } B( {{\mathit X}{(1540)}} \rightarrow {{\mathit \pi}^{+}} {{\mathit \pi}^{-}})}$
$\mathbf {{{\mathit K}_{{{0}}}^{*}{(2600)}}}$ $\mathit I(\mathit J{}^{P}) = 1/2(0{}^{+})$
$\mathbf {{{\mathit X}{(2632)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {B( {{\mathit X}{(2632)}} \rightarrow {{\mathit D}^{0}} {{\mathit K}^{+}})/B( {{\mathit X}{(2632)}} \rightarrow {{\mathit D}_{{{s}}}^{+}} {{\mathit \eta}})}$
$\mathbf {{{\mathit X}{(2680)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit X}{(2710)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(6{}^{+?})$
$\mathbf {{{\mathit X}{(2750)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(7{}^{-?})$
$\mathbf {{{\mathit f}_{{{6}}}{(3100)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{+}(6{}^{++})$
$\mathbf {{{\mathit X}{(3250)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$ 3-Body Decays
$\mathbf {{{\mathit X}{(3250)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$ 4-Body Decays
$\mathbf {{{\mathit X}{(3350)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = ?{}^{?}(?{}^{??})$
$\mathbf {{{\mathit \psi}{(4500)}}}$ $\mathit I{}^{G}(\mathit J{}^{PC}) = 0{}^{-}(1{}^{--})$
FOOTNOTES | 5,631 | 12,405 | {"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} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.208503 |
https://www.fractii.ro/fractions-theory-how-to-subtracting.php?lang=en | 1,566,639,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00197.warc.gz | 819,840,673 | 8,464 | # Fractions, theory: subtracting ordinary math fractions - how to subtract fractions with unlike (different) denominators?
## Tutoring and a practical example: subtracting multiple ordinary (simple, common) fractions
### To subtract fractions with different denominators (unlike denominators), fractions should be first brought to the same denominator. How is it done?
• #### 2. To bring all the fractions to the same denominator, you must calculate the fractions' denominators lowest common multiple, LCM:
• Fractions' denominators must be factored down to their constituent prime factors (denominators prime factorization).
• The denominators' lowest common multiple, LCM, will contain all of the unique prime factors, by the highest powers.
• Whether you don't know how or you'd like to practice the process, go to the address on the numere-prime.ro website: LCM, lowest common multiple of two numbers.
• #### 3. Calculate each fraction's expanding number, the non-zero number that will be used to multiply fraction's both numerator and denominator in order to bring all the fractions to the same common denominator:
• Divide the lowest common multiple, LCM, calculated above, by each fraction's denominator, getting the expanding number; that non-zero number will be used to multiply both the numerator and denominator of each fraction.
• #### 4. Expand fractions - multiply each fraction's both numerator and denominator by its expanding non-zero number.
• At this point, fractions are brought to the same denominator. In order to subtract the fractions simply subtract fractions' numerators. The end fraction will have as a denominator the lowest common multiple calculated above.
### An example of subtracting three ordinary fractions with different denominators (unlike denominators): 6/90 - 16/24 - 30/75
#### Reduce each fraction to lowest terms (simplify):
• Factor each fraction's both numerator and denominator down to their constituent prime factors (numerator and denominator prime factorization), then divide each fraction's both numerator and denominator by their greatest common factor, GCF (also called the greatest common divisor, GCD, or the highest common factor, HCF); GCF has only their unique common prime factors, by the lowest powers.
• Whether you don't know how to calculate the greatest common divisor of two numbers or you just need some refreshment, access this page on numere-prime.ro website: greatest common factor or divisor (GCF, GCD).
• Reduce the 1st fraction to lowest terms (simplify it):
• 6/90 = (2 * 3) / (2 * 32 * 5) = ((2 * 3) ÷ (2 * 3)) / ((2 * 32 * 5) ÷ (2 * 3)) = 1 / (3 * 5) = 1/15
• Reduce the 2nd fraction to lowest terms (simplify it):
• 16/24 = 24 / (23 * 3) = (24 ÷ 23) / ((23 * 3) ÷ 23) = 2/3
• Reduce the 3rd fraction to lowest terms (simplify it):
• 30/75 = (2 * 3 * 5) / (3 * 52) = ((2 * 3 * 5) ÷ (3 * 5)) / ((3 * 25) ÷ (3 * 5)) = 2/5
• At this point all the fractions are reduced (simplified): 6/90 - 16/24 - 30/75 = 1/15 - 2/3 - 2/5
#### Calculate the lowest common multiple, LCM
• Next, we calculate the lowest common multiple LCM of all three fractions' denominators. For that, factor each fraction's denominator down to its constituent prime factors (each fraction's denominator prime factorization); then take ALL the unique denominators' prime factors, by the highest powers.
• First fraction's denominator prime factorization: 15 = 3 * 5
• Second fraction's denominator prime factorization: 3 is already a prime number, it cannot be prime factorized
• Third fraction's denominator prime factorization: 5 is a prime number, it cannot be prime factorized
• The lowest common multiple LCM of all fractions' denominators must contain all of the unique denominators' prime factors by the highest powers:
• LCM (15; 3; 5) = LCM (3 * 5; 3; 5) = 3 * 5 = 15
#### Calculate each fraction's expanding number:
• Calculate each fraction's expanding number - this non-zero number will be used to multiply each fraction's both numerator and denominator. This number is calculated by dividing the lowest common multiple, LCM, by each fraction's denominator:
• first fraction expanding number: 15 ÷ 15 = 1
• second fraction expanding fraction: 15 ÷ 3 = 5
• third fraction expanding fraction: 15 ÷ 5 = 3
#### Bring the fractions to the same denominator:
• To bring the fractions to the same denominator, expand them (multiply each fraction's both numerator and denominator by its corresponding expanding number calculated above):
• the first fraction stays unchanged: 1/15 = (1 * 1)/(1 * 15) = 1/15
• the second fraction becomes: 2/3 = (5 * 2)/(5 * 3) = 10/15
• the third fraction becomes: 2/5 = (3 * 2)/(3 * 5) = 6/15
#### The final result of subtracting the fractions:
• 6/90 - 16/24 - 30/75 = 1/15 - 2/3 - 2/5 = 1/15 - 10/15 - 6/15 = - 15/15 = - 1
• The end fraction is an improper one - the absolute value of the numerator is equal to the absolute value of the denominator. | 1,282 | 4,943 | {"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} | 4.75 | 5 | CC-MAIN-2019-35 | latest | en | 0.880256 |
https://www2.clarku.edu/faculty/djoyce/elements/bookXI/propXI15.html | 1,679,945,716,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00178.warc.gz | 1,175,093,384 | 2,041 | # Proposition 15
If two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel.
Let the two straight lines AB and BC meeting one another be parallel to the two straight lines DE and EF meeting one another not in the same plane.
I say that the plane produced through AB and BC and the plane produced through DE and EF do not meet one another.
XI.11
Draw BG from the point B perpendicular to the plane through DE and EF to where it meets the plane at the point G.
I.31
Draw GH through G parallel to ED, and GK parallel to EF.
XI.Def.3
Now, since BG is at right angles to the plane through DE and EF, therefore it makes right angles with all the straight lines which meet it and lie in the plane through DE and EF.
But each of the straight lines GH and GK meets it and lies in the plane through DE and EF, therefore each of the angles BGH and BGK is right.
And, since BA is parallel to GH, therefore the sum of the angles GBA and BGH equals two right angles.
But the angle BGH is right, therefore the angle GBA is also right. Therefore GB is at right angles to BA. For the same reason GB is also at right angles to BC.
XI.4
Since then the straight line GB is set up at right angles to the two straight lines BA and BC which cut one another, therefore GB is also at right angles to the plane through BA and BC.
XI.14
But planes to which the same straight line is at right angles are parallel, therefore the plane through AB and BC is parallel to the plane through DE and EF.
Therefore, if two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel.
Q.E.D.
## Guide
This proposition is not used in the rest of the Elements. | 410 | 1,838 | {"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-2023-14 | latest | en | 0.94966 |
https://www.scholarify.in/ugc-net-mathematical-reasoning-and-aptitude/ | 1,696,256,598,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00100.warc.gz | 1,051,215,124 | 46,113 | # Mathematical Reasoning and Aptitude | MCQ | Paper 1
##### UGC NET Study Materials for Paper 1 (eBook with MCQ) ⇒ BUY NOW
Mathematical Reasoning and Aptitude is not a new topic of the revised syllabus of UGC NET Paper 1. In an updated syllabus, the unit “Reasoning (Including Mathematical Reasoning)” has been renamed as “ Mathematical Reasoning and Aptitude.” There are no changes in the subtopics under this unit.
You will learn in this unit the basic concept and types of Reasoning, number series, letter series, coding-decoding, blood relationship, etc.
## Free Notes of UGC NET Mathematical Reasoning and Aptitude
We provide free study materials or notes of Mathematical Reasoning and Aptitude for the aspiring candidates. You will learn here according to the new revised syllabus of NTA UGC NET Paper 1.
There are THREE topics of Mathematical Reasoning and Aptitude of UGC NET Paper 1, in which No topic is new. The study materials available on this website are prepared by JRF qualified who are either working as an Assistant Professor or a Ph.D. Scholar.
Click on the link below to read the topics
### Types of Reasoning
Objectives of the topic: This topic covers the concept and types of Reasoning. After learning this topic, you will enhance the analytical skill which will be used in your academic work in the future.
### Number series, Letter series, Codes, and Relationships
Objectives of the topic: In this topic, you will learn only limited reasoning topics which is useful for UGC NET Paper 1 like Number System, Letter Series, Coding & Decoding, Blood relationship, etc. You do not have to go for an in-depth study of Reasoning.
### Mathematical Aptitude (Fraction, Time & Distance, Ratio, Proportion and Percentage, Profit and Loss, Interest and Discounting, Averages, etc.)
Objectives of the topic: This portion will teach you the mathematical sills essential for analytics and calculation. Do not go for an in-depth study of Mathematical aptitude. The topics are limited, like basic calculation related to fraction, ratio, percentage, profit, and loss, etc.
## UGC NET Mathematical Reasoning and Aptitude MCQ
1) Insert the missing number in the following series: 4 16 8 64 ? 256
(A) 16 (B) 24 (C) 32 (D) 20
2) Pointing to a man, a woman said, “His mother is the only daughter of my mother.” How is the woman related to the man?
A) Mother (B) Daughter (C) Sister (D) Grandmother
3) Sahil said, “Siddhesh’s mother is the only daughter of my mother.” How is Sahil related to Siddhesh?
A) Grandfather (B) Father (C) Brother (D) Maternal Uncle
Directions: In each of the following questions, one term in the number series is wrong. Find the wrong term.
4) 125, 126, 124, 127, 123, 129
(A) 123 (B) 124 (C) 126 (D) 129
5) 10, 26, 74, 218, 654, 1946, 5834
(A) 26 (B) 74 (C) 218 (D) 654 | 744 | 2,899 | {"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-2023-40 | longest | en | 0.877376 |
https://kratos-wiki.cimne.upc.edu/index.php?title=How_to_use_the_Spatial_Containers&diff=next&oldid=5880&printable=yes | 1,669,559,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00848.warc.gz | 401,100,271 | 8,702 | # How to use the Spatial Containers
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Revision as of 10:35, 5 May 2012 (view source)Nelson (Talk | contribs) (→Using the Spatial Search Algorithm)← Older edit Revision as of 10:36, 5 May 2012 (view source)Nelson (Talk | contribs) Newer edit → Line 33: Line 33: * Points Searches: Used commonly for searching near points in a radius and can be used for * Points Searches: Used commonly for searching near points in a radius and can be used for − searches contacts for spheric and circle geometries. Use the Kd_tree,Octree and bins for spatial + searches contacts for spheric and circle geometries. Use the Kd_tree,Octree and bins for spatial − decomposition. + decomposition. − * Object Searches: The searches is based in the concept of objects. + * Object Searches: The searches is based in the concept of objects. Is a generic class, thats mean, we can use either objects type even points. For now, use the bins for spatial decomposition. It could be interesting proof hybrid methods would be a − Is a generic class, thats mean, we can use either objects type even points. + future investigation lines. − For now, use the bins for spatial decomposition. + − It could be interesting proof hybrid methods would be a future investigation + − lines. + == The User Configure File == == The User Configure File ==
## Using the Spatial Search Algorithm
Large-scale in Finite Element Methods, Discrete Element Method and Combined Finite-Discrete Element Method simulations involve contact of a large number of separate bodies. Processing contact interaction for all possible contacts would involve a total number of operations proportional to N^2, where N is the total number of separates bodies comprising the problem.
This would be very CPU intensive, and would limit the application that use and need to evaluate the contact forces ,comprising a very small number (a few thousand) of separates bodies. To reduce CPU requirements of processing contact interaction, it is necessary to eliminate couples of discrete elements that are far from each other and are not in contact. A set procedures designed to detect bodies that are close to each other is usually called a contact detection algorithm, or sometimes a contact search algorithm.
With this purpose was created in Kratos Program a contact search algorithm based on spatial decomposition, so it can be used for any application and is unique in that it was implemented in a generic way. Therefore it requires a contact search algorithm having the following characteristics:
• Robust,
• CPU efficient,
• RAM efficient
The space decomposition implemented in Kratos are:
• Trees : Kd-Trees, Octrees
• Bins
There are two kind of searches algorithms in Kratos:
• Points Searches: Used commonly for searching near points in a radius and can be used for
searches contacts for spheric and circle geometries. Use the Kd_tree,Octree and bins for spatial decomposition.
• Object Searches: The searches is based in the concept of objects. Is a generic class, thats mean, we can use either objects type even points. For now, use the bins for spatial decomposition. It could be interesting proof hybrid methods would be a
future investigation lines.
## The User Configure File
It say that is the main file created by the user, which defines the types, and operations contenerdores necessary to make the spatial decomposition and the search for contacts. Is then the parameter that defines Bins.
This consists of four parts:
• Definition of the types and their containers.
• Function that calculates bounding box
• Function of intersection between objects
• Function of intercession between the object and cells.
## An Example Configure File
Suppose that we have a set of discrete elements like a spheres. In our file called "discrete_configure_file.h" need to create the class, types and methods before mentioned. Is useful to create this file using the template of Kratos found in kratos/templates directory. Only we need to do is changes the default names and directives by the name of our file. Normally this file is stored in the custom_utilities directory of the application.
• Change the name of directives
``` #if !defined(KRATOS_DISCRETE_PARTICLE_CONFIGURE_INCLUDED)
#define KRATOS_DISCRETE_PARTICLE_CONFIGURE_INCLUDED
```
• Change the name of class
``` ///@}
///@name Kratos Classes
///@{
template <std::size_t TDimension>
class DiscreteParticleConfigure{
```
• Typing the typenames, objects and container of the particle
``` ///@}
///@name Type Definitions
///@{
typedef Point<Dimension, double> PointType;
typedef std::vector<double>::iterator DistanceIteratorType;
typedef ModelPart::ElementsContainerType::ContainerType ContainerType;
typedef ContainerType::value_type PointerType;
typedef ContainerType::iterator IteratorType;
typedef ModelPart::ElementsContainerType::ContainerType ResultContainerType;
typedef ResultContainerType::value_type ResultPointerType;
typedef ResultContainerType::iterator ResultIteratorType;
typedef ContactPair<PointerType> ContactPairType;
typedef std::vector<ContactPairType> ContainerContactType;
typedef ContainerContactType::iterator IteratorContactType;
typedef ContainerContactType::value_type PointerContactType;
typedef std::vector<PointerType>::iterator PointerTypeIterator;
```
Note: Another types and container different to the model part can be used.
• Computing the bounding box
``` static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint)
{
rHighPoint = rLowPoint = rObject->GetGeometry().GetPoint(0);
double radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
for(std::size_t i = 0; i < 3; i++){
rLowPoint[i] += -radius;
rHighPoint[i] += radius;
}
}
```
Note: It is necessary the name static inline void CalculateBoundingBox is well type, beacause the function is called by the bins.
• The Intersection function between objects.
``` static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2){
array_1d<double, 3> rObj_2_to_rObj_1 = rObj_1->GetGeometry().GetPoint(0) - rObj_2->GetGeometry().GetPoint(0);
double distance_2 = inner_prod(rObj_2_to_rObj_1, rObj_2_to_rObj_1);
const double& radius_1 = rObj_1->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
const double& radius_2 = rObj_2->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
double radius_sum = radius_1 + radius_2;
bool intersect = (distance_2 - radius_sum * radius_sum) <= 0;
return intersect;
}
```
• Intersection cell and objects
``` static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint){
array_1d<double, 3> center_of_particle = rObject->GetGeometry().GetPoint(0);
const double& radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
bool intersect = (rLowPoint[0] - radius <= center_of_particle[0] && rLowPoint[1] - radius <= center_of_particle[1] && rLowPoint[2] - radius <= center_of_particle[2] &&
rHighPoint[0] + radius >= center_of_particle[0] && rHighPoint[1] + radius >= center_of_particle[1] && rHighPoint[2] + radius >= center_of_particle[2]);
return intersect;
}
```
Note: Intersection cell and objects: If we do not have this function a simple return true is enougth.
## Calling the Constructor
Once the configure file is done, we are able to use Contact Search Algorithm, based in a domain decomposition. In the fail where it will be called , we need to add in the directives the corresponding files, those are the configure file and the bins. Maybe is required some redefinition of the typedef defined above in the configure file.
``` static const std::size_t space_dim = 2;
typedef DiscreteConfigure<space_dim> Configure;
typedef Configure::PointType PointType;
typedef PointType::CoordinatesArrayType CoordinatesArrayType;
typedef Configure::ContainerType ContainerType;
typedef Configure::PointerType PointerType;
typedef Configure::IteratorType IteratorType;
typedef Configure::ResultContainerType ResultContainerType;
typedef Configure::ResultPointerType ResultPointerType;
typedef Configure::ResultIteratorType ResultIteratorType;
typedef Configure::ContactPairType ContactPairType;
typedef Configure::ContainerContactType ContainerContactType;
typedef Configure::IteratorContactType IteratorContactType;
typedef Configure::PointerContactType PointerContactType;
typedef Configure::PointerTypeIterator PointerTypeIterator;
typedef ContainerContactType ContainerContactPair;
typedef IteratorContactType IteratorContainerContactPair;
typedef PointerContactType PointerContainerContactPair;
```
After that we call the constructor as follow:
• Using Dynamic Bins
``` IteratorType it_begin = mBoundaryElements.begin();
IteratorType it_end = mBoundaryElements.end();
BinsObjectDynamic<Configure> rBinsObjectDynamic(it_begin, it_end);
```
• Using Static Bins
``` BinsObjectStatic<Configure> Bins(it_begin, it_end);
const std::size_t MaxNumberOfResults = 1000;
std::size_t NumberOfResults = 0;
ResultIteratorType begin;
```
Note: ``` Note that we can use the dynamic or static bins. The difference between them is the dynamic allocation of memory. Naturally, the static bins is more fast because the we know size of containers. ```
• Lastly, we can get the list of pairs of contacts
``` rBins.SearchContact(mPairContacts);
```
• However, is more fast to do a loop and get the bodies that are in contact with a particular element.
``` For(IteratorType it =it_begin; it!=it_end; it++)
rBinsObjectDynamic.SearchObjectsInner(*it, Result);
```
where Result is the list of contact for a particular element.
## References
• Munjiza. Ante, The Combined Finite-Discrete Element Method, John Wiley & Sons, Ltd. 2004 | 2,240 | 10,328 | {"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} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.915087 |
https://www.mersenneforum.org/showpost.php?s=db3fc4823807f0a2f8282f6b94f5491a&p=513749&postcount=22 | 1,604,066,684,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00368.warc.gz | 792,601,426 | 4,622 | View Single Post
2019-04-15, 13:39 #22
MathDoggy
Mar 2019
3716 Posts
Quote:
Originally Posted by MathDoggy Introduction: The twin prime conjecture is a mathematical hypothesis which states that there exists infinitely many prime numbers that differ by 2. Proof by direct method: Let s be the infinite sum of the first n twin prime numbers Let z be the infinite sum of the first n natural numbers Let us assume that there exists a finite amount of twin prime numbers, then by the comparison criterion we will check whether the infinite sum of z diverges or converges because if z diverges then by implication s will also diverge. Now we will prove the divergence of z: z=1+2+3+4+5+6+7.. Now let us take the largest power of 2 which is greater or equal to K, where K is an element of z distinct to 2. Now we have, z=0+1+2+2+2+2+3+3... Obviously this infinite sum diverges then we can conclude that s also diverges therefore there does not exist a finite amount of twin prime numbers. Q.E.D (I have used the same method to prove this conjecture as the Goldbach conjecture) If you want to see the other proof, you can take a look in the Algebraic Number Theory section.
Can somebody point out what parts of the proof are wrong | 304 | 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} | 3.40625 | 3 | CC-MAIN-2020-45 | latest | en | 0.88881 |
https://www.mersenneforum.org/showthread.php?s=6d6d38089bf084ae69d09091a68e1e5c&p=329814 | 1,675,633,623,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00326.warc.gz | 886,552,690 | 8,628 | mersenneforum.org Primes on quadric irreducible polynomials
Register FAQ Search Today's Posts Mark Forums Read
2012-10-06, 14:14 #1 henryzz Just call me Henry "David" Sep 2007 Liverpool (GMT/BST) 37×163 Posts Primes on quadric irreducible polynomials I saw this webpage http://109.90.219.147/devalco/basic_polynoms/ and thought that people might be interested in finding polynomials with higher prime density. He has found some rules on what makes a good poly. Can we find some more?
2013-02-17, 03:12 #2 MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 119210 Posts One quadratic polynomial that merits consideration is h(n) = n^2 + n + 41 It has the property that h(n) is prime for n=0..39. Note that h(40) = 40(40 + 1) + 41. Also, h(n) never has a factor smaller than 40 when n is an integer. I have a proof of this fact. I put some more results on the web at https://sites.google.com/site/mattc1anderson/home-1 I have some new results that I have not included on the internet.
2013-02-17, 05:27 #3 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 235138 Posts This recent prime is relevant to your interest
Similar Threads Thread Thread Starter Forum Replies Last Post Till Factoring 12 2021-08-04 21:01 bhelmes Computer Science & Computational Number Theory 122 2017-08-25 21:09 Nick Puzzles 0 2013-10-29 10:19 chris2be8 FactorDB 1 2012-03-10 16:49 mdettweiler Factoring 15 2010-01-14 21:13
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Sun Feb 5 21:47:03 UTC 2023 up 171 days, 19:15, 1 user, load averages: 0.85, 0.97, 1.03 | 497 | 1,548 | {"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-2023-06 | latest | en | 0.825623 |
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Videos, worksheets, songs and solutions to help Grade 6 students learn how to multiply integers.
In these lessons, we will learn how to use the rules for multiplying integers.
Related Topics:
More Grade 6 Math Lessons, Integer Worksheets, Integer Games
Basic rules for multiplying integers and provides several examples
When multiplying integers:
If the signs are the same, the product is positive
If the signs are the different, the product is negative
Example:
Determine each product
5 × 6 =
-8 × -3 =
-2(9) =
8(-3) =
2(-5)(-6) =
-1 × -3 × -4 × -6=
When three or more integers are multiplied
• The product of an even number of negative numbers is positive.
• The product of an odd number of negative numbers is negative.
Example:
Multiply
a) (-2)(-3)(-5) =
b) (-1)(-5)(-2)(-3) =
How to Multiply and Divide Integers?
If the signs of the two integers are the same, the answer is positive.
If the signs of the two integers are different, the answer is negative.
How to Multiply Integers?
The rules for multiplying integers are:
When we multiply two positive integers, we get a positive integer.
When we multiply two negative integers, we get a positive integer.
When we multiply a positive and a negative integer, we get a negative integer.
Examples:
-3 × -5 =
-4 × 2 =
Multiplication / Multiplying Integers
This video gives some examples dealing with multiplication of integers. The only things we have be really careful of is the sign we get after multiplying. Otherwise, it is the same as multiplying all positive numbers.
Example:
Find the product
a) (-3)(4)(-5)
b) (5)(-1)(8)
Multiply Integers Song
Multiplying and Dividing Integers Song
A song to help you remember the rules for multiplying and dividing integers. Sung to the tune of Are You Sleeping Brother John?
Lyrics:
Multiplying and dividing
Use same sign, you will get a positive
Use different signs, you will get a negative
Multiplying Integers
The rules for multiplying integers are as follows: positive times positive equals positive, positive times negative equals negative, negative times positive equals negative, negative times negative equals positive.
Example:
(-3)(-4) =
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# Need to break out of recursive function
Hello,
I call a function recursively to find an item that exists *anywhere* down
the chain. Let's say I find it five layers deep. Now I've got what I need and
want to break out of that whole stack and continue execution at the point of
the initial call. Is that possible?
Thanks,
Bill
Nov 21 '05 #1
9 12766
Do you mean skip the whole call stack? If that's the case, then AFAIK, you
can't do that. You'll have to follow the call stack back to the point of the
initial call. I believe the reason for this is the for each call to the
recursive method, the current stack frame within the runtime stack has a
pointer only to the previous stack frame from which this call was initiated
and knows nothing about any stack frames before that. As soon as the current
call returns, it returns to the address of the previous stack frame from
which this call was made.
hope that helps..
Imran.
"Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
news:5E**********************************@microsof t.com...
Hello,
I call a function recursively to find an item that exists *anywhere* down
the chain. Let's say I find it five layers deep. Now I've got what I need and want to break out of that whole stack and continue execution at the point of the initial call. Is that possible?
Thanks,
Bill
Nov 21 '05 #2
Bill,
Normally I have the function return a "found" indicator (a Boolean), when I
call it recursively if I find the item I exit the current item.
Something like
Public Function RecusiveFind(value As Object, theCurrentLevel) As
Boolean
For Each item In theCurrentLevel
If item = value Then Return True
If RecusiveFind(value, item.Children) Then Return True
' keep looking at the current level
Next
Return False
End Function
"Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
news:5E**********************************@microsof t.com...
Hello,
I call a function recursively to find an item that exists *anywhere* down
the chain. Let's say I find it five layers deep. Now I've got what I need
and
want to break out of that whole stack and continue execution at the point
of
the initial call. Is that possible?
Thanks,
Bill
Nov 21 '05 #3
Hi Imran,
You can add a global variable. Set it equal to False. Once you hit your
finish point set it to True and Exit Function. Then in the function,
directly after where it calls itself check this variable. If it is True
then Exit Function which will move you right back up the stack and out.
I've never tried using a static variable. I have no idea if that would
work, hmmmm I'll have to try it. Good luck! Ken.
--
Ken Dopierala Jr.
For great ASP.Net web hosting try:
http://www.webhost4life.com/default.asp?refid=Spinlight
If you sign up under me and have problems, email me.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:Og*************@TK2MSFTNGP09.phx.gbl...
Do you mean skip the whole call stack? If that's the case, then AFAIK, you
can't do that. You'll have to follow the call stack back to the point of the initial call. I believe the reason for this is the for each call to the
recursive method, the current stack frame within the runtime stack has a
pointer only to the previous stack frame from which this call was initiated and knows nothing about any stack frames before that. As soon as the current call returns, it returns to the address of the previous stack frame from
which this call was made.
hope that helps..
Imran.
"Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
news:5E**********************************@microsof t.com...
Hello,
I call a function recursively to find an item that exists *anywhere* down the chain. Let's say I find it five layers deep. Now I've got what I
need and
want to break out of that whole stack and continue execution at the
point of
the initial call. Is that possible?
Thanks,
Bill
Nov 21 '05 #4
Why a global variable? Couldn't you just pass your boolean variable as one
of the parameters of the recursive function. In fact, isn't this how you
create a recursive function???
Greg
"Ken Dopierala Jr." <kd*********@wi.rr.com> wrote in message
news:%2****************@TK2MSFTNGP10.phx.gbl...
Hi Imran,
You can add a global variable. Set it equal to False. Once you hit your
finish point set it to True and Exit Function. Then in the function,
directly after where it calls itself check this variable. If it is True
then Exit Function which will move you right back up the stack and out.
I've never tried using a static variable. I have no idea if that would
work, hmmmm I'll have to try it. Good luck! Ken.
--
Ken Dopierala Jr.
For great ASP.Net web hosting try:
http://www.webhost4life.com/default.asp?refid=Spinlight
If you sign up under me and have problems, email me.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:Og*************@TK2MSFTNGP09.phx.gbl...
Do you mean skip the whole call stack? If that's the case, then AFAIK,
you
can't do that. You'll have to follow the call stack back to the point of
the
initial call. I believe the reason for this is the for each call to the
recursive method, the current stack frame within the runtime stack has a
pointer only to the previous stack frame from which this call was
initiated
and knows nothing about any stack frames before that. As soon as the
current
call returns, it returns to the address of the previous stack frame from
which this call was made.
hope that helps..
Imran.
"Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
news:5E**********************************@microsof t.com...
> Hello,
>
> I call a function recursively to find an item that exists *anywhere* down > the chain. Let's say I find it five layers deep. Now I've got what I
need
and
> want to break out of that whole stack and continue execution at the
point
of
> the initial call. Is that possible?
>
> Thanks,
>
> Bill
Nov 21 '05 #5
Right - I understand that. But you're still going to be checking that
variable for each recursive call that you made. Ofcourse, you can avoid the
code after the recursive call within the method. But you're still going to
pass through the entire call stack back to the initial call. There's no way
you can skip the call stack. Or am I missing something?
Imran.
"Ken Dopierala Jr." <kd*********@wi.rr.com> wrote in message
news:%2****************@TK2MSFTNGP10.phx.gbl...
Hi Imran,
You can add a global variable. Set it equal to False. Once you hit your
finish point set it to True and Exit Function. Then in the function,
directly after where it calls itself check this variable. If it is True
then Exit Function which will move you right back up the stack and out.
I've never tried using a static variable. I have no idea if that would
work, hmmmm I'll have to try it. Good luck! Ken.
--
Ken Dopierala Jr.
For great ASP.Net web hosting try:
http://www.webhost4life.com/default.asp?refid=Spinlight
If you sign up under me and have problems, email me.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:Og*************@TK2MSFTNGP09.phx.gbl...
Do you mean skip the whole call stack? If that's the case, then AFAIK, you
can't do that. You'll have to follow the call stack back to the point of
the
initial call. I believe the reason for this is the for each call to the
recursive method, the current stack frame within the runtime stack has a
pointer only to the previous stack frame from which this call was
initiated
and knows nothing about any stack frames before that. As soon as the
current
call returns, it returns to the address of the previous stack frame from
which this call was made.
hope that helps..
Imran.
"Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
news:5E**********************************@microsof t.com...
Hello,
I call a function recursively to find an item that exists *anywhere*
down the chain. Let's say I find it five layers deep. Now I've got what I
need
and
want to break out of that whole stack and continue execution at the
point
of
the initial call. Is that possible?
Thanks,
Bill
Nov 21 '05 #6
You could kill the whole stack by throwing an exception :-)
--
Jonathan Bailey.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:eL***************@TK2MSFTNGP10.phx.gbl...
Right - I understand that. But you're still going to be checking that
variable for each recursive call that you made. Ofcourse, you can avoid the code after the recursive call within the method. But you're still going to
pass through the entire call stack back to the initial call. There's no way you can skip the call stack. Or am I missing something?
Imran.
"Ken Dopierala Jr." <kd*********@wi.rr.com> wrote in message
news:%2****************@TK2MSFTNGP10.phx.gbl...
Hi Imran,
You can add a global variable. Set it equal to False. Once you hit your
finish point set it to True and Exit Function. Then in the function,
directly after where it calls itself check this variable. If it is True
then Exit Function which will move you right back up the stack and out.
I've never tried using a static variable. I have no idea if that would
work, hmmmm I'll have to try it. Good luck! Ken.
--
Ken Dopierala Jr.
For great ASP.Net web hosting try:
http://www.webhost4life.com/default.asp?refid=Spinlight
If you sign up under me and have problems, email me.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:Og*************@TK2MSFTNGP09.phx.gbl...
Do you mean skip the whole call stack? If that's the case, then AFAIK, you can't do that. You'll have to follow the call stack back to the point of the
initial call. I believe the reason for this is the for each call to
the recursive method, the current stack frame within the runtime stack has a pointer only to the previous stack frame from which this call was
initiated
and knows nothing about any stack frames before that. As soon as the
current
call returns, it returns to the address of the previous stack frame from which this call was made.
hope that helps..
Imran.
"Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
news:5E**********************************@microsof t.com...
> Hello,
>
> I call a function recursively to find an item that exists *anywhere*
down
> the chain. Let's say I find it five layers deep. Now I've got what I
need
and
> want to break out of that whole stack and continue execution at the
point
of
> the initial call. Is that possible?
>
> Thanks,
>
> Bill
Nov 21 '05 #7
Well - even when an exception is thrown, the runtime searches up the call
stack to look for an appropriate exception handler which means its still
going to be travelling up the call stack. However, I'm not sure how the
runtime handles exceptions for recursive procedures - is it even aware that
its a recursive call? If so, it could just check once whether the procedure
has an exception handler and if not, it could skip the entire stack sequence
for the recursion and jump right back to the initial calling method. But I
doubt that..I could be wrong though..
Imran.
<jb> wrote in message news:41***********************@news.easynet.co.uk. ..
You could kill the whole stack by throwing an exception :-)
--
Jonathan Bailey.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:eL***************@TK2MSFTNGP10.phx.gbl...
Right - I understand that. But you're still going to be checking that
variable for each recursive call that you made. Ofcourse, you can avoid the
code after the recursive call within the method. But you're still going to
pass through the entire call stack back to the initial call. There's no
way
you can skip the call stack. Or am I missing something?
Imran.
"Ken Dopierala Jr." <kd*********@wi.rr.com> wrote in message
news:%2****************@TK2MSFTNGP10.phx.gbl...
Hi Imran,
You can add a global variable. Set it equal to False. Once you hit your finish point set it to True and Exit Function. Then in the function,
directly after where it calls itself check this variable. If it is True then Exit Function which will move you right back up the stack and out. I've never tried using a static variable. I have no idea if that would work, hmmmm I'll have to try it. Good luck! Ken.
--
Ken Dopierala Jr.
For great ASP.Net web hosting try:
http://www.webhost4life.com/default.asp?refid=Spinlight
If you sign up under me and have problems, email me.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:Og*************@TK2MSFTNGP09.phx.gbl...
> Do you mean skip the whole call stack? If that's the case, then AFAIK, you
> can't do that. You'll have to follow the call stack back to the
point of the
> initial call. I believe the reason for this is the for each call to the > recursive method, the current stack frame within the runtime stack
has
a > pointer only to the previous stack frame from which this call was
initiated
> and knows nothing about any stack frames before that. As soon as the
current
> call returns, it returns to the address of the previous stack frame from > which this call was made.
>
> hope that helps..
> Imran.
>
> "Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
> news:5E**********************************@microsof t.com...
> > Hello,
> >
> > I call a function recursively to find an item that exists
*anywhere* down
> > the chain. Let's say I find it five layers deep. Now I've got what I need
> and
> > want to break out of that whole stack and continue execution at the point
> of
> > the initial call. Is that possible?
> >
> > Thanks,
> >
> > Bill
>
>
Nov 21 '05 #8
I was joking of course :-)
I cant find in the MS website where exceptions are defined but I expect vb
creates a try,catch block around any function which might throw one which
then passes it to the previous caller.
It just 'looks like' you are skipping up the stack in the source code.
--
Jonathan Bailey.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:%2****************@tk2msftngp13.phx.gbl...
Well - even when an exception is thrown, the runtime searches up the call
stack to look for an appropriate exception handler which means its still
going to be travelling up the call stack. However, I'm not sure how the
runtime handles exceptions for recursive procedures - is it even aware that its a recursive call? If so, it could just check once whether the procedure has an exception handler and if not, it could skip the entire stack sequence for the recursion and jump right back to the initial calling method. But I
doubt that..I could be wrong though..
Imran.
<jb> wrote in message news:41***********************@news.easynet.co.uk. ..
You could kill the whole stack by throwing an exception :-)
--
Jonathan Bailey.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:eL***************@TK2MSFTNGP10.phx.gbl...
Right - I understand that. But you're still going to be checking that
variable for each recursive call that you made. Ofcourse, you can avoid
the
code after the recursive call within the method. But you're still
going
to pass through the entire call stack back to the initial call. There's
no
way
you can skip the call stack. Or am I missing something?
Imran.
"Ken Dopierala Jr." <kd*********@wi.rr.com> wrote in message
news:%2****************@TK2MSFTNGP10.phx.gbl...
> Hi Imran,
>
> You can add a global variable. Set it equal to False. Once you hit your
> finish point set it to True and Exit Function. Then in the
function, > directly after where it calls itself check this variable. If it is
True > then Exit Function which will move you right back up the stack and out. > I've never tried using a static variable. I have no idea if that would > work, hmmmm I'll have to try it. Good luck! Ken.
>
> --
> Ken Dopierala Jr.
> For great ASP.Net web hosting try:
> http://www.webhost4life.com/default.asp?refid=Spinlight
> If you sign up under me and have problems, email me.
>
> "Imran Koradia" <no****@microsoft.com> wrote in message
> news:Og*************@TK2MSFTNGP09.phx.gbl...
> > Do you mean skip the whole call stack? If that's the case, then AFAIK, you
> > can't do that. You'll have to follow the call stack back to the point
of
> the
> > initial call. I believe the reason for this is the for each call to the
> > recursive method, the current stack frame within the runtime stack has
a
> > pointer only to the previous stack frame from which this call was
> initiated
> > and knows nothing about any stack frames before that. As soon as
the > current
> > call returns, it returns to the address of the previous stack
frame from
> > which this call was made.
> >
> > hope that helps..
> > Imran.
> >
> > "Bill Borg" <Bi******@discussions.microsoft.com> wrote in message
> > news:5E**********************************@microsof t.com...
> > > Hello,
> > >
> > > I call a function recursively to find an item that exists
*anywhere* > down
> > > the chain. Let's say I find it five layers deep. Now I've got
what I > need
> > and
> > > want to break out of that whole stack and continue execution at the > point
> > of
> > > the initial call. Is that possible?
> > >
> > > Thanks,
> > >
> > > Bill
> >
> >
>
>
Nov 21 '05 #9
lol :)
yeah - I couldn't find anything either but considering how exception
handling works in java, I bet that's how it works.
Imran.
<jb> wrote in message news:41***********************@news.easynet.co.uk. ..
I was joking of course :-)
I cant find in the MS website where exceptions are defined but I expect vb
creates a try,catch block around any function which might throw one which
then passes it to the previous caller.
It just 'looks like' you are skipping up the stack in the source code.
--
Jonathan Bailey.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:%2****************@tk2msftngp13.phx.gbl...
Well - even when an exception is thrown, the runtime searches up the call
stack to look for an appropriate exception handler which means its still
going to be travelling up the call stack. However, I'm not sure how the
runtime handles exceptions for recursive procedures - is it even aware
that
its a recursive call? If so, it could just check once whether the
procedure
has an exception handler and if not, it could skip the entire stack
sequence
for the recursion and jump right back to the initial calling method. But I doubt that..I could be wrong though..
Imran.
<jb> wrote in message news:41***********************@news.easynet.co.uk. ..
You could kill the whole stack by throwing an exception :-)
--
Jonathan Bailey.
"Imran Koradia" <no****@microsoft.com> wrote in message
news:eL***************@TK2MSFTNGP10.phx.gbl...
> Right - I understand that. But you're still going to be checking that > variable for each recursive call that you made. Ofcourse, you can avoid the
> code after the recursive call within the method. But you're still going
to
> pass through the entire call stack back to the initial call. There's
no way
> you can skip the call stack. Or am I missing something?
>
> Imran.
>
> "Ken Dopierala Jr." <kd*********@wi.rr.com> wrote in message
> news:%2****************@TK2MSFTNGP10.phx.gbl...
> > Hi Imran,
> >
> > You can add a global variable. Set it equal to False. Once you hit your
> > finish point set it to True and Exit Function. Then in the function, > > directly after where it calls itself check this variable. If it is True
> > then Exit Function which will move you right back up the stack and
out.
> > I've never tried using a static variable. I have no idea if that
would
> > work, hmmmm I'll have to try it. Good luck! Ken.
> >
> > --
> > Ken Dopierala Jr.
> > For great ASP.Net web hosting try:
> > http://www.webhost4life.com/default.asp?refid=Spinlight
> > If you sign up under me and have problems, email me.
> >
> > "Imran Koradia" <no****@microsoft.com> wrote in message
> > news:Og*************@TK2MSFTNGP09.phx.gbl...
> > > Do you mean skip the whole call stack? If that's the case, then
AFAIK,
> you
> > > can't do that. You'll have to follow the call stack back to the
point
of
> > the
> > > initial call. I believe the reason for this is the for each call to the
> > > recursive method, the current stack frame within the runtime
stack
has
a
> > > pointer only to the previous stack frame from which this call
was > > initiated
> > > and knows nothing about any stack frames before that. As soon as
the > > current
> > > call returns, it returns to the address of the previous stack frame from
> > > which this call was made.
> > >
> > > hope that helps..
> > > Imran.
> > >
> > > "Bill Borg" <Bi******@discussions.microsoft.com> wrote in message > > > news:5E**********************************@microsof t.com...
> > > > Hello,
> > > >
> > > > I call a function recursively to find an item that exists
*anywhere*
> > down
> > > > the chain. Let's say I find it five layers deep. Now I've got
what
I
> > need
> > > and
> > > > want to break out of that whole stack and continue execution
at the
> > point
> > > of
> > > > the initial call. Is that possible?
> > > >
> > > > Thanks,
> > > >
> > > > Bill
> > >
> > >
> >
> >
>
>
Nov 21 '05 #10
### This discussion thread is closed
Replies have been disabled for this discussion. | 5,426 | 21,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | longest | en | 0.880824 |
www.cbtricks.com | 1,516,138,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00187.warc.gz | 400,881,863 | 5,338 | Coax Theory
by Russell E. Clift, AB7IF
Having over the years heard so many differing reasons for what is going on in a coax, concerning SWR, reflections, and other things; I decided to have a short discussion of what is occurring in the line. Here I use a DC voltage, which gets the point across just fine even if we are dealing with RF. In this case I use 50 ohm coax, and assume it takes one microsecond for a voltage wave to travel from end to end from t=o, the moment the switch is closed. I will use the drawing below, first for an open line and second for a shorted (at the output end) line. While I am not an artist, I hope the drawing is ok. We can use lumped components for the coax, in this case I will use 3 sections.
Also, I inserted a 50 ohm resistor in series with the 50 ohm line, to make the numbers easier to be understood. Or at least easier to explain. Note that incident and reflected waves are traveling waves.
(Click on each picture to enlarge the image)
OPEN LINE: When the switch is closed at t=0, 1 amp will flow from the 100 volt source, since the 50 ohm resistor is in series with the 50 ohm impedance of the line, (the resistor is added so the applied voltage to the line will be 50 volts). As the capacitors begin to charge one by one, a voltage wave moves down the line, which we call the incident voltage wave. There is also an incident current wave which propagates with the incident voltage wave. In 1 usec when the incident waves reach the end of the line, and C3 is charged to 50 volts by the current through L3, and there is no more coax to charge so the magnetic field around L3 will collapse, and the collapsing field will keep the current flowing in C3 in the same direction, charging C3 to 100 volts. Since the current in L3 is now gone, the current in L2 collapses, and in the same way charges C2 to 100 volts as well. Likewise now L1's field collapse charges C1 to 100 volts, until the voltage all across the line is 100 volts, and current flow in the line is zero. The source at this time (t=2 microseconds) "sees" an open circuit for the line. The capacitors charging from 50 volts to 100 volts from the end of the line toward the source creates a reflected voltage wave traveling toward the source propagating in phase and in the opposite direction of the incident voltage wave. The inductors, discharging from the open circuit toward the source create a reflected current wave propagating out of phase and in the opposite direction of the incident current wave. In the drawing are shown the voltage and current verses time at both the input and output ends of the line.
SHORTED LINE: The line charges in the same way as the previous circuit but since the voltage must equal zero in a short circuit, when the incident voltage and current waves reach the short circuit (in 1 usec), the charge on C2 must discharge into L3, producing a current of 2 amps in L3. When the voltage of C2 approaches zero, C1 discharges into L2, producing 2 amps of current in it. This goes on until the entire line is at zero volts. The reflected current wave in this case is in phase with the incident current wave, while the voltage reflection in this case is out of phase with the incident voltage wave. At t=2 usec, the source "sees" a shorted line.
The differences for open and shorted lines are:
(1.) The voltage reflection from an open circuit is in phase, and from a shorted line is out of phase with the incident wave.
(2.) The current reflection from an open circuit is out of phase, from a shorted line it is in phase.
When a transmission line is terminated in it's characteristic impedance there is no voltage or current reflection from the line. The electromagnetic fields continue to flow into the termination as if the line were infinitely long. When a mismatch of impedance occurs, reflected waves will be produced and they will interact with the incident waves. The total voltage and current on the line are no longer the result of a single travelling wave from the source to the load. Instead, it is the algebraic sum of two waves travelling in opposite directions. This interaction results in what is known as standing waves. The waves remain in fixed positions along the line while they vary in amplitude and polarity. A wave of any shape can be transmitted along the line without any change of waveshape or magnitude. Looking at the gif below, we see a line driven with a sine wave generator, terminated with a short circuit to maximize the reflection.
(Click on each picture to enlarge the image)
In the the gif on the right the incident waves of current and voltage are shown in a and b, while the standing waves of current and voltage are shown in c and d. In the first jpeg, looking at the waveforms below the circuit, we see a simpler way to visualize the standing waves, by ignoring the polarity and only seeing the magnitude of the waves. This is exactly what an RMS meter would show at each point along the line we measured. The points of zero voltage are called nodes or voltage nodes, while points of maximum voltage are called loops or antinodes. The position of the generator from the load will determine the type of impedance the generator "sees". This is the very cause of so much confusion as to what differing coax lengths do to SWR, or at least what is "seen" as the SWR at the point of meter insertion. I have heard it stated many times by others that the line length does not affect the SWR, but this can only be possible when there is ZERO reflection of energy, a perfect match, something which never happens in the real world. The reason I use a shorted line here is to create maximum reflection to show clearly what is going on. With a termination closer to the line's impedance the resulting reflections would be weaker in amplitude as compared to the incident waves, and the algebraic sum in the standing waves would contain a correspondingly smaller component of reflection amplitude or energy. Normally we want the generator to be at a voltage or current loop, depending upon what we were trying to do. As an example we would want half wave multiples between source and load for a single antenna, and quarter wave or odd multiple's of quarter waves for a dual antenna installation. keep in mind that an antenna is really 36 ohms not 50; so the 14.1 feet of half wave shows 36 ohms towards the radio, and the missing 14 ohms is added by the remaing 3.9 feet of coax in an 18 foot single. This gives us 50 ohms roughly at the radio end and is just exactly why antenna manufacturers spec out 18 feet for the coax run to their antenna's. When a transmission line has standing waves on it the dynamic impedance of the line is no longer equal to the characteristic impedance of the line, unlike the example of only incident travelling waves without any reflections occuring. The reflected waves will alter the input impedance that the generator (your boat anchor radio) "sees".
Going back to the first image, looking at the waveforms below the circuit we see that standing waves go from maximum to minimum in 1/4 wavelength intervals along the line. The open or short circuited line will present to it's source a high or low impedance at multiples of 1/4 wavelength, depending upon the location of the source in reference to the load (an effect you see when you alter coax lengths). At points that are not an exact 1/4 wavelength multiple from the load or antenna, the line acts as a capacitor or an inductor. Current at the generator will either lead or lag the voltage. An open or short circuited line may assume the characteristics of a resonant circuit formed of lumped component elements of inductance and capacitance.
Looking at the last picture we see first an open line and second a shorted line. In either case we see a resistive line at 1/4 wave multiples from the load. Importantly, we see an image of the load impedance every even number of 1/4 waves (1/2 waves), and an inverted or mirrored image at every odd 1/4 wave multiple (back to our dual harness again, in theory we get two 100 ohm or nearly so resistors in parallel letting the radio "see" 50 ohms), and at any other point on the line we see a range of complex impedance which contains inductive or capacitive reactance. This effect is often used to "match" a load/line combination to a generator. In other words, if we have a mismatch between the load (antenna) and the line creating standing waves, then we can use a specific line length to make the generator "see" a good match or SWR. This is the most misunderstood and argued about effect in radio among CB'ers, and we can see from the pictures and explanation not only what is really going on once and for all, but we can also see how to take advantage of these facts to make an antenna system perform as well as is possible, given the particulars involved in any installation. Also, looking at the shorted line picture at each odd 1/4 wave multiple we can see that excepting losses the line appears to vanish to the circuit. What can we use this for you may ask (assuming like me you don't have a life so you actually think about these things); if we take a 1/4 wave line and put a pl 259 on one end, and short the other end, we make a nice trap. Putting a T at the antenna, we connect the shorted 1/4 wave. Now at any frequency other than the resonant frequency of the line, we see a short circuit. We also see a DC short circuit. So we have not only protected our system from static build up and nearby lightning transients, we also have made a very high quality harmonic filter to help keep us out of the neighbors television, one with superior characteristics to a lumped component type of filter, far lower loss due to dissipation, and power handling capability equal to the coax used. As a side benefit, we have also added another radial or counterpoise to the antenna system which helps to improve the overall RF ground and lower the SWR of the system, all while increasing radiation efficiency. This can also be used in a mobile linear installation, as I have done for many years. | 2,224 | 10,091 | {"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-2018-05 | longest | en | 0.946349 |
https://gitter.im/Rule-Based-Integration/Lobby?at=607b844d2e5574669b42d8c0 | 1,642,475,365,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00448.warc.gz | 333,138,361 | 19,142 | ## Where communities thrive
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##### Activity
Miguel Raz Guzmán Macedo
@miguelraz
/MathematicaSyntaxTestSuite-4.16.0/1 Algebraic functions/1.3 Miscellaneous/1.3.2 Algebraic functions.m
┌ Warning: (* {(Sqrt[x]+x)^(2/3), x, 0} *)
/MathematicaSyntaxTestSuite-4.16.0/1 Algebraic functions/1.3 Miscellaneous/1.3.2 Algebraic functions.m
┌ Warning: (* {(-3*x+x^2)^(-1/3), x, 0} *)
/MathematicaSyntaxTestSuite-4.16.0/1 Algebraic functions/1.3 Miscellaneous/1.3.2 Algebraic functions.m
┌ Warning: (* {1/Sqrt[a*x+b*x^3], x, 0} *)
/MathematicaSyntaxTestSuite-4.16.0/1 Algebraic functions/1.3 Miscellaneous/1.3.2 Algebraic functions.m
┌ Warning: (* {E^(a + b*x + c*x^2)/(d + e*x)^2, x, 0} *)
Sorry for the silly parsing questions, but:
1. Why is E^x used instead of Exp[x]?
2. What do the underscores in Int([a_. + b_.*x_), x_Symbol] mean?
Miguel Raz Guzmán Macedo
@miguelraz
The translation to Julia of Rubi rules is almost complete!
Albert D. Rich
@AlbertRich
@miguelraz These problems were probably commented out because they hung earlier versions of Rubi. I restored them and included their optimal antiderivatives as the fourth element of the problem.
The underscore following a variable in Mathematica patterns indicates the variable can stand for any mathematical expression.
To have any hope of successfully porting Rubi 4 to Julia, you need to fully understand how Mathematica's pattern matching works.
Miguel Raz Guzmán Macedo
@miguelraz
@AlbertRich Thanks. Good to know they've been handled.
Miguel Raz Guzmán Macedo
@miguelraz
Also, why do these lowercase functions exist?
4.7.5 Intert Trig Functions.m : "Int[u_.*(a_.*tan[c_. + d_.*x_]^n_. + b_.*sec[c_.+d_.*x_]^n_.)^p_, x_Symbol]",
Miguel Raz Guzmán Macedo
@miguelraz
Any RUBI internals for pattern matching that differ from those in the standard Mathematica documentation?
Albert D. Rich
@AlbertRich
@miguelraz Internally Rubi uses inert trig functions to greatly reduce the number of circular and hyperbolic trig integration rules required.
Before searching for integration rules to apply, Rubi uses the utility routine DeactivateTrig to convert integrands involving the 6 trig and 6 hyperbolic functions to equivalents ones involving just the 3 inert trig functions sin, tan and csc.
For example, Cos[x] is converted to the inert sin[x+Pi/2]. If instead it was converted to the active Sin[x+Pi/2], Mathematica would immediately simplify it back to Cos[x]. Similarly, Sinh[x] is converted to -I sin[I x] and Cosh[x] is converted to sin[I x+Pi/2].
So just 1 rule instead of 4 are required to integrate Sin[x], Cos[x], Sinh[x] and Cosh[x] wrt x. Similar rule consolidations are done for the tangent and secant (circular and hyperbolic) trig functions.
Currently Rubi has over 2000 inert trig integration rules. Without this consolidation there would be over 8000 active trig and hyperbolic integration rules.
Miguel Raz Guzmán Macedo
@miguelraz
Wow, that is clever.
Miguel Raz Guzmán Macedo
@miguelraz
What is the best place to read about those internal design decisions? I just started reading the IntegrationUtilityFunctions.m.
Albert D. Rich
@AlbertRich
@miguelraz That's the best place.
Albert D. Rich
@AlbertRich
@miguelraz Also single step through integrations to see the rules being applied. For example, when integrating Sin[x] (or Sinh[x]) you can see the inactive sin[x_] on the lhs of the rule and the active Cos[x] on the rhs.
Miguel Raz Guzmán Macedo
@miguelraz
Neat!
Why do some rules have If[TrueQ[$LoadShowSteps], Int[...] :=? Is there a reason why Modele[{...}] is used in some places and With[{...}] is used in other places? IIUC, both introduce a local scope to make the calculations less verbose. Miguel Raz Guzmán Macedo @miguelraz 1. Why is E^x used instead of Exp[x]? Bump! Albert D. Rich @AlbertRich @miguelraz When building Rubi with the ability to show steps, a few rules are hard-wired to display more meaningfully when the rule is applied. For your purposes, just assume $LoadShowSteps is False and use the Int rule defined by the fail condition of the If statement.
See the Mathematica online help for the distinction between With, Module and Block.
E^x and Exp[x] are interchangeable since Mathematica automatically simplifies the latter to the former.
Miguel Raz Guzmán Macedo
@miguelraz
Awesome, thanks a lot for that explanation.
Miguel Raz Guzmán Macedo
@miguelraz
Albert, good day.
I've had people asking, so I just want to double check. If we use RUBI as a basis for an integrator, are any licensing issues potentially involved? We know that Mathematica people can be... insistent, so we just want to make sure that we have covered all the bases here. I know that the RUBI repo is MIT-licensed, but are there any grounds for concerns that its uses are compliant with the WolframEngine's terms of use?
Albert D. Rich
@AlbertRich
@miguelraz Sorry, I have not used Wolfram Engine and am unfamiliar with its licensing requirements. Although its program code is written in the syntax of Mathematica's programming language, Rubi was developed independently of Wolfram Research. WRI provides more information at Free Wolfram Engine for Developers.
Miguel Raz Guzmán Macedo
@miguelraz
Thanks as always for the prompt response @AlbertRich . Enjoy the weekend.
Albert D. Rich
@AlbertRich
@miguelraz Yup, I hope to enjoy the weekend by going to my first rock concert in over a year. :))
Miguel Raz Guzmán Macedo
@miguelraz
Awesome! Rock on!
Christian Bauer
@cbauer:matrix.org
[m]
Hi Albert!
Congratulations to successfully pulling off a project like this, highly appreciated! Just recently, I stumbled across an integral for which Mathematica returns
a rather involved yet exact solution. I tried to use Rubi to understand how to solve this integrand by hand. For example, when Rubi applies transformation rule #6851, does this have an exact counterpart in the reference literature or is this simply an internal numbering scheme?
Miguel Raz Guzmán Macedo
@miguelraz
Oh nice to see you here @cbauer:matrix.org !
Albert D. Rich
@AlbertRich
@cbauer:matrix.org I'm glad you find Rubi's integration prowess of value.
When showing integration steps, you and others have requested a way to find for the rules being applied the documentation included with their definition in Rubi's source files.
As a help toward that end, now available on Rubi's website is an Integration Rule Index. It shows the numbers of the rules defined in each of the rule source files. For more on using the index, see the third paragraph of the Rules menu option on the website
Once you know the file a rule is defined in, it is a relatively simple matter to do a text search for it.
Christian Bauer
@cbauer:matrix.org
[m]
Got it, thank you! As an aside, I checked for a few integrals and I must say I am deeply impressed what goes on under the hood for seemingly easy integrands. I never bothered which transformations were applied, how many of them and in which order. Rubi served as an eye opener in that regard.
Momchil Peychev
@mpeychev_gitlab
Hello, hope all is well.
I have a somewhat conceptual (although possibly very stupid) question. If we have that s = Int[e, x], then is it reasonable to expect that D[s, x] - e is equal to 0 (possibly after applying Simplify to the left-hand side)?
This, of course, generally holds, but there are a few cases where this is not the case, for example:
Int[(f^x * x^2) / ((a + b*f^(2*x))^3), x]
Int[x / (b / (f^x) + a * f^x)^3, x]
Can someone explain me what is going on here? I asked a similar question somewhere in SymPy's GitHub issues and the people there explained to me that it is possible to get such discrepancies because of some implicit assumptions, related to the domains for which the integrand and the result of the integration are defined. I am not sure that this is also the case here, though. Am I terribly wrong somewhere? Thanks
Nasser M. Abbasi
@nasser1
Sometimes you need to use FullSimplify
<< Rubi
integrand = (f^x*x^2)/((a + b*f^(2*x))^3);
sol = Int[integrand, x];
FullSimplify[D[sol, x] - integrand]
0
and
integrand = x/(b/(f^x) + a*f^x)^3;
sol = Int[integrand, x];
FullSimplify[D[sol, x] - integrand]
0
rfateman
@rfateman
I see no reason to assume that Simplify, FullSimplify, etc can reduce the D[sol,x]-integrand to zero. The zero-equivalence problem is theoretically undecidable (Richardson proved it). While it may ordinarily work if the expressions are in a simple enough domain, and if it does simplify to zero, that is, at least encouraging, it is not a requirement. There are also others issues that come into play, like is "sol" the smallest, simplest, "most continuous" result? Notions of "simplest" can be controversial. Some people like log(), some people like arctan() and there are lots more ways of fiddling.
Albert D. Rich
@AlbertRich
@mpeychev_gitlab If the derivative of the antiderivative Rubi returns is not equal to the integrand, I consider the antiderivative incorrect and a bug in Rubi. However, the host CAS, in this case Mathematica, may not be able demonstrate expressions equal even using FullSimplify.
The optimal antiderivatives of each of the 70,000+ integrals in Rubi's test-suite has been verified correct. Sometimes this requires manually helping out Mathematica show equality using various techniques.
Momchil Peychev
@mpeychev_gitlab
Thank you @rfateman and @AlbertRich for the insightful replies!
Gavin C
@jg3c:matrix.org
[m]
Are there any update on rubi 5? I'm interested in porting rubi to a different cas system but with no mathematica interpreter it seems like a difficult task.
Albert D. Rich
@AlbertRich
@jg3c:matrix.org The Rubi-5 repository has the latest information on the Rubi 5 Project.
It contains an actual functioning prototype of Rubi 5 that shows the structure of the proposed integrator. Also it provides an example how to compile Rubi 4 pattern matching rules into a Rubi 5 if-then-else decision tree. The repository's README file provides more details.
5 replies
Patrick Scheibe
@halirutan
@/all Hey guys. Since 2020, my workload is too high to properly take care of the Rubi repository (what an understatement). I would really appreciate it if someone would step up to take care of the open pull-requests and the repositories in general. Are there any good candidates who think they could handle the job and have at least some time to spare in the upcoming future?
Sorin S
Hello Patrick, happy to help if that is suitable.
Nasser M. Abbasi
@nasser1
Hello;
fyi, I am getting this error Recursion error on this integral using Rubi 4.16.1 on Mathematica 12.3.1
integrand = -(x^3 + x^6 + 1)/((x^3 + x^5)^(1/4)*(x^6 - 1))
Int[integrand, x]
$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of (If[IntegerQ[#1]||Head[#1]===Rational,Null,Return[False]]&)[-1]$RecursionLimit::reclim2: Recursion depth of 1024 exceeded
during evaluation of AtomQ[-1]||FreeQ[-1,x].
$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of AtomQ[x]||FreeQ[x,x]. Using windows 10. Fyi, This one gives 1/0 error and also recursion error, also using Rubi 4.16.1 on Mathematica 12.3.1. integrand = -(x^6 + 1)/((x^3 + x^5)^(1/4)*(x^6 - 1)) Int[integrand, x]$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of N[300].
$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of Floor[Hold[N[300]]].$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of Message[Simplify::timl,300].
General::stop: Further output of \$RecursionLimit::reclim2 will be suppressed during this calculation.
Power::infy: Infinite expression 1/0 encountered.`
Lachlan Sneff
@lachlansneff:matrix.org
[m]
I've been working on a CAS in rust for a few days and have been planning on converting the RUBI rules to rust when I get around to adding integration support!
Here's the repo if anyone is interested: https://github.com/lachlansneff/das
By the way, what's the format for your rules? Can I add some?
Can rules for definite integration be added?
Albert D. Rich
@AlbertRich
@Anixx The integration rules Rubi uses are defined in the Mathematica source and corresponding pdf files available at https://rulebasedintegration.org/integrationRules.html.
The easiest way to add additional rules is to put them at the end of one of Rubi's existing source files, and then rebuild the system.
Yes, you could even add rules for definite integration.
Miguel Raz Guzmán Macedo
@miguelraz
Hello @AlbertRich , hope all is well. | 3,342 | 12,569 | {"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} | 2.953125 | 3 | CC-MAIN-2022-05 | latest | en | 0.737156 |
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How to Find Area – Regular Polygons Find the length of a side and the length of the apothem (the line segment perpendicular to a side connecting the middle of a side to the center. Multiply the length of the side by the number of sides to get the perimeter of the polygon (p).
HOW TO MEASURE AN AREA ELLIPSE If the geometric shape resembles an ellipse rather than a circle, the formula A = 0.7854 x a x b is used, with a representing the length of the ellipse and b the shorter | 549 | 2,507 | {"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} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.939846 |
https://programs.programmingoneonone.com/2021/02/hackerrank-java-bigdecimal-solution.html | 1,695,296,315,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00020.warc.gz | 529,607,077 | 38,030 | # HackerRank Java BigDecimal problem solution
In this HackerRank Java BigDecimal problem in the java programming language Java's BigDecimal class can handle arbitrary-precision signed decimal numbers. Let's test your knowledge of them!
Given an array, s, of n real number strings, sort them in descending order — but wait, there's more! Each number must be printed in the exact same format as it was read from stdin, meaning that .1 is printed as .1, and 0.1 is printed as 0.1. If two numbers represent numerically equivalent values (e.g., .1 = 0.1), then they must be listed in the same order as they were received as input).
## HackerRank Java BigDecimal problem solution.
```import java.math.BigDecimal;
import java.util.*;
class Solution{
public static void main(String []args){
//Input
Scanner sc= new Scanner(System.in);
int n=sc.nextInt();
String []s=new String[n+2];
for(int i=0;i<n;i++){
s[i]=sc.next();
}
sc.close();
for(int i=1; i<n ; i++){
for(int j=i; j>=1; j--){
if(new BigDecimal(s[j]).compareTo(new BigDecimal(s[j-1]))>0){
String temp = s[j];
s[j] = s[j-1];
s[j-1] = temp;
}else{
break;
}
}
}
//Output
for(int i=0;i<n;i++)
{
System.out.println(s[i]);
}
}
}```
## Second solution in java 8 programming
```import java.io.BufferedReader;
import java.io.IOException;
import java.math.BigDecimal;
import java.util.Arrays;
import java.util.Comparator;
public class Solution
{
public static class V {
public final String s;
public final BigDecimal b;
public V(final String s) {
this.s = s;
this.b = new BigDecimal(s);
}
}
public static void main(String[] args) throws IOException {
final int N = Integer.parseInt(br.readLine().trim(), 10);
final V[] A = new V[N];
for (int i = 0; i < N; i++) {
}
Arrays.sort(A, new Comparator<V>() {
@Override
public int compare(final V o1, final V o2) {
return o1.b.compareTo(o2.b);
}
});
final StringBuilder sb = new StringBuilder();
for (int i = N - 1; i >= 0; i--) {
sb.append(A[i].s).append('\n');
}
System.out.print(sb.toString());
br.close();
br = null;
}
}``` | 525 | 2,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-40 | latest | en | 0.566011 |
https://www.mcqslearn.com/applied/mathematics/linear-equation-in-applied-mathematics.php | 1,560,850,394,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998708.41/warc/CC-MAIN-20190618083336-20190618105336-00483.warc.gz | 813,409,585 | 7,104 | Applied mathematics linear equation MCQs, applied mathematics linear equation quiz answers to learn business courses online. Linear equations multiple choice questions (MCQs), applied mathematics linear equation quiz questions and answers for online mathematics degree. Learn two variable systems of equation, graphical linear equations, linear equations in mathematics, graphing linear equations, linear equations: math, applied mathematics: linear equation test prep for online certifications.
Learn linear equations test MCQs: number of ordered pair values (x,y) to satisfy linear equation ax + by = c are with choices finite, infinte, zero, and rational expression for online mathematics degree. Practice assessment test for scholarships, online learning applied mathematics linear equation quiz questions for competitive assessment in math major.
MCQ: Number of ordered pair values (x,y) to satisfy linear equation ax + by = c are
1. finite
2. infinte
3. zero
4. rational expression
A
MCQ: Set notation which states that 'S' contains (x,y) elements that satisfy linear equation ax + by = c is
1. {(a,b,x,y) | ax+bc = yc}
2. {(x,y) | ax+by = c}
3. {(a,b,c) | ax+by = c}
4. {(x,y,a) | abx+bcy = c}
B
MCQ: Set of (x,y) ordered pair that can satisfy equation is called
1. variable set
2. pair set
3. solution set
4. order set
C
MCQ: Equations in linear system are considered as
1. three degree variable
2. one degree variable
3. second-degree equation
4. first-degree equation
D
MCQ: For solution set, set builder notation is S =
1. {(a,b,c) | ax+by = c}
2. {(x,y,a) | abx+bcy = c}
3. {(a,b,x,y) | ax+ bc = yc}
4. {(x,y) | ax+by = c}
D | 427 | 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} | 4.0625 | 4 | CC-MAIN-2019-26 | latest | en | 0.840806 |
http://www.cfd-online.com/Forums/main/13171-convergence-inviscid-flow-over-sylinder-print.html | 1,477,424,020,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720356.2/warc/CC-MAIN-20161020183840-00299-ip-10-171-6-4.ec2.internal.warc.gz | 358,851,405 | 2,646 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Main CFD Forum (http://www.cfd-online.com/Forums/main/)
- - Convergence for inviscid flow over sylinder (http://www.cfd-online.com/Forums/main/13171-convergence-inviscid-flow-over-sylinder.html)
Rjakk March 21, 2007 06:59
Convergence for inviscid flow over sylinder
Hello,
I have a sylinder which I send an incompressible, inviscid flow over. All walls and sylinder is set to SLIP-condition. First I want to send water over the sylinder with, for instance, 10 m/s, and afterwards increase this to 10.5, and compare the total kinetic energy of the two control-volumes.
But flow over a sylinder isn't steady, is it? I've got great problems with convergence using steady flow. The most important thing is that I compare the two cases for flow at the exact same moment. Can I run a transient analysis for this? Or is there a good way to get convergence using steady flow?
The reason for me wanting to do this, is that I want to find the "added-mass" of my object. Optimally I would want an irrotational flow (potential), but I guess this is not possible with my CFD package (Cd-Adapco Star-CD.)
Mani March 21, 2007 09:24
Re: Convergence for inviscid flow over sylinder
Incompressible, inviscid, rotational (Euler) flow over a cylinder is problematic. There may be a steady solution, but it would be very difficult to obtain numerically for this reason: You cannot avoid numerical dissipation. Numerical dissipation will act somewhat (in a very uncontrolled and arbitrary way) as if you had a small non-zero viscosity. This will result in an effective Reynolds number which may be very large (also depending on free stream velocity and cylinder size). Large Reynolds number means: No steady solution! Small numerical dissipation will really spoil it for you. There could be two ways out of this dilemna: a) make sure you choose a small velocity over a large cylinder diameter, b) increase your artificial dissipation to get a lower effective Reynolds number (this will have a damping effect on the instability). Either way it's going to be an "engineered" solution that depends a lot on numerical dissipation.
Here is what you really should do: Run a potential flow code or apply the well-known potential solution analytically.
Ahmed March 21, 2007 11:53
Re: Convergence for inviscid flow over sylinder
When selecting your viscous model in your code, select the inviscous model (commercial codes have this option). If the code you are using does not have that option, you can always select a viscous model and entering a value of Zero for the viscosity coefficient
All times are GMT -4. The time now is 15:33. | 644 | 2,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-44 | latest | en | 0.904558 |
https://blender.stackexchange.com/questions/51175/script-for-splitting-a-mesh-into-n-components-along-z-axis?noredirect=1 | 1,627,553,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00345.warc.gz | 152,641,469 | 38,175 | # Script for splitting a mesh into n components along z axis
I'm trying to divide a mesh into multiple parts along the z axis. I've followed another tutorial and managed this, which divides the mesh into 2 parts, but I can't seem to divide it into more than two. Quite new to blender scripting so I'm not sure how to fix this. Any help would be appreciated, but the solution does need to be a script. The code I'm using at the moment is below.
import bpy, bmesh
from bpy import context as C
bpy.ops.object.mode_set(mode='EDIT')
bm = bmesh.from_edit_mesh(C.object.data)
edges = []
for i in range(-10, 10, 2):
ret = bmesh.ops.bisect_plane(bm, geom=bm.verts[:]+bm.edges[:]+bm.faces[:], plane_co=(0,0,i), plane_no=(0,0,1))
bmesh.ops.split_edges(bm, edges=[e for e in ret['geom_cut'] if isinstance(e, bmesh.types.BMEdge)])
bmesh.update_edit_mesh(C.object.data)
bpy.ops.mesh.separate(type='LOOSE')
bpy.ops.object.mode_set(mode='OBJECT')
Thanks!
My guess is you have scaled the mesh and haven't allowed for local space for the bmesh operator(s). Eg if you add a prim cylinder and scale it by z 10 fold, it's local space will still range between -1 to 1 for z for each vert coord. Running the script will only give the 0 plane cut.
Multiplying the plane location by the inverse of the objects matrix world will put the cuts into local space.
import bpy, bmesh
from bpy import context as C
from mathutils import Vector
# matrix world inverse
mwi = C.object.matrix_world.inverted()
bpy.ops.object.mode_set(mode='EDIT')
bm = bmesh.from_edit_mesh(C.object.data)
geom = bm.verts[:] + bm.edges[:] + bm.faces[:]
for i in range(-10, 10, 2):
j = mwi @ Vector((0, 0, i)) # Blender <3.90: mwi * Vector((0, 0, i))
ret = bmesh.ops.bisect_plane(bm, geom=geom, plane_co=j, plane_no=(0,0,1))
bmesh.ops.split_edges(bm, edges=[e for e in ret['geom_cut'] if isinstance(e, bmesh.types.BMEdge)])
bmesh.update_edit_mesh(C.object.data)
bpy.ops.mesh.separate(type='LOOSE')
bpy.ops.object.mode_set(mode='OBJECT')
Alternatively apply scale before running splits.
bpy.ops.object.transform_apply(scale=True) | 572 | 2,100 | {"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} | 2.640625 | 3 | CC-MAIN-2021-31 | longest | en | 0.696681 |
http://www.jianping-lai.com/2017/09/05/672-bulb-switcher-ii/ | 1,563,522,656,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00443.warc.gz | 230,427,819 | 12,064 | # 672. Bulb Switcher II
There is a room with `n` lights which are turned on initially and 4 buttons on the wall. After performing exactly `m` unknown operations towards buttons, you need to return how many different kinds of status of the `n` lights could be.
Suppose `n` lights are labeled as number [1, 2, 3 …, n], function of these 4 buttons are given below:
1. Flip all the lights.
2. Flip lights with even numbers.
3. Flip lights with odd numbers.
4. Flip lights with (3k + 1) numbers, k = 0, 1, 2, …
Example 1:
Example 2:
Example 3:
Note: `n` and `m` both fit in range [0, 1000]. | 188 | 592 | {"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-2019-30 | latest | en | 0.857092 |
http://www.stata.com/statalist/archive/2012-06/msg00043.html | 1,500,799,527,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424296.90/warc/CC-MAIN-20170723082652-20170723102652-00488.warc.gz | 550,430,062 | 7,848 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: Comparing multiple means with survey data--revisited
From Rieza Soelaeman To statalist@hsphsun2.harvard.edu Subject Re: st: Comparing multiple means with survey data--revisited Date Fri, 1 Jun 2012 11:56:42 -0500
```OK, got it now. Thank you, Steve for all your help.
Rieza
On Thu, May 31, 2012 at 5:21 PM, Steve Samuels <sjsamuels@gmail.com> wrote:
>
> I still can't make the decision for you, because I don't know the theory and
>
> • Your post states: "I don't think we necessarily can make inferences at the
> population level, given the limited data we have."
>
> This is a misconception: with a well-executed probability sample, one can always
> make inferences about the sampled population. If the sample size is limited,
> then confidence intervals might be wide. But if the sample size is small, then
> the power for tests will be poor. The CIs will provide some information,
> the tests none.
>
> • You want to know if the means are equal. In a finite population, this null
> hypothesis will never be true, so the answer in advance is "No". The real
> question is "How, if at all, different", and this is answered by CIs.
>
>
> If you wish to test a hypothesis of no difference, you need to think in terms of
> a super-population or data-generating model in which the null hypothesis
> could be true. Consider a comparison of two reported disease rates. With 100%
> reporting, the SEs for the rates and their differences would be zero (there's no
> sampling). Yet many practitioners would make inferences on the basis of a
> Poisson model in which the "true" rates do not differ. Hypothesis tests would
> also be a justified if you study the causal influence of the column variable on
> your outcome, since in this case the specific population is of secondary
> interest.
>
> • If you do use the -test- command, specify the mtest(noadjust) option, then
> multiply the p-values for the pairwise tests by 15; multiply the overall 2 d.f.
> p-value in each row by 5. These are the Bonferroni corrections.
>
> • If you compute CIs only (with -lincom-), you are free to add the finite
> population correction (fpc()) option to your -svyset- statement.
>
> • Your colleague grouped the data into wide intervals. Plotting outcomes against
>
> Reference: Levy, P. S., & Lemeshow, S. (2008). Sampling of Populations : Methods
> and Applications (4th ed.). Hoboken, N.J: Wiley.
>
> Steve sjsamuels@gmail.com
>
> On May 30, 2012, at 10:02 PM, Rieza Soelaeman wrote:
>
> Dear Steve,
> Thank you for your reply. The values are negative, because they are
> standardized scores, so individuals that are below the mean have
> negative values, whereas individuals above the mean have positive
> values. Our intention was to see whether the means in each row were
> "equal," hence testing the equality of three means, rather than
> pairwise comparisons (i.e., for individuals where Var_C = 0-9 months,
> are -1.28, -0.57, and -0.36 statistically equal?). Given this
> additional information, is my original -test- statement the correct
> way to asses this?
>
> As far as using -svyset-, I wanted to analyze the data taking into
> consideration the sampling design (2-stage cluster with urban/rural
> stratification), so that the software "knows" to adjust for
> homogeneity within the clusters. I don't think we necessarily can
> make inferences at the population level, given the limited data we
> have.
>
> Thank you also for the tip about using -subset- rather than -if- for
> subsetting the subgroup. I am re-running the means with your code.
>
> Bests,
> Rieza
>
> On Wed, May 30, 2012 at 11:25 AM, Steve Samuels <sjsamuels@gmail.com> wrote:
>
> Correction:
>
> "foreach x of local levels {"
>
> not
>
> "foreach x of local(levels) {"
>
> On May 30, 2012, at 12:05 PM, Steve Samuels wrote:
>
> Rieza:
>
> The means are negative and so don't appear to be "ordinary" descriptive
> statistics. Only you can say whether the purpose of the table is descriptive of
> a population (so that tests are not appropriate) or whether some causal
> hypothesis is in play (eg. "that such-and-such an intervention will show
> stronger effects for higher levels of variable B and for variable A").
>
> The patterns are very clear:
> 1. Means increase with row number.
> 2. In each row, first column means are higher than third column means.
>
> Confidence intervals for differences are okay for descriptive tables, but even
> if
> there is a hypothesis floating around, such intervals would just confuse things
> here.
> There would be a minimum of 15 if you did separate tests in each row, and 105
> if all pairwise comparisons in the table are considered. Note that your -test-
> statement tested equality of three means, not of two.
>
> I do suggest that you add standard errors to the table.
>
> Some alternative code:
>
> *********************
> // convert variable names to lower case for easier typing
> rename VARIABLE_*, lower
>
> svy: mean variable_a, over(variable_b variable_c)
> *********************
>
> For easier copying, you can get the columns of the table with the
> following code.
>
> *************************************************************
> levelsof variable_b, local(levels)
> foreach x of local levels {
> di "variable_b = `x'"
> svy, subpop(if variable_b==`x'): mean variable_a, over(variable_c)
> *****************************************************
>
> For correct standard errors, use the -subpop- option to subset data, not the
> -if- qualifier.
>
> Steve
> sjsamuels@gmail.com
>
> On May 29, 2012, at 11:37 PM, Rieza Soelaeman wrote:
>
> Dear Stata-Lers,
> difference between means in survey data (that is, you can't/shouldn't do
> this, I have copied the thread at the end of this e-mail). I am trying to
> replicate the work of a colleague who left recently. She created a table
> where the rows represent levels of one variable, columns represent the
> levels of another variable, and the cells contain the mean value of a third
> variable for that row/column combination and the number of people in that
> group.
>
> Example:
>
> In cells: Mean of Variable A (n)
>
> --------------------------------------------------------------------------------
> ---------------------
> Variable B (years)
> --------------------------------------------------------------------------------
> ---------------------
> Variable C
> (months) 5-10 11-15 16-20 Total p-value
> --------------------------------------------------------------------------------
> ---------------------
> 0-9 -1.28 (21) -0.57 (60) -0.36 (75) -0.57 (156) 0.032
> 10-18 -1.44 (30) -0.92 (47) -1.00 (54) -1.07 (132) 0.15
> 19-27 -1.95 (64) -1.68 (77) -1.63 (126) -1.72 (268) 0.314
> 28-36 -1.92 (51) -1.83 (52) -1.72 (104) -1.80 (206) 0.652
> 37-45 -1.96 (36) -2.01 (61) -1.65 (54) -1.87 (151) 0.107
> --------------------------------------------------------------------------------
> ---------------------
>
> Usng -svyset-, I was able to get the same means and ns in each cell, but was
> not able to get the same significance level for the difference between the
> means--she used SPSS to get the p-values. I suspect this is because I
> specified the cluster, stratum, and pweights in my -svyset- command, whereas
> the software she used allowed only for the specification of weights (to
> specify a complex sampling design in SPSS requires an extension that costs
>
> For those who are familiar with SPSS, she used the following syntax after
> applying weights, and subsetting for a specific level of VARIABLE_C:
>
> MEANS TABLES= VARIABLE_A BY VARIABLE_B
> /CELLS MEAN COUNT STDDEV
> /STATISTICS ANOVA.
>
> I believe the equivalent in Stata to get the means and p-values is to use
> the following code, but as Steve pointed out in the conversation copied
> below from 2009, this is not theoretically correct:
>
> . svy: mean VARIABLE_A if (VARIABLE_C==4), over(VARIABLE_B)
>
> . test [VARIABLE_A]_subpop_1 = [VARIABLE_A]_subpop_2 = [VARIABLE_A]_subpop_3
>
> My question is whether I should be attempting to compare the means using the
> -svyset-/-test- commands at all (is what I am trying to do
> legitimate), or if I should omit this comparison from my tables?
>
> Thanks,
> Rieza
>
> --------------------------------------------------------------------------------
> ---------------------
>
> Re: st: comparing multiple means with survey data
>
> ________________________________
> From sjsamuels@gmail.com
> To statalist@hsphsun2.harvard.edu
> Subject Re: st: comparing multiple means with survey data
> Date Tue, 23 Jun 2009 12:52:47 -0400
> ________________________________
>
> Your syntax is correct. You don't need the "linearized" option, as it
> is the default for -svy: mean-.
>
> However, hypothesis testing is usually not appropriate for finite
> population studies. See:
> http://www.stata.com/statalist/archive/2009-02/msg00806.html If
> hypothesis testing is appropriate for your situation , then you should
> exclude the finite population correction (fpc) option from your
> -svyset- command.
>
> I'm guessing that you also (or only) want to know how different the
> means in the categories of var2 are. Confidence intervals will
> provide the answer, and you can keep the finite population correction
> in your -svyset- statement if appropriate.
>
> It is poor practice (and cumbersome) to label categories with strings
> like "var2a" "var2b". These are unnecessary as "a", "b", .. have no
> descriptive value. Just make var2 a numeric variable with values 1 2
> 3 4 5. Use -label define- and -label values- to associate the
> numeric values with descriptive text.
>
> Assuming that you do that, the easiest way to to get confidence
> intervals for all pairwise differences after -svy: mean- is to write
> out the 10 statements
>
> lincom _b[1] - _b[2]
> lincom _b[1] - _b[3]
> ...
> lincom _b[4] - _b[5]
>
> For plotting continuous outcomew with groups I recommend -dotplot-
> although it will not take weights.
>
> -Steve
>
> On Mon, Jun 22, 2009 at 4:31 PM, Jean-Gael Collomb <jg@ufl.edu> wrote:
>
> Hello -
> I have been struggling to find a way to compare the means of a different
> categories of one of my variable. I think I have found a way but I wonder
> if there would be a more efficient way to do it. In the following example,
>
> var 2 has five categories (var2a-var2e).
> Here's teh commands I type (after survey setting the data):
>
> svy linearized : mean var1, over(var2)
> test [var1]var2a = [var1]var2b = [var1]var2c = [var1]var2d = [var1]var2e,
> mtest(b)
> test [var1]var2b = [var1]var2c = [var1]var2d = [var1]var2e, mtest(b)
> test [var1]var2c = [var1]var2d = [var1]var2e, mtest(b)
> test [var1]var2d = [var1]var2e, mtest(b)
>
> Is there a better way to do this?
>
> Thanks!
>
> Jean-Gael "JG" Collomb
> PhD candidate
> School of Natural Resources and Environment / School of Forest Resources
> and
> Conservation
> University of Florida
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 3,233 | 11,810 | {"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-2017-30 | latest | en | 0.913542 |
https://curriculum.code.org/documentation/gamelab/greaterThanOperator/ | 1,582,152,608,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00034.warc.gz | 344,885,804 | 6,460 | # Greater than operator
Category:Math
## Greater than operator
Category: Math
Tests whether a value is greater than another value.
Your apps will sometimes need to check the relative size of two values, and then possibly perform some specific action using an if, if-else, or while block. < returns true if the value on the left-hand side of the opertor is strictly greater than the value on the right-hand side of the operator.
### Examples
```// Basic numeric greater than check.
var x = 5;
var y = 4;
console.log(x > 4);
console.log(x > y);
```
Example: Comparing "apples" the "Apples" Basic string greater than check. Case matters for string comparison.
```// Basic string equality check. Case matters for string comparison.
var x = "apples";
var y = "Apples";
console.log(x > "bananas");
console.log(x > y);
```
Example: 2 is greater than "12"? Numeric string to number conversion is automatic in App Lab.
```// Numeric string to number conversion is automatic in App Lab.
var x = 2;
var y = "12";
if(x > y)
{
console.log("less");
}
else
{
console.log("not less");
}
```
### Syntax
```___ > ___
```
### Parameters
Name Type Required? Description
___ any Yes The operands can be a number/string/boolean, or a variable containing a number/string/boolean, or the number/string/boolean returned by a function, or the number/string/boolean result of the evaluation of an expression.
### Returns
Boolean true or false
### Tips
• If you want to test if a value is less than OR equal, you can use the >= operator.
• JavaScript will automatically perform type conversion for you when comparing two values (e.g. the integer 5 will register as equivalent to the string "5").
• When comparing two strings, JavaScript will compare them alphabetically based on character by character comparison left to right. All the upper case letters come before the lower case letters.
• Comparison operators include < <= == > >= !=
Found a bug in the documentation? Let us know at documentation@code.org
## Syntax
```__ > __
```
## Returns
Boolean (true or false)
Found a bug in the documentation? Let us know at documentation@code.org | 495 | 2,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | latest | en | 0.694852 |
https://quizizz.com/admin/quiz/5c49a9552d08e6001a979060 | 1,660,619,930,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00183.warc.gz | 436,975,831 | 9,485 | Image Representation
Save
Edit
Host a game
Live GameLive
Homework
Solo Practice
Practice
• Question 1
30 seconds
Q. (Fill in the blank to complete the sentence) Tiny squares called ..... are the building blocks of all digital images.
elements
squares
pixars
pixels
• Question 2
30 seconds
Q. The digital picture shown has been zoomed in and appears now to be very 'blocky'. What is the correct term for this?
Poor resolution
Digitisation
Pixelation
Granularity
• Question 3
30 seconds
Q. Colour depth is the term used to describe .....
N.B. The term 'bit' means a binary digit e.g. '1' or '0'
how many colours have been used to make the picture
how many colours there in each pixel
none of these
the number of bits used to indicate the colour of a single pixel
• Question 4
30 seconds
Q. Which one of the following answers best describe a limitation of this system
Only a few colours can be displayed
The system could not be used for simple computer games
Only 1 colour can be displayed
Only 2 colors can be displayed
• Question 5
30 seconds
Q. How many colours is the bottom picture showing given that it is using 4 bit colour depth?
16 colours
4 colours
256 colours
5 colours
• Question 6
30 seconds
Q. What will happen to the computer file size as the colour depth is increased?
The file size will get smaller
There will be no change in file size
The file size will get bigger
None of these
• Question 7
30 seconds
Q. Resolution means
how fast can the file be sent
how much detail in an image. higher resolution means more image detail
exactly
change picture so it looks wrong
• Question 8
30 seconds
Q. Colour depth means
the difference in brightness between objects or regions
the relationship between picture width and its height.
the overall lightness or darkness
the number of bits used for each colour of a single pixel.
• Question 9
30 seconds
Q. When we have a 'one-bit' colour, which are the only two binary numbers that are possibilities?
00 and 11
000 and 111
0 and 1
0000 and 1111
• Question 10
30 seconds
Q. How many combinations are there for a '2 bit' colour?
8
4
16
32
• Question 11
30 seconds
Q. Which 'bit' image would give the best picture quality?
24
4
8
2
• Question 12
30 seconds
Q. When you 'compress' a file, what happens? | 581 | 2,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.884183 |
https://webexpertschicago.com/qa/question-what-is-sinking-fund-for-condominium.html | 1,621,245,088,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00466.warc.gz | 617,809,223 | 8,975 | # Question: What Is Sinking Fund For Condominium?
## How do you manage sinking funds?
With a sinking fund, you save up a small amount each month for a certain block of time before you spend.
To determine how much you save, take the total amount to be spent and divide it by the number of months or weeks you have left until you need to make the purchase..
## What is sinking fund in maintenance?
99.co Guides: Sinking Fund, Maintenance Costs and Service Charge. You know what’s a condo maintenance fee, but there’s more to that than meets the eye. … In real estate, a sinking fund is a sum of money periodically set aside by the owners of an estate to cover unexpected emergencies and long-term structural costs.
## What is the difference between a sinking fund and an administrative fund?
The administrative fund is used to pay for maintaining common property and assets, insurance, postage, service contractors such as the body corporate managers or caretakers, and other recurrent spending. The sinking fund is for major expenditure of a capital or non-recurrent nature.
## How much money should be in a strata sinking fund?
If buying into a large strata scheme, you would expect a sinking fund to be hundreds of thousands of dollars. Equally, if you are buying into a block of six, the sinking fund could be reasonable with a balance of only \$60,000, because it is a matter of proportion.
## What are sinking funds example?
An Example Sinking Fund Calculation \$1800 / 8 = \$225 / pay period. For the next four months, you set aside \$225 every time you’re paid. Four months later you have the \$1800 you need to cover the expense! You pay no interest and take on no debt obligations.
## Are sinking funds mandatory?
It is mandatory and highly recommended that a housing society create a Sinking Fund, which it can do by collecting financial contributions at a fixed rate from each of its members on a monthly basis and then accumulating it over the years so that a substantial amount is generated.
## How much is condo maintenance fee in Singapore?
Owners of the condo units pay maintenance fees that form the management committee’s budget. This is typically \$500 to \$700 for most condos, but luxury condos may have costs in excess of \$1,000. The smallest (two-bedroom) unit at Draycott 8, for example, has a maintenance fee of around \$1,070 per month.
## What does condo management do?
Condo Manager is Also a Finance Manager Last but not least, condo property managers are also the managers of financial matters of the condo board. They manage such things as drafting the operational budget, preparing financial statements for each month, preparing tax returns and collecting dues related to condos.
## What is a sinking fund schedule?
Definition of Sinking Fund Schedule Sinking Fund Schedule means a schedule of principal amounts of Bonds to mature or be subject to redemption through the application of Sinking Fund Payments on the specified dates and/or a schedule of principal amounts of Bonds maturing as serial Bonds.
## What is sinking fund in property?
A sinking fund is a long-term savings account that homeowners contribute to every month through service charges. This builds up every year and should pay for any major works that are required over a period of time – such as the painting of communal areas or replacement of a roof.
## What is condo sinking fund Singapore?
Essentially, the sinking fund is used to cover the expenses of the following: Painting the condo or any other structural works necessary to preserve or enhance the appearance of the shared facilities or the condo itself. Renewing or replacing any fixtures of the condo.
## How is sinking fund calculated?
The monthly amount is both the interest to the lender and a deposit into the sinking fund. The interest to the lender is based on an annual rate of 12%. Using the simple interest formula, I = Prt, you have I = 10,000(0.12)(1) = 1,200 per year. … Next, you compute the amount to be deposited in the sinking fund each month.
## What is a sinking fund payment?
A sinking fund is a means of repaying funds borrowed through a bond issue through periodic payments to a trustee who retires part of the issue by purchasing the bonds in the open market.
## Is a reserve fund the same as a sinking fund?
Often used interchangeably with the term reserve fund, the term sinking fund was originally used specifically to refer to money collected to cover the cost of future large individual items of expenditure such as a new roof, or replacement lift.
## What is sinking fund for society?
What is Sinking Fund? Sinking fund can be defined as the fund which is kept in a separate account to be used for major expanses. These expanses include major repairs, new equipment’s and other maintenance activities along with structural changes. As per the by-law No.
## What is sinking fund used for?
A sinking fund is a fund containing money set aside or saved to pay off a debt or bond. A company that issues debt will need to pay that debt off in the future, and the sinking fund helps to soften the hardship of a large outlay of revenue.
## What sinking funds should I have?
Here is a list of 23 different sinking funds:Car repairs.Car insurance.New car savings.Homeowners insurance.Life insurance.Taxes.Doctor / dentist.Medication.More items…• | 1,105 | 5,362 | {"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} | 2.59375 | 3 | CC-MAIN-2021-21 | latest | en | 0.940023 |
https://advancesindifferenceequations.springeropen.com/articles/10.1186/s13662-018-1592-0 | 1,600,810,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206763.24/warc/CC-MAIN-20200922192512-20200922222512-00224.warc.gz | 243,840,557 | 55,062 | # Dynamic behaviors of a discrete-time predator–prey bioeconomic system
## Abstract
Bifurcation and chaotic behavior of a discrete-time singular bioeconomic system are investigated. First, the traditional catch equation is modified after accounting for the handling time of the catch in a singular bioeconomic system. To discover the richer dynamics compared with the continuous form, the proposed system is considered difference scheme. Specially, the tangent space local parameterization condensed method for DAEs is generalized. The new local parameterization method is sufficiently general to be applicable to this type of discrete singular system. Also the dynamic behaviors of the system are investigated, by using normal form theory, center manifold theorem and bifurcation theory, it is shown that the system undergoes a Neimark–Sacker bifurcation and a flip bifurcation, on varying step-size in some range. In addition, numerical simulations are presented not only to illustrate our results with the theoretical analysis, but also to exhibit the complex dynamical behaviors.
## Introduction
The study of bioeconomics has a very long history. Bioeconomics is closely related to the early development of theories in fishery economics. The basic model due to Gordon (1954) and Schaefer (1957) is
$$\frac{dx}{dt}=G(x)-h(x,E),$$
and the instantaneous profit rate is
$$v = ph(x,E) - cE,$$
(1.1)
where the “catch equation”
$$h(x,E)=qEx$$
(1.2)
is called the Schaefer catch–effort relation [1]. x is the biomass at time t. E is the harvesting effort, in the fishery setting, which is typically specified as the number of (standard) vessels actively fishing at time t. The units of harvesting effort would then be Standard Vessel Units (SVU). c is the cost of effort, and p is the ex-vessel price of fish. The parameter q is a constant, called catchability, units of q are $$\mathrm{SVU}^{-1}\times (\mathrm{time}\ \mathrm{unit})^{-1}$$. Thus, q represents the proportion of the current stock x caught by one standard vessel in one time unit (to normalize units, one could let $$q=1$$). To be specific we will now use time units of days (rather than years). Thus $$G(x)$$ now represents the population growth per day and h is the daily harvesting rate, both being measured in tonnes per day.
Predator–prey models with harvesting is always a hot research topic (see [29]). The most general form of the two-dimensional predator–prey model is Kolmogorov’s model [10]:
$$\textstyle\begin{cases} \dot{x}=xf(x,y), \\ \dot{y}=yg(x,y), \end{cases}$$
(1.3)
here x and y denote the prey and predator populations at time t, respectively. In recent years, to analyze this model from an economic point of view, many researchers have studied a class of bioeconomic systems which combine system (1.3) with prey (or predator) harvesting and (1.1), and obtained some excellent results, such as the stability of equilibria, limit cycle, Hopf bifurcation, saddle-node bifurcation, singular induced bifurcation (see [46, 8, 9, 11, 12]). It is given by
$$\textstyle\begin{cases} \dot{x}=xf(x,y)-h(x,E), \\ \dot{y}=yg(x,y), \\ v=ph(x,E)-cE, \end{cases}$$
(1.4)
here E is the effort applied to harvesting the prey at time t. The extended model is a system of differential-algebraic equations (DAEs) with differential variables x, y and algebraic variable E. In these literatures, $$h(x,E)$$ is traditionally taken as $$qEx$$. However, the Schaefer catch–effort relation has some flaws as Ref. [13] points out:
1. (i)
All processes affecting stock productivity (e.g., growth, mortality and recruitment) are subsumed in the effective relationship between effort and catch. This hypothesis has been proved to be not reasonable.
2. (ii)
The catchability coefficient q is not always constant. Improvements in technology and fishing power make q often vary through time.
3. (iii)
The harvesting function did not account for the handling time of the catch and the competition between standard vessels which are utilized for harvesting of resource. On the contrary, evidence for a nonlinear harvest function is quite strong for a number of stocks [14].
In economics there is a tradition of formulating the dynamical systems as processes in discrete time from the outset, as difference equations instead of as differential equations. The cobweb model, the Cournot duopoly model, and the Samuelson–Hicks business cycle model are all examples of this tradition. On the other hand, in population ecology many authors are more focused on discrete-time models for the following reasons: one is that, for nonoverlapping generations, births occur in regular, well-defined “breeding seasons”. This contradicts the assumption that births occur continuously [15]. The other is that the discrete-time models can also govern efficient computing models for continuous ones in order to achieve the numerical emulation. However, as far as we know, there are few studies of the difference scheme of system (1.4) which represents a class of discrete singular systems and no one has considered the general method to be applied to deal with it by now.
The main objective of this paper is to investigate the dynamic behaviors of a discrete-time economic predator–prey model. The model, featuring a more realistic nonlinear harvesting function, will give rise to some interesting dynamic behaviors, as we see in Sect. 4. The paper is organized as follows. We develop a local parameterization method based on the tangent space local parameterization of DAEs in [16] and use it to deal with the aforementioned discrete singular system in the next section. In Sect. 3, we analyze the stability of the fixed point. In addition, we also classify the fixed point based on the geometric properties of trajectories distribution near it. Surprisingly, little study has been done on this aspect for discrete systems. Section 4 is devoted to the main results of this paper, the proposed system is investigated as it undergoes a Neimark–Sacker bifurcation and a flip bifurcation. Section 5 contains a numerical verification of some key results. Conclusion and discussion are presented in Sect. 6.
## Statement of model and local parameterization
Consider the scaled equation, describing a singular bioeconomic system (see [6, 9, 1719]) with nonlinear harvesting function, as
$$\textstyle\begin{cases}\dot{x}=x(a-kx-y)-\frac{xE}{1+mx}, \\ \dot{y}=y(-s+x), \\ v=\frac{pxE}{1+mx}-cE. \end{cases}$$
(2.1)
For the practical significance, the system (2.1) is defined on the set
$$R_{+}^{3}= \bigl\{ (x,y,E)\in R_{+}^{3}\mid x\geq0,y\geq0,E\geq0 \bigr\} ,$$
and is subject to positive initial conditions,
$$x(0)>0,\qquad y(0)>0,\qquad E(0)>0.$$
Equation (2.1) is a semi-explicit autonomous differential-algebraic equations with differential index 1, $$\frac{xE}{1+mx}$$ is the harvesting function with constant m being related to the handling or processing time [17, 19]. Though this nonlinear harvesting function is more realistic, not many researchers have used it in their studies. For $$m=0$$ it reduces to the Schaefer catch–effort equation (1.2).
We put $$(x,y,E)^{T}\triangleq X$$ and use $$X_{0}=(x_{0},y_{0},E_{0})^{T}$$ to denote the equilibria of (2.1). It can be checked that system (2.1) has a unique equilibria,
$$X_{0}= \biggl(s, a-ks-\frac{v}{(p-cm)s-c}, \frac {(1+ms)v}{(p-cm)s-c} \biggr)^{T}.$$
We have $$X_{0}\in R_{+}^{3}$$, in this work, we take parameters $$(a,k,s,p,m,c,v)$$ from the set
$$\pi= \biggl\{ (a,k,s,p,m,c,v)\Bigm| a-ks-\frac {v}{(p-cm)s-c}>0,(p-cm)s-c>0 \biggr\} .$$
For (2.1), if derivatives are replaced by finite difference approximations, then we get the forward Euler difference scheme with step-size δ:
$$\textstyle\begin{cases} x_{n+1}=x_{n}+\delta x_{n} (a-kx_{n}-y_{n}-\frac {E_{n}}{1+mx_{n}} ), \\ y_{n+1}=y_{n}+\delta y_{n}(-s+x_{n}), \\ 0=\frac{px_{n}E_{n}}{1+mx_{n}}-cE_{n}-v, \end{cases}$$
(2.2)
or in map form
$$\textstyle\begin{cases} x\mapsto x +\delta x (a-kx -y -\frac{E }{1+mx } ), \\ y\mapsto y +\delta y (-s+x ), \\ 0=\frac{pxE}{1+mx}-cE-v. \end{cases}$$
(2.3)
We can easily verify that the fixed point of (2.3) is also $$X_{0}$$. Now we will determine the local equivalent parameterization system of (2.3).
We first rewrite (2.1) into the constrained form [16], that is, the differential equations on the manifold,
$$\textstyle\begin{cases} \dot{x}=x(a-kx-y)-\frac{xE}{1+mx}, \\ \dot{y}=y(-s+x), \\ \dot{E}=\delta\tilde{f_{3}}(x,y,E), \\ 0=\frac{pxE}{1+mx}-cE-v. \end{cases}$$
(2.4)
The algebraic equation sets an invariant manifold (denoted by M) in $$R^{3}$$ [16] whose number of independent coordinates is 2. Ė is described through a derivation of the algebraic equation along the solution path with respect to time t and it is not necessary for the concrete formula as we will see later. Accordingly, (2.2) becomes the following difference algebraic equations of the constrained form:
$$\textstyle\begin{cases} x_{n+1}=x_{n}+\delta x_{n} (a-kx_{n}-y_{n}-\frac {E_{n}}{1+mx_{n}} ), \\ y_{n+1}=y_{n}+\delta y_{n}(-s+x_{n}), \\ E_{n+1}=E_{n}+\delta\tilde{f_{3}}(x_{n},y_{n},E_{n}), \\ 0=\frac{px_{n}E_{n}}{1+mx_{n}}-cE_{n}-v. \end{cases}$$
(2.5)
It will be convenient to introduce some notation:
\begin{aligned}& f_{1}(x,y,E)=x+\delta x \biggl(a-kx-y-\frac{E}{1+mx} \biggr), \\& f_{2}(x,y,E)=y+\delta y(-s+x), \\& f_{3}(x,y,E)=E+\delta\tilde{f_{3}}(x,y,E), \\& g(x,y,E)=\frac{pxE}{1+mx}-cE-v, \end{aligned}
and the vector-valued function
$$f=(f_{1},f_{2})^{T}, \qquad \tilde{f}=(f_{1},f_{2},f_{3})^{T}.$$
In terms of the above notations, (2.5) becomes
$$\textstyle\begin{cases} X_{n+1}= \tilde{f}(X_{n}), \\ g(X_{n})=0, \end{cases}$$
(2.6)
or in map form
$$\textstyle\begin{cases} X\mapsto\tilde{f}(X), \\ g(X)=0. \end{cases}$$
(2.7)
Note that the form of (2.6) or (2.7) is generic (representing difference algebraic equations in constrained form). We next generalize the tangent space local parameterization method for general differential-algebraic equations in constrained form to parameterize (2.6) or (2.7). First we give a decomposition for M in $$R^{3}$$ as follows:
$$X-X_{0}=U_{0}Y+V_{0}\widetilde{Y}, \quad \forall X\in M,$$
or
$$X= X_{0}+U_{0}Y+V_{0}\widetilde{Y},\quad \forall X\in M.$$
(2.8)
$$[U_{0},V_{0}]$$ is an arbitrary order-three orthogonal matrix whose columns consist of a basis for $$R^{3}$$ with $$U_{0}$$ is $$3\times2$$ and $$V_{0}$$ is $$3\times1$$. $$Y:=(y_{1},y_{2})^{T}\in R^{2}$$ is the coordinate associated with the projection of $$X-X_{0}$$ onto the column space of $$U_{0}$$, and $$\widetilde{Y} \in R$$ is the coordinate associated with the projection of $$X-X_{0}$$ onto the column space of $$V_{0}$$; note that might be locally represented by Y. Indeed,
$$g(X_{0}+U_{0}Y+V_{0}\widetilde{Y})=0.$$
Let $$G(Y, \widetilde{Y})=g(x_{0}+U_{0}Y+V_{0}\widetilde{Y})$$, therefore
$$\det \bigl(D_{\widetilde{Y}}G(Y, \widetilde{Y}) \bigr)=D_{\widetilde {Y}}g(X_{0}+U_{0}Y+V_{0} \widetilde{Y})=Dg\cdot V_{0}\neq0,$$
where D is a derivation operator. Hence, there exists a smooth map: $$h: R^{2}\rightarrow R$$ guaranteed by the implicit function theorem such that $$\widetilde{Y}=h(Y)$$ locally holds. Putting it in (2.8) gives a local parameterization of the solution manifold with parameter Y as follows:
$$X= X_{0}+U_{0}Y+V_{0}h(Y)\equiv \psi(Y),$$
(2.9)
where
$$\psi(Y)=\left ( \textstyle\begin{array}{@{}c@{}} \psi_{1}(y_{1},y_{2})\\ \psi_{2}(y_{1},y_{2})\\ \psi_{3}(y_{1},y_{2}) \end{array}\displaystyle \right ).$$
Obviously, ψ is related to h.
From (2.9) together with the first equation of system (2.6) we get
$$\psi(Y_{n+1})=\tilde{f}\bigl(\psi(Y_{n})\bigr),$$
which gives
$$X_{0}+U_{0}Y_{n+1}+V_{0}h(Y_{n+1})= \tilde{f}\bigl(\psi(Y_{n})\bigr),$$
multiplying by $$U_{0}^{T}$$ gives
$$U_{0}^{T}X_{0}+Y_{n+1}=U_{0}^{T} \tilde{f}\bigl(\psi(Y_{n})\bigr),$$
that is,
$$Y_{n+1}=U_{0}^{T}\tilde{f}\bigl( \psi(Y_{n})\bigr)-U_{0}^{T}X_{0} ,$$
(2.10)
or in map form
$$Y\mapsto U_{0}^{T}\tilde{f}\bigl(\psi(Y) \bigr)-U_{0}^{T}X_{0},$$
which is the local parameterization of (2.7) near $$X_{0}$$ with parameter Y.
Specifically, if take
$$U_{0}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} 1& 0\\ 0& 1\\ 0& 0 \end{array}\displaystyle \right ),\qquad V_{0}=\left ( \textstyle\begin{array}{@{}c@{}} 0\\ 0\\ 1 \end{array}\displaystyle \right ),$$
then
$$U_{0}^{T}\tilde{f}\bigl(\psi(Y_{n})\bigr)=f \bigl(\psi(Y_{n})\bigr).$$
(2.11)
Here we see that $$\tilde{f_{3}}(X)$$ vanishes, which is why it is not necessary to give the concrete formula of $$\tilde{f_{3}}$$ in Ė of the equation of (2.4). Hence using (2.10) and (2.11), we obtain the following parameterization system:
$$Y_{n+1}=f\bigl(\psi(Y_{n})\bigr)-U_{0}^{T}X_{0},$$
or in map form
$$Y\mapsto f\bigl(\psi(Y)\bigr)-U_{0}^{T}X_{0},$$
that is,
$$Y\mapsto f\bigl(\psi(Y)\bigr)-(x_{0}, y_{0})^{T} .$$
(2.12)
Equation (2.12) is topologically equivalent, near the origin, to the difference algebraic system (2.3) in a neighborhood of $$X_{0}$$.
### Remark 2.1
The fixed point $$X_{0}$$ of (2.3) corresponds to the fixed point $$O(0,0)$$ of (2.12).
### Remark 2.2
Since the form of (2.6) or (2.7) has generality, the new parameterization method is also generic ($$X_{n}$$ can be of arbitrary finite dimensions). Specifically, if we take the tangent space and normal vector space of solution manifold at $$X_{0}$$ as $$U_{0}$$ and $$V_{0}$$, respectively, then the local parameterization (2.9) reduces to the tangent space local parameterization condensed method for DAEs in [16].
## Fixed points classification and stability analysis
Our goal in this section is to examine the local stability of (2.3) based upon the standard linearization technique.
From $$g(\psi(Y))=0$$ we obtain
$$Dg(X)D\psi(Y)=0.$$
(3.1)
Differentiating (2.9) with respect to y and multiplying both sides by $$U_{0}^{T}$$ gives
$$U_{0}^{T}D\psi(Y)=I_{2}.$$
(3.2)
Combine (3.1) and (3.2) to yield
\begin{aligned} D\psi(Y)&=\left ( \textstyle\begin{array}{@{}c@{}} Dg(X)\\ U_{0}^{T} \end{array}\displaystyle \right )^{-1} \left ( \textstyle\begin{array}{@{}c@{}} 0\\ I_{2} \end{array}\displaystyle \right ) \\ &=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} 1& 0 \\ 0& 1 \\ -\frac{Ep}{((p-cm)x-c)(1+mx)}& 0 \end{array}\displaystyle \right ) \end{aligned}
for future use. Note that $$D\psi(Y)$$ has nothing to do with h, so we can give the concrete form of (2.12) by a Taylor formula.
The Jacobian matrix of the system (2.12) evaluated at the origin is given by
\begin{aligned}& D \bigl( f\bigl(\psi(y)\bigr)-(x_{0}, y_{0})^{T} \bigr)\big|_{(0,0)} \\& \quad =Df(X)|_{X_{0}}D\psi(Y)|_{(0,0)} \\& \quad =\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} f_{1x}& f_{1y}& f_{1E}\\ f_{2x}& f_{2y}& f_{2E} \end{array}\displaystyle \right )\Bigg|_{X_{0}} \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} 1& 0 \\ 0& 1 \\ -\frac{E_{0}p}{((p-cm)s-c)(1+ms)}& 0 \end{array}\displaystyle \right ) \\& \quad =\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} 1-\delta s ( k-\frac{vm}{((p-cm)s-c)(1+ms)} )+\frac{vp\delta s}{ ((p-cm)s-c )^{2}(1+ms)}& -\delta s\\ \delta (a-ks-\frac{v}{(p-cm)s-c} )& 1 \end{array}\displaystyle \right ). \end{aligned}
Hence the characteristic equation associated to the Jacobian matrix is given by
$$F(\lambda)=\lambda^{2}+B\lambda+C=0,$$
(3.3)
where
$$\textstyle\begin{cases} B= \delta s (k-\frac{vm}{((p-cm)s-c)(1+ms)} )-\frac{p\delta sv}{((p-cm)s-c)^{2}(1+ms)}-2, \\ C= 1+\frac{p\delta sv}{((p-cm)s-c)^{2}(1+ms)}-\delta s (k-\frac {vm}{((p-cm)s-c)(1+ms)} ) \\ \hphantom{C={}}{}+s\delta^{2} (a-ks-\frac{v}{(p-cm)s-c} ). \end{cases}$$
Putting $$\lambda=1$$ and $$\lambda=-1$$ in $$F(\lambda)$$, respectively, we get
$$F(1)=s\delta^{2}y_{0}=s\delta^{2} \biggl(a-ks- \frac{v}{(p-cm)s-c} \biggr)$$
and
\begin{aligned} F(-1) =&4+\frac{2E_{0}p\delta s}{((p-cm)s-c)(1+ms)^{2}}-2\delta s \biggl(k-\frac{E_{0}m}{(1+ms)^{2}} \biggr)+ \delta^{2}sy_{0} \\ =&4+\frac{2p\delta sv}{((p-cm)s-c)^{2}(1+ms)}-2\delta s \biggl(k-\frac {vm}{((p-cm)s-c)(1+ms)} \biggr) \\ &{}+s\delta^{2} \biggl(a-ks-\frac{v}{(p-cm)s-c} \biggr). \end{aligned}
Let $$\lambda_{1}$$ and $$\lambda_{2}$$ denote the two roots of (3.3), referred to as the eigenvalues (or multipliers) of the fixed point $$O(0,0)$$. We first recall the relevant definitions of topological types for fixed points (see [20]).
### Definition 3.1
$$O(0,0)$$ is called
1. (i)
a hyperbolic fixed point, if the moduli of all eigenvalues do not equal 1; and
2. (ii)
a nonhyperbolic fixed point, if $$|\lambda_{1}|=1$$ or $$|\lambda_{2}|=1$$.
### Definition 3.2
If $$O(0,0)$$ is a hyperbolic fixed point, then it is called
1. (i)
a sink, if $$|\lambda_{1}|<1$$ and $$|\lambda_{2}|<1$$;
2. (ii)
a source, if $$|\lambda_{1}|>1$$ and $$|\lambda_{2}|>1$$; and
3. (iii)
a saddle, if $$\lambda_{1,2}$$ are real, with $$|\lambda_{1}|<1$$ and $$|\lambda_{2}|>1$$ (or $$|\lambda_{1}|>1$$ and $$|\lambda_{2}|<1$$).
Moreover, we can classify fixed points based on the geometric properties of trajectory distribution near them.
### Definition 3.3
$$O(0,0)$$ is called
1. (i)
a saddle, if $$\lambda_{1,2}$$ are real, with $$|\lambda_{1}|<1$$ and $$|\lambda_{2}|>1$$ (or $$|\lambda_{1}|>1$$ and $$|\lambda_{2}|<1$$);
2. (ii)
a node, if $$\lambda_{1,2}$$ are real, with $$|\lambda_{1,2}|<1$$ (or $$|\lambda_{1,2}|>1$$);
3. (iii)
a focus (sometimes called spiral point), if $$\lambda _{1,2}=r(\cos\theta\pm i \sin\theta)$$, $$r\neq1$$; and
4. (iv)
a center, if $$\lambda_{1,2}=\cos\theta\pm i \sin\theta$$.
Thus, we have the following lemma from the relations between roots and coefficients of the quadratic equation [21].
### Lemma 3.1
Suppose that $$F(1)>0$$, then we have
1. (i)
$$|\lambda_{1}|<1$$ and $$|\lambda_{2}|<1$$ iff $$F(-1)>0$$ and $$C<1$$;
2. (ii)
$$|\lambda_{1}|>1$$ and $$|\lambda_{2}|>1$$ iff $$F(-1)>0$$ and $$C>1$$;
3. (iii)
$$|\lambda_{1}|<1$$ and $$|\lambda_{2}|>1$$ (or $$|\lambda_{1}|>1$$ and $$|\lambda_{2}|<1$$) iff $$F(-1)<0$$;
4. (iv)
$$\lambda_{1}$$ and $$\lambda_{2}$$ are complex and $$|\lambda _{1}|=|\lambda_{2}|=1$$ iff $$B^{2}-4C<0$$ and $$C=1$$; and
5. (v)
$$\lambda_{1}=-1$$ and $$|\lambda_{2}|\neq1$$ iff $$F(-1)=0$$ and $$B\neq0,2$$.
Using Lemma 3.1, we obtain the following results.
### Theorem 3.1
Assuming parameters $$(a,k,s,p,m,c,v)\in\pi$$, system (2.3) has a unique positive fixed point $$X_{0}$$ and
1. (i)
it is a sink iff
$$4+2\delta\Lambda_{1}-2\delta\Lambda_{2}+ \delta^{2}\Lambda_{3}>0 \quad \textit{and}\quad \Lambda _{1}-\Lambda_{2}+\delta\Lambda_{3}< 0;$$
2. (ii)
it is a source iff
$$4+2\delta\Lambda_{1}-2\delta\Lambda_{2}+ \delta^{2}\Lambda_{3}>0\quad \textit{and} \quad \Lambda _{1}-\Lambda_{2}+\delta\Lambda_{3}>0;$$
3. (iii)
$$4+2\delta\Lambda_{1}-2\delta\Lambda_{2}+ \delta^{2}\Lambda_{3}< 0;\quad \textit{and}$$
4. (iv)
it is nonhyperbolic if one of the following conditions is satisfied:
1. (iv1)
$$\delta\Lambda_{3}=\Lambda_{2}-\Lambda_{1}$$ and $$(\Lambda _{2}-\Lambda_{1})^{2}-4\Lambda_{3}<0$$;
2. (iv2)
$$4+2\delta\Lambda_{1}-2\delta\Lambda_{2}+\delta^{2}\Lambda _{3}=0$$ and $$\delta\Lambda_{2}-\delta\Lambda_{1}\neq2,4$$.
### Theorem 3.2
With the assumption of Theorem 3.1, $$X_{0}$$ is
1. (i)
a stable node iff
$$\delta(\Lambda_{2}-\Lambda_{1})< 2 \quad \textit{and}\quad 1+\delta\Lambda_{1}-\delta\Lambda _{2}+\delta^{2} \Lambda_{3}>0;$$
2. (ii)
an unstable node iff
$$\Lambda_{2}-\Lambda_{1}< 0;$$
3. (iii)
a stable focus iff
$$(\Lambda_{2}-\Lambda_{1})^{2}-4 \Lambda_{3}< 0 \quad \textit{and}\quad \delta\Lambda_{3}< \Lambda _{2}-\Lambda_{1};$$
4. (iv)
an unstable focus iff
$$(\Lambda_{2}-\Lambda_{1})^{2}-4 \Lambda_{3}< 0 \quad \textit{and}\quad \delta\Lambda_{3}> \Lambda _{2}-\Lambda_{1},$$
where
\begin{aligned}& \Lambda_{1}=\frac{psv}{((p-cm)s-c)^{2}(1+ms)}, \\& \Lambda_{2}=s \biggl(k-\frac{vm}{((p-cm)s-c)(1+ms)} \biggr), \\& \Lambda_{3}=s \biggl(a-ks-\frac{v}{(p-cm)s-c} \biggr). \end{aligned}
## Bifurcation analysis
Based on the analysis of Sect. 3, we discuss a codimension-one bifurcation (Neimark–Sacker bifurcation and flip bifurcation) of system (2.3) at the fixed point.
### Neimark–Sacker bifurcation and invariant curve
First, let $$C=1$$, then we have
\begin{aligned}& s \biggl(a-ks-\frac{v}{(p-cm)s-c} \biggr)\delta^{2}+\frac{psv\delta }{((p-cm)s-c)^{2}(1+ms)} \\& \quad {}-s \biggl(k-\frac{vm}{((p-cm)s-c)(1+ms)} \biggr)\delta=0. \end{aligned}
Solving this equation gives
$$\delta_{1}=\frac{\Lambda_{2}-\Lambda_{1}}{\Lambda_{3}}.$$
We choose δ as a bifurcation parameter to discuss the Neimark–Sacker bifurcation when δ varies in a small neighborhood of $$\delta_{1}$$. If we consider $$\delta^{*}=\delta-\delta_{1}$$, then $$\delta=\delta_{1}$$ is equivalent to $$\delta^{*}=0$$.
The linearization of (2.12) has the characteristic equation at $$O(0,0)$$
$$F(\lambda)=\lambda^{2}+B\bigl(\delta^{*}\bigr)\lambda+C\bigl(\delta^{*} \bigr)=0,$$
where
$$\textstyle\begin{cases} B(\delta^{*})= (\delta^{*}+\delta_{1}) s (k-\frac {vm}{((p-cm)s-c)(1+ms)} )-\frac{(\delta^{*}+\delta _{1})psv}{((p-cm)s-c)^{2}(1+ms)}-2, \\ C(\delta^{*})= 1+\frac{(\delta^{*}+\delta _{1})psv}{((p-cm)s-c)^{2}(1+ms)}-(\delta^{*}+\delta_{1}) s (k-\frac {vm}{((p-cm)s-c)(1+ms)} ) \\ \hphantom{C(\delta^{*})={}}{} +s(\delta^{*}+\delta_{1})^{2} (a-ks-\frac{v}{(p-cm)s-c} ). \end{cases}$$
Let
$$\Omega_{1}=\bigl\{ (a,k,s,p,m,c,v,\delta)\mid (a,k,s,p,m,c,v)\in\pi, 0< \delta =\delta_{1},(\Lambda_{2}-\Lambda_{1})^{2}< 4 \Lambda_{3}\bigr\} .$$
Then, if $$(k,a,m,s,p,c,v,\delta)\in\Omega_{1}$$, the characteristic values are complex conjugate numbers λ and λ̅ with $$|\lambda|=|\overline{\lambda}|=1$$, where
$$\lambda, \overline{\lambda}=-\frac{B(\delta^{*})}{2}\pm\frac{i}{2}\sqrt {4C \bigl(\delta^{*}\bigr)-B^{2}\bigl(\delta^{*}\bigr)},\qquad | \lambda|_{\delta^{*}=0}=\sqrt{C(0)}=1,$$
and
$$\alpha=\frac{d|\lambda(\delta^{*})|}{d\delta^{*}}\bigg|_{\delta^{*}=0}=\frac {1}{2} \frac{\Lambda_{2}-\Lambda_{1}}{\sqrt{1+\Lambda_{1}\delta_{1}-\Lambda _{2}\delta_{1}+\Lambda_{3}\delta_{1}^{2}}}=\frac{\Lambda_{2}-\Lambda_{1}}{2}>0.$$
In addition, it is required that when $$\delta^{*}=0$$, $$\lambda^{m}$$, $${\overline{\lambda}}^{m}\neq1$$ ($$m=1,2,3,4$$), which is equivalent to $$B(0)\neq-2,0,1,2$$, that is, $$\delta_{1}\Lambda_{2}-\delta_{1}\Lambda_{1}\neq 0,2,3,4$$. Note that $$\Lambda_{2}>\Lambda_{1}$$, thus
$$H=\delta_{1}\Lambda_{2}-\delta_{1} \Lambda_{1}\neq2,3,4.$$
(4.1)
If we put
\begin{aligned}& x+\bigl(\delta^{*}+\delta_{1}\bigr)x(a-kx-y)-\frac{xE}{1+mx}-x_{0} \triangleq F_{1}(X), \\& y+\bigl(\delta^{*}+\delta_{1}\bigr)y(-s+x)-y_{0}\triangleq F_{2}(X), \end{aligned}
and
$$(F_{1},F_{2})^{T}\triangleq F_{\delta^{*}},$$
then
$$Y\mapsto F\bigl(\psi(Y)\bigr)-(x_{0}, y_{0})^{T}$$
(4.2)
is a one-parameter map with parameter $$\delta^{*}$$.
We next determine, in order to get the normal form of (4.2), the Taylor expansion of the right-hand side of (4.2) at the origin to order three, that is,
$$\left ( \textstyle\begin{array}{@{}c@{}} y_{1}\\ y_{2} \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{}} a_{1}y_{1}+a_{2}y_{2}+a_{11}y_{1}^{2}+a_{12}y_{1}y_{2}+a_{111}y_{1}^{3}+O((|y_{1}|+|y_{2}|)^{4})\\ b_{1}y_{1}+b_{2}y_{2}+b_{12}y_{1}y_{2}+O((|y_{1}|+|y_{2}|)^{4}) \end{array}\displaystyle \right ),$$
(4.3)
where the terms with 0 coefficients are omitted. By the chain rule, we have the following:
$$\textstyle\begin{cases} a_{1}=\frac{\partial F_{1}(\psi(Y))}{\partial y_{1}} |_{(0,0)}= 1+(\delta^{*}+\delta_{1})s (-k+\frac{E_{0}m}{(1+ms)^{2}}+\frac {E_{0}p}{(1+ms)^{2}((p-cm)s-c)} ), \\ a_{2}=\frac{\partial F_{1}(\psi(Y))}{\partial y_{2}} |_{(0,0)}=-(\delta ^{*}+\delta_{1})s, \\ a_{11}=\frac{\partial^{2} F_{1}(\psi(Y))}{\partial y_{1}^{2}} |_{(0,0)} \\ \hphantom{a_{11}}=\frac{(-c^{2}km^{3}s^{3}+2ckm^{2}ps^{3}-3c^{2}km^{2}s^{2}-kmp^{2}s^{3})(\delta _{1}+\delta^{*})}{(cms-ps+c)^{2}(ms+1)} \\ \hphantom{a_{11}={}}{}+\frac{(4ckmps^{2}-3c^{2}kms-kp^{2}s^{2}+c^{2}E_{0}m)(\delta_{1}+\delta ^{*})}{(cms-ps+c)^{2}(ms+1)} \\ \hphantom{a_{11}={}}{}+\frac{(2ckps-c^{2}k-cE_{0}p)(\delta_{1}+\delta^{*})}{(cms-ps+c)^{2}(ms+1)}, \\ a_{12}=\frac{\partial^{2} F_{1}(\psi(Y))}{\partial y_{1} \partial y_{2}} |_{(0,0)}=-\delta^{*}-\delta_{1}, \\ a_{111}=\frac{\partial^{3} F_{1}(\psi(Y))}{\partial y_{1}^{3}} |_{(0,0)}=-\frac{(c^{2}m^{2}-2cmp+p^{2})E_{0}c(\delta^{*}+\delta _{1})}{(cms-ps+c)^{3}(1+ms)}, \\ b_{1}=\frac{\partial F_{2}(\psi(Y))}{\partial y_{1}} |_{(0,0)}=(\delta ^{*}+\delta_{1})y_{0}, \\ b_{2}=\frac{\partial F_{2}(\psi(Y))}{\partial y_{2}} |_{(0,0)}=1,\qquad b_{12}=\frac{\partial^{2} F_{2}(\psi(Y))}{\partial y_{1} \partial y_{2}} |_{(0,0)}=\delta^{*}+\delta_{1}. \end{cases}$$
Let
$$\sigma=-\frac{B(0)}{2}, \qquad \omega=\frac{\sqrt{4C(0)-B^{2}(0)}}{2},$$
then through the change of basis with
\begin{aligned}& \left ( \textstyle\begin{array}{@{}c@{}} y_{1} \\ y_{2} \end{array}\displaystyle \right )= T\left ( \textstyle\begin{array}{@{}c@{}} z_{1} \\ z_{2} \end{array}\displaystyle \right ), \\& T= \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} a_{2} &0 \\ \sigma-a_{1} &-\omega \end{array}\displaystyle \right ), \\& T^{-1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} \frac{1}{a_{2}} &0 \\ \frac{-\sigma+a_{1}}{a_{2}\omega} &-\frac{1}{\omega} \end{array}\displaystyle \right ), \end{aligned}
we can change system (4.3) into the normal form
$$\left ( \textstyle\begin{array}{@{}c@{}} z_{1} \\ z_{2} \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} \sigma&-\omega\\ \omega &\sigma \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} z_{1} \\ z_{2} \end{array}\displaystyle \right )+\left ( \textstyle\begin{array}{@{}c@{}} \overline{f}_{1}(z_{1},z_{2}) \\ \overline{f}_{1}(z_{1},z_{2}) \end{array}\displaystyle \right ),$$
(4.4)
where the nonlinear terms are
\begin{aligned}& \overline{f}_{1}(z_{1},z_{2})=(a_{2}a_{11}+a_{12} \sigma -a_{12}a_{1})z_{1}^{2}-a_{12} \omega z_{1}z_{2}+a_{111}a_{2}^{2}z_{1}^{3}+O \bigl(\bigl( \vert z_{1} \vert + \vert z_{2} \vert \bigr)^{4}\bigr), \\& \overline{f}_{2}(z_{1},z_{2})= \frac{(\sigma -a_{1})(a_{2}a_{11}-a_{1}a_{12}+a_{12}\sigma)-a_{2}b_{12}(\sigma-a_{1})}{\omega }z_{1}^{2} \\& \hphantom{\overline{f}_{2}(z_{1},z_{2})={}}{}+ \biggl(a_{2}b_{12}-\frac{(\sigma-a_{1})a_{12}\omega}{\omega} \biggr)z_{1}z_{2} \\& \hphantom{\overline{f}_{2}(z_{1},z_{2})={}}{}+\frac{(\sigma-a_{1})a_{111}a_{2}^{2}}{\omega}z_{1}^{3}+O \bigl(\bigl( \vert z_{1} \vert + \vert z_{2} \vert \bigr)^{4}\bigr), \end{aligned}
together with
$$\textstyle\begin{cases} \overline{f}_{1z_{1}}=2(a_{2}a_{11}+a_{12}\sigma-a_{12}a_{1})z_{1}-a_{12}\omega z_{2}+3a_{111}a_{2}^{2}z_{1}^{2},\qquad \overline{f}_{1z_{2}}=-a_{12}\omega z_{1}, \\ \overline{f}_{1z_{1}z_{1}}=2(a_{2}a_{11}+a_{12}\sigma -a_{12}a_{1})+6a_{111}a_{2}^{2}z_{1}, \\ \overline{f}_{1z_{1}z_{2}}=-a_{12}\omega,\qquad \overline{f}_{1z_{2}z_{2}}=0,\qquad \overline{f}_{1z_{1}z_{1}z_{1}}=6a_{111}a_{2}^{2}, \\ \overline{f}_{1z_{1}z_{1}z_{2}}=0,\qquad \overline{f}_{1z_{1}z_{2}z_{2}}=0,\qquad \overline{f}_{1z_{2}z_{2}z_{2}}=0, \\ \overline{f}_{2z_{1}}=\frac{2(\sigma-a_{1})(a_{2}a_{11}-a_{1}a_{12}+a_{12}\sigma )-a_{2}b_{12}(\sigma-a_{1})}{\omega}z_{1}+ (a_{2}b_{12}-(\sigma -a_{1})a_{12} ) z_{2} \\ \hphantom{\overline{f}_{2z_{1}}={}}{}+\frac{3(\sigma-a_{1})a_{111}a_{2}^{2}}{\omega}z_{1}^{2}, \\ \overline{f}_{2z_{2}}= (a_{2}b_{12}-(\sigma-a_{1})a_{12} ) z_{1}, \\ \overline{f}_{2z_{1}z_{1}}=\frac{2(\sigma -a_{1})(a_{2}a_{11}-a_{1}a_{12}+a_{12}\sigma)-a_{2}b_{12}(\sigma-a_{1})}{\omega }+\frac{6(\sigma-a_{1})a_{111}a_{2}^{2}}{\omega}z_{1}, \\ \overline{f}_{2z_{1}z_{2}}=a_{2}b_{12}-(\sigma-a_{1})a_{12}, \qquad \overline {f}_{2z_{2}z_{2}}=0, \qquad \overline{f}_{2z_{1}z_{1}z_{1}}=\frac{6(\sigma -a_{1})a_{111}a_{2}^{2}}{\omega}, \\ \overline{f}_{2z_{1}z_{1}z_{2}}=0,\qquad \overline{f}_{2z_{1}z_{2}z_{2}}=0,\qquad \overline {f}_{2z_{2}z_{2}z_{2}}=0. \end{cases}$$
In order for system (4.4) to undergo a Neimark–Sacker bifurcation at $$O(0,0)$$, we require that the following quantity is not zero [22]:
$$l=-\operatorname{Re} \biggl(\frac{(1-2\lambda)\overline{\lambda}^{2}}{1-\lambda}\gamma _{20} \gamma_{11} \biggr)-\frac{1}{2}|\gamma_{11}|^{2}-| \gamma _{02}|^{2}+\operatorname{Re}(\overline{\lambda} \gamma_{21}),$$
where
\begin{aligned}& \gamma_{20}=\frac{(\overline{f}_{1z_{1}z_{1}}-\overline {f}_{1z_{2}z_{2}}+2\overline{f}_{2z_{1}z_{2}})+i(\overline {f}_{2z_{1}z_{1}}-\overline{f}_{2z_{2}z_{2}}-2\overline{f}_{1z_{1}z_{2}})}{8}, \\& \gamma_{11}=\frac{(\overline{f}_{1z_{1}z_{1}}+\overline {f}_{1z_{2}z_{2}})+i(\overline{f}_{2z_{1}z_{1}}+\overline{f}_{2z_{2}z_{2}})}{4}, \\& \gamma_{02}=\frac{(\overline{f}_{1z_{1}z_{1}}-\overline {f}_{1z_{2}z_{2}}-2\overline{f}_{2z_{1}z_{2}})+i(\overline {f}_{2z_{1}z_{1}}-\overline{f}_{2z_{2}z_{2}}+2\overline{f}_{1z_{1}z_{2}})}{8}, \\& \gamma_{21}=\frac{(\overline{f}_{1z_{1}z_{1}z_{1}}+\overline {f}_{1z_{1}z_{2}z_{2}}+\overline{f}_{2z_{1}z_{1}z_{2}}+\overline{f}_{2z_{2}z_{2}z_{2}}) +i(\overline{f}_{2z_{1}z_{1}z_{1}}+\overline{f}_{2z_{1}z_{2}z_{2}}-\overline {f}_{1z_{1}z_{1}z_{2}}-\overline{f}_{1z_{2}z_{2}z_{2}})}{16}. \end{aligned}
Note that here all the derivatives are taken at $$O(0,0)$$. From the above discussion, we obtain the following result.
### Theorem 4.1
Assume that $$(a,k,s,p,m,c,v,\delta)\in\Omega _{1}$$ and the condition (4.1) holds. If $$l\neq0$$, then the system (2.3) undergoes a Neimark–Sacker bifurcation at the fixed point $$(x_{0},y_{0},E_{0})$$ when the bifurcation parameter δ varies in a small neighborhood of $$\delta_{1}$$. Moreover, if $$l<0$$ (resp. $$l>0$$), then an attracting (resp. repelling) invariant closed curve bifurcates from the fixed point for $$\delta>\delta_{1}$$ (resp. $$\delta<\delta_{1}$$).
### Flip bifurcation and chaos
We proceed to investigate the flip bifurcation. For (3.3), let $$F(-1)=0$$ we obtain
\begin{aligned}& 4+ \frac{2p\delta sv}{((p-cm)s-c)^{2}(1+ms)}-2\delta s \biggl(k-\frac {vm}{((p-cm)s-c)(1+ms)} \biggr) \\& \quad {}+ s\delta^{2} \biggl(a-ks-\frac{v}{(p-cm)s-c} \biggr)=0. \end{aligned}
Solving this equation gives
$$\delta_{2}=\frac{\Lambda_{2}-\Lambda_{1}-\sqrt{(\Lambda_{2}-\Lambda _{1})^{2}-4\Lambda_{3}}}{\Lambda_{3}}.$$
Here we still choose δ as a bifurcation parameter to discuss the flip bifurcation when δ varies in a small neighborhood of $$\delta _{2}$$. In the same way as above, we consider $$\delta^{*}=\delta-\delta_{2}$$, then $$\delta=\delta_{2}$$ is equivalent to $$\delta^{*}=0$$. Since we would like to study the system as we vary $$\delta^{*}$$ near 0, we treat $$\delta ^{*}$$ as a variable (which has no dependence on time) and hence the parameter-free center manifold theorem can easily be extended for a parameter-dependent system.
We begin by putting
$$\overline{X}=\bigl(x,y,E,\delta^{*}\bigr)^{T},\qquad \overline{Y}= \bigl(y_{1},y_{2},\delta^{*}\bigr)^{T}.$$
Accordingly, the local parameterization (2.9) should be rewritten as
$$\overline{X}=\widetilde{\psi}(\overline{Y}),$$
where
\begin{aligned} \begin{aligned} \widetilde{\psi}(\overline{Y})&=\widetilde{\psi}\bigl(y_{1},y_{2}, \delta^{*}\bigr) \\ &=\left ( \textstyle\begin{array}{@{}c@{}} \widetilde{\psi}_{1}(y_{1},y_{2},\delta^{*})\\ \widetilde{\psi}_{2}(y_{1},y_{2},\delta^{*})\\ \widetilde{\psi}_{3}(y_{1},y_{2},\delta^{*})\\ \widetilde{\psi}_{0}(y_{1},y_{2},\delta^{*}) \end{array}\displaystyle \right ), \end{aligned} \end{aligned}
together with
\begin{aligned}& \psi_{1}(y_{1},y_{2})\triangleq\widetilde{ \psi}_{1}\bigl(y_{1},y_{2},\delta^{*}\bigr), \\& \psi_{2}(y_{1},y_{2})\triangleq\widetilde{ \psi}_{2}\bigl(y_{1},y_{2},\delta^{*}\bigr), \\& \psi_{3}(y_{1},y_{2})\triangleq\widetilde{ \psi}_{3}\bigl(y_{1},y_{2},\delta^{*}\bigr), \end{aligned}
and
$$\delta^{*}\triangleq\widetilde{\psi}_{0}\bigl(y_{1},y_{2}, \delta^{*}\bigr).$$
We put
$$F_{1}(\overline{X})=x+\bigl(\delta^{*}+\delta_{2} \bigr)x(a-kx-y)-\frac{xE}{1+mx}-x_{0}$$
and
$$F_{2}(\overline{X})=y+\bigl(\delta^{*}+\delta_{2} \bigr)y(-s+x)-y_{0}, \quad \text{and}\quad F=(F_{1},F_{2})^{T}.$$
Hence we study the three-dimensional extended system
$$\overline{Y}\mapsto \left ( \textstyle\begin{array}{@{}c@{}} F(\widetilde{\psi}(\overline{Y}))-(x_{0}, y_{0})^{T}\\ \delta^{*} \end{array}\displaystyle \right ).$$
(4.5)
As in our study of the Neimark–Sacker bifurcation, we expand the right-hand side of (4.5) in the order-three Taylor polynomial,
\left ( \textstyle\begin{array}{@{}c@{}} y_{1}\\ y_{2}\\ \delta^{*} \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{}} \begin{aligned} &a_{1}y_{1}+a_{2}y_{2}+a_{11}y_{1}^{2}+a_{12}y_{1}y_{2}+a_{13}y_{1}\delta ^{*}+a_{23}y_{2}\delta^{*}+a_{111}y_{1}^{3} \\ &\quad {}+a_{113}y_{1}^{2}\delta^{*}+a_{123}y_{1}y_{2}\delta^{*}+O((|y_{1}|+|y_{2}|+|\delta^{*}|)^{4}) \end{aligned} \\ b_{1}y_{1}+b_{2}y_{2}+b_{12}y_{1}y_{2}+b_{13}y_{1}\delta^{*}+b_{123}y_{1}y_{2}\delta ^{*}+O((|y_{1}|+|y_{2}|+|\delta^{*}|)^{4}) \\ \delta^{*} \end{array}\displaystyle \right ),
(4.6)
where
$$\textstyle\begin{cases} a_{1}=\frac{\partial F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial y_{1}} |_{(0,0,0)}=1+\delta_{2}s (-k+\frac{E_{0}m}{(1+ms)^{2}} )+\frac{\delta_{2}sE_{0}p}{(1+ms)^{2}((p-cm)s-c)}, \\ a_{2}=\frac{\partial F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial y_{2}} |_{(0,0,0)}=-\delta_{2}s,\qquad a_{12}=\frac{\partial^{2} F_{1}(\widetilde {\psi}(\overline{Y}))}{\partial y_{1}\partial y_{2}} |_{(0,0,0)}=-\delta _{2}, \\ a_{11}=\frac{\partial^{2} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial ^{2}y_{1}^{2}} |_{(0,0,0)} \\ \hphantom{a_{11}}=\frac {(-c^{2}km^{3}s^{3}+2ckm^{2}ps^{3}-3c^{2}km^{2}s^{2}-kmp^{2}s^{3})\delta _{2}}{(cms-ps+c)^{2}(ms+1)} \\ \hphantom{a_{11}={}}{}+\frac{(c^{2}E_{0}m+2ckps-c^{2}k-cE_{0}p)\delta_{2}}{(cms-ps+c)^{2}(ms+1)} \\ \hphantom{a_{11}={}}{}+\frac{(4ckmps^{2}-3c^{2}kms-kp^{2}s^{2})\delta_{2}}{(cms-ps+c)^{2}(ms+1)}, \\ a_{13}=\frac{\partial^{2} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial y_{1}\partial\delta^{*}} |_{(0,0,0)}=s (-k+\frac {E_{0}m}{(1+ms)^{2}} )+\frac{sE_{0}p}{(1+ms)^{2}((p-cm)s-c)}, \\ a_{23}=\frac{\partial^{2} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial y_{2}\partial\delta^{*}} |_{(0,0,0)}=-s, \\ a_{111}=\frac{\partial^{3} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial ^{3} y_{1}} |_{(0,0,0)}=-\frac{(c^{2}m^{2}-2cmp+p^{2})E_{0}c\delta _{2}}{(cms-ps+c)^{3}(1+ms)}, \\ a_{113}=\frac{\partial^{3} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial ^{2} y_{1}\partial\delta^{*}} |_{(0,0,0)} \\ \hphantom{a_{113}}=\frac {-c^{2}km^{3}s^{3}+2ckm^{2}ps^{3}-3c^{2}km^{2}s^{2}-kmp^{2}s^{3}}{(cms-ps+c)^{2}(ms+1)} \\ \hphantom{a_{113}={}}{}+\frac{4ckmps^{2}-3c^{2}kms-kp^{2}s^{2}+c^{2}E_{0}m}{(cms-ps+c)^{2}(ms+1)} \\ \hphantom{a_{113}={}}{}+\frac{2ckps-c^{2}k-cE_{0}p}{(cms-ps+c)^{2}(ms+1)}, \\ a_{123}=\frac{\partial^{3} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial y_{1}\partial y_{2}\delta^{*}} |_{(0,0,0)}=-1,\qquad b_{1}=\frac{\partial F_{2}(\widetilde{\psi}(\overline{Y}))}{\partial y_{1}} |_{(0,0,0)}=\delta _{2}y_{0}, \\ b_{2}=\frac{\partial F_{2}(\widetilde{\psi}(\overline{Y}))}{\partial y_{2}} |_{(0,0,0)}=1,\qquad b_{12}=\frac{\partial^{2} F_{2}(\widetilde{\psi }(\overline{Y}))}{\partial y_{1}\partial y_{2}} |_{(0,0,0)}=\delta_{2}, \\ b_{13}=\frac{\partial^{2} F_{2}(\widetilde{\psi}(\overline{Y}))}{\partial y_{1}\partial\delta^{*}} |_{(0,0,0)}=y_{0},\qquad b_{123}=\frac{\partial^{3} F_{1}(\widetilde{\psi}(\overline{Y}))}{\partial y_{1}\partial y_{2}\delta ^{*}} |_{(0,0,0)}=1. \end{cases}$$
Construct an invertible matrix
\begin{aligned}& T=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} t_{11}& t_{12}& 0\\ 1& 1& 0\\ 0& 0& 1 \end{array}\displaystyle \right ), \\& T^{-1}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} \tilde{t}_{11}& \tilde{t}_{12}& 0\\ \tilde{t}_{21}& \tilde{t}_{22}& 0\\ 0& 0& 1 \end{array}\displaystyle \right ), \end{aligned}
where
\begin{aligned}& t_{11}=\frac{a_{2}}{\frac{1}{2}-\frac{1}{2}a_{1}+\frac{1}{2}\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}, \\& t_{12}=\frac{a_{2}}{\frac{1}{2}-\frac{1}{2}a_{1}-\frac{1}{2}\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}, \\& \tilde{t}_{11}=\frac{1}{4}\frac{ (1-a_{1}+\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1} ) (-1+a_{1}+\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1} )}{a_{2}\sqrt{a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}, \\& \tilde{t}_{12}=\frac{1}{2}\frac{1-a_{1}+\sqrt{a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}{\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}, \\& \tilde{t}_{21}=-\frac{1}{4}\frac{ (1-a_{1}+\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1} ) (-1+a_{1}+\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1} )}{a_{2}\sqrt{a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}, \\& \tilde{t}_{22}=\frac{1}{2}\frac{-1+a_{1}+\sqrt {a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}{\sqrt{a_{1}^{2}+4a_{2}b_{1}-2a_{1}+1}}. \end{aligned}
Now, make the translation
$$\left ( \textstyle\begin{array}{@{}c@{}} y_{1}\\ y_{2}\\ \delta^{*} \end{array}\displaystyle \right )= T\left ( \textstyle\begin{array}{@{}c@{}} z_{1}\\ z_{2}\\ \delta^{*} \end{array}\displaystyle \right )$$
to system (4.6), then we get
$$\left ( \textstyle\begin{array}{@{}c@{}} z_{1}\\ z_{2}\\ \delta^{*} \end{array}\displaystyle \right )\rightarrow \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} -1& 0& 0\\ 0& \lambda_{2}& 0\\ 0& 0& 1 \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} z_{1}\\ z_{2}\\ \delta^{*} \end{array}\displaystyle \right )+ \left ( \textstyle\begin{array}{@{}c@{}} \varphi_{1}(z_{1},z_{2},\delta^{*})\\ \varphi_{2}(z_{1},z_{2},\delta^{*})\\ 0 \end{array}\displaystyle \right ),$$
(4.7)
where
$$\textstyle\begin{cases} \varphi_{1}(z_{1},z_{2},\delta^{*})= \alpha_{11}z_{1}^{2}+\alpha_{22}z_{2}^{2}+\alpha _{12}z_{1}z_{2}+\alpha_{13}z_{1}\delta^{*}+\alpha_{23}z_{2}\delta^{*}+\alpha _{123}z_{1}z_{2}\delta^{*}+\alpha_{113}z_{1}^{2}\delta^{*} \\ \hphantom{\varphi_{1}(z_{1},z_{2},\delta^{*})={}}{} +\alpha_{223}z_{2}^{2}\delta^{*}+\alpha_{111}z_{1}^{3}+\alpha_{222}z_{2}^{3}+\alpha _{112}z_{1}^{2}z_{2}+\alpha_{122}z_{1}z_{2}^{2}+\alpha_{133}z_{1}{\delta^{*}}^{2} \\ \hphantom{\varphi_{1}(z_{1},z_{2},\delta^{*})={}}{} +\alpha_{233}z_{2}{\delta^{*}}^{2}+O((|z_{1}|+|z_{2}|+|\delta^{*}|)^{4}), \\ \varphi_{2}(z_{1},z_{2},\delta^{*})= \beta_{11}z_{1}^{2}+\beta_{22}z_{2}^{2}+\beta _{12}z_{1}z_{2}+\beta_{13}z_{1}\delta^{*}+\beta_{23}z_{2}\delta^{*}+\beta _{123}z_{1}z_{2}\delta^{*}+\beta_{113}z_{1}^{2}\delta^{*} \\ \hphantom{\varphi_{2}(z_{1},z_{2},\delta^{*})= {}}{} +\beta_{223}z_{2}^{2}\delta^{*}+\beta_{133}z_{1}{\delta^{*}}^{2}+\beta _{233}z_{2}{\delta^{*}}^{2}+O((|z_{1}|+|z_{2}|+|\delta^{*}|)^{4}), \end{cases}$$
with
$$\textstyle\begin{cases} \alpha_{11}=\tilde{t}_{11}a_{11}t_{11}^{2}+\tilde {t}_{11}a_{12}t_{11}+\tilde{t}_{12}b_{12}t_{11}, \\ \alpha_{12}=2\tilde{t}_{11}a_{11}t_{11}t_{12}+\tilde {t}_{11}a_{12}t_{11}+\tilde{t}_{11}a_{12}t_{12}+\tilde {t}_{12}b_{12}t_{11}+\tilde{t}_{12}b_{12}t_{12}, \\ \alpha_{13}=\tilde{t}_{11}a_{13}t_{11}+\tilde{t}_{12}b_{13}t_{11}+\tilde {t}_{11}a_{23},\qquad \alpha_{22}=\tilde{t}_{11}a_{11}t_{12}^{2}+\tilde {t}_{11}a_{12}t_{12}+\tilde{t}_{12}b_{12}t_{12}, \\ \alpha_{23}=\tilde{t}_{11}a_{13}t_{12}+\tilde{t}_{12}b_{13}t_{12}+\tilde {t}_{11}a_{23},\qquad \alpha_{111}=\tilde{t}_{11}a_{111}t_{11}^{3}, \qquad \alpha _{112}=3\tilde{t}_{11}a_{111}t_{11}^{2}t_{12}, \\ \alpha_{113}=\tilde{t}_{11}a_{113}t_{11}^{2}+\tilde {t}_{11}a_{123}t_{11}+\tilde{t}_{12}b_{123}t_{11}+\tilde {t}_{11}a_{223},\qquad \alpha_{122}=3\tilde{t}_{11}a_{111}t_{11}t_{12}^{2}, \\ \alpha_{123}=2\tilde{t}_{11}a_{113}t_{11}t_{12}+\tilde {t}_{11}a_{123}t_{11}+\tilde{t}_{11}a_{123}t_{12}+\tilde {t}_{12}b_{123}t_{11}+\tilde{t}_{12}b_{123}t_{12}+2\tilde {t}_{11}b_{223}, \\ \alpha_{133}=\tilde{t}_{11}a_{233},\qquad \alpha_{222}=\tilde {t}_{11}a_{111}t_{12}^{3}, \\ \alpha_{223}=\tilde {t}_{11}a_{113}t_{12}^{2}+\tilde{t}_{11}a_{123}t_{12}+\tilde {t}_{12}b_{123}t_{12}+\tilde{t}_{11}a_{223}, \\ \alpha_{233}=\tilde{t}_{11}a_{233}, \end{cases}$$
and
$$\textstyle\begin{cases} \beta_{11}=\tilde{t}_{21}a_{11}t_{11}^{2}+\tilde {t}_{21}a_{12}t_{11}+\tilde{t}_{22}b_{12}t_{11}, \\ \beta_{12}=2\tilde{t}_{21}a_{11}t_{11}t_{12}+\tilde {t}_{21}a_{12}t_{11}+\tilde{t}_{21}a_{12}t_{12}+\tilde {t}_{22}b_{12}t_{11}+\tilde{t}_{22}b_{12}t_{12}, \\ \beta_{13}=\tilde{t}_{21}a_{13}t_{11}+\tilde{t}_{22}b_{13}t_{11}+\tilde {t}_{21}a_{23},\qquad \beta_{22}=\tilde{t}_{21}a_{11}t_{12}^{2}+\tilde {t}_{21}a_{12}t_{12}+\tilde{t}_{22}b_{12}t_{12}, \\ \beta_{23}=\tilde{t}_{21}a_{13}t_{12}+\tilde{t}_{22}b_{13}t_{12}+\tilde {t}_{21}a_{23}, \\ \beta_{113}=\tilde{t}_{21}a_{113}t_{11}^{2}+\tilde {t}_{21}a_{123}t_{11}+\tilde{t}_{22}b_{123}t_{11}+\tilde {t}_{21}a_{223}, \\ \beta_{123}=2\tilde{t}_{21}a_{113}t_{11}t_{12}+2\tilde {t}_{21}a_{223}+\tilde{t}_{21}a_{123}t_{11}+\tilde {t}_{21}a_{123}t_{12}+\tilde{t}_{22}b_{123}t_{11}+\tilde {t}_{22}b_{123}t_{12}, \\ \beta_{133}=\tilde{t}_{21}a_{233},\qquad \beta_{223}=\tilde {t}_{21}a_{113}t_{12}^{2}+\tilde{t}_{21}a_{123}t_{12}+\tilde {t}_{22}b_{123}t_{12}+\tilde{t}_{21}a_{223}, \\ \beta_{233}=\tilde{t}_{21}a_{233}. \end{cases}$$
Having gone through these coordinate changes, the parameter-dependent center manifold of (4.7) can be approximately represented as follows:
\begin{aligned}& W^{c}(0,0,0)= \bigl\{ \bigl(z_{1},z_{2},\delta^{*} \bigr)\in R^{3}\mid z_{2}=h\bigl(z_{1},\delta ^{*} \bigr),h(0,0)=0,Dh(0,0)=0 \bigr\} , \\& h\bigl(z_{1},\delta^{*}\bigr)=c_{0}{\delta^{*}}^{2}+c_{1}z_{1} \delta^{*}+c_{2}z_{1}^{2}+O\bigl(\bigl( \vert z_{1} \vert + \bigl\vert \delta ^{*} \bigr\vert \bigr)^{3}\bigr). \end{aligned}
(4.8)
Since the center manifold is invariant under the flow, one can substitute $$z_{2}=h(z_{1},\delta^{*})$$ into the second formula of (4.7) and obtain
$$h \bigl(-z_{1}+\varphi_{1} \bigl(z_{1},h \bigl(z_{1},\delta^{*}\bigr),\delta^{*} \bigr),\delta ^{*} \bigr)- \lambda_{2}h\bigl(z_{1},\delta^{*}\bigr)-\varphi_{2} \bigl(z_{1},h\bigl(z_{1},\delta ^{*}\bigr),\delta^{*} \bigr)=0.$$
(4.9)
Substituting (4.8) into (4.9), and then comparing the coefficients for $${\delta^{*}}^{2}$$, $$\delta^{*} z_{1}$$ and $$z_{1}^{3}$$, we get
$$c_{0}=0,\qquad c_{1}=-\frac{\beta_{13}}{1+\lambda_{2}},\qquad c_{2}=\frac{\beta _{111}}{1-2\alpha_{11}-\beta_{12}}=0.$$
Therefore, we consider the map which is (4.7) restricted to the center manifold $$W^{c}(0,0)$$,
\begin{aligned} F:z_{1} \rightarrow&-z_{1}+d_{11}z_{1}^{2}+d_{12}z_{1} \delta^{*}+d_{112}z_{1}^{2}\delta ^{*}+d_{122}z_{1}{ \delta^{*}}^{2}+d_{22}{\delta^{*}}^{2}+d_{111}z_{1}^{3}+d_{222}{ \delta ^{*}}^{3} \\ &{}+O\bigl(\bigl( \vert z_{1} \vert + \bigl\vert \delta^{*} \bigr\vert \bigr)^{4}\bigr), \end{aligned}
(4.10)
where
\begin{aligned}& d_{11}=\alpha_{11},\qquad d_{112}= \alpha_{113}+\alpha_{12}c_{1},\qquad d_{22}=\alpha _{33},\qquad d_{222}= \alpha_{333},\qquad d_{12}=\alpha_{13}, \\& d_{122}=\alpha_{133}+\alpha_{23}c_{1}, \qquad d_{111}=\alpha_{111}. \end{aligned}
In order for system (4.10) to undergo a flip bifurcation at the origin, we require that the two quantities
$$\tilde{\alpha}_{1}= \biggl(2\frac{\partial F}{\partial\delta^{*}\partial z_{1}}+\frac{\partial F}{\partial\delta^{*}} \frac{\partial F}{\partial z_{1}} \biggr)_{(0,0)}=2d_{12}$$
and
$$\tilde{\alpha}_{2}= \biggl(\frac{1}{2} \biggl( \frac{\partial^{2}F}{\partial z_{1}^{2}} \biggr)^{2}+\frac{1}{3} \biggl( \frac{\partial^{3}F}{\partial z_{1}^{3}} \biggr) \biggr)_{(0,0)}=2d_{11}^{2}+2d_{111}$$
do not equal zero [22].
Let
$$\Omega_{2}=\bigl\{ (a,k,s,p,m,c,v,\delta)\mid (a,k,s,p,m,c,v)\in\pi, 0< \delta =\delta_{2},\delta\Lambda_{2}-\delta \Lambda_{1}\neq2,4\bigr\} ,$$
then we have the following result.
### Theorem 4.2
Assuming that $$(a,k,s,p,m,c,v,\delta)\in\Omega_{2}$$. If $$\tilde{\alpha}_{2}\neq0$$, then the system (2.3) undergoes a flip bifurcation at the fixed point $$(x_{0},y_{0},E_{0})$$ when the bifurcation parameter δ varies in a small neighborhood of $$\delta _{2}$$. Moreover, if $$\tilde{\alpha}_{2}>0$$ (resp., $$\tilde{\alpha}_{2}<0$$), then the period-2 orbits that bifurcate from $$(x_{0},y_{0},E_{0})$$ are stable (resp., unstable).
## Numerical simulations
In this section, we give two examples to illustrate our theoretical analysis.
### Example 5.1
(Neimark–Sacker bifurcation)
Here we present a numerical analysis of the proposed system (2.3) with the following artificially chosen data: $$a=4$$, $$k=1$$, $$s=2$$, $$p=1$$, $$m=0.1$$, $$c=1$$ and $$v=0.6$$ with $$(a,k,s,p,m,c,v)\in\pi$$. It is easy to verify that the system (2.3) has a unique fixed point $$(2,1.25,0.9)$$ and $$\delta_{1}=0.125$$.
The multipliers of the positive fixed point are $$\lambda_{\pm}\approx 0.9805\pm0.1967i$$ with $$|\lambda_{\pm}|=1$$ and $$\alpha=0.1563>0$$. Since $$H=0.0391\neq2,3,4$$, the condition (4.1) holds. Then, using Theorem 4.1, the system (2.3) undergoes a Neimark–Sacker bifurcation at the fixed point $$(2,1.25,0.9)$$ with $$l=0.0017>0$$.
Figures 1, 2 and 3 reveal that the fixed point $$(2,1.25,0.9)$$ of (2.3) is unstable for $$\delta=0.125+0.0014$$ and becomes stable for $$\delta=0.125-0.035$$, and a repelling limit cycle appears around it at $$\delta=0.125-0.0001$$.
### Example 5.2
(Flip bifurcation)
Fix $$a=16$$, $$k=4$$, $$s=3.9$$, $$p=7$$, $$m=1$$, $$c=1$$ and $$v=1$$ and the fixed point $$(3.9,0.3554,0.2188)$$ with $$(a,k,s,p,m,c,v)\in\pi$$. After computation we get the following quantities:
\begin{aligned}& \delta_{2}=0.1294, \\& \tilde{\alpha}_{1}=-0.1792, \\& \tilde{\alpha}_{2}=0.0022-2\times6.8573\times10^{-6}>0. \end{aligned}
Figures 4(a), 4(b) and 4(c) reveal that the fixed point is stable for $$\delta<0.1294$$, and loses its stability at the flip bifurcation parameter value $$\delta=0.1294$$. We also observe that there is a cascade of period doubling bifurcations. The largest Lyapunov exponents corresponding to Figs. 4(a), 4(b) and 4(c) are computed in Fig. 4(d).
Figure 5 shows the phase portraits which are associated with Figs. 4(a), 4(b) and 4(c). For $$\delta\in(0.12, 0.1616)$$, there are orbits of period $$2, 4, 8, \ldots$$ . When $$\delta=0.1616$$, one can see that the largest Lyapunov exponent is positive, thereby confirming that the system is chaotic. And then for $$\delta\in(0.1616,017)$$ some largest Lyapunov exponents are negative, indicating the existence of periodic windows in the chaos region.
## Conclusion
The present paper is concerned with the dynamics of a discrete-time economic predator–prey system in the presence of a type of nonlinear harvesting function. We find the fixed point and its stability. Most interestingly, we have seen that our results reveal a far richer dynamics of the discrete model compared with the continuous one proposed in [9], including an invariant circle, cascades of period-doubling bifurcation and chaotic sets. We confirm the complexity of the dynamic behavior by computing the largest Lyapunov exponents. This paper extends our previous works (see [8, 9]) and provides a sufficiently general parameterization method for a wide range of discrete singular systems.
However, in the presented harvesting function, we have no concern about the effects of competition between boats which will increase the complexity of the normal form and the amount of calculation. Moreover, for discrete-time systems, Marotto’s theorem [23, 24] is a sufficient criterion for the existence of chaos. If the fixed point of system (2.3) is a snap-back repeller under certain parameter conditions, then one can conclude that the system is chaotic in the Marotto sense. Furthermore, various feedback controls can be implemented for controlling the bifurcation and chaos in the system (2.3), which should be useful for fishery management control and biological conversion. Space prevents that discussion here, but these issues will be the topics of our future research.
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## Acknowledgements
The authors wish to express their gratitude to the editors and reviewers for the helpful comments. This research is supported by National Natural Science Foundation of China under grant No. 71271158.
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The study presented here was carried out in collaboration between all authors. All authors read and approved the final manuscript.
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Correspondence to Weiyi Liu.
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Liu, W., Cai, D. & Shi, J. Dynamic behaviors of a discrete-time predator–prey bioeconomic system. Adv Differ Equ 2018, 133 (2018). https://doi.org/10.1186/s13662-018-1592-0
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• Published:
• 37G10
• 39A28
• 39A30
### Keywords
• Generalized local parameterization method
• Discrete singular system
• Bioeconomic system
• Nonlinear harvesting rate
• Stability
• Codimension-one bifurcation
• Chaotic behavior | 20,712 | 52,637 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.906955 |
https://mathoverflow.net/questions/369130/fuchsian-groups-and-eichlers-result | 1,618,891,731,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039375537.73/warc/CC-MAIN-20210420025739-20210420055739-00351.warc.gz | 474,271,062 | 31,647 | # Fuchsian groups and Eichler's result
Let $$G$$ be a Fuchsian group of first kind contained in $$\text{PSL}_2(\mathbb{R})$$. A result of Eichler says, there exists a finite set $$S\subset G$$ such that any $$\gamma$$ in $$G$$ can be written as a product $$\prod_{i=1}^{k} \gamma_i,$$ where each $$\gamma_i$$ are either in $$S,$$ or power of some parabolic element coming from $$S,$$ for some $$k\ll \log ||\gamma||.$$ Howerver, the original Eichler's paper is in German, (http://matwbn.icm.edu.pl/ksiazki/aa/aa11/aa11111.pdf) and this is giving me trouble to understand the proof. Does anyone know if I can find the proof written in English somewhere else ?
This follows from Theorem 2(i) and Theorem 4 in The structure of words in discrete subgroups of $$\mathrm{SL}(2,\mathbb{C})$$, by Beardon.
Since it isn't explicitly stated, I will roughly summarize/explain how you get the result.
Lets $$D$$ be a convex fundamental polygon for $$G$$, $$S^*$$ the associated generating set, and let $$S$$ be the set $$S^*$$ with the natural parabolics added (generators of maximal parabolic subgroup at each ideal vertex). For example, the natural generating set for $$\mathrm{PSL}_2(\mathbb Z)$$ will have $$S^*=S$$ since the parabolic is already there (guessing Eichler was inspired by Euclidean algorithm). If you have the "usual domain" for a Fuchsian group corresponding to a complete, once punctured, finite volume, hyperbolic torus, $$S^*$$ will be $$\{A,B,A^{-1},B^{-1}\}$$. The fundamental domain has four parabolic vertices, which get identified, so adding $$P=ABA^{-1}B^{-1}$$, its cyclic permutations, and inverses will give $$S$$. In this torus example consider the parabolic vertex corresponding to $$P$$, v, and note that
$$D, AD, (AB)D, (ABA^{-1})D, (ABA^{-1}B^{-1})D=PD$$
all contain $$v$$. More generally you have that $$v \in (P^k W) D$$ where $$W$$ is an initial segment of $$P$$.
Beardon defines decomposition of elements in Fuchsian groups into chunks $$C_i$$ which split into two types: type I are elements which are not longer than some constant $$m$$ and type II are elements longer than $$m$$. Theorem 3 tells you that type II $$C_i$$ are basically parabolics in the sense that there is a parabolic vertex $$v \in D$$ such that
$$v \in D, A_1D,\dots,(A_1 \cdots A_n )D=C_i D$$
where $$A_i \in S^*$$. That means that $$C_i= (P^k W)D$$ where $$P$$ is the parabolic at $$v$$ and $$W$$ is some initial segment of $$P$$.
Theorem 2 tells you that there are log many $$C_i$$ compared to norm and Theorem 4 gives the bounds if you "collapse" the type II pieces/split into parabolic part.
• Hi Paul, thanks for mentioning the reference. Beardon is first considering the Dirichlet polygon as a fundamental domain and then using the elements of side pairing elements to generate words. However, its not clear to me what type of elements they are. Can we say they are finite up to multiplication by some parabolic ? – dragoboy Sep 2 '20 at 16:53
• so you mean any element of $G*$ is either hyperbolic or a parabolic ? no elliptic element is there or I am missing something ? – dragoboy Sep 3 '20 at 15:35
• @dragoboy I am saying you can have any of the types. the PSL2(Z) case has an elliptic in $G^*$ (the order two element) and a parabolic. – user35370 Sep 3 '20 at 15:37
• @dragoboy The parabolic will fix $v$. The $W$ part won't fix $v$, it moves one of the other parabolic vertices of $D$ to $v$. – user35370 Sep 5 '20 at 23:34
• @dragoboy yes, exactly – user35370 Sep 6 '20 at 15:01 | 1,028 | 3,504 | {"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": 41, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-17 | longest | en | 0.84304 |
https://www.sheffield.ac.uk/news/maths-students-have-solution-decorating-perfect-christmas-tree | 1,718,203,007,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00820.warc.gz | 900,448,004 | 15,015 | # Maths students have the solution for decorating the perfect Christmas tree
If you've gone overboard while decorating the Christmas tree this year and it's more gaudy than great then maths students from the University of Sheffield may have the answer.
Members of Sheffield University Maths Society (SUMS) have created a festive formula to ensure just the right ratio of lights, tinsel and baubles are used to give your Christmas tree the perfect look.
Using their 'treegonometry' calculations they say a 152cm (5ft) Christmas tree would need 31 baubles, around 776 cms of tinsel and 478 cms of lights with a 15cm star or angel to top it off.
Their calculations look like this:
• Length of tinsel in cm: (13 x pi) divided by 8, then multiply by tree hight in cm
• Number of baubles: (17 x pi) then multiply by tree hight in cm
• Length of lights in cm: pi multiplied by tree hight in cm
• Height of star in cm: tree height in cm divided by 10
### Formula for the perfect Christmas tree
Number of baubles: 29
Height of star or fairy (cm): 14
Length of tinsel (cm): 715
Length of lights (cm): 440
Sophie Chong, president of SUMS said: "Being interested in maths can be a struggle for a lot of people but this is a brilliant way of showing an applicable side of maths.
"This equation - which was created by two former students and SUMS members a few years ago - is a nice and easy equation to get people involved in maths, as well as a bit of good, festive fun."
## Christmas tree facts
• The world's tallest Christmas tree would, at 2,600ft tall, need more than 16,000 baubles, over 4,000 meters of tinsel, almost 2,500 meters of lights, and an 80 meter tall star.
• The Trafalgar Square tree is usually about 25 meters tall, meaning it needs 515 baubles, over 100 meters of tinsel, 78 meters of lights and a 2.5 meter tall star. | 473 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-26 | latest | en | 0.937953 |
https://nickadamsinamerica.com/tag/math-guide/ | 1,621,281,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992440.69/warc/CC-MAIN-20210517180757-20210517210757-00159.warc.gz | 432,860,852 | 21,869 | # Math Guide
## Phonics Games For KindergartenPhonics Games For Kindergarten
Published at Saturday, March 28th 2020, 07:31:23 AM by Roxane Roussel. Worksheet. When we add numbers that need to be carried over we can only carry over a digit into the right spot. For example, when we add the 4 and 9 from the problem on the board we have to carry over the answer (13) into the correct place value spots.
## Free Printable Worksheets For KindergartenFree Printable Worksheets For Kindergarten
Published at Sunday, March 22nd 2020, 23:16:05 PM by Roesia Vallee. Worksheet. We can’t all be master mathematicians or famous doctors. Who would create the music we listen to while we relax or exercise? Who would paint the artworks we love to admire? Who would write the books we love to read? Each and every child has their own unique talents and abilities. It is so important to encourage these, instead of pushing some ancient notion on your child.
Published at Friday, April 03rd 2020, 18:32:03 PM by Geneva Collet. Worksheet. These time worksheets will produce problems with elapsed days, weeks, months, and years. These time worksheets will produce 10 problems per page. These time worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade.
Published at Tuesday, March 24th 2020, 19:11:18 PM. Worksheet By Yolanda Clement. These Math Flash Cards are great for classroom practice recognizing different shapes. The set will produce thirteen different shapes. These Math Flash Cards are appropriate for Kindergarten, 1st Grade, and 2nd Grade.
## Sample Worksheets For KindergartenSample Worksheets For Kindergarten
Published at Tuesday, March 24th 2020, 18:48:44 PM. Worksheet By Genevre Antoine. We can’t all be master mathematicians or famous doctors. Who would create the music we listen to while we relax or exercise? Who would paint the artworks we love to admire? Who would write the books we love to read? Each and every child has their own unique talents and abilities. It is so important to encourage these, instead of pushing some ancient notion on your child.
Published at Tuesday, March 24th 2020, 18:35:20 PM. Worksheet By Adrianne Devaux. These mixed problems worksheets may be configured for adding and subtracting 2, 3, and 4 digit problems in a vertical format. For the subtraction problems you may select some regrouping, no regrouping, all regrouping, or subtraction across zero. You may select up to 30 mixed problems per worksheet.
Published at Tuesday, March 24th 2020, 18:17:30 PM. Worksheet By Berneen Dumas. These division worksheets are a good introduction for algebra concepts. You may select various types of characters to replace the missing numbers of the division problems on the division worksheets. The format of the division worksheets are horizontal and the answers range from 0 to 99. These division worksheets can be configured to layout the division problems using the division sign or a slash (/) format. You may select between 12 and 30 problems for these division worksheets.
Published at Tuesday, March 24th 2020, 16:54:43 PM. Worksheet By Leala Delannoy. These addition worksheets will produce 12 vertical or horizontal addition problems using dot figures to represent the numbers. You may select the numbers for the addition worksheets to be used from 0 to 10.
###### Kindergarten Daily ScheduleKindergarten Daily Schedule
Published at Tuesday, March 24th 2020, 16:43:16 PM. Worksheet By Pensee Perrier. This section contains all of the graphic previews for the Differentiation Applications Worksheets. We have rate of change, graphing, graph properties, differentials, optimization, Newton’s Method, and related rates worksheets for your use. These Calculus worksheets are a good resource for students in high school.
Published at Tuesday, March 24th 2020, 16:38:47 PM. Worksheet By Voleta Morel. These Math Flash Cards are great for classroom practice recognizing different shapes. The set will produce thirteen different shapes. These Math Flash Cards are appropriate for Kindergarten, 1st Grade, and 2nd Grade.
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# mixed_signal_notes_set_8 - ECE145C/218C notes M Rodwell...
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ECE145C /218C notes, M. Rodwell, copyrighted 2007 High Speed Mixed Signal IC Design notes set 8 Noise in Electrical Circuits: circuit noise analysis Mark Rodwell University of California, Santa Barbara [email protected] 805-893-3244, 805-893-3262 fax
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ECE145C /218C notes, M. Rodwell, copyrighted 2007 Noise in electrical circuits: topics Math: distributions, random variables, expectations, pairs of RV, joint distributions, mean, variance, covariance and correlations. Random processes, description, stationarity, ergodicity, correlation functions, autocorrelation function, power spectral density. Noise models of devices: thermal and "shot" noise. Models of resistors, diodes, transitors, antennas. Circuit noise analysis: network representation. Solution. Total output noise. Total input noise. 2 generator model. En/In model. Noise figure, noise temperature. Signal / noise ratio.
ECE145C /218C notes, M. Rodwell, copyrighted 2007 Noise in electrical circuits: Strategy shown. simply all here are steps key but the detail, more much provide notes noise ece594 The problem. any to extend then can one n, imaginatio much o Without to s. definition and methods analytical develop to this Use problem. meaningful possible simplest Work the limited. is here our time Again,
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ECE145C /218C notes, M. Rodwell, copyrighted 2007 Goal: Computing Signal/Noise Ratio and Sensitivity sin cos clock and data recovery clock and data recovery noise. referred - input equivalent the to magnitude signal input the compare to convenient often is It input. circuit decision at the ratio se signal/noi the find must we y, sensitivit receiver the compute To Receiver Optical Receiver Radio
ECE145C /218C notes, M. Rodwell, copyrighted 2007 Circuit noise analysis: Goals ( ) ? noise referred - output the calculate we do How / where , : s input term fictitious by d represente be can terms noise These load the and amplifier, the generator, the from arises Noise noise. and signal both has output circuit The output noise, input noise, input noise, output noise, load noise, amplifier noise, generator noise, output noise, output noise, v in v in v out in v out A V V V V A V V A V V V V V V V A V = + = + = + + = + =
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ECE145C /218C notes, M. Rodwell, copyrighted 2007 Noise model of this circuit fF 4 . 6 = cbx C fF 3.7 = cbi C = 49 bb R = 3 . 4 ex R = 6500 c R e b m V g ' e b V ' E B C = diff C fF 182 = je C fF 38 = π R 350 4kTRex 4kTRbb 2kT/gm or 2qIc 2kT/Rbb or 2qIb Icbo 2qIcbo 4kT/Rf Rf Rgen 4kTRgen Ree 4kTRee RL 4kTRL Vout Vin Vgen circuit. simple a very with ns calculatio illustrate will We . techniques efficient more learn will We analysis. hand for hard too are methods force - Brute . complexity formidable of is circuits practical most of analysis Noise . generators noise of number large a has circuit The
ECE145C /218C notes, M. Rodwell, copyrighted 2007 Circuit Noise Analysis: 1st Example (a) densities spectral based - Hz sided, - Single : Notation model. noise simplified with amplifier,
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## This note was uploaded on 12/29/2011 for the course ECE 594A taught by Professor Rodwell during the Fall '09 term at UCSB.
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# Problem GCake
Žofka really likes cakes. Her most recent favorite is a rectangular-shaped cake with hard chocolate glazing and white marzipan roses on top. The glazing has been pre-cut in a grid-like pattern at the bakery since this type of glazing is difficult to cut after it hardens. The bakers pride themselves in creating various creations with the roses—they place every rose in the middle of one of the grid squares (at most one rose per square) but the placement differs each time. Žofka invited her friends to share the cake with her—together with her there are as many people as there are roses on the cake as she and each of her friends like a piece with a rose. Žofka can cut only rectangular pieces of the cake along the grid lines. How should she cut the cake so that she and each of her friends gets a rectangular piece with a rose and there is the smallest possible amount of leftover cake?
## Input
The first line contains $p$, $q$, and $n$, where $p \times q$ are the dimensions of the cake and $n$ is the number of roses. Then $n$ lines follow, the $i$-th line contains the coordinates of the $i$-th rose, where the first coordinate is between $1$ and $p$ and the second coordinate is between $1$ and $q$. You may assume that $p$, $q$ and $n$ are at most $10\, 000$.
## Output
The output contains $n + 1$ lines. The first $n$ lines contain four numbers each: the $j$-th line contains $a_ j$, $b_ j$, $c_ j$, $d_ j$, where $1 \leq a_ j \leq c_ j \leq p$ and $1 \leq b_ j \leq d_ j \leq q$ and $(a_ j, b_ j)$ and $(c_ j, d_ j)$ correspond to the grid squares at the opposite corners of the rectangular piece for the $j$-th person. The rectangular pieces have to be non-overlapping. The last line contains a single number, the area of the leftovers. The area of the leftovers should be the smallest possible.
Sample Input 1 Sample Output 1
4 5 4
2 2
3 4
1 4
4 5
1 1 1 5
2 1 2 5
3 1 3 5
4 1 4 5
0
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https://toptipbio.com/category/maths/ | 1,618,659,087,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00299.warc.gz | 662,014,718 | 23,307 | # Maths
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In this article, I will explain how to easily prepare PCR stock primers and how to dilute them into a working primer solution. Mastering qPCR A... | 401 | 1,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-17 | longest | en | 0.867407 |
http://blog.panictank.net/tag/python/ | 1,582,769,446,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00116.warc.gz | 23,069,540 | 27,778 | # The Algorithm Design Manual: Chapter 4
4-1. The Grinch is given the job of partitioning 2n players into two teams of n players each. Each player has a numerical rating that measures how good he/she is at the game. He seeks to divide the players as unfairly as possible, so as to create the biggest possible talent imbalance between team A and team B. Show how the Grinch can do the job in $O(n \log n)$ time.
Solution
First sort the players with an algorithm which runs in $O(n \log n)$ time and afterwards form two teams with the first n player in the first team and the last n players in the second team.
4-3. Take a sequence of 2n real numbers as input.Design an $O(n \log n)$ algorithm that partitions the numbers into n pairs, with the property that the partition minimizes the maximum sum of a pair. For example, say we are given the numbers (1,3,5,9). The possible partitions are ((1,3),(5,9)), ((1,5),(3,9)), and ((1,9),(3,5)). The pair sums for these partitions are (4,14), (6,12), and (10,8). Thus the third partition has 10 as its maximum sum, which is the minimum over the three partitions.
Solution We can minimize the maximum sum if we pair up the lowest and the highest element. But first we have to sort our numbers which takes $O(2n \log n)$ and create the pairs which takes $O(2n)$.
4-6. Given two sets $S_1$ and $S_2$ (each of size n), and a number x, describe an $O(n \log n)$ algorithm for finding whether there exists a pair of elements, one from $S_1$ and one from $S_2$, that add up to x. (For partial credit, give a $\Theta (n^2)$ algorithm for this problem.)
Solution We could sort $S_2$ and then iterate through $S_1$ and calculate the second number $s_1 + s_2 = x \Leftrightarrow s_2 = x - s_1.$ Now we just have to search for $s_2$ in $S_2$ which takes $O(\log n)$ time. Therefore we take $O(2n \log 2n) + O(n \log n) = O(n \log n)$ time.
4-7. Outline a reasonable method of solving each of the following problems. Give the order of the worst-case complexity of your methods.
(a) You are given a pile of thousands of telephone bills and thousands of checks sent in to pay the bills. Find out who did not pay.
(b) You are given a list containing the title, author, call number and publisher of all the books in a school library and another list of 30 publishers. Find out how many of the books in the library were published by each company.
(c) You are given all the book checkout cards used in the campus library during the past year, each of which contains the name of the person who took out the book. Determine how many distinct people checked out at least one book.
Solution (a) We can use quick sort in this scenario which works in expected $O(n \log n)$ time and match the ith bill with the ith check which takes $O(n)$ time.
(b) We can use a dictionary quite easily which counts the books by publisher. This takes $O(k ^cdot n)$ where $k=30$ time.
(c) We can sort the list by name and then iterate through it and count each different name. This takes about $O(n \log n) + O(n) = O(n \log n)$ time.
4-9. Give an efficient algorithm to compute the union of sets A and B, where $n = max(|A|,|B|).$ The output should be an array of distinct elements that form the union of the sets, such that they appear more than once in the union.
(a) Assume that A and B are unsorted. Give an $O(n \log n)$ algorithm for the problem.
(b) Assume that A and B are sorted. Give an O(n) algorithm for the problem.
Solution
(a) We can create one array with every element of A and B. Afterwards we sort this array and iterate through it. In this iteration we just have to check if ith item differs from the (i+1)th item which takes $O(n)$ therefore we need $O(n \log n) + O(n) = O(n \log n).$
(b)
4-12. Devise an algorithm for finding the k smallest elements of an unsorted set of n integers in $O(n + k \log n).$
Solution We can build an “unsorted” heap in $O(n)$, i.e. only bring the smallest item at the top. Now we can extract k times the smallest item which takes $O(k \log n)$ time.
4-13. You wish to store a set of n numbers in either a max-heap or a sorted array. For each application below, state which data structure is better, or if it does not matter. Explain your answers.
(a) Want to find the maximum element quickly.
(b) Want to be able to delete an element quickly.
(c) Want to be able to form the structure quickly.
(d) Want to find the minimum element quickly.
Solution (a) They are both equally fast. In the max-heap it’s the first element in the sorted array the last one.
(b) A max-heap is here better because it takes only $O(\log n)$ instead of $O(n)$ for a sorted array.
(c) Both the heap and the sorted array take $O(n \log n)$ time to be formed.
(d) The minimum element in a sorted array is the first. In a max-heap every leaf could be the minimum element, therefore it takes $O(n).$
4-14. Give an $O(n \log k)$-time algorithm that merges k sorted lists with a total of n elements into one sorted list. (Hint: use a heap to speed up the elementary $O(kn)$– time algorithm).
Solution We can basically do an heap sort on these lists. We iterate from i := 1 to n and select the smallest item from the head of each list which takes $O(1) + O(\log n)$ time. Therefore the algorithm takes $O(n \log n)$ time.
4-15. (a) Give an efficient algorithm to find the second-largest key among n keys. You can do better than $2n - 3$ comparisons.
(b) Then, give an efficient algorithm to find the third-largest key among n keys. How many key comparisons does your algorithm do in the worst case? Must your algorithm determine which key is largest and second-largest in the process?
Solution (a) There’s a faster method for construction heaps which runs in $O(n)$. Afterwards we just have to call find-min two times which takes $O(\log n).$
(b) We can use the same method here as well. The largest second-largest key is implicitly found by constructing the heap.
4-16. Use the partitioning idea of quicksort to give an algorithm that finds the median element of an array of n integers in expected $O(n)$ time.
Solution The partition part of quicksort basically can help us. It determines partitions which are bigger respectively smaller than the pivot element. We just have to find n/2 elements which are smaller or equal to our element m which is then the median. This takes expected $O(n)$ time.
4-19. An inversion of a permutation is a pair of elements that are out of order.
(a) Show that a permutation of n items has at most $n(n-1) / 2$ inversions. Which permutation(s) have exactly $n(n-1) / 2$ inversions?
(b) Let P be a permutation and Pr be the reversal of this permutation. Show that $P$ and $P^r$ have a total of exactly $n(n-1) / 2$ inversions.
(c) Use the previous result to argue that the expected number of inversions in a random permutation is $n(n-1) / 4.$
Solution (a) Let’s take for example the items 1..5. We can get the maximum inversions if we reverse this list, i.e. [5, 4, 3, 2, 1] or more general [n, n-1, …, 1]. Now we can start at the right and count the inversions.
Therefore we got $sum_{i=0}^{n} (i-1) = \frac{n(n-1)}{2}$ inversions at most.
(b) Let’s prove this relation. Assume we got a permutation with n items. Let’s add an other item. The reversal for the old permutation got $\frac{n(n-1)}{2}$ inversions. The (n+1)th item adds n inversions which are:
$(n+1, n), (n+1, n-1), .... (n+1, 1).$ Therefore we get $\frac{n(n-1)}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n(n+1)}{2}$ and we’re done.
(c) A rough approach is to take the smallest and the highest value and assume that inversions are uniformly distribute. Therefore we get $\frac{0 + \frac{n(n-1)}{2}}{2} = \frac{n(n+1)}{4}$
4-24. Let A[1..n] be an array such that the first $n - sqrt{n}$ elements are already sorted (though we know nothing about the remaining elements). Give an algorithm that sorts A in substantially better than $n \log n$ steps.
Solution Merge sort can used quite nicely. We need to sort the remaining $sqrt{n}$ which takes $sqrt{n} \log sqrt{n}$ time. Afterwards we have to use merge on our old and new sorted lists which takes $O(n)$ time.
4-29.
Mr. B. C. Dull claims to have developed a new data structure for priority queues that supports the operations Insert, Maximum, and Extract-Max—all in $O(1)$ worst-case time. Prove that he is mistaken. (Hint: the argument does not involve a lot of gory details—just think about what this would imply about the $Omega(n \log n)$ lower bound for sorting.)
Solution If insert, maximum and extract-max would be possible in $O(1)$ we could use the following algorithm to sort data.
This would sort data in $O(2n) = O(n)$ which is smaller than lower bound $Theta(n \log n).$
4-30. A company database consists of 10,000 sorted names, 40% of whom are known as good customers and who together account for 60% of the accesses to the database. There are two data structure options to consider for representing the database:
• Put all the names in a single array and use binary search.
• Put the good customers in one array and the rest of them in a second array. Only if we do not find the query name on a binary search of the first array do we do a binary search of the second array.
Demonstrate which option gives better expected performance. Does this change if linear search on an unsorted array is used instead of binary search for both options?
Solution Single array, binary search: $\log 10000 = 4$
Good and bad, binary search: $0.6 cdot \log 4000 + 0.4 cdot (\log 4000 + \log 6000) approx 5.11$
The first variant is a bit faster here.
Single array, unsorted: $10000$
Good and bad, unsorted: $0.6 cdot 4000 + 0.4 cdot (4000 + 6000) = 6400$
However in this case the second variant is far superior.
4-32. Consider the numerical 20 Questions game. In this game, Player 1 thinks of a number in the range 1 to n. Player 2 has to figure out this number by asking the fewest number of true/false questions. Assume that nobody cheats.
(a) What is an optimal strategy if n in known?
(b) What is a good strategy is n is not known?
Solution (a) Binary search.
(b) We can start asking if the number is between 1 and two. If not we can double our range from two to four, four and eight, etc.
4-35. Let M be an $n x m$ integer matrix in which the entries of each row are sorted in increasing order (from left to right) and the entries in each column are in increasing order (from top to bottom). Give an efficient algorithm to find the position of an integer x in M, or to determine that x is not there. How many comparisons of x with matrix entries does your algorithm use in worst case?
Solution
This algorithm runs in $O(m\log n)$ time.
4-39. Design and implement a parallel sorting algorithm that distributes data across several processors. An appropriate variation of mergesort is a likely candidate. Mea- sure the speedup of this algorithm as the number of processors increases. Later, compare the execution time to that of a purely sequential mergesort implementation. What are your experiences?
For small to medium sizes the threaded version is a bit slower than the non-threaded one. However, for very big sizes the threaded version works faster.
4-45. Given a search string of three words, find the smallest snippet of the document that contains all three of the search words—i.e. , the snippet with smallest number of words in it. You are given the index positions where these words in occur search strings, such as word1: (1, 4, 5), word2: (4, 9, 10), and word3: (5, 6, 15). Each of the lists are in sorted order, as above.
Solution We can sort these positions with its identifier. Afterwards we iterative through the list and put identifiers on a stack. If we got all of them, we found the snippet. We replace each identifier if we find a nearer one. At the end, we just have to search for the smallest snippet in all snippets.
# The Algorithm Design Manual: Chapter 1
Yeah, new book series! I had laying this book around for about two and an half years and only read about a quarter of it but never worked through it. So, I decided to work through it and post interesting problems and solutions online.
The book consists of two parts. The first part treats the algorithmic basics. The second part is just a reference of different algorithmic problems and ways to solve it. I won’t cover the second part, mainly because there are no exercises.
1-1. Show that a + b can be less than min(a, b).
Solution: 1-1. For any $a, b < 0: a + b < \text{min}(a,b)$. For example:
$a = -5, b = -3. a + b = -8 < \text{min}(-5, -3) = -5.$
1-2. Show that a × b can be less than min(a, b).
Solution For example for $a = -5$ and $b = 3$ the result is $-5 * 3 = -15 < \text{min}(-5, 3) = -5.$
1-5. The knapsack problem is as follows: given a set of integers $S = {s_1, s_2, . . . , s_n}$, and a target number T, find a subset of S which adds up exactly to T. For example, there exists a subset within $S = {1,2,5,9,10}$ that adds up to $T = 22$ but not $T = 23$.
Find counterexamples to each of the following algorithms for the knapsack problem. That is, giving an S and T such that the subset is selected using the algorithm does not leave the knapsack completely full, even though such a solution exists.
(a) Put the elements of S in the knapsack in left to right order if they fit, i.e. the first-fit algorithm.
(b) Put the elements of S in the knapsack from smallest to largest, i.e. the best-fit algorithm.
(c) Put the elements of S in the knapsack from largest to smallest.
Solution:
(a) $S = {1, 2}, T = 2$
(b) $S = {1, 2}, T = 2$
(c) $S = {2, 3, 4}, T = 5$
1-16. Prove by induction that n3 + 2n is divisible by 3 for all n ≥ 0.
Solution: The base case is $n = 0, 0^3 + 2*0 = 0 \text{ mod } 3 = 0$ which is true.
We can assume that this holds up to n. For n + 1 we get:
$(n+1)^3 + 2(n+1) = n^3 + 3n^2 + 3n + 1 + 2n + 1$
$= n^3 + 3n^2 + 5n + 2 = (n^3 + 2n) + (3 (n^2 + n)).$
The first term is parenthesis is our assumption and the second term is obviously divisible by 3, therefore we showed that our assumption is true.
1-26. Implement the two TSP heuristics of Section 1.1 (page 5). Which of them gives better-quality solutions in practice? Can you devise a heuristic that works better than both of them?
Solution:
NearestNeighbour createst shorter paths for small input graphs. For midsize input paths and big input graphs closestPair creates smaller paths. However nearestNeighbour is much faster. In conclusion, it depends on the application which heuristic is more suitable.
1-28. Write a function to perform integer division without using either the / or * operators. Find a fast way to do it.
Solution: A simple way to perform integer division is substracting the divisor and counting each substraction. This only works for positive numbers. (more in the comments)
If we can find a bigger divisor we could speed it up. An easy way is to multipy the divisor by itself as long as it divides the denominator with a rest.
This works quite nice and can speed up the process substantially.
# SPOJ: 1728. Common Permutation
Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation of x that is a subsequence of b.
Solution: That’s short and nice problem. You have to find all letters which are in both strings. I’d actually like to see some other implementations of this. I bet there are some languages which handle this in a smart way.
# SPOJ: 3374. Scavenger Hunt
He does a poor job, though, and wants to learn from Bill’s routes. Unfortunately Bill has left only a few notes for his successor. Bill never wrote his routes completely, he only left lots of little sheets on which he had written two consecutive steps of the routes. […]
This made much sense, since one step always required something from the previous step. George however would like to have a route written down as one long sequence of all the steps in the correct order.
Solution: | 4,216 | 16,010 | {"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": 84, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | longest | en | 0.91017 |
http://jwebsterdesigns.com/docs/archive.php?c81f3a=integrated-brownian-motion | 1,653,608,362,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00663.warc.gz | 32,232,895 | 6,133 | In consequence, only probabilistic models applied to molecular populations can be employed to describe it. is the Dirac delta function. On small timescales, inertial effects are prevalent in the Langevin equation. Δ μ Need help? ω u Equating these two expressions yields a formula for the diffusivity, independent of mg or qE or other such forces: Here the first equality follows from the first part of Einstein's theory, the third equality follows from the definition of Boltzmann's constant as kB = R / NA, and the fourth equality follows from Stokes's formula for the mobility. p You should expect from this that any formula will have an ugly combinatorial factor. Buy direct from Motion Industries! ( {\displaystyle dt} However, when he relates it to a particle of mass m moving at a velocity The multiplicity is then simply given by: and the total number of possible states is given by 2N. = This ratio is of the order of 10−7 cm/s. It follows that can experience Brownian motion as it responds to gravitational forces from surrounding stars. and variance σ Is the space in which we live fundamentally 3D or is this just how we perceive it? f 2 For an arbitrary initial value S0 the above SDE has the analytic solution (under Itô's interpretation): The derivation requires the use of Itô calculus. S 1 log ω T N The first person to describe the mathematics behind Brownian motion was Thorvald N. Thiele in a paper on the method of least squares published in 1880. ) This is an interesting process, because in the Black–Scholes model it is related to the log return of the stock price. Like the random walk, the Wiener process is recurrent in one or two dimensions (meaning that it returns almost surely to any fixed neighborhood of the origin infinitely often) whereas it is not recurrent in dimensions three and higher. More specifically, the fluid's overall linear and angular momenta remain null over time. For example, the assumption that on average occurs an equal number of collisions from the right as from the left falls apart once the particle is in motion. d t u ( expectation of integral of power of Brownian motion, “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, Expectation of an integral w.r.t. = v {\displaystyle v_{\star }} ( The Wiener process Wt is characterized by four facts:[citation needed]. What about if $n\in \mathbb{R}^+$? 5 10 ) is constant. and σ {\displaystyle \tau =Dt} $$\int_0^t \int_0^t s^a u^b (s \wedge u)^c du ds$$ 0 E with $n\in \mathbb{N}$. A geometric Brownian motion (GBM) (also known as exponential Brownian motion) is a continuous-time stochastic process in which the logarithm of the randomly varying quantity follows a Brownian motion (also called a Wiener process) with drift. s \wedge u \qquad& \text{otherwise} \end{cases} 5 , De brownse of browniaanse beweging is een natuurkundig verschijnsel, in 1827 beschreven door de Schotse botanicus Robert Brown bij onderzoek van stuifmeelkorrels in een vloeistof onder de microscoop.Hij merkte op dat de deeltjes, hoewel bestaande uit dode materie, een onregelmatige eigen beweging vertoonden en volgens een toevallig aandoend patroon in alle richtingen weg konden schieten. Birmingham, AL 35210, USA. O In the general case, Brownian motion is a non-Markov random process and described by stochastic integral equations. , denotes the normal distribution with expected value μ and variance σ2. What is obvious though is that $\mathbb{E}[Z_t^2] = ct^{n+2}$ for some constant $c$ depending only on $n$. \end{align} . < {\displaystyle a} Further, assuming conservation of particle number, he expanded the density (number of particles per unit volume) at time This pattern of motion typically consists of random fluctuations in a particle's position inside a fluid sub-domain, followed by a relocation to another sub-domain. j t d B ] 2 so we apply Wick's theorem with $X_i = W_s$ if $i \leq n$ and $X_i = W_u$ otherwise. log / cos ( T t t By introducing the new variables → t Geometric Brownian motion models for stock movement except in rare events. f 2 [citation needed]. d What is $\mathbb{E}[Z_t]$? Two such models of the statistical mechanics, due to Einstein and Smoluchowski are presented below. The expected returns of GBM are independent of the value of the process (stock price), which agrees with what we would expect in reality. [17] But Einstein's predictions were finally confirmed in a series of experiments carried out by Chaudesaigues in 1908 and Perrin in 1909. This implies the distribution of = Description. , will be equal, on the average, to the kinetic energy of the surrounding fluid particle, d lim , V In his original treatment, Einstein considered an osmotic pressure experiment, but the same conclusion can be reached in other ways. μ ∫ f W ρ \\=& \tilde{c}t^{n+2} / ) D [26] Random walks in porous media or fractals are anomalous. 1 2 = {\displaystyle dW_{t}^{2}=O(dt)} Is it too late for me to get into competitive chess? → t \sigma^n (n-1)!! What LEGO piece is this arc with ball joint? T { = can be found from the power spectral density, formally defined as. The Brownian motion model of the stock market is often cited, but Benoit Mandelbrot rejected its applicability to stock price movements in part because these are discontinuous.[9]. 1 Variance of Brownian Motion… By measuring the mean squared displacement over a time interval along with the universal gas constant R, the temperature T, the viscosity η, and the particle radius r, the Avogadro constant NA can be determined. , but its coefficient of variation Brown was studying pollen grains of the plant Clarkia pulchella suspended in water under a microscope when he observed minute particles, ejected by the pollen grains, executing a jittery motion. − S When where E t For a student studying Chinese as a second language, is there any practical difference between the radicals 匚 and 匸? Solutions for you include informative articles, videos, case studies, and more — all under one e-roof! in local coordinates xi, 1 ≤ i ≤ m, is given by ½ΔLB, where ΔLB is the Laplace–Beltrami operator given in local coordinates by. | 1,501 | 6,201 | {"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} | 2.828125 | 3 | CC-MAIN-2022-21 | latest | en | 0.926215 |
https://forum.arduino.cc/t/serial-monitor-gives-data-but-adds-random-letters-to-the-mix/584060 | 1,623,759,981,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00588.warc.gz | 260,853,335 | 5,460 | # Serial Monitor Gives data but adds random letters to the mix
``````#include <math.h>
#define led 3 // the PWM pin the LED is attached to
int full = 255;
double T1 = 293.00; //Temperature Recorded in Kelvin 20C
double T2 = 303.00; //30 C
double T3 = 313.00; //40 C
double R1 = 12470.00;
double R2 = 8416.00;
double R3 = 2972.00;
double L1 = log (R1); //Is actually a ln but this give right values
double L2 = log (R2);
double L3 = log (R3);
double Y1 = 1 / T1;
double Y2 = 1 / T2;
double Y3 = 1 / T3;
double X2 = (Y2 - Y1) / (L2 - L1);
double X3 = (Y3 - Y1) / (L3 - L1);
double C = ((X3 - X2) / (L3 - L2)) * pow ((L1 + L2 + L3), -1);
double B = X2 - C * (L1 * L1 + L1 * L2 + L2*L2);
double A = Y1 - L1 * (B + C*L1*L1);
double temperature43(double Ana) { // Function to convert analog from arduino into temperature values for plot
double temp, VI, Vo, V, r1, r2, b, c, lr2, f;
r1 = 43000; // Resistor of 43K ohms
VI = 5;
Vo = Ana * 5 / 1023;
V = VI / Vo ;
r2 = r1 * (V - 1);
b = B * log(r2);
lr2 = log(r2);
c = C * pow (lr2, 3);
f = A + b + c;
temp = 1 / f;
temp = temp - 273; //into celcius
temp = temp * 9 / 5 + 32; // into farenheight
return temp;
}
double temperature75(double Ana) { // Function to convert analog from arduino into temperature values for plot
double temp, VI, Vo, V, r1, r2, b, c, lr2, f;
r1 = 75000; // Resistor of 43K ohms
VI = 5;
Vo = Ana * 5 / 1023;
V = VI / Vo ;
r2 = r1 * (V - 1);
b = B * log(r2);
lr2 = log(r2);
c = C * pow (lr2, 3);
f = A + b + c;
temp = 1 / f;
temp = temp - 273; //into celcius
temp = temp * 9 / 5 + 32; // into farenheight
return temp;
}
void setup () {
Serial.begin(115200);
pinMode(led, OUTPUT);
}
void loop() {
analogWrite(led, full);
double val1, val2, val3, val4, val5, val6, temp1, temp2, temp3, temp4, temp5, temp6;
temp1 = temperature43(val1); //for board with 43K resitor
temp2 = temperature43(val2); //for board with 43K resitor
temp3 = temperature75(val3); //for board with 75K resitor
temp4 = temperature75(val4); //for board with 75K resitor
temp5 = temperature75(val5); //for board with 75K resitor
temp6 = temperature75(val6); //for board with 75K resitor
Serial.println(temp1);
Serial.println(temp2);
Serial.println(temp3);
Serial.println(temp4);
Serial.println(temp5);
Serial.println(temp6);
Serial.println();
delay(1000);
}
``````
When I run my code it prints the correct values for the temperature most of the time but every once in a while it will print 77.12 as 7rj or something like that. I will attach a screenshot of the serial monitor. Both my Arduino Uno and the Online Monitor are at 115200 baud. I am running Windows 7.
Here is the screen shot of my monitor
The serial Baud rate is probably off slightly. On the standard Arduino, the clock (resonator) is not very accurate.
The problem may go away if you choose a lower Baud rate, so try dropping it in steps, choosing among the allowed possibilities.
If that doesn't fix it, use an Arduino with a genuine crystal as the CPU clock.
When I dropped the Baud rate it worked. Thanks for the help! | 1,037 | 3,038 | {"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} | 2.765625 | 3 | CC-MAIN-2021-25 | latest | en | 0.446848 |
http://tjsct.wikidot.com/usaco-open09-bronze | 1,545,027,596,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00386.warc.gz | 290,290,342 | 10,757 | Usaco Open09 Bronze
Part of USACO Open09
``````**********************************************************************
BRONZE PROBLEMS
**********************************************************************
Four problems numbered 11 through 14
**********************************************************************
Problem 11: Time Cards [Rob Kolstad, 2009]
Farmer John wanted to improve farm productivity, so his cows now
get extra hay if they spend more time at a milking machine. To
implement this plan, he has instituted the use of time cards for
each of the N (1 <= N <= 145) cows conveniently numbered 1..N. When
a cow starts at a milking machine, she enters the start time on the
master time card. Likewise, when she leaves, she notes that on the
master time card, as well. FJ is fortunate to have enough milking
machines that he can milk every cow at the same time.
The time entries are typed into a computer file where each line
includes a cow number C (1 <= C <= N), a keyword ('START' or 'STOP'),
and the time expressed as two space-separated integers HH and MM
(0 <= HH <= 23; 0 <= MM <= 59). Cows never stay at the machine past
midnight. The timecard file is complete in the sense that every
cow's START entry contains a corresponding STOP entry later in the
input file.
Calculate the total time each cow spends at the milking machine.
By way of example, consider a time card file for just two cows. The
file includes not only the number of cows but also the total number
of time card entries, Nlines (1 <= Nlines <= 1,458).
2 6
1 START 9 0
2 START 9 30
1 STOP 10 0
2 STOP 10 15
1 START 17 0
1 STOP 17 42
Cow 1 spent times 9:00-10:00 and 17:00-17:42 at the machine for a
total of one hour and 42 minutes (1:42). Cow 2 spent time 9:30-10:15
at the machine, for a total of 45 minutes.
PROBLEM NAME: timecards
INPUT FORMAT:
* Line 1: Two space-separated integers: N and Nlines
* Lines 2..Nlines+1: Each line contains four space-separated entities:
C, keyword, HH, and MM
SAMPLE INPUT (file timecards.in):
2 6
1 START 9 0
2 START 9 30
1 STOP 10 0
2 STOP 10 15
1 START 17 0
1 STOP 17 42
OUTPUT FORMAT:
* Lines 1..N: Line i contains two space-separated integers that are
respectively the number of hours and minutes that cow i spends
at the milking machine. Of course, the minutes value never
exceeds 59.
SAMPLE OUTPUT (file timecards.out):
1 42
0 45
**********************************************************************
Problem 12: O Those Fads [Jarrah Lacko, 2009]
Like any other teenager, teen cows are occasionally overtaken by
fads. Sometimes it's a hula hoop or a pet rock, other times it's
Counterstrike, Pokemon, Rick Astley, or tribal tattoos on their
udders.
Mathematically, we know that a fad has an initial attractiveness
level L (1 <= L <= 50,000). Cow i has a resistance (0 <= R_i <=
1,000,000) that tells how long she can avoid a fad before having
no alternative but to participate. When a fad's attractiveness level
meets or exceeds a cow's fad resistance, then the cow will want to
Each cow who participates in a fad increases (through peer pressure)
that fad's attractiveness by some value K (1 <= K <= 2,500).
Given a population of N (1 <= N <= 100,000) cows, determine how
many will participate in a fad.
INPUT FORMAT:
* Line 1: Three space-separated integers: N, L, and K
* Lines 2..N+1: Line i+1 contains cow i's a single integer that is fad
resistance: R_i
5 2 3
2
6
12
5
14
INPUT DETAILS:
Five cows with fad resistances 2, 6, 12, 5, and 14. Initial fad
attractiveness is 2; peer pressure adds 3 for each attractiveness
index for each cow that participates.
OUTPUT FORMAT:
* Line 1: A single integer that is the number of cows how ultimately
3
OUTPUT DETAILS:
The initial attraction level of 2 brings in cow #1 (whose resistance
is 2) and raises the attractiveness level to 5, thus attracting cow
#4 (whose resistance is 5) and raising the attractiveness level to
8. This attracts cow #2 (resistance: 6) and raises the attractiveness
level to 11. Neither cow #3 (resistance: 12) or cow #5 (resistance:
14) is sucked into the fad; a total of 3 cows particpate.
**********************************************************************
Problem 13: Good Grass [Rob Kolstad, 2009]
Bessie believes that her favorite pasture has a patch of very special
grass that is the best grass on earth. She thinks it helps cows
produce more milk.
Each of the cows has returned to her assigned grazing spot, which
is at some point in a fully populated rectilinear grid. Consider
the example on the left below where the number of rows, NR (3 <=
NR <= 100) is 6 and the number of columns, NC (3 <= NC <= 100) is
5. The spots for cows are marked with a 'C'.
Bessie actually knows the milk production P_rc (1 <= P_rc <= 100)
of each of the cows in the grid; the cows are tagged with the
production number in the grid on the right, below.
COLUMN COLUMN
1 2 3 4 5 1 2 3 4 5
+-------------- +--------------
1| C C C C C 1| 5 6 7 4 6
R 2| C C C C C R 2| 7 7 8 6 5
O 3| C C C C C O 3| 9 9 8 3 5
W 4| C C C C C W 4| 8 8 7 6 4
5| C C C C C 5| 4 5 2 4 5
6| C C C C C 6| 3 4 2 3 4
Bessie wants to find the location of the special grass. She intends
to do this by finding the 3 x 3 grid of cows whose total milk
production is the largest.
Find the 3 x 3 grid whose nine components sum to the greatest number
and report the value of its upper left corner (first the row, then
the column). In the grid above on the right, the largest sum is 71
found in the grid whose upper left corner (as depicted) is row 2,
column 1.
If two 3 x 3 grids have the same sum, output the one whose row
number is smallest. If more than one grid on that row has the same
sum, output the one with the lowest column number.
PROBLEM NAME: goodgrs
INPUT FORMAT:
* Line 1: Two space-separated integers: NR and NC
* Lines 2..NR+1: Line r+1 contains NC space-separated integers that
represent row r of the pasture's grid.
SAMPLE INPUT (file goodgrs.in):
6 5
5 6 7 4 6
7 7 8 6 5
9 9 8 3 5
8 8 7 6 4
4 5 2 4 5
3 4 2 3 4
OUTPUT FORMAT:
* Line 1: A single integer that is the greatest possible sum in a 3 x
3 square.
* Line 2: Two space-separated integers that are respectively the row
and column of the upper left corner of the 'best' 3 x 3 square
grid.
SAMPLE OUTPUT (file goodgrs.out):
71
2 1
**********************************************************************
Problem 14: Treasure Cave [Rob Kolstad, 2009]
Bessie's grandfather was a pirate who accumulated a great treasure
chest of golden plunder. He hid the treasure chest in a cave that
Bessie has recently discovered right on Farmer John's land! Just
inside the cave's entrance she found a map that told her how to get
the treasure.
The cave has P passages (3 <= P <= 5,000) conveniently numbered
1..P. The entrance is passage 1; the treasure is located in some
reachable passage T (2 <= T <= P), whose value is supplied. Passages
are all approximately the same length; each one leads to a split
where hitherto unexplored numbered passages take the inquisitive
cow deeper underground. No passage appears as the split from more
than one passage, and the map contains a total of NS splits (1 <=
NS <= 5,000).
Bessie wants to know both how far away from the entrance the treasure
lies and also which passage numbers to take to get to the treasure.
Consider the schematic representation of a cave shown below. Passage
numbers are shown close to the passage they name. For this example,
the treasure is at the end of passage number 7:
3/
/
+
/ \ /5
2/ 4\ /
1 / +
----+ 6\ #7 /11
\ \ / /
13\ + +
8\ 10/ \
\ / \12
+
9\
\
Bessie would have to traverse passages 1, 2, 4, 6, and 7 to get to
the treasure, a total distance of 5 (which is simply the passage
count).
The input file includes a set of lines, each with a passage number
N (1 <= N <= P) and the two passages (B1 and B2; 1 <= B1 <= P; 1
<= B2 <= P) that branch off from it. Some line in the input file
will include passage number 1 and its two branches (for our example,
passages 2 and 13; likewise, passage number 8 has two branches: 9
and 10).
Tell Bessie how to get to the treasure.
PROBLEM NAME: tcave
INPUT FORMAT:
* Line 1: Line 1 contains three space-separated integers: P, NS, and T
* Lines 2..NS+1: Each line contains three space-separated integers: N,
B1, and B2
SAMPLE INPUT (file tcave.in):
13 6 7
6 7 8
2 3 4
10 11 12
8 9 10
1 2 13
4 5 6
INPUT DETAILS:
This input describes the sample cave in text.
OUTPUT FORMAT:
* Line 1: The distance D from the entrance to the treasure
* Lines 2..D+1: Line i+1 contains a single integer that is next
passage Bessie takes to get to the treasure.
SAMPLE OUTPUT (file tcave.out):
5
1
2
4
6
7
OUTPUT DETAILS:
As in the text.
**********************************************************************```
```
page revision: 0, last edited: 06 Jun 2009 21:45 | 2,604 | 9,055 | {"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} | 3.609375 | 4 | CC-MAIN-2018-51 | longest | en | 0.857292 |
https://www.cs.purdue.edu/homes/spa/courses/msri04/ait.html | 1,709,156,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00301.warc.gz | 712,654,595 | 2,057 | ## Analysis of Exact Pattern Matching
Math techniques:
1. Combinatorial Calculus (languages)
2. Generating functions
3. Simple complex asymptotics -- poles)
4. Moment analysis
Homework - Problems
1. Details of derivations
2. Limiting distribution
3. Extension to Markov
4. Another example of combinatorial calculus (joint distribution)
### References
1. M. Regnier and W. Szpankowski. On Pattern Frequency Occurrences in a Markovian Sequence, Algorithmica, 22, 631-649, 1998.
2. W. Szpankowski. Average Case Analysis of Algorithms on Sequences . John Wiley & Sons, New York, 2001. (Chap. 7.1 and 8.3).
3. P. Jacquet and W. Szpankowski. Analytic Approach to Pattern Matching. In Applied Combinatorics on Words (eds. Lothaire), Cambridge University Press (Encycl. of Mathematics and Its Applications), 2004.
## Worst Case Minimax Redundancy for memoryless Sources
Math techniques :
1. Shtarkov bound
2. Tree-like generating functions
3. Singularity analysis
Homework - Problems
1. Singularity expansion for T(z) and B(z)
2. Extraction of coefficients for B m (z) -- details
### References:
1. W. Szpankowski. On Asymptotics of Certain Recurrences Arising in Universal Coding. Problems of Information Transmission , 34, No.2, 142-146, 1998.
2. W. Szpankowski. Average Case Analysis of Algorithms on Sequences . John Wiley & Sons, New York, 2001. (Chap. 8.7).
## Error Resilient LZ'77 and Tries/Suffix Trees Analysis
Math techniques
1. Recurrences
2. Poissonization
3. Mellin transform
4. Depoissonization
Homework - Problems :
1. Solve some other trie like recurrences (Chap. 7.6.1)
2. Mellin transform exercises (Chap 9)
3. Depoissonization examples (Chap 10)
4. Limiting distribution of M n .
### References:
1. M. Ward and W. Szpankowski. Analysis of a Randomized Selection Algorithm Motivated by the LZ'77 Scheme. The First Workshop on Analytic Algorithmics and Combinatorics , (ANALCO04), New Orleans, January 10, 2004.
2. W. Szpankowski. Average Case Analysis of Algorithms on Sequences . John Wiley & Sons, New York, 2001. (Chap. 9 and 10). | 573 | 2,067 | {"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-2024-10 | latest | en | 0.573189 |
https://encyclopedia2.thefreedictionary.com/Jacobi+theta+functions | 1,620,945,526,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992514.37/warc/CC-MAIN-20210513204127-20210513234127-00450.warc.gz | 250,434,784 | 10,481 | # Theta Function
(redirected from Jacobi theta functions)
Also found in: Dictionary.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Theta Function
Theta functions are entire functions whose quotients are elliptic functions.
The four principal theta functions are defined by the following rapidly converging series:
θ1(z) = 2q1/4 sin z – 2q9/4 sin 3z + 2q25/4 sin 5z – . . .
θ2(z) = 2q1/4 cos z – 2q9/4 cos 3z + 2q25/4 cos 5z + . . .
θ3(z) = 1 + 2q cos 2z + 2q4 cos 4z + 2q9 cos 6z + . . .
θ4(z) = 1 – 2q cos 2z + 2q4 cos 4z – 2q9 cos 6z + . . .
where ǀqǀ < 1.
When π is added to the argument z, the functions are multiplied by –1, –1, 1, and 1, respectively. When πτ is added to z, the relation between τ and q being given by the equation q = eπiτ the functions are multiplied by the factors – N, N, N, and –N, respectively, where N = q–1e–2ik. It follows that, for example, the quotient θ1(Z)/θ4(Z) is a meromorphic function that does not change when 2π or irr is added to the argument—that is, this quotient is an elliptic function with periods 2π and πτ.
For the representation of automorphic functions, H. Poincaré constructed theta functions that are a generalization of the theta functions described above, which were introduced by K. Jacobi (Jacobi’s notation was somewhat different).
### REFERENCE
Whittaker, E. T., and G. N. Watson. Kurs sovremennogo analiza, 2nd edition, part 2. Moscow, 1963. (Translated from English.) | 473 | 1,513 | {"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} | 3.375 | 3 | CC-MAIN-2021-21 | latest | en | 0.88406 |
https://fr.maplesoft.com/support/help/maplesim/view.aspx?path=Groebner%2Fterminology | 1,726,129,582,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00226.warc.gz | 229,478,559 | 22,999 | Terminology - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
Definitions for leading monomials, coefficients, and terms
Description
• The current terminology used in Groebner is that of Ideals, Varieties and Algorithms by David Cox, John Little, and Donal O'Shea, Springer-Verlag, (1992). In releases of Maple prior to Maple 10, Groebner used a different, conflicting convention. In particular, what is now called a leading monomial used to be called a leading term, and vice versa.
• Note: the old commands Groebner[leadmon] and Groebner[leadterm] still respect the old convention, however they are now deprecated. You should replace them with Groebner[LeadingTerm] or Groebner[LeadingMonomial] respectively.
• The current convention is as follows:
– A monomial is a product of indeterminates from a fixed set X, possibly with repetitions. The coefficient of a monomial is always one.
– A term of a polynomial (with respect to X) is the product of a monomial in X and a coefficient whose degree in X is zero. This coefficient may include other indeterminates not in X. For example, the coefficients may be rational functions in other variables.
– The "leading term" of a polynomial with respect to a monomial order is the term whose monomial is greatest with respect to the order and whose coefficient is non-zero. The coefficient and monomial of this term are called the "leading coefficient" and "leading monomial" of the polynomial, respectively.
• Note that the LeadingTerm command does not actually output terms, but rather the sequence (leading coefficient, leading monomial). This may be changed in a future release of Maple.
Examples
> $\mathrm{with}\left(\mathrm{Groebner}\right):$
> $f≔5{x}^{3}y+{x}^{2}{w}^{2}t+5{x}^{3}yzt-2xz{w}^{3}t+3{y}^{2}{w}^{3}t$
${f}{≔}{-}{2}{}{t}{}{{w}}^{{3}}{}{x}{}{z}{+}{3}{}{t}{}{{w}}^{{3}}{}{{y}}^{{2}}{+}{5}{}{t}{}{{x}}^{{3}}{}{y}{}{z}{+}{t}{}{{w}}^{{2}}{}{{x}}^{{2}}{+}{5}{}{{x}}^{{3}}{}{y}$ (1)
With respect to the definitions above, we will compute the leading coefficient, leading monomial, and leading term of the polynomial f with respect to lexicographic order with x > y > z > w > t.
> $\mathrm{LeadingCoefficient}\left(f,\mathrm{plex}\left(x,y,z,w,t\right)\right)$
${5}$ (2)
> $\mathrm{LeadingMonomial}\left(f,\mathrm{plex}\left(x,y,z,w,t\right)\right)$
${t}{}{{x}}^{{3}}{}{y}{}{z}$ (3)
> $\mathrm{LeadingTerm}\left(f,\mathrm{plex}\left(x,y,z,w,t\right)\right)$
${5}{}{t}{}{{x}}^{{3}}{}{y}{}{z}$ (4)
In releases of Maple prior to Maple 10, Groebner[leadmon] computed what is now returned by LeadingTerm and Groebner[leadterm] computed what is now returned by LeadingMonomial.
> $\mathrm{leadcoeff}\left(f,\mathrm{plex}\left(x,y,z,w,t\right)\right)$
${5}$ (5)
> $\mathrm{leadterm}\left(f,\mathrm{plex}\left(x,y,z,w,t\right)\right)$
${t}{}{{x}}^{{3}}{}{y}{}{z}$ (6)
> $\mathrm{leadmon}\left(f,\mathrm{plex}\left(x,y,z,w,t\right)\right)$
${5}{}{t}{}{{x}}^{{3}}{}{y}{}{z}$ (7) | 940 | 3,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "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} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.850447 |
http://www.solve-variable.com/solve-variable/exponent-rules/8-as-decimal.html | 1,519,359,720,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814393.4/warc/CC-MAIN-20180223035527-20180223055527-00595.warc.gz | 546,236,860 | 11,697 | Free Algebra Tutorials!
Try the Free Math Solver or Scroll down to Tutorials!
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:
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8% as decimal
Related topics:
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Author Message
jrienjeeh
Registered: 05.09.2003
From:
Posted: Saturday 30th of Dec 07:00 Hi math experts . This is my first post in this forum. I struggle a lot with 8% as decimal questions . No matter how hard I try, I just am not able to crack any problem in less than an hour. If things go this way, I reckon I will not be able to pass my math exam.
kfir
Registered: 07.05.2006
From: egypt
Posted: Monday 01st of Jan 10:48 How about some more details about your problem with 8% as decimal? I might be able to give some leads. If you are not able to get a good assistance or some one to sit and sort out your difficulty or if it is too costly , then there might be another way out . There are some good math software that you can explore. I tried them out myself. It came across to me as superb as any tutor can be. I would select Algebrator for the kind of elucidation that you are looking out for . What is attractive about it is that it assists you step by step to the solutions rather than plainly providing the answer. Why not try it out?
Dxi_Sysdech
Registered: 05.07.2001
From: Right here, can't you see me?
Posted: Tuesday 02nd of Jan 18:43 Algebrator is really a good software program that helps to deal with math problems. I remember facing difficulties with logarithms, inverse matrices and factoring. Algebrator gave step by step solution to my algebra homework problem on typing it and simply clicking on Solve. It has helped me through several math classes. I greatly recommend the program.
musicolmidget
Registered: 11.01.2003
From:
Posted: Wednesday 03rd of Jan 07:31 Great! Guys . Only one more problem. Where can I purchase the great software?
Paubaume
Registered: 18.04.2004
From: In the stars... where you left me, and where I will wait for you... always...
Posted: Friday 05th of Jan 07:52 It is available at http://www.solve-variable.com/math-problem-solvinglong-division-face.html and really is the easiest program to get up and running. You can start learning algebra within minutes of downloading the software.
MichMoxon
Registered: 21.08.2001
From:
Posted: Friday 05th of Jan 10:47 I remember having difficulties with subtracting fractions, exponent rules and quadratic inequalities. Algebrator is a really great piece of math software. I have used it through several math classes - Algebra 1, Algebra 2 and Remedial Algebra. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly recommended. | 893 | 3,497 | {"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} | 2.84375 | 3 | CC-MAIN-2018-09 | latest | en | 0.926611 |
http://www.numbersaplenty.com/1380625 | 1,590,821,046,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00332.warc.gz | 191,949,687 | 3,779 | Search a number
1380625 = 54472
BaseRepresentation
bin101010001000100010001
32121010212021
411101010101
5323140000
645331441
714510101
oct5210421
92533767
101380625
11863314
12566b81
1339454c
1427d201
151c411a
hex151111
1380625 has 15 divisors (see below), whose sum is σ = 1762717. Its totient is φ = 1081000.
The previous prime is 1380623. The next prime is 1380629. The reversal of 1380625 is 5260831.
Multipling 1380625 by its sum of digits (25), we get a square (34515625 = 58752).
1380625 divided by its sum of digits (25) gives a square (55225 = 2352).
It can be divided in two parts, 13806 and 25, that added together give a palindrome (13831).
It is a happy number.
The square root of 1380625 is 1175.
It is a perfect power (a square), and thus also a powerful number.
It can be written as a sum of positive squares in 2 ways, for example, as 108241 + 1272384 = 329^2 + 1128^2 .
It is not a de Polignac number, because 1380625 - 21 = 1380623 is a prime.
It is a super-2 number, since 2×13806252 = 3812250781250, which contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (25).
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 1380625.
It is not an unprimeable number, because it can be changed into a prime (1380623) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 14 ways as a sum of consecutive naturals, for example, 29352 + ... + 29398.
21380625 is an apocalyptic number.
1380625 is the 1175-th square number.
1380625 is the 588-th centered octagonal number.
It is an amenable number.
1380625 is a deficient number, since it is larger than the sum of its proper divisors (382092).
1380625 is an frugal number, since it uses more digits than its factorization.
1380625 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 114 (or 52 counting only the distinct ones).
The product of its (nonzero) digits is 1440, while the sum is 25.
The cubic root of 1380625 is about 111.3504332283.
The spelling of 1380625 in words is "one million, three hundred eighty thousand, six hundred twenty-five". | 685 | 2,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-24 | latest | en | 0.90205 |
http://math.stackexchange.com/questions/120992/graph-theory-how-can-i-calculate-the-number-of-vertices-and-edges-if-given-th/120994 | 1,469,457,056,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00050-ip-10-185-27-174.ec2.internal.warc.gz | 148,209,158 | 21,406 | Graph Theory - How can I calculate the number of vertices and edges, if given this example
An algorithm book Algorithm Design Manual has given an description:
Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges, since each vertex neighbors four other vertices and each edge is shared between two vertices.
If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.
Can anyone explain the calculation more easily?
Thanks
-
Each street crossing is a vertex of the graph. An avenue crosses about $200$ streets, and each of these crossings is a vertex, so each avenue contains about $200$ vertices. There are $15$ avenues, each of which contains about $200$ vertices, for a total of $15\cdot 200=3000$ vertices.
To make the description easier, imagine that the avenues all run north-south and the other streets east-west. Then each intersection of an avenue and a street has another intersection due north along the avenue. It also has one due south along the avenue, one due east along the cross street, and one due west along the cross street. That’s a total of four neighboring vertices. There must be an edge from the given vertex to each of those four, so we count $4\cdot 3000=12,000$ edges. But that’s counting each edge twice, once at each end, so there are really only half that many edges, or $6000$.
Now you might object that the vertex (i.e., intersection) in the northwest corner, say, has only two neighboring vertices, one to the east and one to the south, and similarly for the other three corners. You might also worry about the non-corner vertices along the edges, since they seem to have only three neighboring vertices each. But remember, the original figure of $200$ cross streets was only an approximation in the first place, so we might as well ignore these relatively minor edge effects: they probably don’t affect the result much more than the approximation in the $200$ figure already does.
-
Ha, you made the same avenue = north-south and street = east-west connection I made. I wonder if most people would... – El'endia Starman Mar 16 '12 at 16:48
@Brian M. Scott, what means "an avenue crosses streets"? How can an avenue(a large area) cross a number of streets? I'd say "cover" at most. – Jackson Tale Mar 17 '12 at 8:15
@Jackson: In U.S. English avenue is simply one of many terms used for streets. A street whose name is Marietta Avenue, say, might well be a rather short residential street with a curve in it. Or it might be a long, straight street that carries a lot of traffic. When it’s used as a general term, not part of a street name, I expect something called an avenue to be a fairly important street, as is the case in this problem, but it’s still just a street. – Brian M. Scott Mar 17 '12 at 15:47
@BrianM.Scott, ah, ok. Avenue in US means that. I thought avenue means a kind of area. lol. Forgive me for my poor english. – Jackson Tale Mar 19 '12 at 11:33
This is so simple, I mean 15 avenues crossing 200 streets, which means there are 15 * 200 = 3000 crossings, i.e. 3000 nodes.
each nodes have upper, lower, left and right neighbors, so for each node, there are 4 edges connecting to the neighbors. However, each edge has been counted twice since node 1 has an edge connected with node 2, and the node 2 has the same edge connected to node 1
So that is totally 3000*4/2 = 6000 edges
-
If you have 15 vertical and 200 horizontal lines, parallel for each direction, they'll have 15*200 crossings. An edge is a line segment between each crossing. Each crossing connects 4 line segments (ignoring the outer bounds) and each line segment connects two crossings, so it's double the number of crossings.
-
Every junction between an avenue and a street is a vertex. As there are $15$ avenues and (about) $200$ streets, there are (about) $15*200=3000$ vertices. Furthermore, every vertex has an edge along an avenue and an edge along a street that connect it to two other vertices. Hence, there are (about) $2*3000 = 6000$ edges1. Does that answer your question?
1 With regards to edges, a visual way to imagine it would be to imagine that the avenues are going north-south and the streets are going east-west. Start with the junction/vertex in the northwesternmost corner. It is adjacent to two other vertices: one south along the avenue and one east along the street. Similarly, every vertex has a vertex to the south and a vertex to the east (as well as north and west for most of them, but those are irrelevant for this). Hence, there are two edges for every vertex.
-
If you have exactly 15 columns (avenues) of vertices in 200 rows (streets) then basic properties of multiplication give $15 \times 200 = 3000$ vertices.
As for edges, there are 14 edges in each row and 199 edges in each column so there are $14 \times 200 + 199 \times 15 = 5785$ edges. $6000$ was only an approximation.
$4$ of the vertices have two edges; $2\times 198 +2 \times 13 = 422$ vertices have three edges; the other $357$4 vertices have four edges. Each edge has two vertices. So $$4 \times 2 + 422 \times 3 + 5574 \times 4 = 11570 = 2 \times 5785.$$
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# Northwood Company manufactures basketballs. The company has a ball that sells for \$35. At present,...
Northwood Company manufactures basketballs. The company has a ball that sells for \$35. At present, the ball is manufactured in a small plant that relies heavily on direct labor workers. Thus, variable expenses are high, totaling \$24.50 per ball, of which 70% is direct labor cost.
Last year, the company sold 57,000 of these balls, with the following results:
Sales (57,000 balls) \$ 1,995,000 Variable expenses 1,396,500 Contribution margin 598,500 Fixed expenses 493,500 Net operating income \$ 105,000
Required:
1-a. Compute last year's CM ratio and the break-even point in balls. (Do not round intermediate calculations. Round up your final break even answers to the nearest whole number.)
1-b. Compute the the degree of operating leverage at last year’s sales level. (Round your answer to 2 decimal places.)
2. Due to an increase in labor rates, the company estimates that next year's variable expenses will increase by \$3.50 per ball. If this change takes place and the selling price per ball remains constant at \$35.00, what will be next year's CM ratio and the break-even point in balls? (Do not round intermediate calculations. Round up your final break even answers to the nearest whole number.)
3. Refer to the data in (2) above. If the expected change in variable expenses takes place, how many balls will have to be sold next year to earn the same net operating income, \$105,000, as last year? (Do not round intermediate calculations. Round your answer to the nearest whole unit.)
4. Refer again to the data in (2) above. The president feels that the company must raise the selling price of its basketballs. If Northwood Company wants to maintain the same CM ratio as last year (as computed in requirement 1a), what selling price per ball must it charge next year to cover the increased labor costs? (Do not round intermediate calculations. Round your answer to 2 decimal places.)
5. Refer to the original data. The company is discussing the construction of a new, automated manufacturing plant. The new plant would slash variable expenses per ball by 30%, but it would cause fixed expenses per year to increase by 99%. If the new plant is built, what would be the company’s new CM ratio and new break-even point in balls? (Do not round intermediate calculations. Round up your final break even answers to the nearest whole number.)
6 Refer to the data in (5) above.
a. If the new plant is built, how many balls will have to be sold next year to earn the same net operating income, \$105,000, as last year? (Do not round intermediate calculations. Round up your final answer to the nearest whole number.)
b-1. Assume the new plant is built and that next year the company manufactures and sells 57,000 balls (the same number as sold last year). Prepare a contribution format income statement. (Do not round your intermediate calculations.)
b-2. Compute the degree of operating leverage. (Do not round intermediate calculations and round your final answer to 2 decimal places.)
Selling price..................................\$35.00 100%
Variable expenses......................... 24.50 70%
Contribution margin.....................\$10.50 30%
Profit = Unit CM × Q − Fixed expenses
\$ 0 = \$ 10.50 * Q - 493,500
\$ 10.50 Q = 493500
Q = 47000
1B)The degree of operating leverage is:
Degree of operating leverage = Contribution Margin / Net operating income
= 598,500 / 105,000
= 5.70
2)
CM ratio = [35 - (24.5 + 3.50)] / 35
= 7/ 35
= .2 or 20%
Break-even point = 493,500 / (35 - 28)
= 70500
3)
desired sales (balls) = (fixed cost + desired profit) / (contribution per unit)
= (493,000 + 105,000) / 7
= 85428 balls
4)
let selling price be X, contribution = .30 X (X*30%)
Contribution = selling price - variable cost
.30X = X - (24.5 + 3.50)
.70X = 28
X = 28 / .70
= \$40 per unit to be charged
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• The manager for sales contractors wants to construct a control chart for determining whether the 10... | 1,118 | 4,714 | {"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-2023-23 | latest | en | 0.921218 |
https://www.physicsforums.com/threads/how-could-this-group-possibly-have-elements-of-this-order.558261/ | 1,511,439,178,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806771.56/warc/CC-MAIN-20171123104442-20171123124442-00460.warc.gz | 818,300,258 | 15,331 | # How could this group possibly have elements of this order?
1. Dec 8, 2011
### jdinatale
1. The problem statement, all variables and given/known data
It seems to me that every element of Z_45 has order 1, 3, 5, 9, 15, or 45. It seems impossible to have an element of order 2 by Lagrange's Theorem.
Is there another way of looking at this problem?
2. Dec 8, 2011
### micromass
Staff Emeritus
For example, (30,0,0,0) has order 2.
3. Dec 8, 2011
### jdinatale
I see! Do you think a counting argument would be acceptable? For example there are two elements of order 2 in Z_60, (0 and 30), one element of order 2 in Z_45 (0), two elements of order 2 in Z_12 (0 and 6) and two elements of order 2 in Z_36 (0 and 18)
So it seems to me, all of the elements of order 2 in A would be:
(0, 0, 0, 0)
(0, 0, 0, 18)
(0, 0, 6, 0)
(0, 0, 6, 18)
(30, 0, 0, 0)
(30, 0, 6, 0)
(30, 0, 0, 18)
(30, 0, 6, 18)
So 8 elements, correct?
4. Dec 8, 2011
### micromass
Staff Emeritus
(0,0,0,0) does not have order 2.
5. Dec 8, 2011
### jdinatale
Oh thank you, you are correct. Is my argument mathematically correct where I just list out all of the possible elements of order 2?
Also, do you know of a better way to look at the number of subgroups of index 2 in A?
6. Dec 8, 2011
### micromass
Staff Emeritus
Yes, the rest of what you did is absolutely correct!!!! | 493 | 1,359 | {"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-2017-47 | longest | en | 0.887112 |
https://winmundo.com/python-how-to-clamp-an-integer-to-some-range/ | 1,685,855,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00251.warc.gz | 681,882,736 | 15,534 | # python – How to clamp an integer to some range?
## python – How to clamp an integer to some range?
This is pretty clear, actually. Many folks learn it quickly. You can use a comment to help them.
``````new_index = max(0, min(new_index, len(mylist)-1))
``````
``````sorted((minval, value, maxval))[1]
``````
for example:
``````>>> minval=3
>>> maxval=7
>>> for value in range(10):
... print sorted((minval, value, maxval))[1]
...
3
3
3
3
4
5
6
7
7
7
``````
#### python – How to clamp an integer to some range?
many interesting answers here, all about the same, except… which ones faster?
``````import numpy
np_clip = numpy.clip
mm_clip = lambda x, l, u: max(l, min(u, x))
s_clip = lambda x, l, u: sorted((x, l, u))[1]
py_clip = lambda x, l, u: l if x < l else u if x > u else x
``````
``````>>> import random
>>> rrange = random.randrange
>>> %timeit mm_clip(rrange(100), 10, 90)
1000000 loops, best of 3: 1.02 µs per loop
>>> %timeit s_clip(rrange(100), 10, 90)
1000000 loops, best of 3: 1.21 µs per loop
>>> %timeit np_clip(rrange(100), 10, 90)
100000 loops, best of 3: 6.12 µs per loop
>>> %timeit py_clip(rrange(100), 10, 90)
1000000 loops, best of 3: 783 ns per loop
``````
paxdiablo has it!, use plain ol python. The numpy version is, perhaps not surprisingly, the slowest of the lot. Probably because its looking for arrays, where the other versions just order their arguments. | 462 | 1,399 | {"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} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.699827 |
https://ww2.mathworks.cn/help/matlab/learn_matlab/types-of-arrays.html | 1,709,596,403,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00404.warc.gz | 632,600,550 | 22,086 | ## 数组类型
### 多维数组
MATLAB® 环境中的多维数组是具有多个下标的数组。创建多维数组的一种方法是调用具有多个参数的 `zeros``ones``rand``randn`。例如，
`R = randn(3,4,5);`
MATLAB 与丢勒的 4 阶幻方矩阵版本的区别在于交换了两个列。通过交换列，可以生成许多不同的幻方矩阵。语句
`p = perms(1:4);`
```A = magic(4); M = zeros(4,4,24); for k = 1:24 M(:,:,k) = A(:,p(k,:)); end```
```size(M) ans = 4 4 24```
`sum(M,d)`
`sum(M,1)`
`34 34 34 34`
`sum(M,2)`
```34 34 34 34```
`S = sum(M,3)`
```S = 204 204 204 204 204 204 204 204 204 204 204 204 204 204 204 204```
### 元胞数组
MATLAB 中的元胞数组是以其他数组的副本为元素的多维数组。使用 `cell` 函数可以创建空矩阵的元胞数组。但是，更普遍的做法是，通过将其他内容的集合括入花括号 `{}` 中来创建元胞数组。花括号还可以与下标配合使用，以便访问各个元胞的内容。例如，
`C = {A sum(A) prod(prod(A))}`
```C = [4x4 double] [1x4 double] [20922789888000]```
```M = cell(8,1); for n = 1:8 M{n} = magic(n); end M```
```M = [ 1] [ 2x2 double] [ 3x3 double] [ 4x4 double] [ 5x5 double] [ 6x6 double] [ 7x7 double] [ 8x8 double]```
`M{4}`
### 字符与文本
`s = 'Hello'`
`a = double(s)`
```a = 72 101 108 108 111```
`s = char(a)`
`F = reshape(32:127,16,6)';`
```char(F) char(F+128)```
`h = [s, ' world']`
```h = Hello world```
`v = [s; 'world']`
```v = Hello world```
`S = char('A','rolling','stone','gathers','momentum.')`
```S = A rolling stone gathers momentum.```
`C = {'A';'rolling';'stone';'gathers';'momentum.'}`
```C = 'A' 'rolling' 'stone' 'gathers' 'momentum.'```
`C = cellstr(S)`
`S = char(C)`
### 结构体
```S.name = 'Ed Plum'; S.score = 83; S.grade = 'B+'```
```S = name: 'Ed Plum' score: 83 grade: 'B+'```
```S(2).name = 'Toni Miller'; S(2).score = 91; S(2).grade = 'A-';```
```S(3) = struct('name','Jerry Garcia',... 'score',70,'grade','C')```
```S = 1x3 struct array with fields: name score grade```
`S.score`
`S(1).score, S(2).score, S(3).score`
```scores = [S.score] scores = 83 91 70 avg_score = sum(scores)/length(scores) avg_score = 81.3333```
```names = char(S.name) names = Ed Plum Toni Miller Jerry Garcia```
```names = {S.name} names = 'Ed Plum' 'Toni Miller' 'Jerry Garcia'```
```[N1 N2 N3] = S.name N1 = Ed Plum N2 = Toni Miller N3 = Jerry Garcia```
#### 动态字段名称
`structName.(expression)`
`structName.(expression)(7,1:25)`
```function avg = avgscore(testscores, student, first, last) for k = first:last scores(k) = testscores.(student).week(k); end avg = sum(scores)/(last - first + 1);```
```testscores.Ann_Lane.week(1:25) = ... [95 89 76 82 79 92 94 92 89 81 75 93 ... 85 84 83 86 85 90 82 82 84 79 96 88 98]; testscores.William_King.week(1:25) = ... [87 80 91 84 99 87 93 87 97 87 82 89 ... 86 82 90 98 75 79 92 84 90 93 84 78 81];```
```avgscore(testscores, 'Ann_Lane', 7, 22) ans = 85.2500 avgscore(testscores, 'William_King', 7, 22) ans = 87.7500``` | 1,183 | 2,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.325757 |
https://www.weegy.com/?ConversationId=J2OW2LD8&Link=i&ModeType=2 | 1,611,110,923,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00420.warc.gz | 1,073,192,637 | 11,796 | The painting Venus of Urbino, by Titian, is most notable for its A. texture. B. symmetry. C. sensuality. D. symbolism.
The painting Venus of Urbino, by Titian, is most notable for its sensuality.
g
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Updated 193 days ago|7/10/2020 12:51:41 PM
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https://www.mathworks.com/matlabcentral/answers/395514-help-with-updating-slice-plot-in-for-loop | 1,631,860,740,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00193.warc.gz | 926,181,356 | 26,202 | Help with updating slice plot in for loop
8 views (last 30 days)
Jordan Clark on 17 Apr 2018
Commented: Adalric Leung on 5 Jul 2020
I am plotting slices of a block with dimensions x,y,z and using a colour map to represent the temperature of the block, which is described by the 3D-matrix, V, which is the temperature at each data point. The temperature changes over time so I am trying to create an animation using a for loop over time and drawnow.
The problem I have is that I cannot seem to access and change V. I need to implement something to effect of:
if true
[X, Y, Z] = meshgrid([0 x], [0 y], [0 z])
h = slice(X,Y,Z,V,xslice,yslice,zslice)
colorbar
for n = 1:tmax
Vnew = (function that determines the temperature at the next time step)
set(h, V, Vnew)
drawnow
end
end
I then get an error that says "There is no V property on the Surface class."
A workaround I have is to re-plot the whole graph at each step but this is quite slow - any help would be hugely appreciated!
Wick on 2 May 2018
You just need to draw a new slice with the new V_new.
if true
[X, Y, Z] = meshgrid([0 x], [0 y], [0 z]);
h = slice(X,Y,Z,V,xslice,yslice,zslice);
colorbar
for n = 1:tmax
Vnew = (function that determines the temperature at the next time step)
h = slice(X,Y,Z,Vnew,xslice,yslice,zslice);
drawnow
end
end
Adalric Leung on 5 Jul 2020
Hi I'm want to acheive the same thing as the previous person where my block dimensions are:
x = -0.5:dx:0.5; % x-vector (cm)
y = -0.5:dy:0.5; % y-vector (cm)
z = -0.5:dz:0.5; % z-vector (cm)
My temperature matrix is given by a 4D matrix (in x, y, z, time) where the first 3 terms provide the temperature at that location, creating a 3D temperature matrix at a sepcific time. I want to plot the block with those dimensions and then represent the changes in temperature as a colour, but I'm unsure how to interpret the code you wrote since I"m still fairly new to MatLab. More specifically, would the "V" be the 3D temperature matrix at a specific point in time? How would I get the x,y, and z slices?
I've included the data for my 4D temperature matrix if it helps. | 606 | 2,094 | {"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-2021-39 | latest | en | 0.908393 |
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# Count of Nodes whose children gives same remainder when divided by K
Given a binary tree and an integer K. The task is to count the number of nodes having children that give the same remainder when divided by K. Print “-1” if no such node exists.
Examples:
Input: 2 K = 2
/ \
3 5
/ / \
7 8 6
Output: 2
Explanation: Children of 2 are 3 and 5. Both give remainder 1 with 2
Similarly for 5, both children give remainder as 0
Input: 9 K = 5
/ \
7 8
/ \
4 3
Output: -1
Explanation: There is no node having both children with same remainder with K.
Approach: This problem can be solved by simple binary tree traversal. Follow the steps below to solve the given problem.
• Traverse the Binary tree, and for each node, check
• If the node has a left child
• If the node has a right child
• If both children give the same remainder with K.
• Count all such nodes and print their content at the end.
Below is the implementation of the above approach.
## C++
// C++ implementation to print// the nodes having a single child#include using namespace std; // Class of the Binary Tree nodestruct Node { int data; Node *left, *right; Node(int x) { data = x; left = right = NULL; }}; // Function to find the nodes having both// and both of them % K are sameint countNodes(Node* root, int& K, int count){ // Base case if (root == NULL) return count; // Condition to check if the // node is having both child // and both of them % K are same if (root->left != NULL && root->right != NULL && root->left->data % K == root->right->data % K) { count++; } // Traversing the left child count = countNodes(root->left, K, count); // Traversing the right child count = countNodes(root->right, K, count); return count;} // Driver codeint main(){ // Constructing the binary tree Node* root = new Node(2); root->left = new Node(3); root->right = new Node(5); root->left->left = new Node(7); root->right->left = new Node(8); root->right->right = new Node(6); int K = 2; // Function calling cout << countNodes(root, K, 0);}
## Java
// Java code for the above approach import java.io.*;class Node { int data; Node left, right; Node(int data) { this.data = data; left = null; right = null; }}; class GFG { // Function to find the nodes having both // and both of them % K are same static int countNodes(Node root, int K, int count) { // Base case if (root == null) return count; // Condition to check if the // node is having both child // and both of them % K are same if (root.left != null && root.right != null && (root.left.data % K == root.right.data % K)) { count++; } // Traversing the left child count = countNodes(root.left, K, count); // Traversing the right child count = countNodes(root.right, K, count); return count; } public static void main(String[] args) { // Driver code // Constructing the binary tree Node root = new Node(2); root.left = new Node(3); root.right = new Node(5); root.left.left = new Node(7); root.right.left = new Node(8); root.right.right = new Node(6); int K = 2; // Function calling System.out.println(countNodes(root, K, 0)); }}//This code is contributed by Potta Lokesh
## Python3
# Python code for the above approachclass Node: def __init__(self, data): self.data = data; self.left = None; self.right = None; # Function to find the Nodes having both# and both of them % K are samedef countNodes(root, K, count): # Base case if (root == None): return count; # Condition to check if the # Node is having both child # and both of them % K are same if (root.left != None and root.right != None and (root.left.data % K == root.right.data % K)): count += 1; # Traversing the left child count = countNodes(root.left, K, count); # Traversing the right child count = countNodes(root.right, K, count); return count; if __name__ == '__main__': # Driver code # Constructing the binary tree root = Node(2); root.left = Node(3); root.right = Node(5); root.left.left = Node(7); root.right.left = Node(8); root.right.right = Node(6); K = 2; # Function calling print(countNodes(root, K, 0)); # This code is contributed by umadevi9616
## C#
// C# code for the above approachusing System;using System.Collections.Generic; class Node { public int data; public Node left, right; public Node(int data) { this.data = data; left = null; right = null; }}; public class GFG{ // Function to find the nodes having both // and both of them % K are same static int countNodes(Node root, int K, int count) { // Base case if (root == null) return count; // Condition to check if the // node is having both child // and both of them % K are same if (root.left != null && root.right != null && (root.left.data % K == root.right.data % K)) { count++; } // Traversing the left child count = countNodes(root.left, K, count); // Traversing the right child count = countNodes(root.right, K, count); return count; } public static void Main(String[] args) { // Driver code // Constructing the binary tree Node root = new Node(2); root.left = new Node(3); root.right = new Node(5); root.left.left = new Node(7); root.right.left = new Node(8); root.right.right = new Node(6); int K = 2; // Function calling Console.WriteLine(countNodes(root, K, 0)); }} // This code is contributed by 29AjayKumar
## Javascript
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up | 1,743 | 6,037 | {"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} | 3.984375 | 4 | CC-MAIN-2023-23 | latest | en | 0.764129 |
http://primes.utm.edu/curios/page.php?number_id=2951 | 1,529,616,930,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864300.98/warc/CC-MAIN-20180621211603-20180621231603-00245.warc.gz | 267,354,272 | 2,614 | 1423 (another Prime Pages' Curiosity)
Curios: Curios Search: Participate: The smallest prime number that can be formed by concatenating the first four positive integers. [Murthy] >Joy of Thinking: The Beauty and Power of Classical Mathematical Ideas is The Teaching Company's course number 1423. The smallest prime which is sum of composite numbers written with primes repeated twice for each term : 1423 = 22+33+55+1313 = 33+77+1313. [Capelle] A 4-digit prime, which if repeated 4 times after number 4, yields another prime whose sum of digits is 44. [Silva] The smallest 4-digit prime whose sum of the two leftmost digits (1 and 4) equals the sum of the two rightmost digits (2 and 3). [Forouhari] (There is one curio for this number that has not yet been approved by an editor.) Prime Curios! © 2000-2018 (all rights reserved) privacy statement | 222 | 851 | {"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-2018-26 | latest | en | 0.85391 |
https://tex.stackexchange.com/questions/260540/number-equations-and-preserving-alignment-of-equal-signs | 1,653,799,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00405.warc.gz | 622,741,428 | 65,600 | Number equations and preserving alignment of equal signs
So as I have learned from a previous post that the \begin{aligned}... \end{aligned} with the usage of "&" before "=" will help arrange neatly the equations where the equal sign of each equation will be just below the equal sign of the following equation and so on.
My question is if I have a total of 8 equations, how to number each equation and at the same time preserve the alignment of the equal signs.
Note if I use $$...$$, I get only one number for the collection of 8 equations at the very end which is not what I want.
• What about \begin{align}...\end{align} that's well explained in the manual of amsmath? Aug 9, 2015 at 14:56
Here is a small sample of numbering equations.
\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}
\begin{align} % numbers equations automatically
x &= (1-x) + 2x \\ % \\ ends the line
&= (x+y) - y \nonumber \\ % \nonumber removes the equation number
&= \frac{x}{2} + \frac{x}{2} % last line does not have \\
\end{align}
\begin{align*} % use of * removes all numbers in this block
x &= (2x + 5) - (x+5) \\
&= yx - (y-1) x
\end{align*}
\end{document}
This sample produces | 333 | 1,193 | {"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": 2, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-21 | longest | en | 0.832171 |
http://mathoverflow.net/questions/18449/forms-over-finite-fields-and-chevalleys-theorem | 1,469,779,744,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.49/warc/CC-MAIN-20160723071029-00124-ip-10-185-27-174.ec2.internal.warc.gz | 153,503,879 | 18,628 | # Forms over finite fields and Chevalley's theorem
Chevalley's theorem says that if $k$ is a finite field and $f(X_1,...,X_n)$ is a form (homogeneous polynomial) of degree $d < n$, then the equation $f(X_1,...,X_n) = 0$ has a non-trivial solution in $k^n$.
It is known that this result is optimal, in the sense that for each $n$ there exists a form $f(X_1,...,X_n)$ - coming from a norm - of degree $d = n$ which has only the trivial zero. See Brian Conrad's answer below.
I am interested in each form with this property, i.e. each form $f(X_1,...,X_n)$ of degree $d = n$ which has only the trivial zero. What are the known examples/classes of such forms? It would be very nice if we could describe/classify them all...
I am in particular interested in the case of quartic forms.
-
I think that the Fermat quartic in $\mathbb{P}^3$ is an example of a form without zero with $n=d=4$ having no zeros over $\mathbb{F}_5$. If I remember correctly, it is the only diagonal quartic over a finite field admitting no solution. If this is correct, then Martin Bright told me this, otherwise I am wrong! Note that the Fermat quartic is non-singular, and hence it is geometrically irreducible. – damiano Mar 17 '10 at 9:53
## 2 Answers
It is optimal for every $d > 0$. One way to make examples is using extension fields. For any $d > 0$, let $k'/k$ be an extension of degree $d$ and consider the norm map $N:k' \rightarrow k$. Choose a $k$-basis of $k'$, this is expressed as a homogeneous polynomial $f$ of degree $d$ in $d$ variables over $k$. More specifically, if $A$ is any $k$-algebra then the norm map $N_A:k' \otimes_k A \rightarrow A$ is given by evaluation of $f$ on $d$-tuples from $A$. This has no nonzero $k$-rational zeros, since nonzero elements of $k'$ have nonzero norm in $k$.
-
Thanks for the reply. Yes, I know that it is optimal for every $d > 0$, and norms are the obvious examples to show this - I should have mentioned that in the question. But I'm rather interested in "other types" of examples, whatever that may mean. – Wanderer Mar 17 '10 at 1:20
You're asking that the projective hypersurface $(f=0)$ of degree $d$ in projective $(d-1)$-space has no $k$-rational points. If $k$ has sufficiently large size purely in terms of $d$ then the RH bound provides rational points if this hypersurface is smooth and geometrically connected. So if we restrict attention to irreducible $f$ over $k$ then it seems inevitable for it to be a $k^{\times}$-multiple of a norm of a lower-degree homogeneous form over a nontrivial finite extension of degree dividing $d$, or that $(f=0)$ has singularities. – BCnrd Mar 17 '10 at 2:02
@Brian: inevitable only if $\# k \gg d$, you mean, right? Certainly there are smooth counterexamples when $d$ is large compared to $\# k$ and $n$. – Pete L. Clark Mar 17 '10 at 3:45
Yes Pete, I was only considering d=n as in the question, and so I was only pointing out that if we want to make constructions that include most finite fields and use a fixed d then blah-blah-blah... – – BCnrd Mar 17 '10 at 5:40
@Brian: not to belabor the point, but you're not suggesting that this is the case even for $n = d \gg \# k$, right? I ask because (i) this is relevant to the original question and (ii) I am interested myself. – Pete L. Clark Mar 17 '10 at 13:28
Perhaps it is obvious to most readers, but about a year ago I spent several days trying to determine for which pairs (d,n) there existed an anisotropic degree d form in n variables over a finite field $\mathbb{F}_q$. The question was motivated by Exercise 10.16 in Ireland and Rosen's classic number theory text: "Show by explicit calculation that every cubic form in two variables over $\mathbb{F}_2$ has a nontrivial zero."
As many students have discovered over the years, this is false: e.g. take
$f(x_1,x_2) = x_1^3 + x_1^2 x_2 + x_2^3$.
I knew about the existence and anisotropy of norm hypersurfaces for all $n = d$. But what about $n < d$? I confess that I spent some time proving this result in several special cases and even dragged a postdoc into it. Here is a copy of the sheepish email I sent out (in particular to Michael Rosen) later on:
If K is a field, and f(x_1,...,x_n) is an anisotropic form of degree d in n variables, then f(x_1,...,x_{n-1},0) is an anisotropic form of degree d in n-1 variables.
So let K be any field which admits field extensions of every positive degree d. Then for all d there is an anisotropic norm form N in d variables of degree d. For any n < d, setting (d-n) of the variables equal to 0 gives an anisotropic form of degree d in n variables. In particular, this proves "the converse of Chevalley-Warning".
So, not so fascinating after all, then.
I think it is still nontrivial to ask what happens if the hypersurface f is required to be geometrically irreducible. For instance, despite the fact that (q,3,3) is anisotropic, every geometrically irreducible cubic curve over a finite field has a rational point.
AS's question about classifying anisotropic hypersurfaces with $d = n$ is interesting. It may also be interesting to look at the case $d < n$. It is certainly not clear to me that all such anistropic hypersurfaces come from intersecting a norm hypersurface of larger dimension with a linear subspace.
I also want to add that the following generalization seemed less trivial to me (and I still don't know the answer): Chevalley-Warning is also true for sytems of polynomial equations $f_1(x_1,\ldots,x_n) = \ldots = f_r(x_1,\ldots,x_n)$ so long as the sum of the degrees of the $f_i$'s is strictly less than $n$. What kind of counterexamples can we construct here when $d = d_1 + \ldots + d_r \geq n$?
-
The third term of your cubic form should be x_2^3. – Franz Lemmermeyer Mar 17 '10 at 13:37
Thanks. – Pete L. Clark Mar 17 '10 at 14:14
Your last question is interesting, but doesn't the classical norm example still work? Just take norm forms - in disjoint sets of variables - for extensions of degree $d_1,d_2,\,\cdots,d_r$ where $d_1 + d_2 + \cdots + d_r = n$. – Wanderer Mar 17 '10 at 14:19
@AS: Yes, it does. Thanks for pointing this out. Amusingly the case with $d_i = 1$ for all $i$ is given in en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem (and is the only anisotropic example given there), but it did not inspire me to think of the generalization to simultaneous norm equations. Again, looking for geometrically integral examples seems more interesting... – Pete L. Clark Mar 17 '10 at 20:27 | 1,802 | 6,476 | {"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-2016-30 | latest | en | 0.899872 |
https://math.berkeley.edu/~ribet/Math115/99/message1.html | 1,656,108,797,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00699.warc.gz | 435,943,487 | 1,502 | In class next week, we will begin by discussing the material in section 2.1 of the text. Especially, we will prove Theorem 2.15, which characterises those positive integers n that can be expressed in the form a^2 + b^2, where a and b are integers. (For example, 8 is of this form because it is 2^2 + 2^2; so is 16, which is 4^2 + 0^2. The number 7 is not the sum of two squares.)
In preparation for our time together, you might have a look at the recent review of the beautiful "Proofs from the Book," a volume that brings together especially beautiful proofs of interesting results from diverse parts of mathematics. The "Book" in the title is God's notebook of mathematicial proofs, in which the best possible proofs of all theorems are written down. (Paul Erdös invented this concept.) The author of the review lists several highlights of "Proofs from the Book"; one is Don Zagier's proof, first published in 1990, of our Theorem 2.15. The reviewer says that the first published proof of the theorem is that of Euler; he adds that Zagier's proof is so beautiful that it will certainly be the last one to be published! I'll try to understand Don Zagier's argument and then explain it to you. Before discussing Zagier's proof, however, I'll talk about the one in our textbook. Don Zagier is a mathematician in Bonn, Germany who went to high school in Stockton. When you come to my office hours, ask me to point out Zagier in the photos on my wall.
September 11, 1999 | 362 | 1,469 | {"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-2022-27 | latest | en | 0.958824 |
http://roymech.co.uk/Useful_Tables/Beams/Continuous_Beams.html | 1,438,399,578,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988458.74/warc/CC-MAIN-20150728002308-00039-ip-10-236-191-2.ec2.internal.warc.gz | 205,982,143 | 4,358 | Disclaimer: The information on this page has not been checked by an independent person. Use this information at your own risk.
ROYMECH
Continuous Beams
Introduction
Built -in (Encastre) beams are fixed at both ends. Continuous beams, which are beams with more than two supports and covering more than one span, are not statically determinate using the static equilibrium laws.
Nomenclature
e = strain
σ = stress (N/m2)
E = Young's Modulus = σ /e (N/m2)
y = distance of surface from neutral surface (m).
R = Radius of neutral axis (m).
I = Moment of Inertia (m4 - more normally cm4)
Z = section modulus = I/y max(m3 - more normally cm3)
M = Moment (Nm)
w = Distributed load on beam (kg/m) or (N/m as force units)
W = total load on beam (kg ) or (N as force units)
F= Concentrated force on beam (N)
L = length of beam (m)
x = distance along beam (m)
Built in beams
A built in beam is normally considered to be horizontal with both ends built-in at the same level and with zero slope at both ends. A loaded built in beam has a moment at both ends and normally the maximum moments at at one or both of the two end joints.
A built in beam is generally much stronger than a simply supported beam of the same geometry. The bending moment reduces along the beam and changes sign at points of contraflexure between the supports and the load. A typical built-in beam is shown below.
It is not normally possible to determine the bending moments and the resulting stress using static equilibrium. Deflection calculations are often used to enable the moments to be determined.
Repeating section from Beam Theory
It has been proved ref Shear Bending that dM/dx = S and dS/dx = -w = d2M /dx
Where S = the shear force M is the moment and w is the distributed load /unit length of beam. Therefore
Using the above beam as an example...
Using the above equations the bending moment, shear force, deflection, slope can be determined at any point along the beam.
M = EI d 2y/dx 2 = w(- 6x2+6lx -l2)/12
at x = 0 & l then M = -wl2 /12 and at x = l/2 then M = wl2 /24
S = EI d 3y/dx 3 = w(l/2 - x)
at x = 0 then S = w.l/2 at x = l then S = -w.l/2
Continuous Beams
This type of beam is normally considered using the Clapeyron's Theorem ( Three Moments theorem)
The three moments theorem identifies the relationship between the bending moments found at three consecutive supports in a continuous beam. This is achieved by evaluating the slope of of the beam at the end where the two spans join. The slopes are expressed in terms of the three moments and the supported loads which are then equated and the resulting equations solved.
This relationship for spans with supports at the same height and with spans of constant section results in the following expression.
If the beams has a different section for each span then the more general expression applies as shown below.
Example Areas and x 1 value calculations.
Examples using theorem.
Example 1)
This simple example is a two span continuous beam with the ends simple supported,therefore with no moments at the end support points..
1) Bending Moments.
2) Shear Forces
3) Diagrams
Example 2)
This simple example is a three span continuous beam with the ends simple supported,therefore with no moments at the end support points..
The values of A1,x1,A2, and x2 are calculated using the methods above..
Sites Providing Relevant Information Three Moment equation for continous beam analysis...Very advanced paper download Mitcalc...Excel based software including coded beam calculations Granta -Solutions to Standard Problems...Very accessible notes for beams and sections Chaapter 6 Moment Distribution...Chapter of Book inlcuding detailed information | 875 | 3,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2015-32 | longest | en | 0.924055 |
http://www.mrcbaker.com/2015/10/3-act-math-images-circular-farms.html | 1,534,243,134,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209021.21/warc/CC-MAIN-20180814101420-20180814121420-00201.warc.gz | 698,396,296 | 15,583 | ## 07 October 2015
### 3-Act Math Images - Circular Farms
I really like this satellite image from NASA of center-pivot irrigated farm in Kansas - it looks like hundreds of pennies to me.
What questions could we ask with this image?
Image: Dalhart Texas, NASA Visualization
Some of my ideas:
• How much land goes unused in this image because the fields are planted as circles?
• If you could add smaller center-pivot sprinklers in the corners of the fields (and therefore, smaller circles) to cover some of the unused land, how many sprinklers would you need to cover over 98% of a field?
• What proportion of this land is covered in dark farms, and in light colored farms?
• Assuming irrigation costs were the same for both fields, how much more money could a farm make that planted in rectangular plots instead of circular plots of wheat?
• Which setup would best maximize use of the land - one big center pivot sprinkler (one circle on a square plot, two sprinklers that do a half rotation (two hemispheres on a square plot), or four smaller sprinklers that do full rotations (four circles on a square plot)? | 247 | 1,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-34 | longest | en | 0.924821 |
www.attentiveequations.com | 1,627,146,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00668.warc.gz | 658,858,718 | 22,187 | Color Temperature Study of a Figure
Periodically, I visit a studio of a fellow artist and as to be expected it is set up in a way to satisfy his artistic needs. There is a series of sky lights facing northeast and a large window east, south-east. The set-up for the model is quite simple but the dual direction of the light sources is difficult. I struggle in my drawings or paintings with maintaining a sense of a directional light source. Many times, on sunny days, both sources have an equal intensity. Most times, I have chosen one source or the other as the dominant point of light. Another added difficulty is the color temperature of the two light sources- one being more direct and therefore warmer and the other more indirect and therefore cooler (although, this changes as well with the seasons as the sun moves its position in the sky).
I decided to tackle this problem straight on and deal with both light sources equally and study the color temperature differences that occur. This sunny June day found the light as such: The sky lights had a warm directional light because the sun was coming in at an oblique angle directly and this light was to one side of the model and slightly behind; on the other hand the large window was receiving indirect light but was directly opposite the figure creating a definitive coolness on one side of the model. As the model was placed in the space, one side was receiving the warm light from the sky light and the other side was receiving the cool light from the window. Both sources were opposing one another with the core shadow and darker shadow edge running through the middle of the figure. This was really the best position to study these opposing color temperatures.
I began by choosing a gray sheet of Borden Riley charcoal paper of a middle value. I intended to leave the cool gray of the paper as an indication of the cool light source. I sketched the figure with medium vine charcoal, building the armature, making comparisons of relationships and of proportions and drawing the revisions. Next, I built the forms keeping in mind the anatomical landmarks and allowing the rhythm of the forms to relate to the over-all rhythm of the figure as a whole. After completing this first stage of the drawing, I started to plan the pattern of light and shadow and the differences of color temperature between them.
Since the warm light was more direct, I chose this source as the dominant point of light- this would be the light side of the figure. And since the cool light was indirect and not as intense, I chose this secondary light for the shadow side. The division between the two, which was darker, would indicate the core shadow as well as the shadow edge. This is where I began- in the interior of the figure looking carefully at the specific shape of the shadow edge. This created the necessary division between the warm light source and the cool reflected light on the shadow side of the figure.
Next, I chose three pastels to work the light side of the model- a cool white, a warmer white and a warm mid-tone flesh color for the half-tones. I began with the warmer white and used it to indicate the general source of the light- reducing its application as I approached the half-tones, leaving some cool of the paper to show through. Next, I indicated the warmer half-tones near the shadow edge that were of a lower value than the general light. These did not touch the shadow edge, leaving the gray tone of the paper to abut the shadow. Lastly, I applied the cooler white for the highlights.
Next, I worked the shadow side of the figure allowing the cool gray tone of the paper to act as the cool reflected light within the shadow. I did not use any pastel on this cool side. I used the charcoal in varying degrees to indicate the value changes occurring and left the coolness of the paper in areas with a stronger, more intense reflected light. Next, I added dark accents as well as lost and found edges. And lastly, I felt I needed some tone in the hair to off-set it against the flesh and added a warm dark in its mid-tones.
This study allowed me to solve many of the issues I faced in this particular situation and will assist me in future projects there. But, it also showed me how important color temperature is to achieving a sense of form. Robert Henri believed that, “Form can be modeled in black and white, but there are infinitely greater possibilities in modeling through the warmth and coolness of color.” (Henri, The Art Spirit, p.62) Through this experiment, I found that even more true than I first realized.
Author: Judith Reeve
For nearly 30 years I've developed my painting practice in the studio, building on what I leaned from my student days at the Lyme Academy of Fine Art. Along with my daily journey creating images which I write about here on this blog, I am also currently writing a book on the color practice of Robert Henri.
One thought on “Color Temperature Study of a Figure”
1. Sounds like a really difficult lighting situation. I like the drawing very much and I see what you mean about effect of the temperature of the light. It adds another dimension to your description of the forms. A very clear account of the procedure
you used. You make it sound and look easy!
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,092 | 5,357 | {"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-2021-31 | longest | en | 0.974842 |
http://mathhelpforum.com/advanced-statistics/109318-multivariate-statistics-compute-variance.html | 1,481,349,889,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542939.6/warc/CC-MAIN-20161202170902-00071-ip-10-31-129-80.ec2.internal.warc.gz | 175,621,360 | 9,964 | # Thread: multivariate statistics ; compute variance
1. ## multivariate statistics ; compute variance
hi guys,
for X1 = [1,2,4,1]
and X2 = [5,1,2,4]
how do you find the sample means and covariance matrix for the two variables without knowing their distributions ?
2. Originally Posted by masterinex
hi guys,
for X1 = [1,2,4,1]
and X2 = [5,1,2,4]
how do you find the sample means and covariance matrix for the two variables without knowing their distributions ?
The sample mean is just the average of the 4 values.
I guess you want a 2 by 2 sample variance matrix.
Put the sample variances in the diagonal positions and the sample covariance in the off diagonal spot.
NOW do we use n or n-1 in this case?
Not sure if you want the MLEs (n) or the unbiased estimators (n-1). | 207 | 778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-50 | longest | en | 0.898961 |
https://documen.tv/question/in-a-louisiana-chili-cook-off-18-of-the-40-chilis-included-two-types-of-beans-a-photo-of-multipl-24080621-86/ | 1,660,827,757,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00652.warc.gz | 218,425,821 | 15,566 | ## In a Louisiana chili cook-off, 18 of the 40 chilis included two types of beans. A photo of multiple crock pots on a long table i
Question
In a Louisiana chili cook-off, 18 of the 40 chilis included two types of beans.
A photo of multiple crock pots on a long table is shown. The caption says chili cook-off.
What percentage of the chilis did not include two types of beans?
Enter the correct answer in the box.
in progress 0
5 months 2021-09-04T22:20:14+00:00 1 Answers 11 views 0
55%
Step-by-step explanation:
18 out of 40 include chili.
So 22 has no chili
22/40 = 0.55 or 55% | 169 | 591 | {"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} | 3.265625 | 3 | CC-MAIN-2022-33 | latest | en | 0.903362 |
http://staryte.com/publ/homework_assignment/std_5/homework_1_8/23-1-0-258 | 1,524,318,710,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00144.warc.gz | 291,240,999 | 24,712 | Home » Articles » HOMEWORK ASSIGNMENT » STD 5
HOMEWORK 1.8
HOME WORK 1.8 Write the following roman numbers as Arabic Numeral: MCDLII CCXCVII CDIV MMCMXL CMXLVIII Perform the following operation and write the answer in Roman Numeral: 213 x 3 = 876 + 2304 = Perform the following operation and write the answer in Arabic Numeral: LIV + CCCXL VIII = MMCCCXXVIII ÷ XXIV = Find the sum of L.C.M and G.C.F of 16, 32 and 40. Write 48 as the product of prime factors. What are the common factors of 28 and 40? List all prime numbers between 322 and 333. How many odd numbers are there from 31 to 41? List all multiples of 6 which are less than 50. Find the Greatest Common Divisor of 14 and 42 by using listing method. HAVE A NICE DAY: .u-star-rating-16 { list-style:none;margin:0px;padding:0px;width:80px;height:16px;position:relative;background: url('/.s/t/1706/rating.png') top left repeat-x } .u-star-rating-16 li{ padding:0px;margin:0px;float:left } .u-star-rating-16 li a { display:block;width:16px;height: 16px;line-height:16px;text-decoration:none;text-indent:-9000px;z-index:20;position:absolute;padding: 0px;overflow:hidden } .u-star-rating-16 li a:hover { background: url('/.s/t/1706/rating.png') left center;z-index:2;left:0px;border:none } .u-star-rating-16 a.u-one-star { left:0px } .u-star-rating-16 a.u-one-star:hover { width:16px } .u-star-rating-16 a.u-two-stars { left:16px } .u-star-rating-16 a.u-two-stars:hover { width:32px } .u-star-rating-16 a.u-three-stars { left:32px } .u-star-rating-16 a.u-three-stars:hover { width:48px } .u-star-rating-16 a.u-four-stars { left:48px } .u-star-rating-16 a.u-four-stars:hover { width:64px } .u-star-rating-16 a.u-five-stars { left:64px } .u-star-rating-16 a.u-five-stars:hover { width:80px } .u-star-rating-16 li.u-current-rating { top:0 !important; left:0 !important;margin:0 !important;padding:0 !important;outline:none;background: url('/.s/t/1706/rating.png') left bottom;position: absolute;height:16px !important;line-height:16px !important;display:block;text-indent:-9000px;z-index:1 } Category: STD 5 | Added by: Admin (20/Feb/2017) | Author: Yahya Mohamed (Badshah) E W Views: 200 | Rating: 0.0/0 | 669 | 2,167 | {"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-2018-17 | latest | en | 0.469959 |
https://www.nikhef.nl/pub/pr/eSMC.html | 1,544,952,826,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827596.48/warc/CC-MAIN-20181216073608-20181216095608-00509.warc.gz | 997,330,460 | 2,754 | ## SMC: How do quarks spin?
In the Spin Muon Collaboration (SMC), 150 physicists are working in an experiment at CERN to determine the spin direction of quarks in protons and neutrons.
In the front the polarised target set-up is visible with the detectors behind it.
All the electronic equipment is in the yellow huts to the right.
Spin is the quantity of rotation of a particle around its axis. A particle rotating to the left has an upward spin, a particle rotating to the right a downward. Electric charge on such a spinning top causes a current and produces a magnetic dipole. Due to this effect, charged particles with spin behave like compass needles in a magnetic field. The amount of spin direction we call polarisation.
Protons and deuterons (nuclei which are made up from a proton and a neutron) behave like very weak dipole magnets; they have spin. By using different microwave, superconducting and cryogenic techniques, the SMC experiment has succeeded in polarising large quantities of protons and deuterons. During the measurements we looked at the quarks in the protons and deuterons. The question we wanted answered was: Do quark spins have a preferential direction in respect to the spin of the proton or neutron in which they are locked? (figure 1)
To investigate this, muons were shot at targets of polarised protons and deuterons. Muons originate together with neutrinos in the decay of short lived particles which are made with CERN's particle accelerators. Thanks to a mysterious property of neutrinos -which always have left hand spin- the muons are polarised. This polarisation plays a role in the collisions with quarks; chances of collision are higher when a quark and a muon have opposing spin.
Figure 2. The target material is surrounded by a solenoid and a dipole coil (a). When the current through these coils change as shown (b), the direction of the magnetic field is reversed (c). The magnetic dipoles (polarised protons and deuterons follow as indicated by the compass needles (d).
We have investigated if there is a difference in collision chances when we reverse the polarisation direction of the target (figure 2). Such a difference, expressed in the asymmetry A, points at a preferential direction in quark spin in the proton and the deuteron. After the collision, the muons were deflected in a magnetic field and their tracks detected in wire chambers (figure 3). It can be established from the reconstruction of the muons tracks whether the quark with which a muon collided represents a proportionally large or a small amount of mass and impulse of the proton (or neutron). This ratio we call the impulse fraction x of the quark.
The results in figure 4a show we found an asymmetry for the proton: quarks with large x are polarised in the proton spin direction. For quarks with small x this preference disappears. For the deuteron (figure 4b) the asymmetry is smaller because the deutron also contains a neutron in which the roles of quarks with 2/3 and -1/3 charge are reversed. In the proton quarks with 2/3 charge are polarised in the proton spin direction, in the neutron they point at the opposite direction.
Figure 4. The experimentally determined asymmetry of the change of collision with polarisation of beam and target and as a function of the impulse fraction of the quark. The asymmtery is zero by a small x. When x increases, so does the asymmetry. The curve is adjusted for the proton data points. The deuteron data points are clearly below this curve.
When we add up all the quark spin contributions for all values of x, we only find half the value which was predicted by theory 20 years ago. This lack of polarisation has now been irrefutably established in the SMC experiment, and also in experiments at SLAC (Stanford) and DESY (Hamburg). The next step is to research gluons, the carrier particles of the strong nuclear force. At this moment it is not clear how large the spin preference of gluons is. If they are strongly polarised, this would explain the lack of quark polarisation in a natural way. Future experiments will tell us more.
Back to the index. | 910 | 4,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-51 | latest | en | 0.923684 |
http://scienceblogs.com/goodmath/2006/06/13/the-problem-with-irreducibly-c-1/ | 1,444,381,821,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737922497.60/warc/CC-MAIN-20151001221842-00191-ip-10-137-6-227.ec2.internal.warc.gz | 277,921,039 | 44,711 | # The Problem with Irreducibly Complexity (revised post from blogger)
As I mentioned yesterday, I’m going to repost a few of my critiques of the bad math of the IDists, so that they’ll be here at ScienceBlogs. Here’s the first: Behe and irreducibly complexity. This isn’t quite the original blogger post; I’ve made a few clarifications and formatting fixes; but the content remains essentially the same. You can find the original post in my blogger information theory index. The original publication date was March 13, 2006.
Today, I thought I’d take on another of the intelligent design sacred cows: irreducible complexity. This is the cornerstone of some of the really bad arguments used by people like Michael Behe.
To quote Behe himself:
By irreducibly complex I mean a single system composed of several well-matched, interacting parts that contribute to the basic function, wherein the removal of any one of the parts causes the system to effectively cease functioning. An irreducibly complex system cannot be produced directly (that is, by continuously improving the initial function, which continues to work by the same mechanism) by slight, successive modifications of a precursor system, because any precursor to an irreducibly complex system that is missing a part is by definition nonfunctional. An irreducibly complex biological system, if there is such a thing, would be a powerful challenge to Darwinian evolution. Since natural selection can only choose systems that are already working, then if a biological system cannot be produced gradually it would have to arise as an integrated unit, in one fell swoop, for natural selection to have any thing to act on.
Now, to be clear and honest upfront: Behe does not claim that this is a mathematical argument. But that doesn’t mean that I don’t get to use math to shred it.
There are a ton of problems with the whole IC argument, but I’m going to take a different tack, and say that even if those other flaws weren’t there, it’s still a meaningless argument. Because from a mathematical point of view, there’s a critical, fundamental problem with the entire idea of irreducible complexity: you can’t prove that something is irreducible complex.
This is a result of some work done by Greg Chaitin in Algorithmic Complexity Theory. A fairly nifty version of this can be found on Greg’s page.
The fundamental result is: given a system S, you cannot in general show that there is no smaller/simpler system that performs the same task as S.
As usual for algorithmic information theory, the proof is in terms of computer programs, but it works beyond that; you can think of the programs as the instructions to build and/or operate an arbitrary device.
First, suppose that we have a computing system φ, which we’ll treat as a function. So φ(x) = the result of running program x on φ. x is both a program and its input data coded into a single string, so x=(c,d), where c is code, and d is data.
Now, suppose we have a formal axiomatic system, which describes the basic rules that φ operates under. We can call this FAS.
If it’s possible to tell where you have minimal program using the axiomatic system, then you can write a program that examines other programs, and determines if they’re minimal. Even better: you can write a program that will generate a list of every possible minimal program, sorted by size.
Let’s jump aside for just a second to show how you can generate a list of every possible minimum program. Here’s a sketch of the program:
1. First, write a program which generates every possible string of one character, then every possible string of two characters, etc., and outputs them in sequence.
2. Connect the output of that program to another program, which checks each string that it receives as input to see if it’s a syntactically valid program for φ. If it is, it outputs it. If it isn’t, it just discards it.
3. At this point, we’ve got a program which is generating every possible program for φ. Now, remember that we said that using FAS, we can write a program that tests an input program to determine if its minimal. So, we use that program to test our inputs, to see if they’re minimal. If they are, we output them; if they aren’t, we discard them.
Now, let’s take a second, and write out the program in mathematical terms:
Remember that φ is a function modeling our computing system, FAS is the formal axiomatic system. We can describe φ as a function from a combination of program and data to an output: φ(c,d)=result.
In this case, c is the program above; d is FAS. So φ(c,FAS)=a list of minimal programs.
Now, back to the main track.
Using the program that we sketched above, given any particular length, we can easily generate programs larger than that length.
Take our program, c, and our formal axiomatic system, FAS. and compute their
length. Call that l(c,FAS). If we know l(c,FAS), we can run φ(c,FAS) until it generates a string longer than l(c,FAS).
Ok. Now, write a program c’ that for φ that runs φ(c,FAS) until it finds a program K, where the length of the output of φ(K) is larger than l(c,FAS) + length(c’). c’ then outputs the same thing as φ(K).
This is the tricky part. What does this program do? It runs a program which generates a sequence of provably minimal programs. It runs those provably minimal programs until it finds one larger than itself plus all of its data. Then it runs that and emits the output.
So – c’ outputs the same result as a supposedly minimal program K, where K is larger than c’ and its data. But since c’ is a program which emits the same result as K, but is smaller, then K cannot be minimal.
No matter what you do – no matter what kind of formal system you’ve developed for showing that something is minimal, you’re screwed. Godel just came in and threw a wrench into the works. There is absolutely no way that you can show that any system is minimal – the idea of doing it is intrinsically contradictory.
Evil, huh?
But the point of it is quite deep. It’s not just a mathematical game. We can’t tell when something complicated is minimal. Even if we knew every relevant fact about all of the chemistry and physics that affects things, even if the world were perfectly deterministic, we can’t tell when something is as simple as it can possibly be.
So irreducibly complexity is useless in an argument; because we can’t know when something is irreducibly complex.
1. #1 Chris Grant
June 13, 2006
You haven’t fixed the problems with this argument that were explained to you by several people the last time you posted it. To summarize:
(1) “Irreducibly complex” in Behe’s sense is not “minimal” in your sense.
This was explained to you by “Gorobei” in the 5th comment and by “keiths” in the 14th comment.
(2) Being able to show that a particular program is minimal is not the same as having a procedure for determining for every single program P whether P is minimal.
This was explained to you by “viked” in the 7th comment, “Gorobei” in the 8th comment, and “Anonymous” in the 16th comment.
(3) Claiming that there is empirical evidence for irreducible complexity in a system is not the same thing as claiming to have a deductive proof for irreducible complexity in a system.
This was explained to you by “Macht” in the 18th comment.
2. #2 Mark C. Chu-Carroll
June 13, 2006
Chris:
I thought I addressed the complaints in the comments last time; I’ll try again in the comments here. The fundamental point is that the concept of irreducibility is itself flawed. It has nothing to do with how you demonstrate irreducibility; it’s the concept of irreducibility itself that’s the problem.
(1) Behe never gives much in the way of a precise definition of IC, but as far as he defines it, I do not see any reason why it’s fundamentally different than the mathematical notion of minimal: both mean that you can’t remove anything and get the same result. Dembski quite clearly means the information theoretic notion of minimality when he references IC type arguments in his explanatory filter/specified complexity arguments.
(2) The fundamental point of the information theoretic argument is that there’s a threshhold beyond which no proof of minimality is valid. It doesn’t matter how good the proof looks to you. It doesn’t matter how good your allegedly IC system is. Minimality tests fail not because of any flaws in the system, not because there’s anything wrong with the reasoning in your supposed proof of minimality – but because fundamentaly, the notion of minimal complexity is itself flawed: minimal complexity itself lacks meaning.
(3) If you look at the proof I show, there is no problem with the minimality proof of K. And there is no way of knowing if the allegedly irreducible system you’re looking at is beyond the K threshold. The fundamental concept of irreducible systems does not work: It doesn’t matter that you have what looks to be a perfectly valid proof of the irreducibility of a system if the concept of irreducibility itself is flawed. Empricial evidence that a system has a fundamentally meaningless property doesn’t make that property meaningful. Once you cross the complexity threshold where the complexity of your system exceeds that of K in the proof, no proof of minimality has any meaning, because minimality itself has no meaning; and to make matters worse, you can’t determine where that point is!
3. #3 Steve
June 13, 2006
It seems that there is a difference between minimality and what Behe seems to want IC to mean (which is not to say that IC actually makes sense/is valid). It appears to me that minimality is a static concept related a program’s functioning which is invariant to any model you have of how the program was developed.
IC (if it were to be defined rigorously) would have to be a dynamic property embedded in some model of program/object generation – i.e. some partial order that would say that progam X is/could be a child of program Y (given generation method G). Then saying that program X is IC would mean that X’s output is meaningful, while any program Y for which X is a successor to Y (under a specific G, or maybe that there exists a G…) is not meaningful.
4. #4 Macht
June 13, 2006
My point in my comment in the original post was that science doesn’t deal with proof. So the fact that “you can’t prove that something is irreducible complex” is irrelevant to science. I can’t prove “insert any scientific theory here” but as long as it makes testable predictions, is consistent with other theories, etc. the fact that I can’t prove it is true really doesn’t matter.
5. #5 Mark C. Chu-Carroll
June 13, 2006
steve:
I think you’re partially right. The main catch is that Behe does have a kind of minimality constraint. That is, he only allows things to *grow*. That’s a key piece of his argument: you can only produce an IC system by *adding* features – and if you can’t get there by adding features, then it can’t evolve.
So while I think you’re right that he’s arguing for something like a transition system, where X in irreducible if/f any program Y for which X is a successor is non-functional, I think that he also places an additional ordering constraint on it: that Y can only be a predecessor of X if Y is smaller than X.
6. #6 Chris Hyland
June 13, 2006
To be fair to Behe, his original definition ie ‘if you remove any part it fails’ is testable in that you can tell if something is irreducibly complex by sequentially removing parts and seeing if the system still functions. This has absolutely nothing to do with whether or not the system would have evolved though, and you are right that it is no proof the function could not be performed by a simpler system.
7. #7 Mark C. Chu-Carroll
June 13, 2006
Macht:
Sorry, but I think you’re wrong there. While we can’t, for example, prove beyond any reasonable doubt that, for example, gravity behaves the way that relativity says it does, that’s not the kind of concept that underlies IC.
For something like IC, if you can’t prove that something is IC, then it’s a useless argument. If the argument is specifically that a system has the property that it can’t be created incrementally, then saying that you can’t prove things like whether a system has that property means that you don’t have an argument.
If IC is provable – then the information theory argument tears it down.
If it isn’t provable, then it’s got no value whatsoever. IC is an argument that evolution couldn’t exlain certain systems, because they have this specific property that makes them non-evolvable. If we can’t show whether or not any system has this non-evolvable property, then it isn’t an argument at all.
8. #8 Macht
June 13, 2006
I don’t exactly see what distinction you are making. It can’t be that IC is a property, since other areas of science posit specific properties without proving them (e.g., properties of atoms). If it is the word “can’t” that you think makes the difference, I don’t think that is significant either. It certainly hasn’t been proven that nothing can go faster than the speed of light but we have no problem using the word “can’t” in that context.
Maybe you mean something else though.
9. #9 Mark C. Chu-Carroll
June 13, 2006
Macht:
My point is that WRT to irreducible complexity as a theory, if IC isn’t a provable property, then the theory has no meaning whatsoever.
As a theory, what IC says is: If there exists an observed system S with property IC, then evolution can’t have developed S. But if you can’t prove IC, then the IC statement becomes “If there exists an observed system S with a property that I can’t detect, then evolution can’t have developed S.” That’s a totally empty statement.
You can’t have it both ways. Behe wants to use IC as an argument that “proves” evolution can’t explain features of life. If it can’t “prove” that, then it becomes a worthless statement: “I can’t imagine how this system could evolve, therefore it couldn’t have evolved”. On the other hand, if it purports to prove (or demonstrate, if you prefer that term) that evolution couldn’t develop certain features, then the irredicibility of those features must be provable (or demonstrable). But if you try to prove/demonstrate the ICness of a system, then you fall into the IT irreducibility trap.
IC is meaningless as a theory, because it’s built on a fundamentally flawed concept. All that the proof I showed does is demonstrate why that concept (irreducibility) is flawed.
10. #10 secondclass
June 13, 2006
Behe’s original definition of IC: A single system which is composed of several interacting parts that contribute to the basic function, and where the removal of any one of the parts causes the system to effectively cease functioning”.
He admitted that this approach is defective, which it obviously is.
His later definition: An irreducibly complex evolutionary pathway is one that contains one or more unselected steps (that is, one or more necessary-but-unselected mutations). The degree of irreducible complexity is the number of unselected steps in the pathway.
Here IC is a property of an evolutionary pathway rather than a system, which renders it virtually useless.
(Definitions taken from here.)
11. #11 Bronze Dog
June 13, 2006
I’ll try my hand at explaining:
IDer: “Evolution can’t evolve flarschnikit features.”
Mark: “There’s no way to know if a feature is flarschnikit. Here’s why. (Programs) All the features they’ve declared to be flarschnikit are based solely on an inability to think of something more flarschnikity.”
Another angle I like to bring up myself:
1. IC is a negative premise. (This thingy is not reducible)
2. The idea that evolution can’t evolve IC is a negative premise.
3. ID is an affirmative conclusion based on those negative premises.
4. Affirmative conclusions from negative premises are a no-no in logic. Especially if you don’t know all the possibilities.
12. #12 Anonymous
June 13, 2006
There seems to be a serious flaw in the above proof: if you go through running every minimal program to see how long its output is, as soon as you run across *one* non-halting program, c’ gets stuck and never gets past trying to decide how long its output is. Therefore it may not output anything, let alone output the same thing as K. You can’t fix this by examining only programs that halt, because then you’d need a way of deciding which ones those are.
Another flaw: the fact that the OUTPUT of K is longer than the PROGRAM c’ doesn’t prove that the PROGRAM K is longer than the PROGRAM c’. The output of K can easily be larger than K itself.
Ultimately, though, even if your proof can be fixed, it still isn’t applicable to IC. I think you’re slipping a quantifier. You can’t *always* decide minimality in finite time, but that doesn’t mean you can’t *ever* decide it in finite time. Some systems could be provably minimal, others could be provably non-minimal: there are definitely *some* systems whose minimality is undecidable, but not necessarily *all* of them.
You’re going from
~EP: Ax: P(x)=1 if x is minimal, 0 otherwise
(No P can decide minimality of ALL programs)
to
Ax: ~EP: P(x)=1 if x is minimal, 0 otherwise
(No program can decide the minimality of ANY other program)
which is invalid.
IC still has other problems, of course. But this isn’t really one of them.
13. #13 Mark C. Chu-Carroll
June 13, 2006
anon:
You’re making two mistakes. First, you’re scrambling the output of a program and the program itself. And second, you’re assuming that this is a proof that a specific IC system (K) is not minimal.
(1) The program vs the output.
The “IC” system is a program that generates an output. The process of generating an output is the computation representation of the task performed by the IC system.
The process described above doesn’t run any of the programs it’s examining for IC. It’s examining them with a formal axiomatic system to determine if they’re minimal. (If they’re provably minimal, there’s a formal axiomatic system which can prove it, by definition.) It doesn’t run anything until it finds a “minimal system” larger than itself plus its data. That’s an allegedly IC system.
It doesn’t matter how large the output of K is. The point is that K is provably minimal; but something smaller than K can do the same thing as K (because “doing the same thing” means “generating the same output”). So K isn’t minimal.
(2) Proving something about a specific K vs proving a concept.
If you look at the proof, we don’t know what K is. In fact, we can’t know what K is. There is nothing wrong with the proof that K is minimal. The problem is that the concept of minimality itself is flawed. So it doesn’t matter how good the proof of alleged minimality/ICness for a particular system is. The concept is fundamentally flawed: any allegedly IC system might be K for some formal axiomatic system.
14. #14 ArtK
June 13, 2006
Echoing BronzeDog somewhat, my take on IC is that it’s like looking through the wrong end of a telescope. If evolution has a direction (not necessarily “forward,” let’s not go there), then IC is looking in exactly the opposite direction.
I read the concept as this: Given a state, S and all of its possible immediate predecessor states, Pm through Pn, S is IC iff none of the predecessor states has the same function as S.
This condition is not applicable to evolution. Co-option demonstrates that the predecessor’s function need not be the same as the succesor’s and neutral mutation demonstrates that even if non-functional, predecessor states need only be non-lethal. To prove something IC, Behe would have to demonstrate that all of the possible predecessor states were lethal. To disprove that something is IC, it is only necessary to show that there is at least one predecessor state that is non-lethal, which is a pretty low bar.
15. #15 Mark C. Chu-Carroll
June 13, 2006
ArtK:
You’re absolutely right about that – there’s a lot wrong with the IC idea. While I focused on one narrow mathematical point, that’s just one tiny little corner of what’s wrong with the IC concept.
IC as an anti-evolution argument is predicated on the fundamental notion that evolution behaves in a particular way, which has very little to do with how evolution actually behaves.
The IC argument assumes that there’s exactly one way that evolution produces features: via a strictly additive constructive process. No co-option; no reduction; no elimination; addition only.
So according to the IC argument you can’t produce a 3-part device by removing a part from a 4-part device. You can’t create a working device by modifying a different working device. But of course, that’s nonsense.
June 13, 2006
The process described above doesn’t run any of the programs it’s examining for IC. It’s examining them with a formal axiomatic system to determine if they’re minimal. (If they’re provably minimal, there’s a formal axiomatic system which can prove it, by definition.)
Uh, your program requires not only that minimality be always provable, but also non-minimality. I.E. “Is the program s string minimal under FAS?” is decidable, not just partially decidable (provable + disprovable, not just disprovable). Now if non-minimality was provable there’d be no problem; provability of minimality would then imply decidability of minimality. But it’s not obvious that non-minimality is provable.
This is partially fixable: if minimality is provable but not decidable, you can still write a program to enumerate minimal programs until it finds one longer than any given length. But the problem is that either way your proof gives a very weak result, namely that the set of minimal programs is not recursively enumerable. In other words, it is not possible to prove a program is minimal for every minimal program. This is not at all the same thing as for every minimal program being unable to prove that the program is minimal.
Your proof reminds me of the proof that the halting property of Turing machines is not decidable. This doesn’t mean that the halting property lacks meaning; it’s trivially provable for those combinations of programs and inputs that have it, and the set of Turing machines and & inputs that halt is recursively enumerable. Like the set of minimal programs, the set of non-halting programs & input is not recursively enumerable, but it is manifestly possible to find some elements of the latter set.
Similarly, you have not proved that it is impossible to prove a system has Irreducible Complexity. All you have done is proved that it can’t be done for every system.
17. #17 Mark C. Chu-Carroll
June 13, 2006
Andrew:
You’re missing the key bit to that proof.
In the proof about the Halting problem, you’re constructing a specific program for which your halting-detector fails. That is, you’ve identified a property that a program needs to posess to “outwit” a halting-detector, and constructed a program that posesses that property. We can find program for which we can determine whether they halt or not, because we can look for specific properties of the program that guarantee that they will halt. Whether we can determine whether or not a particular program will halt is dependent on the properties of that program.
In the case of minimality, we’re constructing a program which makes the proof of minimality of some arbitrary, unknown provably minimal program invalid.
That’s a huge, important difference. We aren’t constructing a specific non-minimal system which posesses a property that tricks the minimality identifaction process into failing. What we’re doing is constructing a system which undermines the entire concept of proving minimality. The specific system whose minimality proof gets undermined has no special properties at all. The only reason that it’s the one hit by the system is because it was in the right place in the entirely arbitrary ordering of provably minimal systems.
So there is no way of identifying a system whose minimality proof is actually invalid! Because the invalidity of the proof isn’t a property of the program, and it’s not a property of the proof. It’s a property of the concept of minimality.
18. #18 Macht
June 13, 2006
I think you are kind of playing fast and loose with the words “prove” and “demonstrate.” Scientific demonstration isn’t the same as proof. If I say that I’ve demonstrated that some atomic particle has some property, I’m not saying I proved it, I’m saying that I’ve hypothesized it and tested it over and over. My demonstration of this property could turn out to be wrong tomorrow if a better theory comes along.
All you’ve said in this post is that you can’t prove that a system is minimal. You haven’t showed that you can’t hypothesize IC for some system and test it.
Going back to your post, you say:
“The fundamental result is: given a system S, you cannot in general show that there is no smaller/simpler system that performs the same task as S.”
This just isn’t a problem if we are doing science (as opposed to math). You can just hypothesize that there is no simpler system and test it. If it stands up to testing, helps you make useful predictions, etc. then scientists will use the theory. If not, then they will forget it.
19. #19 steve
June 14, 2006
So, the claim is that any claim to demonstrate minimality is invalid, right? Let’s look at a concrete example. Wouldn’t the program that consists of only the STOP symbol be a minimal program to output the empty string? (Or if that doesn’t work, say a two symbol program “output 1” and STOP). How does your proof undermine their minimality? I can’t see how there could be a shorter program with the same output – maybe I’m missing the intuition.
20. #20 Mark C. Chu-Carroll
June 14, 2006
Macht:
I think you’re deliberately missing the point – playing a very Dembski-ish game by playing with meta-issues in order to obscure the point that the entire core of an argument is fundamentally flawed.
I’ve run out of different ways to say it. I’ll just try to restate it in the simplest way I can.
(1) Behe claims to “prove” that certain systems can’t involve, because they have the property that they are irreducibly complex.
(2) If you cannot prove that something is IC, then Behe’s argument degenerates to “Certain systems can’t evolve because Behe says they can’t evolve”.
If you can prove that something is IC, then Behe’s argument hits the information theoretic problem with minimality. If you can’t prove that something is IC, then Behe’s argument becomes nothing but an unsupported bare assertion.
21. #21 Mark C. Chu-Carroll
June 14, 2006
steve:
If you look at the proof, there is a critical threshold, based on the size of the formal axiomatic system (FAS) that proves minimality. The problem with minimality kicks in as soon as the complexity of the system becomes such that the length of the meta-program plus the size of the FAS is smaller than the information-theoretic complexity of the system.
Tiny systems – like a single-instruction program – are probably smaller than that threshold. (I’ll explain the probably bit in a moment.) But we can’t know where that threshold is: because finding the threshold requires finding the minimum size of a FAS that can be used to show mimimality. And that’s the minimality problem biting itself.
The probably thing up there is because of one important thing that we do in information theory. The IT complexity of a program is dependent on the size of the program plus the IT complexity of the machine that it runs on. So a single instruction program can be a very complex beast if it’s a single instruction for a complex machine. To show a trivial example of that, you can imagine a machine which is basically a vonNeumann computer (essentially the mathematical formalism for real computers), but which has one instruction that causes it to output the first 300 digits of the square root of a number. Then computing 300 digits of an irrational number takes only one instruction – but in IT terms, it’s a very complex operation.
June 14, 2006
Wow, this was educational.
First, since I learned a natural consequence of Gödel’s theorems. (I’ve learned to accept the theorems and use them beneficially. But they were not exactly intuitive the first time around. This application is.)
Second, since I learned that IC is as empty as CSI. IC has been special since it isn’t really a one-one correspondence to ID. There has also been plenty of cases were allegedly IC cases has been falsified. So it is satisfying to find that IC is as vacuous as general creationist concepts.
Third, since Bronze and Art also gave new insights.
I find it comical that this creationist concept fails because it algorithmically asks for a perfect solution that isn’t there, meanwhile evolution itself succeeds splendidly while it algorithmically has the decency to ask for good enough (surviving) solutions.
23. #23 SOBrien
June 14, 2006
I don’t think you understand Behe’s argument. He is only talking about evolutionary pairs differing by a single mutation.
The point is this: when a complex system such as blood clotting evolves by a mutation the resulting system is inferior and will lead to extinction of the new species. Mutations are symmetric in time so if it cannot evolve forward then it did not evolve from a precursor.
Proving that there is an unrelated system which does the same thing is irrelevant. You are only allowed to study 2 systems linked by a simple mutation, and that mutation must be a real genetic event.
June 14, 2006
In the case of minimality, we’re constructing a program which makes the proof of minimality of some arbitrary, unknown provably minimal program invalid.
Where I’ve been getting tangled up is in the conditions necessary for constructing such a program. I now see that it’s enough for an infinite number of programs to be provably minimal under FAS for such a program to be constructed. Where I was going wrong was thinking that the program was constructed under the assumption that all minimal programs are provably minimal.
… So there is no way of identifying a system whose minimality proof is actually invalid!
Yes there is: c’ finds such a system. All we have to do is construct and run a slight modification of c’, that outputs K instead of φ(K) to tell us what that system is.
Because the invalidity of the proof isn’t a property of the program, and it’s not a property of the proof. It’s a property of the concept of minimality.
You’ve got bigger problems: the minimality proof isn’t invalid. (c’ is constructed to find to find a system that is provably minimal if your original argument is to make sense: i.e. a valid proof exists). What is happening is that you’re proving something that isn’t true; FAS is inconsistent. (This is the contradiction part of proof by contradiction). If FAS is consistent, then it must not be possible to construct c’.
So why is in not possible to construct c’? Certainly one possibility is that no minimality proofs exist at all. But that’s not the only possibility. c’ is based on the assumption that there is at least one provably minimal program longer than l(c,FAS) + length(c’). This is trivially true if the number of provably minimal programs is infinite, … but not necessarily true if the number of provably minimal programs is finite.
So ultimately, what you’ve proved is that there is at most a finite number of programs that are provably minimal under FAS, and provided an upper bound on their size. While a very interesting result, it’s not a problem for the concept of minimality; perhaps the rest of the minimal programs are simply not provably minimal under FAS, that is, FAS is incomplete. The incompleteness of FAS is not something that Godel hasn’t already proved.
Which gets back to earlier comments:
Once you cross the complexity threshold where the complexity of your system exceeds that of K in the proof, no proof of minimality has any meaning, because minimality itself has no meaning; and to make matters worse, you can’t determine where that point is!
If FAS is consistent, once you cross the threshold of K you won’t be able to find any proofs of minimality for systems.
… There is nothing wrong with the proof that K is minimal. …
… which means that FAS is inconsistent. You had better not be able to find K in the first place. Which means (if FAS is consistent), there are no provably minimal programs at all beyond some threshold.
Where I disagree is that not being able to find that threshold poses a problem: If you can find a proof of minimality you are below the threshold. If you’re greater than l(c,FAS) + length(c’), you’ve over the threshold. If both are true then FAS must be inconsistent and you’ve got bigger problems.
BTW, in your proof, when you say,
… where the length of the output of φ(K) is larger than l(c,FAS) + length(c’).
I think that should be:
… where the length of K is larger than l(c,FAS) + length(c’).
25. #25 Mark C. Chu-Carroll
June 14, 2006
SObrien:
I understand Behe’s argument. It’s a thoroughly bogus argument. I’m just pointing out one mathematical problem with it here; I’m not trying to address all of its problems at once.
The point is this: when a complex system such as blood clotting evolves by a mutation the resulting system is inferior and will lead to extinction of the new species. Mutations are symmetric in time so if it cannot evolve forward then it did not evolve from a precursor.
The problem with this argument is, as I explained in another comment, that it is a strictly additive constructive argument. That is, it assumes that the only way to evolve to a given system is by adding parts in single steps.
The basic IC argument is predicated on several distinct things. Among them:
(1) I can show that for a given system, it has the minimum number of parts necessary to perform its task. This is what the IT argument refutes.
(2) The only way that an IC system could evolve is by a process of adding parts: that is, if my “IC” system has 4 parts, then if it evolved, it must have evolved from a system that had 3 parts in a single step that added one part. This is the part that you’re focusing on. It’s a separate part of the IC argument. It’s *also* bogus, but for a different reason. Real systems evolve in many different ways – evolution is not a directed additive process. Evolution can add parts, copy parts, remove parts, modify parts, and co-opt parts. If parts can be removed, modified, or co-opted, then the argument that IC can’t evolve fails.
Suppose you have a 3-piece system that you claim is minimal and couldn’t have evolved, because if you remove any of the three pieces, it wouldn’t work.
What if there’s a 5-piece system performing some other task that can be made to perform the same task as your 3-piece “IC” system via a single mutation? IC doesn’t address that. Now, what if the 5-piece system can evolve to a 4-piece system in a single mutation? And then the 4-piece to a 3-piece via a single mutation?
That’s evolution of an IC system via co-option and reduction.
The IC argument fails if co-option and reduction are possibilities. And both co-option and reduction are real phenomena that have been observed. (For example, see Tara’s recent piece here.
26. #26 Mark C. Chu-Carroll
June 14, 2006
Andrew:
This is getting beyond the scope of what I think is easily discussed in blog comments. We’re moving from the information theory argument which is an application of a Godel-like proof to the fundamental Godel issues of incompleteness. The answer to your objection lies in the completeness properties of the axiomatic system, which are themselves intrinsically problematic.
I’d suggest looking at one of Greg Chaitin’s books, as I think he’s a lot better at explaining this than I am. “The Limits of Mathematics” is the book where I first saw this proof; the original post also includes a link to Greg’s website where he has his version of this proof. Greg does address the issues you’re asking about. (And he does it with a kind of clarity and flair that I can only dream of.)
27. #27 Steve
June 14, 2006
(1) OK, so given the possibility of a proof below the threshold (and since Andrew said having a proof is sufficient to show that you’re below the threshold) then it seems your proof is demonstrating that if IC were to be feasible to prove, it’d probably only work for simple systems – but is not demonstrating that no system exists for which IC is provable.
(2) More importantly, given Behe’s restriction to additive mutations, I think he’s not requiring full minimality over all smaller programs (which your proof assumes), just minimaility over the subset that are potential precursors. This is a much weaker notion of minimality. Moreover, I think being over the threshold for general minimality needn’t imply that it is similarly impossible to prove minimality over the restricted subset. Proving restricted minimality should be a much easier problem. I think his evolutionary error might get him out of his IT problem.
28. #28 Mark C. Chu-Carroll
June 14, 2006
Steve:
Having a proof isn’t sufficient to show you’re below the threshold. That’s only true if you have an axiomatic system which is both complete and consistent. But per Godel, no axiomatic system is complete and consistent. If you have an incomplete axiomatic system, then the axiomatic system isn’t powerful enough to do minimality proofs. If the axiomatic system is powerful enough to do minimality proofs (i.e., it’s complete), then it’s by definition inconsistent. (See Chaitin for why you need a complete axiomatic system.)
For the other part – yeah, if you accept the invalid restriction that undermines Behe’s entire argument, then you don’t need the information-theory stuff to show that it’s invalid. But if you do that, you’re making the argument moot anyway, because you’re building your argument on a false axiom.
29. #29 Mark C. Chu-Carroll
June 14, 2006
Ok folks, please stop the email that I copped out on Andrew
Andrew’s argument, summarized by me: (I’m using blockquote for formatting; this is not a literal quote of Andrew, but my attempt to summarize; please correct me if I’m summarizing badly):
If you have a consistent axiomatic system, then it won’t return any minimal programs beyond the threshold of the length of the axiomatic system plus the meta-program. So the minimality results that the consistent system generates are correct.
The problems:
1. We don’t know what the threshold is. Regardless of whether the axiomatic system is complete or consistent, it can’t know where that threshold is. To know that threshold is impossible meta-knowledge.
2. An axiomatic system that reasons about programs in an effective computing system encodes the semantics of that computing system in its rules; if the computing system is an effective system, then the axiomatic system is, by definition, complete.
3. If the computing system is not an ECS, then there are reasonable systems that are arbitrarily excluded from the minimality space; a proof of minimality in a non-ECS does not provide that you have a minimal system.
4. The systems of biology – that is, DNA and the processes that it drives in a real biological system – have been shown to be an ECS. (DNA is turing complete – see some of the work IBM has done experimenting with DNA computers.)
30. #30 Steve
June 14, 2006
I agree that restricting to mutation by addition is an incorrect assumption… but wouldn’t the restriction to the subset of predecessor programs from all the kinds of mutations we’ve seen (addition, subtraction, cooptation, etc.) still be a weaker condition than full-blown minimality over every program, thus avoiding the IT problem?
31. #31 Mark Chu-Carroll
June 14, 2006
Steve:
The idea of “restriction to the subset of predecessor programs from all the kinds of mutations we’ve seen (addition, subtraction, cooptation, etc.)” isn’t quite well-defined enough.
You can define “predecessor programs” in a number of different ways. Do you require “predecessor programs” to be a single point-mutation difference from a previous step? Must every “predecessor program” be a fully-functional system performing some function necessary to the survival of the organism?
Those sounds like weaselly questions, but they really aren’t. Real evolution is an amazing process, which has a lot of complexity to it that we don’t normally think of. If we allow the full range of evolutionary modification steps, in the way that they really occur – then the predecessor requirement is not a problem. But one of the tricks that frequently gets played by people like Behe and Dembski is to create artificial restrictions on the process that do not have any basis in reality.
Let me give one example, to try to make clear what I mean. In a real biological system, we frequently see patterns like duplication and cooption, where one mutation produces a duplicate of some element of the genetic code, which is useless but harmless; and then generations later, a mutation coopts one of the duplicates to perform a new function. So you have steps in the process that do not have any intrinsic value to the organism. They’re neutral: they provide no immediate value to the organism, but they also provide no immediate harm. But they open up the possibility of later changes. A lot of ID/IC types would exclude this scenario, because it contains intermediate steps that have no function. They focus on single systems, rather than the organism as a whole, and say “what function did the predecessor of this supposedly IC system perform?”; and if there’s no answer to that question, then they assert that it doesn’t count as a predecessor.
If you make those kinds of restrictions on the transitions, then you wind up with a situation where the IT problem doesn’t really come into play, because you’ve excluded so many paths. But those paths are real – and by excluding them, you’ve created an artificial system that doesn’t model reality. If you allow the full range of mutation and adaptation – then you can find paths to essentially any system; and then the IT problem comes into force.
June 14, 2006
If you have a consistent axiomatic system, then it won’t return any minimal programs beyond the threshold of the length of the axiomatic system plus the meta-program. So the minimality results that the consistent system generates are correct.
That’s an accurate summation.
An axiomatic system that reasons about programs in an effective computing system encodes the semantics of that computing system in its rules; if the computing system is an effective system, then the axiomatic system is, by definition, complete.
Ah, this I think is the heart of the disagreement: I don’t think the axiomatic system will/can be complete.
I think it is time for me to find one of Greg Chaitin’s books; we’re starting to get into technical issues that I’m not really familiar with.
June 14, 2006
SoBrian,
The point is this: when a complex system such as blood clotting evolves by a mutation the resulting system is inferior and will lead to extinction of the new species.
Mark has already addressed the symmetry of mutations in time: It’s just not so. Perhaps Behe is making the common error of assuming the only type of mutations are point-wise mutations? But there is another problem with this argument. If the original mutation resulted in a superior system, then reversing the mutation will result in a inferior system, that will go extinct. But this does not mean that the original (inferior) system could not have existed in the first place, after all it didn’t have the superior system to compete against. And it does not mean that no mutations will result in an even better system, I do not know of what confusion could have lead to that conclusion.
34. #34 Macht
June 14, 2006
Mark,
I’m not “deliberately missing” anything. That kind of talk is really unhelpful and doesn’t get us anywhere. I’ve understood your points since the first time you wrote them and I genuinely think you are wrong.
“1) Behe claims to “prove” that certain systems can’t involve, because they have the property that they are irreducibly complex.”
I really don’t think he does. Where does he claim this?
“(2) If you cannot prove that something is IC, then Behe’s argument degenerates to “Certain systems can’t evolve because Behe says they can’t evolve”.”
No, it then makes his argument “Certain systems can’t evolve because Behe hypothesizes (as opposed to proven) that they are IC.” (Which happens to have been his argument all along.) Behe makes a hypothesis and scientists test that hypothesis and if it holds up, it holds up. If it doesn’t hold up, it doesn’t.
35. #35 Anonymous
June 14, 2006
I agree with Andrew: you’ve proved that the minimality filter c can’t be complete (and in fact can’t prove the minimality of anything past a certain size/complexity threshold), but you haven’t proved that minimality can’t be defined at all. In fact, minimality is very simple to define, even though it’s computationally undecidable *in general* in any given finite system.
c can be accurate in the sense that any program verified as minimal *is* in fact minimal, but by the above proof it can’t prove the minimality of any system larger than some threshold size; therefore, it must miss infinitely many programs that are unprovably minimal.
It doesn’t matter if we know where the threshold is, or if the axiomatic system knows where the threshold is. We know there must be some minimal programs that aren’t provably minimal within the axiomatic system (infinitely many, in fact).
But that doesn’t undermine the validity of the proofs that the system *is* capable of. The provably minimal programs are still a proper subset of all minimal programs. And the minimality or non-minimality of any given program *is* well defined – there are only finitely many shorter programs, each of which is either equivalent or non-equivalent.
It’s just sometimes impossible to determine which in finite time.
P.S. Your original post says that “the length of the output of ?(K) is larger than l(c,FAS) + length(c’).” where I think you mean the length of K itself is larger, which confused me before. You might want to correct that. I don’t know why signing in via TypeKey made my post show up as anonymous – it doesn’t do that on other ScienceBlogs.
P.P.S. Like Andrew, I’m also going to be heading to the library to see if they have any of Chaitin’s books, to pursue this further.
36. #36 Mark C. Chu-Carroll
June 14, 2006
Macht:
If Behe is *not* arguing that the existence of IC systems proves evolution is impossible, then just what do you claim he is arguing?
If his argument *is* that the existence of IC systems proves evolution is impossible, but you can’t ever prove that a system is IC – then what value does his argument have?
We can’t *prove* relativity in the mathematical sense. But there are specific predictions of relativity where we can *prove* that reality matches relativities predictions; if it’s wrong, there will be specific things we can do that will *prove* that relativity is not an accurate description of reality. That’s what makes relativity valuable as a scientific theory.
What value does Behe’s whole IC argument have if it can’t ever be proven or disproven?
Finally, what is Behe’s entire book about, if not proving that evolution can’t explain life?
37. #37 Macht
June 14, 2006
“If Behe is *not* arguing that the existence of IC systems proves evolution is impossible, then just what do you claim he is arguing?”
Ummm, I thought we were talking about proving that something is IC, not proving that evolution is impossible? As I said in my last post, Behe is putting forth a *hypothesis* (not a proof) that he thinks shows that NS+RM isn’t sufficient for certain biological features to come about. But I would still like to know where you got the idea from that Behe is trying to prove anything. You haven’t answered that.
“If his argument *is* that the existence of IC systems proves evolution is impossible,…”
It isn’t, as far as I can tell.
“But there are specific predictions of relativity where we can *prove* that reality matches relativities predictions; if it’s wrong, there will be specific things we can do that will *prove* that relativity is not an accurate description of reality. That’s what makes relativity valuable as a scientific theory.”
See, this is why I said you were playing “fast and loose” with the word “prove.” You are essentially describing falsificationism. But it is well known that in philosophy of science that this neither proves theories right or wrong. Scientists come to accept or reject these theories in light of new evidence or better theories, but this isn’t proof. Here is why. Your logic looks something like this:
If some theory T is true, we would expect to see X.
We see X.
Therefore, theory T is true (that is, proven).
This is, of course, not a valid argument (affirming the consequent).
The part of your logic is:
If some theory T is true, we would expect to see X.
We don’t see X.
Therefore, T isn’t true.
This is, of course, Popper’s famous modus tollens argument for falsificationism. It is valid, but unfortunately it isn’t sound. If we don’t see X, it may be true that T isn’t true, but it could also be that part of our experiment went wrong causing us not to see X, or it could be that some of our background information causes us to think T would expect us to see X when we really wouldn’t, or any number of other things. So, in this case we have’t proven anything, either.
“What value does Behe’s whole IC argument have if it can’t ever be proven or disproven?”
In my very first post I said “proof is for mathematics and alcohol.” What I wrote above explains why science doesn’t deal with proof. So the fact that Behe’s argument can’t be proven or disproven is irrelevant to whether it is valuable as science or not.
“Finally, what is Behe’s entire book about, if not proving that evolution can’t explain life?”
As I’ve said, I read it as a hypothesis or theory about whether evolution can explain life. As I’ve also said, the way to go about dealing with a theory or hypothesis is to test it.
38. #38 Mark C. Chu-Carroll
June 14, 2006
Macht:
This is exactly why I say you’re just playing word games and deliberately avoiding the real issues. This isn’t an argument about falsificationism, Popperism, the philosophy of science, or the things that can go wrong in experimental observations.
My entire point, from the start, has been that Behe’s argument is based on the invalid premise that he can show that some things are irreducibly complex.
By shifting words, and saying things like “Behe’s argument isn’t about trying to prove that evolution can’t evolve IC”, but instead “Behe is trying to propose a hypothesis about whether or not evolution can explain life”, you’re just deliberately obfuscating and avoiding the point.
You can say that “Behe is trying to make an argument about whether or not evolution can explain life. Specifically, his argument is that there are some things that can’t evolve.” Or you can say “Behe is trying to propose a hypothesis about whether or not evolution can explain life. Specifically, his hypothesis is that there are some things that can’t evolve”. In any essential way, how does the second statement differ in meaning from the first? Substituting “argue” for “prove”, or “hypothesis” for “argument” is just word games.
He’s trying to put together a logical argument where if his axioms are true, then his conclusion is true. You can call that an argument, or you can call that a proof. It *does not matter* which words you choose. The basic point is: either he *does* have a argument where the conclusion follows from a set of true axioms via valid logical reasoning; or he does *not*.
One of his axioms is “I can show that some biological systems are irreducibly complex”. If he can’t do that, then the argument fails. Call it argument or hypothesis or proof, call it hypothesis or call it theory; it doesn’t matter.
If “irreducibly complex” is a meaningless term where you can never show whether or not anything is irreducibly complex, then any argument based on being able to show that something is irreducibly complex is nonsense.
39. #39 Troublesome Frog
June 14, 2006
Macht:
I think the point is that Behe’s argument is purely a logical one, not an empirical one. If IC systems exist, it is indeed a problem for evolution. However, a supposedly IC system is only a problem for evolution if it is definitely IC. If it only looks like an IC system, it’s no better than saying, “I can’t explain this, so evolution can’t have done it.” I think that most people intuitively knew that there’s no real way to tell if a system is IC. Mark’s entry is just pointing out that their disbelief can be formalized.
There’s no way of determining empirically whether a system could have evolved. If there is an essentially infinite number of paths to a result, you can’t simply say that because the obvious paths are wrong, there is no way to reach that result. Behe realizes this and posits IC systems as a special case: “Let’s pretend that there are evolutionary results to which all possible evolutionary paths are invalid. If that’s so, evolution fails.” That’s a perfectly fine piece of logical reasoning, but it doesn’t make for much of a book career.
Behe’s major failure is that he spends the rest of his time trying to come up with examples of such systems. The point is, all he can do is come up with systems that LOOK like they’re IC. He could spend his entire life listing the possible evolutionary pathways to one of his IC systems and pointing out how each of those pathways fails, but there’s no reason to believe that he has examined any significant portion of the possible pathways (and good reasons to think that he never could). No useful science comes out of it, so IC is nothing more than a logic game. Calling Mark for staying in logic land for something that can clearly never leave logic land is hardly fair.
40. #40 Macht
June 14, 2006
Mark,
I don’t know how to respond to you because you’ve basically ignored every point I made and then just threw a blanket “you’re obfuscating” retort on it. I point by point showed why your premises are wrong but you haven’t responded to them. I’m not sure what else I can say unless you actually address what I’ve argued.
I clearly explained the difference between a proof and how science works but you just gloss over it without saying why (again, “you’re obfuscating” or “you’re playing word games” isn’t a reason).
I talked about falsificationism and verification because *YOU* brought it up in your relativity example. I clearly showed how these *AREN’T* proofs because they are either invalid or unsound.
You *STILL* haven’t explained where you are getting these arguments from Behe’s work. You are just claiming that he makes them. For example, where does Behe claim that one of his axioms is “I can show that some biological systems are irreducibly complex”? And if you answer my question again with another question like “Well if he isn’t making that arugment, what argument is he making?” I’m going to scream since I’ve repeatedly said what argument he is making.
June 15, 2006
If IC systems exist, it is indeed a problem for evolution.
Solved problem. There are a number of well known examples of structures for which “the basic function” has changed. What is IC for one function isn’t necessarily IC for another. Also, while removing one part from an IC system will cause it to cease functioning (by definition), removing that part from its precursor will not necessarily have that effect: the precursor may have functioned slightly differently, and may well have been more complex. IC just isn’t a useful concept.
42. #42 Mark C. Chu-Carroll
June 15, 2006
Macht:
I think that you’re the one who consistently ignores every point I make, and replies with endless obfuscation and definitional games rather than addressing anything real.
I keep asking, what is Behe’s actual claim, if it isn’t that there are systems that have the property that they’re IC, and evolution can’t produce them?
And you keep dancing around the issue; substituting terms, arguing about whether proof is meaningful in science, and insisting that Behe isn’t arguing that he show that there are IC systems that by virtue of being IC can’t evolve. But all of your statements of what Behe actually says are just ways of rephrasing, from “Behe claims that certain systems can’t evolve because they’re irreducibly complex”, to “Certain systems can’t evolve because Behe hypothesizes (as opposed to proven) that they are IC.”
To me – that’s word games. It’s just substituting the word “hypothesizes” for “proves” or “demonstrates” in a way that lets you hedge and dance around the point, without ever addressing the fundamental issue that Behe’s argument has no value whatsoever.
I don’t care what words you use. The fundamental point remains: Behe’s argument relies on the assertion that some systems can’t have evolved because they are irreducibly complex. If the assertion/hypothesis/whatever that that those things are irreducibly complex, but that assertion/hypothesis/whatever word you like is nothing but a blind assertion, and you can in fact never verify that something is, indeed, IC, then it’s a totally worthless argument.
You can hedge all you want; switch words around all you want; play philosophy-of-science games all you want. But if Behe makes an argument that is, in your words, “Certain systems can’t evolve because Behe hypothesizes (as opposed to proven) that they are IC.” – that is a worthless argument. Because either: IC is a meaningful property, or it isn’t. If IC is meaningful, then he can provide evidence that something is IC – and since IC is a logical/mathematical property, he needs to show it in a mathematical way. If IC isn’t a meaningful property, and he can’t do anything but just assert/hypothesize that things have this IC property, but he can’t ever demonstrate/prove/whatever that anything has the property IC, or that in fact the property IC is actually meaningful in any way – then it’s a totally worthless argument.
43. #43 Bronze Dog
June 15, 2006
IC is a non-issue. Even if you could prove something’s IC.
44. #44 Macht
June 15, 2006
Mark,
The problem is that you seem to think that “prove,” “hypothesize,” “show,” “demonstrate,” and “assert” are all synonyms. They aren’t. What you did in your original post is a proof. Scientists don’t do that. They hypothesize theories and then test them. After a hypothesis has been well tested, then it is asserted to be true. I’m almost positive that you’ve taken enough math courses to give you some idea about what the difference between a “proof” and a “demonstration.” I’m also sure that you understand that demonstrations don’t constitute proof.
I understand that *YOU* don’t care what words I use. That seems to be the entire problem.
45. #45 Mark C. Chu-Carroll
June 15, 2006
Macht:
I’d say that the problem is that you are focusing on distinctions between words, rather than looking at the concepts that those words are talking about.
*It doesn’t matter* what words you use to describe Behe’s IC nonsense. The important point is that either he’s talking about a meaningful concept which he can demonstrate in a meaningful way; or he’s *not* talking about a meaningful concept that he can demonstrate in a meaningful way.
I’m arguing that his theory is based on a fundamentally meaningless concept.
You’re throwing around the distinction between proofs and tests, hypotheses and theory, etc. But if the fundamental concept is invalid, then proof versus test, hypothesis versus theory, and all of the other word games you’ve introduced are all distinctions without a difference.
If you want to play word games, that’s fine. Go ahead. But you’re ignoring the point that my original post is about, and the post that I’ve been arguing through this whole miserable thread: that IC as a theory is nothing but a facade over a meaningless concept.
June 15, 2006
“If his argument *is* that the existence of IC systems proves evolution is impossible,…”
It isn’t, as far as I can tell.
Neither would be quite correct:
A. Yes, if you read Darwin’s Black Box you see that I say the following, “Even if a system is irreducibly complex and could not have been produced directly, however one cannot definitely rule out the possibility of an indirect circuitous route. As the complexity of an interacting system increases though, the likelihood of such an indirect route drops precipitously.”
So here I was arguing well, there’s a big problem for Darwinian theory. These things can’t be produced directly, but nonetheless you can’t rule out an indirect route, but nonetheless building a structure by changing its mechanism and changing its components multiple times is very implausible and the likelihood of such a thing, the more complex it gets, the less likely it appears. So the point is that I was careful in my book to qualify my argument at numerous points, and Professor Miller ignores those qualifications.
So Darwinian theory can explain IC structures, but has a great deal of difficulty doing so. That’s some nice weaseling there as it means his argument cannot be defeated by a single counterexample. And he restricts IC to molecular systems, which we have a much poorer evolutionary record for than macroscopic systems. Poorer, but not non-existant; Darwinian evolution leaves clues in the genes, and in the molecular structure too, by which evolutionary history can sometimes be deduced. Even qualified, his argument has been shown to be wrong.
47. #47 Macht
June 15, 2006
But you haven’t shown that IC is a “fundamentally meaningless concept.” All you’ve done is show that you can’t prove that a system is minimal. That’s not the same as saying it is meaningless. That isn’t even the same as saying that a system can’t be minimal. A system could be minimal despite us not being able to prove that it is. This is exactly why I’m making the distinctions I’m making. A scientist would have no use for proving that something is IC, he would hypothesize that something is IC and then test to see if that hypothesis holds up.
48. #48 Mark C. Chu-Carroll
June 15, 2006
Macht:
Still the same old wordgames. I don’t give a damn if you rephrase it from “proving that something is IC” to “hypothesize something is IC and then test to see if the hypothesis holds up”. The fact remains that no test can ever demonstrate that a real system is IC. A scientist can’t test to see if a hypothesis of IC holds up. It’s a non-testable hypothesis, because the underlying concept is invalid.
No wordgames can change the fact that there IC is a facade with nothing behind it. It’s scientifically useless, because no test can ever confirm its predictions.
You are avoiding that point in favor of endless games of vocabulary shifting and meta-reasoning about word choices. IC is a mathematicaly meaningless and scientifically worthless – because no test can ever confirm its existence.
49. #49 Macht
June 15, 2006
Mark,
You have in no way shown that “the underlying concept is invalid.” All you’ve shown, as I’ve already said, is that you can’t prove a system is minimal. You haven’t shown that a minimal system is a meaningless idea and you haven’t even shown that some system can’t be minimal. The fact that I can’t prove a system to be minimal doesn’t mean it isn’t minimal.
June 15, 2006
SOBrien says:
“The point is this: when a complex system such as blood clotting evolves by a mutation the resulting system is inferior and will lead to extinction of the new species. Mutations are symmetric in time so if it cannot evolve forward then it did not evolve from a precursor.”
Besides being beside the point, as Mark explains, this is also a good illustration of the IT argument.
But first another point. You say erroneously that a mutation should make the system worse. First, there are neutral mutations. Second, we are discussing evolution here. The system is not optimised when it first clots blood. It continues to evolve towards better fitness. In this case it could be optimised clotting. (As good as possible without clogging arteries at the slightest provocation.)
Now to the illustration. The blood clotting system isn’t IC or even minimal as stated, since dolphins and fishes have variations of the same clotting system with fewer components. If you need references I think you find them on talkorigin or in The Pandas Thumb discussion.
51. #51 Mark Chu-Carroll
June 15, 2006
Macht:
Again, you quite deliberately miss the point.
What value does a so-called theory have if it’s relies on an untestable concept?
From the point of view of Behe’s IC theory: what value does IC have if you can never actually confirm that any IC system ever exists anywhere in biology?
Behe claims that the existence of IC systems is a problem for evolution. If we don’t know if the problem actually exists, and we can never confirm that it exists, then is it really a problem at all?
IC comes down to handwaving: *Behe says* that there are IC systems. Therefore, *Behe says* that there’s a problem with the theory of evolution. There’s no evidence for the existence of the problem beyond the fact that *Behe says* that there is. He can’t actually demonstrate its existence.
I can say that Behe must not have written the book that he claims to have written, because that book has the property of frobitzness, and religious people can’t write things that are frobitz. If I can’t tell you how I know that his book is frobitz, and I can’t tell you how to test if his book is frobitz, and I can’t tell you how to test if religious people can write things that are frobitz, then do I have any claim to a valid argument based on the frobitzness of his book?
52. #52 Macht
June 15, 2006
“What value does a so-called theory have if it’s relies on an untestable concept?”
Your post here doesn’t show that the concept is untestable. Your post (and I’m not sure why I’m repeating this again) only shows that you can’t prove it. (Oh yeah, I forgot, this is just a word game.) This is what I said above when I was talking about falsification and confirmation of relativity theory. You said that we can’t prove it in a mathematical sense and then I showed how it can’t be proved (or disproved) in any sense, since any proof will either be invalid or unsound. If you want to claim that IC is untestable, fine. But your post here doesn’t show that it is untestable. Besides, there are plenty people who have looked at IC and have ripped it shreds because the evidence is against it (aka, they’ve tested it and rejected it). So, it is testable.
“*Behe says* that there are IC systems. ”
Yes.
“Therefore, *Behe says* that there’s a problem with the theory of evolution.”
Yes.
“There’s no evidence for the existence of the problem beyond the fact that *Behe says* that there is. ”
Okay, I can deal with that. If there is not evidence, then that means its bad science and it should be rejected. This really has nothing to do with your post, though. In fact I said this a long, long time ago in one of my first comments in my thread:
“This just isn’t a problem if we are doing science (as opposed to math). You can just hypothesize that there is no simpler system and test it. If it stands up to testing, helps you make useful predictions, etc. then scientists will use the theory. If not, then they will forget it.”
And, again, I have to say that you *SHOULD* know the difference between demonstrate and prove, if my assumption is correct that you are fairly well versed in math. That you are confusing these two is just a poor, beginner’s mistake.
You said:
“I can say that Behe must not have written the book that he claims to have written, because that book has the property of frobitzness, and religious people can’t write things that are frobitz. If I can’t tell you how I know that his book is frobitz, and I can’t tell you how to test if his book is frobitz, and I can’t tell you how to test if religious people can write things that are frobitz, then do I have any claim to a valid argument based on the frobitzness of his book?”
See, these are all valid objections to Behe. Obviously, if I put forth a hypothesis and have no way to test the hypothesis, then science will reject it. But, again, this has nothing to do with your original post. Testing something is not the same as proving it. I’ve said this numerous times already, too.
June 15, 2006
Obviously Macht doesn’t get it. If you can’t test for the property in principle, you can’t make observations.
I’m no philosopher but I’m sure falsifiability is stronger than Macht does it. Special relativity would be disproved if we observe speeds faster than light, evolution would be disproved if we observe the fossil of a rabbit in a pre-Cambrian strata. (Ie the theory predicts ‘not X’, we find ‘X’, theory must be wrong.) But since we can’t make a test for IC from the hypotheses, we can’t observe it, so we can’t falsify it. (Not surprisingly, since Mark shows the property is invalidly defined, ie it can’t exist.)
“The systems of biology – that is, DNA and the processes that it drives in a real biological system – have been shown to be an ECS. (DNA is turing complete – see some of the work IBM has done experimenting with DNA computers.)”
Again, wow! The discussion isn’t a waste. I would love to hear more, and how ‘Information is Conserved’ (D*mbski) face up if DNA is turingcomplete, but evolution, say the RM+NS part, isn’t. (Or is it?)
54. #54 Macht
June 15, 2006
Torbjörn,
“(Ie the theory predicts ‘not X’, we find ‘X’, theory must be wrong.)”
I explained why this isn’t the case. Perhaps what is wrong is our belief that the theory predicts “not X.” Perhaps it doesn’t predict that at all. Perhaps what is wrong is “X.” Most modern day scientific “observations” are actually experiments. Experiments always have experimental error, so you can’t rule out that the experiment is wrong. And you can’t rule out that somebody made a mistake in setting up the experiment.
So, no, if theory T predicts “not X” and you see X, you haven’t disproved T. It probably isn’t a theory you would want to pursue anymore, but it isn’t disproven.
55. #55 Christian Albert
June 15, 2006
It would seem that IC is arguing for organized complexity as opposed to complex randomness. Doesn�t organization imply reducibility because organization is basically repeated patterns? If something was irreducible wouldn�t it be completely random apart from being mathematically impossible (Godel incompleteness theorem)?
56. #56 Bronze Dog
June 15, 2006
And you can’t rule out that somebody made a mistake in setting up the experiment.
So, no, if theory T predicts “not X” and you see X, you haven’t disproved T.
It sounds like Macht is banking on all of evolutionary experiments and observations being the result of dumb luck mistakes and bad experimental protocols.
57. #57 Mark C. Chu-Carroll
June 15, 2006
bronze:
Yeah, I think you’re pretty much right. Macht is playing a game of reductionism where nothing can ever be proven or disproven, and we can therefore not ever know anything.
If you’re willing to accept that kind of silly reductionism, then sure, Behe’s garbage is as valid as Einstein – i.e., both become nothing but a pile of unverifiable, untestable rubbish.
58. #58 Macht
June 15, 2006
“Macht is playing a game of reductionism where nothing can ever be proven or disproven, and we can therefore not ever know anything.”
Well, seeing as how I think we can prove/disprove some things and I also think that we can know things that we haven’t proven, you are wrong on both counts.
59. #59 Bronze Dog
June 15, 2006
…I also think that we can know things that we haven’t proven…
Sounds like faith, which is pretty much Behe’s basis for determining which things are IC.
Of course, even if that were a legitimate way of knowing, IC still wouldn’t be a problem. Evolution already has simple ways of making IC structures, and probably lots of clever methods we haven’t even conceived of yet.
60. #60 Macht
June 15, 2006
“It sounds like Macht is banking on all of evolutionary experiments and observations being the result of dumb luck mistakes and bad experimental protocols.”
Seeing as how I haven’t argued against evolutionary theory here (or anywhere, for that matter) and have just been arguing about whether Mark’s proof has any relevance to IC, this is also incorrect.
61. #61 Canuckistani
June 15, 2006
Macht, you wrote:
No, it then makes his argument “Certain systems can’t evolve because Behe hypothesizes (as opposed to proven) that they are IC.” (Which happens to have been his argument all along.) Behe makes a hypothesis and scientists test that hypothesis and if it holds up, it holds up. If it doesn’t hold up, it doesn’t.
Certain systems can’t evolve because Behe hypothesizes they are IC? I’m pretty sure that’s not what you actually want to assert. It’s no wonder to me that Mark thinks you are being deliberately obtuse.
I’m guessing that the assertion you want to make is more along the lines of, “Behe hypothesizes, but does not prove, that certain systems can’t evolve because they are IC.” Mark’s post describes a theorem that asserts, in essence, that no procedure for testing an arbitrary system for the property of IC exists. Macht, how can scientists test Behe’s hypothesis when it’s not possible, even in principle, to tell IC systems from non-IC systems?
June 15, 2006
Macht says:
“”(Ie the theory predicts ‘not X’, we find ‘X’, theory must be wrong.)”
I explained why this isn’t the case.”
No, you were discussing a weaker case.
Regarding the triviality about experimental verification, when we attest an observation or theory beyond reasonable doubt we apply an arbitrary, area contingent, but robust limit.
For example, to verify a phenomena in physics one must have more than 5 sigma certainty observations. What you are discussing is some arbitrary philosophy unreasonable doubt, not science as it is used and understood by practitioners.
63. #63 Bronze Dog
June 15, 2006
“Evolution can’t explain unicorns!”
“But there aren’t any unicorns to explain. At least none that have been shown to exist, yet.”
June 15, 2006
Macht,
Also, you are confused about the common usage of observation and experiment. A series of individual observations is an experiment. You use the observations to establish the experimental error. That is the certainty I spoke about in my earlier comment.
Mistakes and bad experiments (with undiscovered systematic errors, for example) happens, but these are sooner or later found by peer review, replicated experiments or too large discrepancy with theory.
65. #65 Macht
June 15, 2006
“Certain systems can’t evolve because Behe hypothesizes they are IC? I’m pretty sure that’s not what you actually want to assert.”
Yeah, that’s not what I wanted to say. Stupid mistake on my part. You are correct about what I was trying to say.
June 15, 2006
Christian Albert:
It would seem that IC is arguing for organized complexity as opposed to complex randomness. Doesn’t organization imply reducibility because organization is basically repeated patterns? If something was irreducible wouldn’t it be completely random apart from being mathematically impossible (Godel incompleteness theorem)?
Behe doesn’t mean incompressible by irreducible. He means mutual interdependence between all the parts. Which assumes that there is one, unchanging function to the system, because whether the parts are interdependent depends on which function they are performing. He doesn’t like it when others point out this assumption, you can see him do the razzle-dazzle in the transcript I linked to when the attorney did that. He also assumes that (Darwinian) evolution is mainly some sort of linear additive process (just keep adding on parts). (Mark noticed this). I haven’t read the whole transcript so I don’t know if the attorney ever called him on this assumption and forced him to make it explicit. If Behe is honest he states his assumptions upfront. I haven’t seen enough of his work to know if he does this, but I doubt it.
67. #67 Mark C. Chu-Carroll
June 15, 2006
Well Macht, I think we’ve finally managed to figure out where you’re coming from.
Well, seeing as how I think we can prove/disprove some things and I also think that we can know things that we haven’t proven, you are wrong on both counts.
So, just what is it that you purport we can *know* without any proof? I’m willing to bet that I know the answer if you’re honest, but I want to see if you’re willing to come out and admit it.
68. #68 Macht
June 15, 2006
“So, just what is it that you purport we can *know* without any proof?”
Physics, chemistry, biology, etc. There is lots of evidence and well-supported theories in those disciplines, but they don’t really deal proof. What is this, like, the 20th time I’ve said this?
69. #69 Canuckistani
June 15, 2006
Somewhat OT:
FWIW, there is at least one non-religious truth that can be known even in the absence of proof. You can play Godel games in plain English a lot easier than you can in number theory. For example:
Canuckistani cannot prove this statement is true.
It’s obviously true, but you’ll have to prove it yourself, since you can’t take my word for it.
In fact, a portion of Chaitin’s work in algorithmic complexity amounts to putting encoding the Berry paradox into formal number theory.
June 16, 2006
Andrew,
There is a long comment that discusses this at
http://www.pandasthumb.org/archives/2006/06/laudan_demarcat.html#comment-105140
The crux is:
“The long term evolution of most features of life has not been what Behe, or indeed most people, would call direct. And even short term evolution can be indirect in Behe’s terms. So it is surprising to read, on page 40, Behe’s argument against indirect evolution of IC systems. Here is the crux of it:
“Even if a system is irreducibly complex (and thus cannot have been produced directly), however, one can not definitely rule out the possibility of an indirect, circuitous route. As the complexity of an interacting system increases, though, the likelihood of such an indirect route drops precipitously.” (page 40) [Darwin’s Black Box]
He simply asserts that evolution of irreducible complexity by an indirect route is so improbable as to be virtually out of the question, except in simple cases. He makes no special connection between indirect evolution and IC. He offers no evidence. He just asserts that it is too improbable.”
So Mark’s elegant disembowlement of the testability of this concept clarifies Behe’s dishonest treatment further.
June 16, 2006
Falsifiability is a subtle matter. Many of the theories I was taught in physics were known to be wrong. Quantum Chromodynamics? disagrees with experiment on a couple of values. (In something like the sixth place sure, but still not explicable by experimental error). Also fails completely at very small scales and assumes a flat space-time. Nuclear theories (the liquid drop model, the shell model, the alpha particle in a well model)? All false. I believe some of them were even known to be false at the time they were introduced. General Relativity? Doesn’t know how to handle quantum superpositions. Newtonian gravity? Wrong. All “theories in trouble” as the IDers would have it. But not abandoned, because they are useful. They predict things that are true and don’t predict things that are false in some area better (more accurately, or more simply) than all their competitors. Each has a great deal of predictive power (albeit somewhat flawed). It’s not so much that falsifiability allows us to know when to discard a theory as that falsifiability is necessary for a theory to be at all useful in the first place.
June 16, 2006
I’m sorry, but I don’t see the connection between my last comment and yours. I was responding to your comment on Behe’s assumption of interdependence, or “direct” evolution as the Panda’s commenter called it.
If you are discussing my answer to Macht, I agree that the question of falsifiability is subtle. I haven’t studied Popper.
Four things. First, I think my comment to Macht is correct.
Second, the examples of failures you point out seem to me to be mainly theories outside their area of relevance or their singularities. All theories have both, and those doesn’t falsify them.
Third, I’m not sure how strong falsifiability is, but it is subtle as you say. We can look at string theory as an illustration of this.
Some say it hasn’t been tested.
But the theory is concocted by parts and methods that hasn’t yet been solidly mathematically connected with old theories. Still, it has been tested as correct for trivial properties and for nontrivial such as correspondence with semiclassical black hole properties. Further it explained channel properties in some particle experiments before an analogous flux tube method in QCD was invented. Or so they tell me. (And I don’t know enough to see if either the string or the QCD explanation was contorted or natural.)
So some others say it has been tested against falsifiable predictions, both theoretical and experimental.
Finally, when asked what would falsify evolution, the answer in talkorigin from a biologist was a rabbit in preCambrian strata. I think I can say that supraluminal speeds would falsify relativity, not move the area of relevance. For example, gauge theories become unstable in such circumstances ( http://arxiv.org/PS_cache/hep-th/pdf/0606/0606091.pdf )
In summary, I don’t know exactly how strong falsifiability is, but I agree that it seems to have uses. If you follow my earlier link, the discussion has mentioned that demarcation criteria seem to be contingent. Since that is coherent with my view of science as hard to describe workings and possible results off, I can live with that.
June 17, 2006
I’m sorry, but I don’t see the connection between my last comment and yours.
Sorry, yes there is no connection and I screwed up in not given any context at all to my comment. I have nothing further to add to your last comment, it expresses the comprehensive nature of Behe’s wrongness so well. I was indeed discussing your answer to Macht.
Four things. First, I think my comment to Macht is correct.
Yes. However the implications of a theory being falsified are rather different then is often assumed. I used your comment to go off on a tangent of my own that I think is quite important, unfortunately I did not make that clear. A pre-Cambrian rabbit (or rather many pre-Cambrian rabbits, one being possibly a fake) will falsify RM+NS, but by itself it will not cause the theory to be abandoned.
Second, the examples of failures you point out seem to me to be mainly theories outside their area of relevance or their singularities. All theories have both, and those doesn’t falsify them.
That’s a bit circular, the failure points of these theories plays a part is determining what their area of relevance is. And even inside the area of relevance, there generally are some discrepancies with observation. The discrepancies are slight, but Newtonion gravity does not correctly predict the motions of the planets; its original domain. The errors of the nuclear theories are not slight; often being more than an order of magnitute for decay rates. (And more than that the weaknesses of the nuclear theories on theoretical grounds are fairly well understood; slight tweaks will not save them.)
Third, I’m not sure how strong falsifiability is, but it is subtle as you say. We can look at string theory as an illustration of this.
String theory is an illuminative example. Because string theory is not in fact falsifiable; it is much like ID in this regard (Individual string theories may well be falsifiable, and it sounds like there has been very limited but nonetheless exciting progress in making them so.). But string theorists don’t expect their theories to be accepted until they are falsifiable.
Opinions may vary on whether string theory or ID are a promising ideas (and they do), but the string theorists have some hope of creating useful theories if there is something to string theory. No so for the current bunch of clowns promoting ID, who seem either incapeable or uninterested it creating something substantive.
There is one other observation you made that bears on falsifiability: “[string theory] hasn’t yet been solidly mathematically connected with old theories.” Because if/when string theory ever supplants the old theories it will have to explain why the old theories appear to be true. Because while the old theories may be false, they reflect a real pattern in the data. Similarly, the theory of Evolution reflects a real pattern in the data even if pre-Cambrian rabbits are found, and any replacement to RM+NS is going to have to explain that pattern. RM+NS is falsifiable, but the theory is always going to be pretty good at explaining the fossil record. The epicycles of Ptolemy’s successors have been abandoned, but the regularities they described have not disappeared. ID has done a fairly good job of fooling people into thinking some of the regularities described by the tree of life aren’t there, but they are.
I think I can say that supraluminal speeds would falsify relativity, not move the area of relevance. For example, gauge theories become unstable in such circumstances ( http://arxiv.org/PS_cache/hep-th/pdf/0606/0606091.pdf )
Well, the area of relevance would be subluminal speeds, no? I don’t disagree that relativity would be falsified (so long as supraluminal speeds are approprately qualified), but it would nonetheless still have a wide area of applicability.
In summary, I don’t know exactly how strong falsifiability is, but I agree that it seems to have uses. If you follow my earlier link, the discussion has mentioned that demarcation criteria seem to be contingent. Since that is coherent with my view of science as hard to describe workings and possible results off, I can live with that.
Agreed.
74. #74 Anonymous
June 18, 2006
“A pre-Cambrian rabbit (or rather many pre-Cambrian rabbits, one being possibly a fake) will falsify RM+NS, but by itself it will not cause the theory to be abandoned.”
As I said, science, its theory and its demarcation is contingent. This was the opinion of a biologist – if we observe these rabbits, common descent with modification (ie all evolutionary theories) must be wrong.
“That’s a bit circular, the failure points of these theories plays a part is determining what their area of relevance is.”
The failure points (singularities, area of relevance) are predicted by the theories, not observed – no circularity that I can see. They help falsification.
“And even inside the area of relevance, there generally are some discrepancies with observation.”
Experimental error is known and controlled. For your first example, once relativity was known the area of relevancy for Newtonian gravitation (not gravity) was predicted. The theory isn’t falsified but is continued to be used as a convenient approximation within its area of relevance.
The situation you describe for nuclear theories sounds like they aren’t the final explanation, ie the theories aren’t complete, but used until a better theory comes about. This seems like a genuine problem for falsification fantasts – uncomplete theories are allowed. (Ad hocs can’t be a big problem.)
“Because string theory is not in fact falsifiable;”
This is wrong. It is even contested that string theory already has been falsifiable tested. The reason that it is wrong is that it has falsifiable predictions before or latest at Planck scale.
I have noted that philosophers distinguish between falsifiability and testability. Testability is logical falsifiability and practical observability. String theory is logical falsifiable, but the practical observability is contested. Especially since some scientists prefer to allow tests against old theory.
“it is much like ID in this regard”
A contemporary discussion, which is supported by Mark’s observations in this post and elsewhere, is found under the link on demarcation I gave above and centers around the vacuity of ID. Ie while string theory is well defined, connects with old theory and thus old observations, and gives predictions like AdS/CFT, ID does neither of these things.
“Because if/when string theory ever supplants the old theories”
Note that QM and thermodynamics et cetera aren’t going to be supplanted by string theory. String theory will have to make a good connection to them.
“Well, the area of relevance would be subluminal speeds, no?”
Not if relativity predicts subluminal speeds, and it does. Newtonian gravity never made any prediction about light and causality – that was its problem.
June 20, 2006
Sorry, I have no idea why my last comment was anonymous and why i didn’t see that at the time.
June 20, 2006
Rereading my last comment, it is inconsistent of sorts, since a greater use of falsification is used.
Newtonian gravity wasn’t falsified, merely constricted, since the new theory could explain it. This is a contingent use of falsification of course. I agree that when observations of Mercury had problem, the old theory wasn’t falsified outright, but it was understood that it wasn’t complete. Compare with phlogiston that could be falsified outright.
I think Nick Matzke noted an analogous problem on the thread on Laudand and demarcation: “I think that this kind of critique of falsificationism, although very common, is always extremely tenditious because anyone who ever advocated falsifiability recognized that the whole point of it was that many claims could be shown to be false and excluded from the body of science because they were false. The problem that falsificationists were legitimately pointing to was that some claims are “not even wrong”, which sometimes (as with the “God/ID did it” claim) is more pernicious than being wrong.”
77. #77 Anonymous
February 7, 2007
The problem is that you seem to think that “prove,” “hypothesize,” “show,” “demonstrate,” and “assert” are all synonyms. They aren’t.
Sigh. It’s sad how badly the point was missed here. Mark, not being an imbecile, obviously does not think that. What Macht failed to understand is that Behe’s argument — that there are systems that cannot evolve — can go through (ignoring all the other problems with it) if he can prove that there are IC systems. It’s not enough to hypothesize or assert that there are IC systems, because that is simply an opinion, with no logical force. And what could possibly count as a “demonstration” that a system is IC, other than a proof that it is? “system S is IC” is not the sort of proposition that is amenable to empirical demonstration.
What you did in your original post is a proof. Scientists don’t do that.
What ridiculous poppycock. Scientists deal with both empirical claims, established through observation, and analytical claims, established through logical or mathematical proof. But it seems that Macht would have scientists kicked out of the academy for applying modus ponens. Here’s a clue: every time you add two numbers, you perform a proof. How else do you obtain the result? It’s not simply a hypothesis or assertion, nor is it an empirical demonstration. a: first addend = 1 (premise). b: second addend = 2 (premise). c: 1 + 2 = 3 (theorem of arithmetic). d: first addend + second addend = 3 (a, b, c, Liebniz’s Law). Q.E.D.
78. #78 Anonymous
February 7, 2007
can go through
Arrgh! Make that “can only go through”.
February 8, 2007
Macht, if any of the suggestions you make were true, IC can show nothing in regards to evolution. If Behe is “hypthesising” that some things MIGHT be IC, and thus MIGHT be a problem for evolution, then what makes IC any better than, say, “I hypothesise that some biological systems may be magical, and this might be a problem for evolution.”
A bold assertion that something might exist, if there’s no way to tell if it does or doesn’t, or even to accumulate evidence either way, is MEANINGLESS.
80. #80 Xanthir, FCD
June 8, 2007
Okay, someone just linked over here from the Pharyngula comment section, and I read over the section again.
Mark, you know I love you, and have been a long-time reader, but I think you’re a bit off-base here.
Disclaimer: Anything Behe said is probably wrong (ignoring the fact that he’s always wrong, of course). My definitions are my own.
Now, an IC system would simply be a system where, if you knocked out any piece, it stops working. It is not necessarily minimal! There may very well be a smaller system that *does* work. But all the immediately smaller system don’t. In other words, an IC system is a local minimum in complexity
The fact that a system is IC doesn’t say anything about its evolvability, of course. It is completely possible for evolution to build up a larger system and then whittle it down to an IC one, or for evolution to create a system with one task and then coopt it for a different task where it is IC. Anyone who maintains that this is a problem for evolution is an idiot who thinks that mutations are purely additive.
I’m just sayin’ that IC *is* a well-defined and empirically testable quality. | 21,003 | 93,603 | {"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-2015-40 | latest | en | 0.940777 |
https://affairscloud.com/reasoning-questions-direction-sense-set-9/ | 1,701,922,313,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00722.warc.gz | 102,679,782 | 37,502 | # Reasoning Questions : Direction Sense Set 9
Hello Aspirants. Welcome to Online Reasoning Section in AffairsCloud.com.Here we are creating Best question samples from Direction Sense test with explanation, which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
1. A man standing facing north starts walking. After walking for 5 m he took a left turn and walked for 10m. Now he walked for 20 m after turning to his right and again he turned right and finally stopped after walking 10 m. What is the distance from the starting point?
A) 18 m
B) 20 m
C) 15 m
D) 25 m
E) None of these
D) 25 m
Explanation:
2. From a point Sahil starts walking in south-east direction. After walking for 6 m, he turned to west and walks another 10 m. Now he turned towards north-west and walks 6 m and then finally stopped after turning to east and walking 2 m. Find the distance from the starting point.
A) 10 m
B) 8 m
C) 12 m
D) 6 m
E) None of these
B) 8 m
Explanation:
It becomes a parallelogram, so 10-2 = 8
3. From a point Sahil started walked towards south. After walking for 20 m, he took a left turn and walked 40 m. Again he took a left turn and walked 15 m. Now he turned to his right and finally stopped after walking 10 m. How far is he from the starting point?
A) 42.50 m
B) 50.25 m
C) 30.75 m
D) 45 m
E) None of these
B) 50.25 m
Explanation:
√502+52
4. A person started walking in south direction. After walking for 40 m he turned to his left and then walked 20 m. Now he turned to north and walked 30 m. In which direction is he now with respect to starting point?
A) South-east
B) North-east
C) North
D) South
E) North-west
A) South-east
Explanation:
5. Point A is 7 m west of point B. Point C is 5 m north of point B. A point D on AB 2 m away from point A meets BC at point E with DE = √34 m. Find the distance EC.
A) 3 m
B) 2 m
C) 4 m
E) None of these
B) 2 m
Explanation:
BE = √(√34)2 – (25)2
BE = √9 = 3
So EC = 5-3 = 2
6. From a point, Suman walked 10 m towards east, then she took a right turn and walked another 10 m. Then she turned to her left and walked 30 m. Again turning left and after walking 40 m she stopped. How far and in which direction is she from starting point?
A) 50 m, south-east
B) 50 m, north-east
C) 20 m, north-east
D) 50 m, north-west
E) None of these
B) 50 m, north-east
Explanation:
√402+302 = 50
7. Madhuri started walking in north and after walking 5 m she turned to her left. Now she walked 4 m and turned to her left again and walked 5 m. Now finally she turned to her left again and walked 5 m. What is the distance of this point from starting point?
A) 0 m
B) 4 m
C) 2 m
D) 1 m
E) None of these
D) 1 m
Explanation:
8. Point A is 15 m west of point B. Point C is 13 m south of point B and also 5 m east of point E. Point F is 20 m north of point E. What is the distance AF?
A) 49 m
B) 7√49 m
C) 20 m
D) √149 m
E) None of these
D) √149 m
Explanation:
√102+72
9. In above question, what is the direction of point C with respect to point A?
A) West
B) East
C) South
D) South-west
E) South-east
E) South-east
Explanation:
10. Shruti walks 5 m towards south. Now she walks 10 m after turning to her left. Now she turns to her right and walks 2 m, takes a right turn again and walks 4 m. Now finally she takes a right turn and walks 15 m and stops. Find the distance from starting point?
A) 16 m
B) 10 m
C) 10 m
D) 12 m
E) None of these
C) 10 m
Explanation:
PQ = √ 62+82 = √100 = 10
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http://www.cfd-online.com/Forums/openfoam-solving/57870-wing-aerodynamics-fluent-15-comparison-3.html | 1,480,835,834,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00101-ip-10-31-129-80.ec2.internal.warc.gz | 380,225,710 | 27,044 | # Wing Aerodynamics Fluent OF 15 comparison
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February 16, 2009, 10:57 Hi FOAMers! In order to fin #41 Senior Member maddalena Join Date: Mar 2009 Posts: 436 Rep Power: 14 Hi FOAMers! In order to find a proper case set up, I am considering a wing with an AoA = 8°, that it is easier to simulate and has not problem with flow detachment. After trying different schemes combinations, I applied: - gradSchemes: faceMDLimited Gauss linear 0.5 - divSchemes: Gauss upwind everywhere, except for div(phi,U) Gauss limitedLinearV 1 - laplacianSchemes: Gauss linear corrected. A comparison between OF and Fluent is presented in the following pictures: As you can see, two problems are still opened: 1)There is a big gap between OF and Fluent converged cd value: the latter is higher of about 50% in respect to OF. Comparing pressure contours, I have the same general behaviour, but different max and min values. Might it be connected with wallFunctions? I have patch 0 named surf y+: min: 1.27842 max: 52.8088 average: 10.0848 in OF and Yplus max 168.2676 in Fluent... Remember that I am using realizableKE on both. 2)OF cl and cd shows wiggles in their histories, while Fluent values becomes smooth after about 300 iterations. It seems like OF values stop converging after a while. I tried to decrease tolerances for U and p, but no improvement was obtained. Is this problem connected with numerical schemes, relaxation factors and so on or is this more a wrong physical condition? Suggestions are welcome!! Maddalena.
February 16, 2009, 16:09 Hi, I have similar problem #42 New Member Mathias Kvick Join Date: Mar 2009 Location: Stockholm / Istanbul, Sweden / Turkey Posts: 11 Rep Power: 9 Hi, I have similar problems, here are plots of the pressure coefficient at a cross section of the upper and lower surface of my wing. As you can see, the results does not look good at the upper surface, on the lower surface it looks better, but still not good. As I understood it Fluent uses the same wallfunctions as I have in my case.. Let me know if you find out something..
February 16, 2009, 16:34 In Genral different solvers gi #43 Senior Member BastiL Join Date: Mar 2009 Posts: 488 Rep Power: 12 In Genral different solvers give different results. However, 50% difference on drag seems massively to big for me. However I currently have no real explanation for it. It might be possible that the turbulence models behave different in the transition area. I get very simular results on meshes with prism layers (= fewer transition). I never tried it without. Regards BastiL
February 18, 2009, 13:33 Hi FOAMers, In order to under #44 Senior Member maddalena Join Date: Mar 2009 Posts: 436 Rep Power: 14 Hi FOAMers, In order to understand why OF and Fluent converge to such different coefficient values, I analyzed the single forces contribution. Here the results: Fluent pressure force: (-0,4409 0,7928 10,5498) viscous force: (0,4121 -0,0003 0,0159) cl: 0,63170 (converged) cd: 0,08700 (converged) yPlus : min 14,79 max 129,71 OF pressure force: (-0,7441 0,720801 10,7649) viscous force: (0,07900 -0,0011 0,0045) cl: 0,6538 (not converged) cd: 0,0504 (not converged) yPlus: min 1,28 max 52,81 In particular, the contribution along x axis is the main cause of such a difference in the resulting aerodynamic coefficients, and since I have an AoA = 8°, this is more pronounced in cd. In addition, besides I am using the same Hi-Re turbulence model in the two solvers (realizableKE), the same standard wall function including wall roughness and the same mesh, the yPlus values are different. Note that Fluent values are close to experimental ones. In order to have an insight on this points, I run a inviscid case (viscosity → inviscid in Fluent and slip bc at wall in OF) with a A0A = 0°. Here the results: Fluent pressure force: (0,3334 -0,4700 0,5896) viscous force: (0 0 0) (Ok!) cl: 0.035580496 cd: 0.020121824 OF pressure force: (0,1937 -0,5229 0,5089) viscous force: (3,011e-5 -8,3081e-6 -2,4340e-5) (ok!) cl: 0,0307 cd: 0,01169 As you can see, x-component of pressure force is very different in the two solvers, it is the main reason of the 50% difference in cd here, and so it is to the viscous original case. I have been searching in the message board and surfing the web for quite a while now, in order to find an explanation, but with no luck. I have still some open questions: 1)Why there is such a difference between the two solvers in pressure calculation? As reported by Mathias above, the pressure distribution in chord-wise direction is different as well. 2)Why there is such a difference in y+, besides the same mesh, the same wall function and the same grid? 3)Is there something that I am missing (The entire case set up is explained in detail in the previous posts)? 4)Is there any hidden tricks in Fluent that let Fluent simulation converge? Schemes? Boundary conditions? 5)Besides using a prism-layer mesh, is there any other suggestions? Note that I do not use a prism-layer mesh since: - It is difficult to obtain a good mesh with my model - I do not want to end up with millions of cells to have a first estimation of values, - Doing a geometry sensitivities study with a prism-layer mesh will take me more time in creating mesh than running the case itself. Any help will be appreciated!!! Regards, Maddalena
February 19, 2009, 04:38 Hello Maddalena, #45 Member Maruthamuthu Venkatraman Join Date: Mar 2009 Location: Norway Posts: 80 Rep Power: 9 Hello Maddalena, Good work in regards with inviscid computation for comparing pressure forces. 1. I realized you are using Gaussian Node based gradients in Fluent which is recommended for unstructured mesh. In Openfoam are you using something similar type ? Still i am a new to OpenFoam , may be how much this may influence i dont know. 2. To be sure you have to set the kinematic viscosity [m^2/s] in OpenFoam, which is Default.i did this mistake once so just reminding you. 3.The physical distance from the wall to the next cell grid may be different if it is cell centred schemes or its counter part. Make sure that its taking the same physical distance for Y+... 4. There are lots of variables and modeling assumptions involved here, quite difficult to find the exact one for turbulent flows. But looking at your results for potential flow, its seems there may be some better Numerical settings can be used for OpenFoam. Have Fun. Your contribution will end up in some better understanding of OpenFoam.
February 24, 2009, 08:14 Hi Maddalena, Any success #46 New Member Mathias Kvick Join Date: Mar 2009 Location: Stockholm / Istanbul, Sweden / Turkey Posts: 11 Rep Power: 9 Hi Maddalena, Any success in reducing the difference between the two calculations? //Mathias
February 24, 2009, 11:52 Hi Mathias, in the last days #47 Senior Member maddalena Join Date: Mar 2009 Posts: 436 Rep Power: 14 Hi Mathias, in the last days I have been busy with other stuff and I haven't tested anything new. I will try some more settings, included what Maruthamuthu suggested, as soon as I will have some free time. In the meanwhile, please let me know of every updates/news/improvement you have! Cheers, Maddalena
February 27, 2009, 07:39 Hi Maddalena, from your pro #48 New Member Mitja Morgut Join Date: Mar 2009 Location: Trieste, TS, Italy Posts: 1 Rep Power: 0 Hi Maddalena, from your profile I have noticed that you are working in Trieste. I am also from Trieste. Recently I have compared CFX-11 and OF-1.5 for prediction of the flow around NACA0012. If you are interested to contact me, (to arrange a meeting in TS) you can find my e-mail address in the profile. Have a nice day Mitja
March 2, 2009, 04:14 hi Maruthamuthu Venkatraman #49 Senior Member NAVEEN.K.M Join Date: Mar 2009 Location: Bangalore, Karnataka, india Posts: 114 Rep Power: 9 hi Maruthamuthu Venkatraman i am from india,In order to find a proper convergent solution, I am considering a naca 0012 airfoil case with an AoA = 4°, can u please send me how wil u know that whether the solution is converged or not like fluent v wil get at some point v wil get that the solution is converged,but i dont know how v wil get in openfoam 1.4.1.....
March 2, 2009, 04:14 hi Maruthamuthu Venkatraman si #50 Senior Member NAVEEN.K.M Join Date: Mar 2009 Location: Bangalore, Karnataka, india Posts: 114 Rep Power: 9 hi Maruthamuthu Venkatraman sir i am from india,In order to find a proper convergent solution, I am considering a naca 0012 airfoil case with an AoA = 4°, can u please send me how wil u know that whether the solution is converged or not like fluent v wil get at some point v wil get that the solution is converged,but i dont know how v wil get in openfoam 1.4.1.....
March 2, 2009, 04:29 Hello Naveen, #51 Member Maruthamuthu Venkatraman Join Date: Mar 2009 Location: Norway Posts: 80 Rep Power: 9 Hello Naveen, Iam glad to see your progress in using OpenFoam.For your calculation i would recommend you to check the convergence of the drag and lift coeffcient. You need not to address people here in the forum with Sir. Just put your question clearly and if people dont have time to reply you then better try it out... I would recommend you to read Fluid mechanics book by Munsion, Youung and Okashi under the Section "Flow over the immersed bodies". Also Fluid Mechanics by Frank M white. Most of the external aerodynamic theories and expoerimental results are there. You shall make some test case and check your answers... Good luck
March 2, 2009, 23:12 Hi Fomers, i am also facin #52 Member Vishal Jambhekar Join Date: Mar 2009 Location: University Stuttgart, Stuttgart Germany Posts: 90 Blog Entries: 1 Rep Power: 9 Hi Fomers, i am also facing some issues using sonicTurboFoam for Supersonic Flow. i am mainly interested to capture shock boundary layer interaction. Please have a look at this post. Use this link to go directly to the discussion: http://www.cfd-online.com/cgi-bin/Op...ow.cgi?1/11454 __________________ Cheers, Vishal Jambhekar... "Simulate the way ahead......!!!"
March 3, 2009, 08:34 Hi Foamers, when I tried to #53 Senior Member Wolfgang Heydlauff Join Date: Mar 2009 Location: Germany Posts: 136 Rep Power: 10 Hi Foamers, when I tried to simulate an airfoil I realized that to further the boundary borders were, the better the solution got (compared to some windtunnel results). best results I got with a distance of the boundary borders of approx. 160 time the chordlenght (!!). (komegaSST-Model). Since OF has no farfield-correction like "Fluent" or "Flower" I think this is at least necessary. Improving the results a lot I also change the top and bottom boundary type from symmetryPlane to inlet. If far away you get better undisturbed stream flow at the airfoil itself. If symmetryPlane is used, the airflow is changing angle to apply the symmetryPlane conditions. In bad case, the AoA decreases.
April 15, 2009, 04:40 Conclusion #54 Senior Member maddalena Join Date: Mar 2009 Posts: 436 Rep Power: 14 Hello FOAMers, in order to obtain better convergence and results closer to Fluent for my external aerodynamic simulation, I changed one more time my fvSchemes, following some suggestions I had from HRV. Now it looks like this: gradSchemes: faceMDLimited Gauss linear 0.5; divSchemes: Gauss GammaV 1 on div(phi,U); Gauss upwind everywhere else. laplacianSchemes: Gauss linear limited 0.33 on laplacian(DepsilonEff,epsilon), Gauss linear limited 0.5 everywhere else. interpolationSchemes: linear; snGradSchemes: limited 0.5; Moreover, I increased tolerance and relTol for every variable: { ... tolerance 1e-09; relTol 0.01; } This helped me to obtain better numerical convergence. However, a certain difference between OF and Fluent still remain (alpha = 8°): Fluent:CL: 0.6317 CD: 0.0870 pmax:137 pmin:-199 OF:CL: 0.6596 CD: 0.0522 pmax: 115 pmin: -179 I could not find a numerical set-up or domain discretisation that allows me to obtain a smaller gap between the two solvers, so I can say that this is the best I could have with the standard OF distribution for a case similar to mine. Since Fluent results are closer to experimental values, I can conclude that OF 1.5 underestimates aerodynamic coefficients as a consequence of a numerically different pressure field around the wing. Please, feel free to add anything in addition to this. Regards, Maddalena s.m likes this.
April 15, 2009, 11:32 #55 Senior Member Join Date: Mar 2009 Posts: 138 Rep Power: 9 HI Maddalena! HI Foamers! Just by a fluke I found this thread. I am also simulating a flow around an airfoil and I am enthused of the possibility of writing cl cd to file. But I am struggling with some difficulties. Is there anywhere more information about this function?? Best Regards Camoesas Last edited by camoesas; April 16, 2009 at 03:12.
April 20, 2009, 08:20 #56 Senior Member Join Date: Mar 2009 Posts: 138 Rep Power: 9 HI Everybody! I still have not found more information about the forceCoeffs function. Altough it is working fine. But in the end I get this error message: *** glibc detected *** simpleFoam: corrupted double-linked list: 0x0000000000cc1ae0 *** Inconsistency detected by ld.so: dl-open.c: 260: dl_open_worker: Assertion `_dl_debug_initialize (0, args->nsid)->r_state == RT_CONSISTENT' failed! Do I have to worry? I have already searched this board but without results. Any ideas? Anyone gets this message too? Best Regards Camoesas
June 10, 2009, 02:55 solving multielement airfoil case in openfoam #58 Senior Member NAVEEN.K.M Join Date: Mar 2009 Location: Bangalore, Karnataka, india Posts: 114 Rep Power: 9 does any one know how to solve multielement airfoil case in openfoam which is imported from gambit...and suggest me how to set boundary conditions for mach no 0.15..
June 11, 2009, 02:03 Multielement airfoil analysis #59 Senior Member NAVEEN.K.M Join Date: Mar 2009 Location: Bangalore, Karnataka, india Posts: 114 Rep Power: 9 HI EVERYBODY Does any one know how to solve mutielement airfoil in openfoam 1.4.1 or 1.5...if any one knows suggest me how to set boundary conditions which is imported from gambit..
June 11, 2009, 02:18 #60 Senior Member NAVEEN.K.M Join Date: Mar 2009 Location: Bangalore, Karnataka, india Posts: 114 Rep Power: 9 hi maruthamuthu_venkatraman my self NAVEEN from INDIA...i hav completed my naca 0012 airfoil analysis case in openfoa 1.4.1 and 1.5....i got good results in both the versions...thanks for your suggestions....now i am doing MTech project on multielement airfoil analysis with double slotted flaps and i hav completed my grid in gambit(c-grid) and now i need to solve in openfoam...i am getting some errors like this during solving in openfoam 1.4.1 #0 Foam::error:rintStack(Foam::Ostream&) in "/home/openfoam14/OpenFOAM/OpenFOAM-1.4.1/lib/linux64GccDPOpt/libOpenFOAM.so" #1 Foam::sigFpe::sigFpeHandler(int) in "/home/openfoam14/OpenFOAM/OpenFOAM-1.4.1/lib/linux64GccDPOpt/libOpenFOAM.so" #2 ?? in "/lib64/libc.so.6" #3 Foam::PCG::solve(Foam::Field&, Foam::Field const&, unsigned char) const in "/home/openfoam14/OpenFOAM/OpenFOAM-1.4.1/lib/linux64GccDPOpt/libOpenFOAM.so" #4 Foam::fvMatrix::solve(Foam::Istream&) in "/home/openfoam14/OpenFOAM/OpenFOAM-1.4.1/lib/linux64GccDPOpt/libfiniteVolume.so" #5 main in "/home/openfoam14/OpenFOAM/OpenFOAM-1.4.1/applications/bin/linux64GccDPOpt/simpleFoam" #6 __libc_start_main in "/lib64/libc.so.6" #7 Foam::regIOobject::readIfModified() in "/home/openfoam14/OpenFOAM/OpenFOAM-1.4.1/applications/bin/linux64GccDPOpt/simpleFoam" can u give me some suggestions whare i am going wrong and also suggest how to set boundary conditions.....
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