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https://math.stackexchange.com/questions/3617696/can-you-suggest-some-good-optimization-books | # Can you suggest some good optimization books?
I am looking for optimization books. Can you suggest some good materials?
First, I started with Convex Optimization by Stephen Boyd & Lieven Vandenberghe, but I don't like it because they don't give examples of proofs and techniques, only theory and talking. I need some classical books, for example I like books such as: Zorich, Kreyszig, Kolmogorov and Fomin. Please suggest books that have a similar style. Especially I need books to pursue research in Reinforcement Learning.
• Optimisation is a pretty wide area, if you are more specific you might get better answers. Apr 9, 2020 at 18:30
• I'm not sure what you mean by "only theory" because Boyd and Vandenberghe is filled with applications. Apr 10, 2020 at 0:24
• You might be interested in reading Reinforcement Learning and Optimal Control by Bertsekas:amazon.com/Reinforcement-Learning-Optimal-Control-Bertsekas/dp/… Apr 10, 2020 at 0:27
Since you've mentioned you wish to take up research in reinforcement learning, I'm assuming by "optimization" you mean both convex and non-convex optimization. I'd suggest you the following depending on your level of understanding:
1) Introduction to Linear Optimization by Bertsimas and Tsitsiklis: A good starting book on linear optimization that will prepare you for convex optimization.
2) Introductory Lectures on Convex Optimization by Yurii Nesterov: Nesterov is a living legend in the field of convex optimization. You might have heard his name from the famous Nesterov Momentum technique. Be warned, this book isn't light in its usage of mathematics. An even math-heavier version of this book titled Lectures on Convex Optimization is also published by Springer.
3) Optimization for Machine Learning by Sra, Nowozin and Wright: While I haven't taken a look at this book, I'm told it is high quality and covers both convex and non-convex optimization.
4) Non-Convex Optimization for Machine Learning by Jain and Kar: This monograph is a big gun, when it comes to non-convex optimization, and is freely available online. I'm not sure how much of the material would be helpful to you, but it should serve as a good reference. There are not many books which specifically cater to non-convex optimization, this book is one of the first dedicated texts I could find. But then again, this isn't exactly an easy read.
Happy learning.
Optimization by Vector Space Methods by Luenbeger, and Non-Linear Programming by Bertsekas.
• I was curious... Is this one by Luenbeger still relevant? How does it compare to his newer one "Linear and Nonlinear Programming"? Thank you. Jun 16 at 1:05
Try these books $$(1)$$ N. S. Kambo, Mathematical Programming Techniques, East West Press, 1997 and $$(2)$$ E.K.P. Chong and S.H. Zak, An Introduction to Optimization, 2nd Ed., Wiley, 2010.
• thanks, i started (2), and really loved it.I am also interested in solving problems, but i don't have solutions for those problems.Do you know any source that provides solutions? Apr 15, 2020 at 13:33
Belegundu and Chandrupatla, Optimization Concepts and Applications in Engineering. | 2022-06-29 05:55:12 | {"extraction_info": {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42808452248573303, "perplexity": 643.6316907814388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00515.warc.gz"} |
http://www.mathics.net/doc/reference-of-built-in-symbols/assignment/nvalues/ | # NValues
• NValues[a]
• N[a] = 3;
• NValues[a]
You can assign values to NValues:
• NValues[b] := {N[b, MachinePrecision] :> 2}
• N[b]
Be sure to use SetDelayed, otherwise the left-hand side of the transformation rule will be evaluated immediately, causing the head of N to get lost. Furthermore, you have to include the precision in the rules; MachinePrecision will not be inserted automatically:
• NValues[c] := {N[c] :> 3}
• N[c]
Mathics will gracefully assign any list of rules to NValues; however, inappropriate rules will never be used:
• NValues[d] = {foo -> bar};
• NValues[d]
• N[d] | 2017-09-21 06:41:02 | {"extraction_info": {"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, "math_score": 0.3487080931663513, "perplexity": 1730.8982540760176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687702.11/warc/CC-MAIN-20170921063415-20170921083415-00026.warc.gz"} |
https://codegolf.stackexchange.com/questions/97436/alphabet-diamond/97505 | # Alphabet Diamond
You've seen the amazing alphabet triangle, the revenge of the alphabet triangle and now it's time for the revenge of the revenge of the alphabet triangle!
Introducing...
THE ALPHABET DIAMOND!
Your task is to output this exact text, lowercase/uppercase does not matter, though newlines do:
bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb
cdefghijklmnopqrstuvwxyzabazyxwvutsrqponmlkjihgfedc
defghijklmnopqrstuvwxyzabcbazyxwvutsrqponmlkjihgfed
efghijklmnopqrstuvwxyzabcdcbazyxwvutsrqponmlkjihgfe
fghijklmnopqrstuvwxyzabcdedcbazyxwvutsrqponmlkjihgf
ghijklmnopqrstuvwxyzabcdefedcbazyxwvutsrqponmlkjihg
hijklmnopqrstuvwxyzabcdefgfedcbazyxwvutsrqponmlkjih
ijklmnopqrstuvwxyzabcdefghgfedcbazyxwvutsrqponmlkji
jklmnopqrstuvwxyzabcdefghihgfedcbazyxwvutsrqponmlkj
klmnopqrstuvwxyzabcdefghijihgfedcbazyxwvutsrqponmlk
lmnopqrstuvwxyzabcdefghijkjihgfedcbazyxwvutsrqponml
mnopqrstuvwxyzabcdefghijklkjihgfedcbazyxwvutsrqponm
nopqrstuvwxyzabcdefghijklmlkjihgfedcbazyxwvutsrqpon
opqrstuvwxyzabcdefghijklmnmlkjihgfedcbazyxwvutsrqpo
pqrstuvwxyzabcdefghijklmnonmlkjihgfedcbazyxwvutsrqp
qrstuvwxyzabcdefghijklmnoponmlkjihgfedcbazyxwvutsrq
rstuvwxyzabcdefghijklmnopqponmlkjihgfedcbazyxwvutsr
stuvwxyzabcdefghijklmnopqrqponmlkjihgfedcbazyxwvuts
tuvwxyzabcdefghijklmnopqrsrqponmlkjihgfedcbazyxwvut
uvwxyzabcdefghijklmnopqrstsrqponmlkjihgfedcbazyxwvu
vwxyzabcdefghijklmnopqrstutsrqponmlkjihgfedcbazyxwv
wxyzabcdefghijklmnopqrstuvutsrqponmlkjihgfedcbazyxw
xyzabcdefghijklmnopqrstuvwvutsrqponmlkjihgfedcbazyx
yzabcdefghijklmnopqrstuvwxwvutsrqponmlkjihgfedcbazy
zabcdefghijklmnopqrstuvwxyxwvutsrqponmlkjihgfedcbaz
abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba
bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb
abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba
zabcdefghijklmnopqrstuvwxyxwvutsrqponmlkjihgfedcbaz
yzabcdefghijklmnopqrstuvwxwvutsrqponmlkjihgfedcbazy
xyzabcdefghijklmnopqrstuvwvutsrqponmlkjihgfedcbazyx
wxyzabcdefghijklmnopqrstuvutsrqponmlkjihgfedcbazyxw
vwxyzabcdefghijklmnopqrstutsrqponmlkjihgfedcbazyxwv
uvwxyzabcdefghijklmnopqrstsrqponmlkjihgfedcbazyxwvu
tuvwxyzabcdefghijklmnopqrsrqponmlkjihgfedcbazyxwvut
stuvwxyzabcdefghijklmnopqrqponmlkjihgfedcbazyxwvuts
rstuvwxyzabcdefghijklmnopqponmlkjihgfedcbazyxwvutsr
qrstuvwxyzabcdefghijklmnoponmlkjihgfedcbazyxwvutsrq
pqrstuvwxyzabcdefghijklmnonmlkjihgfedcbazyxwvutsrqp
opqrstuvwxyzabcdefghijklmnmlkjihgfedcbazyxwvutsrqpo
nopqrstuvwxyzabcdefghijklmlkjihgfedcbazyxwvutsrqpon
mnopqrstuvwxyzabcdefghijklkjihgfedcbazyxwvutsrqponm
lmnopqrstuvwxyzabcdefghijkjihgfedcbazyxwvutsrqponml
klmnopqrstuvwxyzabcdefghijihgfedcbazyxwvutsrqponmlk
jklmnopqrstuvwxyzabcdefghihgfedcbazyxwvutsrqponmlkj
ijklmnopqrstuvwxyzabcdefghgfedcbazyxwvutsrqponmlkji
hijklmnopqrstuvwxyzabcdefgfedcbazyxwvutsrqponmlkjih
ghijklmnopqrstuvwxyzabcdefedcbazyxwvutsrqponmlkjihg
fghijklmnopqrstuvwxyzabcdedcbazyxwvutsrqponmlkjihgf
efghijklmnopqrstuvwxyzabcdcbazyxwvutsrqponmlkjihgfe
defghijklmnopqrstuvwxyzabcbazyxwvutsrqponmlkjihgfed
cdefghijklmnopqrstuvwxyzabazyxwvutsrqponmlkjihgfedc
bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb
## This is code-golf, lowest bytecount wins.
Rules:
1. Standard loopholes are disallowed.
2. a must be the center of the alphabet diamond.
• Nice challenge. I personally think it would make more sense if the corners were a and the center was z – ETHproductions Oct 25 '16 at 20:30
• @ETHproductions I wanted to make it harder for the golfing languages who can push the alphabet ;). Though I doubt it's much harder. – Magic Octopus Urn Oct 25 '16 at 20:32
• @carusocomputing A left rotate would do it. For example, .< would do it in Pyth: .<S5 1 would result in [2, 3, 4, 5, 1]. It is the same as .<[1 2 3 4 5)1. Not sure for the diamond, though. – Erik the Outgolfer Oct 26 '16 at 17:01
• @EriktheGolfer I realize, there are some smarter ways to do it too, like collapsing the shift in a do-while. The diamond itself should require shifting, so there are ways to combine the initial shift in the actual diamond iterations as well. It was an intentional part of the challenge to see who optimized their looping. – Magic Octopus Urn Oct 26 '16 at 17:05
• @carusocomputing I meant about b being the initial letter instead of a. Of course shifting is vital too. – Erik the Outgolfer Oct 26 '16 at 17:08
# 05AB1E, 13 12 bytes
A27FÀDûˆ}¯û»
Try it online!
Explanation
A # push alphabet
27F # 27 times do
À # rotate alphabet left
Dû # create a palendromized copy
ˆ # add to global list
} # end loop
¯ # push global list
û # palendromize list
» # merge list on newline
# implicit output
• Could probably save a byte using registers and the g to calc the length of the alphabet instead of duping it. – Magic Octopus Urn Oct 25 '16 at 23:43
• @carusocomputing: Unless I'm doing it wrong (feel free to educate me if that is the case) I ended up at 13 doing that. I did manage to save a byte with the global list though. – Emigna Oct 26 '16 at 6:20
• A©gF®À©û})û» is what I was thinking. By no means are you ever doing it wrong! I've learned all I know from watching you heh. The global list is the same basic idea. – Magic Octopus Urn Oct 26 '16 at 14:49
• Now you're tied with me. :3 – Oliver Ni Oct 26 '16 at 19:28
• @Oliver: Indeed! And with 2 different approaches in the same language :) – Emigna Oct 26 '16 at 19:35
:se ri|h<_<cr>jjYZZPqqx$pYpq26@qdjyH:g/^/m0<cr>VP:%norm DPA<C-v><C-r>"<C-v><esc>x<cr> Drawing on inspiration from Lynn's awesome vim answer to take the idea of stealing the alphabet from the help docs. You can watch it happen in real time as I struggle to remember the right sequence of keystrokes! Note that this gif is slightly outdated because it produces the wrong output, and I haven't gotten around to re-recording it yet. • I tried this program and got this. – LegionMammal978 Jun 15 '17 at 13:03 • Pls rerecord ty – ASCII-only Sep 7 '17 at 12:18 # MATL, 14 bytes 2Y226Zv27Zv!+) Try it online! 2Y2 % Push string 'abc...z' 26Zv % Push symmetric range [1 2 ... 25 26 25 ... 2 1] 27Zv % Push symmetric range [1 2 ... 25 26 27 26 25 ... 2 1] ! % Transpose into a column + % Addition with broadcast. Gives a matrix of all pairwise additions: % [ 2 3 ... 26 27 26 ... 3 2 3 4 ... 27 28 27 ... 4 3 ... 27 28 ... 51 52 51 ... 28 27 28 29 ... 52 53 52 ... 29 28 27 28 ... 51 52 51 ... 28 27 ... 2 3 ... 26 27 26 ... 3 2 ] ) % Index modularly into the string. Display implicitly # PHP, 129 Bytes for(;++$i<27;)$o.=($s=($f=substr)($r=join(range(a,z)),$i,26-$i)).$t.strrev($s.$t=$f($r,0,$i))."\n";echo$o.$f($o,0,51).strrev($o);
• syntax error, unexpected '(' on line 1 Which php version? – Tschallacka Oct 26 '16 at 9:19
• @Tschallacka PHP > 7 before you can write for($f=substr; and $f($r=join(range(a,z)),$i,26-$i)) instead of ($f=substr)($r=join(range(a,z)),$i,26-$i)) to avoid the error – Jörg Hülsermann Oct 26 '16 at 9:44 ## Haskell, 75 bytes g=(++)<*>reverse.init unlines$g$g.take 26.($cycle['a'..'z']).drop<$>[1..27] How it works: g=(++)<*>reverse.init -- helper function that takes a list and appends the -- reverse of the list with the first element dropped, e.g. -- g "abc" -> "abcba" <$>[1..27] -- map over the list [1..27] the function that
drop -- drops that many elements from
($cycle['a'..'z']) -- the infinite cycled alphabet and take 26 -- keeps the next 26 chars and g -- appends the reverse of itself -- now we have the first 27 lines g -- do another g to append the lower half unlines -- join with newlines # Jelly, 13 bytes Øaṙ1ṭṙJ$ŒBŒḄY
Try it online!
## Explanation
Øaṙ1ṭṙJ$ŒBŒḄY Main link. No arguments Øa Get the lowercase alphabet ṙ1 Rotate left by 1 ṭ Append to$ Monadic chain
J Indices of the alphabet [1, 2, ..., 26]
ṙ Rotate the alphabet by each
ŒB Bounce each rotation
ŒḄ Bounce the rotations
Y Join with newlines and print implicitly
• Øaṙ'JŒBŒḄY for 10 :) – Jonathan Allan Oct 25 '16 at 22:36
• @JonathanAllan Thanks but that's missing the middle part which is why I had to do that ṙ1ṭ bit. Also ØaṙJŒBŒḄY is fine, you don't need the quick since it vectorizes on the right to 0 – miles Oct 25 '16 at 23:04
• Totally missed that the diamond was not perfect! Oh well... – Jonathan Allan Oct 25 '16 at 23:09
# C, 76 bytes
Function, to be called as below. Prints capital letters.
f(i){for(i=2756;--i;)putchar(i%52?90-(abs(i%52-26)+abs(i/52-26)+25)%26:10);}
//call like this
main(){f();}
Simple approach, add the x and y distances from the centre of the square, plus an offset of 25 for the a in the middle, take modulo 26 and subract from 90, the ASCII code for Z. Where i%52==0 a newline ASCII 10 is printed.
• Your offset of +25 is the same as -1 in modulo 26 – Karl Napf Oct 25 '16 at 22:42
• @KarlNapf C doesn't implement modulo of negative numbers the way mathematicians do. -1%26 in C is -1, not 25. The result is a [ in the centre instead of the expected A. Thanks anyway, you would have been correct in a language such as Ruby where -1%26 does equal 25. – Level River St Oct 25 '16 at 22:55
# R, 71 bytes
cat(letters[outer(c(1:27,26:1),c(0:25,24:0),"+")%%26+1],sep="",fill=53)
outer creates a matrix with the indices of the letters, letters[...] then creates a vector with the correct letters in. cat(...,sep="",fill=53) then prints it out with the desired formatting.
• Nice one! Somehow I had forgotten about the fill option for cat. Great way of printing formatted matrices. – Billywob Oct 26 '16 at 8:35
# Jelly, 11 bytes
27RØaṙŒḄŒBY
Explanation:
27R range of 1...27
Øa the alphabet
ṙ rotate
ŒḄŒB bounce in both dimensions
Y join on newline
# Python 2, 96 85 bytes
Printing the uppercase version (saves 1 byte).
R=range
for i in R(53):print''.join(chr(90-(abs(j-25)+abs(i-26)-1)%26)for j in R(51))
previous solution with help from muddyfish
s="abcdefghijklmnopqrstuvwxyz"*3
for i in range(53):j=min(i,52-i);print s[j+1:j+27]+s[j+25:j:-1]
• isn't just typing the alphabet fewer chars? – Blue Oct 25 '16 at 20:56
## Perl, 77 bytes
Requires -E at no extra cost.
Pretty standard approach... I don't like the calls to reverse I think there's likely a more maths based approach to this, will see how I get on!
@a=((a..z)x3)[$_..$_+26],$a=pop@a,(say@a,$a,reverse@a)for 1..26,reverse 1..25
### Usage
perl -E '@a=((a..z)x3)[$_..$_+26],$a=pop@a,(say@a,$a,reverse@a)for 1..26,reverse 1..25'
• You can save 1 byte by removing the whitespace after reverse in reverse 1..25. The for needs it though. – simbabque Oct 26 '16 at 14:31
• @simbabque maybe it's a Perl version thing, but reverse1..25 results in 0..25. I'm running 5.18.2... – Dom Hastings Oct 26 '16 at 15:24
• You're right. Because the bareword reverse1 is undefined. Makes sense. – simbabque Oct 26 '16 at 15:39
# JavaScript (ES6), 97 96 bytes
Saved 1 byte thanks to @user81655
R=(n,s=25,c=(n%26+10).toString(36))=>s?c+R(n+1,s-1)+c:c
C=(n=1,r=R(n))=>n<27?r+
${C(n+1)} +r:r Two recursive functions; C is the one that outputs the correct text. Try it here: R=(n,s=25,c=(n%26+10).toString(36))=>s?c+R(n+1,s-1)+c:c C=(n=1,r=R(n))=>n<27?r+${C(n+1)}
+r:r
console.log(C())
• @user81655 I always forget about string interpolation when newlines are involved :P – ETHproductions Oct 26 '16 at 13:13
# Python 3, 119 bytes
I tried to exploit the two symmetry axes of the diamond, but this ended up more verbose than Karl Napf's solution.
A='abcdefghijklmnopqrstuvwxyz'
D=''
for k in range(1,27):
D+=A[k:]+A[:k]
D+=D[-2:-27:-1]+'\n'
print(D+D[:51]+D[::-1])
• A good solution nonetheless! You can cut down 3 bytes by writing the for loop in 1 line: for k in range(1,27):D+=A[k:]+A[:k];D+=D[-2:-27:-1]+'\n' – FlipTack Oct 30 '16 at 10:55
• shortened again: replace A with 'bcdefghijklmnopqrstuvwxyza' and replace range(1,27) with range(26). My byte count is now 114 – FlipTack Oct 30 '16 at 11:33
unlines[[toEnum$mod(-abs j-abs i)26+97|j<-[-25..25]]|i<-[-26..26]] ## C, 252 bytes #define j q[y] #define k(w,v) (v<'z')?(w=v+1):(w=97) char q[28][52],d;main(){int y,x=1;y=1;j[0]=98;j[50]=98;for(;y<27;y++){for(;x<26;x++){(x<1)?(k(d,q[y-1][50])):(k(d,j[x-1]));j[50-x]=d;j[x]=d;}x=0;j[51]=0;puts(j);}strcpy(j,q[1]);for(;y;y--)puts(j);} Formatted, macro-expanded version, which is hopefully more intelligible: #define j q[y] #define k(w,v) (v<'z')?(w=v+1):(w=97) char q[28][52],d; main(){ int y,x=1; y=1; q[1][0]=98;q[1][50]=98; //98 takes one less byte to type than the equivalent 'b' for(;y<27;y++){ for(;x<26;x++){ (x<1)? (k(d,q[y-1][50])) :(k(d,q[y][x-1])); q[y][50-x]=d; q[y][x]=d; } x=0; q[y][51]=0; puts(q[y]); } strcpy(q[y],q[1]); for(;y;y--)puts(q[y]); } I know this can't win, but I had fun trying. This is my first ever attempt at code golf. • Welcome to code golf, it's addicting haha ;). – Magic Octopus Urn Oct 27 '16 at 13:28 ## Batch, 255 bytes @echo off set l=abcdefghijklmnopqrstuvwxyz set r=yxwvutsrqponmlkjihgfedcba for /l %%i in (1,1,27)do call:u for /l %%i in (1,1,25)do call:l :l set r=%r:~2%%l:~-1%. set l=%l:~-2%%l:~0,-2% :u set r=%l:~-1%%r:~0,-1% set l=%l:~1%%l:~0,1% echo %l%%r% Explanation: The subroutine u rotates the alphabet outwards by one letter from the centre, which is the pattern used in the top half of the desired output. The subroutine l rotates the alphabet inwards by two letters. It then falls through into the u subroutine, achieving an effective single letter inward rotation. Finally the last line is printed by allowing the code to fall through into the l subroutine. ## Pyke, 2322 21 bytes GV'th+jj_t+j)K]0n'JOX Try it here! GV ) - repeat 26 times, initially push alphabet. 'th+ - push tos[1:]+tos[0] j - j = tos j - push j _t+ - push pop+reversed(pop)[1:] j - push j K - pop ]0 - list(stack) n'JOX - print "\n".join(^), - splat ^[:-1] # Charcoal, 24 21 bytes -3 bytes thanks to ASCII-only. Ｆ²⁷«Ｐ✂×β³⊕ι⁺ι²⁷↓»‖Ｂ↓→ Try it online! Link is to verbose version. ...I need to work on my Charcoal-fu. :P • some easy bytes off – ASCII-only Sep 7 '17 at 8:34 • Now I find out that the fourth argument to Slice is optional. >_> – totallyhuman Sep 7 '17 at 10:47 • All (yes, all four) arguments are optional – ASCII-only Sep 7 '17 at 10:49 • wat, what does niladic Slice even do? – totallyhuman Sep 7 '17 at 10:50 • Oh wait nvm yeah the first argument is required :P – ASCII-only Sep 7 '17 at 10:52 # PowerShell, 64 bytes 1..27+26..1|%{$y=$_ -join(0..25+24..0|%{[char](97+($_+$y)%26)})} Try it online! ## C++, 191179166165 139 bytes -12 bytes thanks to Kevin Cruijssen -14 bytes thanks to Zacharý -26 bytes thanks to ceilingcat #import<cstdio> #define C j;)putchar(97+j main(){for(int i=0,j,d=1;i<53;d+=i++/26?-1:1){for(j=d;26+d>C++%26);for(j--;d<=--C%26);puts("");}} • You can save 12 bytes like this: #include<iostream> int main(){for(int i=0,j,d=1;i<53;d+=i++/26?-1:1){for(j=0;j<26;)std::cout<<char((j+++d)%26+97);for(j=24;j>=0;)std::cout<<char((j--+d)%26+97);std::cout<<'\n';};} – Kevin Cruijssen Sep 4 '17 at 11:08 • Dang it Kevin, ya ninja'd me. – Adalynn Sep 4 '17 at 13:57 • @KevinCruijssen Thanks you for the help. By the way, your code in the comment contained unprintable unicode characters. – HatsuPointerKun Sep 4 '17 at 14:35 • @HatsuPointerKun That's something that happens automatically in the comments I'm afraid. It's indeed pretty annoying when you try to copy-paste code from it.. :) – Kevin Cruijssen Sep 4 '17 at 14:57 • You can abuse macros by adding this: #define C j;)std::cout<<char(97+(d+j, and then changing the last line to this: int main(){for(int i=0,j,d=1;i<53;d+=i++/26?-1:1){for(j=0;26>C++)%26);for(j=24;0<=C--)%26);std::cout<<'\n';};} – Adalynn Sep 4 '17 at 15:28 # JavaScript (ES6), 128115 114 bytes a='abcdefghijklmnopqrstuvwxyz' for(o=[i=27];i--;)o[26-i]=o[26+i]=(a=(p=a.slice(1))+a[0])+[...p].reverse().join o # Groovy - 103 97 bytes I realise there are cleverer ways of doing this but... {t=('a'..'z').join();q={it[-2..0]};c=[];27.times{t=t[1..-1]+t[0];c<<t+q(t)};(c+q(c)).join('\n')} When run the result of the script is the requested answer. (Thanks to carusocomputing for the tip on saving 7 bytes). Updated example accordingly on: • Instead of the for loop you can use 27.times(){} and save 7 bytes ;). – Magic Octopus Urn Oct 28 '16 at 13:39 ## Racket 293 bytes (let*((ls list->string)(rr reverse)(sr(λ(s)(ls(rr(string->list s))))))(let p((s(ls(for/list((i(range 97 123)))(integer->char i)))) (n 0)(ol'()))(let*((c(string-ref s 0))(ss(substring s 1 26))(s(string-append ss(string c)(sr ss))))(if(< n 53)(p s(+ 1 n)(cons s ol)) (append(rr ol)(cdr ol)))))) Ungolfed: (define (f) (define (sr s) ; sub-fn reverse string; (list->string (reverse (string->list s)))) (let loop ((s (list->string (for/list ((i (range 97 123))) (integer->char i)))) (n 0) (ol '())) (define c (string-ref s 0)) (define ss (substring s 1 26)) (set! s (string-append ss (string c) (sr ss))) (if (< n 53) (loop s (add1 n) (cons s ol)) (append (reverse ol) (rest ol))))) Testing: (f) Output: '("bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb" "cdefghijklmnopqrstuvwxyzabazyxwvutsrqponmlkjihgfedc" "defghijklmnopqrstuvwxyzabcbazyxwvutsrqponmlkjihgfed" "efghijklmnopqrstuvwxyzabcdcbazyxwvutsrqponmlkjihgfe" "fghijklmnopqrstuvwxyzabcdedcbazyxwvutsrqponmlkjihgf" "ghijklmnopqrstuvwxyzabcdefedcbazyxwvutsrqponmlkjihg" "hijklmnopqrstuvwxyzabcdefgfedcbazyxwvutsrqponmlkjih" "ijklmnopqrstuvwxyzabcdefghgfedcbazyxwvutsrqponmlkji" "jklmnopqrstuvwxyzabcdefghihgfedcbazyxwvutsrqponmlkj" "klmnopqrstuvwxyzabcdefghijihgfedcbazyxwvutsrqponmlk" "lmnopqrstuvwxyzabcdefghijkjihgfedcbazyxwvutsrqponml" "mnopqrstuvwxyzabcdefghijklkjihgfedcbazyxwvutsrqponm" "nopqrstuvwxyzabcdefghijklmlkjihgfedcbazyxwvutsrqpon" "opqrstuvwxyzabcdefghijklmnmlkjihgfedcbazyxwvutsrqpo" "pqrstuvwxyzabcdefghijklmnonmlkjihgfedcbazyxwvutsrqp" "qrstuvwxyzabcdefghijklmnoponmlkjihgfedcbazyxwvutsrq" "rstuvwxyzabcdefghijklmnopqponmlkjihgfedcbazyxwvutsr" "stuvwxyzabcdefghijklmnopqrqponmlkjihgfedcbazyxwvuts" "tuvwxyzabcdefghijklmnopqrsrqponmlkjihgfedcbazyxwvut" "uvwxyzabcdefghijklmnopqrstsrqponmlkjihgfedcbazyxwvu" "vwxyzabcdefghijklmnopqrstutsrqponmlkjihgfedcbazyxwv" "wxyzabcdefghijklmnopqrstuvutsrqponmlkjihgfedcbazyxw" "xyzabcdefghijklmnopqrstuvwvutsrqponmlkjihgfedcbazyx" "yzabcdefghijklmnopqrstuvwxwvutsrqponmlkjihgfedcbazy" "zabcdefghijklmnopqrstuvwxyxwvutsrqponmlkjihgfedcbaz" "abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba" "bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb" "cdefghijklmnopqrstuvwxyzabazyxwvutsrqponmlkjihgfedc" "defghijklmnopqrstuvwxyzabcbazyxwvutsrqponmlkjihgfed" "efghijklmnopqrstuvwxyzabcdcbazyxwvutsrqponmlkjihgfe" "fghijklmnopqrstuvwxyzabcdedcbazyxwvutsrqponmlkjihgf" "ghijklmnopqrstuvwxyzabcdefedcbazyxwvutsrqponmlkjihg" "hijklmnopqrstuvwxyzabcdefgfedcbazyxwvutsrqponmlkjih" "ijklmnopqrstuvwxyzabcdefghgfedcbazyxwvutsrqponmlkji" "jklmnopqrstuvwxyzabcdefghihgfedcbazyxwvutsrqponmlkj" "klmnopqrstuvwxyzabcdefghijihgfedcbazyxwvutsrqponmlk" "lmnopqrstuvwxyzabcdefghijkjihgfedcbazyxwvutsrqponml" "mnopqrstuvwxyzabcdefghijklkjihgfedcbazyxwvutsrqponm" "nopqrstuvwxyzabcdefghijklmlkjihgfedcbazyxwvutsrqpon" "opqrstuvwxyzabcdefghijklmnmlkjihgfedcbazyxwvutsrqpo" "pqrstuvwxyzabcdefghijklmnonmlkjihgfedcbazyxwvutsrqp" "qrstuvwxyzabcdefghijklmnoponmlkjihgfedcbazyxwvutsrq" "rstuvwxyzabcdefghijklmnopqponmlkjihgfedcbazyxwvutsr" "stuvwxyzabcdefghijklmnopqrqponmlkjihgfedcbazyxwvuts" "tuvwxyzabcdefghijklmnopqrsrqponmlkjihgfedcbazyxwvut" "uvwxyzabcdefghijklmnopqrstsrqponmlkjihgfedcbazyxwvu" "vwxyzabcdefghijklmnopqrstutsrqponmlkjihgfedcbazyxwv" "wxyzabcdefghijklmnopqrstuvutsrqponmlkjihgfedcbazyxw" "xyzabcdefghijklmnopqrstuvwvutsrqponmlkjihgfedcbazyx" "yzabcdefghijklmnopqrstuvwxwvutsrqponmlkjihgfedcbazy" "zabcdefghijklmnopqrstuvwxyxwvutsrqponmlkjihgfedcbaz" "abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba" "bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb" "abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba" "zabcdefghijklmnopqrstuvwxyxwvutsrqponmlkjihgfedcbaz" "yzabcdefghijklmnopqrstuvwxwvutsrqponmlkjihgfedcbazy" "xyzabcdefghijklmnopqrstuvwvutsrqponmlkjihgfedcbazyx" "wxyzabcdefghijklmnopqrstuvutsrqponmlkjihgfedcbazyxw" "vwxyzabcdefghijklmnopqrstutsrqponmlkjihgfedcbazyxwv" "uvwxyzabcdefghijklmnopqrstsrqponmlkjihgfedcbazyxwvu" "tuvwxyzabcdefghijklmnopqrsrqponmlkjihgfedcbazyxwvut" "stuvwxyzabcdefghijklmnopqrqponmlkjihgfedcbazyxwvuts" "rstuvwxyzabcdefghijklmnopqponmlkjihgfedcbazyxwvutsr" "qrstuvwxyzabcdefghijklmnoponmlkjihgfedcbazyxwvutsrq" "pqrstuvwxyzabcdefghijklmnonmlkjihgfedcbazyxwvutsrqp" "opqrstuvwxyzabcdefghijklmnmlkjihgfedcbazyxwvutsrqpo" "nopqrstuvwxyzabcdefghijklmlkjihgfedcbazyxwvutsrqpon" "mnopqrstuvwxyzabcdefghijklkjihgfedcbazyxwvutsrqponm" "lmnopqrstuvwxyzabcdefghijkjihgfedcbazyxwvutsrqponml" "klmnopqrstuvwxyzabcdefghijihgfedcbazyxwvutsrqponmlk" "jklmnopqrstuvwxyzabcdefghihgfedcbazyxwvutsrqponmlkj" "ijklmnopqrstuvwxyzabcdefghgfedcbazyxwvutsrqponmlkji" "hijklmnopqrstuvwxyzabcdefgfedcbazyxwvutsrqponmlkjih" "ghijklmnopqrstuvwxyzabcdefedcbazyxwvutsrqponmlkjihg" "fghijklmnopqrstuvwxyzabcdedcbazyxwvutsrqponmlkjihgf" "efghijklmnopqrstuvwxyzabcdcbazyxwvutsrqponmlkjihgfe" "defghijklmnopqrstuvwxyzabcbazyxwvutsrqponmlkjihgfed" "cdefghijklmnopqrstuvwxyzabazyxwvutsrqponmlkjihgfedc" "bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb" "abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba" "zabcdefghijklmnopqrstuvwxyxwvutsrqponmlkjihgfedcbaz" "yzabcdefghijklmnopqrstuvwxwvutsrqponmlkjihgfedcbazy" "xyzabcdefghijklmnopqrstuvwvutsrqponmlkjihgfedcbazyx" "wxyzabcdefghijklmnopqrstuvutsrqponmlkjihgfedcbazyxw" "vwxyzabcdefghijklmnopqrstutsrqponmlkjihgfedcbazyxwv" "uvwxyzabcdefghijklmnopqrstsrqponmlkjihgfedcbazyxwvu" "tuvwxyzabcdefghijklmnopqrsrqponmlkjihgfedcbazyxwvut" "stuvwxyzabcdefghijklmnopqrqponmlkjihgfedcbazyxwvuts" "rstuvwxyzabcdefghijklmnopqponmlkjihgfedcbazyxwvutsr" "qrstuvwxyzabcdefghijklmnoponmlkjihgfedcbazyxwvutsrq" "pqrstuvwxyzabcdefghijklmnonmlkjihgfedcbazyxwvutsrqp" "opqrstuvwxyzabcdefghijklmnmlkjihgfedcbazyxwvutsrqpo" "nopqrstuvwxyzabcdefghijklmlkjihgfedcbazyxwvutsrqpon" "mnopqrstuvwxyzabcdefghijklkjihgfedcbazyxwvutsrqponm" "lmnopqrstuvwxyzabcdefghijkjihgfedcbazyxwvutsrqponml" "klmnopqrstuvwxyzabcdefghijihgfedcbazyxwvutsrqponmlk" "jklmnopqrstuvwxyzabcdefghihgfedcbazyxwvutsrqponmlkj" "ijklmnopqrstuvwxyzabcdefghgfedcbazyxwvutsrqponmlkji" "hijklmnopqrstuvwxyzabcdefgfedcbazyxwvutsrqponmlkjih" "ghijklmnopqrstuvwxyzabcdefedcbazyxwvutsrqponmlkjihg" "fghijklmnopqrstuvwxyzabcdedcbazyxwvutsrqponmlkjihgf" "efghijklmnopqrstuvwxyzabcdcbazyxwvutsrqponmlkjihgfe" "defghijklmnopqrstuvwxyzabcbazyxwvutsrqponmlkjihgfed" "cdefghijklmnopqrstuvwxyzabazyxwvutsrqponmlkjihgfedc" "bcdefghijklmnopqrstuvwxyzazyxwvutsrqponmlkjihgfedcb") # Pyth, 21 19 bytes j+PKm+PJ.<Gd_JS27_K Try it online! Explanation: j+PKm+PJ.<Gd_JS27_K expects no input j joins on new line + + joins two strings P P prints everything but the last element K initialize K and implicitly print m for...in loop, uses d as iterator variable J initialize J and implicitly print .< cyclically rotate G initialized to the lowercase alphabet d iterating variables of m _ _ reverse J call J S27 indexed range of 27 K call K # SOGL V0.12, 10 bytes zl{«:}«¹╬, Try it Here! Explanation: z push the lowercase alphabet l{ } repeat length times « put the 1st letter at the end : duplicate « put the 1st letter at the end (as the last thing called is duplicate) ¹ wrap the stack in an array ╬, quad-palindromize with 1 X and Y overlap # Kotlin, 106 bytes {(0..52).map{i->(0..50).map{j->print((90-((Math.abs(j-25)+Math.abs(i-26)-1)+26)%26).toChar())} println()}} ## Beautified { (0..52).map {i-> (0..50).map {j-> print((90 - ((Math.abs(j - 25) + Math.abs(i - 26) - 1)+26) % 26).toChar()) } println() } } ## Test var v:()->Unit = {(0..52).map{i->(0..50).map{j->print((90-((Math.abs(j-25)+Math.abs(i-26)-1)+26)%26).toChar())} println()}} fun main(args: Array<String>) { v() } TryItOnline Port of @Karl Napf's answer VBA (Excel) , 116 Bytes Sub a() For i=-26To 26 For j=-25To 25 b=b & Chr(65+(52-(Abs(j)+Abs(i))) Mod 26) Next Debug.Print b b="" Next End Sub Following Sir Joffan's Logic. :D # VBA, 109 105 78 Bytes Anonymous VBE immediate window function that takes no input and outputs the alphabet diamond to the VBE immediate window. For i=-26To 26:For j=-25To 25:?Chr(65+(52-(Abs(j)+Abs(i)))Mod 26);:Next:?:Next # uBASIC, 86 bytes 0ForI=-26To26:ForJ=-25To25:?Left$(Chr\$(65+(52-(Abs(J)+Abs(I)))Mod26),1);:NextJ:?:NextI
Try it online!
# MY-BASIC, 89 bytes
Anonymous function that takes no input and outputs to the console.
For i=-26 To 26
For j=-25 To 25
Print Chr(65+(52-(Abs(j)+Abs(i)))Mod 26)
Next
Print;
Next
Try it online! | 2021-05-08 05:07:11 | {"extraction_info": {"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, "math_score": 0.1712559014558792, "perplexity": 4067.5758365242073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988837.67/warc/CC-MAIN-20210508031423-20210508061423-00319.warc.gz"} |
https://www.semanticscholar.org/paper/TensorFlow%3A-Large-Scale-Machine-Learning-on-Systems-Abadi-Agarwal/9c9d7247f8c51ec5a02b0d911d1d7b9e8160495d | • Corpus ID: 5707386
# TensorFlow: Large-Scale Machine Learning on Heterogeneous Distributed Systems
@article{Abadi2016TensorFlowLM,
title={TensorFlow: Large-Scale Machine Learning on Heterogeneous Distributed Systems},
author={Mart{\'i}n Abadi and Ashish Agarwal and Paul Barham and Eugene Brevdo and Z. Chen and Craig Citro and Gregory S. Corrado and Andy Davis and Jeffrey Dean and Matthieu Devin and Sanjay Ghemawat and Ian J. Goodfellow and Andrew Harp and Geoffrey Irving and Michael Isard and Yangqing Jia and Rafal J{\'o}zefowicz and Lukasz Kaiser and Manjunath Kudlur and Josh Levenberg and Dandelion Man{\'e} and Rajat Monga and Sherry Moore and Derek Gordon Murray and Christopher Olah and Mike Schuster and Jonathon Shlens and Benoit Steiner and Ilya Sutskever and Kunal Talwar and Paul A. Tucker and Vincent Vanhoucke and Vijay Vasudevan and Fernanda B. Vi{\'e}gas and Oriol Vinyals and Pete Warden and Martin Wattenberg and Martin Wicke and Yuan Yu and Xiaoqiang Zheng},
journal={ArXiv},
year={2016},
volume={abs/1603.04467}
}
TensorFlow is an interface for expressing machine learning algorithms, and an implementation for executing such algorithms. A computation expressed using TensorFlow can be executed with little or no change on a wide variety of heterogeneous systems, ranging from mobile devices such as phones and tablets up to large-scale distributed systems of hundreds of machines and thousands of computational devices such as GPU cards. The system is flexible and can be used to express a wide variety of…
8,938 Citations
TensorFlow: A system for large-scale machine learning
The TensorFlow dataflow model is described and the compelling performance that Tensor Flow achieves for several real-world applications is demonstrated.
Operator Vectorization Library – A TensorFlow Plugin
TensorFlow is an interface for implementing machine learning applications that can be accelerated by using Graphics Processing Units (GPUs). It is rapidly becoming a standard tool in this space.
Improving the Performance of Distributed TensorFlow with RDMA
This work presents a RDMA-capable design of TensorFlow, which shows a great scalability among the training scale and gets nearly 6$$\times$$× performance improvements over the original distributed Tensor Flow, based on gRPC.
TensorX: Extensible API for Neural Network Model Design and Deployment
• Computer Science
ArXiv
• 2020
TensorX is a Python library for prototyping, design, and deployment of complex neural network models in TensorFlow, aiming to make available high-level components like neural network layers that are, in effect, stateful functions, easy to compose and reuse.
SingleCaffe: An Efficient Framework for Deep Learning on a Single Node
SingleCaffe is presented, a DL framework that can make full use of hardware equipped with high computing power and improve the computational efficiency of the training process and the experimental results show that SingleCaffe can improve training efficiency well.
A Tour of TensorFlow
This paper reviews TensorFlow and puts it in context of modern deep learning concepts and software, its basic computational paradigms and distributed execution model, its programming interface as well as accompanying visualization toolkits and comment on observed use-cases of Tensor Flow in academia and industry.
DLVM : A MODERN COMPILER INFRASTRUCTURE FOR DEEP LEARNING
Many current approaches to deep learning make use of high-level toolkits such as TensorFlow, Torch, or Caffe. Toolkits such as Caffe have a layer-based programming framework with hard-coded gradients
Increasing Portable Machine Learning Performance by Application of Rewrite Rules on Google Tensorflow Data Flow Graphs
Machine Learning is an important field that is usually limited by execution performance. The approach commonly used to solve this problem is to make use of parallelism provided by hardware such as
Scalability Study of Deep Learning Algorithms in High Performance Computer Infrastructures
This project show how the training of a state-of-the-art neural network for computer vision can be parallelized on a distributed GPU cluster, Minotauro GPU cluster from Barcelona Supercomputing Center with the TensorFlow framework.
In-Database Machine Learning: Gradient Descent and Tensor Algebra for Main Memory Database Systems
• Computer Science
BTW
• 2019
This work aims to incorporate gradient descent and tensor data types into database systems, allowing them to handle a wider range of computational tasks, and implements tensor algebra and stochastic gradient descent using lambda expressions for loss functions as a pipelined operator in a main memory database system.
## References
SHOWING 1-10 OF 67 REFERENCES
Project Adam: Building an Efficient and Scalable Deep Learning Training System
• Computer Science
OSDI
• 2014
The design and implementation of a distributed system called Adam comprised of commodity server machines to train large deep neural network models that exhibits world-class performance, scaling and task accuracy on visual recognition tasks and shows that task accuracy improves with larger models.
Large Scale Distributed Deep Networks
This paper considers the problem of training a deep network with billions of parameters using tens of thousands of CPU cores and develops two algorithms for large-scale distributed training, Downpour SGD and Sandblaster L-BFGS, which increase the scale and speed of deep network training.
Caffe: Convolutional Architecture for Fast Feature Embedding
Caffe provides multimedia scientists and practitioners with a clean and modifiable framework for state-of-the-art deep learning algorithms and a collection of reference models for training and deploying general-purpose convolutional neural networks and other deep models efficiently on commodity architectures.
An introduction to computational networks and the computational network toolkit (invited talk)
The computational network toolkit (CNTK), an implementation of CN that supports both GPU and CPU, is introduced and the architecture and the key components of the CNTK are described, the command line options to use C NTK, and the network definition and model editing language are described.
Building high-level features using large scale unsupervised learning
• Quoc V. Le, +5 authors A. Ng
• Computer Science, Mathematics
2013 IEEE International Conference on Acoustics, Speech and Signal Processing
• 2013
Contrary to what appears to be a widely-held intuition, the experimental results reveal that it is possible to train a face detector without having to label images as containing a face or not.
Multilingual acoustic models using distributed deep neural networks
• G. Heigold, +4 authors J. Dean
• Computer Science
2013 IEEE International Conference on Acoustics, Speech and Signal Processing
• 2013
Experimental results for cross- and multi-lingual network training of eleven Romance languages on 10k hours of data in total show average relative gains over the monolingual baselines, but additional gain from jointly training the languages on all data comes at an increased training time of roughly four weeks.
cuDNN: Efficient Primitives for Deep Learning
A library similar in intent to BLAS, with optimized routines for deep learning workloads, that contains routines for GPUs, and similarly to the BLAS library, could be implemented for other platforms.
On rectified linear units for speech processing
This work shows that it can improve generalization and make training of deep networks faster and simpler by substituting the logistic units with rectified linear units.
Sequence to Sequence Learning with Neural Networks
• Computer Science
NIPS
• 2014
This paper presents a general end-to-end approach to sequence learning that makes minimal assumptions on the sequence structure, and finds that reversing the order of the words in all source sentences improved the LSTM's performance markedly, because doing so introduced many short term dependencies between the source and the target sentence which made the optimization problem easier.
Batch Normalization: Accelerating Deep Network Training by Reducing Internal Covariate Shift
• Computer Science
ICML
• 2015
Applied to a state-of-the-art image classification model, Batch Normalization achieves the same accuracy with 14 times fewer training steps, and beats the original model by a significant margin. | 2022-01-22 18:36:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.402389794588089, "perplexity": 3044.142396842096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00591.warc.gz"} |
http://aoi-okami-09.deviantart.com/ | # aoi-okami-09
Joey
## Favourites
blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah, blah blah blah blah blah blah blah blah blah blah blah blah blah, blah blah, blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah.
• Mood: Amused
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United Kingdom
Quizilla - quizilla.teennick.com/my/profi…
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## Groups
Admin of 1 Group Member of 42 Groups
Mar 1, 2014 Student Filmographer
Thanks for the watch! ^^
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No problem, I really like your artwork
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Hi there! Can I get ur attention 4 a moment please? I've posted a mouse character of mine called Chibi, I've given him 3 different colours n was wonderful which 1 colour u like best.
Dec 6, 2013 Student Artist
white, I don't know what it is but a white mouse looks cute.
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thank you very much for the
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No problem! =3
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Thanks for the fav
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no prob =3
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https://en.wikisource.org/wiki/Elementary_Principles_in_Statistical_Mechanics/Chapter_XIV | # Elementary Principles in Statistical Mechanics/Chapter XIV
CHAPTER XIV.
DISCUSSION OF THERMODYNAMIC ANALOGIES.
If we wish to find in rational mechanics an a priori foundation for the principles of thermodynamics, we must seek mechanical definitions of temperature and entropy. The quantities thus defined must satisfy (under conditions and with limitations which again must be specified in the language of mechanics) the differential equation
$d\epsilon = T d\eta - A_1 da_1 - A_2 da_2 - \mathrm{etc.} ,$ (482)
where $\epsilon$, $T$, and $\eta$ denote the energy, temperature, and entropy of the system considered, and $A_1 da_1$ etc., the mechanical work (in the narrower sense in which the term is used in thermodynamics, i. e., with exclusion of thermal action) done upon external bodies.
This implies that we are able to distinguish in mechanical terms the thermal action of one system on another from that which we call mechanical in the narrower sense, if not indeed in every case in which the two may be combined, at least so as to specify cases of thermal action and cases of mechanical action.
Such a differential equation moreover implies a finite equation between $\epsilon$, $\eta$, and $a_1$, $a_2$, etc., which may be regarded as fundamental in regard to those properties of the system which we call thermodynamic, or which may be called so from analogy. This fundamental thermodynamic equation is determined by the fundamental mechanical equation which expresses the energy of the system as function of its momenta and coördinates with those external coördinates ($a_1$, $a_2$, etc.) which appear in the differential expression of the work done on external bodies. We have to show the mathematical operations by which the fundamental thermodynamic equation, which in general is an equation of few variables, is derived from the fundamental mechanical equation, which in the case of the bodies of nature is one of an enormous number of variables.
We have also to enunciate in mechanical terms, and to prove, what we call the tendency of heat to pass from a system of higher temperature to one of lower, and to show that this tendency vanishes with respect to systems of the same temperature.
At least, we have to show by a priori reasoning that for such systems as the material bodies which nature presents to us, these relations hold with such approximation that they are sensibly true for human faculties of observation. This indeed is all that is really necessary to establish the science of thermodynamics on an a priori basis. Yet we will naturally desire to find the exact expression of those principles of which the laws of thermodynamics are the approximate expression. A very little study of the statistical properties of conservative systems of a finite number of degrees of freedom is sufficient to make it appear, more or less distinctly, that the general laws of thermodynamics are the limit toward which the exact laws of such systems approximate, when their number of degrees of freedom is indefinitely increased. And the problem of finding the exact relations, as distinguished from the approximate, for systems of a great number of degrees of freedom, is practically the same as that of finding the relations which hold for any number of degrees of freedom, as distinguished from those which have been established on an empirical basis for systems of a great number of degrees of freedom.
The enunciation and proof of these exact laws, for systems of any finite number of degrees of freedom, has been a principal object of the preceding discussion. But it should be distinctly stated that, if the results obtained when the numbers of degrees of freedom are enormous coincide sensibly with the general laws of thermodynamics, however interesting and significant this coincidence may be, we are still far from having explained the phenomena of nature with respect to these laws. For, as compared with the case of nature, the systems which we have considered are of an ideal simplicity. Although our only assumption is that we are considering conservative systems of a finite number of degrees of freedom, it would seem that this is assuming far too much, so far as the bodies of nature are concerned. The phenomena of radiant heat, which certainly should not be neglected in any complete system of thermodynamics, and the electrical phenomena associated with the combination of atoms, seem to show that the hypothesis of systems of a finite number of degrees of freedom is inadequate for the explanation of the properties of bodies.
Nor do the results of such assumptions in every detail appear to agree with experience. We should expect, for example, that a diatomic gas, so far as it could be treated independently of the phenomena of radiation, or of any sort of electrical manifestations, would have six degrees of freedom for each molecule. But the behavior of such a gas seems to indicate not more than five.
But although these difficulties, long recognized by physicists,[1] seem to prevent, in the present state of science, any satisfactory explanation of the phenomena of thermodynamics as presented to us in nature, the ideal case of systems of a finite number of degrees of freedom remains as a subject which is certainly not devoid of a theoretical interest, and which may serve to point the way to the solution of the far more difficult problems presented to us by nature. And if the study of the statistical properties of such systems gives us an exact expression of laws which in the limiting case take the form of the received laws of thermodynamics, its interest is so much the greater.
Now we have defined what we have called the modulus ($\Theta$) of an ensemble of systems canonically distributed in phase, and what we have called the index of probability ($\eta$) of any phase in such an ensemble. It has been shown that between the modulus ($\Theta$), the external coördinates ($a_1$, etc.), and the average values in the ensemble of the energy ($\epsilon$), the index of probability ($\eta$), and the external forces ($A_1$, etc.) exerted by the systems, the following differential equation will hold:
$d\overline\epsilon = - \Theta d\overline\eta - \overline A_1 da_1 - \overline A_2 da_2 - \mathrm{etc.}$ (483)
This equation, if we neglect the sign of averages, is identical in form with the thermodynamic equation (482), the modulus ($\Theta$) corresponding to temperature, and the index of probability of phase with its sign reversed corresponding to entropy.[2]
We have also shown that the average square of the anomalies of $\epsilon$, that is, of the deviations of the individual values from the average, is in general of the same order of magnitude as the reciprocal of the number of degrees of freedom, and therefore to human observation the individual values are indistinguishable from the average values when the number of degrees of freedom is very great.[3] In this case also the anomalies of $\eta$ are practically insensible. The same is true of the anomalies of the external forces ($A_1$, etc.), so far as these are the result of the anomalies of energy, so that when these forces are sensibly determined by the energy and the external coördinates, and the number of degrees of freedom is very great, the anomalies of these forces are insensible.
The mathematical operations by which the finite equation between $\overline\epsilon$, $\overline\eta$, and $a_1$, etc., is deduced from that which gives the energy ($\epsilon$) of a system in terms of the momenta ($p_1\ldots p_n$) and coördinates both internal ($q_1\ldots q_n$) and external ($a_1$, etc.), are indicated by the equation
$e^{-\frac{\psi}{\Theta}} = \mathop{\int\ldots\int}^{\rm all}_{\rm phases}\,dq_1\ldots dq_n \, dp_1 \ldots dp_n, ,$ (484)
where
$\psi = - \Theta \overline\eta + \overline\epsilon .$
We have also shown that when systems of different ensembles are brought into conditions analogous to thermal contact, the average result is a passage of energy from the ensemble of the greater modulus to that of the less,[4] or in case of equal moduli, that we have a condition of statistical equilibrium in regard to the distribution of energy,[5]
Propositions have also been demonstrated analogous to those in thermodynamics relating to a Carnot's cycle,[6] or to the tendency of entropy to increase,[7] especially when bodies of different temperature are brought into contact.[8]
We have thus precisely defined quantities, and rigorously demonstrated propositions, which hold for any number of degrees of freedom, and which, when the number of degrees of freedom ($n$) is enormously great, would appear to human faculties as the quantities and propositions of empirical thermodynamics.
It is evident, however, that there may be more than one quantity denned for finite values of $n$, which approach the same limit, when $n$ is increased indefinitely, and more than one proposition relating to finite values of $n$, which approach the same limiting form for $n = \infty$. There may be therefore, and there are, other quantities which may be thought to have some claim to be regarded as temperature and entropy with respect to systems of a finite number of degrees of freedom.
The definitions and propositions which we have been considering relate essentially to what we have called a canonical ensemble of systems. This may appear a less natural and simple conception than what we have called a microcanonical ensemble of systems, in which all have the same energy and which in many cases represents simply the time-ensemble, or ensemble of phases through which a single system passes in the course of time.
It may therefore seem desirable to find definitions and propositions relating to these microcanonical ensembles, which shall correspond to what in thermodynamics are based on experience. Now the differential equation
$d\epsilon = e^{-\phi} V \, d\log V - \overline{A_1}|_{\epsilon} \, da_1 - \overline{A_2}|_{\epsilon} \, da_2 - \mathrm{etc.} ,$ (485)
which has been demonstrated in Chapter X, and which relates to a microcanonical ensemble, $\overline{A_1}|_{\epsilon}$ denoting the average value of $A_1$ in such an ensemble, corresponds precisely to the thermodynamic equation, except for the sign of average applied to the external forces. But as these forces are not entirely determined by the energy with the external coördinates, the use of average values is entirely germane to the subject, and affords the readiest means of getting perfectly determined quantities. These averages, which are taken for a microcanonical ensemble, may seem from some points of view a more simple and natural conception than those which relate to a canonical ensemble. Moreover, the energy, and the quantity corresponding to entropy, are free from the sign of average in this equation.
The quantity in the equation which corresponds to entropy is $\log V$, the quantity $V$ being defined as the extension-in-phase within which the energy is less than a certain limiting value ($\epsilon$). This is certainly a more simple conception than the average value in a canonical ensemble of the index of probability of phase. $\operatorname{Log} V$ has the property that when it is constant
$d\epsilon = - \overline{A_1}|_{\epsilon} \, da_1 - \overline{A_2}|_{\epsilon} \, da_2 - \mathrm{etc.} , ,$ (486)
which closely corresponds to the thermodynamic property of entropy, that when it is constant
$d\epsilon = - A_1 \, da_1 - A_2 \, da_2 - \mathrm{etc.} ,$ (487)
The quantity in the equation which corresponds to temperature is $e^{-\phi}V$, or $d\epsilon/d\log V$. In a canonical ensemble, the average value of this quantity is equal to the modulus, as has been shown by different methods in Chapters IX and X.
In Chapter X it has also been shown that if the systems of a microcanonical ensemble consist of parts with separate energies, the average value of $e^{-\phi}V$ or any part is equal to its average value for any other part, and to the uniform value of the same expression for the whole ensemble. This corresponds to the theorem in the theory of heat that in case of thermal equilibrium the temperatures of the parts of a body are equal to one another and to that of the whole body. Since the energies of the parts of a body cannot be supposed to remain absolutely constant, even where this is the case with respect to the whole body, it is evident that if we regard the temperature as a function of the energy, the taking of average or of probable values, or some other statistical process, must be used with reference to the parts, in order to get a perfectly definite value corresponding to the notion of temperature.
It is worthy of notice in this connection that the average value of the kinetic energy, either in a microcanonical ensemble, or in a canonical, divided by one half the number of degrees of freedom, is equal to $e^{-\phi}V$, or to its average value, and that this is true not only of the whole system which is distributed either microcanonically or canonically, but also of any part, although the corresponding theorem relating to temperature hardly belongs to empirical thermodynamics, since neither the (inner) kinetic energy of a body, nor its number of degrees of freedom is immediately cognizable to our faculties, and we meet the gravest difficulties when we endeavor to apply the theorem to the theory of gases, except in the simplest case, that of the gases known as monatomic.
But the correspondence between $e^{-\phi}V$ or $d\epsilon/d\log V$ and temperature is imperfect. If two isolated systems have such energies that
$\frac{d\epsilon_1}{d\log V_1} = \frac{d\epsilon_2}{d\log V_2} ,$
and the two systems are regarded as combined to form a third system with energy
$\epsilon_{12} = \epsilon_1 + \epsilon_2 ,$
we shall not have in general
$\frac{d\epsilon_{12}}{d\log V_{12}} = \frac{d\epsilon_1}{d\log V_1} = \frac{d\epsilon_2}{d\log V_2} ,$
as analogy with temperature would require. In fact, we have seen that
$\frac{d\epsilon_{12}}{d\log V_{12}} = \overline{\frac{d\epsilon_1}{d\log V_1}}\bigg|_{\epsilon_{12}} = \overline{\frac{d\epsilon_2}{d\log V_2}}\bigg|_{\epsilon_{12}} ,$
where the second and third members of the equation denote average values in an ensemble in which the compound system is microcanonically distributed in phase. Let us suppose the two original systems to be identical in nature. Then
$\epsilon_1 = \epsilon_2 = \overline{\epsilon_1}|_{\epsilon_{12}} = \overline{\epsilon_2}|_{\epsilon_{12}} .$
The equation in question would require that
$\frac{d\epsilon_{1}}{d\log V_{1}} = \overline{\frac{d\epsilon_1}{d\log V_1}}\bigg|_{\epsilon_{12}} ,$
i. e., that we get the same result, whether we take the value of $d\epsilon_1/d\log V_1$ determined for the average value of $\epsilon_1$ in the ensemble, or take the average value of $d\epsilon_1/d\log V_1$. This will be the case where $d\epsilon_1/d\log V_1$ is a linear function of $\epsilon_1$. Evidently this does not constitute the most general case. Therefore the equation in question cannot be true in general. It is true, however, in some very important particular cases, as when the energy is a quadratic function of the $p$'s and $q$'s, or of the $p$'s alone.[9] When the equation holds, the case is analogous to that of bodies in thermodynamics for which the specific heat for constant volume is constant.
Another quantity which is closely related to temperature is $d\phi/d\epsilon$. It has been shown in Chapter IX that in a canonical ensemble, if $n>2$, the average value of $d\phi/d\epsilon$ is $1/\Theta$, and that the most common value of the energy in the ensemble is that for which $d\phi/d\epsilon = 1/\Theta$. The first of these properties may be compared with that of $d\epsilon/d\log V$, which has been seen to have the average value $\Theta$ in a canonical ensemble, without restriction in regard to the number of degrees of freedom.
With respect to microcanonical ensembles also, $d\phi/d\epsilon$ has a property similar to what has been mentioned with respect to $d\epsilon/d\log V$. That is, if a system microcanonically distributed in phase consists of two parts with separate energies, and each with more than two degrees of freedom, the average values in the ensemble of $d\phi/d\epsilon$ for the two parts are equal to one another and to the value of same expression for the whole. In our usual notations
$\overline{\frac{d\phi_1}{d\epsilon_1}}\bigg|_{\epsilon_{12}} = \overline{\frac{d\phi_2}{d\epsilon_2}}\bigg|_{\epsilon_{12}} = \frac{d\phi_{12}}{d\epsilon_{12}}$
if $n_1 > 2$, and $n_2 > 2$.
This analogy with temperature has the same incompleteness which was noticed with respect to $d\epsilon/d\log V$, viz., if two systems have such energies ($\epsilon_1$ and $\epsilon_2$) that
$\frac{d\phi_1}{d\epsilon_1} = \frac{d\phi_2}{d\epsilon_2} ,$
and they are combined to form a third system with energy
$\epsilon_{12} = \epsilon_1 + \epsilon_2 ,$
we shall not have in general
$\frac{d\phi_{12}}{d\epsilon_{12}} = \frac{d\phi_1}{d\epsilon_1} = \frac{d\phi_2}{d\epsilon_2} .$
Thus, if the energy is a quadratic function of the $p$'s and $q$'s, we have[10]
$\frac{d\phi_1}{d\epsilon_1} = \frac{n_1 - 1}{\epsilon_1}, \qquad \frac{d\phi_2}{d\epsilon_2} = \frac{n_2 - 1}{\epsilon_2} ,$
$\frac{d\phi_{12}}{d\epsilon_{12}} = \frac{n_{12} - 1}{\epsilon_{12}} = \frac{n_1 + n_2 - 1}{\epsilon_1 + \epsilon_2} ,$
where $n_1$, $n_2$, $n_{12}$, are the numbers of degrees of freedom of the separate and combined systems. But
$\frac{d\phi_1}{d\epsilon_1} = \frac{d\phi_2}{d\epsilon_2} = \frac{n_1 + n_2 - 2}{\epsilon_1 + \epsilon_2} .$
If the energy is a quadratic function of the $p$'s alone, the case would be the same except that we should have $\tfrac 12 n_1$, $\tfrac 12 n_2$, $\tfrac 12 n_{12}$, instead of $n_1$, $n_2$, $n_{12}$. In these particular cases, the analogy between $d\epsilon/d\log V$ and temperature would be complete, as has already been remarked. We should have
$\frac{d\epsilon_{1}}{d\log V_{1}} = \frac{\epsilon_1}{n_1}, \qquad \frac{d\epsilon_{2}}{d\log V_{2}} = \frac{\epsilon_2}{n_2} ,$
$\frac{d\epsilon_{12}}{d\log V_{12}} = \frac{\epsilon_{12}}{n_{12}} = \frac{d\epsilon_{1}}{d\log V_{1}} = \frac{d\epsilon_{2}}{d\log V_{2}} ,$
when the energy is a quadratic function of the $p$'s and $q$'s, and similar equations with $\tfrac 12 n_1$, $\tfrac 12 n_2$, $\tfrac 12 n_{12}$, instead of $n_1$, $n_2$, $n_{12}$, when the energy is a quadratic function of the $p$'s alone.
More characteristic of $d\phi/d\epsilon$ are its properties relating to most probable values of energy. If a system having two parts with separate energies and each with more than two degrees of freedom is microcanonically distributed in phase, the most probable division of energy between the parts, in a system taken at random from the ensemble, satisfies the equation
$\frac{d\phi_1}{d\epsilon_1} = \frac{d\phi_2}{d\epsilon_2} ,$ (488)
which corresponds to the thermodynamic theorem that the distribution of energy between the parts of a system, in case of thermal equilibrium, is such that the temperatures of the parts are equal.
To prove the theorem, we observe that the fractional part of the whole number of systems which have the energy of one part ($\epsilon_1$) between the limits $\epsilon_1'$ and $\epsilon_1''$ is expressed by
$e^{-\phi_{12}} \int_{\epsilon_1'}^{\epsilon_1''} e^{\phi_1 + \phi_2} \, d\epsilon_1 ,$
where the variables are connected by the equation
$\epsilon_1 + \epsilon_2 = \mathrm{constant} = \epsilon_{12} .$
The greatest value of this expression, for a constant infinitesimal value of the difference $\epsilon_1'' - \epsilon_1'$, determines a value of $\epsilon_1$, which we may call its most probable value. This depends on the greatest possible value of $\phi_1 + \phi_2$. Now if $n_1 > 2$, and $n_2 > 2$, we shall have $\phi_1 = -\infty$ for the least possible value of $\epsilon_1$, and $\phi_2 = -\infty$ for the least possible value of $\epsilon_2$. Between these limits $\phi_1$ and $\phi_2$ will be finite and continuous. Hence $\phi_1 + \phi_2$ will have a maximum satisfying the equation (488).
But if $n_1 \leq 2$, or $n_2 \leq 2$, $d\phi_1/d\epsilon_1$ or $d\phi_2/d\epsilon_2$ may be negative, or zero, for all values of $\epsilon_1$ or $\epsilon_2$, and can hardly be regarded as having properties analogous to temperature.
It is also worthy of notice that if a system which is microcanonically distributed in phase has three parts with separate energies, and each with more than two degrees of freedom, the most probable division of energy between these parts satisfies the equation
$\frac{d\phi_1}{d\epsilon_1} = \frac{d\phi_2}{d\epsilon_2} = \frac{d\phi_3}{d\epsilon_3} .$
That is, this equation gives the most probable set of values of $\epsilon_1$, $\epsilon_2$, and $\epsilon_3$. But it does not give the most probable value of $\epsilon_1$, or of $\epsilon_2$, or of $\epsilon_3$. Thus, if the energies are quadratic functions of the $p$'s and $q$'s, the most probable division of energy is given by the equation
$\frac{n_1 - 1}{\epsilon_1} = \frac{n_2 - 1}{\epsilon_1} = \frac{n_3 - 1}{\epsilon_3} .$
But the most probable value of $\epsilon_1$ is given by
$\frac{n_1 - 1}{\epsilon_1} = \frac{n_2 + n_3 - 1}{\epsilon_2 + \epsilon_3} ,$
while the preceding equations give
$\frac{n_1 - 1}{\epsilon_1} = \frac{n_2 + n_3 - 2}{\epsilon_2 + \epsilon_3} .$
These distinctions vanish for very great values of $n_1$, $n_2$, $n_3$. For small values of these numbers, they are important. Such facts seem to indicate that the consideration of the most probable division of energy among the parts of a system does not afford a convenient foundation for the study of thermodynamic analogies in the case of systems of a small number of degrees of freedom. The fact that a certain division of energy is the most probable has really no especial physical importance, except when the ensemble of possible divisions are grouped so closely together that the most probable division may fairly represent the whole. This is in general the case, to a very close approximation, when $n$ is enormously great; it entirely fails when $n$ is small.
If we regard $d\phi/d\epsilon$ as corresponding to the reciprocal of temperature, or, in other words, $d\epsilon/d\phi$ as corresponding to temperature, $\phi$ will correspond to entropy. It has been defined as $\log(dV/d\epsilon)$. In the considerations on which its definition is founded, it is therefore very similar to $\log V$. We have seen that $d\phi/d\log V$ approaches the value unity when $n$ is very great.[11]
To form a differential equation on the model of the thermodynamic equation (482), in which $d\epsilon/d\phi$ shall take the place of temperature, and $\phi$ of entropy, we may write
$d\epsilon = \bigg(\frac{d\epsilon}{d\phi}\bigg)_a\, d\phi + \bigg(\frac{d\epsilon}{da_1}\bigg)_{\phi,a}\, da_1 + \bigg(\frac{d\epsilon}{da_2}\bigg)_{\phi,a}\, da_2 + \mathrm{etc.} ,$ (489)
or
$d\phi = \frac{d\phi}{d\epsilon}\,d\epsilon + \frac{d\phi}{da_1}\,da_1 + \frac{d\phi}{da_2}\,da_2 + \mathrm{etc.}$ (490)
With respect to the differential coefficients in the last equation, which corresponds exactly to (482) solved with respect to $d\eta$, we have seen that their average values in a canonical ensemble are equal to $1/\Theta$, and the averages of $A_1/\Theta$, $A_2/\Theta$, etc.[12] We have also seen that $d\epsilon/d\phi$ (or $d\phi/d\epsilon$) has relations to the most probable values of energy in parts of a microcanonical ensemble. That $(d\epsilon/da_1)_{\phi,a}$, etc., have properties somewhat analogous, may be shown as follows.
In a physical experiment, we measure a force by balancing it against another. If we should ask what force applied to increase or diminish $a_1$ would balance the action of the systems, it would be one which varies with the different systems. But we may ask what single force will make a given value of $a_1$ the most probable, and we shall find that under certain conditions $(d\epsilon/da_1)_{\phi,a}$, a represents that force.
To make the problem definite, let us consider a system consisting of the original system together with another having the coördinates $a_1$, $a_2$, etc., and forces $A_1'$, $A_2'$ etc., tending to increase those coördinates. These are in addition to the forces $A_1$, $A_2$, etc., exerted by the original system, and are derived from a force-function ($-\epsilon_q'$) by the equations
$A_1' = -\frac{d\epsilon_q'}{da_1} , \quad A_2' = -\frac{d\epsilon_q'}{da_2} , \quad \mathrm{etc.}$
For the energy of the whole system we may write
$E = \epsilon + \epsilon_q' + \tfrac 12 m_1 \dot a_1^2 + \tfrac 12 m_2 \dot a_2^2 + \mathrm{etc.} ,$
and for the extension-in-phase of the whole system within any limits
$\int \ldots \int dp_1 \ldots dq_n \, da_1 \, m_1 \, d\dot a_1 \, da_2 \, m_2 \, d\dot a_2 \ldots$
or
$\int e^\phi \, d\epsilon \, da_1 \, m_1 \, d\dot a_1 \, da_2 \, m_2 \, d\dot a_2 \ldots ,$
or again
$\int e^\phi \, dE \, da_1 \, m_1 \, d\dot a_1 \, da_2 \, m_2 \, d\dot a_2 \ldots ,$
since $d\epsilon = dE$, when $a_1$, $\dot a_1$, $a_2$, $\dot a_2$, etc., are constant. If the limits are expressed by $E$ and $E + dE$, $a_1$ and $a_1 + da_1$, $\dot a_1$ and $a_1 + d\dot a_1$, etc., the integral reduces to
$e^\phi \, dE \, da_1 \, m_1 \, d\dot a_1 \, da_2 \, m_2 \, d\dot a_2 \ldots$
The values of $a_1$, $\dot a_1$, $a_2$, $\dot a_2$, etc., which make this expression a maximum for constant values of the energy of the whole system and of the differentials $dE$, $da_1$, $d\dot a_1$, etc., are what may be called the most probable values of $a_1$, $\dot a_1$, etc., in an ensemble in which the whole system is distributed microcanonically. To determine these values we have
$de^\phi = 0 ,$
when
$d(\epsilon + \epsilon_q' + \tfrac 12 m_1 \dot a_1^2 + \tfrac 12 m_2 \dot a_2^2 + \mathrm{etc.}) = 0 .$
That is,
$d\phi = 0 ,$
when
$\bigg(\frac{d\epsilon}{d\phi}\bigg)_a\, d\phi + \bigg(\frac{d\epsilon}{da_1}\bigg)_{\phi,a}\, da_1 - A_1' da_1 + \mathrm{etc.} + m_1 \dot a_1 d\dot a_1 + \mathrm{etc.} = 0 .$
This requires
$\dot a_1 = 0, \quad \dot a_2 = 0, \quad \mathrm{etc.} ,$
and
$\bigg(\frac{d\epsilon}{da_1}\bigg)_{\phi,a} = A_1', \quad \bigg(\frac{d\epsilon}{da_2}\bigg)_{\phi,a} = A_2', \quad \mathrm{etc.}$
This shows that for any given values of $E$, $a_1$, $a_2$, etc. $\Big(\frac{d\epsilon}{da_1}\Big)_{\phi,a}$, $\Big(\frac{d\epsilon}{da_2}\Big)_{\phi,a}$, etc., represent the forces (in the generalized sense) which the external bodies would have to exert to make these values of $a_1$, $a+2$, etc., the most probable under the conditions specified. When the differences of the external forces which are exerted by the different systems are negligible,—$(d\epsilon/da_1)_{\phi,a}$, etc., represent these forces.
It is certainly in the quantities relating to a canonical ensemble, $\overline\epsilon$, $\Theta$, $\overline\eta$, $\overline A_1$, etc., $a_1$, etc. that we find the most complete correspondence with the quantities of the thermodynamic equation (482). Yet the conception itself of the canonical ensemble may seem to some artificial, and hardly germane to a natural exposition of the subject; and the quantities $\epsilon$, $\frac{d\epsilon}{d\log V}$, $\log V$, $\overline{A_1}|_{\epsilon}$, etc., $a_1$, etc., or $\epsilon$, $\frac{d\epsilon}{d\phi}$, $\phi$, $\Big(\frac{d\epsilon}{da_1}\Big)_{\phi,a}$, etc., $a_1$, etc., which are closely related to ensembles of constant energy, and to average and most probable values in such ensembles, and most of which are defined without reference to any ensemble, may appear the most natural analogues of the thermodynamic quantities.
In regard to the naturalness of seeking analogies with the thermodynamic behavior of bodies in canonical or microcanonical ensembles of systems, much will depend upon how we approach the subject, especially upon the question whether we regard energy or temperature as an independent variable.
It is very natural to take energy for an independent variable rather than temperature, because ordinary mechanics furnishes us with a perfectly defined conception of energy, whereas the idea of something relating to a mechanical system and corresponding to temperature is a notion but vaguely defined. Now if the state of a system is given by its energy and the external coördinates, it is incompletely defined, although its partial definition is perfectly clear as far as it goes. The ensemble of phases microcanonically distributed, with the given values of the energy and the external coördinates, will represent the imperfectly defined system better than any other ensemble or single phase. When we approach the subject from this side, our theorems will naturally relate to average values, or most probable values, in such ensembles.
In this case, the choice between the variables of (485) or of (489) will be determined partly by the relative importance which is attached to average and probable values. It would seem that in general average values are the most important, and that they lend themselves better to analytical transformations. This consideration would give the preference to the system of variables in which $\log V$ is the analogue of entropy. Moreover, if we make $\phi$ the analogue of entropy, we are embarrassed by the necessity of making numerous exceptions for systems of one or two degrees of freedom.
On the other hand, the definition of $\phi$ may be regarded as a little more simple than that of $\log V$, and if our choice is determined by the simplicity of the definitions of the analogues of entropy and temperature, it would seem that the $\phi$ system should have the preference. In our definition of these quantities, $V$ was defined first, and $e^\phi$ derived from $V$ by differentiation. This gives the relation of the quantities in the most simple analytical form. Yet so far as the notions are concerned, it is perhaps more natural to regard $\phi$ as derived from $e^\phi$ by integration. At all events, $e^\phi$ may be defined independently of $V$, and its definition may be regarded as more simple as not requiring the determination of the zero from which $V$ is measured, which sometimes involves questions of a delicate nature. In fact, the quantity $e^\phi$ may exist, when the definition of $V$ becomes illusory for practical purposes, as the integral by which it is determined becomes infinite.
The case is entirely different, when we regard the temperature as an independent variable, and we have to consider a system which is described as having a certain temperature and certain values for the external coördinates. Here also the state of the system is not completely defined, and will be better represented by an ensemble of phases than by any single phase. What is the nature of such an ensemble as will best represent the imperfectly defined state?
When we wish to give a body a certain temperature, we place it in a bath of the proper temperature, and when we regard what we call thermal equilibrium as established, we say that the body has the same temperature as the bath. Perhaps we place a second body of standard character, which we call a thermometer, in the bath, and say that the first body, the bath, and the thermometer, have all the same temperature.
But the body under such circumstances, as well as the bath, and the thermometer, even if they were entirely isolated from external influences (which it is convenient to suppose in a theoretical discussion), would be continually changing in phase, and in energy as well as in other respects, although our means of observation are not fine enough to perceive these variations.
The series of phases through which the whole system runs in the course of time may not be entirely determined by the energy, but may depend on the initial phase in other respects. In such cases the ensemble obtained by the microcanonical distribution of the whole system, which includes all possible time-ensembles combined in the proportion which seems least arbitrary, will represent better than any one time-ensemble the effect of the bath. Indeed a single time-ensemble, when it is not also a microcanonical ensemble, is too ill-defined a notion to serve the purposes of a general discussion. We will therefore direct our attention, when we suppose the body placed in a bath, to the microcanonical ensemble of phases thus obtained.
If we now suppose the quantity of the substance forming the bath to be increased, the anomalies of the separate energies of the body and of the thermometer in the microcanonical ensemble will be increased, but not without limit. The anomalies of the energy of the bath, considered in comparison with its whole energy, diminish indefinitely as the quantity of the bath is increased, and become in a sense negligible, when the quantity of the bath is sufficiently increased. The ensemble of phases of the body, and of the thermometer, approach a standard form as the quantity of the bath is indefinitely increased. This limiting form is easily shown to be what we have described as the canonical distribution.
Let us write $\epsilon$ for the energy of the whole system consisting of the body first mentioned, the bath, and the thermometer (if any), and let us first suppose this system to be distributed canonically with the modulus $\Theta$. We have by (205)
$\overline{(\epsilon - \overline \epsilon)^2} = \Theta^2 \frac{d\overline\epsilon}{d\Theta} ,$
and since
$\overline\epsilon_p = \frac n2 \Theta ,$
$\frac{d\overline\epsilon}{d\Theta} = \frac n2 \frac{d\overline\epsilon}{d\overline\epsilon_p} .$
If we write $\Delta \epsilon$ for the anomaly of mean square, we have
$(\Delta \epsilon)^2 = \overline{(\epsilon - \overline \epsilon)^2} .$
If we set
$\Delta \Theta = \frac{d\Theta}{d\overline \epsilon}\Delta \epsilon ,$
$\Delta \Theta$ will represent approximately the increase of $\Theta$ which would produce an increase in the average value of the energy equal to its anomaly of mean square. Now these equations give
$(\Delta \Theta)^2 = \frac{2\Theta^2}{n} \frac{d\overline\epsilon_p}{d\overline\epsilon} ,$
which shows that we may diminish $\Delta \Theta$ indefinitely by increasing the quantity of the bath.
Now our canonical ensemble consists of an infinity of microcanonical ensembles, which differ only in consequence of the different values of the energy which is constant in each. If we consider separately the phases of the first body which occur in the canonical ensemble of the whole system, these phases will form a canonical ensemble of the same modulus. This canonical ensemble of phases of the first body will consist of parts which belong to the different microcanonical ensembles into which the canonical ensemble of the whole system is divided.
Let us now imagine that the modulus of the principal canonical ensemble is increased by $2\Delta \Theta$, and its average energy by $2\Delta \epsilon$. The modulus of the canonical ensemble of the phases of the first body considered separately will be increased by $2\Delta \Theta$. We may regard the infinity of microcanonical ensembles into which we have divided the principal canonical ensemble as each having its energy increased by $2\Delta \epsilon$. Let us see how the ensembles of phases of the first body contained in these microcanonical ensembles are affected. We may assume that they will all be affected in about the same way, as all the differences which come into account may be treated as small. Therefore, the canonical ensemble formed by taking them together will also be affected in the same way. But we know how this is affected. It is by the increase of its modulus by $2\Delta \Theta$, a quantity which vanishes when the quantity of the bath is indefinitely increased.
In the case of an infinite bath, therefore, the increase of the energy of one of the microcanonical ensembles by $2\Delta \epsilon$, produces a vanishing effect on the distribution in energy of the phases of the first body which it contains. But $2\Delta \epsilon$ is more than the average difference of energy between the microcanonical ensembles. The distribution in energy of these phases is therefore the same in the different microcanonical ensembles, and must therefore be canonical, like that of the ensemble which they form when taken together.[13] As a general theorem, the conclusion may be expressed in the words:—If a system of a great number of degrees of freedom is microcanonically distributed in phase, any very small part of it may be regarded as canonically distributed.[14]
It would seem, therefore, that a canonical ensemble of phases is what best represents, with the precision necessary for exact mathematical reasoning, the notion of a body with a given temperature, if we conceive of the temperature as the state produced by such processes as we actually use in physics to produce a given temperature. Since the anomalies of the body increase with the quantity of the bath, we can only get rid of all that is arbitrary in the ensemble of phases which is to represent the notion of a body of a given temperature by making the bath infinite, which brings us to the canonical distribution.
A comparison of temperature and entropy with their analogues in statistical mechanics would be incomplete without a consideration of their differences with respect to units and zeros, and the numbers used for their numerical specification. If we apply the notions of statistical mechanics to such bodies as we usually consider in thermodynamics, for which the kinetic energy is of the same order of magnitude as the unit of energy, but the number of degrees of freedom is enormous, the values of $\Theta$, $d\epsilon/d\log V$, and $d\epsilon/d\phi$ will be of the same order of magnitude as $1/n$, and the variable part of $\overline \eta$, $\log V$, and $\phi$ will be of the same order of magnitude as $n$.[15] If these quantities, therefore, represent in any sense the notions of temperature and entropy, they will nevertheless not be measured in units of the usual order of magnitude,—a fact which must be borne in mind in determining what magnitudes may be regarded as insensible to human observation.
Now nothing prevents our supposing energy and time in our statistical formulae to be measured in such units as may be convenient for physical purposes. But when these units have been chosen, the numerical values of $\Theta$, $d\epsilon/d\log V$, $d\epsilon/d\phi$, $\overline \eta$, $\log V$, $\phi$, are entirely determined,[16] and in order to compare them with temperature and entropy, the numerical values of which depend upon an arbitrary unit, we must multiply all values of $\Theta$, $d\epsilon/d\log V$, $d\epsilon/d\phi$ by a constant ($K$), and divide all values of $\overline \eta$, $\log V$, and $\phi$ by the same constant. This constant is the same for all bodies, and depends only on the units of temperature and energy which we employ. For ordinary units it is of the same order of magnitude as the numbers of atoms in ordinary bodies.
We are not able to determine the numerical value of $K$ as it depends on the number of molecules in the bodies with which we experiment. To fix our ideas, however, we may seek an expression for this value, based upon very probable assumptions, which will show how we would naturally proceed to its evaluation, if our powers of observation were fine enough to take cognizance of individual molecules.
If the unit of mass of a monatomic gas contains $\nu$ atoms, and it may be treated as a system of $3\nu$ degrees of freedom, which seems to be the case, we have for canonical distribution
$\overline\epsilon_p = \tfrac 32 \nu \Theta ,$
$\frac{d\overline\epsilon_p}{d\Theta} = \tfrac 32 \nu .$ (491)
If we write $T$ for temperature, and $c_v$ for the specific heat of the gas for constant volume (or rather the limit toward which this specific heat tends, as rarefaction is indefinitely increased), we have
$\frac{d\epsilon_p}{dT} = c_v ,$ (492)
since we may regard the energy as entirely kinetic. We may set the $\epsilon_p$ of this equation equal to the $\overline\epsilon_p$ of the preceding, where indeed the individual values of which the average is taken would appear to human observation as identical. This gives
$\frac{d\Theta}{dT} = \frac{2c_v}{3\nu} ,$ (X)
whence
$\frac 1K = \frac{2c_v}{3\nu} .$ (493)
a value recognized by physicists as a constant independent of the kind of monatomic gas considered.
We may also express the value of $K$ in a somewhat different form, which corresponds to the indirect method by which physicists are accustomed to determine the quantity $c_v$. The kinetic energy due to the motions of the centers of mass of the molecules of a mass of gas sufficiently expanded is easily shown to be equal to
$\tfrac 32 pv ,$
where $p$ and $v$ denote the pressure and volume. The average value of the same energy in a canonical ensemble of such a mass of gas is
$\tfrac 32 \Theta \nu ,$
where $\nu$ denotes the number of molecules in the gas. Equating these values, we have
$pv = \Theta \nu ,$ (494)
whence
$\frac 1K = \frac \Theta T = \frac{pv}{\nu T} .$ (495)
Now the laws of Boyle, Charles, and Avogadro may be expressed by the equation
$pv = A \nu T ,$ (496)
where $A$ is a constant depending only on the units in which energy and temperature are measured. $1/K$, therefore, might be called the constant of the law of Boyle, Charles, and Avogadro as expressed with reference to the true number of molecules in a gaseous body.
Since such numbers are unknown to us, it is more convenient to express the law with reference to relative values. If we denote by $M$ the so-called molecular weight of a gas, that is, a number taken from a table of numbers proportional to the weights of various molecules and atoms, but having one of the values, perhaps the atomic weight of hydrogen, arbitrarily made unity, the law of Boyle, Charles, and Avogadro may be written in the more practical form
$p v = A' T \frac mM,$ (497)
where $A'$ is a constant and m the weight of gas considered. It is evident that $1~K$ is equal to the product of the constant of the law in this form and the (true) weight of an atom of hydrogen, or such other atom or molecule as may be given the value unity in the table of molecular weights.
In the following chapter we shall consider the necessary modifications in the theory of equilibrium, when the quantity of matter contained in a system is to be regarded as variable, or, if the system contains more than one kind of matter, when the quantities of the several kinds of matter in the system are to be regarded as independently variable. This will give us yet another set of variables in the statistical equation, corresponding to those of the amplified form of the thennodynamic equation.
1. See Boltzmann, Sitzb. der Wiener Akad., Bd. LXIIL, S. 418, (1871).
2. See Chapter IV, pages 44, 45.
3. See Chapter VII, pages 73-75.
4. See Chapter XIII, page 160.
5. See Chapter IV, pages 35-37.
6. See Chapter XIII, pages 162, 163.
7. See Chapter XII, pages 143-151.
8. See Chapter XIII, page 159.
9. This last case is important on account of its relation to the theory of gases, although it must in strictness be regarded as a limit of possible cases, rather than as a case which is itself possible.
10. See foot-note on page 93. We have here made the least value of the energy consistent with the values of the external coördinates zero instead of $\epsilon_a$, as is evidently allowable when the external coördinates are supposed invariable.
11. See Chapter X, pages 120, 121.
12. See Chapter IX, equations (321), (327).
13. In order to appreciate the above reasoning, it should be understood that the differences of energy which occur in the canonical ensemble of phases of the first body are not here regarded as vanishing quantities. To fix one's ideas, one may imagine that he has the fineness of perception to make these differences seem large. The difference between the part of these phases which belong to one microcanonical ensemble of the whole system and the part which belongs to another would still be imperceptible, when the quantity of the bath is sufficiently increased.
14. It is assumed—and without this assumption the theorem would have no distinct meaning—that the part of the ensemble considered may be regarded as having separate energy.
15. See equations (124), (288), (289), and (314); also page 106.
16. The unit of time only affects the last three quantities, and these only by an additive constant, which disappears (with the additive constant of entropy), when differences of entropy are compared with their statistical analogues. See page 19. | 2016-05-06 14:57:09 | {"extraction_info": {"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": 322, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8862664699554443, "perplexity": 228.54111058866582}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461861831994.45/warc/CC-MAIN-20160428164351-00052-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://quantumprogress.wordpress.com/2011/09/12/telling-the-full-story-of-a-number/ | I tell my students every number in science tells a story—the story of a measurement. The way you write the number tells the reader how it was measured, and the units you use tell the reader what it was that you measured.
I use a number of stories to get students to think about this. I suggest that they think of writing a number down with the same effort it takes to spell the number out in cursive on a check—One thousand, one hundred and fifteen (antiquated, I know). We talk about how arabic numerals really are a tremendous gift for allowing us to communicate very complex concepts $6.02\times 10^{23}$ atoms in a mole, in a tiny number of marks on the page, and to imagine how much harder it would be to express ideas like this using Roman Numerals, or even hash marks.
It is really hard for students to engrain this habit of fully telling the stories of the numbers they write. In the era of cheap calculation, it’s far too easy to write $\frac{1}{3}$, $0.\bar{3}$ or $0.3333333333$, even if the latter requires 10 times more ink than the most thoughtful answer. We sort of make a big deal out of it—we call numbers without units naked numbers, and I hide my eyes whenever I see them.
To further help them with the habit, I tell my students that on any concept involving a calculation, they haven’t shown full understanding until they solve the problem while telling the full story of every number, even those in intermediate steps. Even if a student writes $4\times 5=20$ off in the corner, I consider this less than full understanding if they don’t clothe their numbers with units. I do this, because I think it’s a lot easier to build the habit of “I will tell the story of every number I write” than it is to build the habit of “I will tell the story of the last number I write,” and when students get in the habit of writing the complete story of their numbers, they’ll see that it is one of the most powerful aids in problem solving they have.
Two problems have come up with this approach however. The first is that too many of my students see their problems as “writing naked numbers” rather than telling the full story of the number. This leads me to believe they aren’t getting the big picture.
The second problem with this is that we haven’t really gotten a great way to illustrate the value of precision in measurements the lab. Our work with buggies really doesn’t require great precision, and when they find that their prediction is off from the actual result, they chalk it up to “human error” (my pet peeve) or the fact that the buggies don’t move in straight lines. What I really want is an activity early on that pushes them to see just how precisely they can predict something. And I realized too late that we really have that with Dan Meyer’s boat in the river problem. If you don’t measure the time very precisely, by stopping the video and scrubbing to the moment when Dan reaches the top of the stairs, you’ll end up with a time that’s wildly off.
Still this isn’t quite what I want from the lab—I’d like for my students to see two things in working with measurement:
• Improving precision of measurements can allow us to discriminate previously unseen things. For instance, it was improvements in our ability to measure time using atomic clocks that gave us experimental proof of special relativity.
• Improved prevision can’t be had for free, it comes with costs, usually in terms of time and cost. Nobel prizes have often been won for adding a few decimal places on to measuring one quantity or another, with atomic clocks being just one example.
That’s were I am now. Even though I am telling my students to tell the full story of the numbers they write, I don’t think I’m giving them an opportunity to appreciate the full story for themselves, and so for now, these guidelines become simple must-do’s to avoid losing points on assessments, which is totally where I don’t want to be. So maybe some of my readers will have some insight on how to move this idea forward.
1. September 13, 2011 6:27 am
John:
Thanks for the post. Numbers do tell stories and sometimes we don’t spend enough time telling them, and the history surrounding the number. Students need to understand that they don’t materialize out of thin air, especially numbers like Avogadro’s number or Planck’s constant. Their is an elegance associated with many numbers, even those that are simple. By the way, Avogadro’s number is 6.02 x 10^23 atoms per mole. And then what is a 1 mole. Well, it is defined by Avogardo’s number, as well as the number of grams equivalent to the atomic or molecular weight. Thanks for the post and maybe correct your number so all the chemists of the world don’t inundate you.
Thanks!
Bob
• September 13, 2011 6:54 am
Thanks Bob,
I should have caught that, but what’s a power of 10 between friends?🙂
2. September 13, 2011 8:08 am
I personally shy away from being the unit police, especially when student work represents them working something out for themselves (and I just happened to collect it). I want the issue come up in discussion or in writing when there’s a real audience (i.e. not teacher) who might be confused or might need to know. For example, last week students were working on my goal-less video problem with the speedometer; One groups got like 6,000 mph per hour and another group got like 1.67 mph / s and another group got like .0005 miles per second per second. We talked about how and why they got different numbers and if they meant the same thing or if they meant a different thing? It was a struggle to talk about what 6,000 mph per hour meant. A minute later, another group was asking for help, because they solved for the distance two ways and got different answers. Quickly another student realized that while they had converted the acceleration into miles/s/s, they had left their speeds in terms of mph. They spotted the unit problem before I did.
• September 13, 2011 9:31 am
I totally hear you. I don’t want to be the unit police, and I really don’t want to be the sig fig police (especially since sig figs is a word I never use). But I’ve found that the act of writing a number is something students to almost without thinking, that they need incentive to really stop and think about the number they are writing. So far, I’m sad to admit the only incentive I’ve found is the “you don’t have full understanding until you set up a habit of telling the story with every number” and the grade stick that implies.
• September 13, 2011 9:52 am
I suppose what I see is that we both want the need to both care and be careful with units to be authentic –not because “our teacher makes us” and not because “naked numbers are bad”. The first is problematic for me because of the authority-driven nature. The second one is problematic for me because its divorced from authentic epistemological concerns.
• September 13, 2011 9:57 am
I’m totally with you. I think I wrote the original post I wanted a way to get past both “teacher forces me” and “naked numbers” to an intrinsic motivation to tell the full story of a number. And I’ve had lots of moments like what you describe where students find the value of telling the full story of a number through a lab or discussion, and these are the moments I strive for, but so far, I find they are not enough to promote the day-to-day attention to this important detail.
3. September 13, 2011 9:01 am
when i taught physics, i could never seem to motivate units/sig figs really well, but this seems like a perfect way to frame the discussion and a perfect mantra for the students to keep in their head. every time they write down a number thinking that they are telling a story. thanks for sharing, i’ll apply a modified idea to word problems in calc this year
4. September 13, 2011 4:32 pm
Not quiiittteee what this topic was about, but if you missed it I wanted to share Geoff’s link here http://pedagoguepadawan.net/124/measurementuncertaintyactivities/ about establishing measurement uncertainty in his class. I dig this a lot. Instead of arbitrary lopping off of numbers, his students figure out what the best measurement they can reasonably get in that class with those tools.
• September 17, 2011 12:08 pm
Jason,
Thanks for reminding of this awesome activity from Geoff. I’d seen it before, but forgotten about it. | 2016-09-28 05:08:05 | {"extraction_info": {"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": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41664189100265503, "perplexity": 678.3349308075442}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661311.64/warc/CC-MAIN-20160924173741-00198-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2347733/intersections-between-a-cubic-b%C3%A9zier-curve-and-a-line | # Intersections Between a Cubic Bézier Curve and a Line
If one end and a control point of a cubic Bézier curve is connected by a straight line, is there a simple way to find out whether this straight line intersects the Bezier curve? If it intersects then what will be the corresponding Bézier curve parameter's value?
Cubic Bezier curve and a straight line intersection tells two end point as a line, but I need one end point and one control point.
The calculations are roughly the same regardless of what line is used.
Suppose the line has equation $ax+by+cz = d$. In vector form, this is $$\mathbf{A}\cdot \mathbf{X} = d,$$ where $\mathbf{A} = (a,b,c)$. Also, suppose the Bézier curve has equation $$\mathbf{X}(t) = (1-t)^3\mathbf{P}_0 + 3t(1-t)^2\mathbf{P}_1 + 3t^2(1-t)\mathbf{P}_2 + t^3\mathbf{P}_3$$ Substituting this into the previous equation, we get $$(1-t)^3(\mathbf{A} \cdot \mathbf{P}_0) + 3t(1-t)^2(\mathbf{A} \cdot \mathbf{P}_1) + 3t^2(1-t)(\mathbf{A} \cdot \mathbf{P}_2) + t^3(\mathbf{A} \cdot \mathbf{P}_3) - d = 0$$ This is a cubic equation that you need to solve for $t$. The $t$ values you obtain (at most 3 of them) are the parameter values on the Bézier curve at the intersection points. If any of these $t$ values is outside the interval $[0,1]$, you'll probably want to ignore it.
In your particular case, you know that the line passes through either the start-point or the end-point of the Bézier curve, so either $t=0$ or $t=1$ is a solution of the cubic equation, which makes things easier. Suppose $t=0$ is a solution. Then the cubic above can be written in the form $$t(at^2 + bt + c) =0$$ Now you only have to find $t$ values in $[0,1]$ that satisfy $at^2 + bt +c =0$, which is easy.
• And you may need to require that $t \in [0,1]$. – lhf Jul 8 '17 at 13:24
• @lhf -- yes, good point. I'll add a comment about that. Thanks. – bubba Jul 8 '17 at 13:53
this can be solved analytically, by solving a cubic function within a cubic function, one rather complicated cubic function.
getting a points closest euclidean [distance to a cubic bezier] is solved on shadertoy.com
doing that for all points along a line gives you a cubic function with 1 or 2 local minima (or a line if all 3 CV are colinear AD line is parallel to the linear bezier).
for a quick estimation, cubic beziers are constrained to a triangular bounding volume (or line if the CV are colinear), this allows you to skip solving a big cubic function for many cases. | 2019-06-16 14:35:20 | {"extraction_info": {"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, "math_score": 0.8976481556892395, "perplexity": 208.17588451909631}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998250.13/warc/CC-MAIN-20190616142725-20190616164725-00503.warc.gz"} |
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# Rescue. Your friend (mass 65.0 kg) is standing on the ice
ISBN: 9780321675460 31
## Solution for problem 90P Chapter 6
University Physics | 13th Edition
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Problem 90P
Rescue?. Your friend (mass 65.0 kg) is standing on the ice in the middle of a frozen pond There is very little friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 m/s while you remain at rest. What is the average power supplied by the force you applied?
Step-by-Step Solution:
Solution 90P Step 1: Her mass is 65 kg. Initially she was at rest and by pulling her final velocity reached to 6 m/s. The change in kinetic energy is, K.E = K.E K.E final initial 2 2 = 1/2 × m × (V f V i 2 = 1/2 × 65 × (6 0) = 0.5 × 65 × 36 = 1170 J .
Step 2 of 3
Step 3 of 3
## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Inverse Trigonometric Functions: Integration
?In Exercises 1-20, find the indefinite integral. $$\int \frac{1}{4+(x-3)^{2}} d x$$
#### Related chapters
Unlock Textbook Solution | 2022-08-08 22:09:55 | {"extraction_info": {"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, "math_score": 0.23751042783260345, "perplexity": 1974.3031004517063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00584.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-p-prerequisites-chapter-p-review-exercises-page-76/29 | ## College Algebra 7th Edition
Published by Brooks Cole
# Chapter P, Prerequisites - Chapter P Review - Exercises - Page 76: 29
#### Answer
$\frac{1}{6}$
#### Work Step by Step
We convert the negative exponent to a fraction and the fraction exponent to a root: $216^{-1/3}=\displaystyle \frac{1}{216^{1/3}}=\frac{1}{\sqrt[3]{216}}=\frac{1}{6}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-09-20 18:53:34 | {"extraction_info": {"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, "math_score": 0.6353259682655334, "perplexity": 4315.015792861718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156554.12/warc/CC-MAIN-20180920175529-20180920195929-00499.warc.gz"} |
https://labs.tib.eu/arxiv/?author=J.%20Rozynek | • ### An example of the interplay of nonextensivity and dynamics in the description of QCD matter(1606.09033)
March 20, 2019 hep-ph
Using a simple quasi-particle model of QCD matter, presented some time ago in the literature, in which interactions are modelled by some effective fugacities z, we investigate the interplay between the dynamical content of fugacities z and effects induced by nonextensivity in situations when this model is used in a nonextensive environment characterized by some nonextensive parameter q $\neq$ 1 (for the usual extensive case q=1). This allows for a better understanding of the role of nonextensivity in the more complicated descriptions of dense hadronic and QCD matter recently presented (in which dynamics is defined by a lagrangian, the form of which is specific to a given model).
• Substantial experimental and theoretical efforts worldwide are devoted to explore the phase diagram of strongly interacting matter. At LHC and top RHIC energies, QCD matter is studied at very high temperatures and nearly vanishing net-baryon densities. There is evidence that a Quark-Gluon-Plasma (QGP) was created at experiments at RHIC and LHC. The transition from the QGP back to the hadron gas is found to be a smooth cross over. For larger net-baryon densities and lower temperatures, it is expected that the QCD phase diagram exhibits a rich structure, such as a first-order phase transition between hadronic and partonic matter which terminates in a critical point, or exotic phases like quarkyonic matter. The discovery of these landmarks would be a breakthrough in our understanding of the strong interaction and is therefore in the focus of various high-energy heavy-ion research programs. The Compressed Baryonic Matter (CBM) experiment at FAIR will play a unique role in the exploration of the QCD phase diagram in the region of high net-baryon densities, because it is designed to run at unprecedented interaction rates. High-rate operation is the key prerequisite for high-precision measurements of multi-differential observables and of rare diagnostic probes which are sensitive to the dense phase of the nuclear fireball. The goal of the CBM experiment at SIS100 (sqrt(s_NN) = 2.7 - 4.9 GeV) is to discover fundamental properties of QCD matter: the phase structure at large baryon-chemical potentials (mu_B > 500 MeV), effects of chiral symmetry, and the equation-of-state at high density as it is expected to occur in the core of neutron stars. In this article, we review the motivation for and the physics programme of CBM, including activities before the start of data taking in 2022, in the context of the worldwide efforts to explore high-density QCD matter.
• ### Nuclear Entalpies(1311.3591)
May 3, 2014 nucl-th
In a compressed Nuclear Matter (NM) an increasing pressure between the nucleons starts to increase the ratio of a nucleon Fermi to average single particle energy and in accordance with the Hugenholtz-van Hove theorem the longitudinal Momentum Sum Rule (MSR) is broken in a Relativistic Mean Field (RMF) approach. We propose to benefit from the concept of enthalpy in order to show how to fulfill the MSR above a saturation density with pressure corrections. As a result a nucleon mass can decrease with density, making the Equation of State (EoS) softer. The course of the EoS in our modified RMF model is close to a semi-empirical estimate and to results obtained from extensive DBHF calculations with a Bonn A potential, which produce the EoS stiff enough to describe neutron star properties (mass-radius constraint), especially the most massive known neutron star. The presented model has proper saturation properties, including good values of a compressibility and a spin-orbit term.
• ### Finite Pressure Corrections to Nucleon Structure Function Inside a Nuclear Medium(1104.0093)
May 2, 2012 hep-ph, nucl-th
Our model calculations performed in the frame of the Relativistic Mean Field (RMF) approach show how important are the modifications of nucleon Structure Function (SF) in Nuclear Matter (NM) above the saturation point. They originated from the conservation of a parton longitudinal momenta - essential in the explanation of the EMC effect at the saturation point of NM. For higher density the finite pressure corrections emerge from the Hugenholtz -van Hove (HH) theorem valid for NM and asks to modify the nucleon SF. The density evolution of the nuclear SF seems to be stronger for higher densities. Here we show that the course of Equation o State (EoS) in our modified Walecka model is very close to that obtained from extensive DBHF calculations with a Bonn A potential. The nuclear compressibility decreases. Our model - a nonlinear extension of nuclear RMF, has no additional parameters.
• ### Nonextensive critical effects in relativistic nuclear mean field models(1102.4497)
Feb. 22, 2011 hep-ph, nucl-th
We present a possible extension of the usual relativistic nuclear mean field models widely used to describe nuclear matter towards accounting for the influence of possible intrinsic fluctuations caused by the environment. Rather than individually identifying their particular causes we concentrate on the fact that such effects can be summarily incorporated in the changing of the statistical background used, from the usual (extensive) Boltzman-Gibbs one to the nonextensive taken in the form proposed by Tsallis with a dimensionless nonextensivity parameter $q$ responsible for the above mentioned effects (for $q \rightarrow 1$ one recovers the usual BG case). We illustrate this proposition on the example of the QCD-based Nambu - Jona-Lasinio (NJL) model of a many-body field theory describing the behavior of strongly interacting matter presenting its nonextensive version. We check the sensitivity of the usual NJL model to a departure from the BG scenario expressed by the value of $| q - 1|$, in particular in the vicinity of critical points.
• ### The optical model potential of the $\Sigma$ hyperon in nuclear matter(0911.3288)
Nov. 17, 2009 nucl-th
We present our attempts to determine the optical model potential $U_\Sigma = V_\Sigma -iW_\Sigma$ of the $\Sigma$ hyperon in nuclear matter. We analyze the following sources of information on $U_\Sigma$: $\Sigma N$ scattering, $\Sigma^-$ atoms, and final state interaction of $\Sigma$ hyperons in the $(\pi,K^+)$ and $(K^-.\pi)$ reactions on nuclear targets. We conclude that $V_\Sigma$ is repulsive inside the nucleus and has a shallow a tractive pocket at the nuclear surface. These features of $V_\Sigma$ are consistent with the Nijmegen model F of the hyperon-nucleon interaction.
• ### The Single Particle Sum Rules in the Deep-Inelastic Region(nucl-th/0406045)
May 12, 2005 hep-ph, nucl-th
We have modelled parton distribution in nuclei using a suitably modified nuclear Fermi motion. The modifications concern the nucleon rest energy which changes the Bj\"orken x in nuclear medium. We also introduce final state interactions between the scattered nucleon and the rest of the nucleus. The energy-momentum sum rule is saturated. Good agreement with experimental data of the EMC effect for $x> 0.15$ and nuclear lepton pair production data has been obtained.
• ### The nuclear pions and quark distributions in deep inelastic scattering on nuclei(nucl-th/0110036)
Oct. 13, 2001 astro-ph, hep-ph, nucl-th
We propose simple Monte Carlo method for calculating parton distribution in nuclei. The EMC effect is automatically included by means of two parameters, which characterize the change of nuclear pion field. Good agreement with experimental data in the broad range of variable x is obtained.
• ### A Model for the Parton Distribution in Nuclei(hep-ph/9910401)
Dec. 10, 1999 hep-ph, nucl-th
We have extended recently proposed model of parton distribution i nucleons to the case of nucleons in nuclei. | 2019-10-22 05:31:34 | {"extraction_info": {"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, "math_score": 0.7080720663070679, "perplexity": 1239.1538734914998}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987798619.84/warc/CC-MAIN-20191022030805-20191022054305-00509.warc.gz"} |
http://tex.stackexchange.com/questions/61586/tikz-picture-fitted-to-page-edge | # TiKZ-picture fitted to page edge
Using TiKZ, I'm trying to create an image that fits the margin of odd-sided pages (a water stamp of sort) for an A6 booklet. My idea was at first to create an image that is exactly the same size as an A6 page, compile to PDF and then use eso-pic to set it as the background of the booklet document.
However, I've realized I want to use commands such as \thesection to show information in the margin, so I'd rather do this in a way where everything fits in the same document. An example of what I want to accomplish can be seen here, where the document shows an A4 with eight A6 pages, the first one being odd-numbered.
I can't figure out how to scale the tikzpicture to fit the page exactly. Some parts of the image should be fitted to the page edge, so I need it to scale exactly, without margins. You could say I'm trying to do this, only the opposite.
-
\documentclass[10pt]{scrartcl}
\usepackage[margin=20mm,a6paper]{geometry}
\usepackage{tikz}
\usepackage{xifthen}
\usetikzlibrary{fit,calc}
\usepackage{lipsum}
\usepackage{everypage}
\parindent0mm
\newcommand{\currentsidemargin}{%
\ifodd\value{page}%
\oddsidemargin%
\else%
\evensidemargin%
\fi%
}
\ifthenelse{\isodd{\thepage}}
{ \begin{tikzpicture}[overlay, remember picture]
\path (current page.north west) ++(\hoffset, -\voffset) node[anchor=north west, shape=rectangle, inner sep=0, minimum width=\paperwidth, minimum height=\paperheight] (pagearea) {};
\path (pagearea.north west) ++(1in+\currentsidemargin,-1in-\topmargin-\headheight-\headsep) node[anchor=north west, shape=rectangle, inner sep=0, minimum width=\textwidth, minimum height=\textheight] (textarea) {};
\node[inner sep=0,fit=(pagearea.north east)(textarea.north east)(pagearea.south east)] (marginbox) {};
\draw (marginbox.south east) rectangle (marginbox.north west);
\draw (marginbox.south east) -- (marginbox.north west);
\end{tikzpicture}
}{}
}
\begin{document}
\lipsum[1-10]
\end{document}
-
This is fantastic! Will be tried out immediately! =) – Tomas Lycken Jun 28 '12 at 22:01
I was able to modify this to suit my needs perfectly. The key to success was nesting tikzpictures, with the node anchored to current page.north east. Thanks a lot! – Tomas Lycken Jun 28 '12 at 23:29
I don't exactly understand the question but you can produce your pages rather bluntly and then combine them via for example Can I convert a 16 page pdf into an 8x2 matrix in LaTeX?
For the multiple page file you can use something like
\documentclass{article}
\usepackage[paperheight=148mm,paperwidth=105mm]{geometry}
\usepackage{tikz,lipsum}
\begin{document}
\thispagestyle{empty}
\begin{tikzpicture}[remember picture,overlay]
\draw[line width=5mm,red!20]
([shift={(-0.5\pgflinewidth,-0.5\pgflinewidth)}]current page.north east)
rectangle
([shift={(0.5\pgflinewidth,0.5\pgflinewidth)}]current page.south west);
\node[scale=3,opacity=0.2,rotate=60] at (current page.center) {\textsf{First Watermark}};
\end{tikzpicture}
\lipsum[2]
\newpage
\thispagestyle{empty}
\begin{tikzpicture}[remember picture,overlay]
\draw[line width=5mm,blue!20]
([shift={(-0.5\pgflinewidth,-0.5\pgflinewidth)}]current page.north east)
rectangle
([shift={(0.5\pgflinewidth,0.5\pgflinewidth)}]current page.south west);
\end{tikzpicture}
\lipsum[3]
\end{document}
Note that you have to run twice for remember picture,overlay to work properly.
EDIT Only for the side of the page,
\documentclass{article}
\usepackage[paperheight=148mm,paperwidth=105mm]{geometry}
\usepackage{tikz,lipsum}
\begin{document}
\begin{tikzpicture}[remember picture,overlay]
\draw[line width=5mm,blue!20]
([shift={(-0.5\pgflinewidth,\pgflinewidth)}]current page.north east)
--
([shift={(-0.5\pgflinewidth,0)}]current page.south east) node[circle,anchor=east,pos=0.2,fill,draw,text=black] {\Large $\Gamma$};
\end{tikzpicture}
\lipsum[3]
\end{document}
-
This is great, but it doesn't really answer to my main problem: how to fit the watermark to the edge of the page. Whatever I try, the picture is just enlarged to the left instead of moved to the right... – Tomas Lycken Jun 28 '12 at 19:08
Those rectangles are also watermarks. You can use the (current page) node to position your picture. Can you provide more details about the watermark even make a compilable example so that we can have a look at the positioning? – percusse Jun 28 '12 at 19:13 | 2016-05-25 17:19:16 | {"extraction_info": {"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, "math_score": 0.8693023324012756, "perplexity": 2890.642928158489}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049275181.24/warc/CC-MAIN-20160524002115-00195-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://alpheratz.net/dda-2015-recent-dynamical-evolution-of-mimas-and-enceladus/ | DDA 2015 – Recent dynamical evolution of Mimas and Enceladus
This is one of a series of notes taken during the 2015 meeting of the AAS Division on Dynamical Astronomy, 3-7 May, at CalTech. An index to this series (all the papers presented at the meeting) is here.
Session: Moon Formation and Dynamics III
Maja Cuk (SETI Institute)
Abstract
Mimas and Enceladus are the smallest and innermost mid-sized icy moons of Saturn. They are each caught in a 2:1 orbital resonance with an outer, larger moon: Mimas with Tethys, Enceladus with Dione. This is where the similarities end. Mimas is heavily cratered and appears geologically inactive, while Enceladus has a young surface and high tidal heat flow. Large free eccentricity of Mimas implies low tidal dissipation, while Enceladus appears very dissipative, likely due to an internal ocean. Their resonances are very different too. Mimas is caught in a 4:2 inclination type resonance with Tethys which involves inclinations of both moons. Enceladus is in a 2:1 resonance with Dione which affects only Enceladus’s eccentricity. The well-known controversy over the heat flow of Enceladus can be solved by invoking a faster tidal evolution rate than previously expected (Lainey et al. 2012), but other mysteries remain. It has been long known that Mimas has very low probability of being captured into the present resonance, assuming that the large resonant libration amplitude reflects sizable pre-capture inclination of Mimas. Furthermore, Enceladus seems to have avoided capture into a number of sub-resonances that should have preceded the present one. An order of magnitude increase in the rate of tidal evolution does not solve these problems. It may be the time to reconsider the dominance of tides in the establishment of these resonances, especially if the moons themselves may be relatively young. An even faster orbital evolution due to ring/disk torques can help avoid capture into smaller resonances. Additionally, past interaction of Mimas with Janus and Epimetheus produce some of the peculiarities of Mimas’ current orbit. At the meeting I will present numerical integrations that confirm the the existence of these problems, and demonstrate the proposed solutions.
Notes
• tidal rates $\dfrac{1}{a}\dfrac{d a}{d t}$: Mimas = 59, Enceladus = 23
• numerical integrations — brute force
• artificial migration
• slow
• the trouble with Mimas
• Mimas and Tethys in inclination-type 4:2 MMR
• inclination of both moons affected by the resonance
• libration amp. of resonance is large, ~100 deg $\rightarrow$ primordial Mimas inclination — doesn’t work
• eccentricity of Tethys has complex effects
• Mimas-Tethys evolution rate: $\dfrac{da_{moon}}{dt} \propto \dfrac{R^5_{planet}}{a^{3/2}}$
• introduce ad hoc ring torques — artificial torque on Prometheus
• gives Tethys resonance a kick
• $\therefore$ don’t take Mimas-Tethys resonance too seriously | 2019-07-15 22:15:36 | {"extraction_info": {"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, "math_score": 0.4543374180793762, "perplexity": 5584.846922146798}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524254.28/warc/CC-MAIN-20190715215144-20190716001144-00193.warc.gz"} |
https://proj.org/operations/transformations/helmert.html | # Helmert transform¶
New in version 5.0.0.
The Helmert transformation changes coordinates from one reference frame to another by means of 3-, 4-and 7-parameter shifts, or one of their 6-, 8- and 14-parameter kinematic counterparts.
Alias helmert Domain 2D, 3D and 4D Input type Cartesian coordinates (spatial), decimalyears (temporal). Output type Cartesian coordinates (spatial), decimalyears (temporal). Input type Cartesian coordinates Output type Cartesian coordinates
The Helmert transform, in all its various incarnations, is used to perform reference frame shifts. The transformation operates in cartesian space. It can be used to transform planar coordinates from one datum to another, transform 3D cartesian coordinates from one static reference frame to another or it can be used to do fully kinematic transformations from global reference frames to local static frames.
All of the parameters described in the table above are marked as optional. This is true as long as at least one parameter is defined in the setup of the transformation. The behavior of the transformation depends on which parameters are used in the setup. For instance, if a rate of change parameter is specified a kinematic version of the transformation is used.
The kinematic transformations require an observation time of the coordinate, as well as a central epoch for the transformation. The latter is usually documented alongside the rest of the transformation parameters for a given transformation. The central epoch is controlled with the parameter t_epoch. The observation time is given as part of the coordinate when using PROJ’s 4D-functionality.
## Examples¶
Transforming coordinates from NAD72 to NAD83 using the 4 parameter 2D Helmert:
proj=helmert convention=coordinate_frame x=-9597.3572 y=.6112 s=0.304794780637 theta=-1.244048
Simplified transformations from ITRF2008/IGS08 to ETRS89 using 7 parameters:
proj=helmert convention=coordinate_frame x=0.67678 y=0.65495 z=-0.52827
rx=-0.022742 ry=0.012667 rz=0.022704 s=-0.01070
Transformation from ITRF2000 to ITRF93 using 15 parameters:
proj=helmert convention=position_vector
x=0.0127 y=0.0065 z=-0.0209 s=0.00195
dx=-0.0029 dy=-0.0002 dz=-0.0006 ds=0.00001
rx=-0.00039 ry=0.00080 rz=-0.00114
drx=-0.00011 dry=-0.00019 drz=0.00007
t_epoch=1988.0
## Parameters¶
Note
All parameters are optional but at least one should be used, otherwise the operation will return the coordinates unchanged.
+convention=coordinate_frame/position_vector
New in version 5.2.0.
Indicates the convention to express the rotational terms when a 3D-Helmert / 7-parameter more transform is involved. As soon as a rotational parameter is specified (one of rx, ry, rz, drx, dry, drz), convention is required.
The two conventions are equally popular and a frequent source of confusion. The coordinate frame convention is also described as an clockwise rotation of the coordinate frame. It corresponds to EPSG method code 1032 (in the geocentric domain) or 9607 (in the geographic domain) The position vector convention is also described as an anticlockwise (counter-clockwise) rotation of the coordinate frame. It corresponds to as EPSG method code 1033 (in the geocentric domain) or 9606 (in the geographic domain).
This parameter is ignored when only a 3-parameter (translation terms only: x, y, z) , 4-parameter (3-parameter and theta) or 6-parameter (3-parameter and their derivative terms) is used.
The result obtained with parameters specified in a given convention can be obtained in the other convention by negating the rotational parameters (rx, ry, rz, drx, dry, drz)
Note
This parameter obsoletes transpose which was present in PROJ 5.0 and 5.1, and is forbidden starting with PROJ 5.2
+x=<value>
Translation of the x-axis given in meters.
+y=<value>
Translation of the y-axis given in meters.
+z=<value>
Translation of the z-axis given in meters.
+s=<value>
Scale factor given in ppm.
+rx=<value>
X-axis rotation in the 3D Helmert given arc seconds.
+ry=<value>
Y-axis rotation in the 3D Helmert given in arc seconds.
+rz=<value>
Z-axis rotation in the 3D Helmert given in arc seconds.
+theta=<value>
Rotation angle in the 2D Helmert given in arc seconds.
+dx=<value>
Translation rate of the x-axis given in m/year.
+dy=<value>
Translation rate of the y-axis given in m/year.
+dz=<value>
Translation rate of the z-axis given in m/year.
+ds=<value>
Scale rate factor given in ppm/year.
+drx=<value>
Rotation rate of the x-axis given in arc seconds/year.
+dry=<value>
Rotation rate of the y-axis given in arc seconds/year.
+drz=<value>
Rotation rate of the y-axis given in arc seconds/year.
+t_epoch=<value>
Central epoch of transformation given in decimalyear. Only used spatiotemporal transformations.
+exact
Use exact transformation equations.
See (5)
+transpose
Deprecated since version 5.2.0: (removed)
Transpose rotation matrix and follow the Position Vector rotation convention. If +transpose is not added the Coordinate Frame rotation convention is used.
## Mathematical description¶
In the notation used below, $$\hat{P}$$ is the rate of change of a given transformation parameter $$P$$. $$\dot{P}$$ is the kinematically adjusted version of $$P$$, described by
(1)$\dot{P}= P + \hat{P}\left(t - t_{central}\right)$
where $$t$$ is the observation time of the coordinate and $$t_{central}$$ is the central epoch of the transformation. Equation (1) can be used to propagate all transformation parameters in time.
Superscripts of vectors denote the reference frame the coordinates in the vector belong to.
### 2D Helmert¶
The simplest version of the Helmert transform is the 2D case. In the 2-dimensional case only the horizontal coordinates are changed. The coordinates can be translated, rotated and scale. Translation is controlled with the x and y parameters. The rotation is determined by theta and the scale is controlled with the s parameters.
Note
The scaling parameter s is unitless for the 2D Helmert, as opposed to the 3D version where the scaling parameter is given in units of ppm.
Mathematically the 2D Helmert is described as:
(2)\begin{split}\begin{align} \begin{bmatrix} X \\ Y \\ \end{bmatrix}^B = \begin{bmatrix} T_x \\ T_y \\ \end{bmatrix} + s \begin{bmatrix} \hphantom{-}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{bmatrix} \begin{bmatrix} X \\ Y \\ \end{bmatrix}^A \end{align}\end{split}
(2) can be extended to a time-varying kinematic version by adjusting the parameters with (1) to (2), which yields the kinematic 2D Helmert transform:
(3)\begin{split}\begin{align} \begin{bmatrix} X \\ Y \\ \end{bmatrix}^B = \begin{bmatrix} \dot{T_x} \\ \dot{T_y} \\ \end{bmatrix} + s(t) \begin{bmatrix} \hphantom{-}\cos \dot{\theta} & \sin \dot{\theta} \\ -\sin\ \dot{\theta} & \cos \dot{\theta} \\ \end{bmatrix} \begin{bmatrix} X \\ Y \\ \end{bmatrix}^A \end{align}\end{split}
All parameters in (3) are determined by the use of (1), which applies the rate of change to each individual parameter for a given timespan between $$t$$ and $$t_{central}$$.
### 3D Helmert¶
The general form of the 3D Helmert is
(4)\begin{align} V^B = T + \left(1 + s \times 10^{-6}\right) \mathbf{R} V^A \end{align}
Where $$T$$ is a vector consisting of the three translation parameters, $$s$$ is the scaling factor and $$\mathbf{R}$$ is a rotation matrix. $$V^A$$ and $$V^B$$ are coordinate vectors, with $$V^A$$ being the input coordinate and $$V^B$$ is the output coordinate.
In the Position Vector convention, we define $$R_x = radians \left( rx \right)$$, $$R_z = radians \left( ry \right)$$ and $$R_z = radians \left( rz \right)$$
In the Coordinate Frame convention, $$R_x = - radians \left( rx \right)$$, $$R_z = - radians \left( ry \right)$$ and $$R_z = - radians \left( rz \right)$$
The rotation matrix is composed of three rotation matrices, one for each axis.
\begin{split}\begin{align} \mathbf{R}_X &= \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos R_x & -\sin R_x \\ 0 & \sin R_x & \cos R_x \end{bmatrix} \end{align}\end{split}
\begin{split}\begin{align} \mathbf{R}_Y &= \begin{bmatrix} \cos R_y & 0 & \sin R_y\\ 0 & 1 & 0\\ -\sin R_y & 0 & \cos R_y \end{bmatrix} \end{align}\end{split}
\begin{split}\begin{align} \mathbf{R}_Z &= \begin{bmatrix} \cos R_z & -\sin R_z & 0\\ \sin R_z & \cos R_z & 0\\ 0 & 0 & 1 \end{bmatrix} \end{align}\end{split}
The three rotation matrices can be combined in one:
\begin{align} \mathbf{R} = \mathbf{R_X} \mathbf{R_Y} \mathbf{R_Y} \end{align}
For $$\mathbf{R}$$, this yields:
(5)$\begin{split}\begin{bmatrix} \cos R_y \cos R_z & -\cos R_x \sin R_z + & \sin R_x \sin R_z + \\ & \sin R_x \sin R_y \cos R_z & \cos R_x \sin R_y \cos R_z \\ \cos R_y\sin R_z & \cos R_x \cos R_z + & - \sin R_x \cos R_z + \\ & \sin R_x \sin R_y \sin R_z & \cos R_x \sin R_y \sin R_z \\ -\sin R_y & \sin R_x \cos R_y & \cos R_x \cos R_y \\ \end{bmatrix}\end{split}$
Using the small angle approxition the rotation matrix can be simplified to
(6)\begin{split}\begin{align} \mathbf{R} = \begin{bmatrix} 1 & -R_z & R_y \\ Rz & 1 & -R_x \\ -Ry & R_x & 1 \\ \end{bmatrix} \end{align}\end{split}
Which allow us to express the most common version of the Helmert transform, using the approximated rotation matrix:
(7)\begin{split}\begin{align} \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^B = \begin{bmatrix} T_x \\ T_y \\ T_z \\ \end{bmatrix} + \left(1 + s \times 10^{-6}\right) \begin{bmatrix} 1 & -R_z & R_y \\ Rz & 1 & -R_x \\ -Ry & R_x & 1 \\ \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^A \end{align}\end{split}
If the rotation matrix is transposed, or the sign of the rotation terms negated, the rotational part of the transformation is effectively reversed. This is what happens when switching between the 2 conventions position_vector and coordinate_frame
Applying (1) we get the kinematic version of the approximated 3D Helmert:
(8)\begin{split}\begin{align} \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^B = \begin{bmatrix} \dot{T_x} \\ \dot{T_y} \\ \dot{T_z} \\ \end{bmatrix} + \left(1 + \dot{s} \times 10^{-6}\right) \begin{bmatrix} 1 & -\dot{R_z} & \dot{R_y} \\ \dot{R_z} & 1 & -\dot{R_x} \\ -\dot{R_y} & \dot{R_x} & 1 \\ \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^A \end{align}\end{split}
The Helmert transformation can be applied without using the rotation parameters, in which case it becomes a simple translation of the origin of the coordinate system. When using the Helmert in this version equation (4) simplifies to:
(9)\begin{split}\begin{align} \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^B = \begin{bmatrix} T_x \\ T_y \\ T_z \\ \end{bmatrix} + \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^A \end{align}\end{split}
That after application of (1) has the following kinematic counterpart:
(10)\begin{split}\begin{align} \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^B = \begin{bmatrix} \dot{T_x} \\ \dot{T_y} \\ \dot{T_z} \\ \end{bmatrix} + \begin{bmatrix} X \\ Y \\ Z \\ \end{bmatrix}^A \end{align}\end{split} | 2021-01-17 08:56:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 12, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000096559524536, "perplexity": 1559.395800272855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00012.warc.gz"} |
http://math-mprf.org/journal/articles/id1102/ | Gibbs Measures for Self-Interacting Wiener Paths
#### M. Gubinelli
2006, v.12, №4, 747-766
ABSTRACT
We study a class of specifications over $d$-dimensional Wiener measure which are invariant under translation of the paths. We address the problem of existence and uniqueness of the Gibbs measures and prove a central limit theorem for the rescaled process. These results apply to the study of the ground state of the Nelson model of a quantum particle interacting with a scalar boson field.
Keywords: Gibbs measures,Nelson model,scaling limits | 2017-06-27 18:58:41 | {"extraction_info": {"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, "math_score": 0.6845089793205261, "perplexity": 454.167238558741}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00710.warc.gz"} |
https://www.aimsciences.org/article/doi/10.3934/dcds.2010.28.1469 | Article Contents
Article Contents
# New insights into the classical mechanics of particle systems
• The classical mechanics of particle systems is developed on the basis of the precept that, kinematics and the notion of force aside, power expenditures are of the foremost importance. The essential properties of forces between particles follow from requiring that the net power expended within any subsystem of particles be frame-indifferent. Furthermore, requiring that the net power expended on any subsystem of particles by external agencies be frame-indifferent yields force, moment, and power balances. These balances account for inertia but hold relative to any frame-of-reference, inertial or noninertial. Assuming that each particle possesses an interaction energy that embodies the extent to which it is attracted or repelled by other particles leads to the proposition of an interaction-energy inequality that serves as a purely mechanical statement of the second law of thermodynamics. In combination with the power balance, this inequality provides an avenue to ensure that constitutive equations do not permit a violation of thermodynamics. This inequality is used to develop the simplest class of constitutive equations that account for both conservative and dissipative particle-particle interactions.
Mathematics Subject Classification: Primary: 70A05, 70F25; Secondary: 37J60.
Citation:
• on this site
/ | 2023-01-27 08:31:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.48260530829429626, "perplexity": 781.8416641733963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00718.warc.gz"} |
https://www.r-bloggers.com/2016/02/rbokeh-version-0-3-4-released/ | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
The rbokeh package version 0.3.4 was recently released.
First of all, if you are new to rbokeh, check out the documentation here to get familiar with the package.
install.packages("rbokeh", repos = "http://packages.tessera.io")
Since this is the first post about a release, we’ll talk about what’s new from when we hit version 0.3. There have been many updates, but most are either internal refactoring, bug fixes, or minor feature additions. See the comlpete NEWS file for full details.
Non-standard evaluation
The most major interal aspect of the bump to version 0.3 is a major overhaul of the logic for non-standard evaluation to make specification of arguments more flexible and robust. Major thanks to Barret Schloerke for his work on this and for Hadley’s lazy_eval package. We supported this prior to 0.3 but it is now much more robust.
Bokeh 0.11.0
We have updated the BokehJS library to version 0.11.0. The major user-facing change that this brings is that the mouse wheel zoom tool is turned off by default. For those who have been using rbokeh for a while, this may take some getting used to, but I think it is for the better, especially when embedding multiple plots in a scrollable page. Prior to this change, rbokeh plots could hijack the page scrolling resulting in unwanted zooming on the plot.
Another minor user-facing change on the R side is that we have greatly reduced the verbosity of the functions.
There are several other features that have not been covered thoroughly in the documentation and we’ll be spending some more time posting about those soon.
Also, we are working hard on several enhancements, mostly to do with interactivity and shiny integration, and we are excited to share these soon! | 2021-04-13 16:32:21 | {"extraction_info": {"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, "math_score": 0.23681138455867767, "perplexity": 2088.1919876618363}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00265.warc.gz"} |
https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-measurement-topic/cc-4th-area-and-perimeter/e/area_of_squares_and_rectangles | # Area of squares and rectangles
Find the area of rectangles and squares when given side lengths. Find the side length of a square when given the area.
### Problem
The area of a square is 49 square meters. How long is each side?
meters
Get first 3 correct, or 5 in a row. | 2016-07-24 12:58:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7692227363586426, "perplexity": 677.2255252751632}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824037.46/warc/CC-MAIN-20160723071024-00313-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathoverflow.net/revisions/71434/list | MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
4 made proof slightly better
Let $Y$ be a member of the Turing degree $[Y\hspace{.04 in}]$. $\;$ Define $canhalt : \omega \times \omega \to \{\text{false},\text{true}\}$ by
$canhalt(s,t) \iff$
there exists an $s$-state $Y$-oracle machine that runs exactly $t$ steps if started on a blank tape
Define $pair : \omega \times \omega \to \omega$ to be the Cantor pairing function. $\; \; pair$ has a graph and is a bijection.
There are only finitely many $m$-state $Y$-oracle machines, and these are easily enumerated,
so define $\langle S_0,S_1,S_2,S_3,...\rangle$ by
$((2\cdot n)\in S_{pair(s,t)}) \iff n=s$
and
$(((2\cdot n)+1)\in S_{pair(s,t)}) \iff (t\lt n$ and $canhalt(s,n))$
and note that for all $s$, $\{t : canhalt(s,t)\}$ is finite.
Define $bb_Y : \omega \to \omega$ by $bb_Y(s) = \operatorname{max}(\{t : canhalt(s,t)\})$. $\;$ ($bb_Y$ does not necessarily have a graph)
Define $E = \{n : n\, \text{ is even} \}$. $\;$ By construction, for all members $n$ of $E$, $\; n\in S_{pair(m,bb_Y(m))} \subseteq E \;$.
Assuming the Union Principle, let $I$ be a subset of $\omega$ such that $\; \; \; \displaystyle\bigcup_{i\in I} \; S_i \; \; = \; \; X \; \; \;$.
By the construction of $\langle S_0,S_1,S_2,S_3,...\rangle$ and $I$, for all $s$ there exists $t$ such that $pair(s,t)\in I$,
and for all $s$ and $t$ if $pair(s,t)\in I$ then $bb_Y(s) \leq t$.
Let $\langle mach_0,mach_1,mach_2,mach_3,...\rangle$ be an a reasonable enumeration of the $Y$-oracle machines such that with . $\;$ Define $states : \omega \to \omega$ defined by $\; states(m) =$ the number of states in $mach_m \;$, ;$. Since the enumeration is reasonable,$states$is computablehas a graph. For all$m$and$t$, if$pair(states(m),t)\in I$then$mach_m$halts within$t$steps if started on a blank tape$\impliesmach_m$halts if started on a blank tape$\impliesmach_m$runs exactly a member of $\{t : canhalt(states(m),t)\}$ steps if started on a blank tape$\impliesmach_m$halts within$bb_Y(states(m))$steps if started on a blank tape$\impliesmach_m$halts within$t$steps if started on a blank tape Now, since the enumeration is reasonable, define $H = \{m : mach_m\; \text{halts within}\; t\; \text{steps when started on a blank tape, where}\; pair(states(m),t)\in I \}$. By the above,$[Y\hspace{.04 in}]' = [Y\hspace{.02 in}'] = [H\hspace{.02 in}]$exists.$\; $This works for all Turing degrees, so (RCA0 + Union Principle) proves all of ACA0.$\; $Clearly ACA0 proves the Union principle, and ACA0 is stronger than RCA0. Therefore the Union Principle is equivalent to ACA0 over RCA0. 3 fixed another typo Let$Y$be a member of the Turing degree$[Y\hspace{.04 in}]$.$\; $Define $canhalt : \omega \times \omega \to \{\text{false},\text{true}\}$ by$canhalt(s,t) \iff$there exists an$s$-state$Y$-oracle machine that runs exactly$t$steps if started on a blank tape Define$pair : \omega \times \omega \to \omega$to be the Cantor pairing function. There are only finitely many$m$-state$Y$-oracle machines, and these are easily enumerated, so define$\langle S_0,S_1,S_2,S_3,...\rangle$by$((2\cdot n)\in S_{pair(s,t)}) \iff n=s$and$(((2\cdot n)+1)\in S_{pair(s,t)}) \iff (t\lt n$and$canhalt(s,n))$and not note that for all$s$, $\{t : canhalt(s,t)\}$ is finite. Define$bb_Y : \omega \to \omega$by $bb_Y(s) = \operatorname{max}(\{t : canhalt(s,t)\})$.$\; $Define $E = \{n : n\, \text{ is even} \}$. By construction, for all members$n$of$E$,$\; n\in S_{pair(m,bb_Y(m))} \subseteq E \;$. Assuming the Union Principle, let$I$be a subset of$\omega$such that$\; \; \; \displaystyle\bigcup_{i\in I} \; S_i \; \; = \; \; X \; \; \; $. By construction of$\langle S_0,S_1,S_2,S_3,...\rangle$and$I$, for all$s$there exists$t$such that$pair(s,t)\in I$, and for all$s$and$t$if$pair(s,t)\in I$then$bb_Y(s) \leq t$. Let$\langle mach_0,mach_1,mach_2,mach_3,...\rangle$be an enumeration of the$Y$-oracle machines such that with$states : \omega \to \omega$defined by$\; states(m) =$the number of states in$mach_m \;$,$states$is computable. For all$m$and$t$, if$pair(states(m),t)\in I$then$mach_m$halts within$t$steps if started on a blank tape$\impliesmach_m$halts if started on a blank tape$\impliesmach_m$runs exactly a member of $\{t : canhalt(states(m),t)\}$ steps if started on a blank tape$\impliesmach_m$halts within$bb_Y(states(m))$steps if started on a blank tape$\impliesmach_m$halts within$t$steps if started on a blank tape Now, define $H = \{m : mach_m\; \text{halts within}\; t\; \text{steps when started on a blank tape, where}\; pair(states(m),t)\in I \}$. By the above,$[Y\hspace{.04 in}]' = [Y'] Y\hspace{.02 in}'] = [H\hspace{.02 in}]$exists.$\; $This works for all Turing degrees, so (RCA0 + Union Principle) proves all of ACA0.$\; $Clearly ACA0 proves the Union principle, and ACA0 is stronger than RCA0. Therefore the Union Principle is equivalent to ACA0 over RCA0. 2 fixed typo Let$Y$be a member of the Turing degree$[Y]$. [Y\hspace{.04 in}]$. $\;$ Define $canhalt : \omega \times \omega \to \{\text{false},\text{true}\}$ by
$canhalt(s,t) \iff$
there exists an $s$-state $Y$-oracle machine that runs exactly $t$ steps if started on a blank tape
Define $pair : \omega \times \omega \to \omega$ to be the Cantor pairing function.
There are only finitely many $m$-state $Y$-oracle machines, and these are easily enumerated,
so define $\langle S_0,S_1,S_2,S_3,...\rangle$ by
$((2\cdot n)\in S_{pair(s,t)}) \iff n=s$
and
$(((2\cdot n)+1)\in S_{pair(s,t)}) \iff (t\lt n$ and $canhalt(s,t))$ canhalt(s,n))$and not that for all$s$, $\{t : canhalt(s,t)\}$ is finite. Define$bb_Y : \omega \to \omega$by $bb_Y(s) = \operatorname{max}(\{t : canhalt(s,t)\})$.$\; $Define $E = \{n : n\, \text{ is even} \}$. By construction, for all members$n$of$E$,$\; n\in S_{pair(m,bb_Y(m))} \subseteq E \;$. Assuming the Union Principle, let$I$be a subset of$\omega$such that$\; \; \; \displaystyle\bigcup_{i\in I} \; S_i \; \; = \; \; X \; \; \; $. By construction of$\langle S_0,S_1,S_2,S_3,...\rangle$and$I$, for all$s$there exists$t$such that$pair(s,t)\in I$, and for all$s$and$t$if$pair(s,t)\in I$then$bb_Y(s) \leq t$. Let$\langle mach_0,mach_1,mach_2,mach_3,...\rangle$be an enumeration of the$Y$-oracle machines such that with$states : \omega \to \omega$defined by$\; states(m) =$the number of states in$mach_m \;$,$states$is computable. For all$m$and$t$, if$pair(states(m),t)\in I$then$mach_m$halts within$t$steps if started on a blank tape$\impliesmach_m$halts if started on a blank tape$\impliesmach_m$runs exactly a member of $\{t : canhalt(states(m),t)\}$ steps if started on a blank tape$\impliesmach_m$halts within$bb_Y(states(m))$steps if started on a blank tape$\impliesmach_m$halts within$t$steps if started on a blank tape Now, define $H = \{m : mach_m\; \text{halts within}\; t\; \text{steps when started on a blank tape, where}\; pair(states(m),t)\in I \}$. By the above,$[Y]' [Y\hspace{.04 in}]' = [Y'] = [H]$H\hspace{.02 in}]$ exists. This works for all Turing degrees, so (RCA0 + Union Principle) proves all of ACA0. Clearly ACA0 proves the Union principle, and ACA0 is stronger than RCA0.
Therefore the union principle Union Principle is equivalent to ACA0 over RCA0.
1 | 2013-06-19 23:08:59 | {"extraction_info": {"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, "math_score": 0.9989750385284424, "perplexity": 5817.550739241462}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368709805610/warc/CC-MAIN-20130516131005-00038-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/four-potential-of-moving-charge.792215/ | # Four-potential of moving charge
Tags:
1. Jan 14, 2015
### unscientific
1. The problem statement, all variables and given/known data
(a) Find the four-vector potential of a moving charge
(b) Find source time and z-component of electric field
(c) Find electric potential to first order of x and hence electric field
2. Relevant equations
3. The attempt at a solution
Part(a)
$$\phi = \frac{q}{4\pi \epsilon_0 r}$$
$$\vec A = 0$$
Consider world line of aritrarily moving charge, where the vector potential is only dependent on what charge is doing at source event. The relevant distance is source-field distance $r_{sf}$. We represent this in 4-vector displacement $R = (ct, \vec r)$ where $t = \frac{r_{sf}}{c}$. Consider $R \cdot U = \gamma(-rc + \vec r \cdot \vec v)$. It gives the right form in rest frame. Thus:
$$A = \frac{q}{4\pi \epsilon_0} \frac{ \frac{U}{c} }{-R \cdot U}$$
Part (b)
$$t_s = \frac{r_{sf}}{c} = \frac{\sqrt{z^2+a^2}}{c}$$
$$R \cdot U = \gamma (-ct + \vec r \cdot \vec v) = -\gamma r c = -\gamma c \sqrt{z^2 + a^2}$$
Thus the four-potential is
$$A = \frac{q}{4\pi \epsilon_0 c^2 \sqrt{z^2 + a^2}} (c, \vec v_s) = (\frac{\phi}{c}, \vec A)$$
To find electric field,
$$E_z = -\frac{\partial \phi}{\partial z} - \frac{\partial A_z}{\partial t}$$
z-component of $\vec A$ is zero, so
$$E_z = -\frac{\partial \phi}{\partial z} = -\frac{qz}{4\pi \epsilon_0 \left( z^2 + a^2 \right)^{\frac{3}{2}}} \hat k$$
Part (c)
$$r = \sqrt{ z^2 + \left[ a cos(\omega t) - \Delta x \right]^2 + a^2 sin^2(\omega t) }$$
$$r = \approx \sqrt{ z^2 + a^2 - 2a \Delta x cos(\omega t) } \approx \sqrt{z^2 + a^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{z^2 + a^2} \right]$$
For $R \cdot U$, we have
$$R \cdot U = \gamma(-rc + \vec r \cdot \vec v) = \gamma \left[ -cr + a\omega \Delta x sin (\omega t) \right]$$
$$\phi = -\frac{qc}{4\pi \epsilon_0} \frac{1}{ c\sqrt{a^2 + z^2} \left[ 1 - \frac{a \Delta x cos(\omega t)}{a^2 + z^2} \right] + a\omega \Delta x sin(\omega t) }$$
$$\phi \approx \frac{-qc}{4\pi \epsilon_0} \frac{1}{c\sqrt{a^2+z^2} + \Delta x \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2}} \right] }$$
$$\phi \approx \frac{-qc}{4\pi \epsilon_0} \left[ 1 - \frac{a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2+z^2}} }{c\sqrt{a^2+z^2}} \Delta x \right]$$
$$E_x = -\frac{\partial \phi}{\partial x} = \frac{qc}{4\pi \epsilon_0 c^2 (a^2 + z^2) \left[ a\omega sin(\omega t) - \frac{ac cos(\omega t)}{\sqrt{a^2 + z^2 }} \right] }$$
Can't seem to match the final expression - what worries me is the lack of $\omega^2$ in the answer.
2. Jan 14, 2015
### Orodruin
Staff Emeritus
What happened to the x-component of the vector potential? | 2017-10-17 10:29:59 | {"extraction_info": {"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, "math_score": 0.696042537689209, "perplexity": 1999.5945039631854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00098.warc.gz"} |
https://onlinejudge.org/board/viewtopic.php?f=9&t=9148 | ## 10060 - A hole to catch a man
All about problems in Volume 100. If there is a thread about your problem, please use it. If not, create one with its number in the subject.
Moderator: Board moderators
10153EN
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### 10060 - A hole to catch a man
I want to ask if there's mistakes in the sample input? It states that the line of input is in the form of T X0 Y0 X1 Y2 ... X0 Y0
However, for the sample input:
2
2 0 0 0 10 5 15 12 10 10 0 0 0
5 0 0 5 100 100 0 0 0
5 3
1
2 0 0 10 0 10 10 0 10
5 2
0
The second last line isn't of this format. Do I understand wrongly? If not, can anyone give the actual sample input data for this case? Thx.
10153EN
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Posts: 148
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Problem solved. Just to add the first point at the end of the line~
ljb
New poster
Posts: 2
Joined: Wed Oct 09, 2002 10:03 am
### 10060 is there any mistake in the sample input?
the problem say:
Ti X0 Y0 X1 Y1 X2 Y2 … … Xn Yn X0 Y0
Where Ti is the thickness of the sheet, and Xi Yi are the coordinates of corner points. The line ends with co-ordinate of the first point.
but the second input does not end with the x0 y0
1
2 0 0 10 0 10 10 0 10 <-should be 0?
5 2
i got wrong answer ,55555555555
cytse
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I think that line should be
2 0 0 10 0 10 10 0 10 0 0
jpfarias
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Location: Brazil
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### 10060 It's frustating
Man, I thought I've got this problem, but it appears that something is wrong...
My approach to solve this problem was summing up all the volumes of the pieces and then dividing by the volume of the man hole. What's wrong?
Thanks, JP!
Whinii F.
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It's likely you rounded off your result, but you must floor() it to get AC.
JongMan @ Yonsei
CodeMaker
Experienced poster
Posts: 183
Joined: Thu Nov 11, 2004 12:35 pm
### 10060
Hi.......really this one is tough any help plz? i m getting WRONG ANSWER. A lot actually
Code: Select all
#include<cstdio>
#include<cmath>
struct point
{
double x,y;
};
point p[10];
int n;
double area(point a,point b,point c)
{
double area;
area=a.x*(b.y-c.y)+b.x*(c.y-a.y)+c.x*(a.y-b.y);
return 0.5*fabs(area);
}
int main()
{
double sum,pt,r,t,total,pi=2*acos(0.0);
int k;
// freopen("in.in","r",stdin);
while(scanf("%d",&n)==1 && n)
{
total=0;
for(k=0;k<n;k++)
{
scanf("%lf%lf%lf%lf%lf",&pt,&p[0].x,&p[0].y,&p[1].x,&p[1].y);
sum=0;
while(scanf("%lf%lf",&p[2].x,&p[2].y) && (p[2].x!=p[0].x || p[2].y!=p[0].y))
{
sum+=area(p[0],p[1],p[2]);
p[1]=p[2];
}
total+=pt*sum;
}
scanf("%lf%lf",&r,&t);
printf("%.0lf\n",floor(total/(pi*r*r*t)));
}
return 0;
}
Jalal : AIUB SPARKS
shahriar_manzoor
System administrator & Problemsetter
Posts: 399
Joined: Sat Jan 12, 2002 2:00 am
### hmm
The formula u r using is only acceptable for convex polygon. On the other hand the input polygons may be concave. For concave polygons some or the triangle areas must be negative if you want to get the true result.
CodeMaker
Experienced poster
Posts: 183
Joined: Thu Nov 11, 2004 12:35 pm
Thank you, Now I have solved this problem and also thanks to my friend Dip who helped me to find the precision error.
Jalal : AIUB SPARKS
asif_rahman0
Experienced poster
Posts: 209
Joined: Sun Jan 16, 2005 6:22 pm
### 10060
Hello I'm facing problem to understand the problem 10060.Can anybody help me?
daveon
Experienced poster
Posts: 229
Joined: Tue Aug 31, 2004 2:41 am
Hi,
The problem is basically this:
Given a volume L1,L2,...,LN and a volume of a man hole MH,
how many times can MH go into L1+L2+...+LN or
answer = floor( (L1+L2+...+LN) / MH );
mohiul alam prince
Experienced poster
Posts: 120
Joined: Sat Nov 01, 2003 6:16 am
Location: Dhaka (EWU)
### 10060 A Hole to Catch a Man WA
Hi
I am getting wa in this problem but don't know why. I have trying to solve
this problem by this equation
Code: Select all
(x1y2 + x2y3 + x3y4 + ............. + xny1) - (y1x2 + y2x3 + y3x4 +.......... ynx1)
can any body help me.
Here is my code
Code: Select all
#include <stdio.h>
#include <string.h>
#include <math.h>
#define pi acos(-1)
char Table[100024];
double x[20005];
double y[20005];
int main() {
int i, j;
int N;
//freopen("D:\\in.txt", "r", stdin);
while (scanf("%d", &N) == 1) {
if (N == 0) break;
gets(Table);
double ans1 = 0, ans2 = 0, ans = 0;
for (j = 1; j <= N; j++) {
gets(Table);
char *p;
p = strtok(Table, " ");
int n = 1;
int m = 1;
double t;
sscanf(p, "%lf", &t);
while (1) {
p = strtok(NULL, " ");
if (p == NULL) break;
sscanf(p, "%lf", &x[n++]);
p = strtok(NULL, " ");
sscanf(p, "%lf", &y[m++]);
}
ans1 = ans2 = 0.0;
for (i = 1; i < n - 1; i++) {
ans1 += ((x[i] * y[i + 1]) - (y[i] * x[i + 1]));
}
ans1 += ((x[n] * y[1]) - (y[n] * x[1]));
ans += fabs(0.5 * (ans1) * t);
}
double area;
double r, s;
scanf("%lf %lf", &r, &s);
area = 1.0 * s * pi * r * r;
printf("%.0lf\n", floor(ans / area + .5));
}
return 0;
}
Thanks
MAP
sohel
Guru
Posts: 856
Joined: Thu Jan 30, 2003 5:50 am
Location: New York
Why are you using gets to take input..
each polygon may be given in more than one line !!
mohiul alam prince
Experienced poster
Posts: 120
Joined: Sat Nov 01, 2003 6:16 am
Location: Dhaka (EWU)
Sohel
My solution has other problems i have fixed this problem and got
AC.
MAP.
chops
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Posts: 9
Joined: Sat Jan 29, 2005 10:48 pm
Location: dhaka
Contact:
please give me some I/O. | 2019-11-16 01:26:21 | {"extraction_info": {"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, "math_score": 0.4856712818145752, "perplexity": 7081.663116801086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668716.69/warc/CC-MAIN-20191116005339-20191116033339-00354.warc.gz"} |
http://math.stackexchange.com/questions/248390/what-is-the-difference-between-a-poisson-process-and-a-poisson-point-process | # What is the difference between a poisson process and a poisson point process?
What is the difference between a poisson process and a poisson point process???? in terms of properties and definition ? If possible, please also provide an explanation in layman terms.
-
At time t the former counts the points of the latter located in the interval (0,t). – Did Dec 1 '12 at 11:03
A process $(N_t)_{t\geq 0}$ is called a Poisson process with parameter $\lambda>0$ if it is a Lévy process and
1) $N_t-N_s\sim\mathrm{po}(\lambda(t-s))$ for all $0\leq s<t$.
2) $t\mapsto N_t(\omega)$ is non-decreasing with values in $\mathbb{N}_0$ for almost all $\omega$.
Being a Lévy process means that $N_0=0$, the sample paths are càdlàg and that the process has independent increments. The sample paths looks something like this ($\lambda=1$):
When talking about Poisson point processes there is, first of all, no time parameter in question. Let's consider $\mathbb{R}^n$ and the family of all closed subsets $$\mathcal{F}=\{F\subseteq \mathbb{R}^n\mid F\text{ is closed in }\mathbb{R}^n\}.$$ Now, one equips $\mathcal{F}$ with the topology of closed convergence and then we can talk about the Borel sets $\mathcal{B}(\mathcal{F})$ of $\mathcal{F}$. Let $(\Omega,\mathcal{A},P)$ be a probability space.
Definition: A mapping $Z:\Omega\to\mathcal{F}$ is called a random closed set if $Z$ is $(\mathcal{A},\mathcal{B}(\mathcal{F}))$-measurable. The measure $P\circ Z^{-1}$ is called the distribution of $Z$.
In order to introduce point processes in general, we need to introduce the family of all locally finite sets in $\mathbb{R}^n$: $$\mathcal{F}_{\text{lf}}=\{F\in\mathcal{F}\mid \#(F\cap C)<\infty \text{ for all compact subsets }C\}.$$
Definition: A random closed set $X$ with $P_X(\mathcal{F}_{\text{lf}})=1$ is called a point process in $\mathbb{R}^n$. The function $\Theta$ on $\mathcal{B}(\mathbb{R}^n)$ given by $$\Theta(A)=E[\#(X\cap A)],\quad A\in \mathcal{B}(\mathbb{R}^n)$$ is called the intensity measure of $X$.
With this in hand a Poisson point process is defined as:
Definition: Let $X$ be a point process on $\mathbb{R}^n$ with intensity measure $\Theta$. Then $X$ is called a Poisson point process if $\#(X\cap A)\sim \mathrm{po}(\Theta(A))$ for all $A$ with $\Theta(A)<\infty$.
-
med = with? – Did Dec 1 '12 at 11:04
@did: Thanks, a little Danish slipped in there :) – Stefan Hansen Dec 1 '12 at 11:46
Indeed. Chassez le naturel, il revient au galop, as they say. – Did Dec 1 '12 at 12:23 | 2015-05-06 15:37:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9131674766540527, "perplexity": 159.1726266050857}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430458866468.44/warc/CC-MAIN-20150501054106-00042-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://en.wikibooks.org/wiki/Introduction_to_Physical_Science/A.1 | # Introduction to Physical Science/A.1
= Appendix A1: Practice Problems Set 1
Example problems
• There is a article of handicraft,it has a V=10cm^3 and M=157g. Is it pure gold?(Note: D Of pure gold is 19.3g/cm^3)
Answer: NO
Di=M/V=157/10=15.7g/cm^3
because 15.7g/cm^3<19.3g/cm^3,so we can judge this article of handicraft is not pure gold.
• What are the SI unit base symbols and what do they mean?
Answer [m]-length
[kg]-Mass
[s]-time
[a]-electrical current
[k]-temp
[mol]-amount of substance
[cd]- liminous intensity
## Practice Problems and Solutions
Students - please insert your questions and solutions in the appropriate categories below and make sure you sign your problems. Use a # sign to number the problem and put a space before each line in your answer to put it in a box. --J.M.Pearce (talk) 12:55, 11 February 2008 (UTC)
### Displacement
• You are on a trip to visit your grandparents in Myrtle Beach (719 mile trip) for spring break. You drive half way there and realize that you forgot your toothbrush. So you turn around and drive back to get it. Then you complete the drive and spend the week with your grandparents and then go home. What is your displacement?
Solution. Your displacement is 0 because displacement is final position minus
initial position.
• Dawn is driving to her new job today. She left at 8am this morning. She drove 10 miles from her house in the small town of Cranston to the town of Meadenrock. From Meadenrock, she drove 20 miles to Richport. From Richport, She drove 40 miles back through Meadenrock and Cranston to Porterville. When she reached Porterville she realized she was very lost. What is Dawn's total displacement?
Displacement = total change in position
Start - Cranston
20mi - Richport (this is the farthest point before she makes a wrong turn)
40mi backwards - Porterville
30 - 40 = -10
so...her total displacement is only 10 miles Ten miles will equal the quality of light years from the units of space and volume which = 206.
### Force
• A linebacker runs towards a Quarterback who is standing in the pocket looking for a receiver. The linebacker tackles the stationary quarterback in the pocket. What is the linebackers acceleration if we know the Linebackers weight is 312kg and the force of the impact on the QB is 342 Newtons.
Solution: F=ma
First you must manipulate the equation for this particular problem
A=f/m
Once the equation has been manipulated, then you simply plug your information in.
A=f/m
A=342kg/312N
A=1.096m/s^2
• A 10.0 kg log was pulled along the ground (assuming with no friction) by a force of 200 N. What is the acceleration?
We know:
F=ma
F= 200 N
m= 10 kg
200 N = 10.0 kg (a)
200 N/10.0 kg = a
20 m/s^2= a
• What is the force of a 6kg baby sitting in your lap?
Force (F) = Mass (m) x Accleration (a)
Here the a is due to gravity = 9.8 m/s^2 or about 10 m/s^2
So F = ma = 6kg x 10 m/s^2 = 60 N
• If a person who weighs 180lbs falls out of a tree, with how much force does he hit the ground with.
You know that F=ma
1 lbs = 0.4356 kg
180 * .4356 = 78.41 Kg
a= force of gravity 9.8 m/s^2 so...
F= 78.41Kg (9.8m/s^2)
F=768.42 N
• A hockey puck is sent flying over the ice. If the air resistance and friction of the hockey puck are ignored, what is the force required to keep that puck moving at a constant velocity?
The solution would be zero newtons.
• A 32.1 kg log is rasied off the ground with a rope. The acceleration of the gravity is 9.8m/s^2. If the upward acceleration of the log is 3.9m/s^2, find the force excerted by the rope on the log. Please use units of newtons.
F=MA
F=(32.1kg)(9.8m/s^2+ 3.9m/s^2)
F=(32.1kg)(13.7m/s/s)
F= 439.77 N
• A 15.0 kg mass pulled along a frictionless surface by a horizontal force of 100 N will have what acceleration?
Use F=ma
(100N)=(15.0kg)a
a=6.67m/s^2
• An object sits on a frictionless surface. There is a 16 N force applied to an object parallel to the surface and its acceleration is at 2m/s/s. What is its mass?
Use F=ma
16N=m(2m/s^2)
m=8kg
• The pressure inside of a bottle is 7 atm, and the pressure outside of is 3 atm. The inner radius of the bottle is 2cm. so what is the frictional force on the cork?
P=P(inside)-P(outside)=4 atm
F=(3.14r^2)P=3.14x(0.02)^2x1.013x10^5 x4 atm =508.9N
• there is an object of mass 2kg on the table. and there is a force pushes it to move forward and this force is 5N, but the object is still there. it doesn't move at all. find out what forces apply on this object.
answer: G(gravity)=mg=2kgx9.8n/kg=19.6N
F=G=mg=19.6N
F(push)=5N
Friction force=F(push)=5N
and G and F are opposite directions, F(push) and friction are opposite directions.
• What is the force due to gravity of a hamburger sitting on a table? Assume that each bun weighs 1/4 kg and the beef weighs 1/2 kg.
F=ma
a=g=9.8s/m^2
m= 1/4 + 1/4 + 1/2
m= 1 kg
F= (1kg)(9.8s/m^2)
F= 9.8 N
mmmm tasty :)
• A constant force is applied to an object, causing the object to accelerate at 8.0m/s^2. What will the acceleration be if the objects mass is doubled?
answer: 4m/s^2
a=f/m
• A baseball player is sliding into 2nd base, identify the forces on the player.
Weight, normal force by ground, kinetic friction force by ground
• Suzan pulled a wagon that weighed 5 kg along the street at a force of 100 N, what is the acceleration? Assume she was pulling the wagon directly in the direction of motion (no angle).
F=ma
a=f/m
a=100N/5kg
a=20N/kg
• What is the force needed to pull a car that weighs 67,500 kg at an acceleration of 5m/s^2?
F=ma
F=(67,500 kg)(5m/s^2)
F=337,500 N
• If you have a box that is pulled at an acceleration of 10 m/s^2 with a force of 98 N. What is the boxes mass?
F=ma
m=F/a
m= 98 N / 10 m/s^2
m= 9.80 kg
• Jose pushes a boulder that weighs 60 kg and exerts a force of 60 N. What is the acceleration of the boulder?
F=ma
60 N= 60kg x a
a= 1 m/s^2
• Which Law states that if no force acts on an object, then its speed and direction do not change?
Newton's 1st Law
### Velocity
• If you are given the maximum velocity of 20m/s for a car. What would be its acceleration if it requires 500 seconds to reach max speed?
To solve you would first see that acceleration is $change in V/change in T$ = $Vf-Vi/Tf-Ti$
Therefore, the acceleration would be $20m/s-0m/s/500s-0s$ = 0.04m/s2
• A car is traveling down the Highway and it takes the car 20 minutes to travel from Exit 1 to Exit 2. The car was traveling 5 mph over the posted speed limit which is 55 mph. What is the distance between these two exits in miles.
V=d/t
t= 20min/60min/hr = 1/3 hour -- To solve for time you must first convert minutes into hours
v= 60 mph
d= vt = 60 mph x 1/3 hour - Once you have found time and distance all you have to do is plug those answers into the equation d=vt, and then solve.
The final answer is d=20 miles
• If the change in the velocity of vector A = 5m/s to the right and the change in the velocity of vector B = 8m/s to the right, then what is the total change in velocity to the right?
The correct answer is 13 m/s.
Remember to add vectors from head to tail. Therefore, Vector A + Vector B = Total change in velocity.
5 m/s + 8 m/s = 13 m/s
• A.Given that a van has a height of 1.1m and 1.8 is the track width, what is the SSF?(static stability folder)
The correct answer is 0.81. Remember to plug the numbers into the SSF formula. t/2h=v^2/rg.
• Part B. Given R=15m, what is the fastest you can take a corner without tipping over the van?
The correct answer is V=10.91m/s. Remember to use part one's answer in place of t/2h. Also remember that G= 9.8m/s.
1. Joe drove 885 meters in 11 minutes at a constant velocity. At what velocity was Joe traveling?
Answer: 1.34 m/s
v = d/t
v = 885 meters/11 minutes
11min.x60sec.=660seconds
v = 885 meters/660 seconds
v = 1.34 meters/second
• Sarah has drive 318 meters to the grocery store to get ingredients to bake a cake for her mom's birthday. When she left the store she realized she forgot to buy milk. So she stopped at the gas station conveinent store on the way home. The conveinent store is located exctly halfway between the grocery store and her house. What was her displacement when she stopped at the conveinent store?
Answer: 159 meters
d = df-di
* 318 meters / 2
d = 318 meters - 159 meters
d = 159 meters
• A professional baseball pitcher throws a curveball to the number four hitter of the opposing team. The curveball travels 60ft to homeplate. The batter hits the curveball for a homerun which went 409ft to dead center field. What is the total displacement of the baseball from home plate?
The correct answer would be 409ft
Remember that displacement is the Δx
therefore the equation is Δx= xf - xi Δx= 409-60 Δx= 369ft
• A football quarterback throws a touch pass of 25 meters and it takes the ball 1.75 seconds to reach the receiver. How fast did the QB throw the ball.
the correct answer is 14.29 m/s
v= d/t
v= 25 meters/ 1.75 seconds
v= 14.29 m/s
• A toddler pushes "Block A" to the right toward "block B"(which is at rest). "Block A"(which is 1kg), has a velocity of 6 m/s moving to the right toward "Block B". "Block B" is 2kg. Now "Block A" is at rest and "Block B" moves to the right at an unknown velocity. Find this unknown velocity.
∑ MV0 = ∑ MVf
(1kg)(6m/s)+(2kg)(0m/s)=1kg
6kgm/s=2kgV
2kg 2kg
V=3m/s
#### Speed Limits
• An Olympic runner ran the a 800 m in 86 s. What was his average speed?
The runner's average speed is given by:
average speed= distance/time= 800/86= 9.3 m/s
• A car is going 94.5 km/h. A deer runs out infront of the car, causing it to come to a complete stop in 7.8 s. What is the acceleration of the car?
We know that acceleration is a change in velocity over a given changed in time A=V/T
So we plug 95 km/h in for V, but V is measured in m/s so we must convert. 94.5 km * 1000m/km
= 95,000km and 1 hr * (60m/1hr) * (60s/1m)=3,600 s so after converting V=95,000m/3,600s
V= 26.39 m/s then we plug in T which is 8s so A=(26.39 m/s)/ 7.8s A=3.3 m/s2
• Julie leaves Clarion at 8:00 am to drive to Pittsburgh, 100 miles aways. She travels at a constant speed of 50mph. Matt leaves Clarion at 8:00am and travels a constant speed of 60mph. How long does the first person to arrive have to wait for the second person?
Using the equation V=d/t we can manipulate it to solve for time for each person.
So using the equation t=d/v we plug in the numbers for Julie's trip t= 100mi/50mph.
We get t=2hrs for Julie's time. Now we use the same equation to solve for Matt's time.
t=d/v. t=100mi/60mph. Matt's time equals 1.67hrs. So if you take Julie's time and
subtract Matt's time from it 2hrs-1.67hrs=0.33. Therefore, Matt has to wait ~20 mins for
Julie to arrive in Pittsburgh.
• A speedboat with a velocity of 50 m/s only has 15 m until it crosses the finish line. How long will it take the boat to cross the finish line?
ΔV = ΔX/Δt (change in velocity equals the change in displacement divided by the change in time)
Δt = Δx/Δv
15m/ 50m/s = 0.3 s
• What is the correct equation for velocity?
a) changeV= changeX /changet = Xf - Xi / tf - ti b) V=2(pi)r / T
correct answer : a)
• What is the speed of a vehicle that travels 800 meters in 20 seconds?
v=d/t
v=800 meters /20 seconds
v=40 meters per second
• How long will your trip take (in hours) 425 km at a speed of 70 km per hour?
t=d/v
t=425,000 meters /70,000 meters
t=6.07 hours
• One car is traveling 23.8 m/s. Another car traveled 89 meters in 2.7 seconds. Which car has a higher velocity?
V=change in distance/change in time
V of the first car = 23.8 (given)
V of second car = 89m/2.7s
V of second car = 32.96m/s
Therefore, the second car has a higher velocity than the first.
### Circular Motion
• A baseball with a 3 in diameter is rolled on the ground with no slipping and it revolves 50 times how far does it go?
The correct answer is ~ 471 inches
Distance Traveled = 50*C
Diameter/(d)= 3in
C=d(pie)
pie= 3.14
50*C
50d(pie)
50(3in)(3.14)= ~ 471 inches
• If a baseball is rolled on the ground and revolves seven times, how far does it go in feet? Given the diameter is six inches.
The correct answer is 131.88in or 10.99 feet
C=2TTR
d=2r
C=dTT
distance=7c
7dTT
TT is equal to pi or 3.14.
7(6in)TT=42inTT=42*3.14=131.88in or 10.99ft.
• If a child rolls a ball on the ground and resolves ten times. How far does the ball travel? The diameter of the ball is 4 inches.
Formula: C=2(TT)(r)
D=2(r)
C=(D)(TT)
=(10)(D)(3.14)
=(10)(4)(3.14)
Solution: 125.6 inches or 10.5 feet
• A regular bowling path is 19.15m long; the diameter of a bowling ball is 21.5cm. If the bowling ball goes the whole way straightly in 8seconds, what is its frequency in this situation?
Circumference of the bowling ball: 21.5*3.14=67.51cm
Length of the bowling path in cm: 19.15*100=1915cm
Rounds of the bowling ball rolled:1915/67.51=28.37 rounds
Frequency: 28.37/8= 3.55 r/sec
• If a ferris wheel takes 3 minutes to go one revolution and you want to pay the man running it to get an extra long ride with your significant other. He tells you that he will make your ride longer for each dollar you give him by 1 revolution. You only have 5 dollars in your wallet. How long will your ride be?
1 revolution takes 3 minutes. 5 dollars times 1 revolution = 5 revolutions. 5
revolutions times 3 minutes a revolution = 15 minutes.
• A basketball is 9.07 inches in diameter. It rolls across the court 7 times before a player stops it. How far did it roll?
D=7C
C=2TTrsquared= TTd
D=7(TTx9.07in)=199.46 in
The distance traveled by the ball was approximately 199.46 in.
### Static Stability Factor
• If you are driving a Jeep Laredo with a track width (t) of 2.2 m, with a height (h) of .95 m, what is the static stability the factor (SSF)?
The formula for (SSF)= t/2h
t= 2.2 m
h= .95 m
2.2 m / 2 (.95 m)
2.2 m / 1.9 m
SSF = 1.16
• If a vehicle is 1.2m high, to the center of gravity, and the distance between the tires is 2.3, at what velocity can you take a curve with the radius being 12m.
t=2.3 m
h=1.2 m
r=12 m
g=9.8 m/s²
v=sq. root of rgt/2h
v=12(9.8m/s²)(2.3)/2.4
v=sq.root of 112.7m/s²
v= 10.61m/s²
• Given that a racecar has a height of 2.5m and the width of the racetrack is 2.8m. What is the static stability folder (SSF)?
Formula: t/2h=v^2/rg
2.5/2(2.8)=2.5/5.6
Solution: .45m
Part B. Given R=25m. Coming into turn three, what is the fastest that the racecar can take the turn without tipping over?
Formula: t/2h=v^2/rg (Substitute t/2h with the previous answer of .45m)
.45m=v^2/rg (Substitute rg with 25 (being r) and 9.8 (being g)
.45m=v^2/(25)(9.8)
Multiply both sides by (25)(9.8)
.45(25)(9.8)=v^2
Take the square root of .45(25)(9.8)
Solution: 10.5m
• A Hummer with a SSF of 1.25 is rounding a curve with r=12 m. How fast can the Hummer go around the curve without flipping?
Remember tipping point is t/2h=v2/rg
Rearrange so that v2=trg/2h
SSF=t/2h
v2=(SSF)rg
v2=(1.25)(12 m)(9.8 m/s2)
v2=147 m/s2
v=12.1 m/s
A Hummer can go around the curve r=12 m at a speed of 12.1 m/s without flipping.
• A mini cooper has a track width (t) of 57.7 inches, with the central mass height (h) of 25 inches. Suppose it is running with 215 km/h, what would be the max radius of the curve can this mini copper turn?
57.7inches= 1.47m
25inches= 0.64m
215 km/h= 59.72m/second
t/2h=v^2 /rg
1.47/ (2*0.64) = 59.72^2/ r(9.8)
1.15=3655.48/r(9.8)
11.27= 3655.48/r
3566.48/11.27=r
r=316.45m
• If a vehicle has a track width of 1.5m and a SSF of 1.25, what is the height of its center of mass?
t=1.5m
SSF=1.25
SSF=t/2h
solve the SSF=t/2h equation for "h"--
(2h)SSF=t
Divide both sides by (2)(SSF)
h=t/2(SSF)
Plug the known values of "t" and the SSF into this equation to get the height.
h=1.5m/2(1.25)
h=0.6m
The height of the vehicle's center of mass is 0.6 meters.
• If we know that the track width of a van is 1.5m, the height is .75m, what is the SSF?
the equation for SSF is
SSF=t/2h
SSF=1.5(2*.75)
SSF= 2.25
### Torque
• Torque depends on what two things?
The magnitude of the force, and where the force is applied or the point of application.
• What is the measure of effectiveness of a force to turn or twist an object?
Torque
• A pinwheel is turned with the force of 6N at a distance of 10cm from the rotation. What is the torque?
Torque is T=rF(perpendicular), which is measured in Nm.
Torque= (.10m)(6)= 0.6Nm
• A child of mass of 50 kg is sitting on a seesaw 4m from the pivot- What is the magnitude of the torque created by his weight?
Remember acceleration of gravity= 9.8 m/s2
F=ma=(50 kg)(9.8N/kg)
T=Fr=(50 kg)(9.8N/kg)(4m)
T= 1960 Nm
• A mother and child are playing on a see-saw. The child is sitting 3 m from the pivot. The magnitude of torque created by the child is 999.6 Nm. How much does the child weigh?
The equation for torque is T=rF. Remember F=ma. So, T=rma. Solve equation for m, m=T/ra.
m=T/ra
m=(999.6 Nm)/(3 m * 9.8 N/kg)
m=(999.6)/(29.4 kg)
m=34 kg
A child sitting 3 m from the pivot, creating a torque of 999.6 Nm, weighs 34 kg.
• If somebody is using a wrench with the F of 30N at a distance of 18cm from the rotation axis, what is the magnitude of the torque?
The equation for torque is T=Fr. Remember to change cm to m.
T=Fr
T=(30N)(.18m)
T=5.4Nm
• A circular silo is a used to store grain for feeding cattle. A man is trying to turn a silo unloader by hand. Each arm of the unloader is parallel to the axis of the unloader. How much Force should he use if the silo has a circumference of 20 feet and is using a torque of 740 Nm?
C=2pir
20=2pir
10/pi=r
3.18 ft=r
feet must be changed to meters so 3.18(30.5)= 96.99 cm= .9699 m
T=rF(parallel)
740 Nm= .9699m(F)
F= 762.9N
• A little girl of 46Kg is being pushed on a merry go round counter clockwise. She is 2.4m from the center. What is the magnitude of the Torque?
T=+ rF(perpendicular)
-
(46kg)x(9.8N/kg)<--Gravity
(46kg)x(9.8N/kg)x(2.4)=1081Nm
• Phil took his car to get his tires aligned. The mechanic elevated the car and spun the tire with a force of 30 N. The force he exerts is 62 cm from the rotational axis. What is the magnitue of the torque?
T=Fr
T=30N x 62 cm / 100cm/m
T=18.6 Nm
• A door is pushed open with a force of 15N. The force is applied perpendicular to the door at a distance of 85 cm from the rotation axis. What is the magnitude of the torque?
T=Fr
T=15N x 85cm / 100cm/m
T=12.75Nm
### States of Matter
• When talking about Pascal's Principal, a change in WHAT? at any point in a confined fluid is transmitted everywhere throughout the fluid?
It is a change in pressure.
• Which of the three states of matter (solid, liquid, and gas) has the strongest bond?
Solids, they have a rigid/fixed shape, and the strongest bond of the states of matter.
• All of the states of matter,(solids, liquids, and gases), are described by what? And what is the equation for the description?
All states of matter are described by their DENSITY. This is mass per unit volume.
The equation for this is D=m/V.
• There are three states of matter (solid, liquid, and gas) two of them fall into a category that has no fixed shape and is easily deformed. What is that category and what two fall into that category.
The category is FLUIDS and the two states of matter that are in FLUIDs are gases and
liquids.
• Gold has a density of 19.3g/cm^3. What is the volume of a gold bar that has a mass of 78g?
V= md
V= (78g)(1 cm^3/19.3g)
V= 4.04 cm^3
• The volume of a mystery fluid is 12 cubic meters with a mass of 37kg. What is it's density?
D=m/v
D=37kg/12m^3
D= 3.083kg/m^3
• Which out of the three states of matter [Solid, Liquid, Gas]is the only one able to have a change in volume?
Gas, it is compressible, and when heated, its molecules moved farther apart.
• If the pressure on the box is 3atm, the area of box on the surface is 12cm^2.What is the force on the box?
F=P*A
P=3*1.013*10^5=3.039*10^5Pa
F=3.039*10^5pa*12*(1/1000)m^2=364.68N
### Elasticity and Oscillations
• What is the definition of elasticity?
Means that a object returns to it's orginal shape and size after it has been deformed.
• Steve picks up a rubber band, which at rest measures 2.5 inches long. He then streches the rubber band across two fingers, the rubber band is now 3.5 inches longer than it was at rest. What is the strain Steve is placing on the rubber band?
2.5in=6.35cm, 6in=15.24cm
Strain= change in length/length
(15.24cm-6.36cm)/6.35cm = Strain
Strain = 1.4cm
• The original length of a spring is .20 m long. When the spring is stretched to .30 m long with the stress=3.5, what is the stiffness of the spring?
stress/strain = stiffness
strain = ΔL/L
3.5 / .30-.20/.20 = 7 [Pa]
• What are the Four Fundamental Forces?
1. Force of gravity- graviton
2. Strong Nuclear Force- holds protons and neutrons together in an atom
3. Weak Nuclear Force- radioactive decay
4. Electromagnetic Force- contact forces
• Debbie is exercising with an elastic cord which when at rest measures .5m. When she pulls it it stretches to a length of 1.2m. What is the strain on the cord?
strain=ΔL/L
strain=1.2-.5/.5
strain=.7/.5
strain=1.4
1. A Frisbee is 25.4 cm in diameter. If it is thrown, not caught, and hits a wall the new diameter is 24.6 cm. What is the strain on the Frisbee?
Strain=deltaL/L
=-0.08/ 25.4 cm= -0.03 | 2014-10-21 05:31:10 | {"extraction_info": {"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": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46254363656044006, "perplexity": 2006.6552269044553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443901.32/warc/CC-MAIN-20141017005723-00330-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1886801/show-that-the-one-point-compactification-of-mathbbq-is-not-hausdorff | Show that the one point compactification of $\mathbb{Q}$ is not Hausdorff
https://en.wikipedia.org/wiki/Alexandroff_extension
Definition: one point compactification
Let $X$ be any topological space, and let $\infty$ be any object which is not already an element of $X$. Put $X^{*}=X\cup \{\infty \}$, and topologize $X^*$ by taking as open sets all the open subsets $U$ of $X$ together with all subsets $V$ which contain $\infty$ and such that $X\setminus V$ is closed and compact
Show that the one point compactification of $\mathbb{Q}$ which is $\mathbb{Q}^*$ is Not Hausdorff.
What to do? What...
My Attempt:
Suppose to the contrary $\mathbb{Q}^*$ is Hausdorff, then we take two points $x$, $\infty \in \mathbb{Q}^*$, $x \neq \infty$, and produce disjoint open sets $U,V$ such that $x \in U$ and $\infty \in V$
By definition, we know that $\mathbb{Q} \backslash V$ is a closed and compact space such that $x \in U \subseteq \mathbb{Q} \backslash V$
We wish to produce a contradiction such that $\mathbb{Q} \backslash V$ is not closed, or not compact. But we know that $\mathbb{Q} \backslash V$ has to be closed since $V$ is open, therefore we need to show $\mathbb{Q} \backslash V$ is not compact.
Let $\mathcal{U}$ be an open cover of $\mathbb{Q} \backslash V$. Since $\mathbb{Q} \backslash V$ is claimed to be compact, then $\mathcal{U}$ has a finite subcover $\{U_i|i \in F\}$, $F$ is finite in $\mathbb{Q} \backslash V$. Then for all $x \in \mathbb{Q}\backslash V$, $\exists i \in F$ s.t. $x \in U_i$ (...Ugh everything seems fine...)
Can someone provide me with some help as to how to go on with this proof? Thanks a bunch.
• – user348749 Aug 9 '16 at 8:51
• More generally, one can show that $X^*$ is Hausdorff iff $X$ is Hausdorff and locally compact. – Alex Provost Aug 9 '16 at 17:24
Given your open sets $U$ and $V$ in $\Bbb Q^*$, you can continue the argument as follows.
$U$ is an ordinary open nbhd of $x$ in $\Bbb Q$, so it contains an open nbhd of $x$ of the form $(a,b)\cap\Bbb Q$ for some irrational $a,b\in\Bbb R$. Let $W=(a,b)\cap\Bbb Q$; clearly $W\cap V=\varnothing$, so in particular $\infty\notin\operatorname{cl}_{\Bbb Q^*}W$. Thus,
$$\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q=W\;.$$
But this is clearly impossible, since $W$ is not compact.
If you want an explicit example of an open cover of $W$ with no finite subcover, let $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ be sequences in $(a,b)$ such that
• $\langle a_n:n\in\Bbb N\rangle$ is strictly decreasing and converges to $a$,
• $\langle b_n:n\in\Bbb N\rangle$ is strictly increasing and converges to $b$, and
• $a_0<b_0$.
Let $U_n=(a_n,b_n)\cap\Bbb Q$ for $n\in\Bbb N$; then $\{U_n:n\in\Bbb N\}$ is the desired cover.
• Hi, how do you reach the conclusion that: $\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q=W\;.$? I understand why $\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q$, but then how does that equal to $W$? And what do you mean by $W$ is not compact? Do you just mean that we defined $W = (a,b) \cap \Bbb{Q}$ but later reached the conclusion $W = [a,b] \cap \Bbb{Q}$...do you mean the latter $[a,b] \cap \Bbb{Q}$ is compact? How? Thanks for ur help – Olórin Aug 9 '16 at 22:45
• @John: $a$ and $b$ are irrational, so $[a,b]\cap\Bbb Q=(a,b)\cap\Bbb Q$. \\ I mean that $W$ is not compact. I even exhibited an open cover with no finite subcover. – Brian M. Scott Aug 10 '16 at 0:10
• I agree, I just wish to understand your top proof. So we let $W = (a,b) \cap \mathbb{Q}$ which is open and not compact, then we found that $\text{cl}_\mathbb{Q}^* W = W$, which implies it is closed and compact... so we reach a contradiction. I think what I am not getting is why is $(a,b) \cap \mathbb{Q}$ not compact in $\mathbb{Q}^*$, and why if we show that $\text{cl}_\mathbb{Q}^* W = W$ is closed, then it is compact. After all we are not working in $\mathbb{R}$, so the intuition about closed and bounded = compact doesn't hold. – Olórin Aug 10 '16 at 0:25
• @John: Every closed subset of a compact space is compact. Since $W$ is closed in $\Bbb Q^*$, it must be compact. But in fact it's just $(a,b)\cap\Bbb Q$ with its usual topology, which is not compact. This contradiction shows that $x$ and $\infty$ cannot in fact be separated by disjoint open sets. – Brian M. Scott Aug 10 '16 at 0:35
Let $C$ be a compact subset of $\mathbb Q$. Then because the inclusion map $\mathbb Q\to\mathbb R$ is continuous it follows then $C$ is a compact subset of $\mathbb R$, so it is bounded and has all its limit points.
Let $a,b\in\mathbb R$ with $a<b$ such that $(a,b)\cap\mathbb Q\subseteq C$, then the closure of $C$ in $\mathbb R$ includes $(a,b)$. So by contradiction $C$ has empty interior. Therefore every open set containing $\infty$ is dense.
Finally let $x\in \mathbb Q$ and let $U,V$ be open sets with $x\in U$ and $\infty\in V$, then $V$ is dense so $U\cap V\ne\emptyset$. | 2019-08-24 19:47:37 | {"extraction_info": {"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, "math_score": 0.9788143634796143, "perplexity": 77.1986526990053}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00357.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-derivative-of-f-x-6-5x-1 | # How do you find the derivative of f(x)=(6-5x)^-1?
Apr 8, 2016
$f ' \left(x\right) = \frac{5}{6 - 5 x} ^ 2$
#### Explanation:
Use the chain rule, where if
$f \left(x\right) = g \left(h \left(x\right)\right)$
then
$f ' \left(x\right) = h ' \left(x\right) g ' \left(h\right)$.
If we say then that $h \left(x\right) = 6 - 5 x$ and $g \left(x\right) = {h}^{-} 1$,
$h ' \left(x\right) = - 5$
$g ' \left(h\right) = - \frac{1}{h} ^ 2 = - \frac{1}{6 - 5 x} ^ 2$
then
$f ' \left(x\right) = \frac{5}{6 - 5 x} ^ 2$ | 2019-06-24 16:17:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9621525406837463, "perplexity": 2315.30287459303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999615.68/warc/CC-MAIN-20190624150939-20190624172939-00250.warc.gz"} |
https://physics.stackexchange.com/questions/458563/power-absorbed-by-electron-in-plane-electromagnetic-wave | # Power absorbed by electron in plane electromagnetic wave
How can the power (in Watts) absorbed by the electron be calculated, knowing the incident electric field amplitude $$E_0$$, wavelength $$\lambda$$, and the electron momentum relaxation time $$\tau$$ in the medium ?
The units seem to check out for $$\frac {|q_e| E_0 \lambda }\tau$$ (result is in Watts), but the correct answer is off by several orders of magnitude. What is missing?
I guess you're also doing the Plasmonics course of edX. (Great course by the way, although I find the exercises quite difficult.)
I solved the task using the formula for power that was shown in the lecture video:
$$P = <\cfrac{\vec{F}+\vec{F}^*}{2}\cdot \cfrac{\vec{v}+\vec{v}^*}{2}> = \cfrac{1}{2}\Re (\vec{F}^* \cdot \vec{v})$$
The velocity is time derivative of position $$\vec{r}$$. To get $$\vec{r}$$ let's solve the equation of motion that is accounting for collisions (this was done in the edX course):
$$m_e\ddot{\vec{r}}=-\gamma m_e \dot{\vec{r}} - |e|\vec{E_0}e^{-i\omega t}$$ which yields the solution:
$$\vec{r}(t)=\cfrac{|e|}{m_e\, \omega \,(\omega + i\gamma)}\vec{E_0}e^{-i\omega t}$$ $$\vec{F}$$ is Lorentz force due to electric field:
$$\vec{F} = |e| \vec{E_0}e^{-i\omega t}$$
So that gives me: $$P = \cfrac{1}{2}\Re (\vec{F}^* \cdot \vec{v}) = \cfrac{|e|^2}{2} \Re \left(E_0^2 \cfrac{-i\omega}{m_e \omega(\omega + i\gamma)} \right) = \cfrac{E_0^2\,e^2 \gamma}{2 m_e \,(\omega^2 + \gamma^2)}$$
Now we now that $$\lambda = \frac{2 \pi c}{\omega}$$, from this we can calculate $$\omega$$ knowing the wavelength. And following again the defitions from edX course $$\gamma = \frac{1}{\tau}$$.
As you can see, the unit is still OK although the expression changed a lot.
I would like to know if there is a simpler way but I cannot think of any right now.
• Incredibly clear and straightforward explanation. Thank you so much. – WantsToLearn Feb 4 at 21:54 | 2019-06-20 07:23:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8876069188117981, "perplexity": 429.19145874243054}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999163.73/warc/CC-MAIN-20190620065141-20190620091141-00299.warc.gz"} |
https://electronics.stackexchange.com/questions/123482/why-isnt-homemade-hardware-a-thing | # Why isn't homemade hardware a thing [duplicate]
What prevents someone with some funding from creating and selling their own hardware? Let's ignore the legal side of things and assume the only limit is technology.
Could someone create something akin to a 1995-2000 time period processor without specialized equipment? What about slightly earlier, like the "computer pioneers" did?
## marked as duplicate by The Photon, Passerby, PeterJ, Scott Seidman, Nick Alexeev♦Jul 31 '14 at 1:29
• Making your own PCB's with off-the-shelf chips is a thing. Lots of people do it. Making your own chips costs 10's of thousands to millions of dollars, so not many people do it as a hobby. – The Photon Jul 31 '14 at 0:09
• – The Photon Jul 31 '14 at 0:11
• @ThePhoton not many people is an understatement. I'd venture to say its around 0. Maybe 10 max. – Funkyguy Jul 31 '14 at 0:49
• OK, so this is going a wee bit further back than, say, 1995 but I can't resist adding this link for an example of 'homemade' hardware: youtube.com/watch?v=EzyXMEpq4qw – Alfred Centauri Jul 31 '14 at 2:29
• If you're permitted to consider an FPGA as raw material, then a 486-era processor is definitely feasible in terms of resources and logic. In terms of effort, of course, cloning a 486 would be a lot of work, to say the least! – Brian Drummond Jul 31 '14 at 8:47
If you're talking about fabricating your own IC like maybe an Intel '386 or '486 processor "without specialized equipment" you can forget it. However, the computer pioneers built the first computers with vacuum tubes, and there's nothing preventing you from buying a boatload of them and wiring them together to duplicate what they did. Or, you could get a bunch of TTL NAND gates and build a processor with them. If you just want to learn about digital logic you don't have to go that far, you could experiment with some simple circuits, then take a course in Verilog or VHDL. At that point you could implement a simple processor in an FPGA rather than wiring together a bunch of discrete gates.
• "... there's nothing preventing you from buying a boatload of them ..." Except availability. runs – Ignacio Vazquez-Abrams Jul 31 '14 at 0:53
• They still make them in China and Russia mostly for audio application, but you can buy all you want :) – John D Jul 31 '14 at 2:25
• Oh, I know you can still get them, I've seen them in surplus stores over here by the hundreds. But getting a hold of 10000 of them will still take some time (and some serious scratch). – Ignacio Vazquez-Abrams Jul 31 '14 at 2:26
• LOL yes, absolutely, but of course it wasn't a serious suggestion anyway :) Though I know a guy who built a switching power supply with vacuum tubes just to see how it would work. – John D Jul 31 '14 at 2:50
• There are guys making their own valves. In most cases though, why bother making/developing a new processor or other device when there's so many readily available for very low cost that fit the bill. Joining them together & programming them to your requirement is easily done & commonly done. – John U Jul 31 '14 at 10:14
Could someone create something akin to a 1995-2000 time period processor without specialized equipment?
Lol no.
What about slightly earlier, like the "computer pioneers" did?
Still no. The first computer pioneers were government funded, while the personal computer pioneers used commercially available parts to build their own computers. Could you recreate say the Apple Mac (Mac 128k) by hand with the right ICs and wiring? Yes. People have. But its not practical let alone trivial.
There are some experiments in tediousness where simple computers are hand made, but often still use mass produced ICs.
Intel’s fabrication plants can churn out hundreds of thousands of processor chips a day. But what does it take to handcraft a single 8-bit CPU and a computer? Give or take 18 months, about \$1,000 and 1,253 pieces of wire.
Steve Chamberlin, a Belmont, California, videogame developer by day, set out on a quest to custom design and build his own 8-bit computer. The homebrew CPU would be called Big Mess of Wires or BMOW. Despite its name, it is a painstakingly created work of art.
Examples in pointless endeavors, for brag points. 4-bit transistor and passives processor. | 2019-07-17 12:48:58 | {"extraction_info": {"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, "math_score": 0.17353220283985138, "perplexity": 2188.1876137140935}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525187.9/warc/CC-MAIN-20190717121559-20190717143559-00167.warc.gz"} |
http://www.openwetware.org/wiki/Physics307L_F09:People/Mahony/Rough | # Physics307L F09:People/Mahony/Rough
SJK 13:53, 2 December 2009 (EST)
13:53, 2 December 2009 (EST)
I think this is a very good rough draft. There is a lot of work needed (see all comments below), but I think you've written a really good foundation to make that very do-able. My guess is that the most substantial work will be required in expanding the results and conclusions section. I gave you some ideas for that, and also you'll naturally be able to expand it due to your "extra data" next week. OK, Good luck!
# The Speed of Light
SJK 11:19, 30 November 2009 (EST)
11:19, 30 November 2009 (EST)
Author & Contact info is missing. Also, a more descriptive title is needed. As it is now, a reader may assume your article is a vast review of all topics related to the speed of light, whereas a more descriptive title could indicate that you're measuring speed of light in a specific way, etc.
## Abstract
SJK 11:31, 30 November 2009 (EST)
11:31, 30 November 2009 (EST)
You are homing in on a short and to-the-point abstract, compared with an abstract that gives more motivation and conclusion. That's OK, if done precisely, which you can probably do. In terms of getting a bit more practice, though, I'd probably prefer that you: (1) expanded the first sentence a bit, if you can add a bit of "why" it's important. (2) Add a second sentence saying that there's a variety of extremely precise methods for measuring the speed of light. (3) then in your next sentence, preface it by saying you're using the time of flight method. (4) your sentence "by positioning the LED..." can be broken up into two sentences -- first that you created fixed distances and measured the change in time delays for each; next that you used linear regression to produce ___ m/s. (5) Finally, add a conclusion sentence--is it consistent w/ accepted value and any other comments.
The speed of light is an important fundamental constant in physics. In this experiment we use a Time-Amplitude Converter to measure the delay between an LED and a photomultiplier tube. By positioning the LED at different distances from the PMT, and using the conversion ratio of voltage to time, we were able to fit our data to a line, with the slope corresponding to our measured speed of light of 2.941(15)*10^8 m/s.
SJK 11:25, 30 November 2009 (EST)
11:25, 30 November 2009 (EST)
LED is not defined and probably should be. PMT is almost defined, you just need to put (PMT) after the "photomultiplier tube." It's good to define almost all acronyms, although you can use your judgement on some of them, such as DNA, e.g.
## Introduction
SJK 11:49, 30 November 2009 (EST)
11:49, 30 November 2009 (EST)
This is an excellent foundation for the introduction, but it's too succinct, I think. It's so concise that it could almost serve as half an abstract! In some cases, such as when page limits are very short, this could be OK. In this case, though, I'd like you to expand each of your ideas. I think you can do this relatively easily based on what you have. Just build sentences around what you already have. For example, what was the community focusing on after 1887? Since you mention it, how was it that special relativity gained wide acceptance? References to key experiments? More important are the "many experiments done to more accurately measure the speed." What were the nature of these experiments? What are the different methods? Can you link to a few specific research papers that reported a value for the speed? Finally, your last sentence is the only sentence about what you're reporting here. You can expand this to say that you're using time of flight, and that you will compare your result with the accepted value, or you could say that the result you produce does compare favorably. And / or that this experiment is a satisfying way to explore methods for measuring the speed of light.
With the refutation of the aether theory by the Michelson-Morley experiment in 1887, light became a subject of focus in the scientific community.[1] Equally controversial was Einstein's theory of special relativity, published in 1905, which proposed that the speed at which light propagated was a fundamental constant, invariant of the speed of the reference frame in which it was observed.[2] This theory became widely accepted, and throughout the rest of the 20th century, many experiments were done to more accurately measure this speed.[3] In 1983 the meter was redefined by the CGPM as the distance traveled by light in 1/299,792,458 seconds, giving the speed of light the exact value of 299,792,458 meters/second.[4] In our experiment, we set out to measure this speed to see if our experimental data matched the accepted value.
## Methods
SJK 12:01, 30 November 2009 (EST)
12:01, 30 November 2009 (EST)
Your style for the methods section is great. And the information you have included is very good with some minor additions needed (as noted). Some details are missing: did you average the signals? How much averaging? How did you define "intensity" on the oscilloscope? Also, in your results section, you talk about "in the 1st trial," but the methods does not mention trials, or differences between them (as far as I can see).
More important is that data analysis methods are completely missing. You may talk about this sufficiently below (I haven't looked closely yet), but in any case, it belongs in the methods section. Just like you do for the equipment, you'll want to cite the software that you use, and you'll want to spell out any algorithms you developed. It's up to you whether you want to include specific formulas or link to them in an appendix or elsewhere.
Figure 1: Time Walk Effect- Different amplitudes of a signal cause a time shift in a trigger signal.
SJK 11:50, 30 November 2009 (EST)
11:50, 30 November 2009 (EST)
Is there a model number for LED or PMT? If not, can you describe them somehow? (pulse rate, color, etc.)
We positioned a photomultiplier tube (PMT) powered by a Bertran 313B Power Supply on one end of a carboard tube. We placed a LED in the other end, powered by a Harrison Laboratories 6207A PSU. We measured the time difference between the LED's pulse and the photomultiplier's response with a Ortec 567 TAC/SCA Module plugged into a Harshaw NQ-75 NIM Bin. We placed a Canberra 2058 Delay Module between the PMT and the TAC to guarantee the response pulse would be received by the TAC after the triggering pulse from the LED. SJK 11:56, 30 November 2009 (EST)
11:56, 30 November 2009 (EST)
All figures used in a report will be referred to in the text. You can do that here by saying something like "See Figure 2 for experimental setup." Or work it into a sentence as you did for the time walk figure. Probably you'll need to renumber those so that the first one you refer to is Figure 1. Also, figure captions need a bit more detail: For time walk figure: tell reader what the x and y axes represent, and what the typical scales would be (nanoseconds/ volts). For Figure 2, I'm not sure how to expand it, but probably should, even if just saying "with key equipment labeled" instead of "labels." I do love the panorama, BTW.
We measured the TAC's voltage using a Tektronix TDS 1002 Oscilloscope. This voltage corresponded to the time between the LED trigger pulse and the PMT response pulse with the LED at different positions, all 10 cm apart. As the LED got closer to the PMT, the intensity would increase, and this would cause error due to "time walk." The oscilloscope displays a signal by triggering when the signal reaches some threshold. The "time walk" effect is the change in time of this trigger signal due to a change in amplitude of the input signal. In this experiment, a change in intensity of the LED signal causes the oscilloscope and the TAC to trigger at a different time, and the TAC will produce a different voltage. To minimize the error due to time walk (see Figure 1) we used a set of polarizers placed on the PMT and the LED to keep the intensity of the LED pulses constant. We measured the intensity of the LED when it was at its maximum distance from the PMT, and then we rotated the PMT with the polarizer attached so that the intensity of the LED signal remained constant for every measurement with the LED in a closer position.
Figure 2: Panorama of the setup with labels
## Results and Discussion
SJK 13:40, 2 December 2009 (EST)
13:40, 2 December 2009 (EST)
I really like your plot: it clearly conveys a lot of information very clearly. It's definitely a good way of presenting your net result. I think, though, some other figures should be included to make the results section more substantial. To do this, ask yourself what the reader may want to see after looking at the figure you have. In my mind, this would be individual linear regression fits for some or all of the data sets. If you did that as a preceding figure, I think you'd have enough to talk about. As it is now, there is very little text in your results and discussion section, which is not standard. So, I think it'd be good by starting out, even the way you have, "In the first trial..." and then saying, "Trial one and the linear regression fit is displayed in Figure X.A. Subsequent trials (Figure X.B-E) included 11 positions, separated by __ cm. As seen in Figure X, no noticeable trend was seen in terms of uncertainty of individual measurements, or on residuals (you could even plot all the residuals on another graph, which could be interesting"
In the first trial, we measured the voltage of the TAC with the LED in 10 different positions. For every subsequent trial, we measured the voltage with the LED at 11 positions. After the first trial, we used the averaging function on the oscilloscope. This function took a time average of a signal, which reduced the noise, so we could better measure its voltage. The TAC was set to produce a 10V signal for a 100 ns delay. We used this ratio of 1V/10ns to convert our measured voltages into times.
I used the chi-square minimization technique to fit the data with a line. The slope of the line and standard error were used in a weighted average to compute the measured speed of light. This value was:
$2.941(15)\cdot 10^{8} m/s$
The exact speed of light is approximately:
$2.998\cdot 10^{8} m/s$
The calculated speed of light was 4 sigma away. Assuming only normally distributed random error, the probability of measuring the same value we did is 0.006%.SJK 13:35, 2 December 2009 (EST)
13:35, 2 December 2009 (EST)
A couple comments about this. First, this is definitely on the right track for how to compare your measurement to a precise accepted value. As you've written it, though, your wording should be, "the probability of measuring 4 sigma or more away from the mean (in either direction) is 0.006%." That is, you're reporting 1 minus the error integral from -4sigma to + 4sigma. Furthermore, you should add a comment, "Thus, it is almost certain that substantial systematic error remained in our measurements." (You can make a goal of eliminating or identifying this systematic error next week.)
I would be pretty much OK with this kind of comparison. But it's worth noting that when you're out on those tails of the distribution, the estimate of the standard deviation of the mean is pretty important. In reality, we usually don't care so much about whether it's 0.006% or 0.3% : in either case we'd be pretty certain there's substantial systematic error. But if you are reporting those kinds of numbers and asserting confidence, then I think what you need to do is use "Student's t-test," which would account for how many degrees of freedom were used in estimating the standard error. It's worth reading, just to find out that the guy who invented it was using the pseudonum "Student" because his employer, Guinness, didn't want him revealing the trade secret that the brewery was using statistics to improve its processes: http://en.wikipedia.org/wiki/T_test#History
Figure 3: Trials 1-6, the accepted speed of light, and the calculated value from the measurements, along with their corresponding uncertainties, are shown.
SJK 13:42, 2 December 2009 (EST)
13:42, 2 December 2009 (EST)
Usually, the figure legend is not included (probably due to space limitations?). However, it works well for you, so I'd leave it. Nevertheless, I think more description in the figure legend is needed to explain what the various symbols and error bars refer to. E.g. green triangle represents the mean of all 6 trials and the error bars represent one standard error of the mean.
The supplementary data and analysis can be seen here.SJK 13:44, 2 December 2009 (EST)
13:44, 2 December 2009 (EST)
Hyperlinks are usually not included in the text, but rather you'd have a numbered "reference" (endnote) here and in the reference would include the link. For example, "Raw data and source code are freely available[7]."
## Conclusions
SJK 13:46, 2 December 2009 (EST)
13:46, 2 December 2009 (EST)
Aside from the revised language regarding probability as I mentioned above, I think this is a pretty good conclusion. I think investigating the systematic error next week would be good, including trying out the DAQ card if you'd like.
The probability of measuring the same value we did is 0.006%. Assuming only normally distributed random error, the likelihood of this happening again is quite low. I conclude that the experimental data deviated from the accepted value due to systematic error. I believe the cause of this error was inadequate minimization of the time walk effect caused by the reliance on human judgment in determining when the intensity of the LED pulse signal matched the original signal. This error might be reduced by the use of a computer to measure the LED signal, rather than using the screen of an oscilloscope. This method is far more quantitative, and I believe it would yield more accurate results.
## Acknowledgements
SJK 13:48, 2 December 2009 (EST)
13:48, 2 December 2009 (EST)
You can revise this to make it more typical. "I thank Ryan Long for help with electronic lab notebook, instrumentation setup, data acquisition, and data analysis...I thank A. Barron for his open access lab report which provided guidance on bib tex references and manuscript style (this is not something you'd read in a peer-reviewed report, but I really like that you give him props, so I think you should leave it in and just try to make it sound "formal."
Thanks my lab partner Ryan for his help with running the lab, taking data, and finishing up the lab notebook with me. I'd also like to thank Dr. Koch for his helpful explanations of various parts of the setup.
Thanks to A. Barron, who I referred to for help in formatting citations as well as getting a general idea of what I needed to write.[5]
## References
SJK 11:43, 30 November 2009 (EST)
11:43, 30 November 2009 (EST)
These first two are the kind of original peer-reviewed research I'm looking for--good!
1. Michelson, Albert Abraham & Morley, Edward Williams (1887), "On the Relative Motion of the Earth and the Luminiferous Ether", American Journal of Science 34: 333–345 [Michelson]
2. Albert Einstein (1905) "Zur Elektrodynamik bewegter Körper", Annalen der Physik 17: 89 SJK ~~~~~
~~~~~
[SR]
3. Measurement of the speed of light. T. G. Blaney C. C. Bradley G. J. Edwards B. W. Jolliffe D. J. E. Knight W. R. C. Rowley K. C. Shotton & P. T. Woods Nature 251, 46 (1974) | doi:10.1038/251046a0. http://www.nature.com/nature/journal/v251/n5470/pdf/251046a0.pdf
[Blaney]
4. Base unit definitions: Meter. Nov 15 2009. http://physics.nist.gov/cuu/Units/meter.html
[NIST]
5. A. Barron's Final Report
[Barron] | 2014-07-25 22:21:32 | {"extraction_info": {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6342419981956482, "perplexity": 843.4643918586146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997894799.55/warc/CC-MAIN-20140722025814-00156-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://studybay.com/math/tags/andersen/ | # Tag
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### Chen Prime
A Chen prime is a prime number for which is either a prime or semiprime. Chen primes are named after Jing Run Chen who proved in 1966 that there are infinitely many such primes (Chen's theorem).The first Chen primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... (OEIS A109611). The first primes that are not Chen primes are 43, 61, 73, 79, 97, 103, 151, ... (OEIS A102540).The lesser of any twin prime is always a Chen prime. Apart from twin prime records, the largest known Chen prime known as of Oct. 2005 was(https://primes.utm.edu/primes/page.php?id=75857),which has 70301 digits.There are infinitely many cases of 3 Chen primes in arithmetic progression (Green and Tao 2005). The following 3074-digit case produces Chen primes for , 1, 2, where denotes the primorial:
### Palindromic Prime
A palindromic prime is a number that is simultaneously palindromic and prime. The first few (base-10) palindromic primes are 2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, ... (OEIS A002385; Beiler 1964, p. 228). The number of palindromic primes less than a given number are illustrated in the plot above. The number of palindromic numbers having , 2, 3, ... digits are 4, 1, 15, 0, 93, 0, 668, 0, 5172, 0, ... (OEIS A016115; De Geest) and the total number of palindromic primes less than 10, , , ... are 4, 5, 20, 20, 113, 113, 781, ... (OEIS A050251). Gupta (2009) has computed the numbers of palindromic primes up to .The following table lists palindromic primes in various small bases. OEISbase- palindromic primes2A11769711, 101, 111, 10001, 11111, 1001001, 1101011, ...3A1176982, 111, 212, 12121, 20102, 22122, ...4A1176992, 3, 11, 101, 131, 323, 10001, 11311, 12121, ...5A1177002, 3, 111, 131, 232, 313, 414, 10301, 12121,..
### Cunningham Chain
A sequence of primes is a Cunningham chain of the first kind (second kind) of length if () for , ..., . Cunningham primes of the first kind are Sophie Germain primes.It is conjectured there are arbitrarily long Cunningham chains. The longest known Cunningham chains are of length 17, with the first examples found corresponding to (first kind; J. Wroblewski, May 2008) and (second kind; J. Wroblewski, Jun. 2008).The smallest prime beginning a complete Cunningham chain of the first kind of lengths , 2, ... are 13, 3, 41, 509, 2, 89, 1122659, 19099919, 85864769, 26089808579, ... (OEIS A005602).The smallest prime beginning a complete Cunningham chain of the second kind of lengths , 2, ... are 11, 7, 2, 2131, 1531, 33301, 16651, 15514861, 857095381, 205528443121, ... (OEIS A005603)...
### Cousin Primes
Pairs of primes of the form (, ) are called cousin primes. The first few are (3, 7), (7, 11), (13, 17), (19, 23), (37, 41), (43, 47), (67, 71), ... (OEIS A023200 and A046132).A large pair of cousin (proven) primes start with(1)where is a primorial. These primes have 10154 digits and were found by T. Alm, M. Fleuren, and J. K. Andersen (Andersen 2005).As of Jan. 2006, the largest known pair of cousin (probable) primes are(2)which have 11311 digits and were found by D. Johnson in May 2004.According to the first Hardy-Littlewood conjecture, the cousin primes have the same asymptotic density as the twin primes,(3)(4)where (OEIS A114907) is the twin primes constant.An analogy to Brun's constant, the constant(5)(omitting the initial term ) can be defined. Using cousin primes up to , the value of is estimated as(6)..
### Sexy Primes
Sexy primes are pairs of primes of the form (, ), so-named since "sex" is the Latin word for "six.". The first few sexy prime pairs are (5, 11), (7, 13), (11, 17), (13, 19), (17, 23), (23, 29), (31, 37), (37, 43), (41, 47), (47, 53), ... (OEIS A023201 and A046117). As of November 2005, the largest known sexy prime pair starts with(1)where is a primorial. These primes have 10154 digits and were found by M. Fleuren, T. Alm, and J. K. Andersen (Andersen 2005).Sexy constellations also exist. The first few sexy triplets (i.e., numbers such that each of is prime but is not prime) are (7, 13, 19), (17, 23, 29), (31, 37, 43), (47, 53, 59), ... (OEIS A046118, A046119, and A046120). As of October 2005, the largest known sexy triplet starts with(2)These primes have 5132 digit digits and were found by Davis (2005).The first few sexy quadruplets are (11, 17, 23, 29), (41, 47, 53, 59), (61, 67, 73, 79), (251, 257, 263, 269),..
### Prime Gaps
A prime gap of length is a run of consecutive composite numbers between two successive primes. Therefore, the difference between two successive primes and bounding a prime gap of length is , where is the th prime number. Since the prime difference function(1)is always even (except for ), all primes gaps are also even. The notation is commonly used to denote the smallest prime corresponding to the start of a prime gap of length , i.e., such that is prime, , , ..., are all composite, and is prime (with the additional constraint that no smaller number satisfying these properties exists).The maximal prime gap is the length of the largest prime gap that begins with a prime less than some maximum value . For , 2, ..., is given by 4, 8, 20, 36, 72, 114, 154, 220, 282, 354, 464, 540, 674, 804, 906, 1132, ... (OEIS A053303).Arbitrarily large prime gaps exist. For example, for any , the numbers , , ..., are all composite (Havil 2003, p. 170). However, no general method..
### Prime Arithmetic Progression
An arithmetic progression of primes is a set of primes of the form for fixed and and consecutive , i.e., . For example, 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089 is a 10-term arithmetic progression of primes with difference 210.It had long been conjectured that there exist arbitrarily long sequences of primes in arithmetic progression (Guy 1994). As early as 1770, Lagrange and Waring investigated how large the common difference of an arithmetic progression of primes must be. In 1923, Hardy and Littlewood (1923) made a very general conjecture known as the k-tuple conjecture about the distribution of prime constellations, which includes the hypothesis that there exist infinitely long prime arithmetic progressions as a special case. Important additional theoretical progress was subsequently made by van der Corput (1939), who proved than there are infinitely many triples of primes in arithmetic progression, and Heath-Brown (1981),..
### Integer Sequence Primes
Just as many interesting integer sequences can be defined and their properties studied, it is often of interest to additionally determine which of their elements are prime. The following table summarizes the indices of the largest known prime (or probable prime) members of a number of named sequences.sequenceOEISdigitsdiscoverersearch limitcommentsalternating factorialA00127259961260448M. Rodenkirch (Sep. 18, 2017)100000 (M. Rodenkirch, Dec. 15, 2017)finite sequence; largest certified prime has index 661; the rest are probable primesApéry-constant primeA119334141141E. W. Weisstein (May 14, 2006)9089 (E. W. Weisstein, Mar. 22, 2008)status unknownApéry number A092825662410136E. W. Weisstein (Mar. 2004) (E. W. Weisstein, Mar. 2004)probable primeApéry number 87E. W. Weisstein..
Check the price | 2020-09-23 08:40:47 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8917551040649414, "perplexity": 526.1644141769946}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400210616.36/warc/CC-MAIN-20200923081833-20200923111833-00461.warc.gz"} |
https://www.sarthaks.com/8663/equilibrium-constant-decomposition-phosphorus-pentachloride-decomposition-depicted | # At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3 . If decomposition is depicted as,
+1 vote
3.7k views
At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3 . If decomposition is depicted as,
∆rH° = 124.0 kJmol–1
a) Write an expression for Kc for the reaction.
b) What is the value of Kc for the reverse reaction at the same temperature?
c) What would be the effect on Kc if
(i) more PCl5 is added
(ii) pressure is increased?
(iii) The temperature is increased?
+1 vote
by (128k points)
selected by
(a)
(b) Value of Kc for the reverse reaction at the same temperature is:
(c) (i) Kc would remain the same because in this case, the temperature remains the same.
(ii) Kc is constant at constant temperature. Thus, in this case, Kc would not change.
(iii)In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased. | 2022-10-06 23:50:00 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8665622472763062, "perplexity": 2846.474073459601}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337889.44/warc/CC-MAIN-20221006222634-20221007012634-00681.warc.gz"} |
http://cjcp.ustc.edu.cn/html/hxwlxb_en/2016/6/cjcp1605105.htm | Chinese Journal of Chemical Physics 2016, Vol. 29 Issue (6): 703-709
#### The article information
Li-zhao Xie, Le-chen Chen, Mo-zhen Wang, Qi-chao Wu, Xiao Zhou, Xue-wu Ge
In-situ Enhanced Toughening of Poly (ethylene terephthalate)/elastomer Blends via Gamma-Ray Radiation at Presence of Trimethylolpropane Triacrylate
Chinese Journal of Chemical Physics, 2016, 29(6): 703-709
http://dx.doi.org/10.1063/1674-0068/29/cjcp1605105
### Article history
Accepted on: June 6, 2016
In-situ Enhanced Toughening of Poly (ethylene terephthalate)/elastomer Blends via Gamma-Ray Radiation at Presence of Trimethylolpropane Triacrylate
Li-zhao Xiea, Le-chen Chena, Mo-zhen Wanga, Qi-chao Wub, Xiao Zhoub, Xue-wu Gea
Dated: Received on May 12, 2016; Accepted on June 6, 2016
a. CAS Key Laboratory of Soft Matter Chemistry, Department of Polymer Science and Engineering, University of Science and Technology of China, Hefei 230026, China;
b. Guangdong Tianan New Material Co., Ltd., Foshan 528000, China
*Author to whom correspondence should be addressed. Mo-zhen Wang, E-mail:pstwmz@ustc.edu.cn; Xue-wu Ge, E-mail:xwge@ustc.edu.cn, Tel:+86-551-63600843
Abstract: Gamma-ray radiation has always been a convenient and effective way to modify the interfacial properties in polymer blends. In this work, a small amount of trimethylolpropane triacrylate (TMPTA) was incorporated into poly (ethylene terephthalate) (PET)/random terpolymer elastomer (ST2000) blends by melt-blending. The existence of TMPTA would induce the crosslinking of PET and ST2000 molecular chains at high temperatures of blending, resulting in the improvement in the impact strength but the loss in the tensile strength. When the PET/ST2000 blends were irradiated by gamma-ray radiation, the integrated mechanical properties could be enhanced significantly at a high absorbed dose. The irradiated sample at a dose of 100 kGy even couldn't be broken under the impact test load, and at the same time, has nearly no loss of tensile strength. Based on the analysis of the impactfractured surface morphologies of the blends, it can be concluded that gamma-ray radiation at high absorbed dose can further in situ enhance the interfacial adhesion by promoting the crosslinking reactions of TMPTA and polymer chains. As a result, the toughness and strength of PET/ST2000 blend could be dramatically improved. This work provides a facial and practical way to the fabrication of polymer blends with high toughness and strength.
Ⅰ. INTRODUCTION
With the rapid development of modern polymer industry, polymer blends have attracted considerable scientific and industrial interest since their properties can be finely tuned by varying the composition and the type of the components [1-4]. However, due to the unfavorable enthalpy of mixing, macro-phase separation occurs in most polymer blends, which leads to the deterioration in the integrated mechanical properties [5-7]. Therefore, many methods have been developed to manipulate the interface properties of polymer blends to realize the exceptional properties that the blending can offer [8-13].
In general, the introduction of macromolecular compatibilizers, such as graft, block or star copolymers, can lower the interfacial tension between the partially miscible or immiscible phases so as to improve the interfacial affinity [14-16]. In our previous work [17], poly (acrylic acid) (PAA) grafted poly (ethylene terephthalate) (PET) resins (PET-g-PAA) prepared by γ-ray radiation was introduced in PET/ethylene-methyl acrylate-glycidyl methacrylate random terpolymer (ST2000) blend (PET/ST2000) as the compatibilizer. The ternary PET/PET-g-PAA/ST2000 blend with only 6wt% of PET-g-PAA has an improved impact strength, twice that of PET/ST2000. PET resins grafted with other polymers such as methyl acrylate [18], could also have the similar effect on improving the mechanical properties of PET/ST2000 blends. However, the disadvantages of this method are obvious. First, the copolymeric compatibilizer themselves have little contribution to the strength and stiffness of polymer blends because they are generally composed of "soft" chain segments. In some cases, their introduction even brings in the fatal loss of the strength of the blends [19, 20]. Second, the preparation and introduction of these copolymeric compatibilizers will inevitably increase the synthetic and processing cost of the blends [21-23].
In this work, we focused on the effect of γ-ray radiation on the mechanical properties of widely-used PET/ST2000 blends at the presence of TMPTA. The change in the morphologies of PET/ST2000 blends and mechanical properties with different content of TMPTA was then investigated in detail. It was found that the toughening effect of ST2000 on PET can be further in-situ enhanced under γ-ray radiation at the presence of TMPTA. The toughness mechanism was also discussed with the micro-voiding and plastic deformation theories.
Ⅱ. EXPERIMENTS
PET resin (CB651, [η]=0.75 dL/g) and ethylene-methyl acrylate-glycidyl methacrylate random terpolymer (ST2000, 1.5wt% of glycidyl methacrylate) were purchased from Far Eastern Industry (Shanghai, China) and Shanghai Xiuhu Chemical Co., Ltd., respectively. Analytical reagents, including phenol, tetrahydrofuran (THF), and tetrachloroethane, were purchased from Shanghai Chemical Reagents Co., Ltd. TMPTA (95%, technical grade) was supplied by Laiyu Chemical (Shangdong, China).
Before the preparation of PET/ST2000 blends, the raw PET granules were dried at 90 ℃ for 24 h and ST2000 granules were dried at 50 ℃ for 12 h. The dried PET and ST2000 were firstly premixed together with TMPTA in a homogenizer (WJ-30) at room temperature. The premixed blends were then added into a feeding device and transported to a co-rotating twin-screw extruder (TE-35, China) with a screw diameter of 35 mm and an overall L/D of 37. The feed rate was 300 r/min. The temperatures of the first to the seventh regions were set as 140, 200, 260, 260, 260, 260, and 260 ℃. The die temperature was also 260 ℃. The screw speed was 300 r/min. The extrudates were cooled in water. After being pelletized, the extrudates were dried at 90 ℃ for 24 h, then injection-moulded into the standard specimens for the tensile and notched Izod impact strength measurement using an injection-moulding machine (HTF80X1, China). The temperatures of the first to the sixth regions were set as 265, 260, 260, 260, 255, and 25 ℃, respectively. The injection pressure was 80 MPa. The screw rate was 20 r/min. The retention time was 35 s. The weight content of ST2000 in all samples was fixed as 20%. The as-prepared standard specimens were then thermally sealed into plastic bags filled with nitrogen gas and exposed in the radiation field of 60Co γ-ray at a dose rate of 83.3 Gy/min. The 60Co source with a radioactivity of 1.37×1015 Bq is located in University of Science and Technology of China. The absorbed dose ranged from 10 kGy to 150 kGy.
The gel fractions of the radiated samples for impact tests were measured by the solvent extraction method. The samples (0.2-1.0 g) wrapped with nickel mesh were extracted in a Soxhlet extractor with THF at 66 ℃ for 24 h. The THF extracted samples were immersed in 30 mL of phenol/tetrachloroethane mixed solvent (1/2, W/W) under magnetic stirring for 24 h at 110 ℃. Finally, the extracted samples were taken out and dried in vacuum oven at 100 ℃ till a constant weight. The gel fraction, G, was calculated by the following equation:
$G/\% = \frac{{{W_1}}}{{{W_0}}} \times 100$ (1)
where W0 and W1 are the weights of the samples before and after extraction, respectively.
Fourier-transform infrared (FT-IR) spectra of the solvent-extracted samples were recorded on a Nicolet-8700 infrared spectrometer (Thermo Scientific Instrument Co., USA) at a resolution of 1 cm-1. The samples were prepared by mixing the grinded gel with KBr and pressing into a thin film.
The notched Izod impact strengths of all samples were tested on a Memory Impact Test machine (JJ-20, Intelligent Instrument Equipment Co., Ltd.) at room temperature according to GB/T 1843-2008 (ISO 180: 2000). The tensile properties of all samples were conducted on an electronic universal testing machine (WSM-20KB, Intelligent Equipment Co., Ltd.) at room temperature. The dumbbell-shaped specimens were stretched until they were broken at a crosshead rate of 50 mm/min according to GB/T1040.2-2006 (ISO 527-2: 1993). A minimum of five tensile and impact specimens were tested for each reported value.
The microstructure of the samples before and after the impact test was observed by field-emission scanning electron microscopy (SEM，JEOL JSM-6700, Japan, 5 kV). The samples before impact test were observed after being fractured in liquid nitrogen. After the fractured surfaces were etched by THF at 66 ℃ for 12 h to remove the elastomer component, the samples were then dried in a vacuum oven at 50 ℃ for 12 h and sputter coated with gold.
Dynamic mechanical analysis of the samples were performed on Pyris Diamond DMS 6100 DMTA (Perkin-Elmer) at a heating rate of 5 ℃/min and a frequency of 1 Hz using the bending model. The temperature was scanned from -90 ℃ until the sample became too soft to be tested.
Ⅲ. RESULTS AND DISCUSSION A. Effect of γ-ray radiation on the mechanical properties of PET/ST2000 blends at the presence of TMPTA
The impact strength, tensile strength, and elongation at break of PET blends with different content of TMPTA before and after γ-ray radiation at various absorbed doses are shown in Fig. 1. It is seen from Fig. 1(a) that the introduction of a little amount of TMPTA (1-2wt%) can slightly increase the impact strength of PET/ST2000 blend. However, the impact strength of PET/ST2000 blend with more content of TMPTA (>2wt%) falls instead. At the same time, the elongation at break increases slightly with the content of TMPTA (Fig. 1(c)), also indicating the improvement in the toughness of PET/ST2000 blends. But the tensile strength drops with the content of TMPTA (Fig. 1(b)). TMPTA molecule has three active C=C bonds. The grafting or crosslinking of polymer chains by TMPTA will occur during the melt-blending process at high temperatures [31]. When the content of TMPTA is low, a little amount of copolymer composed of PET and ST2000 chains connected by TMPTA will be produced during the melt blending process, and can act as the compatibilizer to enhance the interfacial interaction between PET and ST2000, resulting in the increase of the toughness of the PET/ST2000 blend. But excessive TMPTA may lead to a high crosslinking degree in the blends, and weaken the integrated mechanical property of PET/ST2000 blends.
FIG. 1 (a) Impact strength, (b) tensile strength, and (c) elongation at break of PET/ST2000 blends with different content of TMPTA before and after being irradiated by γ-ray at various absorbed doses. The impact samples of PET/ST2000 blends with 2wt% TMPTA irradiated at 100 and 150 kGy didn’t break under the same impact test condition.
After being irradiated by γ-ray radiation at an absorbed dose above 10 kGy, all of the PET/ST2000 blends exhibit improved impact strength. The relationships between the impact strength and the content of TMPTA for the irradiated PET/ST2000 blends are similar to that for the un-irradiated PET/ST2000 blend, i.e. 2wt% of TMPTA can reach the highest impact strength for all the irradiated PET/ST2000 blends. However, the rate of increase in the impact strength is dramatically high when the absorbed dose is above 100 kGy as listed in Table Ⅰ since the PET/ST2000 blends with 2wt% of TMPTA irradiated at 100 and 150 kGy even cannot break under the impact test condition. But the tensile strengths for these two samples almost have little change, which indicates the γ-ray radiation produces an in situ enhanced toughening and strengthening effect on PET/ST2000 blend.
Table Ⅰ The impact strength of PET/ST2000 blends with different content of TMPTA at various absorbed doses.
The storage modulus is a measure of the stiffness for polymer [32], thus DMTA was carried out to evaluate the mechanical properties, as exhibited in Fig. 2. All the irradiated blends have a higher storage modulus than the un-irradiated sample. The glass transition temperature (Tg) can be determined as the peak temperature on the tanδ curves. It was obvious that the Tg of PET in all studied PET/ST2000 blends have little change due to the macro-phase separation between PET and ST2000. And the Tg of ST2000 in irradiated samples is hardly distinguished, which should be attributed to the crosslinking of ST2000 under γ-ray radiation or at high temperatures.
FIG. 2 DMTA curves of PET/ST2000 blends with 2wt% of TMPTA at different absorbed doses.
Since γ-ray radiation can produce free radicals randomly on any polymer chains and TMPTA molecules, it can be expected that the crosslinking between polymer chains can be in-situ enhanced under γ-ray radiation. The gel fractions in the raw PET (GPET), raw ST2000 (GST), and the PET/ST2000 (GPET/ST) blends with 2wt% TMPTA treated with the same melting blend condition were measured and are shown in Fig. 3. It can be seen that ST2000 could be fully crosslinked at high temperatures at the presence of 2wt% TMPTA, while only a little part of PET (15 wt%) can be crosslinked at the same condition without γ-ray radiation. The result is easy to understand as shown in many thermoplastic elastomers-based dynamic vulcanization systems [33]. There are about 35% of gel in PET/ST2000 blend incorporated with 2wt% of TMPTA, which makes a remarkable increase in the impact strength of the blend. When all the samples were exposed under the γ-ray radiation, the GPET and GPET/ST slightly increased with the absorbed doses till a constant above 100 kGy, indicating that the PET and ST2000 chains at the phase interface can be further crosslinked by TMPTA so as to enhance the affinity of the phase interface resulting in the improvement of the impact strength of the blends.
FIG. 3 Gel fractions of the PET, ST2000, and PET/ST2000 blends with 2wt% TMPTA before and after being irradiated at different absorbed doses. All the samples were treated with the same blending condition.
The FTIR spectrum of the gel extracted from PET/ST2000 blends is shown in Fig. 4, compared with those of pure PET, ST2000, and TMPTA. As the ST2000 could be fully crosslinked either at high temperatures or under γ-ray radiation, so the gel extracted from PET/ST2000 blends must contain the ST2000 component. In the spectrum of PET, the characteristic peaks of benzene rings in the PET chains could be found at 727 and 875 cm-1 which were assigned to the aromatic ring C-H out-of-plane bending vibrations and deformation vibrations, respectively [34, 35]. The peaks at 1577 and 1506 cm-1 were due to the in-plane aromatic ring vibrations [24]. All of these characteristic peaks of PET could not be found in the spectra of TMPTA and ST2000 but in the gel. From the gelation analysis and FTIR characterization, we confirmed the existence of both ST2000 and PET chains network formed after radiation. Due to the high mobility, TMPTA acted as a bridge connecting PET matrix and ST2000 phase, leading to the enhanced interfacial interaction.
FIG. 4 FTIR spectra of neat PET, ST2000, TMPTA, and the gel of PET/ST2000 blends with 2wt% of TMPTA irradiated at a dose of 100 kGy.
B. Morphologies and toughening mechanism of PET/ST2000 blends at the presence of TMPTA
Figure 5 exhibits the morphologies of the cryofractured samples of PET/ST2000 blends with 2wt% of TMPTA before and after γ-ray radiation. There are no observable changes in the size and size distributions of the dispersed ST2000 particles after the samples were irradiated by γ-ray radiation.
FIG. 5 SEM images of PET/ST2000 with 2wt% of TMPTA at different absorbed doses. (a) 0 kGy, (b) 10 kGy, (c) 30 kGy, (d) 50 kGy, (e) 100 kGy, and (f) 150 kGy. The weight content of ST2000 is 20wt% for all the samples.
FIG. 6 SEM images of the impact-fractured surface at zone A, B, and C of PET and PET/ST2000 blends with 2wt% of TMPTA before (0 kGy) and after being irradiated at different doses (10 and 100 kGy).
Scheme 1 The in-situ enhanced toughening mechanism of PET/ST2000 blends at the presence of TMPTA under γ-ray radiation.
Ⅳ. CONCLUSION
Ⅴ. ACKNOWLEDGMENTS
This work was supported by the National Natural Science Foundation of China (No.51173175), the Foshan Scientific and Technological Innovation Team Project (No.2013IT100041), and the Foshan University-City Cooperation Project (Scientific and Technological Innovation Project, No.2014HK100291).
We thank Prof. Yuan Hu and Dr. Bi-bo Wang of the University of Science and Technology of China for their helpful advice and assistance.
Reference | 2019-08-22 04:35:01 | {"extraction_info": {"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, "math_score": 0.5668562650680542, "perplexity": 5497.433454804832}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00156.warc.gz"} |
http://mathhelpforum.com/statistics/218132-when-would-you-multiply-permutations.html | # Thread: when would you multiply permutations?
1. ## when would you multiply permutations?
hey all,
so, I was practicing some permutation problems and this one came up.
'A company is trying to fill some vacancies. four of five accountants will be selected and two of four lawyers will be selected. how many different ways can they be selected?'
I found out both permutations, but then I was stuck on what to do. The answer key said I should multiply them because the different permutations are independent of one another. Is this true? if so, why? and what happens if the permutations are dependent?
2. ## Re: when would you multiply permutations?
Hey hellothisismyname.
Its easier to show this in terms of probability. We can relate probabilities to frequencies by noting that a probability of something P(A) = #Frequency of A occuring / #Total Number of Things.
In probability we have a theorem that says if A and B are independent then P(A and B) = P(A)P(B).
The proof of this is based on conditional probability. We define the probability of A given B as P(A|B) = P(A and B)/P(B).
Lets consider two random events A and B. If they are independent, then it means that knowing one of them won't affect the probability of the other one. In other words lets say you had the probability of selecting a lawyer and the probability of selecting an accountant: in this case the two events would be independent if knowing one result didn't have any effect on the other.
Mathematically we write this as P(A|B) = P(A) and P(B|A) = P(B). In other words, any extra information about anything outside of what you are looking at doesn't have any impact on the probability which means that there is no causal effect between two things.
If things were not independent then changing one thing would affect the other thing. But if they are independent then changing one thing doesn't change the other. That is the intuitive idea behind independence.
So P(A|B) = P(A and B)/P(B), if P(A|B) = P(A) then re-arranging gives us:
P(A|B) = P(A) = P(A and B)/P(B) so this implies P(A and B) = P(A)P(B) if P(B) != 0.
Since probabilities are relative frequencies, you can use this result for permutations and combinations if you are dealing with independent things.
If for example you have a total number of possibilities as N then you can convert a probability to a frequency by multiplying it by N. In other words:
Freq_A = P(A)*N.
The above should give you an idea of why it works (i.e. the multiplication rule).
3. ## Re: when would you multiply permutations?
Thanks, that really cleared it up for me. If the two variables are dependent, is their any way you can use permutations/combinations in a similar manner? or is it simply not feasible to do so? | 2017-03-29 02:39:17 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.802610456943512, "perplexity": 291.43040446059155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218190134.67/warc/CC-MAIN-20170322212950-00043-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-geometry/139196-complex-analysis-entire-function.html | # Thread: complex analysis entire function
1. ## complex analysis entire function
Show that an entire function whose real part is positive throughout the complex plane must be a constant.
Can I get some help please?
2. Originally Posted by dori1123
Show that an entire function whose real part is positive throughout the complex plane must be a constant.
Can I get some help please?
Suppose $\displaystyle f(z)$ satisfies the properties stated above, let's look at $\displaystyle \frac{1}{f(z)}$. This function is also entire since $\displaystyle f(z) \neq 0$.
Since $\displaystyle f(z) \neq 0, \; 0<a\leq |f(z)|$ for some $\displaystyle a\neq0$. Thus $\displaystyle |f(z)|$ is bounded from below by a non zero real number.
So $\displaystyle \left|\frac{1}{f(z)}\right|\leq\frac1a$ or in other words, $\displaystyle \left|\frac{1}{f(z)}\right|$ is bounded from above.
Thus by Liouville's Theorem $\displaystyle \frac{1}{f(z)}$ is (a non zero) constant, which implies $\displaystyle f(z)$ is constant too. | 2018-05-26 10:40:47 | {"extraction_info": {"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, "math_score": 0.9614067077636719, "perplexity": 216.7514534650215}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00483.warc.gz"} |
https://www.stat.berkeley.edu/~aldous/RWG/Book_Ralph/Ch8.S2.html | # 8.2 Improved bounds on $L^{2}$ distance
The central theme of the remainder of this chapter is that norms other than the $L^{1}$ norm (and closely related variation distance) and $L^{2}$ norm can be used to improve substantially upon the bound
$\|P_{i}(X_{t}\in\cdot)-\pi(\cdot)\|_{2}\leq\pi^{-1/2}e^{-t/\tau_{2}}.$ (8.26)
## 8.2.1 $L^{q}$ norms and operator norms
Our discussion here of $L^{q}$ norms will parallel and extend the discussion in Chapter 2 Section yyy:6.2 of $L^{1}$ and $L^{2}$ norms. Given $1\leq q\leq\infty$, both the $L^{q}$ norm of a function and the $L^{q}$ norm of a signed measure are defined with respect to some fixed reference probability distribution $\pi$ on $I$, which for our purposes will be the stationary distribution of some irreducible but not necessarily reversible chain under consideration. For $1\leq q<\infty$, the $L^{q}$ norm of a function $f:I\to{\bf R}$ is
$\|f\|_{q}:=\left(\sum_{i}\pi_{i}|f(i)|^{q}\right)^{1/q},$
and we define the $L^{q}$ norm of a signed measure $\nu$ on $I$ to be the $L^{q}$ norm of its density function with respect to $\pi$:
$\|\nu\|_{q}:=\left(\sum_{j}\pi^{1-q}_{j}\,|\nu_{j}|^{q}\right)^{1/q}.$
For $q=\infty$, the corresponding definitions are
$\|f\|_{\infty}:=\max_{i}|f(i)|$
and
$\|\nu\|_{\infty}:=\max_{j}(|\nu_{j}|/\pi_{j}).$
Any matrix ${\bf A}:=(a_{ij}:i,j\in I)$ operates on functions $f:I\to{\bf R}$ by left-multiplication:
$({\bf A}f)(i)=\sum_{j}a_{ij}f(j),$ (8.27)
and on signed measures $\nu$ by right-multiplication:
$(\nu{\bf A})_{j}=\sum_{i}\nu_{i}a_{ij}.$ (8.28)
For (8.27), fix $1\leq q_{1}\leq\infty$ and $1\leq q_{2}\leq\infty$ and regard ${\bf A}$ as a linear operator mapping $L^{q_{1}}$ into $L^{q_{2}}$. The operator norm $\|{\bf A}\|_{q_{1}\to q_{2}}$ is defined by
$\|{\bf A}\|_{q_{1}\to q_{2}}:=\sup\{\|{\bf A}f\|_{q_{2}}:\|f\|_{q_{1}}=1\}.$ (8.29)
The sup in (8.29) is always achieved, and there are many equivalent reexpressions, including
$\displaystyle\|{\bf A}\|_{q_{1}\to q_{2}}$ $\displaystyle=$ $\displaystyle\max\{\|{\bf A}f\|_{q_{2}}:\|f\|_{q_{1}}\leq 1\}$ $\displaystyle=$ $\displaystyle\max\{\|{\bf A}f\|_{q_{2}}/\|f\|_{q_{1}}:f\neq 0\}.$
Note also that
$\|{\bf B}{\bf A}\|_{q_{1}\to q_{3}}\leq\|{\bf A}\|_{q_{1}\to q_{2}}\|{\bf B}\|% _{q_{2}\to q_{3}},\ \ \ 1\leq q_{1},q_{2},q_{3}\leq\infty.$ (8.30)
For (8.28), we may similarly regard ${\bf A}$ as a linear operator mapping signed measures $\nu$, measured by $\|\nu\|_{q_{1}}$, to signed measures $\nu{\bf A}$, measured by $\|\nu{\bf A}\|_{q_{2}}$. The corresponding definition of operator norm, call it $\||{\bf A}\||_{q_{1}\to q_{2}}$, is then
$\||{\bf A}\||_{q_{1}\to q_{2}}:=\sup\{\|\nu{\bf A}\|_{q_{2}}:\|\nu\|_{q_{1}}=1\}.$
A brief calculation shows that
$\||{\bf A}\||_{q_{1}\to q_{2}}=\|{\bf A}^{*}\|_{q_{1}\to q_{2}},$
where ${\bf A}^{*}$ is the matrix with $(i,j)$ entry $\pi_{j}a_{ji}/\pi_{i}$, that is, ${\bf A}^{*}$ is the adjoint operator to ${\bf A}$ (with respect to $\pi$).
Our applications in this chapter will all have ${\bf A}={\bf A}^{*}$, so we will not need to distinguish between the two operator norms. In fact, all our applications will take ${\bf A}$ to be either ${\bf P}_{t}$ or ${\bf P}_{t}-{\bf E}$ for some $t\geq 0$, where
${\bf P}_{t}:=(p_{ij}(t):i,j\in I)$
xxx notation ${\bf P}_{t}$ found elsewhere in book?
and ${\bf E}=\lim_{t\to\infty}{\bf P}_{t}$ is the transition matrix for the trivial discrete time chain that jumps in one step to stationarity:
${\bf E}=(\pi_{j}:i,j\in I),$
and where we assume that the chain for $({\bf P}_{t})$ is reversible. Note that ${\bf E}$ operates on functions essentially as expectation with respect to $\pi$:
$({\bf E}f)(i)=\sum_{j}\pi_{j}f(j),\ \ \ i\in I.$
The effect of ${\bf E}$ on signed measures is to map $\nu$ to $(\sum_{i}\nu_{i})\pi$, and
${\bf P}_{t}{\bf E}={\bf E}={\bf E}{\bf P}_{t},\ \ \ t\geq 0.$ (8.31)
## 8.2.2 A more general bound on $L^{2}$ distance
The following preliminary result, a close relative to Chapter 3, Lemmas yyy:21 and 23, is used frequently enough in the sequel that we isolate it for reference. It is the simple identity in part (b) that shows why $L^{2}$-based techniques are so useful.
###### Lemma 8.10
(a) For any function $f$,
$\frac{d}{dt}\|{\bf P}_{t}f\|^{2}_{2}=-2\mbox{{\cal E}}({\bf P}_{t}f,{\bf P}_% {t}f)\leq-\frac{2}{\tau_{2}}{{\rm var}}_{\pi}\,{\bf P}_{t}f\leq 0.$
(b)
$\|{\bf P}_{t}-{\bf E}\|_{2\to 2}=e^{-t/\tau_{2}},\ \ \ t\geq 0.$
Proof. (a) Using the backward equations
$\frac{d}{dt}p_{ij}(t)=\sum_{k}q_{ik}\,p_{kj}(t)$
we find
$\frac{d}{dt}({\bf P}_{t}f)(i)=\sum_{k}q_{ik}[({\bf P}_{t}f)(k)]$
and so
$\displaystyle\frac{d}{dt}\|{\bf P}_{t}f\|^{2}_{2}$ $\displaystyle=$ $\displaystyle 2\sum_{i}\sum_{k}\pi_{i}[({\bf P}_{t}f)(i)]q_{ik}[({\bf P}_{t}f)% (k)]$ $\displaystyle=$ $\displaystyle-2\mbox{{\cal E}}({\bf P}_{t}f,{\bf P}_{t}f)\mbox{\ \ \ by % Chapter~{}3 yyy:(70)}$ $\displaystyle\leq$ $\tau_{2}$
(b) From (a), for any $f$ we have
$\frac{d}{dt}\|({\bf P}_{t}-{\bf E})f\|^{2}_{2}=\frac{d}{dt}\|{\bf P}_{t}(f-{% \bf E}f)\|^{2}_{2}\leq-\frac{2}{\tau_{2}}\|({\bf P}_{t}-{\bf E})f\|^{2}_{2},$
which yields
$\displaystyle\|({\bf P}_{t}-{\bf E})f\|^{2}_{2}$ $\displaystyle\leq$ $\displaystyle\|({\bf P}_{0}-{\bf E})f\|^{2}_{2}\,e^{-2t/\tau_{2}}=({{\rm var}}% _{\pi}\,f)e^{-2t/\tau_{2}}$ $\displaystyle\leq$ $\displaystyle\|f\|^{2}_{2}\,e^{-2t/\tau_{2}}.$
Thus $\|{\bf P}_{t}-{\bf E}\|_{2\to 2}\leq e^{-t/\tau_{2}}$. Taking $f$ to be the eigenvector
$f_{i}:=\pi^{-1/2}_{i}u_{i2},\ \ \ i\in I,$
of ${\bf P}_{t}-{\bf E}$ corresponding to eigenvalue $\exp(-t/\tau_{2})$ demonstrates equality and completes the proof of (b).
The key to all further developments in this chapter is the following result.
###### Lemma 8.11
For an irreducible reversible chain with arbitrary initial distribution and any $s,t\geq 0$,
$\|P(X_{s+t}\in\cdot)-\pi(\cdot)\|_{2}\leq\|P(X_{s}\in\cdot)\|_{2}\|{\bf P}_{t}% -{\bf E}\|_{2\to 2}=\|P(X_{s}\in\cdot)\|_{2}\,e^{-t/\tau_{2}}.$
Proof. The equality is Lemma 8.10(b), and
$\|P(X_{s+t}\in\cdot)-\pi(\cdot)\|_{2}=\|P(X_{s}\in\cdot)({\bf P}_{t}-{\bf E})% \|_{2}\leq\|P(X_{s}\in\cdot)\|_{2}\|{\bf P}_{t}-{\bf E}\|_{2\to 2}$
proves the inequality.
We have already discussed, in Section 8.1, a technique for bounding $\tau_{2}$ when (as is usually the case) it cannot be computed exactly. To utilize Lemma 8.11, we must also bound $\|P(X_{s}\in\cdot)\|_{2}$. Since
$\|P(X_{s}\in\cdot)\|_{2}=\|P(X_{0}\in\cdot){\bf P}_{s}\|_{2}$ (8.32)
xxx For NOTES: By Jensen’s inequality (for $1\leq q<\infty$), any transition matrix contracts $L^{q}$ for any $1\leq q\leq\infty$.
and each ${\bf P}_{t}$ is contractive on $L^{2}$, i.e., $\|{\bf P}_{t}\|_{2\to 2}\leq 1$ (this follows, for example, from Lemma 8.10(a); and note that $\|{\bf P}_{t}\|_{2\to 2}=1$ by considering constant functions), it follows that
$1$ (8.33)
and the decrease is strictly monotone unless $P(X_{0}\in\cdot)=\pi(\cdot)$. From (8.32) follows
$1\leq q^{*}\leq\infty$ (8.34)
and again
$1$ (8.35)
The norm $\|{\bf P}_{s}\|_{q^{*}\to 2}$ decreases in $q^{*}$ (for fixed $s$) and is identically $1$ when $q^{*}\geq 2$, but in applications we will want to take $q^{*}<2$. The following duality lemma will then often prove useful. Recall that $1\leq q,q^{*}\leq\infty$ are said to be (Hölder-)conjugate exponents if
$\frac{1}{q}+\frac{1}{q^{*}}=1.$ (8.36)
###### Lemma 8.12
For any operator ${\bf A}$, let
${\bf A}^{*}=(\pi_{j}a_{ji}/\pi_{i}:i,j\in I)$
denote its adjoint with respect to $\pi$. Then, for any $1\leq q_{1},q_{2}\leq\infty$,
$\|{\bf A}\|_{q_{1}\to q_{2}}=\|{\bf A}^{*}\|_{q^{*}_{2}\to q^{*}_{1}}.$
In particular, for a reversible chain and any $1\leq q\leq\infty$ and $s\geq 0$,
$\|{\bf P}_{s}\|_{2\to q}=\|{\bf P}_{s}\|_{q^{*}\to 2}.$ (8.37)
Proof. Classical duality for $L^{q}$ spaces (see, e.g., Chapter 6 in [303]) asserts that, given $1\leq q\leq\infty$ and $g$ on $I$,
$\|g\|_{q^{*}}=\max\{|\langle f,g\rangle|:\|f\|_{q}=1\}$
where
$\langle f,g\rangle:=\sum_{i}\pi_{i}f(i)g(i).$
Thus
$\displaystyle\|{\bf A}^{*}g\|_{q^{*}_{1}}$ $\displaystyle=$ $\displaystyle\max\{|\langle f,{\bf A}^{*}g\rangle|:\|f\|_{q_{1}}=1\}$ $\displaystyle=$ $\displaystyle\max\{|\langle{\bf A}f,g\rangle|:\|f\|_{q}=1\},$
and also
$|\langle{\bf A}f,g\rangle|\leq\|{\bf A}f\|_{q_{2}}\|g\|_{q^{*}_{2}}\leq\|{\bf A% }\|_{q_{1}\to q_{2}}\|f\|_{q_{1}}\|g\|_{q^{*}_{2}},$
so
$\|{\bf A}^{*}g\|_{q^{*}_{1}}\leq\|{\bf A}\|_{q_{1}\to q_{2}}\|g\|_{q^{*}_{2}}.$
Since this is true for every $g$, we conclude $\|{\bf A}^{*}\|_{q^{*}_{2}\to q^{*}_{1}}\leq\|{\bf A}\|_{q_{1}\to q_{2}}$. Reverse roles to complete the proof.
As a corollary, if $q^{*}=1$ then (8.34) and (8.37) combine to give
$\|P(X_{s}\in\cdot)\|_{2}\leq\|{\bf P}_{s}\|_{1\to 2}=\|{\bf P}_{s}\|_{2\to\infty}$
and then
$\|P(X_{s+t}\in\cdot)-\pi(\cdot)\|_{2}\leq\|{\bf P}_{s}\|_{2\to\infty}\,e^{-t/% \tau_{2}}$
from Lemma 8.11. Thus
$\sqrt{{\hat{d}}(2(s+t))}\leq\|{\bf P}_{s}\|_{2\to\infty}\,e^{-t/\tau_{2}}.$ (8.38)
Here is a somewhat different derivation of (8.38):
###### Lemma 8.13
For $0\leq s\leq t$,
$\displaystyle\sqrt{{\hat{d}}(2t)}=\|{\bf P}_{t}-{\bf E}\|_{2\to\infty}$ $\displaystyle\leq$ $\displaystyle\|{\bf P}_{s}\|_{2\to\infty}\|{\bf P}_{t-s}-{\bf E}\|_{2\to 2}$ $\displaystyle=$ $\displaystyle\|{\bf P}_{s}\|_{2\to\infty}\,e^{-(t-s)/\tau_{2}}.$
Proof. In light of (8.31), (8.30), and Lemma 8.10(b), we need only establish the first equality. Indeed, $\|P_{i}(X_{t}\in\cdot)-\pi(\cdot)\|_{2}$ is the $L^{2}$ norm of the function $(P_{i}(X_{t}\in\cdot)/\pi(\cdot))-1$ and so equals
$\max\left\{\left|\sum_{j}(p_{ij}(t)-\pi_{j})f(j)\right|:\|f\|_{2}=1\right\}=% \max\{|(({\bf P}_{t}-{\bf E})f)(i)|:\|f\|_{2}=1\}.$
Taking the maximum over $i\in I$ we obtain
$\displaystyle\sqrt{{\hat{d}}(2t)}$ $\displaystyle=$ $\displaystyle\max\left\{\max_{i}|(({\bf P}_{t}-{\bf E})f)(i)|:\|f\|_{2}=1\right\}$ $\displaystyle=$ $\displaystyle\max\{\|({\bf P}_{t}-{\bf E})f\|_{\infty}:\|f\|_{2}=1\}$ $\displaystyle=$ $\displaystyle\|{\bf P}_{t}-{\bf E}\|_{2\to\infty}.~{}\ \ \rule{4.3pt}{4.3pt}$
Choosing $s=0$ in Lemma 8.11 recaptures (8.26), and choosing $s=0$ in Lemma 8.13 likewise recaptures the consequence (8.5) of (8.26). The central theme for both Nash and log-Sobolev techniques is that one can improve upon these results by more judicious choice of $s$.
## 8.2.3 Exact computation of $N(s)$
The proof of Lemma 8.13 can also be used to show that
$N(s):=\|{\bf P}_{s}\|_{2\to\infty}=\max_{i}\|P_{i}(X_{s}\in\cdot)\|_{2}=\max_{% i}\sqrt{\frac{p_{ii}(2s)}{\pi_{i}}},$ (8.39)
as at (8.9). In those rare instances when the spectral representation is known explicitly, this gives the formula
xxx Also useful in conjunction with comparison method—see Section 3.
xxx If we can compute this, we can compute ${\hat{d}}(2t)=N^{2}(t)-1$. But the point is to test out Lemma 8.13.
$N^{2}(s)=1+\max_{i}\pi_{i}^{-1}\sum_{m=2}^{n}u^{2}_{im}\exp(-2\lambda_{m}s),$ (8.40)
and the techniques of later sections are not needed to compute $N(s)$. In particular, in the vertex-transitive case
$N^{2}(s)=1+\sum_{m=2}^{n}\exp(-2\lambda s).$
The norm $N(s)$ clearly behaves nicely under the formation of products:
$N(s)=N^{(1)}(s)N^{(2)}(s).$ (8.41)
###### Example 8.14
The two-state chain and the $d$-cube.
For the two-state chain, the results of Chapter 5 Example yyy:4 show
$N^{2}(s)=1+\frac{\max(p,q)}{\min(p,q)}e^{-2(p+q)s}.$
In particular, for the continuized walk on the $2$-path,
$N^{2}(s)=1+e^{-4s}.$
By the extension of (8.41) to higher-order products, we therefore have
$N^{2}(s)=(1+e^{-4s/d})^{d}$
for the continuized walk on the $d$-cube. This result is also easily derived from the results of Chapter 5 Example yyy:15. For $d\geq 2$ and $t\geq\frac{1}{4}d\log(d-1)$, the optimal choice of $s$ in Lemma 8.13 is therefore
$s={\textstyle\frac{1}{4}}d\log(d-1)$
and this leads in straightforward fashion to the bound
${\hat{\tau}}\leq{\textstyle\frac{1}{4}}d(\log d+3).$
While this is a significant improvement on the bound [cf. (8.12)]
${\hat{\tau}}\leq{\textstyle\frac{1}{4}}(\log 2)d^{2}+{\textstyle\frac{1}{2}}d$
obtained by setting $s=0$, i.e., obtained using only information about $\tau_{2}$, it is not
xxx REWRITE!, in light of corrections to my notes.
as sharp as the upper bound
${\hat{\tau}}\leq(1+o(1)){\textstyle\frac{1}{4}}d\log d$
in (8.13) that will be derived using log-Sobolev techniques.
###### Example 8.15
The complete graph.
For this graph, the results of Chapter 5 Example yyy:9 show
$N^{2}(s)=1+(n-1)\exp\left(-\frac{2ns}{n-1}\right).$
It turns out for this example that $s=0$ is the optimal choice in Lemma 8.13. This is not surprising given the sharpness of the bound in (8.7) in this case. See Example 8.32 below for further details.
###### Example 8.16
Product random walk on a $d$-dimensional grid.
Consider again the benchmark product chain (i.e., the “tilde chain”) in Example 8.5. That chain has relaxation time
$\tau_{2}=d\left(1-\cos\left(\frac{\pi}{m}\right)\right)^{-1}\leq\frac{1}{2}dm^% {2},$
so choosing $s=0$ in Lemma 8.13 gives
$\displaystyle{\hat{\tau}}$ $\displaystyle\leq$ $\displaystyle\frac{d}{1-\cos(\pi/m)}\left(\frac{1}{2}\log n+1\right)$ (8.42) $\displaystyle\leq$ $\displaystyle{\textstyle\frac{1}{4}}dm^{2}(\log n+2).$
This bound can be improved using $N(\cdot)$. Indeed, if we first consider continuized random walk on the $m$-path with self-loops added at each end, the stationary distribution is uniform, the eigenvalues are
$\lambda_{l}=1-\cos(\pi(l-1)/m),\ \ \ 1\leq l\leq m,$
and the eigenvectors are given by
$u_{il}=(2/m)^{1/2}\cos(\pi(l-1)(i-{\textstyle\frac{1}{2}})/m),\ \ \ 0\leq i% \leq m-1,\ \ \ 2\leq l\leq m.$
According to (8.40) and simple estimates, for $s>0$
$N^{2}(s)-1\leq 2\sum_{l=2}^{m}\exp[-2s(1-\cos(\pi(l-1)/m)]\leq 2\sum_{l=1}^{m-% 1}\exp(-4sl^{2}/m^{2})$
and
$\displaystyle\sum_{l=2}^{m-1}\exp(-4sl^{2}/m^{2})$ $\displaystyle\leq$ $\displaystyle\int_{x=1}^{\infty}\exp(-4sx^{2}/m^{2})\,dx$ $\displaystyle=$ $\displaystyle m\left(\frac{\pi}{4s}\right)^{1/2}P\left(Z\geq\frac{2(2s)^{1/2}}% {m}\right)$ $\displaystyle\leq$ $\displaystyle\left(\frac{\pi/4}{4s/m^{2}}\right)^{1/2}\exp(-4s/m^{2})$ $\displaystyle\leq$ $\displaystyle(4s/m^{2})^{-1/2}\exp(-4s/m^{2})$
when $Z$ is standard normal; in particular, we have used the well-known (xxx: point to Ross book exercise) bound
$P(Z\geq z)\leq{\textstyle\frac{1}{2}}e^{-z^{2}/2},\ \ \ z\geq 0.$
Thus
$N^{2}(s)\leq 1+2[1+(4s/m^{2})^{-1/2}]\exp(-4s/m^{2}),\ \ \ s>0.$
Return now to the “tilde chain” of Example 8.5, and assume for simplicity that $m_{1}=\cdots=m_{d}=m$. Since this chain is a slowed-down $d$-fold product of the path chain, it has
$N^{2}(s)\leq\left[1+2\left(1+\left(\frac{4s}{dm^{2}}\right)^{-1/2}\right)\exp% \left(-\frac{4s}{dm^{2}}\right)\right]^{d},\ \ \ s>0.$ (8.43)
In particular, since ${\hat{d}}(2t)=N^{2}(t)-1$, it is now easy to see that
${\hat{\tau}}\leq Km^{2}d\log d=Kn^{2/d}d\log d$ (8.44)
for a universal constant $K$.
xxx We’ve improved on (8.42), which gave order $m^{2}d^{2}\log m$.
xxx When used to bound $d(t)$ at (8.4), the bound (8.44) is “right”: see Theorem 4.1, p. 481, in [118].
The optimal choice of $s$ in Lemma 8.13 cannot be obtained explicitly when (8.43) is used to bound $N(s)=\|{\bf P}_{s}\|_{2\to\infty}$. For this reason, and for later purposes, it is useful to use the simpler, but more restricted, bound
$0\leq s\leq dm^{2}/16$ (8.45)
To verify this bound, simply notice that $u+2ue^{-u^{2}}+2e^{-u^{2}}\leq 4$ for $0\leq u\leq\frac{1}{2}$. When $s=dm^{2}/16$, (8.45), and $\tau_{2}\leq dm^{2}/2$ are used in Lemma 8.13, we find
${\hat{\tau}}\leq{\textstyle\frac{3}{4}}m^{2}d^{2}(\log 2+{\textstyle\frac{3}{4% }}d^{-1}).$
xxx Improvement over (8.42) by factor $\Theta(\log m)$, but display following (8.43) shows still off by factor $\Theta(d/\log d)$. | 2018-09-24 04:17:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 244, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.973771333694458, "perplexity": 582.5377721778673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160142.86/warc/CC-MAIN-20180924031344-20180924051744-00358.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/142433/ | # Upstream Tracking and the Decay $B^{0} \to K^{+}\pi^{-}\mu^{+}\mu^{-}$ at the LHCb Experiment
Bowen, Espen. Upstream Tracking and the Decay $B^{0} \to K^{+}\pi^{-}\mu^{+}\mu^{-}$ at the LHCb Experiment. 2017, University of Zurich, Faculty of Science. | 2021-07-29 18:57:17 | {"extraction_info": {"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, "math_score": 0.9713956713676453, "perplexity": 7737.104821735604}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00603.warc.gz"} |
https://aviation.stackexchange.com/questions/35076/what-happens-if-i-activate-vfr-flight-plan-later-than-filed-etd | # What happens if I activate VFR flight plan later than filed ETD?
Let's say I filed a VFR flight plan with ETD (1500), en route (2 hours), ETA (1700)
If I depart and activate the plan at 1600, now the new ETA will be 1800 which is an hour later the the filed ETA. Is the ETA changed automatically on the system? or do I have to contact FSS and change to prevent initiation of the search and rescue?
• When you call FSS to activate your flight plan you give them your time of departure. Unless your estimated time en route changes, the ETA will be amended correctly. – JScarry Jan 28 '17 at 21:14
• Are you asking about a specific country or jurisdiction? – Pondlife Jan 29 '17 at 16:48
Assuming that you've submitted a Delay Message in order to prevent your flightplan from being deleted you have to do nothing at all.
In your flightplan you only state your ETD (Expected Time of Departure) and ETE (Estimated Time Enroute). Your ETA (Estimated Time of Arrival) will be calculated and relayed to all stations in need of the ETA by AIS (Aeronautical Information Service).
Filed ETD: 1500 | 2019-10-16 13:43:22 | {"extraction_info": {"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, "math_score": 0.19609157741069794, "perplexity": 4766.991358305829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00304.warc.gz"} |
https://openreview.net/forum?id=SJl7DsR5YQ | ## ReNeg and Backseat Driver: Learning from demonstration with continuous human feedback
Sep 27, 2018 Blind Submission readers: everyone Show Bibtex
• Abstract: Reinforcement learning (RL) is a powerful framework for solving problems by exploring and learning from mistakes. However, in the context of autonomous vehicle (AV) control, requiring an agent to make mistakes, or even allowing mistakes, can be quite dangerous and costly in the real world. For this reason, AV RL is generally only viable in simulation. Because these simulations have imperfect representations, particularly with respect to graphics, physics, and human interaction, we find motivation for a framework similar to RL, suitable to the real world. To this end, we formulate a learning framework that learns from restricted exploration by having a human demonstrator do the exploration. Existing work on learning from demonstration typically either assumes the collected data is performed by an optimal expert, or requires potentially dangerous exploration to find the optimal policy. We propose an alternative framework that learns continuous control from only safe behavior. One of our key insights is that the problem becomes tractable if the feedback score that rates the demonstration applies to the atomic action, as opposed to the entire sequence of actions. We use human experts to collect driving data as well as to label the driving data through a framework we call Backseat Driver'', giving us state-action pairs matched with scalar values representing the score for the action. We call the more general learning framework ReNeg, since it learns a regression from states to actions given negative as well as positive examples. We empirically validate several models in the ReNeg framework, testing on lane-following with limited data. We find that the best solution in this context outperforms behavioral cloning has strong connections to stochastic policy gradient approaches.
• TL;DR: We introduce a novel framework for learning from demonstration that uses continuous human feedback; we evaluate this framework on continuous control for autonomous vehicles.
• Keywords: learning from demonstration, imitation learning, behavioral cloning, reinforcement learning, off-policy, continuous control, autonomous vehicles, deep learning, machine learning, policy gradient
0 Replies | 2020-01-18 05:34:28 | {"extraction_info": {"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, "math_score": 0.5171042084693909, "perplexity": 1416.3829854388978}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250592261.1/warc/CC-MAIN-20200118052321-20200118080321-00402.warc.gz"} |
https://exceptionshub.com/python-ensure-relative-path-is-used-in-cmd-exe-when-running-batch-files-from-excel.html | Home » excel » python – Ensure Relative Path is Used in cmd.exe when Running Batch Files From Excel
# python – Ensure Relative Path is Used in cmd.exe when Running Batch Files From Excel
Questions:
I have an Excel file with a button that calls a Shell command in VB:
Shell(ThisWorkbook.Path & "\python_script.bat", vbNormalFocus)
The Shell command calls the above batch file that runs a Python script:
python python_script.py
All the pertinent files (the Excel file, the batch file, the data files, the Python file) are all located in the same directory, call this sample_program, because I am building this for someone else and I intend for them to simply unzip it and run it.
In Excel, when testing this, I click the button and then get this error:
C:\Users\<user_name>\Documents>python python_script.py
python: can't open file 'python_script.py': [Errno 2] No such file or directory
For some reason, although all these files are in the same location, the cmd.exe is running from my user directory: C:\Users\<user_name>\Documents
I don’t want cmd.exe to use this path; I want it to use the path\to\sample_program directory.
How do I get this to use relative paths so when I transfer this folder to someone else, and they place it anywhere, it will work as a self-contained unit?
Try this, if path\to\sample_program is on the same drive as your home drive
Shell "cmd.exe /k cd " & ThisWorkbook.Path & "&&python_script.bat"
or this if path\to\sample_program is not on the same drive as your home drive, or you dont know in advance
Shell "cmd.exe /k " & Left(ThisWorkbook.Path, 2) & "&&cd " & ThisWorkbook.Path & "&&python_script.bat"
You can use Environ("username") to get the user logon name, so for your sample (which I presume is Windows 7)
Sub GetDir()
MsgBox "C:\Users\" & Environ("username") & "\Documents"
End Sub
You can also automatically retrieve certain locations regardless of OS using SpecialFOlders, ie
Sub GetPath()
Set wshShell = CreateObject("WScript.Shell")
Documents = wshShell.SpecialFolders("MyDocuments")
MsgBox Documents
End Sub | 2021-04-11 00:44:09 | {"extraction_info": {"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, "math_score": 0.4846700429916382, "perplexity": 7935.277411863027}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038060603.10/warc/CC-MAIN-20210411000036-20210411030036-00516.warc.gz"} |
https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Structural_Mechanics_(Wierzbicki)/07%3A_Energy_Methods_in_Elasticity/7.02%3A_Equivalence_of_the_Minimum_Potential_Energy_and_Principle_of_Virtual_Work | 7.2: Equivalence of the Minimum Potential Energy and Principle of Virtual Work
The concept of virtual displacement $$\delta u_i$$ is the backbone of the energy methods in mechanics. The virtual displacement is a small hypothetical displacement which satisfy the kinematic boundary condition. The virtual strains $$\delta \epsilon_{ij}$$ are obtained from the virtual displacement by
$\delta \epsilon_{ij} = \frac{1}{2} (\delta u_{i,j} + \delta u_{j,i})$
The increment of stress $$\delta \sigma_{ij}$$ corresponding to the increment of strain is obtained from the elasticity law
$\sigma_{ij} = C_{ijkl}\epsilon_{kl}$
$\delta \sigma_{ij} = C_{ijkl}\delta \epsilon_{kl}$
Therefore, by eliminating $$C_{ijkl}$$
$\sigma_{ij} \delta \epsilon_{ij} = \epsilon_{ij}\delta \sigma_{ij} \label{8.15}$
The total strain energy of the elastic system $$\prod$$ is the sum of the elastic strain energy stored and the work of external forces
$\prod = \int_{V} \frac{1}{2} \sigma_{ij} \epsilon_{ij} dv − \int_{S} T_iu_i ds$
In the above equation the surface traction are given and considered to be constant. The stresses $$\sigma_{ij}$$ are not considered to be constant because they are related to the variable strains. For equilibrium the potential energy must be stationary, $$\delta \prod = 0$$ or
$\delta \int_{V} \frac{1}{2} \sigma_{ij} \epsilon_{ij} dv − \delta \int_{S} T_i u_i ds \\ = \frac{1}{2} \int_{V} \delta (\sigma_{ij} \epsilon_{ij}) dv − \int_{S} T_i \delta u_i ds \\ = \frac{1}{2} \int_{V} (\delta \sigma_{ij} \epsilon_{ij} + \sigma_{ij} \delta \epsilon_{ij}) dv − \int_{S} T_i \delta u_i ds = 0 \label{8.17}$
The two terms in the integrand of the volume integral are equal in view of Equation \ref{8.15}. Therefore, Equation \ref{8.17} can be written in the equivalent form
$\int_{V} \sigma_{ij} \delta \epsilon_{ij} dv = \int_{S} T_i \delta u_i ds$
which is precisely the principle of virtual work. The above proof goes also in the opposite direction. Assuming the principle of virtual work one can show that the stationarity of the total potential energy holds.
This page titled 7.2: Equivalence of the Minimum Potential Energy and Principle of Virtual Work is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Tomasz Wierzbicki (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 2022-08-16 12:30:03 | {"extraction_info": {"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, "math_score": 0.9174172878265381, "perplexity": 310.32855761581794}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00058.warc.gz"} |
https://www.macajournal.com/article_693796.html | Document Type : Original Article
Author
Department of Mathematics, Payame Noor University(PNU), P. O. Box 19395-4697, Tehran, Iran
10.30495/maca.2022.1952804.1049
Abstract
The objectives of this article are to present the notions of anti complex fuzzy subalgebras and anti complex fuzzy ideals of Lie algebras under $S$-norms and we prove that the level subset of them are also subalgebras and ideals of Lie algebras, respectively. Next, we introduce the union and summation of them and we prove that the union and summation of them are also anti complex fuzzy subalgebras and anti complex fuzzy ideals of Lie algebras under $S$-norms, respectively. Finally, we investigate the homomorphic image (pre-image) of them under Lie homomorphisms.
Keywords | 2022-12-06 10:46:15 | {"extraction_info": {"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, "math_score": 0.7895973920822144, "perplexity": 1473.899839336324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711077.50/warc/CC-MAIN-20221206092907-20221206122907-00078.warc.gz"} |
https://www.studysmarter.us/textbooks/math/fundamentals-of-differential-equations-and-boundary-value-problems-9th/series-solutions-of-differential-equations/q16e-in-problems-15-17solve-the-given-initial-value-problem-/ | Suggested languages for you:
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Found in: Page 453
### Fundamentals Of Differential Equations And Boundary Value Problems
Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069
# In Problems 15-17,solve the given initial value problem x2y"+5xy'+4y=0.y(1) =3 and y'(1) = 7
The solution of the given initial value problem is y=3x-2+13x-2 ln x.
See the step by step solution
## Define Cauchy-Euler equations:
In mathematics, a Cauchy problem is one in which the solution to a partial differential equation must satisfy specific constraints specified on a hypersurface in the domain.
An initial value problem or a boundary value problem is an example of the Cauchy problem.
The equation will be in the form of, ax2y"+bxy'+cy=0.
## Find the general solution:
The given equation is,
x2y"+5xy'+4y=0
Let L be the differential operator defined by the left-hand side of equation, that is,
L [x] (t) = x2y"+5xy'+4y
And let.
w (r,t) = xr
Substituting the w(r,t) in place of x(t), you get
L [w] (t)=x2 (xr)"+5x (xr)'+4 (xr)
=x2 (r (r-1)) xr-2 +5x (r) xr-1 +4xr
= (r2-r) xr +5rxr +4xr
=(r2+4r+4) xr
Solving the indicial equation,
r2+4r+4 =0
(r+2)2 =0
There are repeated roots at r= -2
Thus there are two linearly independent solutions given by
y1=c1x-2 and y2 = c2x-2 lnx
The general solution for the equation will be
y=c1x-2 +c2x-2 lnx
## Determine the initial value:
For the given initial conditions,
y(1)=3 and y'(1)=7
y(x)=c1 x-2+c2 x-2 lnx
y(1) = c1
c1=3
y'(x)=c1 (-2) x-3 +c2 [(-2) x-3 lnx+x-2 (1/x)]
y'(1) = -2c1+c2
-2c1+c2=7
Solving the two simultaneous equations (1) and (2), you get the values of two constants c1 and c2 as,
c1 = 3 and c2 = 13
Thus the solution of the given initial value problem is y=13 x-2+13x-2 ln x. | 2023-03-30 01:26:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8596259951591492, "perplexity": 4957.573242335316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00554.warc.gz"} |
https://mathoverflow.net/questions/327869/which-are-good-algorithms-for-finding-hamiltonian-path-not-necessarily-a-circle | Which are good algorithms for finding Hamiltonian path (not necessarily a circle) up to now?
I am not expertise in graph theory. So have to ask this question here. The term "good" means that the algorithms should be efficient for general undirected simple connected graphs with a higher success probability.
I also read the question Efficient Hamiltonian cycle algorithms for graph classes.
But it does answer my concern.
• You will need to be more specific for a helpful answer. Here is a list of resources you might want to consult: web.archive.org/web/20100324030526/http://alife.ccp14.ac.uk/… – Carlo Beenakker Apr 12 '19 at 12:19
• Thanks a lot! But I cannot open this link after some times trying ... Is this link correct? Thanks again! – Licheng Wang Apr 12 '19 at 12:37
• it works for me... – Carlo Beenakker Apr 12 '19 at 12:49
• Does "needless a circle" mean that the Hamiltonian path need not be a cycle? If so, probably "not necessarily" is clearer than "needless". – LSpice Apr 12 '19 at 14:57
• Yes! Thanks! I have edited the title just now. – Licheng Wang Apr 13 '19 at 1:11
The Hamiltonian path problem is NP-complete in general, so only heuristics could exist if P≠NP.
The rotation-extension heuristic may be the simplest heuristic:
Input: undirected graph G with size n
Let P be a path
Let e be a random edge of G
P:=[e]
Loop:
If Extension(G,P)≠∅:
P:=Extension(G,P)
Goto Loop
Let Π be the family of the Posa extensions¹ of P in G
For π in Π:
If Extension(G,π)≠∅:
P:=Extension(G,π)
Goto Loop
{Remark: The heuristic is not able to extend the path, so we must stop}
If P is a hamiltonian path, return P, otherwise stop without returning anything.
Subprocedure: Extension
Input: undirected graph G and a path P⊆G
For x in vertices of G:
if x is connected with one of P's endpoints p:
Return P+(p,x)
Return ∅
1: as defined in https://www.sciencedirect.com/science/article/pii/S0012365X06005097
In other words, the program finds extensions and extensions after rotations until there're none, and return a hamiltonian path if there is one.
For more sophiscated heuristics, one can use methods from the Flinders Hamiltonian Cycle Project. | 2020-01-28 00:23:05 | {"extraction_info": {"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, "math_score": 0.6741132736206055, "perplexity": 1463.2225018149568}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00398.warc.gz"} |
https://docs.openmc.org/en/stable/pythonapi/generated/openmc.deplete.LEQIIntegrator.html | openmc.deplete.LEQIIntegrator¶
class openmc.deplete.LEQIIntegrator(operator, timesteps, power=None, power_density=None)[source]
Deplete using the LE/QI CFQ4 algorithm.
Implements the LE/QI Predictor-Corrector algorithm using the fourth order commutator-free integrator.
“LE/QI” stands for linear extrapolation on predictor and quadratic interpolation on corrector. This algorithm is mathematically defined as:
\begin{aligned} y' &= A(y, t) y(t) \\ A_{last} &= A(y_{n-1}, t_n - h_1) \\ A_0 &= A(y_n, t_n) \\ F_1 &= \frac{-h_2^2}{12h_1} A_{last} + \frac{h_2(6h_1+h_2)}{12h_1} A_0 \\ F_2 &= \frac{-5h_2^2}{12h_1} A_{last} + \frac{h_2(6h_1+5h_2)}{12h_1} A_0 \\ y_p &= \text{expm}(F_2) \text{expm}(F_1) y_n \\ A_1 &= A(y_p, t_n + h_2) \\ F_3 &= \frac{-h_2^3}{12 h_1 (h_1 + h_2)} A_{last} + \frac{h_2 (5 h_1^2 + 6 h_2 h_1 + h_2^2)}{12 h_1 (h_1 + h_2)} A_0 + \frac{h_2 h_1)}{12 (h_1 + h_2)} A_1 \\ F_4 &= \frac{-h_2^3}{12 h_1 (h_1 + h_2)} A_{last} + \frac{h_2 (h_1^2 + 2 h_2 h_1 + h_2^2)}{12 h_1 (h_1 + h_2)} A_0 + \frac{h_2 (5 h_1^2 + 4 h_2 h_1)}{12 h_1 (h_1 + h_2)} A_1 \\ y_{n+1} &= \text{expm}(F_4) \text{expm}(F_3) y_n \end{aligned}
It is initialized using the CE/LI algorithm.
Parameters: operator (openmc.deplete.TransportOperator) – Operator to perform transport simulations timesteps (iterable of float) – Array of timesteps in units of [s]. Note that values are not cumulative. power (float or iterable of float, optional) – Power of the reactor in [W]. A single value indicates that the power is constant over all timesteps. An iterable indicates potentially different power levels for each timestep. For a 2D problem, the power can be given in [W/cm] as long as the “volume” assigned to a depletion material is actually an area in [cm^2]. Either power or power_density must be specified. power_density (float or iterable of float, optional) – Power density of the reactor in [W/gHM]. It is multiplied by initial heavy metal inventory to get total power if power is not speficied. operator (openmc.deplete.TransportOperator) – Operator to perform transport simulations chain (openmc.deplete.Chain) – Depletion chain timesteps (iterable of float) – Size of each depletion interval in [s] power (iterable of float) – Power of the reactor in [W] for each interval in timesteps
__call__(bos_conc, bos_rates, dt, power, i)[source]
Perform the integration across one time step
Parameters: conc (numpy.ndarray) – Initial concentrations for all nuclides in [atom] rates (openmc.deplete.ReactionRates) – Reaction rates from operator dt (float) – Time in [s] for the entire depletion interval power (float) – Power of the system in [W] i (int) – Current depletion step index proc_time (float) – Time spent in CRAM routines for all materials in [s] conc_list (list of numpy.ndarray) – Concentrations at each of the intermediate points with the final concentration as the last element op_results (list of openmc.deplete.OperatorResult) – Eigenvalue and reaction rates from intermediate transport simulation
__iter__()
Return pairs of time steps in [s] and powers in [W]
__len__()
Return integer number of depletion intervals
integrate()
Perform the entire depletion process across all steps | 2020-03-31 10:25:29 | {"extraction_info": {"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, "math_score": 0.9999992847442627, "perplexity": 9202.592847854034}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00307.warc.gz"} |
https://byjus.com/question-answer/abc-is-right-angled-at-a-ad-is-perpendicular-to-bc-ab-5cm-bc-13-cm-and-ac-12-cm-find-the-area-of-abc-also-find-the-length-of-ad/ | Question
# $\Delta ABC$ is right-angled at $A$.$AD$ is perpendicular to $BC$, $AB=5cm,BC=13cm$and $AC=12cm$. Find the area of$\Delta ABC$. Also, find the length of$AD$.
Open in App
Solution
## Step1: Finding the area of$\Delta ABC$ $∆ABC$is right–angled at $A$.$AB=5cm,BC=13cm$ and $AC=12cm$now, area of triangle$\begin{array}{rcl}∆ABC& =& \frac{1}{2}×altitude×base\\ & =& \frac{1}{2}×AB×AC\\ & =& \frac{1}{2}×5cm×12cm\\ & =& 30cm²\end{array}$Hence, area of$∆ABC=30cm²$Step2:Finding length of $AD$Area of triangle$\begin{array}{rcl}A∆ABC& =& \frac{1}{2}×BC×AD\\ & ⇒& 30=\frac{1}{2}×13×AD\\ & ⇒& \frac{30×2}{13}=AD\\ & ⇒& AD=\frac{60}{13}\\ & ⇒& AD=4.61cm\end{array}$Therefore the area of$∆ABC=30cm²$ and length of $\begin{array}{rcl}AD& =& 4.61cm\end{array}$
Suggest Corrections
17 | 2023-02-03 00:59:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8593698143959045, "perplexity": 4072.8672447195163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500041.2/warc/CC-MAIN-20230202232251-20230203022251-00752.warc.gz"} |
http://math.stackexchange.com/questions/261319/wave-equation-on-the-half-line/261603 | # Wave equation on the half-line
Let $u(x,t)$ be solution of initial boundary value problem $$u_{tt}=u_{xx},\quad 0<x<\infty,\quad t>0$$ $$u(x,0)=\cos\left(\frac{\pi x}{2}\right),\quad 0<x<\infty$$ $$u_{t}(x,0)=0,\quad 0<x<\infty$$ $$u_{x}(0,t)=0,\quad t\geq0$$ then
1. $u(2,2)=?$
2. $u(1/2,1/2)=?$
I am stuck on this problem . Can anyone help me please........
I can't solve it with separation of variable........ I don't know where to begin?
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how can we apply d'Alambert's formula, not getting sir.......... – pankaj Dec 18 '12 at 12:36
The problem is posed in the positive real line. The boundary condition suggests to consider the initial value problem in the while real line extending the initial data to be even. Once you have that, use d'Alambert's formula.
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how can we apply d'Alambert's formula, not getting sir.......... – pankaj Dec 18 '12 at 12:16
One approach is to assume that you can split $u(x,t)=X(x)\cdot T(t)$. I don't know the english expression, but it should be something like "separation approach". Using this for the equation $u_{tt}=u_{xx}$, we get \begin{align} \frac{d^2T}{dt^2} \cdot X = T \cdot \frac{d^2X}{dx^2} \\ \Leftrightarrow \frac{d^2T}{dt^2} \cdot T^{-1} = X^{-1} \frac{d^2X}{dx^2}X \equiv -\lambda^2 \end{align} Where $\lambda \in C$ is an arbitrary constant. The reason, why the expression are constant, is that the left side only depends on $t$ while the right side only depends on $x$. If now the i.e. the left side would not be constant, we couldn't reproduce that change on the right side as it is only depend on $x$. The reason why I choose $\lambda^2$ and not $-\lambda$ or $+\lambda^2$ becomes clear in the next step. From the above it follows \begin{align} &\frac{d^2T}{dt^2} = -\lambda^2 T \Rightarrow T(t) = a \sin(\lambda t) + b \cos(\lambda t) \\& \frac{d^2X}{dt^2} = -\lambda^2 X \Rightarrow X(x)=c \sin(\lambda x) + d \cos(\lambda x) \end{align} If we had chosen $\lambda^2$ instead of $-\lambda^2$ we would have got another solution to the two ODEs, i.e. the $\exp()$. But as the initial condition suggest trigonometric functions, $\sin()$ and $\cos()$ fit better. Okay. So we have $u(x,t) =X(x)T(t)$ Now we check for initial condition. \begin{align} u(x,0) = X(x)\cdot T(0) = \cos(\frac{\pi x}{2}) \end{align} This suggests $c=0,\lambda=\frac{\pi}{2}$ and with $T(0) = b$ it follows $d\cdot b=1$ and $X(x) =d\cdot \cos(\frac{\pi x}{2})$.
Next initial condition is $u_t(x,0)=0$. \begin{align} u_t(x,0) &= T'(t)X(0) =(\frac{\pi}{2} a \cos(\frac{\pi}{2} 0) -\frac{\pi}{2} b \sin(\frac{\pi}{2} 0)) \cdot d\cos(\frac{\pi x}{2}) = \\ &=(\frac{\pi}{2} a)\cdot d\cos(\frac{\pi x}{2})=0 \end{align} As $d\neq0$, it follows $a=0$ and therefore $T(t) = b\cos(\frac{\pi t}{2})$. Combining this, we have \begin{align} u(x,t) = b\cos(\frac{\pi t}{2})\cdot d \cos(\frac{\pi x}{2})=e\cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2}) \end{align} With $e=b\cdot d$. But as $b\cdot d = 1$ as stated above, the solution becomes \begin{align} u(x,t) = \cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2}) \end{align} Now we still have to check the last initial condition $u_x(0,t)=0$. \begin{align} u_x(0,t) = X'(x)T(0) = \frac{\pi}{2} \sin(0)T(0) = 0 \end{align} So this one also holds. Therefore we have the solution to the PDE as $u(x,t) = \cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2})$.
Note: In my opinion we could have gone a lot faster. Why? Except for the first initial condition the problem looks quite symmetric. So thought that maybe also $T(t) = \cos(\frac{\pi t}{2})$. And if you check this with the first initial condition it holds.
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if there is any shortcut please explain sir......... – pankaj Dec 18 '12 at 12:49
Sorry, what do you mean exactly? – cmmndy Dec 18 '12 at 13:18
Note that when without the condition $u_x(0,t)=0$ this is in fact a just-determining problem and the solution can be expressed by using D’Alembert’s formula $u(x,t)=\dfrac{1}{2}\left(\cos\dfrac{\pi(x+t)}{2}+\cos\dfrac{\pi(x-t)}{2}\right)=\cos\dfrac{\pi x}{2}\cos\dfrac{\pi t}{2}$
Check for $u_x(0,t)$ :
$u_x(x,t)=-\dfrac{\pi}{2}\sin\dfrac{\pi x}{2}\cos\dfrac{\pi t}{2}$
$u_x(0,t)=-\dfrac{\pi}{2}\sin\dfrac{\pi\times0}{2}\cos\dfrac{\pi t}{2}=0$
$\therefore u(x,t)=\cos\dfrac{\pi x}{2}\cos\dfrac{\pi t}{2}$ is already an exact solution of this problem, we don't need to use the result in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf#page=2
Hence $u(2,2)=\cos\dfrac{\pi\times2}{2}\cos\dfrac{\pi\times2}{2}=\cos\pi\cos\pi=1$
Hence $u\left(\dfrac{1}{2},\dfrac{1}{2}\right)=\cos\dfrac{\pi\times\dfrac{1}{2}}{2}\cos\dfrac{\pi\times\dfrac{1}{2}}{2}=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{4}=\dfrac{1}{2}$
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This is a wave equation on the half-line with Neumann boundary condition. You can use even extension(fortunately, everything you need to extend is automatically even) and d'Alambert's Formula to solve it.
I think the answer should be $u(x,t) =\frac { 1 }{ 2 } \left[ cos\frac { \pi (x+t) }{ 2 } +cos\frac { \pi (x-t) }{ 2 } \right]$ with x and t positive real numbers. Plugin $(2,2)$ and $( \frac{1}{2} , \frac{1}{2} )$ you'll get 1 and 1/2
- | 2015-11-30 17:09:56 | {"extraction_info": {"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": 7, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9990240335464478, "perplexity": 470.0245641208749}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398462709.88/warc/CC-MAIN-20151124205422-00060-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://pure.mpg.de/pubman/faces/ViewItemFullPage.jsp?itemId=item_3335362_3&view=EXPORT | English
# Item
ITEM ACTIONSEXPORT
GWTC-2.1: Deep Extended Catalog of Compact Binary Coalescences Observed by LIGO and Virgo During the First Half of the Third Observing Run
The LIGO Scientific Collaboration, the Virgo Collaboration, Abbott, R., Abbott, T. D., Acernese, F., Ackley, K., et al. (in preparation). GWTC-2.1: Deep Extended Catalog of Compact Binary Coalescences Observed by LIGO and Virgo During the First Half of the Third Observing Run.
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Creators:
The LIGO Scientific Collaboration, Author
the Virgo Collaboration, Author
Abbott, R., Author
Abbott, T. D., Author
Acernese, F., Author
Ackley, K., Author
Affeldt, C.1, Author
Agarwal, D., Author
Agathos, M., Author
Agatsuma, K., Author
Aggarwal, N., Author
Aguiar, O. D., Author
Aiello, L., Author
Ain, A., Author
Ajith, P., Author
Albanesi, S., Author
Affiliations:
1Laser Interferometry & Gravitational Wave Astronomy, AEI-Hannover, MPI for Gravitational Physics, Max Planck Society, ou_24010
2Astrophysical and Cosmological Relativity, AEI-Golm, MPI for Gravitational Physics, Max Planck Society, ou_1933290
3Binary Merger Observations and Numerical Relativity, AEI-Hannover, MPI for Gravitational Physics, Max Planck Society, ou_2461691
4Astrophysical Relativity, AEI-Golm, MPI for Gravitational Physics, Max Planck Society, ou_24013
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Free keywords: General Relativity and Quantum Cosmology, gr-qc
Abstract: The second gravitational-wave transient catalog, GWTC-2, reported on 39 compact binary coalescences observed by the Advanced LIGO and Advanced Virgo detectors between 1 April 2019 15:00 UTC and 1 October 2019 15:00 UTC. Here, we present GWTC-2.1, which reports on a deeper list of candidate events observed over the same period. We analyze the final version of the strain data over this period, which is now publicly released. We employ three matched-filter search pipelines for candidate identification, and estimate the probability of astrophysical origin for each candidate event. While GWTC-2 used a false alarm rate threshold of 2 per year, we include in GWTC-2.1, 1201 candidates that pass a false alarm rate threshold of 2 per day. We calculate the source properties of a subset of 44 high-significance candidates that have a probability of astrophysical origin greater than 0.5, using the default priors. Of these candidates, 36 have been reported in GWTC-2. If the 8 additional high-significance candidates presented here are astrophysical, the mass range of candidate events that are unambiguously identified as binary black holes (both objects $\geq 3M_\odot$) is increased compared to GWTC-2, with total masses from $\sim 14M_\odot$ for GW190924_021846 to $\sim 184M_\odot$ for GW190426_190642. The primary components of two new candidate events (GW190403_051519 and GW190426_190642) fall in the mass gap predicted by pair-instability supernova theory. We also expand the population of binaries with significantly asymmetric mass ratios reported in GWTC-2 by an additional two events ($q \lt 0.61$ and $q \lt 0.62$ at $90\%$ credibility for GW190403_051519 and GW190917_114630 respectively), and find that 2 of the 8 new events have effective inspiral spins $\chi_\mathrm{eff} > 0$ (at $90\%$ credibility), while no binary is consistent with $\chi_\mathrm{eff} \lt 0$ at the same significance.
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Dates: 2021-08-02
Publication Status: Not specified
Pages: 6 figures, 5 tables
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Rev. Type: -
Identifiers: arXiv: 2108.01045
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show | 2021-11-28 23:23:37 | {"extraction_info": {"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, "math_score": 0.7609474658966064, "perplexity": 8422.562477912985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00167.warc.gz"} |
https://caddellprep.com/january-2016-geometry-regents/ | 1. William is drawing pictures of cross sections of the right circular cone below.
Which drawing can not be a cross section of a cone?
2. An equation of a line perpendicular to the line represented by the equation $y = -\dfrac{1}{2}x - 5$ and passing through ($6, -4$) is
(1) $y = -\dfrac{1}{2}x + 4$
(2) $y = -\dfrac{1}{2}x - 1$
(3) $y = 2x + 14$
(4) $y = 2x - 16$
3. In parallelogram $QRST$ shown below, diagonal $\overline{TR}$ is drawn, $U$ and $V$ are points on $\overline{TS}$ and $\overline{QR}$, respectively, and $\overline{UV}$ intersects $\overline{TR}$ at $W$.
If $m\angle S = 60^{\circ}, m\angle SRT = 83^{\circ}$, and $m\angle TWU = 35^{\circ}$, what is $m\angle WVQ$?
(1) $37^{\circ}$
(2) $60^{\circ}$
(3) $72^{\circ}$
(4) $83^{\circ}$
4. A fish tank in the shape of a rectangular prism has dimensions of $14$ inches, $16$ inches, and $10$ inches. The tank contains $1680$ cubic inches of water. What percent of the fish tank is empty?
(1) $10$
(2) $25$
(3) $50$
(4) $75$
5. Which transformation would result in the perimeter of a triangle being different from the perimeter of its image?
(1) $(x, y) \rightarrow (y,x)$
(2) $(x, y) \rightarrow (x, -y)$
(3) $(x, y) \rightarrow (4x, 4y)$
(4) $(x, y) \rightarrow (x + 2, y - 5)$
6. In the diagram below, $\overleftrightarrow{FE}$ bisects $\overline{AC}$ at $B$, and $\overleftrightarrow{GE}$ bisects $\overline{BD}$ at $C$.
Which statement is always true?
(1) $\overline{AB} \cong \overline{DC}$
(2) $\overline{FB} \cong \overline{EB}$
(3) $\overleftrightarrow{BD}$ bisects $\overline{GE}$ at $C$
(4) $\overleftrightarrow{AC}$ bisects $\overline{FE}$ at $B$
7. As shown in the diagram below, a regular pyramid has a square base whose side measures $6$ inches.
If the altitude of the pyramid measures $12$ inches, its volume, in cubic inches, is
(1) $72$
(2) $144$
(3) $288$
(4) $432$
8. Triangle $ABC$ and triangle $DEF$ are graphed on the set of axes below.
Which sequence of transformations maps triangle $ABC$ onto triangle $DEF$?
(1) a reflection over the $x$-axis followed by a reflection over the $y$-axis
(2) a $180^{\circ}$ rotation about the origin followed by a reflection over the line $y = x$
(3) a $90^{\circ}$ clockwise rotation about the origin followed by a reflection over the $y$-axis
(4) a translation $8$ units to the right and $1$ unit up followed by a $90^{\circ}$ counterclockwise rotation about the origin
9. In $\triangle ABC$, the complement of $\angle B$ is $\angle A$. Which statement is always true?
(1) $tan \angle A = tan \angle B$
(2) $sin \angle A = sin \angle B$
(3) $cos \angle A = tan \angle B$
(4) $sin \angle A = cos \angle B$
10. A line that passes through the points whose coordinates are ($1,1$) and ($5,7$) is dilated by a scale factor of $3$ and centered at the origin. The image of the line
(1) is perpendicular to the original line
(2) is parallel to the original line
(3) passes through the origin
(4) is the original line
11. Quadrilateral $ABCD$ is graphed on the set of axes below.
When $ABCD$ is rotated $90^{\circ}$ in a counterclockwise direction about the origin, its image is quadrilateral $A'B'C'D'$. Is distance preserved
under this rotation, and which coordinates are correct for the given vertex?
(1) no and $C'(1,2)$
(2) no and $D'(2,4)$
(3) yes and $A'(6,2)$
(4) yes and $B'(-3,4)$
12. In the diagram below of circle $O$, the area of the shaded sector $LOM$ is $2\pi cm^{2}$.
If the length of $\overline{NL}$ is $6$ cm, what is $m\angle N$?
(1) $10^{\circ}$
(2) $20^{\circ}$
(3) $40^{\circ}$
(4) $80^{\circ}$
13. In the diagram below, $\triangle ABC \sim \triangle DEF$.
If $AB = 6$ and $AC = 8$, which statement will justify similarity by SAS?
(1) $DE = 9, DF = 12,$ and $\angle A \cong \angle D$
(2) $DE = 8, DF = 10,$ and $\angle A \cong \angle D$
(3) $DE = 36, DF = 64,$ and $\angle C \cong \angle F$
(4) $DE = 15, DF = 20,$ and $\angle C \cong \angle F$
14. The diameter of a basketball is approximately $9.5$ inches and the diameter of a tennis ball is approximately $2.5$ inches. The volume of the basketball is about how many times greater than the volume of the tennis ball?
(1) $3591$
(2) $65$
(3) $55$
(4) $4$
15. The endpoints of one side of a regular pentagon are ($1,4$) and ($2,3$). What is the perimeter of the pentagon?
(1) $\sqrt{10}$
(2) $5\sqrt{10}$
(3) $5\sqrt{2}$
(4) $25\sqrt{2}$
16. In the diagram of right triangle $ABC$ shown below, $AB = 14$ and $AC = 9$.
What is the measure of $\angle A$, to the nearest degree?
(1) $33$
(2) $40$
(3) $50$
(4) $57$
17. What are the coordinates of the center and length of the radius of the circle whose equation is $x^{2} + 6x + y^{2} - 4y = 23$?
(1) ($3,-2$) and $36$
(2) ($3,-2$) and $6$
(3) ($-3,2$) and $36$
(4) ($-3,2$) and $6$
18. The coordinates of the vertices of $\triangle RST$ are $R(-2, -3), S(8,2)$, and $T(4,5)$. Which type of triangle is $\triangle RST$?
(1) right
(2) acute
(3) obtuse
(4) equiangular
19. Molly wishes to make a lawn ornament in the form of a solid sphere. The clay being used to make the sphere weighs $.075$ pound per cubic inch. If the sphere’s radius is $4$ inches, what is the weight of the sphere, to the nearest pound?
(1) $34$
(2) $20$
(3) $15$
(4) $4$
20. The ratio of similarity of $\triangle BOY$ to $\triangle GRL$ is $1:2$. If $BO = x + 3$ and $GR = 3x - 1$, then the length of $\overline{GR}$ is
(1) $5$
(2) $7$(3) $10$
(4) $20$
21. In the diagram below, $\overline{DC}$, $\overline{AC}$, $\overline{DOB}$, $\overline{CB}$, and $\overline{AB}$ are chords of circle $O$, $\overleftrightarrow{FDE}$ is tangent at point $D$, and radius $\overline{AO}$ is drawn. Sam decides to apply this theorem to the diagram: “An angle inscribed in a semi-circle is a right angle.”
Which angle is Sam referring to?
(1) $\angle AOB$
(2) $\angle BAC$
(3) $\angle DCB$
(4) $\angle FDB$
22. In the diagram below, $\overline{CD}$ is the altitude drawn to the hypotenuse $\overline{AB}$ of right triangle $ABC$.
Which lengths would not produce an altitude that measures $6\sqrt{2}$?
(1) $AD = 2$ and $DB = 36$
(2) $AD = 3$ and $AB = 24$
(3) $AD = 6$ and $DB = 12$
(4) $AD = 8$ and $AB = 17$
23. A designer needs to create perfectly circular necklaces. The necklaces each need to have a radius of $10$ cm. What is the largest number of necklaces that can be made from $1000$ cm of wire?
(1) $15$
(2) $16$
(3) $31$
(4) $32$
24. In $\triangle SCU$ shown below, points $T$ and $O$ are on $\overline{SU}$ and $\overline{CU}$, respectively. Segment $OT$ is drawn so that $\angle C \cong \angle OTU$.
If $TU = 4, OU = 5$, and $OC = 7$, what is the length of $\overline{ST}$?
(1) $5.6$
(2) $8.75$
(3) $11$
(4) $15$ | 2020-08-08 05:46:32 | {"extraction_info": {"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": 200, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8189263343811035, "perplexity": 244.35542172230822}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737289.75/warc/CC-MAIN-20200808051116-20200808081116-00039.warc.gz"} |
http://jacobw.xyz/posts/005_shuffle/ | Jacob Wood
Objective
Identify a small experiment and take it through the process of gathering data, modeling, and analyzing.
Folk wisdom claims 7 shuffles is sufficient to thoroughly mix up a deck of cards. This claim originates from a paper published in 1986 by David Aldous and Persi Diaconis and summarized in the New York Times in 1990.
Does my mediocre shuffling reflect this common wisdom? This post is an attempt to investigate that question.
# Collecting Data
The first step to assessing my shuffling performance is to investigate a few real world shuffles. To start I recorded 5 the results of 5 shuffles by hand to get a sense of the variation in the distribution. Each run involved shuffling the deck with a riffle shuffle but not aligning the left and right hand piles:
The resulting order was then recorded as a string of 1s (card from left hand) and 2s (card from right hand). The 5 initial shuffles were:
1112212122121212121212121212121212121212121212111221
1222212212212121221212121212121212222112112211221122
2221121212211212112121221211221212121122222112212122
2222122121221121212121122112211222121222211222112211
2221221221122221212112221122121212212211122122121221
Eyeballing here shows quite a bit of variation from shuffle to shuffle. I set out to record the results of 100 independent shuffles to start to quantify the distribution.
The most straightforward approach seemed to be recording the cards falling during the shuffling sequence in slow motion (phone camera can capture at 240fps). The video was recorded in a semi-reproducible environment to ease processing:
A sample of the resulting video looks like:
Even at 240fps some of the faster cards aren't captured in transit, and the ones that are captured are blurry and tough to detect with any sort of definable edge or feature. We can address this by instead capturing the face of the shuffle pile as it changes (the time constant between dropping new cards is much larger than the one associated with the drop itself).
## Shuffling - Angled Down
The new and improved data collection environment involved even more cardboard:
Initially, the video collected during a shuffle looks promising. It is easy to identify transitions between shuffled cards and which hand is dropping the card:
However, when we look a little closer we don't notice any funny business:
Which is a problem because there was indeed some funny business. The 3 of clubs was tucked between the 9 and 4 of spades.
You can make out the edge of the card in the video, but we never see the face because the shuffle isn't perfect and I release both the 3 of clubs and the 4 of spades at the same time.
This doesn't happen in every shuffle, but it does happen occasionally. We might argue that we can just throw out shuffles where we don't process 52 different cards to avoid the missed card measurement error. However, that would introduce a systematic bias into the measurement - we would be removing all the worst shuffles from the dataset and our resulting impression of our shuffling performance would be better than it should be.
## Post Shuffle Conveyor Belt
We had a lot of cardboard to spare and I had to try getting the power tools involved somehow. The gravity-fed rubber band and drill feeder worked, but was more trouble than it was worth.
## Post Shuffle Riffle
One way to address the systematic bias mentioned above is to decouple the event of the shuffle from the recording of it. That way, if we were to make an error that invalidates the recording, we have no reason to expect the error would alter the perceived distribution. One way to do this is to record the order of the cards after the shuffle. If we record the order before and after each shuffle we can back out the 1s and 2s that make up our riffle model.
If we take the cards and riffle them in front of the camera we can then look for cards in each frame and record the order.
This strategy is also prone to missing a card, but we can address the problem this time. We can record two different riffles and compare the resulting orders, using the knowledge from the second recording to fill in gaps or resolve discrepancies. If we are unable to order the cards with sufficient confidence we can drop that shuffle from the dataset without introducing bias.
### Transfer Learning Post Processing
Of course, we won't be processing the video by hand. I was hoping to use this project to do some in-the-wild machine learning so we will leverage that here. There are a few projects on Github doing playing card detection but they are generally looking for whole cards on specific backdrops. We will train a new model and make use of the controlled environment it is being deployed in.
#### Data Labelling
We'll start off by labelling a bit of the data we collected to train on. We'll use Python and OpenCV to flip through the video files and write frames:
import cv2
Most of the models we will be considering want small and square images as input. Fortunately, we riffled in a very consistent fashion so we can crop all the video frames in the same small square. We'll use 224x224 as our image size for now.
def prep_frame(f):
return cv2.resize(f[0:800, 700:1500], (224,224))
We can then run through the video frame by frame and label each image. To make things easier we will consider two different classification problems for each frame independently - suit and rank.
1. Read a new frame from a video
2. Crop the frame as dictated above
3. Show the frame and wait for a keypress
4. Upon keypress:
• If "c", "s", "h", "d", or "0" (for clubs, spades, hearts, diamonds, none) save to that directory with unique ID generated from frame counter
• If "q" then quit the program
• If other key then skip the frame and go to the next
1. Repeat until the end of the video is reached
cap = cv2.VideoCapture("INPUT.MOV")
i = 0
while True:
i+=1
if not ret:
print("End of video")
break
f = prep_frame(f)
cv2.imshow('frame', f)
key = cv2.waitKey(0)
if chr(key) in 'cshd0':
cv2.imwrite("suit_data/" + chr(key) + "/" + str(i) + ".jpg", f)
elif chr(key) in 'q':
break
cap.release()
cv2.destroyAllWindows()
After running through all the video frames we should have generated a bit of training data for each suit.
And we can verify the classifications look decent:
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import ImageGrid
import numpy as np
fig = plt.figure(figsize=(10., 6.))
grid = ImageGrid(fig, 111, nrows_ncols=(3, 5), axes_pad=0.1, share_all=True)
grid[0].get_yaxis().set_ticks([])
grid[0].get_xaxis().set_ticks([])
grid[0].set_title("C")
grid[1].set_title("D")
grid[2].set_title("H")
grid[3].set_title("S")
grid[4].set_title("0")
ims = []
for _ in range(3):
for s in "cdhs0":
im_file = random.choice(os.listdir("suit_data/"+s))
ims.append(im[...,::-1])
for ax, im in zip(grid, ims):
ax.imshow(im)
plt.show()
Rinse and repeat for the ranks of each card.
#### Model Training
Now that we have a subset of our data labelled we can train an algorithm to label the rest of the videos we will take.
There are plenty of models out there that do a great job of parsing images and learning their requisite features. We would like to take a model that is good at understanding images and have it learn to classify our specific images. This is the definition of transfer learning.
We'll follow along with the TensorFlow example detailed here for the most part.
First bring in a few packages we will need:
import os
import matplotlib.pylab as plt
import numpy as np
import tensorflow as tf
import tensorflow_hub as hub
The we can browse through the available models on TensorFlow Hub for a suitable classification model. Our problem is pretty easy, and we will have no trouble using a small MobileNet model. This model takes in 224x224 images (which is what we saved our training data at, so no need to resize).
model_name = "mobilenet_v2_100_224"
image_size = (224,224)
batch_size = 16
data_dir = "suit_data"
We'll use Keras as our framework. Keras provide a convenient API to split up our dataset:
def build_dataset(subset):
return tf.keras.preprocessing.image_dataset_from_directory(
data_dir,
validation_split=.20,
subset=subset,
label_mode="categorical",
seed=123,
image_size=image_size,
batch_size=1)
Which we can use to generate a training and a validation dataset:
train_ds = build_dataset("training")
val_ds = build_dataset("validation")
And record a few pieces of information before we change the datasets:
train_size = train_ds.cardinality().numpy()
val_size = val_ds.cardinality().numpy()
class_names = tuple(train_ds.class_names)
We will want to pre-process our image inputs and introduce some artificial warping to flesh out the data set.
First, the MobileNet model expects inputs from 0 to 1 and our current images contain pixel values from 0 to 255. We can fix that with a normalization layer that scales the data by 1/255. We'll want to do this to both the training and the validation data.
Next, we can move the images around a bit to resemble changes we might see in new data. These changes could consist of:
• rotation
• vertical translation
• horizontal translation
• zoom
• contrast
We will only want to apply these mutations to the training data.
normalization_layer = tf.keras.layers.Rescaling(1. / 255)
preprocessing_train = tf.keras.Sequential([
normalization_layer,
tf.keras.layers.RandomRotation(0.1),
tf.keras.layers.RandomTranslation(0, 0.2),
tf.keras.layers.RandomTranslation(0.2, 0),
tf.keras.layers.RandomZoom(0.2, 0.2),
tf.keras.layers.RandomContrast(0.1),
])
preprocessing_val = tf.keras.Sequential([
normalization_layer
])
Then we can take our data, stick it into batches, and apply the preprocessing described above to get it all ready to go. Note - we need to add a repeat() call to our training data to ensure we can make enough data during the training runs (this seems weird to me?).
train_ds = train_ds.unbatch().batch(batch_size)
train_ds = train_ds.repeat()
train_ds = train_ds.map(lambda images, labels:(preprocessing_train(images), labels))
val_ds = val_ds.unbatch().batch(batch_size)
val_ds = val_ds.map(lambda images, labels:(preprocessing_val(images), labels))
With the data ready to go we can prepare the model we are going to train. We need to add a few things to the main MobileNet model specified earlier to link everything together.
First, we need to add an input layer that is compatible with our RGB image size (224x224x3).
That can feed into the MobileNet model which gets downloaded from TensorFlow Hub.
We probably also want to include some dropout to prevent overtraining. 20% is a standard value to start with.
Finally, we need a Dense layer that will act as the classifier for the 5 different classes we have: none, clubs, diamonds, hearts, and spades. We will include some L2 regularization in this Dense layer to keep the kernel weights in check.
model = tf.keras.Sequential([
tf.keras.layers.InputLayer(input_shape=image_size + (3,)),
hub.KerasLayer(model_handle, trainable=True),
tf.keras.layers.Dropout(rate=0.2),
tf.keras.layers.Dense(len(class_names),
kernel_regularizer=tf.keras.regularizers.l2(0.0001))
])
Next we need to define how we want to train the model. This is done with model.compile() and some definitions.
• Optimizer: Stochastic Gradient Descent with a small learning rate should work just fine, but feel free to poke around
• Loss Function: Categorical Crossentropy is recommended when there are two or more labels that are one-hot encoded
• from_logits = True must be used when the outputs are not normalized (as is the case when we have not soft-maxed the outputs)
• label_smoothing takes our one-hot encoded outputs and smooths out the confidence a bit. Instead of encoding a club as [0,1,0,0,0] we encode it as, say, [0.01,0.96,0.01,0.01,0.01]. This can help regularization a bit.
• Metrics: We will just monitor accuracy here
model.compile(
optimizer=tf.keras.optimizers.SGD(learning_rate=0.005),
loss=tf.keras.losses.CategoricalCrossentropy(from_logits=True, label_smoothing=0.1),
metrics=['accuracy'])
Now we are ready to hit go on the training. We will train the model for 10 standard epochs and see how we're doing. This might take a few minutes.
steps_per_epoch = train_size // batch_size
validation_steps = val_size // batch_size
hist = model.fit(
train_ds,
epochs=10,
steps_per_epoch=steps_per_epoch,
validation_data=val_ds,
validation_steps=validation_steps).history
Epoch 1/10
45/45 [==============================] - 44s 908ms/step - loss: 1.1458 - accuracy: 0.5792 - val_loss: 0.9850 - val_accuracy: 0.6420
Epoch 2/10
45/45 [==============================] - 39s 884ms/step - loss: 0.7227 - accuracy: 0.8610 - val_loss: 0.6280 - val_accuracy: 0.9205
Epoch 3/10
45/45 [==============================] - 40s 879ms/step - loss: 0.5971 - accuracy: 0.9368 - val_loss: 0.4976 - val_accuracy: 0.9886
Epoch 4/10
45/45 [==============================] - 40s 878ms/step - loss: 0.5762 - accuracy: 0.9579 - val_loss: 0.5069 - val_accuracy: 0.9943
Epoch 5/10
45/45 [==============================] - 39s 875ms/step - loss: 0.5431 - accuracy: 0.9691 - val_loss: 0.4847 - val_accuracy: 0.9943
Epoch 6/10
45/45 [==============================] - 39s 877ms/step - loss: 0.5381 - accuracy: 0.9705 - val_loss: 0.4756 - val_accuracy: 1.0000
Epoch 7/10
45/45 [==============================] - 40s 878ms/step - loss: 0.5272 - accuracy: 0.9803 - val_loss: 0.4778 - val_accuracy: 0.9886
Epoch 8/10
45/45 [==============================] - 40s 879ms/step - loss: 0.5193 - accuracy: 0.9803 - val_loss: 0.4799 - val_accuracy: 0.9886
Epoch 9/10
45/45 [==============================] - 39s 878ms/step - loss: 0.5227 - accuracy: 0.9789 - val_loss: 0.4841 - val_accuracy: 0.9830
Epoch 10/10
45/45 [==============================] - 39s 873ms/step - loss: 0.5221 - accuracy: 0.9733 - val_loss: 0.4808 - val_accuracy: 0.9943
Tough to beat that. We probably could have gotten away with only 5 epochs, but oh well.
Finally, we can save the trained model:
model.save("suit_predictor")
#### Deploying the Model
Now we need to use our models to identify all the cards in a riffle as it goes by and back out the shuffled string of 1s and 2s.
We'll start by loading in our models and the classes the predictions represent:
ranks = ('0', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'j', 'k', 'q', 'z')
suits = ('0', 'c', 'd', 'h', 's')
image_size = (224, 224)
We also need to make sure we do the same preprocessing on new images that we did on previous ones. That means cropping, scaling, and normalizing:
def prep_frame(f):
return cv2.resize(f[0:800, 700:1500], (224,224))
normalization_layer = tf.keras.layers.Rescaling(1. / 255)
preprocessing_model = tf.keras.Sequential([normalization_layer])
def prep_input(inp):
f = prep_frame(inp)
arr = np.array([f[...,::-1].astype(np.float32)])
return preprocessing_model(arr)
It is also going to be helpful to label the images directly as we monitor some results. We will borrow the draw_text function from this stackoverflow answer:
def draw_text(img, text, font=cv2.FONT_HERSHEY_PLAIN, pos=(0, 0), font_scale=3, font_thickness=2, text_color=(0, 0, 0), text_color_bg=(255, 255, 255)):
x, y = pos
text_size, _ = cv2.getTextSize(text, font, font_scale, font_thickness)
text_w, text_h = text_size
cv2.rectangle(img, pos, (x + text_w, y + text_h), text_color_bg, -1)
cv2.putText(img, text, (x, y + text_h + font_scale - 1), font, font_scale, text_color, font_thickness)
Now we can see how we did by making predictions on new video frames. To prevent spurious classifications we will only record a card if we ID it two frames in a row.
cap = cv2.VideoCapture("INPUT2.MOV")
last_card = ""
card_order = []
viz = True
while True:
if not ret:
print("No frame")
break
inp = prep_input(frame)
preds = suit_model.predict(inp)
suit_pred = suits[np.argmax(preds)]
preds = rank_model.predict(inp)
rank_pred = ranks[np.argmax(preds)]
if rank_pred!="z" and suit_pred!="0":
card = rank_pred + suit_pred
if card == last_card and (len(card_order)==0 or card_order[-1] != card):
card_order.append(card)
if card == last_card and viz:
draw_text(frame, rank_pred + suit_pred, pos=(800, 500))
last_card = card
if viz:
cv2.imshow('frame', frame)
if cv2.waitKey(3) == ord('q'):
break
cv2.destroyAllWindows()
If the video we processed above has two independent riffles in it we should end up with a card_order vector that is, ideally, 104 elements long with 52 unique elements. That probably won't be the case. Instead, we probably get something that looks like:
0d
8c
jc
0h
jh
9s
9d
kh
.
.
.
0s
9h
9d
as
0d
8c
jc
0h
jh
8s
9s
9d
kh
.
.
.
0s
9h
Our two most robust measurements should be the last card of the first riffle and the last card of the second riffle. We can use these to split the vector in half and start lining things up (we could also detect the break between riffle 1 and riffle 2 using the timestamps in the video). In the example above the last card would be the 9h.
After 9h split and align:
9d
as
0d
0d 8c
8c jc
jc 0h
0h jh
jh 8s
9s 9s
9d 9d
kh kh
. .
. .
. .
0s 0s
9h 9h
Now we can crawl through the two vectors and look for similar neighbors. On the first pass we will just note what is missing and mark it with an empty character:
num_el = len(set(card_order))
last_card = card_order[-1]
r1 = card_order[:card_order.index(last_card)+1]
r2 = card_order[card_order.index(last_card)+1:]
i = 0
while i < num_el:
if r1[i] == r2[i] or r1[i]==" " or r2[i]==" ":
i += 1
continue
elif i > len(r1):
r1.append(" ")
elif i > len(r2):
r2.append(" ")
elif r1[i] == r2[i+1]:
r1.insert(i," ")
elif r2[i] == r1[i+1]:
r2.insert(i," ")
elif r1[i] == r2[i+2]:
r1.insert(i," ")
elif r2[i] == r1[i+2]:
r2.insert(i," ")
i = 0
[print(i,j) for i,j in zip(r1,r2)]
After alignment first pass:
9d
as
0d 0d
8c 8c
jc jc
0h 0h
jh jh
8s
9s 9s
9d 9d
kh kh
0s 0s
9h 9h
Then we can crawl through again and determine what to do with the empty characters. We either toss them or infill based on how many other entries for that particular card we see elsewhere:
ord = []
for (e1,e2) in zip(r1, r2):
if e1 == " ":
if card_order.count(e2) > 1:
continue
else:
ord.append(e2)
elif e2 == " ":
if card_order.count(e1) > 1:
continue
else:
ord.append(e1)
elif e1 == e2:
ord.append(e1)
as
0d
8c
jc
0h
jh
8s
9s
9d
kh
0s
9h
Almost there. Our last step is to link two deck orders and back out the shuffle sequence. For example, we might see these two deck orders in a row:
as 8s
0d as
8c 0d
jc 9s
0h 8c
jh 9d
8s kh
9s jc
9d 0h
kh jh
0s 0s
9h 9h
To back out the order we first find where the cut happened and then iterate through the resulting deck to see which hand each card came out of:
def get_shuffle(o1, o2):
i = 0
for e in o2:
if e==o1[i]:
i+=1
else:
cut_loc = i
lh = o1[:cut_loc]
rh = o1[cut_loc:]
o = ""
il = 0
ir = 0
for e in o2:
if il<len(lh) and e==lh[il]:
o += "1"
il += 1
elif e==rh[ir]:
o += "2"
ir += 1
else:
ValueError("not feasible shuffle result")
return o
o1 = ["as","0d","8c","jc","0h","jh","8s","9s","9d","kh","0s","9h"]
o2 = ["8s","as","0d","9s","8c","9d","kh","jc","0h","jh","0s","9h"]
get_shuffle(o1,o2)
'211212211122'
Whew. A lengthy process but the result is a decent data collection pipeline.
# Building a Shuffling Model
A recording of 100 of my shuffles can be found here.
This part of the project is all done in Julia.
## Data Exploration
Bring in a few packages
using Plots, StatsPlots
using Random
using Distributions
using StatsBase
using KernelDensity
using Trapz
And read in the data from the recorded file into a vector of vectors of Int.
S_rec = [[parse(Int,c) for c in l] for l in readlines("rec.txt")]
One way we can "score" a shuffle is by counting the number of card runs there were. A perfect shuffle, with two piles of 26 cards interwoven one at a time, would score 52 on this metric.
function score_shuffle(S)
score = 1
for i in 2:52
if S[i] != S[i-1]
score += 1
end
end
score
end
scores = score_shuffle.(S_rec)
plot(scores, st=:scatter, legend=false, smooth=true,
xlabel="Row", ylabel="Score")
Ooph. A large range of performance and definitely got tired over time. But is this any good? It is tough to evaluate without context. The Gilbert-Shannon-Reeds (GSR) model dates back to 1955 and is the de-facto distribution used to model a riffle shuffle. The model has two steps:
1. Split the deck into left and right piles ($A$ and $B$) according to a binomial distribution
2. Drop the cards one at a time with probability $P(A) = A/(A+B)$
This can be implemented in Julia:
function gsr_shuffle()
cut = rand(Binomial(52),1)[1]
nl = cut
nr = 52 - cut
out = zeros(Int,52)
for i in 1:52
pl = nl / (nl + nr)
if rand() < pl
out[i] = 1
nl -= 1
else
out[i] = 2
nr -= 1
end
end
out
end
Now we can make a bunch of GSR-shuffled decks and see how our scores compare.
S_scores = score_shuffle.(S_vec)
GSR_vec = [gsr_shuffle() for _ in 1:10000]
GSR_scores = score_shuffle.(GSR_vec)
histogram(scores, label="Me", bins=15:2:52, lw=2, alpha=0.5, norm=:probability)
histogram!(GSR_scores, label="GSR", bins=15:2:52, lw=2, alpha=0.5, norm=:probability,
xlabel="Score", ylabel="Fraction")
Looks like we outperform the shuffling model by quite a bit on this metric! Our mean score turns out to be 33 vs 27 for the GSR shuffle. This indicates we should probably build our own model instead of assuming the GSR.
## Building a Model
### Deck Split
The first action in shuffling is splitting the deck into two halves. Ideally you end up with two stacks of 26 cards. The standard approach to modeling this action (the approach taken in the GSR model) is to draw the card split from a binomial distribution. Let's see if that looks decent for us.
split_vec = [count(s.==1) for s in S_vec]
histogram(split_vec, bins=16.5:1:35.5, xticks=17:35, normalize=:pdf, label="observed")
plot!(Binomial(52), st=:line, xlims=(17,35), label = "expected (binomial)",
xlabel="Number of Cards in Left Hand", ylabel="Fraction of Cases")
The binomial distribution is not a great fit here. 84/100 cases I ended up with <26 cards in my left hand and the distribution is much sharper (centered around 24) than the binomial would indicate. We should fit a different distribution to this action.
Our distribution is discrete and univariate over the integers from (let's say) 16 to 36. We can come up with a few ways of generating the underlying distribution:
• Use the measured histogram
• Fit a known discrete distribution to the data
• Fit a known continuous distribution to the data and round
• Use the data to generate a kernel density estimator
#### Measured Histogram
The most straightforward way to translate our measurements into a discrete probability distribution is to assume the data directly describes the underlying distribution. This assumption is most likely to hold true when you have a lot of measured data that has borne out the entirety of the underlying distribution. This assumption may appear to be true when the measured histogram is smooth and well-shaped.
The data we have shows a 1% chance of drawing a 30, and a 0% chance of drawing a 29. This is a small discrepancy but seems unlikely to me to be true.
The resultant PMF would be identical to the histogram of measurements:
#### Known Discrete Distribution
The Distributions.jl package provides a collection of distributions and the ability to fit them to experimental data using (usually) maximum likelihood estimation. The distributions available for this functionality are shown below.
• Binomial(52) - would result from drawing each card as either right or left with a fixed probability
• Binomial(20) - would result from only randomly drawing the 20 middle cards with a fixed probability (and assuming the other 32 are split 16 left, 16 right)
• Discrete Uniform - would result if every outcome in the possible solution space has the same probability
• Geometric - usually interpreted as the number of trials needed for a single outcome to materialize. Tough to apply in this case.
• Poisson - the probability of a given number of independent events occurring in a specific period of time. We don't expect card dropping events to be independent.
Among these distributions we expect the binomial (especially the binomial with a smaller number of chance draws) to perform the best. And indeed that is true, the Binomial(20) distribution (which resulted in a 39% chance that each of the 20 randomly drawn card ends up in my left hand) fits the data decently well as seen in the chart below.
One thing we might (definitely) want to do is truncate the distribution - we certainly don't ever want our split to be <0 or >52, and we probably wouldn't shuffle the cards if we had <16 or >36 cards in one hand, it just doesn't feel right. We'll apply 16-36 truncation in all the results going forward.
split_vec = [count(s.==1) for s in S_vec]
histogram(split_vec, bins=15.5:36.5, norm=:pdf, label="Measured", alpha=0.5)
f = fit_mle(Binomial, 52, split_vec)
f = truncated(f,16,36)
plot!(f, st=:line, label = "Binomial(52)", lw=2)
x = 16:36
f = fit_mle(Binomial, x[end]-x[1], split_vec.-x[1])
plot!(x, st=:line, pdf.(f, x.-x[1]), label = "Binomial(20)", lw=2)
distributions = [
(DiscreteUniform, "DiscreteUniform")
(Geometric, "Geometric")
(Poisson, "Poisson")
]
for d in distributions
f = fit_mle(d[1], split_vec)
f = truncated(f,16,36)
plot!(f, st=:line, marker=false, label = d[2], lw=2)
end
plot!(xlabel="Number of Cards in Left Hand", ylabel="Fraction of Cases", xlims=(16,36), xticks=16:36)
Note - discrete distributions are plotted here as continuous for ease of viewing.
#### Known Continuous Distribution
Similarly to the discrete options, Distributions.jl provides a host of continuous distributions that can be easily fit to our experimental data.
distributions = [
(Exponential, "Exponential")
(LogNormal, "LogNormal")
(Normal, "Normal")
(Gamma, "Gamma")
(Laplace, "Laplace")
(Pareto, "Pareto")
(Poisson, "Poisson")
(Rayleigh, "Rayleigh")
(InverseGaussian, "InverseGaussian")
(Uniform, "Uniform")
(Weibull, "Weibull")
]
histogram(split_vec, bins=15.5:36.5, norm=:pdf, label="observed", alpha=0.5)
for d in distributions
f = fit_mle(d[1], split_vec)
f = truncated(f,16,36)
plot!(f, st=:line, label = d[2], lw=2)
end
plot!(xlabel="Number of Cards in Left Hand", ylabel="Fraction of Cases", xlims=(16,36), xticks=16:36)
We can discount quite a few of these right off the bat and then we are left with the LogNormal, Normal, Gamma, and InverseGaussian distributions.
• Log Normal - usually the result of an event which is the product of multiple independent random variables
• Normal - usually the result of an event which is the sum of multiple independent random variables
• Gamma - can be used to model wait times - such as when will the nth event occur?
• Inverse Gaussian - if a normal distribution describes possible values of a random walk process at a fixed time, the inverse gaussian describes the possible times at which we might see a specific value of a random walk process.
In order to round these to the proper domain we can integrate them over the rounding range of each integer. In the end these all look practically identical:
distributions = [
(LogNormal, "LogNormal")
(Normal, "Normal")
(Gamma, "Gamma")
(InverseGaussian, "InverseGaussian")
]
histogram(split_vec, bins=15.5:36.5, norm=:pdf, label="observed", alpha=0.5)
for d in distributions
f = fit_mle(d[1], split_vec)
f = truncated(f,16,36)
#Integrate
x_ep = 15.5:36.5
x_cp = 16:36
y_cp = [cdf(f, x_ep[i+1]) - cdf(f, x_ep[i]) for i in 1:length(x_ep)-1]
plot!(x_cp, y_cp, st=:line, label = d[2], lw=2)
end
plot!(xlabel="Number of Cards in Left Hand", ylabel="Fraction of Cases", xlims=(16,36), xticks=16:36)
You might be able to make a case for any of these. They all can fit the data we have pretty well. If pressed I might argue the normal distribution makes the most sense here so that's what I'll assume going forward.
#### Kernel Density Estimate
The other option we have when generating a distribution here is to take the measured histogram and apply some smoothing via kernel density estimation. This is a great choice if we ever just want to make sure we are drawing from something very close to the measured data. That might be nice if the underlying mechanisms are poorly understood or too complex to attempt to model.
The Julia package KernelDensity.jl makes this process easy. The main parameter of interest is the kernel bandwidth we use to smooth with - larger values will smooth over spurious measurements at the expense of pulling down peaks. Smaller measurements won't do much smoothing work and may capture more fine structure than you would hope.
This is also a continuous estimator - so we need to integrate over the relevant region to get a probability mass function over the integers.
histogram(split_vec, bins=15.5:36.5, norm=:pdf, label="observed", alpha=0.5)
for bw = [0.2, 0.5, 1.0, 2.0]
k = kde(split_vec, boundary=(26-10, 26+10), bandwidth=bw)
x = 16:0.01:36
y = pdf(k, x)
x_ep = 15.5:36.5
x_cp = 16:36
y_cp = [trapz(x[(x.>x_ep[i]) .& (x.<x_ep[i+1])], y[(x.>x_ep[i]) .& (x.<x_ep[i+1])]) for i in 1:length(x_ep)-1]
plot!(x_cp, y_cp, lw=2, label="BW = " * string(bw))
end
plot!(xlabel="Number of Cards in Left Hand", ylabel="Fraction of Cases", xlims=(16,36), xticks=16:36)
A bandwidth of 0.5 smooths over the spurious 29-30 bump while retaining most of the rest of the structure, it probably wins the eye test here.
#### Winning Distribution
We can compare the leading solutions from each of the previous 4 sections:
At this point it comes down to either our understanding of the underlying mechanism that is driving the distribution or aesthetic preference. Let me know if you come up with a good causal model. In the meantime, I'll use the normal distribution with a mean of 23.8 and a standard deviation of 1.8.
### Dropping Cards
After splitting the deck in two we start to drop cards from either our left or right hand. The GSJ model predicts that this will happen probabilistically, with the probability of dropping from either the left or the right according solely to the fraction of remaining total cards that are currently in that hand. Is this a good model for us?
#### GSR Calibration
We can view GSR as making a prediction each time we are about to drop a card. We can then look at what actually happened, and see if things that were supposed to happen 25% of the time actually happened 25% of the time. This kind of analysis is commonly referred to as model calibration.
We don't have perfect point estimates here - so we need to bucket the probabalities. Anything within the blue squares below is considered likely "calibrated". For example, the bottom left point says that whenever the model predicted an event to happen between 0-10% of the time, it actually happened 0% of the time.
The points are sized roughly corresponding to how many predictions they represent. We notice that the model does a pretty good job overall - the only time it is off by more than 10% is the 10-20% bucket which is only comprised of 50 observations. Not bad.
probs = Float64[]
outcs = Bool[]
cards = Int[]
for S in S_vec
for (i,s) in enumerate(S)
nl = count(S[i:end].==1)
nr = count(S[i:end].==2)
prob = nl/(nl+nr)
outc = (s == 1)
push!(probs, prob)
push!(outcs, outc)
push!(cards, nl+nr)
end
end
edges = 0:0.1:1.0
binC = 0.05:0.1:0.95
h = fit(Histogram, probs, edges)
binindices = StatsBase.binindex.(Ref(h), probs)
outF = Float64[]
dps = Int[]
for i in 1:length(binC)
idx = binindices .== i
push!(outF, count(outcs[idx]) / count(idx))
push!(dps, count(idx))
end
X = vcat([[x-0.05,x-0.05,x+0.05,x+0.05,x-0.05,NaN] for x in binC]...)
Y = vcat([[x-0.05,x+0.05,x+0.05,x-0.05,x-0.05,NaN] for x in binC]...)
plot(X, Y, fill=true, label="Perfectly Calibrated", aspect_ratio=1, size=(500,500),
xlabel = "Predicted Probability",
ylabel = "Observed Probability",
xticks = edges, yticks=edges)
outF[dps.<10] .= 0
dps[dps.<10] .= 0
binX = [mean(probs[binindices.==i]) for i in 1:length(dps)]
scatter!(binX, outF, markersize=log.(dps/maximum(dps)).+9, label="Actual", legend=:topleft, xlims=(-0.05,1.05), ylims=(-0.05,1.05))
#### Differences from GSR
What are some ways in which we might differ from the GSR model? Perhaps we systematically favor the right or left hand at certain times:
side_ev = [mean([s[i] for s in S_vec]) for i in 1:52] .- 1
plot(side_ev, label="observed", ylims=(0, 1), ylabel="← Left Hand Right Hand →", xlabel="Card Number", lw=2)
GSR_side_ev = [mean([s[i] for s in GSR_vec]) for i in 1:52] .- 1
plot!(GSR_side_ev, label="GSR model", lw=2)
It looks like we do systematically favor the right hand at the beginning and end of the shuffle.
Of particular interest in shuffling efficiency is runs of cards, do we have more or longer runs than the GSR model would predict?
[IN WORK]
# Ingredients
• pngquant
• Gimp
• ffmpeg
• jpegoptim
• iPhone 11
• cardboard
• Bicycle playing cards
• Python 3.9.7
• OpenCV 4.5.4
• Matplotlib 3.3.4
• Numpy 1.19.5
• Tensorflow 2.7.0
• Tensorflow_hub 0.12.0
• Julia 1.7.1
• Distributions.jl | 2022-09-27 06:52:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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, "math_score": 0.42282170057296753, "perplexity": 3064.6049922091224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334992.20/warc/CC-MAIN-20220927064738-20220927094738-00343.warc.gz"} |
https://cs.stackexchange.com/questions/108712/the-barbers-paradox-first-order-logic-formalization/108716 | # The barbers paradox first order logic formalization
I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.
Sentence: there exists a barber who shaves all the people that don't shave themselves.
My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"
And now it seems to be easier to substitute with variables so:
$$\exists x(\lnot S(x,x) \rightarrow S(b,x))$$
where $$S(x,x)$$ is shaving(verb), $$x$$ are the persons (who don't shave themselves), and $$b$$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.
First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $$x$$ such that if $$x$$ does't shave themself, the barber does." Instead, what you want to say is that "For all $$x$$ it holds that if $$x$$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.
Second, instead of using a constant name $$b$$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.
Whith this, the sentence becomes
There is a thing $$y$$ which is a barber and such that for all people $$x$$ who don't shave themselves, $$y$$ shaves $$x$$.
which translates as
$$\exists y (B(y) \land \forall x (\neg S(x,x) \to S(y,x)))$$
• thank you for your answer @lemontree. before marking this answer, i am just wondering about the difference between both answers, could you please the difference between using the verb B(y) in comparison to saying exists b like in the other answer? thank you very much for your explaining – hps13 May 3 at 6:10
• $b$ is just a (more or less arbitrary) name for an object that the formula says nothing about other than that this object shaves everyone who doesn't shave themselves. But we additionally want to express that this object is a barber, which we do by $B(b)$. – lemontree May 3 at 10:28
• The other queston is whether to use a constant name $b$ (without a quantifier) or a vaiable with an existential introduction ($\exists y$/$\exists b$; the name of the variable does not matter in this case.) It is better to use $\exists$, because the English sentence says "There is a barber", while using a constant name $b$ just talks about "the barber", which excludes the possibility that there might be more than one such barber. – lemontree May 3 at 10:29
You're not being asked to do any inference; just to express something.
there exists a barber who shaves all the people that don't shave themselves
translates directly as
$$\exists b\, \forall x\,\big(\neg S(x,x) \rightarrow S(b,x)\big)\,.$$
(There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)
Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $$x$$ be that person, so $$\neg S(x,x)$$ is false, so $$\neg S(x,x)\rightarrow\text{anything}$$ is true.
• thank you so much for your explanation! – hps13 May 3 at 6:09 | 2019-08-21 03:09:03 | {"extraction_info": {"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6293142437934875, "perplexity": 466.5600770352139}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00441.warc.gz"} |
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Analyzing Equations and Inequalities
Top
Sub Topics While analyzing equations and inequalities we focus on some points. For analyzing equations and inequalities we follow some points: 1. First solve the expression. 2. Then understand and use the Properties of Real Numbers. 3. Then simplify the equation and inequalities. 4. At last find the absolute values equation and inequality. If equal sign is present in the equations, and equation is used to find the unknown variables. One method for solving the equation is substitution method. Suppose we have an equation p + q = 2p – 1; We know that equation is used to find the unknown variable. Here in this equation we solve for unknown variable ‘p’; On solving the equation for unknown variable ‘p’ we get: ⇒ p + q = 2p – 1; ⇒p = q + 1; Now we use substitution method: Now substitute the value of ‘p’ in the equation: ⇒ p + q = 2p – 1; Put p = q + 1; ⇒ (q + 1) + q = 2(q + 1) – 1; Now we solve the value of unknown variable ‘q’. ⇒q + 1 + q = 2q + 2 – 1; Now combine like terms: ⇒2q – 2q = 1 – 1; ⇒0; This is how we solve the equation by substitution method. And in case of inequality, if we have two variables ‘m’ and ‘n’, then there are conditions for analyzing the inequality equations. Suppose ‘m’ is less than equal to ‘n’; ⇒ m ≤ n; Inequality present; ‘m’ is greater than equal to n then we can write: ⇒m ≥ n; And if ‘m’ is greater than n then we can write; ⇒m > n; And if ‘m’ is smaller than ‘n’ then we can write; ⇒m < n; These all are the properties of inequality. If in any equation these symbols are present then we can say there is an inequality in the expression.
Expressions and Formulas
An expression is defined as the Combination of two or more terms and can be written in the form of Relations between variables and constant values. The only difference between the expression and the equation is that the sign of equal to is missing in the expression, which exists in the equation.
Now, let us learn how to solve expressions and formulas. In order to solve the expressions and formulas, we will put the value of the variable‘s’ in the expression and thus get the numerical value of the expression. It will be clearer with the following example.
Let us consider the expression 2ab + 4c, where we have the value of ‘a’ as 2, b = 3 and c = 4.
Now we will place these numerical values in the given expression, so it will become:
2 * 2 * 3 + 4 * 4
= 12 + 16 = 28.
Sometimes the expressions are in form of powers of the variables too. In such case, we proceed as follows:
In case we have 2x^2 + 4y as an expression, where x = 2 and y = 3, then we will solve the expression as follows :
2 * 2^2 + 4 * 3
= 2 * 4 + 12
= 8 + 12 = 20
Now we will look at the formulas. Formulas are used to find the values if the variables are known. In case, if length and breadth of the Rectangle are known and we need to find the values of the perimeter of the rectangle and area of the rectangle, then for this we will apply the formula as follows:
Perimeter = 2 * (length + breadth)
This is called solving the values using the variables.
Solving Equations
When we have to solve equations, we Mean that we are finding the value of the variable, which satisfies the equations. The equations can be Linear Equations of one variable or the pair of Linear Equations with Two Variables. Firstly we are solving equations with one variable.
In order to solve the equation, we need to use different methods. First method will be to get the value of the variable by hit and trial method. According to this method, we will find the value of the variable by trying different values for the variables in the equation. On solving if we get the left side of the equation equal to the right side of the equation, then we say that the particular value satisfies the equation. As we go on trying different values which satisfies the equations, so we call it hit and trial method.
Another method to solve the equation is by the Algebra method. According to algebra method, we say that the variables of the equation are taken to one side of the equation and the constants are taken to another side of the equation. Thus we say that on solving we get the value for the variable of the equation.
Now if we take the pair of linear equations with two variables, these equations can be solved either by the graphical method or by the method of algebra.
In case we solve the pair of equations and get the value of the two variables in the equation, we say that the two values of the variables will satisfy both the equations. If we draw the graph for the two linear equations, we will get the solution of the equation by finding the meeting Point of the equation. The coordinates of the meeting point is the solution to the given equation.
Solving Absolute Value Equations
By the term absolute value, we Mean that we are going to find the value of the equation, which is always a positive number. All absolute values have magnitude, but no direction and the absolute value is represented by | | sign, which we call as the modulus sign.
When we say the term solving absolute value equations, we mean that we are going to find the value of the variable in the form of magnitude only. Suppose we are given the equation as follows:
Y = | x|
Here we observe that whatever be the value of x, either a positive value or the negative value, the result will be a positive value, as the sign of modulus is placed with it. Let us take x as a positive number first, suppose 5. In such a situation, we say that |5| is a positive value. Thus the value of y = 5, for x = 5
Now let us take the value of x = -5, in case the value of x = -5, the value of y = | -5 |, y = 5
So we observe that either the value of x is positive or it is a negative Integer, we come to the conclusion that the value of y will be the modulus value of x, thus it will be a positive number.
So for each two values of x, we will get one value of y.
Still we have one situation, where for one value of x, we have one value of y and this value of x is 0. Since we know that zero either a positive number or a negative number will always be the same value, which is represented by the origin. The graph so plotted for this function will be in the form of English letter V, which passes through the origin.
Properties of Real Numbers
The series of Real Numbers include the numbers which can be both rational and Irrational Numbers. These numbers are called real numbers as they are not imaginary. Let us look at the properties of real numbers one by one:
First we look at the closure property of real numbers, according to which we say that if we have two real numbers say a and b, then we say that the sum of two real numbers is also a real number. It means that the closure property holds true for the addition of real numbers.
Similarly, if we take the two real numbers ‘a’ and ‘b’, then the difference of two real numbers is also a real number. It means that the closure property holds true for the subtraction of the real numbers.
On the other hand, if we take the two real numbers ‘a’ and ‘b’, then the multiplication of two real numbers is also a real number. It means that the closure property holds true for the product of the real numbers. Similarly, if we take the two real numbers ‘a’ and ‘b’, then the division of two real numbers is also a real number. It means that the closure property holds true for the quotient of the real numbers.
Further looking at the properties of real numbers, we say that the associative and the Commutative Property of the addition and multiplication of real numbers holds true. But these two properties do not hold true for the subtraction and division of the real numbers. It means if we have the real numbers a, b and c, then by commutative property of real numbers, we have :
(a + b) = (b + a) and a * b = b * a
But a – b < > b – a and a/ b < > b / a
Solving Inequalities Algebraically
Talking into consideration Solving Inequalities algebraically such as : x + 3 > 0, was a simple and easy expression and we must remember that when the expression is multiplied by any of the negative number, then the sign of inequality changes.
To solve inequalities, we Mean finding all of its solutions. To solve the inequality, we need to find the value of the variable. So we say that a solution of an inequality is a number which when substituted for the variable, makes the existence of the inequality a true statement. Let us take the following inequality:
X – 2 > 5
Now if we substitute 8 for x, the inequality becomes 8 – 2 > 5. So we say that x = 8 is a solution of the inequality. On the other hand, we will substitute -2 for the value of x gives the false statement, which can be expressed as:
(-2) – 2 > 5.
So we say that x = -2 is not a true statement. So we say that x = -2 is not the solution to the inequality. We must remember that inequality always has many solutions. In order to solve the equations, there are certain manipulations of the inequality which does not change the solutions. We must remember certain rules for the solution of inequalities:
1. On adding or subtracting the same number on both sides, makes the inequality exactly same. So we say that the equation x – 3 > 5 is same as x – 3 + 3 > 5 + 3
Or x > 8 is the solution to the inequality.
2. In case we change the position of the left and the right side of the inequality, the sign of inequality changes, thus if we have:
X + 2 > 4,
Or 4 < x + 2 .
Solving Absolute Value Inequalities
Absolute value can be defined as measure of how further a number is from 0. For example: ‘10’ is 10 away from zero and -50 is 50 away from zero. Absolute value of 0 is 0 and absolute value of 5 is 5. These all are examples of Absolute Value Function. Negative Numbers are not included in absolute value function. Suppose we have any negative number then it is necessary to remove negative sign from number. So we take only positive values and zero in case of absolute function. Absolute value function is represented by the symbol '|'. This symbol is known as bar. If we put any negative number between this symbol then we get positive number outside this symbol. Now we will see process of Solving Absolute Value Inequalities.
If (<,> <, >) these symbols are present in any expression then we can say that equation contains inequality in it. Let's see how to solve absolute value inequalities. Suppose we have | 2p + 3 | < 6, absolute value inequality.
Solution: Given inequality | 2p + 3 | < 6, now first we will solve linear inequality. So we can write given inequality as:
=> - 6 < 2p + 3 < 6, it means 2p + 3 is greater than -6 and less than 6. Now subtract 3 from inequality. On subtracting we get:
=> - 6 – 3 < 2p + 3 – 3 < 6 – 3, on further solving we get:
=> -9 < 2p < 3,
Now divide whole inequality by 2, on dividing we get:
=> - 9 / 2 < p < 3 / 2,
On solving | 2p + 3 | < 6 we get –9/ 2 < p < 3 / 2. This is how we solve absolute value inequalities. | 2017-03-29 09:18:47 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8109905123710632, "perplexity": 174.54645100251886}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218190236.99/warc/CC-MAIN-20170322212950-00266-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/9795/what-is-the-most-natural-definition-of-the-weak-hypercharge-coupling-constant-if | # What is the most natural definition of the weak hypercharge coupling constant if grand unification is wrong?
A tricky question. Here is the famous graph of the running of the three coupling constants in the standard model: http://www-ekp.physik.uni-karlsruhe.de/~deboer/html/Forschung/unification_eng.eps .
The graph shows, in its top curve, the running of the coupling constant alpha_1. This is the coupling of the weak hypercharge coupling constant for the weak hypercharge group U(1)_Y, which is one of the three gauge groups of the standard model of particle physics.
But there is a tricky detail. In that curve, alpha_1 is multiplied with 5/3. This factor 5/3 comes from the assumption that GUTs are valid. The factor ensures that the various group traces of U(1)_Y, SU(2) and SU(3) are normalized in the correct way when they form the SU(5), SO(10) or any other grand unification gauge group.
In the case that grand unification is wrong, the factor 5/3 cannot be deduced. Which factor would be natural in this case?
Clarification added, after remarks by Lubos Motl: it is assumed in the question that the usual definition of the weak hypercharge is used, Y_W=2(Q-T_3), in which left-handed quarks have hypercharge 1/3.
-
Of course, if you assume a normalization with no relationship between any groups where the hypercharge $U(1)$ is normalized just like any other $U(1)$ would be, and your goal is to make the formulae as simple as possible on the paper (which is not really a physical criterion), then $5/3$ is replaced by $1$. You just omit the $5/3$ factor. But this is a kind of vacuous statement because one may only compare the fine-structure constants of the different group factors if there is some relationship between them which is either grand unification or plays the same role.
One more preemptive comment: at low energies, it is not true that the hypercharge fine-structure constant renormalized by another simple factor such as $5/3$ yields the electromagnetic fine-structure constant. At the GUT scale, similar relations exist but the electromagnetic fine-structure constant is not well-defined there. | 2013-05-21 19:09:38 | {"extraction_info": {"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, "math_score": 0.8959905505180359, "perplexity": 252.66437060251585}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368700438490/warc/CC-MAIN-20130516103358-00090-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://scikit-learn.org/dev/modules/generated/sklearn.metrics.precision_recall_curve.html | # sklearn.metrics.precision_recall_curve¶
sklearn.metrics.precision_recall_curve(y_true, probas_pred, *, pos_label=None, sample_weight=None)[source]
Compute precision-recall pairs for different probability thresholds.
Note: this implementation is restricted to the binary classification task.
The precision is the ratio tp / (tp + fp) where tp is the number of true positives and fp the number of false positives. The precision is intuitively the ability of the classifier not to label as positive a sample that is negative.
The recall is the ratio tp / (tp + fn) where tp is the number of true positives and fn the number of false negatives. The recall is intuitively the ability of the classifier to find all the positive samples.
The last precision and recall values are 1. and 0. respectively and do not have a corresponding threshold. This ensures that the graph starts on the y axis.
Read more in the User Guide.
Parameters
y_truendarray of shape (n_samples,)
True binary labels. If labels are not either {-1, 1} or {0, 1}, then pos_label should be explicitly given.
probas_predndarray of shape (n_samples,)
Target scores, can either be probability estimates of the positive class, or non-thresholded measure of decisions (as returned by decision_function on some classifiers).
pos_labelint or str, default=None
The label of the positive class. When pos_label=None, if y_true is in {-1, 1} or {0, 1}, pos_label is set to 1, otherwise an error will be raised.
sample_weightarray-like of shape (n_samples,), default=None
Sample weights.
Returns
precisionndarray of shape (n_thresholds + 1,)
Precision values such that element i is the precision of predictions with score >= thresholds[i] and the last element is 1.
recallndarray of shape (n_thresholds + 1,)
Decreasing recall values such that element i is the recall of predictions with score >= thresholds[i] and the last element is 0.
thresholdsndarray of shape (n_thresholds,)
Increasing thresholds on the decision function used to compute precision and recall. n_thresholds <= len(np.unique(probas_pred)).
plot_precision_recall_curve
Plot Precision Recall Curve for binary classifiers.
PrecisionRecallDisplay
Precision Recall visualization.
average_precision_score
Compute average precision from prediction scores.
det_curve
Compute error rates for different probability thresholds.
roc_curve
Compute Receiver operating characteristic (ROC) curve.
Examples
>>> import numpy as np
>>> from sklearn.metrics import precision_recall_curve
>>> y_true = np.array([0, 0, 1, 1])
>>> y_scores = np.array([0.1, 0.4, 0.35, 0.8])
>>> precision, recall, thresholds = precision_recall_curve(
... y_true, y_scores)
>>> precision
array([0.66666667, 0.5 , 1. , 1. ])
>>> recall
array([1. , 0.5, 0.5, 0. ])
>>> thresholds
array([0.35, 0.4 , 0.8 ]) | 2021-06-23 18:54:17 | {"extraction_info": {"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, "math_score": 0.49530333280563354, "perplexity": 2905.257038486518}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00019.warc.gz"} |
https://physics.stackexchange.com/questions/483143/initial-speed-is-zero-and-so-is-power | # Initial speed is zero and so is power?
If I want to accelerate something from standstill to max speed, with a constant force (acceleration and mass don't change), the equation P = F * v would say that in the beginning we use 0 W power. How is that possible? Since power is the rate of transference of energy to the body (J/s), it would appear that that rate should be constant.
Why is it that a fast moving body requires more power to accelerate than a still body? And if we would like to know how much power we need to accelerate that body, should we then use the maximum speed?
• Do you know calculus? – David White May 29 '19 at 16:08
If you want to accelerate a body from $$v$$ to $$v+\Delta v$$, the associated change in energy is
$$\Delta E=E\left(v+\Delta v\right)-E\left(v\right)=\dfrac{1}{2}m\left(v+\Delta v\right)^{2}-\dfrac{1}{2}mv^{2}\approx mv \Delta v$$
You can see that for larger $$v$$ it costs more energy for the same increment $$\Delta v$$. Essentially the power is
$$\dfrac{\Delta E}{\Delta t}=\underbrace{m\dfrac{\Delta v}{\Delta t}}_{ma=F}v=Fv$$
in the beginning we use 0 power. How is that possible?
We apply a force, but, at the very beginning, the power is 0, and is small as long as the speed is small.
Since power is the rate of transference of energy to the body (J/s), it would appear that that rate should be constant.
It is not constant, for the specific case of constant acceleration.
Why is it that a fast moving body requires more power to accelerate than a still body?
This is more a matter of fact: the same force, acting on increasing speed, correspond to more power. I can try to connect it to another concept (although this does not represent a more explicit answer): the kinetic energy is proportional to $$v^2$$ (square of the speed), thus, as the object moves faster, we need more and more power to give the necessary increase in kinetic energy.
And if we would like to know how much power we need to accelerate that body, should we then use the maximum speed of 2 m/s?
We need an increasing power from the beginning to the moment at which the speed is 2 m/s. So there is not a single value of power. We can evaluate the total energy needed to accelerate to $$v=$$2 m/s: it equals the final kinetic energy, i.e. $$E=1/2 m v^2$$ (where $$m$$ is the mass of the object and $$v$$ the final speed). Dividing this value by the time required for the acceleration, i.e. by $$\Delta t=v/a$$ (where $$a$$ is the acceleration), we get the average power $$E/\Delta t$$.
$$\frac{dv}{dE} = \frac{d}{dE} [(2E/m)^{\frac 1 2}]$$ $$\frac{dv}{dE} = \sqrt{\frac 2 {mE}}$$
which diverges at $$E=0$$ (i.e. $$v=0$$), so the velocity is infinitely sensitive to energy for a body at rest.
You can understand it if you think that the power is the rate of doing work on the accelerated object. This is the rate in time. As the speed increases the distance traveled in the same time interval (1 second for example) increases. But the work done is force times displacement (or distance). So in the same time interval the distance traveled is larger so the work done is higher so the power is higher.
Bear in mind that power as a function of time is the rate at which work is done as a function of time, or
$$P(t)=\frac{dW(t)}{dt}$$
For a constant force $$F$$ acting through a distance $$d$$ the work done $$W$$ is
$$W=Fd$$
If the force is constant, so is the acceleration. The distance $$d$$ a mass travels at constant acceleration given zero initial velocity is given by
$$d=\frac{at^2}{2}$$
That means the work done as a function of time is
$$W(t)=F\frac{at^2}{2}$$
Then power as a function of time, $$P(t)$$ which we gave at the beginning, is given by
$$P(t)=\frac{dW(t)}{dt}=Fat$$
This tells us for a constant force$$F$$, at $$t=0$$ with zero initial velocity, the power is zero and increases linearly with time thereafter.
Hope this helps.
This is a really good question!
Let me phrase it in a slightly different way, because it's the same problem according to calculus. The problem is, if I let go of a ball at rest, then the velocity is zero, so surely it's not moving, so why does it ever move? We would say “yeah, its velocity is zero for an instant, but it is increasing because it is in a state of acceleration.”
Similarly you might say: as we know, the speed is related to the kinetic energy and the change in kinetic energy over time is due to power exerted on that object, but it is at rest: nothing can exert any power on it; so why does it ever move? And the answer is the same: the power exerted is zero for an instant, but it is increasing because it is in a state of acceleration.
In other words when we are looking at a curve at a point, we have a tangent line which is very important, but there are also deviations from those tangent lines which are also important: they are just more important over longer timescales whereas the tangent line is only important over the shortest timescales. We speak of the "order" of the zeros, the ball at rest has a position which is constant to first order in time, but then to second order in time it is not constant. This means that its position is $$y = y_0 + \frac12 a~(t - t_0)^2$$ for some constants $$y_0, t_0, a$$.
Mathematics actually furnishes us with some even weirder examples. For example you might want a curve $$y(x)$$ which is constant to all orders in some variable $$x$$, but is not the constant curve $$y(x) = C$$. Mathematics says that yeah, that's certainly doable, and a great example is the function $$y(x) = \begin{cases} e^{-1/x^2}& x\ne 0\\ 0& x = 0 \end{cases}$$at the point $$x=0$$. Every derivative of this thing at that point is well-defined and zero, but the function still manages to “crawl out of the abyss.”
This decomposition of a function into its behavior over various “orders” in time or space is called a Taylor series, and each Taylor series only approximates a function within a certain “radius of convergence,” which in the above case happens to be 0. If you are doing physics then you generally assume (until you get into stochastic mathematics—in other words, until you start explicitly modeling noise) that the world is perfect and smooth and every function that you're using to model the world has some Taylor series that is valid for a nice big interval.
So that is why without thinking about it we expand to first order, second order, third order... so your teachers have been surreptitiously getting you to think in terms of first-order changes. But to your question, what do we do when the first-order change is zero?, the answer is always going to be well then we need to know more about the second order change. What if that's zero, too? Well, then the third order. And so on.
In you context you would have an average power of $$P_{avg} =F\Delta v$$ where $$\Delta v$$ is the change in velocity. This way, if you accelerate at a constant force $$P_{avg} =F \times 2m/s$$. With no other information about the force or the time you want to spend accelerating the object you can't get a value for $$P_{avg}$$.
• The equation is OK. And the interpretation too. The power varies with velocity so it will increase if the force is constant. $F \Delta v$ is the average power, or $\frac{\Delta E}{\Delta t}$ for the entire interval. – nasu May 29 '19 at 15:44
• The equation is $P=F v$, where $v$ is the speed, not the "change". This is connected to the non-linear increase of kinetic energy, $1/2 m v^2$. – Doriano Brogioli May 29 '19 at 15:48
• OK I'm sorry, I'll edit that... :/ apparently I was the one that misunderstood the question – Ballanzor May 29 '19 at 15:53 | 2020-01-24 17:55:40 | {"extraction_info": {"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, "math_score": 0.8267232775688171, "perplexity": 173.59486966719714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00248.warc.gz"} |
https://math.libretexts.org/Bookshelves/Algebra |
# Algebra
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ | 2019-04-20 16:51:17 | {"extraction_info": {"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, "math_score": 0.44280099868774414, "perplexity": 193.479040514725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578529898.48/warc/CC-MAIN-20190420160858-20190420182858-00243.warc.gz"} |
https://deepai.org/publication/dynamical-neural-network-information-and-topology | # Dynamical Neural Network: Information and Topology
A neural network works as an associative memory device if it has large storage capacity and the quality of the retrieval is good enough. The learning and attractor abilities of the network both can be measured by the mutual information (MI), between patterns and retrieval states. This paper deals with a search for an optimal topology, of a Hebb network, in the sense of the maximal MI. We use small-world topology. The connectivity γ ranges from an extremely diluted to the fully connected network; the randomness ω ranges from purely local to completely random neighbors. It is found that, while stability implies an optimal MI(γ,ω) at γ_opt(ω)→ 0, for the dynamics, the optimal topology holds at certain γ_opt>0 whenever 0≤ω<0.3.
## Authors
• 1 publication
• 2 publications
• 1 publication
• 4 publications
• ### Memory Retrieval in the B-Matrix Neural Network
This paper is an extension to the memory retrieval procedure of the B-Ma...
03/14/2011 ∙ by Prerana Laddha, et al. ∙ 0
• ### Emergent Criticality Through Adaptive Information Processing in Boolean Networks
We study information processing in populations of Boolean networks with ...
04/20/2011 ∙ by Alireza Goudarzi, et al. ∙ 0
• ### Learning Topology and Dynamics of Large Recurrent Neural Networks
Large-scale recurrent networks have drawn increasing attention recently ...
10/05/2014 ∙ by Yiyuan She, et al. ∙ 0
• ### Optimal Machine Intelligence Near the Edge of Chaos
It has long been suggested that living systems, in particular the brain,...
09/11/2019 ∙ by Ling Feng, et al. ∙ 0
• ### Neighbor connectivity of k-ary n-cubes
The neighbor connectivity of a graph G is the least number of vertices s...
10/27/2019 ∙ by Tomáš Dvořák, et al. ∙ 0
• ### Iterative Deep Learning for Network Topology Extraction
This paper tackles the task of estimating the topology of filamentary ne...
12/04/2017 ∙ by Carles Ventura, et al. ∙ 0
• ### Deep learning architectures for inference of AC-OPF solutions
We present a systematic comparison between neural network (NN) architect...
11/06/2020 ∙ by Thomas Falconer, et al. ∙ 0
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## 1 Introduction
The collective properties of attractor neural networks (ANN), such as the ability to perform as an associative memory, has been a subject of intensive research in the last couple of decades[2], dealing mainly with fully-connected topologies. More recently, the interest on ANN has been renewed by the study of more realistic architectures, such as small-world [3],[5] or scale-free [4],[16] models. The storage capacity and the overlap with the memorized patterns are the most used measures of the retrieval ability for the Hopfield-Hebb networks[6],[7]. Comparatively less attention has been paid to the study of the mutual information (MI) between stored patterns and the neural states[8][9], although neural networks are information processing machines.
A reason for this relatively low interest is twofold: on the one hand, it is easier to deal with the global parameter , than with
, a function of the conditional probability of neuron states
given the patterns . This can be solved for the so called mean-field networks
which satisfy the law of large numbers, hence MI is a function only of the macroscopic parameters
, and the load rate (where is the number of uncorrelated patterns, and is the neuron connectivity). On the other hand, the load is enough to measure the information if the overlap is close to , since in this case the information carried by any single binary neuron is almost 1 bit. It is true for a fully-connected (FC) network, for which the critical [6], with (with a sharp transition to for larger ): in this case, the information rate is about , as can be seen in the left panel of Fig.1. There we show the overlap (upper) and information for several architectures. However, in the case of diluted networks the transition is smooth. In particular, the random extremely diluted (RED) network has load capacity [11] but the overlap falls continuously to , which yields null information at the transition, , as seen in right panel of Fig.1 (dashed line). Such indetermination shows that one must search for the value of corresponding to the maximal information , instead of .
We address the problem of searching for the optimal topology, in the sense of maximizing the mutual information. Using the graph framework [4], one can capture the main properties of a wide range of neural systems, with only 2 parameters: , which is the average rate of links per neurons, where is the network size, and , which controls the rate of random links (among all neighbors). When is large, the clustering coefficient is large () and the mean-length-path between neurons is small (), whatever is. When is small, then if is too small, and , but if it is about , the network behaves again as if , with and . This region, called small-world (SW), is rather usefull when one is interested to built networks where the information transmition is fast and efficient, with high capacity in presence of significant noise, but do not wants to spent too much wiring [18]. Small-world networks may model many biological systems [15]. For instance, in a brain local connections dominate in intracortex, while there are a few intercortical connections [14].
In Fig.1 we show the overlap (upper) and information for several architectures. In the left panel, it is seen that the maximum information rate, , of FC network is about , while in the right panel, we show extremely-diluted networks (ED). The RED network () has . The right panel of Fig.1 plot also the overlap and the information for the local extremely diluted network (LED, ), with , and a small-world extremely diluted network (SED, ), with . We see that the ED transitions are smooth. The central panel of Fig.1 plot moderately diluted (MD) networks, which are commented later. Theoretical results fit well with the simulations, except for small , where theory underestimate it. Previous works about small-world attractor neural networks [13] studied only the overlap , so no result about information were known.
Our main goal in this work is to solve the following question: how does the maximal information, behaves with respect to the network topology? To our knowledge, up to now, there were no answer to this question. We will show that, near to the stationary retrieval states, for every value of the randomness , the extremely-diluted network, performs the best, . However, regarding the attractor basins, starting far from the patterns, the optimal topology holds for moderate . For instance, if transients are taken in account, values of lead to an optimal with .
The structure of the paper is the following: in the next section we review the information measures used in the calculations; in Sec.3, we define the topology and neuro-dynamics model. The results are shown in Sec.4, where we study retrieval by theory and simulation (with random patterns and with images); conclusions are drawn in last section.
## 2 The Information Measures
### 2.1 The Neural Channel
The network state at a given time is defined by a set of binary neurons, . Accordingly, each pattern
, is a set of site-independent random variables, binary and uniformly distributed:
. The network learns a set of independent patterns .
The task of the neural channel is to retrieve a pattern (say, ) starting from a neuron state which is inside its attractor basin, , i.e.: . This is achieved through a network dynamics, which couples neighbor neurons by the synaptic matrix with cardinality .
### 2.2 The Overlap
For the usual binary non-biased neurons model, the relevant order parameter is the between the neural states and a given pattern:
mμtN≡1N∑iξμiσti, (1)
at the time step . Note that both positive and negative patterns, carry the same information, so the absolute value of the overlap measures the retrieval quality: means a good retrieval. Alternatively, one can measure the error in retrieving using the Hamming distance: .
Together with the overlap, one needs a measure of the load, which is the rate of pattern bits per synapses used to store them. Since the synapses and patterns are independent, the load is given by
.
We require our network to have long-range interactions. Therefore, we regard a mean-field network (MFN), the distribution of the states is site-independent, so every spatial correlation such as can be neglected, which is reasonable in the asymptotic limit . Hence the condition of the law of large numbers, are fulfilled. At a given time step of the dynamical process, the network state can be described by one particular overlap, let say . The order parameters can thus be written, when , as
. The brackets represent average over the joint distribution
, for a single neuron (we can drop the index ). This macroscopic variable describes the information processing of the network, at a given time step of the dynamics. Along with this signal parameter, the residual microscopic overlaps yield the cross-talk noisy, its statistics complete the network macro-dynamics.
### 2.3 Mutual Information
For a long-range system, it is enough to observe the distribution of a single neuron in order to know the global distribution [9]. This is given by the conditional probability of having the neuron in a state , at each (unspecified) time step , given that in the same site the pattern being retrieved is . For the binary network we are considering, , [10] where the overlap is .
The joint distribution of is interpreted as an ensemble distribution for the neuron states and inputs . In the conditional probability, , all type of noise in the retrieval process of the input pattern through the network (both from environment and over the dynamical process itself) is enclosed.
With the above expressions and , we can calculate the MI [9], a quantity used to measure the prediction that an observer at the output () can do about the input () (we drop the time index ). It reads , where is the entropy and is the conditional entropy. We use binary logarithms to measure the information in bits. The entropies are [10]:
S[σ|ξ]=−1+m2log21+m2−1−m2log21−m2, S[σ]=1[bit]. (2)
We define the information rate as
i(α,m)=MI[→σ|{→ξμ}]/|J|≡αMI[σ;ξ], (3)
since for independent neurons and patterns, . When the network approaches its saturation limit , the states can not remain close to the patterns, then is usually small. So, while the number of patterns increase, the information per pattern decreases. Therefore, information is a non-monotonic function of the overlap and load rate, see Fig.1, which reaches its maximum value at some value of the load .
## 3 The Model
### 3.1 The Network Topology
The synaptic couplings are , where the connectivity matrix has a local and a random parts, , and W are synaptic weights. The local part connects the nearest neighbors, , with in the asymmetric case, on a closed ring. The random part consists of independent random variables , distributed with probability , and otherwise, with , where is the mean number of random connections of a single neuron. Hence, the neuron connectivity is . The network topology is then characterized by two parameters: the connectivity ratio, defined as , and the randomness ratio, . The plays the role of rewiring probability in the small-world model (SW) [3]. Our model was proposed by Newman and Watts [20], which has the advantage of avoiding disconneting the graph.
Note that the topology can be defined by an adjacency list connecting neighbors, , with . So the storage cost of this network is . Hence, the information is , Eq.(3), where the load rate is scaled as . The learning algorithm updates W, according to the Hebb rule
Wμij=Wμ−1ij+1Kξμiξμj. (4)
The network starts at , and after learning steps, it reaches a value . The learning stage is a slow dynamics, being stationary-like in the time scale of the much faster retrieval stage, we define in the following.
### 3.2 The Neural Dynamics
The neural states, , are updated according to the stochastic parallel dynamics:
σt+1i=sign(hti+Tx),hti≡∑jJijσtj,i=1...N (5)
where is a normalized random variable and is the temperature-like environmental noise. In the case of symmetric synaptic couplings, , an energy function can be defined, whose minima are the stable states of the dynamics Eq.(5).
In the present paper, we work out the asymmetric network by simulation (no constraints ). The theory was carried out for symmetric networks. As it is seen in Fig.1, theory and simulation shows similar results, except for local networks (theory underestimate , where the symmetry may play some role. We restrict our analysis also for the deterministic dynamics (). The stochastic macro-dynamics comes from the extensive number of learned patterns, .
## 4 Results
We studied the information for the stationary and dynamical states of the network were studied as a function of the topological parameters, and . A sample of the results for simulation and theory is shown in Fig.1, where the stationary states of the overlap and information are plotted for the FC, MD and ED arquitetures. It can be seen that information increases with dilution and with randomness of the network. A reason for this behavior is that dilution decreases the correlation due to the interference between patterns. However, dilution also increases the mean-path-length of the network, thus, if the connections are local, the information flows slowly over the network. Hence, the neuron states can be eventually trapped in noisy patterns. So, is small for even if .
### 4.1 Theory: Stationary States
Following to the Gardner calculations[11], at temperature T=0 the MFN approximation gives the fixed point equations:
m=erf(m/√rα), (6) χ=2φ(m/√rα)/√rα; (7) r=∞∑k=0ak(k+1)χk,ak=γTr[(C/K)k+2] (8)
with , . The parameter is the probability of existence of cycle of length in the connectivity graph. The can be calculated either by using Monte Carlo [17], or by an analytical approach, which gives , where
is the Fourier transform of the probability of links,
. For an RED and FC networks one recover the known results for and respectively [2].
The theoretical dependence of the information on the load, for FC, MD and ED networks, with local, small-world and random connections, are plotted in the fat lines in Fig.1. A comparison between theory and simulation is also given in Fig.1. It can be seen that both results agree for most , but theory fails for . One reason is that theory uses symmetric constraint, while simulation was carried out with asymmetric synapsis. Figure 3 shows their maxima vs. the parameters . It is seen that the optimal is at . This implies that the best topology for information (stationary states) is the extreme diluted network, with purely random connectivity.
### 4.2 Simulation: Attractors and Transients
We have studied the behavior of the network varying the range of connectivity and randomness . We used Eq.(5). Both local and random connections are asymmetric. The simulation was carried out with synapses, storing an adjacency list as data structure, instead of . For instance, with , we used . In [13] the authors use , which is far from asymptotic limit.
We studied the network by searching for the stability properties and transients of the neuron dynamics. To look for stability, we started the network at some pattern (with initial overlap ), and wait until it stays or leave it after a flag time step (unless it converges to a fixed point before ). When we check transients, we start with , and stop the dynamics at the time . Usually, parallel (all neurons) updates is a large enough delay for retrieval. Indeed in most case far before the saturation, after the network end up in a pattern, however, near , even after the network has not yet relaxed.
In first place, we checked for the stability properties of the network: the neuron states start precisely at a given pattern (which changes at each learned step ). The initial overlap is , so, after time steps in retrieving, the information for final overlap is calculated. We plot it as a function of , and its maximum is evaluated. We averaged over a window in the axis of , usually . This is repeated for various values of the connectivity ratio and randomness parameters. The results are in the upper panels of Fig.4.
Second, we checked for the retrieval properties: the neuron states start far from a learned pattern, but inside its basin of attraction, . The initial configuration is chosen with distribution: , for all neurons (so we avoid a bias between local/random neighbors). The initial overlap is now , and after steps, the information is calculated. The results are in the lower panels of Fig.4. The first observation now is that the maximal information increases with dilution (smaller ) if the network is more random, , while it decreases with dilution if the network is more local, .
The comparison between upper () and lower parts of Fig.4, shows that the non-monotonic behavior of the information with dilution and randomness, is stronger for the retrieval () than for the stability properties (). One can understand this in terms of the basins of attraction. Random topologies have very deep attractors, specially if the network is diluted enough, while regular topologies almost lose their retrieval abilities with dilution. However, since the basins becomes rougher with dilution, then network takes longer to reach the attractor. Hence, the competition between depth-roughness is won by the more robust MD networks.
Each maximal in Fig.4 is plotted in Fig.5. We see that, for intermediate values of the randomness parameter there is an optimal information respect to the dilution , if dynamics is truncated. We observe that the optimal is shifted to the left (stronger dilution) when the randomness of the network increases.
For instance, with , the optimal is at while with , it is . This result does not change qualitatively with the flag time, but if the dynamics is truncated early, the optimal , for a fixed , is shifted to more connected networks. However, the behavior depends strongly on the initial condition: respect to , where the maximal are pronounced, with , the dependence on the topology becomes almost flat. We see also that for there is no intermediate optimal topology. It is worth to note that the simulation converges to the theoretical results if when .
### 4.3 Simulation with Images
The simulations presented so far use artificial patterns randomly generated. In order to check if our results are robust against possibly correlations existent in realistic patterns, we test the algorithm with images. We see that the same non-monotonic behavior for is observed here.
We have checked the results by using data derived from the Waterloo image database. We are working with square shaped patches. In order to use Hebb-like non-sparse code binary network and still preserve the structure of the image we process the images preserving the edges, by applying edge filter. Each pixel of the patch represents a different neuron. The number of connections is up to and the feasible connectivities (more than 3 patterns) are .
Note that the procedure, strictly speaking, does not guarantee the conditions for the distribution of , because neither is uniform (due to the threshold in large blocks), nor are uncorrelated (due to image edges).
We are choosing at random the origin of the patch and the image to be used from the available 12 images. The topology of the network is a ring with small world topology. The results of the simulation, using Chen filter, are shown in Fig.3. The optimal connectivity with and is found to be . The fluctuation now are much larger than with random patterns, due to correlation and small network size. In the stationary states, , the optimal connectivity remains at , with . The results agree qualitatively with simulation for random patterns, Fig.4, where the initial overlaps are and (in Fig.3 it is always ).
## 5 Conclusions
In this paper we have studied the dependence of the information capacity with the topology for an attractor neural network. We calculated the mutual information for a Hebb model, for storing binary patterns, varying the connectivity () and randomness () parameters, and obtained the maximal respect to , . Then we look at the optimal topology, in the sense of the information, . We presented stationary and transient states. The main result is that larger always leads to higher information .
From the stability calculations, the stationary optimal topology, is the extremely diluted (RED) network. Dynamics shows, however, that this is not true: we found there is an intermediate optimal , for any fixed . This can be understood regarding the shape of the attractors. The ED waits much longer for the retrieval than more connected networks do, so the neurons can be trapped in spurious states with vanishing information. We found there is an intermediate optimal , whenever the retrieval is truncated, and it remains up to the stationary states.
Both in nature and in technological approaches to neural devices, dynamics is an essential issue for information process. So, an optimized topology holds in any practical purpose, even if no attemption is payed to wiring or other energetic costs of random links [18]. The reason is a competition between the broadness (larger storage capacity) and roughness (slower retrieval speed) of the attraction basins.
We believe that the maximization of information respect to the topology could be a biological criterium (where non-equilibrium phenomena are relevant) to build real neural networks. We expect that the same dependence should happens for more structured networks and learning rules.
Acknowledgments Work supported by grants TIC01-572, TIN2004-07676-C01-01, BFI2003-07276, TIN2004-04363-C03-03 from MCyT, Spain.
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• [5] Masuda, N. and Aihara, K. Biol. Cybernetics, 90: 302 (2004)
• [6] Amit, D., Gutfreund, H., Sompolinsky, H.: Phys. Rev. A 35 (1987) 2293
• [7] Okada, M.: Neural Network 9/8 (1996) 1429
• [8] Perez-Vicente, C., Amit, D.: J. Phys. A, 22 (1989) 559
• [9] Dominguez, D., Bolle, D.: Phys. Rev. Lett 80 (1998) 2961
• [10] Bolle, D., Dominguez, D., Amari, S.: Neural Networks 13 (2000) 455
• [11] Canning, A. and Gardner, E. Partially Connected Models of Neural Networks, J. Phys. A, 21, 3275-3284, 1988
• [12] Kupermann, M. and Abramson, G. Phys. Rev. Lett. 86: 2909, 2001
• [13] McGraw, P.N. and Menzinger, M. Phys. Rev. E 68: 047102-1, 2003
• [14] Rolls, E., Treves, A., Neural Network and Brain Function. Oxford University Press, 2004
• [15] Sporns, O. et al., Cognitive Sciences, 8(9): 418-425, 2004
• [16] Torres, J. et al., Neurocomputing, 58-60: 229-234, 2004
• [17] Dominguez, D., Korutchev, K.,Serrano, E. and Rodriguez F.B., LNCS 3173: 14-29, 2004
• [18] Adams, R., Calcraft, L., Davey, N.: ICANNGA05, preprint (2005)
• [19] Li, C., Chen, G.: Phys. Rev. E 68 (2003) 52901
• [20] Newman, M.E.J., Watts, D.J: Phys. Rev. E 60 (1999) 7332
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https://pysm3.readthedocs.io/en/latest/api/pysm.DecorrelatedModifiedBlackBody.html | # DecorrelatedModifiedBlackBody¶
class pysm.DecorrelatedModifiedBlackBody(map_I=None, map_Q=None, map_U=None, freq_ref_I=None, freq_ref_P=None, map_mbb_index=None, map_mbb_temperature=None, nside=None, pixel_indices=None, mpi_comm=None, correlation_length=None)[source] [edit on github]
See parent class for other documentation.
Parameters
correlation_length: float
This number set the scale in logarithmic space for the distance in freuqency past which the MBB emission becomes decorrelated. For frequencies much much closer than this distance, the emission is well correlated.
Methods Summary
get_emission(self, freqs) Function to calculate the emission of a decorrelated modified black body model.
Methods Documentation
get_emission(self, freqs)[source] [edit on github]
Function to calculate the emission of a decorrelated modified black body model. | 2019-12-10 23:51:42 | {"extraction_info": {"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, "math_score": 0.6536183953285217, "perplexity": 14390.326338166666}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00408.warc.gz"} |
https://ieeexplore.ieee.org/xpl/tocresult.jsp?reload=true&isnumber=3714 | # IEEE Transactions on Power Electronics
## Filter Results
Displaying Results 1 - 21 of 21
• ### PWM-controlled SRC with inductive output filter at constant switching frequency
Publication Year: 1992, Page(s):289 - 295
Cited by: Papers (8)
| | PDF (524 KB)
By using the PWM control scheme in the series resonant power converter (SRC) with inductive output filter, the converter can be operated at a constant frequency. This converter has lower switching loss than the PWM converter and better control characteristics than the ordinary SRC. Since the peak current in the present converter equals the load current, it has the lowest possible peak current stre... View full abstract»
• ### Modeling and improved current control of series resonant converter with nonperiodic integral cycle mode
Publication Year: 1992, Page(s):280 - 288
Cited by: Papers (12)
| | PDF (792 KB)
A dynamic modeling and an improved current control technique for a series resonant power converter with nonperiodic integral cycle mode are proposed to overcome the disadvantages of an integral cycle mode-controlled series resonant converter. The internal operational characteristics, are investigated in detail and an improved current control technique is developed based on this analysis. Using the... View full abstract»
• ### Compensated carrier PWM synchronization: a novel method to achieve self-regulation and AC unbalance compensation in AC fed converters
Publication Year: 1992, Page(s):342 - 348
Cited by: Papers (10)
| | PDF (624 KB)
The compensated carrier PWM synchronization (CCPS) method for AC-fed PWM power converters is presented. The method provides a solution to PWM converters fed by industrial power systems (IPSs). Such environments usually present unbalances and magnitude fluctuations of AC voltages. Those circumstances impair standard PWM techniques because low-order harmonics are produced and DC-link regulation is p... View full abstract»
• ### A discrete adaptive field-oriented induction motor drive
Publication Year: 1992, Page(s):411 - 419
Cited by: Papers (15)
| | PDF (564 KB)
A discrete model reference adaptive controller (MRAC) is designed and implemented. This MRAC makes the performance of the field-oriented induction motor drives insensitive to parameter changes. Only the information of the reference model and the plant output are required. Hence, the proposed controller is easy to implement practically. For designing the proposed adaptive controller, the dynamic mo... View full abstract»
• ### A generic soft switching converter topology with a parallel nonlinear network for high-power application
Publication Year: 1992, Page(s):324 - 331
Cited by: Papers (6)
| | PDF (472 KB)
A soft switching concept that derives from the resonant link and resonant pole power converters and combines the best features of resonant and hard switching converters is applied to a phase arm. The inductor of the resonant LC is designed to saturate, thus having effectively two inductance values: a very large value during the conduction period and a small value that is only active durin... View full abstract»
• ### A rugged soft commutated PWM inverter for AC drives
Publication Year: 1992, Page(s):385 - 392
Cited by: Papers (15) | Patents (2)
| | PDF (656 KB)
A novel DC-DC power converter for variable-speed AC power drives using the zero-voltage switching technique is described. This converter combines the advantages of soft commutated inverters and those of conventional pulsewidth modulated (PWM) inverters. In the proposed scheme, the soft commutation reduces the constraints on the switches, and the PWM enables simple and efficient regulation of the p... View full abstract»
• ### Design of transcutaneous energy transmission system using a series resonant converter
Publication Year: 1992, Page(s):261 - 269
Cited by: Papers (54) | Patents (1)
| | PDF (700 KB)
A full-bridge zero-voltage-switched series resonant power converter that transfers 12-48 W of regulated power across a large, variable air gap has been designed and built. This converter can power an artificial heart through intact skin by utilizing a transformer with an air gap of 1-2 cm between the primary and the secondary. The secondary winding would be implanted under the skin, and the primar... View full abstract»
• ### A parallel resonant converter with postregulators
Publication Year: 1992, Page(s):296 - 303
Cited by: Papers (4) | Patents (5)
| | PDF (688 KB)
A parallel resonant power converter (PRC) with postregulator(s) power synchronized to the primary switching frequency is investigated for discontinuous-conduction mode operation. It is analyzed with a magnetic amplifier, which blocks the rising edge of the secondary voltage, and with a synchronized buck regulator, which blocks part of the trailing edge of the secondary voltage waveform under the s... View full abstract»
• ### Optimal feed-forward compensation for PWM DC/DC converters with linear' and quadratic' conversion ratio
Publication Year: 1992, Page(s):349 - 355
Cited by: Papers (40) | Patents (6)
| | PDF (440 KB)
An analytical procedure to optimize the feedforward compensation for any PWM DC/DC power converters is described. Achieving zero DC audiosusceptibility was found to be possible for the buck, buck-boost, Cuk, and SEPIC cells; for the boost converter, however, only nonoptimal compensation is feasible. Rules for the design of PWM controllers and procedures for the evaluation of the hardware-introduce... View full abstract»
• ### Multikilojoule inductive modulator with solid-state opening switches
Publication Year: 1992, Page(s):420 - 424
Cited by: Papers (18)
| | PDF (368 KB)
A four-stage inductive storage XRAM (inverse MARX) circuit was used to amplify current and power. By means of switching action, individual inductors are charged in series and discharged in parallel to the load. A microprocessor controls the timing of the switches, which results in a predetermined load current waveform. The use of present asymmetric GTOs at much higher currents than rated allowed t... View full abstract»
• ### A unity power factor current-regulated SPWM rectifier with a notch feedback for stabilization and active filtering
Publication Year: 1992, Page(s):356 - 363
Cited by: Papers (25)
| | PDF (856 KB)
A stand-alone, unity power factor, current-regulated sinusoidal pulsewidth modulated (SPWM) rectifier is described. The topology is based on the series connection of three four-valve single-phase bridges, which allows the conventional two-stage logic SPWM strategy to be used without interphase interference. The problems of stability and low harmonic waveform distortion are identified. Solutions ar... View full abstract»
• ### Digital control of an induction motor drive by a stochastic estimator and airgap magnetic flux feedback loop
Publication Year: 1992, Page(s):393 - 403
Cited by: Papers (5)
| | PDF (692 KB)
An approach for the problem of airgap flux estimation in induction motors is presented. The Kalman filter algorithm is developed to provide an estimated state vector containing flux linkage components. The estimated fluxes are then used to implement a direct flux control loop through an inverter-fed AC drive scheme. The overall control system is developed around a digital unit based on a 16 bit mi... View full abstract»
• ### Frequency-domain analysis of series resonant converter for continuous conduction mode
Publication Year: 1992, Page(s):270 - 279
Cited by: Papers (32)
| | PDF (748 KB)
Most previous analyses of a series resonant DC-DC power converter (SRC) have been performed in the time domain. A comprehensive frequency-domain analysis of the SRC for steady-state operation, along with experimental results is presented. Simple analytical design equations are derived for the basic performance parameters of the converter operating in the continuous conduction mode (CCM) using Four... View full abstract»
• ### Characteristics of load resonant converters operated in a high-power factor mode
Publication Year: 1992, Page(s):304 - 314
Cited by: Papers (57) | Patents (7)
| | PDF (816 KB)
The performance of the parallel resonant power converter and the combination series/parallel resonant power converter (LCC converter) when operated above resonance in a high power factor mode are determined and compared for single phase applications. When the DC voltage applied to the input of these converters is obtained from a single phase rectifier with a small DC link capacitor, a relatively h... View full abstract»
• ### A discretized current control technique with delayed input voltage feedback for a voltage-fed PWM inverter
Publication Year: 1992, Page(s):364 - 373
Cited by: Papers (39)
| | PDF (576 KB)
A current control technique for a voltage-fed PWM inverter is presented. The discretized state equation of an inverter and a load independent of operating conditions with the delayed input voltage feedback has been derived using the averaging concept. The discretized current controller is proposed to reduce the current error as fast as possible using the deadbeat control strategy and to stabilize ... View full abstract»
• ### Nonlinear capacitors in snubber circuits for GTO thyristors
Publication Year: 1992, Page(s):425 - 429
Cited by: Papers (9)
| | PDF (296 KB)
Many applications with GTO thyristors require snubber circuits, at least for, turn-off. In achieving high switching frequencies there are conflicting design criteria for the snubber capacitor-losses versus timing conditions. Improvements to this situation are expected by employing nonlinear capacitors. Features of snubber circuits with nonlinear capacitors are described, and experimental results o... View full abstract»
• ### Rectifier for minimum line-current harmonics and maximum power factor
Publication Year: 1992, Page(s):332 - 341
Cited by: Papers (55) | Patents (2)
| | PDF (884 KB)
Rectifier line current harmonics interfere with proper power system operation, reduce rectifier power factor, and limit the power available from a given service. The rectifier's output filter inductance determines the rectifier line current waveform, the line current harmonics, and the power factor. Classical rectifier analysis usually assumes a near-infinite output filter inductance, which introd... View full abstract»
• ### A pulse frequency modulated PWM inverter for induction motor drives
Publication Year: 1992, Page(s):404 - 410
Cited by: Papers (17)
| | PDF (468 KB)
A PWM pulse pattern optimization method using pulse frequency modulation (PFM) is described. In conventional PWMs the pulse frequency is kept constant. In the proposed PFM, however, the pulse frequency is adjusted. The PFM technique is intended to not only reduce the magnetic acoustic noises of driven motors but also to improve the performance of sinusoidal inverters. The PWM pulse patterns are ba... View full abstract»
• ### Quasi-linear modeling and control of DC-DC converters
Publication Year: 1992, Page(s):315 - 323
Cited by: Papers (26)
| | PDF (640 KB)
A quasi-linear approach is proposed for modeling and control of DC-DC power converters. The method presented is derived by perturbing an approximate large signal equation around a varying operating point in a reduced variable space. This differs from the usual practice of applying the perturbation technique around a fixed operating condition. In the proposed algorithm, the control equation and the... View full abstract»
• ### Techniques for the practical application of duality to power circuits
Publication Year: 1992, Page(s):374 - 384
Cited by: Papers (31)
| | PDF (788 KB)
Techniques that simplify the application of duality to power circuits are presented. Duals are derived for ideal switches (including two- and four-quadrant switches) and semiconductor-controlled rectifiers (SCRs), transformers and coupled inductors, and capacitors. A collection of common topological structures and corresponding duals is also given. This library of dual relations allows even compli... View full abstract»
• ### Design considerations for p-i-n thyristor structures
Publication Year: 1992, Page(s):430 - 435
Cited by: Papers (2) | Patents (2)
| | PDF (508 KB)
An analysis of a high-voltage gate turn-off (GTO) thyristor structure with a double-layered n base (p-i-n structure) is presented. From integration of Poisson's equation, an expression for the forward-blocking voltage at the onset of avalanche breakdown is obtained. Simple design criteria are developed to calculate the optimal thickness and doping density of the n base of a conventional pnpn struc... View full abstract»
## Aims & Scope
IEEE Transactions on Power Electronics covers fundamental technologies used in the control and conversion of electric power.
Full Aims & Scope | 2018-05-27 19:56:11 | {"extraction_info": {"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, "math_score": 0.2644149661064148, "perplexity": 9723.430631134082}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00248.warc.gz"} |
https://www.wptricks.com/question-category/plugin-gravity-forms/ | ## Gravity Forms: Create fields programmatically
Question
Is there any possiblity to create and manage gravity fields dynamically? I have form with ID 4 with couple fields. I want to add some fields of type checkbox with my own custom external data which can change in time and ...
0
20 hours 0 Answers 4 views
## Trying to modify the background of a form on a specific page
Question
Here's my CSS: .page-id-9399 body #gform_wrapper_3 { background-color: #363c42; } This is related to a Gravity Forms (contact form submission). I'd like the background of the FORM to be the color as above. I thought the above code would work because ...
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6 days 0 Answers 13 views
## Assigning Gravity Form Vendor Submission to Logged in Vendor
Question
First & foremost, apologies if I am missing any particular information - I'll do my best to be as thorough, concise & specific to help communicate this stupid problem I'm just stuck on. I explain it here: https://www.loom.com/share/f2505d2f81c04ce5a37cdf552ee85f77 But to summarize it ...
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2 weeks 0 Answers 26 views
## Make “Other” the first option for a radio button field in Gravity Forms
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I am looking to change the order of the radio fields for a single specific field on a specific form in Gravity Forms. I would like the "Other" field to display first - before the pre-provided radio options. How it currently ...
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2 weeks 0 Answers 22 views
## Modify gform_other_choice_value for specific form and specific field in Gravity Forms
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I am looking to change the "other" text for a single specific field on a specific form in Gravity Forms. It seems gform_other_choice_value does not work the way some of the other Gravity Forms filters work in that adding the form ...
0
2 weeks 0 Answers 20 views
## How can I get the total average quiz results of the Gravity Forms Quiz Addon?
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I'm using the Quiz Addon for Gravity Forms, and I want to show the average total quiz results on the front end so that people taking the quiz can see how they stack up against everyone else. I was going to ...
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2 weeks 0 Answers 19 views
## Paypal standard add-on don’t work
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I've a site realized with WordPress. I've installed Gravity Forms v. 1.8.20 and the Gravity Forms PayPal Standard Add-On v. 2.3 I've two modules where the content is a bit difference about text information, but either has a submit button with PayPal ...
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2 weeks 0 Answers 32 views
## How to use third-party SendGrid Email Validation API in Gravity Forms?
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Please let me know how to use SendGrid Email Validation API in Gravity Forms, using gform_field_validation filter? Gravity Forms filter example: add_filter( 'gform_field_validation', function ( $result,$value, $form,$field ) { if ( $field->get_input_type() === 'email' &&$result['is_valid'] ) { ...
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3 weeks 0 Answers 30 views
## Pass created post id from gform_after_submission action to gform_confirmation filter?
Question
I am trying to create a post when my Gravity Form submits and redirect to the post permalink on form confirmation. See below the function which creates my post/order when my form submits using the
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4 weeks 0 Answers 44 views
## Gravity form is broken after javascript defer
Question
I want to optimise my website javascript, but i'm also using Gravity form and if I deferred js the form doesn't show up, there will be an error in the console saying jQuery is not a function. I have tried these ...
0
1 month 0 Answers 40 views | 2020-06-03 05:15:29 | {"extraction_info": {"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, "math_score": 0.1706051528453827, "perplexity": 3640.3389989458037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347432237.67/warc/CC-MAIN-20200603050448-20200603080448-00507.warc.gz"} |
https://math.meta.stackexchange.com/questions/28131/should-the-block-design-tag-be-removed | # Should the (block-design) tag be removed?
I noticed for the first time today that there is a (block-design) tag, with no description and 12 questions. It seems that this tag is completely redundant, as there is already a (combinatorial-designs) tag (which does have a description, 154 questions, and is the exact same thing). Should this tag be removed/merged?
(This is my first post on meta so I don't know if I'm phrasing this question correctly. Any feedback is appreciated.)
## 1 Answer
I introduced that tag.
Reviewing the description for "combinatorial designs" in detail, I agree that "block design" is a specialization of the same topic and could be merged in without loss.
In other words, yes, let's remove and replace it with .
• I don't think removing is a bad idea, although we might also consider making block-design a synonym of combinatorial-design -- as you said, the term "block design" is common, and it's not unrealistic to expect it to be a tag. Here is the page to do so, but I'm not sure if it should be done -- quid has stated there can be downsides to synonyms, but I'm not sure I see one here. – pjs36 Mar 31 '18 at 1:04
• I absolutely think that merging as a synonym is a good idea. Both are very common terms, and have an identical meaning. I also agree that the downsides pointed out by quid do not seem to apply in this situation (I think it's unlikely either of these two tags would be used unintentionally by people mistaking their meaning). – Morgan Rodgers Mar 31 '18 at 1:12
• It sounds like we will make a synonym then. However, the convention seems to be using plural forms of nouns in tag names, in which case I guess we would remove all block-design tags, and create the synonym block-designs $\to$ combinatorial-designs, to keep with that convention. (I can't decide if I should suggest this in an answer, or just keep the conversation under this post) – pjs36 Mar 31 '18 at 1:24
• @psj36 I agree that this is a reasonable synonym. I'll give it some more time but intend to implement the proposed action(s) rather soon. (Including the plural, though I am personally not a fan of this; but yeah, it's tradition here like this and a relatively harmless one at that although [rant suppressed] ;-) – quid Mar 31 '18 at 15:38
• @quid I’m not a fan of “block designs” in the plural, either. – Wildcard Mar 31 '18 at 18:29
• Statisticians using block designs may distinguish between "blocks" and "treatments", doing something to which combinatorialists are strangers. I don't know that that's an occasion for a tag on m.s.e. – Michael Hardy Apr 7 '18 at 21:49 | 2020-01-18 20:45:41 | {"extraction_info": {"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, "math_score": 0.6170920133590698, "perplexity": 1095.1314999735796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00462.warc.gz"} |
https://physics.stackexchange.com/questions/162788/can-one-proton-attract-two-electrons | # Can one proton attract two electrons?
Suppose that in an empty space there is only one proton. This proton would have created a field of positive charge which should attract possible negative electrons, so now we add two electrons on opposite sides of the proton in a similar distance. What happens then?
Are both electrons attracted to proton with full/half the force before the one that was a little bit closer reaches the proton first and forms a bond and the other one now glides past neutral hydrogen atom?
• Um...the proton will attract each electron equally strongly, and it will attract one of two as strongly as one of hundred. What changes when you add electrons is that they also repel each other. – ACuriousMind Feb 1 '15 at 16:26
• You might ask the chemists about the stability of the ion $\mathrm{H}^-$ ... – dmckee --- ex-moderator kitten Feb 1 '15 at 16:31 | 2021-01-18 01:39:45 | {"extraction_info": {"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, "math_score": 0.46693116426467896, "perplexity": 553.1161289265642}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00348.warc.gz"} |
http://accessmedicine.mhmedical.com/content.aspx?bookid=331§ionid=40726725 | Chapter 16
Body temperature is controlled by the hypothalamus. Neurons in both the preoptic anterior hypothalamus and the posterior hypothalamus receive two kinds of signals: one from peripheral nerves that transmit information from warmth/cold receptors in the skin and the other from the temperature of the blood bathing the region. These two types of signals are integrated by the thermoregulatory center of the hypothalamus to maintain normal temperature. In a neutral temperature environment, the metabolic rate of humans produces more heat than is necessary to maintain the core body temperature in the range of 36.5–37.5°C (97.7–99.5°F).
A normal body temperature is maintained ordinarily, despite environmental variations, because the hypothalamic thermoregulatory center balances the excess heat production derived from metabolic activity in muscle and the liver with heat dissipation from the skin and lungs. According to studies of healthy individuals 18–40 years of age, the mean oral temperature is 36.8° ± 0.4°C (98.2° ± 0.7°F), with low levels at 6 a.m. and higher levels at 4–6 p.m. The maximum normal oral temperature is 37.2°C (98.9°F) at 6 a.m. and 37.7°C (99.9°F) at 4 p.m.; these values define the 99th percentile for healthy individuals. In light of these studies, an a.m. temperature of >37.2°C (>98.9°F) or a p.m. temperature of >37.7°C (>99.9°F) defines a fever. The normal daily temperature variation is typically 0.5°C (0.9°F). However, in some individuals recovering from a febrile illness, this daily variation can be as great as 1.0°C. During a febrile illness, the diurnal variation usually is maintained, but at higher, febrile levels. The daily temperature variation appears to be fixed in early childhood; in contrast, elderly individuals can exhibit a reduced ability to develop fever, with only a modest fever even in severe infections.
In women who menstruate, the a.m. temperature is generally lower in the 2 weeks before ovulation; it then rises by ˜0.6°C (1°F) with ovulation and remains at that level until menses occur. Body temperature can be elevated in the postprandial state. Pregnancy and endocrinologic dysfunction also affect body temperature.
### Fever
Fever is an elevation of body temperature that exceeds the normal daily variation and ...
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This div only appears when the trigger link is hovered over. Otherwise it is hidden from view. | 2015-03-03 08:52:28 | {"extraction_info": {"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, "math_score": 0.19896793365478516, "perplexity": 4377.424327791018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936463165.18/warc/CC-MAIN-20150226074103-00255-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://mathoverflow.net/revisions/79250/list | ## Return to Question
Post Closed as "off topic" by Andrew Stacey, HW, Mark Sapir, Igor Rivin, Bill Johnson
2 edited tags
1
# Question about the proof of the fact that IR is not quasi-isomtric to IR^2
Hello. Yesterday we proofed, that $\mathbb{R}$ is not quasi-isometric to $\mathbb{R}^2$ (both endowed with the standard Euclidean metric).
Step 1.: $\mathbb{R}$ is q.i. $\mathbb{Z}$ and $\mathbb{R}^2$ is q.i. to $\mathbb{Z}^2$, so we only need to show $\mathbb{Z}$ is q.i. to $\mathbb{Z}^2$.
This is clear.
Step 2.: Indeed suppose that $f:\mathbb{Z}\mapsto \mathbb{Z}^2$ is a $(\lambda,C)$-quasi-isometry for some $\lambda\ge1$ and $C\ge0$. As $f$ is $(\lambda,C)$-quasi-isometric embedding it follwows that $\frac{1}{\lambda}d_X(x,y)-C\le d_Y(f(x),f(y))$ for all $x,y\in X$. This implies that for any $x,y\in X$ we have $d_X(x,y)\le\lambda(d_Y(f(x),f(y))+C)$. Chosing $x=0$ the implies that
$f(X)\cap N_r(f(0))\subset f(N_{\lambda(r+C)}(0))$
As $f$ is further C-quasi-surjective it follows that:
$N_{r-C}(f(0))\subset N_C(f(X)\cap N_r(f(0))\subset N_C(f(N_{\lambda(r+C)}(0))$
This is also clear.
Step 3.: Now $\mid N_{r-C}(f(0))\mid$ grows quadratically in r while
$\mid N_C(f(N_{\lambda(r+C)}(0))\mid\le\mid N_C(f(0))\mid\cdot\mid f(N_{\lambda(r+C)}(0))\mid\le\mid N_C(f(0))\mid\cdot\mid N_{\lambda(r+C}(0)\mid$
grows ato most linearly in $r$. Thus for large$r$ we have
$\mid N_{r-C}(f(0))\mid >\mid N_C(f(N_{\lambda(r+C)}(0))\mid$
Now my questions are:
1. What is $\mid N_{r-C}(f(0))\mid$? Ist it the area of an circle of radius $r-C$ and mddle point $f(0)$ or what does this absolute value brackets mean in this thense?
2. If it is. Is $\mid N_C(f(N_{\lambda(r+C)}(0))\mid$ also a cirlce? I don't think so, because $f(N_{\lambda(r+C)}(0))$ does not have to be connected. Right? But anyway we are looking at the area of this thing under these "absolut value brackets", right?
3. And why does the second inequalitiy of Step 3. holds: I mean why does $\mid f(N_{\lambda(r+C)}(0))\mid\le\mid N_{\lambda(r+C)}(0)\mid$???
Thaks for help! | 2013-05-21 21:01:33 | {"extraction_info": {"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, "math_score": 0.9332391023635864, "perplexity": 619.0173274510969}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368700563008/warc/CC-MAIN-20130516103603-00058-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.ias.ac.in/listing/bibliography/pram/SHI-HUA_CHEN | • SHI-HUA CHEN
Articles written in Pramana – Journal of Physics
• Effect of electric field and temperature on the binding energy of bound polaron in an anisotropic quantum dot
The ratio between the confinement lengths in the $xy$-plane and the $z$ direction plays an important role in determining the properties of anisotropic quantum dot. Within a variational approach of Pekar type, we investigated theoretically the effects of electric field and temperature on the ground-state binding energies of hydrogenic impurity polarons in KBr anisotropic quantum dot. The obtained results illustrate that the binding energies increase with the electric field strength and temperature but decrease with the position of the impurity when considering different confinement lengths in the $xy$-plane and the $z$ direction and present the properties of the anisotropic quantum dot.
• Influence of magnetic field and Coulomb field on the Rashba effect in a triangular quantum well
The influence of magnetic field and Coulomb field on the Rashba spin–orbit interaction in a triangular quantum well was studied using Pekar variational method. We theoretically derived the expression of the boundmagnetopolaron ground-state energy. The energy of the bound magnetopolaron splits under the influence of the Rashba effect. From this phenomenon, it is concluded that the effects of orbital and spin interactions on the polaron energy in different directions must be considered. Because of the contribution of the magnetic field cyclotron resonance frequency to the Rashba spin–orbit splitting, the energy spacing becomes larger as the magnetic field cyclotron resonance frequency increases. Compared to the bare electron, the bound polaron is more stable, and the energy of bound polaron split is more stable.
• # Pramana – Journal of Physics
Volume 95, 2021
All articles
Continuous Article Publishing mode
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019 | 2021-09-25 23:36:00 | {"extraction_info": {"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, "math_score": 0.5018190741539001, "perplexity": 790.4542511852679}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00681.warc.gz"} |
https://socratic.org/questions/591bc49a7c01496f347a5c43 | How many sets of quantum numbers specify all the 2p orbitals?
Jul 30, 2017
Well, there are three sets... The $2 p$ orbitals are triply-degenerate, i.e. $2 l + 1 = 3$, and as such, three sets of quantum numbers specify such orbitals.
$\left(n , l , {m}_{l}\right) = \left(2 , 1 , \left\{- 1 , 0 , + 1\right\}\right)$
Do you spot three here?
Recall:
• The principal quantum number $n$ specifies what energy level one is in: $n = 1 , 2 , 3 , . . .$
• The angular momentum quantum number $l$ describes the shape of the orbital.
$l = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , . . . , n - 1 \leftrightarrow s , p , d , f , g , h , i , k , . . .$.
If $l = 0$, the orbital is sharp (spherical), an $s$ orbital.
If $l = 1$, the orbital is principal (polarized), an $p$ orbital.
If $l = 2$, the orbital is diffuse, an $d$ orbital.
If $l = 3$, the orbital is "fundamental", an $f$ orbital.
• The magnetic quantum number ${m}_{l}$ corresponds to each orbital "orientation", and describes a separate orbital orthogonal to the rest. ${m}_{l} = \left\{- l , - l + 1 , . . . , 0 , . . . , l - 1 , l\right\}$.
These are all three quantum numbers required to describe an orbital. If you want to describe orbital occupations, you'll need to specify the electron spin, ${m}_{s}$. | 2021-01-17 09:59:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "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, "math_score": 0.9649366736412048, "perplexity": 1323.9872183251434}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00731.warc.gz"} |
http://openstudy.com/updates/4f1c9600e4b04992dd23b1fb | Ladybug870 3 years ago Solve the equation be sure to show your work all final answers must be in simplest form. n-1/10=10
$\frac{n-1}{10} = 10$ $10(\frac{n-1}{10}) = 10*10$ $n-1 = 100$ $(n-1)+1=100+1$ $n=101$ now lets check $n=101$ $\frac{101-1}{10}=10$ $10=10$ | 2015-11-25 14:37:10 | {"extraction_info": {"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, "math_score": 0.5986045598983765, "perplexity": 468.13442493259635}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398445208.17/warc/CC-MAIN-20151124205405-00115-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/126529/a-criterion-for-freeness-over-a-local-ring | Let $A=K[[X_1,\dots,X_n]]$ where $K$ is a field. Let $M$ be a finitely generated torsion-free $A$-module, such that
1. for all $k$, the $A[1/X_k]$-module $M[1/X_k]$ is free of rank $d$;
2. for every $i \neq j$, we have $M = M[1/X_i] \cap M[1/X_j]$.
Does this imply that $M$ is free?
It certainly does if $n=1$ (easy), and also if $n=2$ (reduce $M$ modulo $X$), but things seem trickier if $n \geq 3$.
This question comes up when trying to prove that some $(\varphi,\Gamma)$-modules over rings in several variables are free.
-
No, this is false as soon as $n ≥ 3$. A second syzygy $M$ of the residue field $K$ gives a counterexample: each $M[1/X_i]$ is projective, hence free, and it is reflexive, so the second condition is satisfied. On the other hand the projective dimension of $K$ as a module is $n$, so $M$ can not be free.
I think the answer is no. Suppose $M = B$ is a finite ring extension of $A$, say with an isolated singularity, which is both Cohen-Macaulay except over the closed point of $A$ and normal.
Then $M$ cannot be free, since free modules are Cohen-Macaulay. On the other hand, it is free away from the closed point by the Cohen-Macaulay hypothesis. The second hypothesis follows from the S2 assumption. | 2014-10-26 02:07:55 | {"extraction_info": {"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, "math_score": 0.9551827907562256, "perplexity": 104.24046878053903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119653672.23/warc/CC-MAIN-20141024030053-00268-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1180395/how-to-find-union-and-intersection-of-these-relations | # How to find union and intersection of these relations?
Problem:
Let $R_1$ and $R_2$ be the "divides" and "is the multiple of " relations on the set of all positive integers respectively. That is, $R_1 = \{(a,b) | a \text{ divides }b\}$ and $R_2 = \{(a,b) | a \text{ is a multiple of }b\}$. Find
a. $R_1 \cup R_2$
b. $R_1 \cap R_2$
My work:
For a, I understand the concept of a union. I am trying to apply this example, say you have a relation $R_1$ with ordered pairs $\{(1,1), (2,2), (3,3)\}$ and a relation $R_2$ with ordered pairs $\{(1,1),(1,2),(1,3),(1,4)\}$. The union $R_1 \cup R_2$ would consist of the ordered pairs $\{(1,1),(1,2),(1,3),(1,4),(2,2), (3,3)\}$.
Applying that idea here, I got that the union $R_1 \cup R_2$ relation would be $\{(a,b) | a \text{ divides } b \text{ or } a \text{ is a multiple of }b\}$ or $\{(a,b) | a = cb \text{ or } b = ak \text{ for some integers }k \text{ and }c\}$.
Is there someway to simpify this into one expression?
For b , going off the last example, the intersection $R_1 \cap R_2$ would consist of ordered pairs $\{(1,1)\}$. Now applying that idea here, $R_1 \cap R_2$ relation would be $\{(a,b) | a \text{ divides } b \text{ or } a \text{ is a multiple of }b\}$ or $\{(a,b) | a = cb \text{ or } b = ak \text{ for some integers }k \text{ and }c\}$.
Investigating this further, I found that $a = cb$ and $b = ak$ is only true when $k=c= 1$ so overall the relation would consist of $\{(a,b) | a = b\}$. Is that correct?
a) You could say that $R_1\cap R_2=\{(a,b): a \text{ divides } b \text{ or } b \text{ divides } a\}$ as well.
b) Yup. $a=cb$ and $b=ka$ implies $a=c(ka)$, so $ck=1$. Since $c,k$ are integers, $c=k=1$.
• Anything else would obfuscate things I think. Technically, $\max(\frac{a}{b},\frac{b}{a})\in\mathbb Z$ does the trick as well, but I can't come up with anything more clever. – StevenClontz Mar 8 '15 at 5:48
• On that note, I guess $\min(a,b)$ divides $\max(a,b)$ might be what you're looking for. – StevenClontz Mar 8 '15 at 5:49
• Is another way of looking at a be evaluating $\frac ab$ = m and $\frac ba$ = k and saying that the only way $\frac ab$ and $\frac ba$ evaluates to integers is if a = b? – committedandroider Mar 9 '15 at 2:09 | 2020-01-20 09:17:18 | {"extraction_info": {"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, "math_score": 0.7828611731529236, "perplexity": 222.58206218925454}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00215.warc.gz"} |
https://socratic.org/questions/57ea75a711ef6b405443285c | # Question #3285c
Sep 27, 2016
$x = 25 g$
#### Explanation:
The initial mass of the solution was 100g
As 50% (w/w) solute was present in 100g solution , we can say the mass of the solute in initial solution was 50g
After addition of x g solute in 100g solution the changed mass of solution became (x+100)g and total amount of solute became (x+50)g.
So the percentage of solute in final solution became 60%.
So
$\frac{x + 50}{x + 100} \times 100 = 60$
$\implies \frac{x + 50}{x + 100} = \frac{60}{100} = 0.6$
$\implies x + 50 = 0.6 \left(x + 100\right)$
$\implies x + 50 = 0.6 x + 60$
$\implies x - 0.6 x = 60 - 50$
$\implies 0.4 x = 10$
$\implies x = \frac{10}{0.4} = 25$ | 2019-11-19 20:47:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5120914578437805, "perplexity": 5031.340829798745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670255.18/warc/CC-MAIN-20191119195450-20191119223450-00208.warc.gz"} |
https://escholarship.org/uc/item/12h952p6 | Open Access Publications from the University of California
## Published Web Location
https://doi.org/10.1007/s00021-015-0207-8
Abstract
The second-grade fluid equations are a model for viscoelastic fluids, with two parameters: α > 0, corresponding to the elastic response, and $${\nu > 0}$$ν>0, corresponding to viscosity. Formally setting these parameters to 0 reduces the equations to the incompressible Euler equations of ideal fluid flow. In this article we study the limits $${\alpha, \nu \to 0}$$α,ν→0 of solutions of the second-grade fluid system, in a smooth, bounded, two-dimensional domain with no-slip boundary conditions. This class of problems interpolates between the Euler-α model ($${\nu = 0}$$ν=0), for which the authors recently proved convergence to the solution of the incompressible Euler equations, and the Navier-Stokes case (α = 0), for which the vanishing viscosity limit is an important open problem. We prove three results. First, we establish convergence of the solutions of the second-grade model to those of the Euler equations provided $${\nu = \mathcal{O}(\alpha^2)}$$ν=O(α2), as α → 0, extending the main result in (Lopes Filho et al., Physica D 292(293):51–61, 2015). Second, we prove equivalence between convergence (of the second-grade fluid equations to the Euler equations) and vanishing of the energy dissipation in a suitably thin region near the boundary, in the asymptotic regime $${\nu = \mathcal{O}(\alpha^{6/5})}$$ν=O(α6/5), $${\nu/\alpha^{2} \to \infty}$$ν/α2→∞ as α → 0. This amounts to a convergence criterion similar to the well-known Kato criterion for the vanishing viscosity limit of the Navier-Stokes equations to the Euler equations. Finally, we obtain an extension of Kato’s classical criterion to the second-grade fluid model, valid if $${\alpha = \mathcal{O}(\nu^{3/2})}$$α=O(ν3/2), as $${\nu \to 0}$$ν→0. The proof of all these results relies on energy estimates and boundary correctors, following the original idea by Kato.
Many UC-authored scholarly publications are freely available on this site because of the UC's open access policies. Let us know how this access is important for you. | 2022-11-29 21:33:58 | {"extraction_info": {"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, "math_score": 0.714485764503479, "perplexity": 608.5851222011681}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00332.warc.gz"} |
https://g4rud4.gitlab.io/2021/Heist-InCTF-Internationals-2021/ | tl;dr
• Finding default browser and the top visited website.
• Extract timestamp, ID, Hostname of the TeamViewer FileTransfer session.
Challenge Points: 913
Challenge Solves: 15
Challenge Author: g4rud4
Challenge Description
MD5 Hashes:
• A916E26016180D2C5189061D652DC9E1 Heist.7z
• 31f23c78ff99142bad2a778db6a64163 Heist.E01
Initial Analysis
We are given with a disk dump, lets use add the data source to Autopsy. We can see that, the given dump is from Windows 10 Home system, and Owner of the system is Danial Benjamin.
By checking all installed applications, we can see there are 3 web browsers(Google Chrome, Mozilla Firefox, Brave), Slack, Voice Modulator, TeamViewer, AnyDesk etc.
1. What is the default browser that the Heist leader is using on the device?
As there were 3 browsers being installed on the system. Let us check which is the default browser that user is using.
We can find that by checking the following registry key.
NTUSER.DAT: Software\Microsoft\Windows\Shell\Associations\UrlAssociations\{http|https}\UserChoice
From the above highlighting, We get it as ChromeHTML, which means user is using Chrome as his default browser.
So the answer will be Chrome or Google_Chrome.
1. What is the top-visited website in the leader’s system on the default browser?
Now we know Chrome is the default browser. Chrome stores it user data in the following folder Users/Danial Benjamin/AppData/Local/Google/Chrome/User Data/Default/.
In that directory we can find a SQlite database named Top Sites. This database provides us a list of top-visited websites by the user and gives a url_rank for each site. Sorting the table(top_sites) according to url_rank. We can get the top visited website user visited.
As highlighted above, we can see for url_rank - 0, we have https://www.ebay.com/. Converting it to the given format, we get ebay.com
1. When was the latest file transfer session initiated in TeamViewer?
We need to find when the file transfer session initiated in TeamViewer.
TeamViewer stores its user data, in these following locations:
1. C:\Program Files\TeamViewer\
2. C:\Users\<User Profile>\AppData\Roaming\TeamViewer\
Main files of interest would be Connection_incoming.txt & Connections.txt. These files store the incoming and outgoing connections from TeamViewer.
Here is an example representation for data found in Connection_incoming.txt.
Img src: mii-cybersec
On comparing both files, we can have only one file transfer session in Connection_incoming.txt. From that we can get the time initiated, TeamViewer ID, and the Hostname of the remote PC.
As highlighted above, we can see that TeamViewer ID is 920981533, Remote PC’s Hostname is DESKTOP-S34NLCJ, and time file transfer session initiated is 20-07-2021 07:48:50 in UTC.
But for this we need only the time initiated for this question. Which is 20-07-2021_07:48:50.
1. What is the ID, Hostname of that file-transfer session?
We found the TeamViewer ID and hostname of the Remote PC from the previous question.
So the TeamViewer ID and Hostname are 920981533 & DESKTOP-S34NLCJ.
Converting the answer in the given format we get, 920981533_DESKTOP-S34NLCJ.
Flag
Combining all answers we can get the flag. | 2023-01-30 05:50:20 | {"extraction_info": {"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, "math_score": 0.27287641167640686, "perplexity": 8040.731417761188}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00085.warc.gz"} |
https://scipost.org/submissions/2112.09336v2/ | # Triplet character of 2D-fermion dimers arising from $s$-wave attraction via spin-orbit coupling and Zeeman splitting
### Submission summary
As Contributors: Joachim Brand · Ulrich Ebling Arxiv Link: https://arxiv.org/abs/2112.09336v2 (pdf) Date accepted: 2022-05-02 Date submitted: 2022-04-06 02:35 Submitted by: Brand, Joachim Submitted to: SciPost Physics Academic field: Physics Specialties: Atomic, Molecular and Optical Physics - Theory Condensed Matter Physics - Theory Approach: Theoretical
### Abstract
We theoretically study spin-$1/2$ fermions confined to two spatial dimensions and experiencing isotropic short-range attraction in the presence of both spin-orbit coupling and Zeeman spin splitting - a prototypical system for developing topological superfluidity in the many-body sector. Exact solutions for two-particle bound states are found to have a triplet contribution that dominates over the singlet part in an extended region of parameter space where the combined Zeeman- and center-of-mass-motion-induced spin-splitting energy is large. The triplet character of dimers is purest in the regime of weak $s$-wave interaction strength. Center-of-mass momentum is one of the parameters determining the existence of bound states, which we map out for both two- and one-dimensional types of spin-orbit coupling. Distinctive features emerging in the orbital part of the bound-state wave function, including but not limited to its $p$-wave character, provide observable signatures of unconventional pairing.
Published as SciPost Phys. 12, 167 (2022)
We thank both referees for their detailed comments and constructive suggestions that have helped to improve the readability of our manuscript. Following the referees’ comments, and also noticing inconsistencies in the treatment of the bound state threshold in the original manuscript, we have revised the manuscript substantially. The resulting changes to the numerical results and also to formulas, e.g. for the bound state thresholds, now show an enlarged parameter regime for the existence of bound states as well as enlarged parameter regimes where the triplet character of bound states could be observed. Thus the new results confirm and, at times, strengthen the conclusions of the original manuscript.
We elaborate further below on how we have incorporated the referees’ suggestions into the revised version. For convenience, we repeat passages from each individual report before commenting on them. At the very end of this reply, we discuss additional revisions made in the manuscript that were not prompted by the referee reports.
Response to Report 1 and associated changes:
We are grateful to Prof Schmelcher for the positive evaluation of our work, his stimulating thoughts, and the many concrete suggestions for improving the readability of our manuscript. The itemized Main points and Minor comments from the report have been addressed as follows:
Main points:
a. “I think that already at this stage it should be made clear to the reader that the matrix M possesses the property M M^T = 1 and hence q^2 = p^2 in the case of 2D spin-orbit coupling…”
We thank the referee for this helpful suggestion, which we have picked up as detailed below.
a.1.: "A brief discussion in section 2 should be added about the above-mentioned properties of the M matrix."
Below Eq. (4), we have added the sentence ‘The fact that ... diag(1,0).’ to discuss these properties of the M matrix and consequences for 2D-type vs. 1D-type spin-orbit couplings.
a.2.: "An explanation, in the beginning of section 3, which addresses that M M^T = 1 leads to the bound-state energies being independent of the precise form of the 2D spin-orbit coupling would be useful. In addition, it should be explained that the same is not true regarding the wavefunctions which are sensitive to the form of M."
The preamble of section 3 has been extended to discuss these important points in detail. Specifically, we added the part ‘because they give ... [see Eqs. (35)].’
a.3.: "On a related note, in section 5 the authors write ‘It is possible to transform these results into those applicable to other 2D-type spin-orbit couplings by swapping and/or mirroring the relative-momentum axes according to the structure of M-matrices given in Table. 1’. However, no explicit formula is given. Therefore given these issues, the authors should provide an additional appendix where they demonstrate how the wavefunction transforms for the different 2D spin-orbit coupling types."
We found that it wasn’t necessary to relegate these details to an additional appendix, as they can be written down quite compactly. Hence, we have added the detailed description asked for by the referee directly in section 5 as the new penultimate paragraph ‘It is straightforward ... spin-orbit couplings.^2’, including the explanatory footnote 2.
b. "In Eq. (5) [sic: (9)?] the term B_P.Sigma is introduced which is then briefly discussed. Considering the results at the end of section 3.2, I think a more to-the-point discussion regarding the properties of this term would be helpful. As a guide to the authors, in the following, I provide a possible interpretation of the results of Fig. 4(b), based on Eq. (5) [sic: (9)?], which they can verify or falsify based on their data and then update the discussion in sections 2 and 3.2 accordingly."
We thank the referee for sharing his ideas about how the form of the spin-splitting vector B_P could be interpreted, and how this interpretation may explain numerical results for the triplet-state admixture in dimer states. After thinking this through, however, we ultimately decided that treating the Q-dependent in-plane part of B_P as a Rabi coupling is not useful in the general case, especially if the Zeeman energy h is small or absent. Also, the interplay of the Zeeman term B_P.Sigma and the term proportional to lambda q [see new Eq. (12), which was Eq. (9) in the previous version] appears to be less trivial than envisioned by the referee. This is particularly apparent from the new results for the triplet admixture in dimers [new Fig. 5(b)] obtained after correcting an issue with the threshold energy of two-particle bound states (see detailed discussion below). While these considerations have prevented us from adopting the referee’s proposed viewpoint, we decided to follow the suggestion to discuss effects arising from B_P in greater detail right after this quantity is introduced [new Eq. (13)]; specifically in the passage "The last term … or h vanish.". As part of this discussion, we foreshadow similarities and differences between how finite h and finite P affect bound-state properties.
c.: "The derivations in section 2.2 are confusing in my opinion. This particularly applies to those relating to Eq. (23). More specifically, one can easily verify that c_P V_0 is the normalization factor of |psi_b>> ... It is important that the authors comment on this. … The same change also makes clear to the reader that |psi_b(p)> are not orthonormal, a fact which is not mentioned in the current version of the manuscript. In addition, it lifts the ambiguity on how the orbital wavefunctions are calculated in section 5."
We have substantially revised the explanations and derivations in section 2.2 to clarify all the issues raised by the referee in this point. See especially the text and new equations between Eqs. (29) and (34). In particular, the normalization factor is now explicitly and properly defined with the new symbol N_P (for what was previously c_P V_0), and the nonorthonormality of the spin kets |psi_b(p)> is mentioned explicitly [just below Eq. (25)]. Corrections have also been made in equations outside section 2.2 as necessary.
d.: "For the derivation of Eq. (31) the completeness of the two-body helicity spin-basis <Eq. (2) given in the report> has to be employed. A property that is not mentioned in the current form of the manuscript. ... To make the derivation more transparent this property should be added in section 2.1 where the two-body helicity basis is first introduced. This comment should be subsequently referred to in the discussion regarding Eq. (31)."
We have added the completeness relation given as Eq. (2) in the report as new Eq. (19) in section 2.1, introduced by the sentence ‘For fixed P ... this subspace;’. As suggested by the referee, this relation is then referred to in the discussion just above Eq. (41) [the previous Eq. (31)] ‘Employing the resolution (19) ... ‘.
1.: "In Eq. (2) h is introduced but is not defined as denoting the Zeeman energy. The authors should make sure that this quantity is defined in the updated version of their manuscript. In addition, they might consider changing the notation from h to Delta to avoid any possible confusion with the Planck constant."
We now define the Zeeman energy h explicitly after Eq. (2). We decided to stick with the symbol h, as it is widely adopted in the related literature and to avoid a possible confusion of the symbol Delta with an energy gap or pair potential, e.g., in a Bogoliubov-deGennes Hamiltonian.
2.: "In the discussion following Eq. (3), it might be useful to note that M provides a transformation from the configuration to spin space. Moreover, the authors might consider amending the expression regarding lambda(p) by including the expansion of the matrix multiplications <mathematical expression in report>. This change will make the distinction between spatial coordinates and spin coordinates clear."
We found this remark really useful and have included its message in the new paragraph above Eq. (4) [previous Eq. (3)]. In particular, we added the mathematical expression suggested by the referee as our new Eq. (3), and we now refer to q as the ‘vector of momentum-dependent spin splittings’ to distinguish it from the real-space orbital vector p.
3.: "Regarding Eq. (10) it should be commented that alpha_1 and alpha_2 take values from -1 to 1 and describe the helicity states."
We have added the requested information below Eq. (14) [previously Eq. (10)].
4.: "In Eq. (23) it might be preferable to denote c_P as a function c(P), or even as N(P), where the latter points out its role as the normalization constant."
As part of our revisions in section 2.2, we have chosen to absorb c_P into the new normalization factor N_P. See, e.g., Eq. (31) [previously Eq. (22)].
5.: "The term ‘secular equation’ might be unfamiliar to some physicists, it might be a good idea to replace it with some more familiar term characteristic or eigenvalue equation."
We have replaced ‘secular’ by ‘characteristic’.
6.: "In Eq. (29) the transformation from momentum to real space is only done for the relative coordinate while keeping the COM momentum constant, I think that also the latter fact should be mentioned."
We have added a footnote (footnote 1) below Eq. (39) [previously Eq. (29)] to clarify this point.
7.: "During the discussion regarding the reduction to dimensionless quantities, it should be noted that this procedure results in the various quantities mentioned thereafter being measured in the characteristic units of the spin-orbit coupling strength, i.e. ... ."
We have added the passage ‘Introducing characteristic units ... define the dimensionless quantities’ below Eq. (46) to convey the suggested information. We also decided to display the definitions of the rescaled energies more explicitly in the new Eqs. (47).
8.: "When the dimensionless threshold energy \tilde{E}_th is introduced in section 3.1, Eq. (15), which defines this quantity, should also be mentioned as a reminder."
In the revised manuscript, an extended discussion of the energy threshold is now provided at the beginning of section 2.2, which is referenced when introducing \tilde{E}_th in section 3.1 [below Eq. (49)].
9.: "Additionally, at the beginning of section 3.2 the ‘quantity appearing on the r.h.s. of Eq. (37)’ should be replaced by the concrete specification of the quantity, namely, F_P(epsilon_0, h) for P<>0."
We have made that replacement as recommended.
10.: "The quantity \tilde{P} is undefined within the main text, with its definition given only in the caption of Fig. 3. The authors should also define it where it first appears in the main text."
In the revised manuscript, \tilde{P} is now introduced already in the second paragraph of section 3.2.
11.: "During the discussion in section 4, it remains unresolved how the P_y component of the COM affects the results and one has to trace back to Eq. (8) and (9) to verify that it contributes to a ... shift of the bound state energy. The authors should briefly comment on this in section 4."
We have added the sentence ‘The other COM momentum component ... [see Eq. (21)].’ in the first paragraph of section 4 to point out where P_y really matters.
12.: “In the end of section 4, to emphasize the fact that ‘This is important for experiments, as 1D-type spin-orbit coupling is easier to realize, but tuning the interactions to a sufficiently weak strength (e.g. near the zero-crossing of a Feshbach resonance) might prove challenging.’, it would be important to include the appropriate citations.”
We have reworded the corresponding sentence as the revised results discussed below (and seen in the new Fig. 6b) imply less stringent requirements for the interaction strength to be weak. Instead, we point to the potential experimental challenges in tuning to a narrow parameter regime where the binding energy is small and point to Ref. [1] as a general reference to the experimental requirements for 1D-type spin-orbit coupling. This change affects the last sentence of Sec. 4 starting with “While 1D-type spin-orbit coupling …”.
13.: A brief discussion about the envisaged experimental probes should be included in the introduction of the manuscript.
We have added the sentence ‘The polarized spin-triplet character …’ discussing experimental probes to the second-to-last paragraph of the introduction, as suggested.
1. In section 6, which describes the possible experimental detection, a criterion referring to a characteristic momentum scale ... is derived. The authors should briefly comment on how this momentum scale compares to the resolution that is experimentally achievable considering the typical temperatures in state-of-the-art ultracold atom experiments.
The issue of experimental resolution has been addressed in the slightly rewritten section 6, with reference to the experimental work cited as Ref. [44]. Specifically, the changes affect the new sentences ‘As the square root … experiments’ after the formula for the momentum scale and ‘Due to the single-particle resolution … post selection’.
Response to Report 2 and associated changes:
We thank the Referee for their very supportive report. The requested changes/clarifications have been implemented in the revised manuscript as follows:
1.: “The parameter b defined below Eq. (5) gives the relative strength of the Zeeman and SO interaction seems to have the dimension of momentum. Does this parameter then adequately reflects the relative strength of these Zeeman and SO interactions. Would it not be better to use a dimensionless ratio of interaction strengths?”
The appropriate dimensionless parameter measuring the relative strengths of Zeeman splitting and spin-orbit coupling is \tilde{h}=h/(m lambda^2) defined in new Eq. (47b) in the revised manuscript. In contrast, the quantity b=h/lambda corresponds to the momentum scale where Zeeman splitting and spin-orbit spin splitting coincide. After pondering the Referee’s question, we decided that it would be best to eliminate b from the formalism so as to avoid any confusion about its meaning. Hence, we redefined the quantity Z(p) as in new Eq. (6) and made all necessary adjustments in formulae. Now h appears wherever b used to be, which makes physical interpretations much clearer.
2.: “Equations (9) establish that the relative motion of two particles in the COM frame depends on the COM momentum P. This is an unusual situation because the reason one introduces the COM and relative coordinates in the first place is to decouple the Hamiltonian into two commuting parts (the COM and internal Hamiltonians). Hence, two points should be clarified. First, is there really any practical advantage to using the coordinates in Eq. (6)? Second, it would be helpful to mention exactly which interactions are responsible for the coupling between the external and internal degrees of freedom. Are these the SO coupling terms of the kind \sigma_x p_y?”
Yes, the transformation to relative and COM coordinates is helpful because it reduces the dimensionality of the problem from four degrees of freedom (two particles in two spatial dimensions) to two separate problems with two degrees of freedom each (the COM motion being that of a free particle). While the COM momentum still enters the relative-motion problem, it does so as parameter only, which can be considered to have a fixed value. We have added a discussion of the benefits of COM-motion separation in the last paragraph before the start of Sec. 2.1 (‘While the coordinate … generate such terms.’) where we also explain that all forms of spin-orbit coupling considered in this work generate the COM-momentum-dependent terms in the relative-motion problem.
3.: “In Section 2.2 the authors use the Green function approach to obtain the bound states of s-wave interacting particles. Is this the only approach that can be used? It would also be helpful for the general reader to understand how this approach works using a simple example. A reference to the approach being applied to, e.g., two interacting particles in the absence of the SO interaction, would be helpful.
We have added a new paragraph at the beginning of Sec. 2.2 to point the reader to the relevant previous studies of 2D-fermion bound states upon which our work builds, including the seminal work on pairing in 2D without spin-orbit coupling referred to in new Ref. [47].
We felt it would be useful to better describe the origin and value of the energy threshold below which stable bound states exist. We also realized that our previous application of the energy threshold was incorrect in some parameter regimes. To this end, we have made the following changes:
• new Eq. (8) gives the minimum of the single-particle energy dispersion,
• new Eq. (18) gives the minimum of the relative-motion contribution to the two-particle energy, and
• the second paragraph of section 2.2 now contains an extensive discussion about energy thresholds.
In particular, we now explain that two different energy thresholds need to be distinguished, an absolute one [E_th^abs, given in new Eq. (21)] that constitutes the absolute minimum of energy available to two unbound particles, and the threshold E_th^rel given in new Eq. (22) that is the minimum of energy available to two unbound particles with their COM momentum fixed at a given value. We elucidate the relationship between these two thresholds and their different physical meanings to motivate our definition of the bound-state energy using E_th^rel [see new Eq. (23)]. As the two thresholds differ when the COM momentum is finite, we include the critical line corresponding to the more stringent E_th^abs in the plot shown in new Fig. 4(a).
In the process of reconsidering the threshold energy, we realized that our previous definition for E_th does not apply for \tilde{h}>1. As a result of using the wrong E_th, we had excluded a region in parameter space where a stable bound state also exists; this is the region between the dashed black and the solid blue curves in the new Fig. 1(a). We rectified this error, derived the new critical boundary given in Eq. (51) and illustrate its form in the new Fig. 2. We furthermore recalculated the triplet-state admixture for the extended region in parameter space where a bound state exists and find that also the region of triplet-dominated dimers is enlarged. See new Fig. 3. Thus, the new results strengthen our conclusions regarding triplet character of bound states and extend the region of experimental accessibility. We have also recalculated Figs. 4, 5 and 6 (previously Figs. 3, 4 and 5) to implement the correct threshold for the other situations considered in our article (finite COM momentum or 1D-type spin-orbit coupling). The discussion of these results in the main text has been adjusted as needed. All main conclusions remain the same, with the only change being the overall greater prevalence of triplet-dominated dimers. Figures 7, 8 and 9 (previously Figs. 6, 7 and 8) remain unchanged.
The discussion of mathematical details presented in Appendices C and D has been adjusted and extended to account for the corrected form of the energy threshold.
The new references [50] and [51] have been included when discussing features of the new critical boundary Eq. (51), in particular, how it differs from the two-particle version of the Chandrasekhar-Clogston criterion.
### List of changes
Extensive changes were made to the text and figures in response to the referees’ comments and due to the need of distinguishing the two separate dissociation thresholds (where some the previous formulas and figures were incorrect), as detailed in the authors’ response section.
### Submission & Refereeing History
#### Published as SciPost Phys. 12, 167 (2022)
Resubmission 2112.09336v2 on 6 April 2022
Submission 2112.09336v1 on 20 December 2021
## Reports on this Submission
### Report
The authors have addressed all my suggestions in my previous report satisfactorily. I therefore recommend publication in SciPost.
• validity: high
• significance: high
• originality: high
• clarity: high
• formatting: -
• grammar: -
### Strengths
See my previous report
### Weaknesses
See my previous report
### Report
The authors have addressed all of my previous questions and revised the manuscript accordingly. I am happy to recommend the revised version for publication in Scipost Physics.
• validity: high
• significance: high
• originality: high
• clarity: high
• formatting: excellent
• grammar: excellent | 2022-06-27 14:04:38 | {"extraction_info": {"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, "math_score": 0.7447566986083984, "perplexity": 912.0892807182795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00188.warc.gz"} |
https://www.projecteuclid.org/euclid.bjps/1442513451 | Brazilian Journal of Probability and Statistics
Generalizations of some probability inequalities and $L^{p}$ convergence of random variables for any monotone measure
Abstract
This paper has three specific aims. First, some probability inequalities, including Hölder’s inequality, Lyapunov’s inequality, Minkowski’s inequality, concentration inequalities and Fatou’s lemma for Choquet-like expectation based on a monotone measure are shown, extending previous work of many researchers. Second, we generalize some theorems about the convergence of sequences of random variables on monotone measure spaces for Choquet-like expectation. Third, we extend the concept of uniform integrability for Choquet-like expectation. These results are useful for the solution of various problems in machine learning and made it possible to derive new efficient algorithms in any monotone system. Corresponding results are valid for capacities, the usefulness of which has been demonstrated by the rapidly expanding literature on generalized probability theory.
Article information
Source
Braz. J. Probab. Stat., Volume 29, Number 4 (2015), 878-896.
Dates
Accepted: May 2014
First available in Project Euclid: 17 September 2015
https://projecteuclid.org/euclid.bjps/1442513451
Digital Object Identifier
doi:10.1214/14-BJPS251
Mathematical Reviews number (MathSciNet)
MR3397398
Zentralblatt MATH identifier
1328.60051
Citation
Agahi, Hamzeh; Mohammadpour, Adel; Mesiar, Radko. Generalizations of some probability inequalities and $L^{p}$ convergence of random variables for any monotone measure. Braz. J. Probab. Stat. 29 (2015), no. 4, 878--896. doi:10.1214/14-BJPS251. https://projecteuclid.org/euclid.bjps/1442513451
References
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• Bassan, B. and Spizzichino, F. (2005). Relations among univariate aging, bivariate aging and dependence for exchangeable lifetimes. J. Multivariate Anal. 93, 313–339.
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• Choquet, G. (1954). Theory of capacities. Ann. Inst. Fourier (Grenoble) 5, 131–295.
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• Hoeffding, W. (1963). Probability inequalities for sums of bounded random variables. J. Amer. Statist. Assoc. 58, 13–30.
• Huber, P. J. and Strassen, V. (1973). Minimax tests and the Neyman–Pearson theorem for capacities. Ann. Statist. 1, 251–263.
• Huber, P. J. and Strassen, V. (1974). Correction to Minimax tests and Neyman Pearson theorem for capacities. Ann. Statist. 2, 223–224.
• Maccheroni, F. and Marinacci, M. (2005). A strong law of large numbers for capacities. Ann. Probab. 33, 1171–1178.
• Maslov, V. P. and Samborskij, S. N. (1992). Idempotent Analysis. Providence: American Mathematical Society.
• Matkowski, J. (1996). On a characterization of $L^{p}$-norm and a converse of Minkowski’s inequality. Hiroshima Math. J. 26, 277–287.
• Mesiar, R. (1995). Choquet-like integrals. J. Math. Anal. Appl. 194, 477–488.
• Mesiar, R., Li, J. and Pap, E. (2010). The Choquet integral as Lebesgue integral and related inequalities. Kybernetika (Prague) 46, 1098–1107.
• Ouyang, Y. and Mesiar, R. (2009). On the Chebyshev type inequality for seminormed fuzzy integral. Appl. Math. Lett. 22, 1810–1815.
• Pap, E. (2002). Pseudo-additive measures and their applications. In Handbook of Measure Theory, Vol. II (E. Pap, ed.) 1403–1465. Amsterdam: Elsevier.
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• Shilkret, N. (1971). Maxitive measure and integration. Indag. Math. (N.S.) 8, 109–116.
• Suárez García, F. and Gil Álvarez, P. (1986). Two families of fuzzy integrals. Fuzzy Sets and Systems 18, 67–81.
• Sugeno, M. (1974). Theory of fuzzy integrals and its applications. Ph.D. thesis, Tokyo Institute of Technology.
• Wasserman, L. and Kadane, J. (1990). Bayes’ theorem for Choquet capacities. Ann. Statist. 18, 1328–1339.
• Zhu, L. and Ouyang, Y. (2011). Hölder inequality for Choquet integral and its applications. Mohu Xitong yu Shuxue 25, 146–151. | 2019-10-21 11:42:11 | {"extraction_info": {"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, "math_score": 0.6640821695327759, "perplexity": 4622.099327390009}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00343.warc.gz"} |
https://scifi.stackexchange.com/questions/22602/was-snape-romantically-attracted-to-mulciber?noredirect=1 | # Was Snape romantically attracted to Mulciber?
From Slytherincess quoting JKR interview about why Snape hung out with Death Eaters even if it cost him Lily:
He [Snape] wanted Lily and he wanted Mulciber too.
As several people (myself included) noted, a teenage boy would do anything to get the girl he wants; including ditching any social group he is with. Especially a nerdy geek like Severus who doesn't have a pick of all the cheerleaders.
Which leaves only two options - either JKR didn't think this through, never having met a nerd guy in school - OR, she was hinting at something unusual, ala her Albus/Gellert later reveal.
What would be one reason for Snape to be torn between the girl he has feelings for and his buddies? That's right, if he ALSO had feelings for one of the buddies. Which brings the exact quote - He wanted Lily and he wanted Mulciber - back into focus in a new light.
Thus, the question: Did Severus Snape have romantic feelings for fellow wannabe Death Eater Mulciber?
Please note that i'm only looking for canon/JKR information, not random guessing. This means someone may need to ask JKR directly since I don't recall any such details in her interviews.
• My fan fic spidey senses just went off ... ;) – Slytherincess Aug 26 '12 at 15:48
• I think this is a case of ignoring the context of a quote and coming out with some unusual subtext. My comments on @Slytherincess' answer elaborate on my interpretation of the quote. I can reproduce them as a developed answer here, but since I have no new direct quotes from JKR, only interpretation of current quotes, I'm not sure it fits your requirements. – Gabe Willard Aug 26 '12 at 17:22
• @DVK generalizing much? And who is to say that Snape is a "normal geek"? – Dason Aug 26 '12 at 22:08
• @DVK And all this time I thought it was just me that snarked in my basement while brewing stuff. – dlanod Aug 26 '12 at 23:39
• @DVK your premise about geek behavior is flawed. I was a teenage geek as well, and there were a great many things I wouldn't stop doing for any girl, no matter how big the crush. Habit very often defeats teenage love ;) – Andres F. Aug 27 '12 at 2:31
I do not think that is really what Rowling was implying with that statement. The full quote is as follows:
Nithya: Lily detested mulciber,averyif[sic] snape really loved her,why didnt he sacrifice their company for her sake
J.K. Rowling: Well, that is Snape’s tragedy. Given his time over again he would not have become a Death Eater, but like many insecure, vulnerable people (like Wormtail) he craved membership of something big and powerful, something impressive.
J.K. Rowling: He wanted Lily and he wanted Mulciber too. He never really understood Lily’s aversion; he was so blinded by his attraction to the dark side he thought she would find him impressive if he became a real Death Eater.
From the context of the quote, it seems to be saying that Snape wanted what Lily could offer, but he also wanted what Mulciber (and the other Death Eaters) could offer; love and power respectively.
Consider Snape's life and perspectives as a boy/teen/young man. At that age, most guys simply have little concept of the long term consequences of their actions. One theme of Rowling's work is unconditional love. We see it in Lily's sacrifice for Harry, we see it in Harry's sacrifice for his friends. There's a big difference between love and desire. Snape desired Lily, but did he truly love her yet? He loved her presence and her kindness, certainly, but was he truly willing to live his life with and for her instead of just for himself? I don't think so. I think that's why Rowling calls it Snape's greatest tragedy. He chose the Death Eaters' companionship for what they could give him: power. All the people who mocked "Sniveling Snape" would certainly fear Snape the Death Eater. Snape had a clear choice to make: he could love Lily, and live his life with her in peace and quiet, or he could choose the Death Eaters, and finally end his years of torture at the hands of his enemies. As a child who grew up in a loveless world, I don't think Snape truly understood love. He understood fear and power far more. He thought he could choose power, and Lily would be drawn to him because of it. I think Snape began to really understand love as he saw Voldemort's hatred and evil unfold, at exact odds with the Lily he had feelings for from boyhood. When he found out that Lily was a target of Voldemort, it was then that Snape chose to love her more than his own position. He knew he would be largely hated and mistrusted by both Death Eaters and most of the Order from then on, but if by doing so he could protect Lily, then so be it.
We are not told if Snape had a relationship with Mulciber in any way other than as friends and coworkers under Voldemort, but we are told that Snape never stopped having feelings for Lily. His Patronus, an expression of his innermost being, reflected her. After she died, he dedicated his whole life to her memory. When considering the full context of that quote, and comparing what we are clearly told of Snape's life, I don't think we find any canonical support for a Snape/Mulciber relationship.
• +1. I don't agree with some of the premises, but it's a great answer on the whole. As noted, I won't accept it since it is merely inference and not JKR's explicit quote, but that's not at all your fault. – DVK-on-Ahch-To Aug 27 '12 at 10:26
• @DVK I think the combination of the full quote I provided and the summary overview of Snape's life at the end is fairly conclusive canon evidence, honestly. I doubt we'll get better, unless JKR goes really in depth on Snape's inevitable Pottermore entry for book 7. – Gabe Willard Aug 27 '12 at 19:02 | 2019-10-16 09:49:24 | {"extraction_info": {"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, "math_score": 0.18605807423591614, "perplexity": 4476.203403973962}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00344.warc.gz"} |
https://de.maplesoft.com/support/help/maple/view.aspx?path=Student%2FLinearAlgebra%2FMinor | Student[LinearAlgebra] - Maple Programming Help
Home : Support : Online Help : Education : Student Packages : Linear Algebra : Computation : Standard : Student/LinearAlgebra/Minor
Student[LinearAlgebra]
Minor
compute a minor of a Matrix
Calling Sequence Minor(A, r, c, options)
Parameters
A - Matrix r - integer; row to omit c - integer; column to omit options - (optional) parameters; for a complete list, see LinearAlgebra[Minor]
Description
• The Minor(A, r, c) command, where A is an $mxm$ (square) Matrix, returns the determinant of the $\left(m-1\right)x\left(m-1\right)$ submatrix found by omitting the rth row and the cth column of A.
Examples
> $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{LinearAlgebra}\right]\right):$
> $A≔⟨⟨a|b|c⟩,⟨d|e|f⟩,⟨g|h|i⟩⟩$
${A}{≔}\left[\begin{array}{ccc}{a}& {b}& {c}\\ {d}& {e}& {f}\\ {g}& {h}& {i}\end{array}\right]$ (1)
> $\mathrm{Minor}\left(A,3,3\right)$
${a}{}{e}{-}{b}{}{d}$ (2)
> $B≔⟨⟨1,2,3,-4⟩|⟨5,6,-7,8⟩|⟨9,-10,11,12⟩|⟨-13,14,15,16⟩⟩$
${B}{≔}\left[\begin{array}{cccc}{1}& {5}& {9}& {-13}\\ {2}& {6}& {-10}& {14}\\ {3}& {-7}& {11}& {15}\\ {-4}& {8}& {12}& {16}\end{array}\right]$ (3)
> $\mathrm{Minor}\left(B,2,3\right)$
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https://www.nature.com/articles/s41598-020-67669-0?error=cookies_not_supported&code=150ec629-a864-403a-a5a6-cd8d5316db39 | # An expanded repertoire of intensity-dependent exercise-responsive plasma proteins tied to loci of human disease risk
## Abstract
Routine endurance exercise confers numerous health benefits, and high intensity exercise may accelerate and magnify many of these benefits. To date, explanatory molecular mechanisms and the influence of exercise intensity remain poorly understood. Circulating factors are hypothesized to transduce some of the systemic effects of exercise. We sought to examine the role of exercise and exercise intensity on the human plasma proteome. We employed an aptamer-based method to examine 1,305 plasma proteins in 12 participants before and after exercise at two physiologically defined intensities (moderate and high) to determine the proteomic response. We demonstrate that the human plasma proteome is responsive to acute exercise in an intensity-dependent manner with enrichment analysis suggesting functional biological differences between the moderate and high intensity doses. Through integration of available genetic data, we estimate the effects of acute exercise on exercise-associated traits and find proteomic responses that may contribute to observed clinical effects on coronary artery disease and blood pressure regulation. In sum, we provide supportive evidence that moderate and high intensity exercise elicit different signaling responses, that exercise may act in part non-cell autonomously through circulating plasma proteins, and that plasma protein dynamics can simulate some the beneficial and adverse effects of acute exercise.
## Introduction
Physical activity, including structured exercise, is associated with numerous health benefits including enhanced cognition1, reduction in cardiovascular disease (CVD)2, improved cancer outcomes3, and decreased mortality4. Cardiovascular benefits from exercise training have been ascribed to improvements in lipid profiles, blood pressure, and insulin sensitivity and reductions in inflammation, but a substantial portion of the observed cardiovascular benefit remains unexplained by conventional risk factor reductions5. Routine exercise accordingly holds a central place in guideline-directed care for the promotion of cardiovascular and neurological health6,7,8, with current physical activity guidelines proposing moderate and vigorous activity exercise as comparable alternatives for preventing CVD and promoting overall health7. However, mounting clinical evidence suggests that different exercise intensities may confer distinct physiologic and health benefits9, while exercise at high intensity has also been associated with a discrete adverse health risks both acutely10 and over the longer term11. At present, however, the biological mechanisms by which exercise confers beneficial and adverse health effects and the degree to which these mechanisms vary as a function of exercise intensity remain incompletely understood12.
Prior work has explored the impact of acute exercise on cardiac structure13, DNA methylation14, circulating metabolites15, and microRNAs16. Data defining the impact of exercise on the plasma-based proteome and the degree to which the proteome responds differentially to variable exercise intensities are comparatively limited. Several prior studies have examined protein changes in specific tissues (e.g. cardiac or skeletal muscle) using rodent models or human skeletal muscle biopsies17,18, while characterization of circulating proteins in exercise has largely been limited to select cytokines, myokines, and lipokines and focused studies of extracellular vesicle-bound proteins19,20. Plasma-based proteins play fundamental roles in numerous biological processes including growth, repair, and signaling in both disease and health21 and may facilitate exercise-induced cellular, metabolic, and physiologic changes12.
We hypothesized that the human plasma proteome would demonstrate distinct intensity-dependent responses to a single session of exercise and that these acute changes, when integrated over time, might contribute to the beneficial and adverse effects of chronic moderate and vigorous intensity exercise. To address these hypotheses, we employed a well-validated aptamer-based proteomics platform21,22 to measure plasma concentrations of 1,305 circulating proteins before and after acute exercise at intensities chosen to approximate the moderate and vigorous options proposed by clinical guidelines. We then identified genetic loci simultaneously associated with circulating protein levels (protein quantitative trait loci, pQTLs) and with important clinical phenotypes from genome wide association studies (GWAS) to estimate the predicted effect of exercise on relevant human traits.
## Results
### Exercise physiology
Subjects had an average age of 21 ± 1 years, normal body mass index (22.8 ± 2 kg/m2), no known medical conditions (Table 1) and reported similar levels of habitual physical activity (4–6 days/week of exercise and 20–30 miles/week of running). Baseline cardiopulmonary exercise testing demonstrated maximal oxygen consumption of 62 ± 5 ml/kg/min at peak achieved heart rate (HR) of 195 ± 7 beats per minute (100 ± 4% of age-predicted maximum), with ventilatory threshold HR of 182 ± 10 beats per minute (Table 1; Fig. 1a). In a cross-over design, participants subsequently completed 5-mile treadmill runs at both moderate intensity (6 m.p.h) and high intensity (maximal effort) on separate weeks (see study design schematic in Fig. 1a). Average heart rate over the final mile was 150 ± 16 bpm (82% of the ventilatory threshold HR) during the moderate intensity and 187 ± 7 bpm (102% of the ventilatory threshold HR) during the high intensity run (Fig. 1b). All participants experienced a decline in plasma cortisol following moderate intensity exercise (Fig. 1c) and an increase in plasma cortisol following high intensity exercise (Fig. 1d), consistent with prior reports of discordant cortisol responses to these different intensities of exercise23.
### The human plasma proteome responds to acute exercise differentially relative to exercise intensity
Plasma concentrations of 1,305 proteins (see SI Table S1 for complete list) were measured before and immediately following moderate and high-intensity 5-mile treadmill runs. A total of 623 different proteins (48% of measured proteome) were dynamically regulated by acute exercise. Of these, 25 and 439 proteins were uniquely responsive to moderate and high intensity exercise respectively, while 159 changed at both exercise intensities (Supplemental Data File). Overall 184 distinct proteins were responsive to moderate-intensity exercise (14% of measured proteome) (Fig. 1e), while 598 proteins changed with high-intensity exercise (46% of measured proteome) (Fig. 1f), representing a > 3-fold increase in the number of exercise-responsive proteins at high intensity effort.
To further evaluate the impact of exercise intensity, we focused on the 159 proteins modulated by both moderate and high intensity exercise. Comparing fold change at moderate intensity (FCM) to fold change at high intensity (FCH), we observed a range of intensity dependence, with the most intensity-dependent group (n = 22; SI Table S2) increasing by at least 25% more during high intensity than moderate intensity exercise (Fig. 2). In contrast, the least intensity-dependent group of proteins (n = 44) changed to a nearly equivalent degree at moderate and high intensity exercise (FCH within 5% of FCM). All proteins that changed significantly with both types of exercise did so concordantly. Specifically, proteins that decreased during moderate intensity exercise also decreased during high intensity while proteins that increased did so after both exercise intensities. No proteins changed in opposite directions analogous to plasma cortisol.
### Distinct exercise-relevant functional processes are enriched at moderate and high intensity exercise
Enrichment for gene ontology (GO) curated sets was performed to identify functional pathways altered by moderate and high intensity exercise (Table 2). At moderate intensity, the top two processes with the highest enrichment included bone ossification and lipophagy. Proteins related to multiple pathways relevant to the inflammatory response were additionally enriched, including neutrophil, granulocyte, and monocyte chemotaxis and inflammatory cell migration. At high intensity, the top positively enriched protein sets notably included multiple neurologic pathways including both canonical and non-canonical Wnt signaling and neuronal axonogenesis (collateral sprouting). Other pathways enriched with high intensity exercise included the free radical generation, the inflammatory response (monocyte migration, T-cell cytokine production), and vascular smooth muscle cell migration.
### Inferred tissue contribution of exercise-responsive human plasma proteome
In order to discern which tissues might be contributing to plasma proteins, we used a probabilistic model of transcriptional inference to map likely tissue sources for the set of proteins increasing in the plasma with exercise (n = 120 at moderate intensity and n = 250 at high intensity, representing 261 total unique proteins). We observed a 2.1-fold increase in dynamic elevated protein species at high as compared to moderate intensity. At both exercise intensities, proteomic contribution to the plasma involved nearly all organ systems (Fig. 3a), with the most prominent absolute donor tissues being the nervous, cardiovascular, and gastrointestinal systems. Skeletal muscle appeared to be a relatively minor tissue source of donor protein. However, when adjusted for platform representation (Fig. 3b), proteins inferred to derive from skeletal muscle were overrepresented and in contrast proteins derived from the collective gastrointestinal system were relatively underrepresented. Other protein sources enriched relative to the overall platform include blood, cardiovascular, and nervous tissue.
### Exercise-regulated proteins are genetically tied to and simulate observed effects on exercise-associated traits
Out of the 623 exercise-regulated proteins, the plasma abundance of 273 (44%) has previously been linked to 272 protein quantitative trait loci (pQTLs). 55% of identified exercise-regulated pQTL-associated proteins were under cis genetic control, 61% under trans genetic control and 16% under both cis and trans genetic control. Chromosomal positions of the pQTLs and associated exercise responsive proteins (Fig. 4a) reveal widespread involvement across the entire genome. Protein-associated pQTLs were linked to a diverse group of phenotypic traits; the numbers of pQTLs and associated proteins across cardiovascular (coronary artery disease (CAD), blood pressure, and dyslipidemia), neurologic, and oncologic phenotypes are highlighted (Fig. 4b). Finally, pQTLs linked to CAD (Fig. 4c) and blood pressure (Fig. 4d) are shown along with associated exercise-responsive proteins and annotated to depict the proteomic response to exercise. Notably, the simulated impact of exercise on CAD risk loci is heterogeneous, with multiple proteins appearing to contribute in both directions towards increasing and decreasing risk, with 6 of 15 total proteins moving in a direction suggesting benefit. In contrast, the simulated impact on blood pressure showed that the exercise responsive dynamic of protein-pQTL combinations was more concordant and associated with an improved risk profile (12 of 14 proteins moving in a direction consistent with lower blood pressure).
## Discussion
Using an analytical approach that physiologically defines exercise intensity in individual human subjects and leverages the hypothesis-neutral measurement of 1,305 plasma proteins, we characterized and compared the human plasma proteomic response to acute bouts of moderate and high intensity exercise. Our principal findings can be summarized as follows. First, the human plasma proteome is responsive to acute exercise in an intensity-dependent manner. Second, moderate and high intensity exercise stimulate distinct plasma protein changes that correspond to distinct functional biological pathways. Third, the acute proteomic response to exercise appears to derive from systemic tissue sources with enriched contributions from the cardiovascular, neurological and skeletal muscle systems. Finally, by integrating the proteomic response to exercise with available pQTL and GWAS data we were able to simulate the effects of acute exercise with regard to variable impacts on coronary artery disease risk and acute decreases in blood pressure. Taken together, we provide supportive evidence that exercise in part may act through circulating factors, and we highlight the role that circulating plasma proteins may play in mediating some of exercise’s established effects on CAD and blood pressure.
### Distinct proteomic responses to moderate and high intensity aerobic exercise
Endurance exercise is among the most potent health interventions for the primary and secondary prevention of a wide range of adverse health conditions. To date, however, mechanistic underpinnings of how exercise confers its health benefits remain incompletely understood, and the potential role of the plasma proteome remains largely unexplored. We therefore conducted this study with the goals of performing a minimally biased assessment of whether the plasma proteome responds to acute bouts of exercise and to what degree this response varies as a function of exercise intensity. Results of this effort provide several novel insights into the proteomic response to acute exercise. First, we demonstrate that the plasma proteome is responsive to a short bout of exercise, with almost half of ~ 1,300 measured proteins changing significantly at one of the two studied exercise intensities. Second, we found that the proteomic response was consistently bi-directional, with both up- and down-regulation of distinct protein species, suggesting that the observed protein changes were non-random. Third, we show that the human plasma proteomic response varies in an intensity- or ‘dose’-dependent manner and that circulating protein changes at high intensity exercise are greater in both number and magnitude than those observed at moderate intensity.
### Distinct functional pathways are enriched with moderate and high intensity aerobic exercise
The application of established gene sets to the group of exercise-regulated proteins allows insight into biologic processes relevant to exercise and differentially impacted by exercise intensity. Two well established effects of exercise include its ability to prevent osteoporosis24 and to improve lipid profiles25. At moderate intensity exercise we observed that our top two enriched pathways involved the promotion of bone growth and enhanced lipophagy which concerns the degradation and metabolism of lipids. In both cases, exercise-induced proteomic changes are concordant with clinically-observed impacts of exercise training, suggesting that these pathways may be one way through which moderate intensity exercise, when repeated over time, might act (through the plasma proteome) to exert its impacts, in this case to improve bone health and to reduce lipid-associated metabolic risk. Pathway enrichment at high intensity exercise was particularly notable for the prominent role of neurologic processes, highlighting the close interplay of exercise and the nervous system26. Several of the top enriched pathways pertained to Wnt signaling, which is known to play essential roles in the regulation of central nervous system angiogenesis27 and hippocampal neurogenesis28. An additional mechanistic link is suggested by the enrichment of pathways related to neuronal adaptation and axonogenesis (collateral sprouting), which is in line with clinical observations associating aerobic exercise with neurogenesis and synaptic plasticity29.
Notably, the collateral sprouting gene set includes the well-studied brain-derived neurotrophic factor (BDNF). BDNF has been hypothesized to mediate the improvements in cognition and mood observed with exercise, and prior work has documented changes in circulating levels of BDNF with both acute and regular exercise30. However, all of these studies examined moderate intensity exercise or did not report intensity. Our data confirm the exercise responsiveness of BDNF, with levels rising significantly after both moderate and high intensity exercise, and extend the current literature to show that BDNF also appears to be intensity responsive, with high intensity exercise in our platform producing a nearly 30% increase in BDNF levels relative to moderate intensity (Fig. 2; Supplemental Data File). These findings are particularly salient given the rising prevalence of dementia, the absence of efficacious therapies, and links between sedentary behavior and memory loss31.
### Cardiovascular, neurological, and muscular enrichment in the acute plasma proteome
Aerobic exercise requires integrated multi-organ system function, and we expected transcriptional inference to reveal multiple source tissues contributing to the plasma proteomic response. The cardiovascular system experiences workload-dependent increases in pressure and volume stress during exercise, and its role as a major source of circulating proteins was unsurprising. In contrast, the findings of enriched expression of exercise-responsive proteins in the nervous system was unexpected. This novel finding suggests that the nervous system, which is not classically viewed as playing a key role in endurance exercise physiology, responds to exercise and may play mechanistic roles in transducing its health benefits. Further elucidation of the precise protein sources within the neurological (i.e. peripheral versus central neurons or glial cells) system coupled with clarification of downstream effects represent critical areas for future work.
### The proteomic exercise response simulates acute exercise effects on exercise-relevant traits
A pQTL is a genetic locus strongly associated with circulating plasma levels of a given protein. These same genetic loci may be strongly associated with a particular clinical phenotype as documented via a GWAS. Although not proof of causality, integration of these data permits a simulation of the impact of exercise-associated protein changes on a given trait. For example, high levels of a given protein under genetic control may be associated with an adverse trait, and exercise might reduce the level of this protein and thus provide therapeutic benefit. This framework highlights how plasma proteins might integrate the influences of a given gene product with the environment and with behaviors like exercise. Although we found pQTLs across a number of clinical strata (Fig. 4a), we focused these analyses on the acute effects of exercise on CAD and systemic blood pressure given exercise’s well-established and clinically relevant impacts on these traits.
Post exercise hypotension, first described by Hill in 189737, refers to the protracted attenuation in resting blood pressure that lasts for several hours after exercise. The precise mechanism by which this occurs and why humans evolved to have this response remain open questions. Nevertheless, this reduction in blood pressure from a single session of exercise is a reliable post-exercise finding thought to confer some of the beneficial effects of exercise. We observe that the simulated impact of exercise-responsive proteins on blood pressure regulation appears broadly concordant in a beneficial direction, suggesting that the proteomic response appears to reproduce known post-exercise physiology in line with well-established clinical acute and chronic observations38. Taken in sum, these data suggest that exercise may modulate both CAD risk and blood pressure in part through non-cell autonomous mechanisms by influencing circulating proteins tied to risk-conferring genetic loci independent of traditional risk markers. Such a proteomic basis for risk transduction raises the intriguing possibility of therapeutic targets for people with elevated polygenic risk or burdensome disease.
Several limitations of this study are noteworthy. First, while data presented here are among the most comprehensive characterizations to date of how acute exercise perturbs the plasma proteome, we acknowledge that our use of a commercially-available proteomics platform introduces bias, samples only a portion of the vast proteome, and does not represent a complete characterization of exercise’s impact on circulating proteins. Second, we studied a small group of young, healthy, fit males. Future study of females, older participants of both sexes, less aerobically-fit individuals, and patients with established CVD is warranted, and we acknowledge that exercise protein regulation may differ in these populations. While the training status and objective fitness was similar across the study population, we could not evaluate for heterogeneity in proteomic response based on these factors. Further, while our study population did include individuals from different racial backgrounds, we are limited by sample size in our ability to parse racial or ethnic differences in the proteomic effects of exercise. We additionally do not have measures of plasma volume pre and post-exercise, so we cannot ascertain to what extent changes in this may have impacted our results. However, the bi-directional changes in protein concentration suggest that our results were not simply due to hemoconcentration. Finally, our experimental design was not linked to clinical outcomes and was limited to short-duration exercise, with the exact exercise “volume” (running distance) held constant. The majority of exercise’s clinical health benefits is observed in those who transition from sedentary to moderate activity. We thus cannot exclude, and consider it probable, that some of the differences between moderate and high intensity exercise observed in longer-term epidemiologic studies stem from differences in volume; the corollary of this possibility is that some of the proteomic differences described in this study may be attenuated at increased volumes of exercise. The extent to which chronic high volume moderate-intensity exercise approximates lower volume high-intensity exercise remains to be seen. Future work aimed at examining longer- and varying-duration endurance exercise and alternative forms of exercise including strength training represent logical future areas of investigation.
In conclusion, we provide the first comprehensive characterization of how the human plasma proteome responds to acute moderate and high intensity aerobic exercise, and in doing so we expand the repertoire of known exercise-responsive proteins. Functional analyses suggest that distinct proteomic responses translate into the distinct biological functions that underlie numerous exercise-associated traits integral in human health and disease. Overlaying genomic data onto observed protein changes with exercise, we find explanatory congruence between estimated effects drawn from the high intensity proteomic response and clinical observations surrounding CAD risk and post-exercise hypotension. These data support the concept that exercise may confer its beneficial and adverse effects by influencing plasma proteins and signaling through a non-cell autonomous mechanism. These findings set the stage for future work to deconstruct the specific signaling networks through which exercise transduces its benefits and exerts its harms and to determine how the human proteome might be manipulated to promote human health.
## Methods
We conducted a prospective, repeated-measures study examining the physiologic effects of varied exercise intensity. Twelve healthy adult males without known CVD (age 19–24 years) participated in treadmill running sessions at varied intensities (treadmill speed), with blood samples collected before and immediately after each exercise bout for proteomic profiling. We evaluated samples from before and after 5-mile runs at 6 m.p.h. (moderate intensity) and at maximal volitional effort (high intensity). The Institutional Review Board of Massachusetts General Hospital approved this study. Accordingly all elements of the research were performed in accordance with relevant guidelines and regulations and informed consent was obtained from all participants.
### Participant recruitment and exercise testing
Subject recruitment has previously been described16. Inclusion criteria included male sex and ages 18–30 years. Six participants identified as Caucasian. Exclusion criteria included known heart, liver, or kidney disease or a viral illness within the preceding 2 weeks. Informed consent was obtained from all participants. Baseline data included demographics, medical and athletic history, and basic anthropometrics. Each participant underwent a maximal, effort-limited cardiopulmonary exercise test on a treadmill ergometer as previously described16. Subjects were then randomly assigned to complete variable intensity exercise sessions in varying order, with each exercise session completed 1 week apart. Participants abstained from all exercise above and beyond activities of daily living for a minimum of 48 h prior to each exercise session and arrived for exercise sessions following an overnight fast (excepting water). Exercise session start time (09:00), ambient room temperature (69–72 degrees Fahrenheit), and humidity (20–30%) were held constant across visits. Plasma cortisol is known to decline following low-to-moderate intensity exercise and increase following high intensity exercise23. To confirm that prescribed exercise intensities in this study were physiologically distinct, we measured plasma cortisol before and after both exercise bouts. Demographic and exercise data are reported as mean ± standard deviation.
### Aptamer-based proteomic profiling
Profiling methods have been previously described21,22. Venous blood was collected immediately before and after treadmill running from a superficial upper extremity vein using standard phlebotomy techniques. Samples were drawn into standard anticoagulant ethylenediaminetetraacetic acid (EDTA) treated vacutainer tubes (BD, Franklin Lakes, NJ) and spun at 2,700–2,800 RCF in a Medilite Centrifuge (Thermo Scientific, Waltham, MA) for 12 min to separate plasma. Plasma aliquots (400 μL) were frozen and stored at – 80 °C for analysis. Quantitative levels in plasma samples were assayed by the SOMAscan platform (SomaLogic, Boulder, Colorado). Samples were assayed in one single batch (n = 48). A total of 1,305 proteins were assayed.
### Protein source inference
Among proteins whose levels increased in the plasma after acute exercise, we hypothesized this was unlikely reflective of de novo synthesis, given the short timeframe of exercise (approximately 30 min), and more likely represented translocation or active secretion of proteins from tissues into plasma. We sought to infer the tissue sources from which dynamically-regulated proteins most likely derived. To do so, we devised a computational method of transcriptional inference to derive a probabilistic map of likely donor tissue sources. We made two a priori assumptions: (1) a given protein most likely derives from a tissue in which its messenger transcript is found and (2) the probability that a given tissue is the source of a given protein is proportional to the relative expression of the protein’s corresponding gene in that tissue. Sequencing data from the Genotype-Tissue Expression (GTEx) database39 were used to assign a tissue source probability to each protein that increased with acute exercise. For each protein, we assigned a probabilistic weight based on gene expression as reflected in RNA-seq transcript quantification expressed in transcripts per kilobase million (TPM) and recovered from next generation sequencing of human cadaveric tissues (n = 46 distinct human tissues). We defined this relationship as follows:
$$p\left( {Tissue\,a} \right) = k\left( {\frac{{TPM_{a} }}{{TPM_{a} + TPM_{b} + TPM_{c} \ldots TPM_{n} }}} \right)$$
Here p(tissue) is the probability that a given protein derives from a certain source tissue (a), where (a), (b) … (n) represent all the available sites of tissue expression. We defined these probabilities individually for each protein in question and subsequently aggregated across all proteins, with each protein weighted equally and grouped by organ system (SI Table S3).
### Functional annotation and enrichment analysis
To characterize the functional pathways present at moderate and high intensity exercise, we performed gene ontology (GO) analysis on the upregulated proteins derived from respective intensities using an open source tool with expanded curation of functional sets40,41. The GO database was accessed 12/01/2019, with data analyzed using the PANTHER Overrepresentation Test. The Binomial test was used, and p values are reported after Bonferroni correction to adjust for multiple comparisons.
### Trait-based protein annotation
We used previously-reported protein quantitative trait loci (pQTLs)22 to map exercise-regulated proteins to strongly-associated cis and trans sentinel genetic sequence variants. To estimate the anticipated effects of protein changes we queried our sentinel variants against genome-wide association studies (GWAS) data using Phenoscanner42, with GWAS results filtered for genome wide signifigant variants, p < 5 × 10–8. Results were manually filtered to identify cardiovascular and neurological traits. Published beta coefficients were used to estimate the directional effect of exercise regulation.
### Statistical analyses
All proteins were examined for differential abundance analysis using the LIMMA package in R43. Relative changes in protein abundance between resting and post-exercise samples were analyzed using a paired analysis with a Benjamani-Hochberg false-discovery rate of 5% to limit type 1 error and multiplicity44. Statistical analysis was performed in R 3.5 (R Foundation for Statistical Computing, Vienna, Austria). Full results are provided in Supplemental Data File, and individual participant data may be made available upon reasonable request to the corresponding author.
## Abbreviations
Coronary artery disease
CVD:
Cardiovascular disease
GWAS:
Genome wide association study
HR:
Heart rate
pQTL:
Protein quantitative trait locus
V̇O2 :
Oxygen consumption
VT:
Ventilatory threshold
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## Acknowledgements
J.S.G. was supported by the MGH NIH T32 Training Grant (HL007208), the John S. LaDue Memorial Fellowship, the MGH Physician Scientist Development Program, and the AHA-Harold Amos Medical Faculty Development Program. A.R. is funded by the NIH (R01AG061034) and the AHA (16SFRN31720000). A.L.B. is funded by the NIH/NHLBI (Ja).
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Guseh, J.S., Churchill, T.W., Yeri, A. et al. An expanded repertoire of intensity-dependent exercise-responsive plasma proteins tied to loci of human disease risk. Sci Rep 10, 10831 (2020). https://doi.org/10.1038/s41598-020-67669-0
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# Direct atomic scale determination of magnetic ion partition in a room temperature multiferroic material
## Abstract
The five-layer Aurivillius phase Bi6TixFeyMnzO18 system is a rare example of a single-phase room temperature multiferroic material. To optimise its properties and exploit it for future memory storage applications, it is necessary to understand the origin of the room temperature magnetisation. In this work we use high resolution scanning transmission electron microscopy, EDX and EELS to discover how closely-packed Ti/Mn/Fe cations of similar atomic number are arranged, both within the perfect structure and within defect regions. Direct evidence for partitioning of the magnetic cations (Mn and Fe) to the central three of the five perovskite (PK) layers is presented, which reveals a marked preference for Mn to partition to the central layer. We infer this is most probably due to elastic strain energy considerations. The observed increase (>8%) in magnetic cation content at the central PK layers engenders up to a 90% increase in potential ferromagnetic spin alignments in the central layer and this could be significant in terms of creating pathways to the long-range room temperature magnetic order observed in this distinct and intriguing material system.
## Introduction
Multi-state memory devices based on single-phase multiferroic materials (containing both ferroelectric (FE) and ferromagnetic (FM) memory states) have been road-mapped1 as promising architectures for memory scaling beyond current technologies. However, there are relatively few2,3,4,5 materials demonstrating FE and FM properties in a single-phase at room temperature (RT), and consequently no such devices currently exist.
Due to conflicting electronic structure requirements for ferroelectricity and ferromagnetism it has been considered for some time that the two properties tend to be mutually exclusive6. However, recently there has been considerable interest in the engineering of layer-structured materials to accommodate both FE and FM cations within the same structure4, 5, 7. The Aurivillius (Bi2O2(A m−1 B mO3m+1))8 FEs offer a naturally layer structured system in which layers of m perovskite (PK) blocks are sandwiched between (Bi2O2)2+ layers with a structure that approximates that of fluorite. The PK layers readily accommodate magnetic cations (Fe, Mn, Co) on the PK B-sites, and compositions in the system Bim+1Fem−3Ti3O3m+3 (BTFO) have been shown to exist with values of m ranging from 4 to 99 with evidence of antiferromagnetic, weak ferromagnetic and magnetoelectric behaviour10,11,12. Recently, evidence has been reported for exchange-bias and spin cluster glass effects, together with weak ferromagnetism at ca100K in the m = 9 BTFO system13. The inclusion of other magnetic ions such as Co or Mn on the B-site, together with Fe has been pursued with the intention of improving the magnetic properties of the BTFO system, and strong FM and FE has been reported in m = 4 Bi5Fe0.5Co0.5Ti3O15 14. However, it is now well-established that very small (<0.1%) amounts of FM impurities such as CoFe2O4 can be seriously misleading when seeking FM behaviour in these systems15. FE/FM multiferroicity was demonstrated in the single-phase Aurivillius system: Bi6TixFeyMnzO18 (B6TFMO) where x = 2.80–3.04, y = 1.32–1.52, z = 0.54–0.645, 15, 16. Thin films of these materials possess spontaneous polarisation (e.g. Ps~30 μC/cm2 in m = 3 BTFO)17 and saturation magnetisation (MS; up to 215 emu/cm3 in B6TFMO)16 and demonstrate magnetic-field-induced FE domain switching at RT. This is the only material to date in which the multiferroicity has been demonstrated to originate from the Aurivillius phase at a defined confidence level (>99.5%)15. The B6TFMO system is therefore an exciting candidate for potential use in multiferroic, magnetoelectric devices which could potentially meet future industry requirements in multi-state memory applications18, 19.
RT FE/FM multiferroicity in B6TFMO and the other similar compounds is unexpected, given the relatively dilute level of magnetic cations available to undergo FM exchange interactions. There are five PK layers per half-unit cell (m = 5) in this structure, giving three symmetrically-distinct B-site locations over which the magnetic cations (Mn and Fe) can be distributed20 (Fig. 1(a))21. Our knowledge to-date on magnetic cation partitioning9, 22,23,24,25 has mainly depended on inference from structural refinement and statistical methods, which cannot provide information on the localisation of cations for example around defect sites. Previous reports of magnetic cation partitioning within the perovskite (PK) layers in Aurivillius structures, were determined by the Rietveld refinement of X-ray and neutron powder diffraction data. In the m = 3 compound Bi2Sr2Nb2MnO12−δ, the Mn ions were observed to order into the centre layer, away from the (Bi2O2)2+ layers22. In Bi5Ti3CrO15 (m = 4), Giddings et al.23 presented evidence from Reitveld analysis of neutron diffraction data that the Cr ions partition partially into the central two layers. Lomanova et al.9 used Mössbauer spectroscopy to study the BTFO system for 3.5 ≤ m ≤ 9, and showed that for m < 7, the Fe partitions away from the PK blocks adjacent to the (Bi2O2)2+ layers, but that for m ≥ 7 the Fe and Ti are randomly distributed over the PK B-sites. However, Hevoches et al.24 were unable to detect any departure from a random distribution of Fe ions in the compound Bi5Ti3FeO15, by either Rietveld refinement of neutron diffraction data, or Mossbauer spectroscopy. By contrast, partial-partitioning of the Fe away from the PK blocks adjacent to the (Bi2O2)2+ layers, and into the central three PK layers was observed by Rietveld refinement of X-ray diffraction data in the m = 5 compound Bi6Ti3Fe2O18 20. Statistical analysis of extended X-ray absorption fine structure (EXAFS) data for this compound has also indicated a preference for the Fe3+ cations to occupy the ‘inner’ sites within the PK layers25. However, EXAFS is a statistical technique whereby experimental information on unique octahedral sites is refined to theoretically calculated structural models, and no information can be derived on the extent of cation partitioning. Recently, there has been evidence of Fe3+ ion partitioning presented by Huang et al.13 from HAADF-EELS analysis in the m = 9 BTFO system. Here, Fe3+ was observed to partition away from the central layers of the PK blocks and towards the (Bi2O2)2+ layers. This is an interesting result which contrasts with the earlier work9 that indicated a tendency for the Fe to partition into the centres of the PK blocks layers or, for compositions with m ≥ 7, to be randomly distributed. Huang et al.13 do not remark upon this dispartity between their observations and the earlier work, but make the important observation that the structural partitioning of magnetic cations may be implicated in the magnetic behaviour of this type of material. We can note that the disparity between different authors’ observations of Fe3+ cation partitioning in different BTFO systems may be a consequence of differences between sample preparation methodologies. Prior to the work by Huang et al.13, while there had been studies at atomic resolution of PK multi-layer structures26, PK BiFeO3 layers27 and Ruddelsden-Popper faults in perovskite PK BaSnO3 layers28, there had been no direct observations of B-site cation distributions within Aurivillius phase structures, either generally or specifically in the multiferroic B6TFMO system29, 30. As Huang et al.13 have observed, information on magnetic cation distributions is important to further our understanding of multiferroic systems. Nearest neighbour (NN) and next-nearest-neighbour (NNN) interactions between magnetic cations are key in determining the occurrence of ferromagnetism and antiferromagnetism in oxides. For the Bi5Ti3FeO15 (m = 4 BTFO) system, density functional theory (DFT) predicts rather strong couplings (JNN40–50 meV) for Fe3+ cations in NN positions31. However for Fe3+ cations distributed in NNN positions, the coupling is relatively weak (JNNN 1–2 meV) and becomes negligible among second-nearest neighbours. The juxtaposition and arrangement of Mn-O-Fe and Mn-O-Mn linkages is important in determining the existence of ferromagnetism, or antiferromagnetism, depending on the valence states of the ions concerned and the bond lengths32,33,34. Therefore, as both Birenbaum and Ederer31 and Huang et al.13 have pointed-out, knowledge of the B-site distribution of the magnetic cations in the Aurivillius multiferroics in general is important in understanding their ferromagnetism. We consider here specifically the distribution of Mn/Fe cations in B6TFMO for the same reason.
Here we use atomic-resolution aberration-corrected high-angle annular dark-field scanning transmission electron microscopy (HAADF-STEM) combined with EDX (energy dispersive X-ray analysis) and EELS (electron energy loss spectroscopy) analytical techniques to differentiate between closely-packed cations with similar atomic number (Z) and distinguish Ti/Mn/Fe cations dominating at certain B-site locations. We obtain direct atomic-scale visualisation of magnetic cation partitioning and demonstrate a clear preference for Mn cations to partition into the central PK layer, which we infer is due to strain- and electrostatic-energy considerations. We reveal further changes in the relative proportions of magnetic ions in the region of stacking fault defects and Out-of-Phase Boundaries (OPBs). We believe this evidence to be central in explaining pathways to long-range magnetic order and the distinct RT multiferroic properties of this tantalising material system.
## Results
### Atomic Resolution Imaging and Chemical Mapping
Two samples, B6TFMO-A and B6TFMO-B, of similar composition, but from different growth runs, were selected for imaging experiments. Both samples are FE and FM at RT with MS values up to 215 emu.cm−35, 16. By sampling through the depth of the B6TFMO samples along a column of atoms, atomic-resolution HAADF-STEM images of differing magnification were collected (Fig. 1(b) and Supplementary Figure S1). Based on intensity and cation radii differences, the heavier (Z = 83) and larger (radius ≥ 1.11 Å)35 Bi cations at the fluorite-type layers and at the A-sites of the PK layers can clearly be distinguished from the transition metal cations at the B-sites of the PK layers. This data, in conjunction with X-ray diffraction and transmission electron microscopy (TEM) images5, 15, 16, confirms that the majority of the samples are m = 5. By uncorrected STEM-EDX/EELS mapping, it is impossible to distinguish between Ti/Mn/Fe (Z = 22, 25 and 26, respectively) cations in a closely-packed material at atomic-scale. The NION is equipped with a 100 mm2 windowless EDX detector combined with a probe corrected cold field emission electron source which allows atomic-resolution mapping with high sensitivity, making it possible to chemically identify not only the Bi cations but also between the Ti, Mn and Fe cations dominating at B-site locations in B6TFMO. Notably, from the Ti maps in both samples (Fig. 2(d)) and Supplementary Figure S2(d)), we observe an increased signal for Ti within the columns of atoms at the outer (O) PK layers, compared with the intermediate (M) and centre (C) PK layers. On the other hand, we observe a striking decrease in signal for Mn within the columns of atoms at the O PK layers and note an obvious preference for Mn to partition into the M and C PK layers (Fig. 2(e) and Supplementary Figure S2(e)). Atomic-resolution EELS images (Fig. 2(g) and Supplementary Figure S2(g)) confirm these observations. The average relative B-site proportions of Ti, Mn and Fe at the O, M and C layers of samples B6TFMO-A and B6TFMO-B are presented in Supplementary Table S1 and in the histograms in Fig. 3. The B-site proportion (expressed in terms of percentage of B-site occupancy) of Ti decreases considerably (approx. 20%) from the O layers to the C layers. From the spread in the measured proportions made by averaging along 12 nm lengths of four separate PK layers of each type in Sample A and 6 nm lengths of three separate PK layers of each type in Sample B (see the Methods section below), and from the two (unrelated) samples A and B, we estimate the uncertainty in these proportions as being ca 3%. There is a significant (approx. 17 to 20%) increase in the B-site proportion of Mn from the O layers to the C layers. Overall, there appears to be a slight preference (3 to 5% increase) for Fe to partition into the inner PK sites rather than the O sites, but this preference is not significant. The preference for Mn to partition into the C PK sites and for Ti to partition to the O PK sites leads to a significant (8 to 9%) increase in the proportion of magnetic cations at the C layer sites compared with a fully a disordered distribution of B-site cations.
Some regions of the samples display OPB defects where there is displacement by a fraction of a lattice parameter (c/x) between two adjacent B6TFMO layers parallel to the z-direction. These are a common occurrence in materials of high structural anisotropy, such as the Aurivillius phases, although the type and density of these defects can vary widely from sample to sample and from grain to grain within samples36, 37. The OPB density for B6TFMO-A and B6TFMO-B is not high enough in this case to cause peak splitting in conventional X-ray diffraction analysis16, 36; but the atomic structure of the OPBs can easily be seen at the local scale by this optimised HAADF-STEM technique and appear as steps as viewed edge-on in Fig. 4. These can form when the c-axis is inclined with respect to the substrate surface allowing propagation of OPBs through the film, as delineated by the arrows in Fig. 4(a) or when adjacent, out-of-phase nuclei meet during growth as indicated by arrows A/B/C in Fig. 4(b) 37. OPBs can be accompanied by an insertion of an ‘extra’ perovskite block, giving regions of different m-layers, as evident by m = 6 regions in both figures. Stacking faults can also be visible in Fig. 4(b) (highlighted by blue rectangles), where there are sub-unit-cell intergrowths of different m-layers (e.g. m = 6, m = 4) and can terminate in an OPB (arrows D/E/G). It is obvious from these images that physical interruption of the lattice is compensated by a deficiency or excess of bismuth atoms. This is further demonstrated by atomic-resolution EDX (Fig. 5, and the data in Tables 1 and 2) and EELS (Figure S3(i–p)). Atomic-resolution chemical mapping of a stacking fault region (blue rectangle A in Fig. 5(a)) with m = 6 and m = 4 intergrowths reveals that Ti content is increased and Mn content is drastically decreased locally at the defect (Layer 6/region A in Fig. 5(a)) compared with the defect-free average (Supplementary Table S1). The overall effect of the stacking fault on the average proportion of Mn in layers 1–10 (Fig. 5(a and c) and Tables 1 and 2) is only a ~2% decrease compared with the defect-free average, which is not significant.
The considerable effect that an OPB has on the local stoichiometry in the vicinity of the OPB is demonstrated by the green ellipsoid B in Fig. 5(a) and the EDX data in Table 1 (region B). A substantial increase (~30%) in Ti compensates for the deficiency in Bi, while the Fe concentration is depleted (~17%) and there is virtually no Mn present (Fig. 5(a,b), Table 1) at the OPB defect region. On the other hand, the presence of OPBs does not have a significant effect on the larger regions. For example in layers 1–11 in the region between OPBs C and D in Fig. 5(c), the amounts of Ti, Fe and Mn are not changed significantly (Table 2) compared with the defect-free average (Supplementary Table S1). It is important to note that there is a significant ~13% (~8%Mn and ~5%Fe) increase in magnetic cation partitioning to the C PK layers (layers 3/4/9) for this region between OPBs, compared with a ~9% (~6%Mn and ~3%Fe) increase in partitioning to the C PK layers for defect-free regions. It therefore appears that disruptions of the Aurivillius phase lattice by OPBs may result in further partitioning of magnetic cations to the C PK layers of B6TFMO.
OPBs and structural defects can have considerable impact upon material properties38. Changes in local chemistry at the defects and corresponding changes to magnetic cation distribution in the defect regions could be significant in determining magnetic behaviour in B6TFMO and its sample-to-sample variability. As seen in this work and in previous TEM studies16, the defects vary strongly from grain-to-grain and even vary from region-to-region within the same grain. We have also seen that the extent of magnetoelectric switching varies from sample-to-sample and from grain-to-grain in B6TFMO5, 16. Correlating magnetic and magnetoelectric properties with structural defects is the subject of on-going/future work.
## Discussion
### Mechanisms driving cation partitioning in the BT6FMO Structure
The observation that Mn partitions preferentially into the C layers of the 5-layer Aurivillius layer structure of B6TFMO is extremely interesting in terms of its implications for magnetic properties and has not been directly observed before. The chemical vapour deposition technique used to synthesise the B6TFMO samples16 does not impose any layer-by-layer ordering of the precursors during the deposition process, which must therefore occur as a natural part of the fabrication process. There are three possible mechanisms for cation partitioning of the cations.
Garcia-Guaderrama et al. start with Kikuchi’s model39 for Aurivillius phase materials. This model, which follows Armstrong and Newnham40, uses ionic radius arguments to deduce that the PK blocks are under compression, and the (Bi2O2)2+ layers under tension, providing a strain energy to drive cations larger than Ti4+ away from the layers closest to the (Bi2O2)2+ layers. Giddings et al.23 propose two reasons for the partitioning of Cr3+ way from the outer layers in the m = 4 Bi5Ti3CrO15. The first is the preference for Ti4+ to occupy more-distorted octahedra (a reason also proposed by Lomanova et al.9 in an analysis of several different Bi4Ti3O12−BiFeO3 compounds) with the degree of octahedral distortion determined by the Shannon site distortion parameter41 Δn (see Supplementary Information). The second is based upon Madelung energy calculations indicating an electrostatic preference for the more-highly-charged Ti4+ cation to partition close to the outer layer of oxygens in the (Bi2O2)2+ layers. Thus the three potential mechanisms to drive magnetic cation partitioning are: strain-energy considerations, the preference of Ti4+ to occupy more-highly-distorted octahedra and electrostatic energy considerations.
Calculation of Δn values for RT B6TFMO based on the previously-published structure20 indicates values of 0.70, 0.46 and 0.05 respectively for the O, M and C layer octahedra (Fig. 1(a)). These are much smaller than the values reported for Bi5Ti3CrO15, and would be further reduced at the synthesis temperature, although the M and C sites would probably become more similar at higher temperatures23. These results indicate that site asymmetry is not a strong driver for cation partitioning in B6TFMO, leaving strain energy considerations39 and electrostatic interactions as being the most likely mechanisms. Of these, the calculations of Birenbaum and Ederer31, 42 indicate that the strain energy factor is more important.
Consider the high-spin31 ionic radii (in 6-fold coordination) for the B-site ions in the B6TFMO system (Supplementary Table S2)34. The ionic radii for Fe3+ and Mn3+ are both slightly bigger than Ti4+. According to Kikuchi39, that would drive these ions away from (Bi2O2)2+, and as both are trivalent, the electrostatic argument for partitioning would also apply. However, the ionic radii for Fe3+ and Mn3+ are very similar (and the charges are identical), so these mechanisms by themselves would not explain why Mn is preferentially pushed into the C PK blocks. However, Mn2+ is much bigger than both Ti4+ and Fe3+.
We know (see Supplementary Information) that it is energetically much easier to reduce Mn3+ to Mn2+ than Fe3+ to Fe2+ at the B6TFMO synthesis temperature (1100 K), so a substantial part of the Mn in the system will be present as Mn2+. Mn2+ is 33% bigger than Ti4+, but Fe3+ is only 6% bigger than Ti4+, so there is a much larger strain-energy (Kikuchi)39 impetus towards partitioning of Mn2+ than Fe3+. Similarly, we would expect that the divalent Mn2+ would have a stronger preference than Fe3+ for partitioning away from the outer oxygens of the (Bi2O2)2+ layers. On the other hand Fe3+ would still, as a trivalent ion slightly bigger than Ti4+, have some drive for partitioning. This would explain our observation of a 15% increase in the B-site proportion of Mn at the C layers, compared with only a 3–5% increase in Fe partitioning to the C layers. This only slight preference for Fe to partition into the C sites is consistent with Mössbauer data9 and DFT calculations for the 4-layered Bi5Ti3FeO15 31.
We can use the Boltzmann equation (see Supplementary Information) to model the expected distribution of the Mn and Fe within the three PK layers:
$$C(x)=C(0)\exp \{\frac{-b{x}^{2}}{2\sigma }\}$$
(1)
Which is a Gaussian function, with
$$\sigma =\sqrt{\frac{kT}{2b}},{\rm{or}}\,{b}=\frac{kT}{2{\sigma }^{2}}$$
(2)
The experimental values of C(x) (Supplementary Table S1) have been fitted using this equation and the modelled values are plotted for both B6TFMO-A and B6TFMO-B in Fig. 3. It can be seen that the fit is a good one, and the two values of σ are 0.51 ± 0.01α and 1.6 ± 0.1α for Mn and Fe respectively, corresponding to the values of b being 0.182 eV/nm2 and 0.0185 eV/nm2 for Mn and Fe respectively. This means that the energy difference between the three available sites varies ten times faster for manganese relative to iron. The energy differences between the C-to-M and C-to-O PK B-sites are given in Table 3.
As the temperature is reduced during sample cooling, we would expect the Gibbs free energy ΔG for Mn reduction to become positive at approximately 1060 K. Below this temperature, Mn2+ will probably start to convert back to Mn3+ in the oxidising atmosphere. Hitherto there has been no direct determination of the oxidation state of Mn in the Aurivillius bismuth titanates, although it has been inferred from detailed stoichiometry studies5 that there are variable amounts of Mn3+ and Mn4+ in different grains of B6TFO thin films. In pure perovskites such as lead zirconate titanate43, 44 and barium titanate45, 46 ceramics it has been shown by Electron Paramagnetic Resonance (EPR) that Mn exists as Mn2+, Mn3+ and Mn4+, depending on the partial pressure of the oxygen in the firing atmosphere, with a tendency to form Mn4+ as the atmosphere becomes more oxidising, or a tendency to form Mn2+ in the presence of donor dopants such as La3+ or Nb5+. In Mn-doped Pb(Mg0.33Nb0.67)O3-PbTiO3 single crystals, the Mn has been shown by X-ray Absorption Spectroscopy47 to be mainly present as Mn4+. In the case of the B6TFO studied here, which is fired in an oxidising atmosphere, with no donor dopants presence, we would expect any Mn2+ that is present at the synthesis temperature to convert back to Mn3+ and Mn4+, and for oxygen to diffuse back into the lattice to compensate. The activation energy for an oxygen vacancy to diffuse in PKs is approximately 0.7 eV, while that for a B-site cation is between 10 and 15 eV47. Hence, we would expect that the oxygen stoichiometry in the system would easily normalise to compensate for any change in Mn oxidation state, but the cations would tend to stay locked in the positions determined during the film annealing process. The direct determination of Mn oxidation state in this system is an aspect of considerable interest because of its potential importance for magnetic properties, and will be the subject of future study.
The results from the localised compositional analysis at defects agrees with these observations. Region A in Fig. 5(a), for example, is a short region (seven octahedra) of PK units that is both adjacent to, and bounded on each edge by, (Bi2O2)2+. It would be reasonable to expect this region to be under greater compressive stress as a consequence, and indeed there is no Mn and very little Fe in this region. The localised 6-layer regions between OPBs are also richer in magnetic cations (especially Mn) in their C PK layers than the average 5-layer structure. Again, this would be expected from the non-linear energy model expounded above.
### Cation Partitioning and Magnetic Properties
The B-site magnetic cation content (47%) for disordered B6TFMO is above the percolation threshold (31% of magnetic sites)42 to permit NN coupling interactions. However, at the threshold, the number of NN super-exchange interactions is often too low to allow for RT magnetisation. An increase in the percentage of magnetic cations above the percolation threshold at the C B-sites (>8.5% on average and greater in the region of OPBs) would be expected to lead to an increase in the magnetic ordering temperature4. The DFT calculations performed by Birenbaum and Ederer31 have predicted the magnetic coupling strengths between NN and NNN interactions for the m = 4 BTFO system and shown that the strength of NN couplings for magnetic cations located within inner sites (J NN ~43 to 46 meV) is more than twice that for adjacent inner to outer site (J NN ~12 to 20 meV) NN coupling interactions, stressing the potential importance of the site-site interactions in the inner PK layers in determining magnetic behaviour. They also predict that in this system (with Fe3+ as the only magnetic cation), while for NN couplings, AFM coupling has lower energy than FM, (as would be expected from the Goodenough-Kanamori (GK) rules)32, there is little energy difference between the AFM and FM orientations for NNN and greater interactions. Moving to the possibilities within the B6TFMO system, according to the GK rules, Fe3+-O-Mn4+ or Mn3+-O-Mn4+ interactions are expected to show FM coupling via double-exchange mechanisms33. Furthermore, FM coupling of Mn3+-O-Mn3+ is also possible via semicovalent exchange e.g. for the longer Mn3+-O-Mn3+ bonds arranged perpendicular to the (100) planes in the perovskite-type manganites33. A decrease in the c/a ratio of the Aurivillius structure31 (and an increase in the in-plane Mn3+-O-Mn3+ bond lengths) as a result of increased partitioning of the larger magnetic cations to the C PK layers may lead to an increase in FM semicovalent interactions (see Supplementary Information). Hence, magnetic cation partitioning to the C sites in B6TFMO could be significant in permitting pathways to long-range RT magnetic order through an increase in the probability of potentially-ferromagnetic Fe-O-Mn or Mn-O-Mn NN interactions in the C and I layers, which are in any case going to be more important in mediating FM orientations than the O layers31. We can estimate the magnitude of this effect by using the relative proportions of the magnetic ions on the B-sites in the C layer. The probability of there being a potentially-ferromagnetic Fe-O-Mn or Mn-O-Mn NN interaction for an ion sitting in a C layer B-site is given by (see Supplementary Information):
$${P}_{F}=2{P}_{Fe}^{C}{P}_{Mn}^{C}+\,{P}_{Mn}^{C}{P}_{Mn}^{C}$$
(3)
where $${P}_{X}^{C}$$ is the probability of finding the ion X on a C layer B-site.
We find that P F = 0.1 for a B6TFO lattice with randomly distributed Fe and Mn ions, but is increased by 67% to P F = 0.17 for the partitioned proportions reported in sample A and by 88% to P F = 0.19 for the proportions reported in sample B. The corresponding increase for the I layer averages 50% for the two samples. It seems likely that this will have a positive effect on the percolation of magnetic spins in the system, and thus the probability of large-scale ferromagnetic effects emerging. We note that in this system, magnetic percolation is likely to take place through the spins on the Fe3+, Mn3+ and Mn4+ ions in both the C and I layers, with the potential for coupling both within and between the layers. Although the calculation of the percolation thresholds for this system are complex and beyond the scope of this work, we note that a value of P F = 0.19 is of the same order as the percolation threshold that has been calculated for 2D and 3D systems when higher than NN interactions are considered49, 50.
It is recognized that in complex structures such as this, it would be interesting to have 3D compositional information, and atomic resolution imaging along other zone axes will be the subject of further work. The potential effects of defects such as stacking faults and OPBs may also be important in affecting the distribution of the magnetic cations. We note from this work that the Mn content of the lattice is depleted in the immediate region of an OPB, and therefore must be increased to some extent in the rest of the lattice. Hence, a high density of OPBs, as has been observed in some Aurivillius systems36, 38 and can be deliberately introduced through non-stoichiometry51, may be effective in increasing the degree of magnetic spin percolation, although more work is required to investigate this. Further work, perhaps using EPR, is also needed to determine the valence states of the magnetic ions in this B6TFMO system and this, coupled with more detailed theoretical studies should help engender a better understanding of the ferromagnetism that has been observed experimentally.
## Conclusions
We have shown that aberration corrected HAADF-STEM combined with EDX and EELS is capable of distinguishing between Ti/Mn/Fe of similar Z in a closely-packed material at the atomic-scale and this has provided clear atomic-scale demonstration of magnetic cation partitioning between the perovskite layers in in a multiferroic Aurivillius system, with some further localisation due to defects. In our samples, we have observed a strong preference for Mn to partition to the central PK layers, but only a slight tendency for the Fe to partition. The clear preference for Mn cations to partition into the C PK layers of B6TFMO is attributed to a combination of relieving the compressive stress at the (Bi2O2)2+/outer PK layer interface and to facilitate preferential occupation of the outer sites with cations having a larger effective electronic charge (Ti4+). A model has been presented to explain this partitioning based on the partial reduction of Mn3+ to the significantly-larger Mn2+ ion at the film synthesis temperature, an effect which is not expected to occur for Fe3+ in oxidizing conditions at this temperature. A Boltzmann-distribution-based model has also been presented that accurately describes the cation partitioning profiles, and indicates that the driving energy for Mn partitioning is ca 10x greater than that for Fe. The magnetic cation partitioning results in a notable increase in B-site magnetic cation composition from ~47% (~13%Mn) when considering a disordered structure to >54% (>19%Mn) with the observed partitioning at the C sites. We noted in the introduction that there is contradictory evidence for Fe ion partitioning ion the BTFO system, and this may well be due to differences in sample preparation methodologies (synthesis temperatures, atmospheres etc). Such differences could be expected to change the degree to which Fe3+ will reduce to Fe2+ during synthesis, and according to the model presented here, this would change the degree to which it would partition. Higher synthesis temperatures and/or more reducing firing atmospheres would have the effect of increasing the amount of Fe2+ during synthesis. Previous modelling of cation partitioning in the Aurivillius system31, 42 has been undertaken at room temperature, while the cation distribution necessarily takes place at the oxide synthesis temperature. We suggest that consideration of the cation oxidation states and lattice strains at these much higher temperatures may be a fruitful area for future theoretical research on cation partitioning in this system, and this insight may aid the experimental synthesis of materials in which higher magnetic spin percolation densities can be obtained.
Atomic-resolution chemical mapping reveals that the atomic structure at the defect regions deviates from stoichiometric composition and that disruptions to the lattice by e.g. OPBs results in further partitioning >59% (>20%Mn) of magnetic cations to the C PK layers of B6TFMO. The consequences and control of defect densities in this system may well be important in determining magnetic properties and requires further study. We have noted that the partitioning of Mn into the central PK layers will increase the probability of the potentially-ferromagnetic Fe-Mn and Mn-Mn nearest neighbour (NN) interactions. We calculate that this will be a nearly 90% increase in the central layer for the proportions observed here, and correspondingly reduce the probability of antiferromagnetic Fe-Fe NN interactions. We feel that these observations should strongly encourage further theoretical studies of this type of system, which hitherto31 have not considered the consequences of including Mn as a magnetic ion. Further experimental work is also required to determine the oxidation states of the Mn in this system.
## Methods
B6TFMO-A and B6TFMO-B were synthesized at separate times and in different growth runs, on c-sapphire substrates by liquid injection chemical vapour deposition (LI-CVD) methods11 and post-annealed at 850 °C (1123 K). The average stoichiometries, as determined by HR-SEM (high resolution scanning electron microscopy) with EDX, were Bi6Ti3.04Fe1.42Mn0.54O18 (B6TFMO-A) and Bi6Ti2.99Fe1.46Mn0.55O18 (B6TFMO-B). Analysis of B6TFMO A and B6TFMO B (over sample areas of 25 × 25 µm2 up to 1.2 × 1.2 mm2) was conducted using the FEI Helios Nanolab HR-SEM mode and EDX equipped with an X-Max 80 detector and AZTec analysis software from Oxford Instruments. Cross-sections of the B6TFMO films were prepared using a FEI Dual Beam Helios NanoLab 600i Focused Ion Beam (FIB) (final thinning @ 93 pA, 3 kV, final polish 2 kV, 28 pA) and were mounted on TEM grids. Thin coatings of Au were deposited on the sample surfaces to avoid possible surface charging effects. The samples were cleaned using a Fischione 1020 plasma cleaner prior to STEM imaging and EDX/EELS analysis. Imaging and analysis was performed on a NION UltraSTEM 200 operating at 200 kv. EDX analysis was performed using Bruker 100 mm2 windowless EDX detector. EDX acquisition and analysis was performed using Bruker Esprit 2.0. A Gatan Enfinium and GMS 2.0 was used for EELS acquisition and analysis. Energy filtered images acquired at 300 kV on an FEI Titan TEM with Gatan Tridiem Energy Filtering system demonstrated that thicknesses of the regions used for imaging were <35 nm. Further experimental details of the microscopy are given in the Supplementary Information. Images were taken along the [110] crystallographic zone axis. The EDX signal intensities for Ti, Fe and Mn were used as a proxy for the relative proportions of each atom on each of the five PK layers in the structure, normalized to 100% B-site occupancy. It has been noted52, 53 that inelastic scattering of electrons can lead to substantial differences between EDX measurements of composition and actual values. The inelastic scattering is proportional to the atomic numbers Z of the elements concerned, increases with sample thickness and is dependent upon the degree of electron channeling. In these measurements, the Z values of the atoms concerned are similar (Z Ti = 22, Z Mn = 25 and Z Fe = 26), so any corrections for inelastic scattering would also be expected to be similar. In addition, it is the variation in the relative proportions of the magnetic ions across the PK layers that are of greatest interest in this work, and Fe and Mn differ only by one electron. The measurements were only taken along the [110] zone axis in each sample, so sample-to-sample effects due to channeling variations do not need to be taken into account. In addition, the sample was <35 nm thick and EDX intensity data was collected from, and averaged across, several different regions of different thicknesses of each sample. For these reasons, we believe that it is a fair assumption that the EDX intensities represent a good proxy for the relative atomic proportions of each atom. The fact that two unrelated samples of different thicknesses gave similar results, also gives confidence that this assumption is fair. A degree of signal-to-noise improvement was obtained by summing the EDX signal information for each ion along each layer for a total length of about 48 nm (or ca 360 unit cell lengths) for sample A and for 18 nm (or ca135 unit cells lengths) for sample B (see the green lines in Supplementary Figure S4(a and b)). This summation had the effect of integrating-out spatial noise.
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## Acknowledgements
This publication has emanated from research conducted with the financial support of the Royal Society and Science Foundation Ireland (SFI) University Research Fellowship UF 140263 and SFI funded centre AMBER (SFI/12/RC/2278). The authors are grateful to Ahmad Faraz for growing one of the samples used in this study.
## Author information
Authors
### Contributions
L.K. designed the materials, conceived the HAADF-STEM project, took part in the HAADF-STEM experiments and analysed the data. C.D. performed the HAADF-STEM experiments and compiled the data, while V.N. supervised this work. M.S. performed HR-SEM average compositional analysis of the samples and prepared ultra-thin cross-sections of the samples for subsequent HAADF-STEM analysis. R.W.W. conceived the original project to seek multiferroic behavior in this system, and undertook the theoretical analysis of cation partitioning and its effects on magnetic percolation. M.P. managed the AMBER centre at Tyndall/UCC. L.K. and R.W.W. drafted the manuscript. All authors discussed the results and contributed to the writing of the manuscript.
### Corresponding author
Correspondence to Roger W. Whatmore.
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Keeney, L., Downing, C., Schmidt, M. et al. Direct atomic scale determination of magnetic ion partition in a room temperature multiferroic material. Sci Rep 7, 1737 (2017). https://doi.org/10.1038/s41598-017-01902-1
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• & Roger W. Whatmore | 2021-04-19 07:43:09 | {"extraction_info": {"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, "math_score": 0.7013112306594849, "perplexity": 4291.214753808966}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00360.warc.gz"} |
https://yutsumura.com/tag/back-substitution/ | Tagged: back-substitution
Problem 654
Suppose $M$ is an $n \times n$ upper-triangular matrix.
If the diagonal entries of $M$ are all non-zero, then prove that the column vectors are linearly independent.
Does the conclusion hold if we do not assume that $M$ has non-zero diagonal entries? | 2018-03-21 20:43:23 | {"extraction_info": {"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, "math_score": 0.9417348504066467, "perplexity": 218.56668688263412}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647692.51/warc/CC-MAIN-20180321195830-20180321215830-00572.warc.gz"} |
http://people.du.ac.in/~dksingh1/ | Mac OS X Personal Web Sharing
Dr. Dhiraj Kumar Singh Assistant Professor Department of Mathematics Zakir Husain Delhi College (University of Delhi) Jawaharlal Nehru Marg Delhi 110002 INDIA Emails: dhiraj426@rediffmail.com : dksingh@zh.du.ac.in Webpage:people.du.ac.in/~dksingh1/ Mobile Number: 09868322749 List of Research Publications: 1. P. Nath and D.K. Singh, The general solutions of a functional equation related to information theory, Ratio Mathematica (Italy), 15(2005), 1 - 24. 2. P. Nath and D.K. Singh, On a multiplicative type sum form functional equation and its role in information theory, Applications of Mathematics (Czech Republic), 51 (5) (2006), 495 - 516. 3. P. Nath and D.K. Singh, On a sum form functional equation and its role in information theory, Proc. Of National Conference on 'Information Technology: Setting Trends in Modern Era' and 8th Annual Conference of Indian Society of Information Theory and Applications (March 18-20, 2006, N.C. College of Engineering, Israna (Panipat), Haryana, India), (2008), 88 - 94.pdf 4. P. Nath and D.K. Singh, A sum form functional equation related to various entropies in information theory, Glasnik Matematicki (Crotia), 43 (1) (2008), 159 - 178. 5. P. Nath and D.K. Singh, A sum form functional equation and its relevance in information theory, The Australian Journal of Mathematical Analysis and Applications (Australia), 5(1) Article 9 (2008), 1 - 18. 6. P. Nath and D.K. Singh, Some sum form functional equations containing at most two unknown mappings, Proceedings: 9th National Conference of Indian Society of Information Theory & Applications (ISITA) on "Information Technology and Operational Research Applications" December 8 - 9, 2007, Graduate School of Business & Administration, Greater Noida, Uttar Pradesh, India) (2008), 54 - 71.pdf 7. P. Nath and D.K. Singh, On a sum form functional equation related to entropies and some moments of a discrete random variable, Demonstratio Mathematica (Poland), 42 (1) (2009), 83 - 96. 8. P. Nath and D.K. Singh, On a sum form functional equation and its applications in information theory and statistics, Aequtiones Mathematicae, 81 (1-2) (2011), 77-95. 9. P. Nath and D.K. Singh, On a functional equation containing an indexed family of unknown mappings, Functional Equations in Mathematical Analysis (Editor: Themistocles Rassias and Janusz Brzdek), Springer Optimization and Its Applications 52 (2011), 671-687. 10. P. Nath and D.K. Singh, On a sum form functional equation and its relevance in information theory and cryptanalysis, Novi Sad Journal of Mathematics (Serbia), 43 (1) (2013), 59-71. 11. P. Nath and D.K. Singh, On a sum form functional equation related to entropies of type (\alpha, \beeta), Asian-European Journal of Mathematics, 6 (2) (2013), (13 pages). 12. P. Nath and D.K. Singh, On a sum form functional equation containing three unknown mappings, Annales Universitatis Paedagogicae Cracoviensis Studia Mathematica, 12, 2013, 69-81. 13. P. Nath and D.K. Singh, On a sum form functional equation related to various nonadditive entropies in information theory, Tamsui Oxford Journal of Information and Mathematical Sciences, 30(November), 2014, 23-43. 14. P. Nath and D.K. Singh, On a functional equation arising in statistics, Kragujevac Journal of Mathematics, 39(2), 2015, 141-148. 15. P. Nath and D.K. Singh, On a functional equation and its corresponding sum form, Mathematical Modeling, Optimization and Information Technology (Editor: Om Parkash), Lambert Academic Publishers (2015), 211-226. 16. P. Nath and D.K. Singh, Some functional equations and their corresponding sum forms, Sarajevo Journal of Mathematics, To apper 2015. Memberships: 1. Indian Science Congress Association 2. Ramanujan Mathematical Society 3. Indian Society of Information Theory and Application 4. Forum for Interdisciplinary Mathematics 5. Academic of Discrete Mathematics and Applications | 2017-10-19 05:38:37 | {"extraction_info": {"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, "math_score": 0.39418742060661316, "perplexity": 5279.324968564325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00719.warc.gz"} |
https://ec.gateoverflow.in/1807/gate-ece-1991-question-13 | 5 views
In the radiation pattern of a $3$-element array of isotropic radiators equally spaced at distances of $\frac{\lambda}{4}$ it is required to place a null at an angle of $33.56$ degrees off the end-fire direction. Calculate the progressive phase shifts to be applied to the elements. Also calculate the angle at which the main beam is placed for this phase distribution. | 2022-10-04 04:27:28 | {"extraction_info": {"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, "math_score": 0.35693854093551636, "perplexity": 398.53628622220714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337473.26/warc/CC-MAIN-20221004023206-20221004053206-00652.warc.gz"} |
https://zbmath.org/?q=an:07192947 | # zbMATH — the first resource for mathematics
Multiplicity results for $$(p,q)$$ fractional elliptic equations involving critical nonlinearities. (English) Zbl 1442.35500
Summary: In this paper, we prove the existence of infinitely many nontrivial solutions for the class of $$(p,q)$$ fractional elliptic equations involving concave-critical nonlinearities in bounded domains in $$\mathbb{R}^N$$. Further, when the nonlinearity is of convex-critical type, we establish the multiplicity of nonnegative solutions using variational methods. In particular, we show the existence of at least $$cat_{\Omega}(\Omega)$$ nonnegative solutions.
##### MSC:
35R11 Fractional partial differential equations 35J20 Variational methods for second-order elliptic equations 49J35 Existence of solutions for minimax problems 47G20 Integro-differential operators 45G05 Singular nonlinear integral equations
Full Text: | 2021-09-19 04:48:10 | {"extraction_info": {"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, "math_score": 0.47891032695770264, "perplexity": 819.9196591047541}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00420.warc.gz"} |
https://mathoverflow.net/questions/311592/algorithm-to-compute-the-voronoi-diagram-of-points-line-segments-and-triangles | # Algorithm to compute the Voronoi diagram of points, line segments and triangles in $\mathbb{R}^3$
Is there a known algorithm to compute the (generalized) Voronoi diagram of a set of points, line segments and triangles in $$\mathbb{R}^3$$? If yes, are there any available implementations?
I know that there are two methods with available code for line segments and points on the plane, described in the following papers:
I am also aware of a method that computes voxelizations for a closely related 3D problem, where the sites/objects are surfaces: http://sci.utah.edu/~jedwards/research/gvd/
But I'm not being able to find an algoithm for the case when the sites are vertices, line segments and triangles in space. The output should be the Voronoi diagram vertex coordinates and the (possibly curved) bisector descriptions.
• As background personal history: I spent one week writing software for this in Mathematica in 1999 for a single figure in my book Pattern classification (2nd ed.), which, as far as I know, is the first book to contain such an exact three-dimensional Voronoi tessellation. (Alas, I cannot find that code right now.) – David G. Stork Sep 27 '18 at 23:31
• Thanks for your answer, David. What algorithm did you use to generate the exact three-dimensional tesselation for your book? – Leonardo Sacht Sep 28 '18 at 1:15
• I defined a separating plane for each pair of point, then computed the intersection points, then formed the cells based on those points. – David G. Stork Sep 28 '18 at 2:45
Voronoi diagrams of points in $$R^3$$ are now implemented in several software libraries and can be computed, for example, in a few lines of Python code. This has not always been the case, so it still isn't trivial.
However, moving from Voronoi diagrams of points to diagrams of more complicated objects shifts the difficulty from asymptotical complexity issues (i.e., devising asymptotically efficient algorithms for point sets) to the difficulty of representing and computing with curved bisectors.
Furthermore, implementing algorithms that work on curved bisectors encounters inherent robustness difficulties. As an example, the vertex of a 3D Voronoi diagram is, by definition, a single intersection point of four edge curves in $$R^3$$, or equivalently of six face surfaces.
Robustness issues in geometric computing are a known research problem (see, for example, here, here or here), with many papers presenting problems even for simpler objects such as points and lines (for example, this one). Handling these degenerate cases in general and for curved objects, in particular, requires both theoretical and practical sophisticated methods. The book "Effective Computational Geometry for Curves and Surfaces" presents some of these methods for many known problems and in the context of your question, chapter 2 is especially relevant and can refer you to additional papers.
Even for the limited case of line segments in $$R^2$$, the two excellent implementations you cited apply advanced methods to handle these problems. The CGAL implementation adopts the Exact Geometric Computation (EGC) paradigm (including exact computations with square roots) and applies advanced geometric and arithmetic filtering for acceleration. The VRONI implementation, on the other hand, uses a more engineering approach for floating point computations, which combines the relaxation of epsilon thresholds with a multi-level recovery process. Both implementations are very impressive in their achievement. However, the problem in $$R^2$$ is relatively easy compared to the 3D problem in your question. In $$R^2$$, the bisectors are just parabola segments and there is no need to handle surfaces in $$R^3$$ and their intersection curves. In the Voronoi diagram of triangles, segments, and points in $$R^3$$, the bisector surfaces are quadric surfaces and the curves of the Voronoi edges can be polynomials of degree 4.
The only implementation I am aware of, which addresses these difficulties, is the one presented in this paper. They use a tracing algorithm (similar to the one in this previous paper) and apply exact arithmetic to handle the inherent robustness issues. In fact, they developed a special exact algebraic library (MAPC - library for manipulating algebraic points and curves) in order to be able to handle these curves and surfaces. The papers describing the implementation (short and long version) are a good reference to the algorithmic and practical difficulties of computing the Voronoi diagram of polyhedrons in $$R^3$$.
While their method was implemented in the past, I'd be surprised if the Voronoi code is still maintained. At the time, I managed to compile and run the MAPC library, but even compiling just the library on its own wasn't an easy task (and required dependencies on several outside libraries).
All this leads me to the conclusion that, unfortunately, it is unlikely that there are any available exact implementations to your problem. However, depending on your application, you may be able to use one of the non-exact solutions that exist. One direction, as you mentioned, is approximate solutions (such as this one) or voxelization methods, which can be accelerated using GPUs. Another practical approach, which may apply to you, is based on the Voronoi diagrams of points in $$R^3$$, for which there are software implementations as mentioned above. In these methods (mentioned in the book above), you sample the input objects, perform 3D Voronoi diagram computations on the sample points and then prune the output from unwanted faces.
• Thanks a lot for your answer! I read the paper by Culver you suggested and indeed they describe a complete solution for the problem. They also present the details and how to face the algorithmic difficulties of the problem. Unfortunately I couldn't find their code anywhere but at least now I have a description of a solution and can implement it myself. I will also consider the non-exact solutions you suggested for a relaxed version of my problem. – Leonardo Sacht Feb 6 '19 at 22:33
This is close, and may lead (through its references and using Google scholar to trace its future citations) to what you seek. This is an algorithm to compute the Voronoi diagram within a triangulated polyhedron. The medial axis in the title (and figure) is a subset of the Voronoi diagram.
Culver, Tim, John Keyser, and Dinesh Manocha. "Exact computation of the medial axis of a polyhedron." Computer Aided Geometric Design 21, no. 1 (2004): 65-98.
"Our algorithm computes the portion of the generalized Voronoi diagram that lies within the polyhedron. As a post-process, it then removes certain sheets, leaving the medial axis."
• I see after posting this that @IddoHanniel's "this paper" links to the same paper. – – Joseph O'Rourke Feb 6 '19 at 0:59
• Thanks for your suggestion. This solves the problem, though I couldn't find any implementation available. – Leonardo Sacht Feb 6 '19 at 22:35
for line segments, de berg et al. wrote a chapter about voronoi diagrams(chapter 7) in their book computational geometry. They also explain how to adapt fortunes algorithm to work with disjoint line segments, but they don't list a new algorithm in pseudocode. | 2021-05-06 15:08:36 | {"extraction_info": {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5953332185745239, "perplexity": 515.7460722700533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00030.warc.gz"} |
https://www.nature.com/articles/s41598-022-24949-1?error=cookies_not_supported&code=906c3c8b-dbd9-4e35-a3c4-fec0a28a8a56 | ## Introduction
In a fin-type or nanosheet field effect transistor (FET) of a logic semiconductor device, it has been proposed to use metal gate materials, for examples, metal carbides (TiC, TiAlC) and metal nitrides (TiN, TaN, AlN, TiAlN)1,2,3,4,5. Ternary metal compound such as TiAlC belongs to high-melting point, high-hardness, and high-wear resistance materials1,6,7. Conventionally, the TiAlC film and TiC film in semiconductor devices are etched by wet etching using H2O2 mixtures5,8,9,10. However, a poor metal removability in wet etching requires a prolonged etching time to fully remove the target metals, and in the worst case, the metal gate can be damaged5. In order to fabricate the next generation FETs in semiconductor industries, it is strongly demanded to develop an etching method that enables to control the selective and isotropic removal of metal carbides (TiAlC, TiC, and AlC) at an atomic layer level5,9,10. No dry etching (plasma etching) of ternary material TiAlC with atomic level control has been developed yet.
Atmospheric pressure plasma (APP) and medium-pressure plasma techniques with a large difference in chemical kinetics compared to low-pressure plasma are able to miniaturize equipment size, fabrication cost, and energy consumption11,12,13,14,15,16,17. Medium-pressure plasma (0.2–50 kPa) can produce higher plasma density compared to vacuum plasma and larger plasma volume compared to APP18,19,20. In order to improve the plasma density at a remote region where the substrate is placed, we have inserted a long floating metal wire inside the discharge tube to enhance the electric field not only near the copper coil region, but also at a remote region. Therefore, a rich radical source (1014 cm−3) near the sample surface that is far from the coil region can be obtained15. The generated rich radical source can produce a large amount of etchant or co-reactant species to enhance the reaction rate with sample surface. This radical-rich environment plays an important role in controlling isotropic etching of 3D multilayer semiconductor devices.
Here we have first developed a new dry etching method for metal carbides such as ternary material TiAlC by using a floating wire (FW)-assisted vapor plasma of Ar gas mixed with vapor sources of NH4OH-based mixtures at medium pressure. Although the mechanism of wet etching and dry etching can be different because the formed compounds are dissolved in solutions in wet etching, and for plasma etching, the formed compounds should be volatile in the gas phase, the wet etching brings lots of useful ideas to develop new etching chemistries for traditional materials or dry etching methods for new materials. In this study, we aim to develop a new etching method (wet-dry etching or wet-like plasma etching) that can combine the advantages of wet etching (high isotropy and selectivity) and dry etching (high controllability) for new materials or hard-to-etch materials. Surface reaction plays a key role in developing atomic layer etching (ALE) processes, which normally are proceeded in multi-steps including surface modification to reduce the surface energy of sample surface in the first step, and then removal of modified layer in the next step. Here the surface modification, such as hydrogenation and methylamination, of the TiAlC film, was obtained by controlling the active radicals, such as NH, H, and OH. The treated TiAlC surface can be removed via the formation of modifed layers. Lastly, the mechanism for plasma etching of metal carbides (TiAlC, TiC, AlC) is proposed. The new etching method will be explored for etching metals and metal compounds such as nitrides, carbides, and oxides to determine if the selectivity and isotropy commonly seen in wet etching will also occur in dry etching.
## Materials and methods
### Sample preparation
TiAlC films were prepared on Si wafer by vacuum evaporation with the Ti, Al sources, and C2H2 gas. TiAlC/Si samples were prepared with the size of 15 mm × 15 mm (for wet etching) and 15 mm × 20 mm (for dry etching). TiAlC films were analyzed by using an ellipsometry (M-2000, J.A. Woolam Co.) with a Xe arc light source (FLS-300). A model was used for spectral fitting of ellipsometric data of TiAlC sample including a top layer (native oxide, deposited layer, or modified layer), a TiAlC layer, an interface layer, and Si substrate, as shown in Fig. 1a. The pristine TiAlC film on Si substrate (TiAlC/Si) has a thickness of around 35 nm.
The dielectric function of the TiAlC film is expressed by the Gen-Osc model involving one Drude and three Lorentz oscillators21. The Gen-Osc model with the best-fit parameters is shown in Table 1.
Dispersions of optical constants including refractive index (n) and extinction coefficient (k) of TiAlC film as functions of the wavelength ranged from 245 to 1000 nm were obtained by the Gen-Osc model, as shown in Fig. 1b. The n and k values at 633 nm are 2.58 and 0.83, respectively. Cross-sectional microstructure and film thickness of samples were characterized by a cold field-emission scanning electron microscope (FE-SEM, SU-8230, Hitachi).
### Wet etching of TiAlC
To develop the new etching chemistries for TiAlC, both halogen-containing mixtures and non-halogen mixtures were used to test the potential etching chemistries. The 35 nm-TiAlC/Si samples (15 mm × 15 mm) were immersed in liquid mixtures such as peroxide solution that is a mixture of a 30% by weight water-based solution of hydrogen peroxide H2O2 (H2O2, 30 wt%), hydrochloric acid solution (HCl, 36 wt%), ammonium hydroxide solution (NH4OH, 29 wt%), and deionized water H2O. Table 2 shows a list of four experimental conditions (L1, L2, L3, and L4) of liquid mixtures for wet etching of TiAlC film.
### Dry etching of TiAlC by FW-assisted vapor plasma
A dry etching method of TiAlC was developed by using FW-assisted vapor plasma. A long floating metal wire was placed inside the discharge tube to enhance the plasma density, that provides a rich radical source at a remote region14,15. The FW-assisted plasma can generate high densities of radicals, electronically excited particles, and photons in the visible and UV range, in which radicals are able to travel long distances. The charged particles from atmospheric-pressure plasmas have very short lifetime and are difficult to reach the substrate surface at a far distance12,22. In order to improve this, the FW-plasma is designed to assist the short-lived particles to reach the substrate surface placed far away from the plasma source.
The FW-assisted plasma was connected with a process chamber, a vacuum dry pump, and a heating unit, as shown in Fig. 2a. The FW-assisted plasma consists of a 500-mm-high discharge quartz tube (inner diameter of 6 mm) with a three-turn Cu coil connected with a very high-frequency (VHF) power of 100 MHz. A polytetrafluoroethylene (PTFE) seal was used to connect the discharge tube with the chamber. The FW made of metal wire is coated by a protective material to avoid the chemical reaction with plasma species. The distance between the center of copper coil and the sample surface is 140 mm. The distance between the sample surface and the discharge tube is 2 mm. Working pressure can be controlled from atmospheric pressure to medium pressure by using a rough valve (RV) and a fine valve (FV) between a dry pump (Kashiyama, NeoDry 15E) and the process chamber. The value of working pressure was recorded by a pressure gauge (Baratron MKS, 628B). In this study, the working pressure was controlled at 0.64 kPa.
The vapor was flowed from upstream with Ar gas to generate remote FW-Ar/upstream vapor plasma along the discharge tube. Liquid mixtures including deionized water H2O and two ammonium hydroxide solutions including (NH4OH, 28 wt%) and (NH4OH, 17 wt%) were prepared.
In order to generate FW-Ar/vapor plasma, the Ar gas of 1.5 standard liter per minute (slm) was used. The saturated vapor pressures of 100%, 28%, and 17% ammonia hydroxide at 25 °C are 1007 kPa, 83 kPa, and 30 kPa, respectively23, that are higher than the working pressure of 0.64 kPa used in this study. A list of three experimental conditions (P1, P2, and P3) to generate FW-Ar/vapor plasmas by various liquid mixtures injected from upper vapor line is shown in Table 3. The FW-Ar/vapor plasmas were generated at 100 W, and the temperature of the liquid canister (Tcan) was controlled at 70 °C. The substrate temperature (Tsub) obtained from plasma discharge was measured by a thermocouple (around 150 °C, no additional heater was used).
### Plasma diagnostics
The optical emission spectra (OES) of the FW-Ar/vapor plasmas such as the emissions of Ar, OH, NH, Hβ, Hα, O were detected by using a spectrometer (Ocean Photonics, HR4000CG-UV-NIR) with the wavelength from 200 to 900 nm. The measured point was set on the TiAlC film surface (Fig. 2b). The distance between the sample center and the head of optical fiber is 150 mm.
### Material characterization
In order to analyze the surface modification of the TiAlC surface, X-ray photoelectron spectra (XPS) were obtained using a spectrometer (ESCALAB 250; Vacuum Generator, UK) equipped with an Al Kα (photon energy = 1486.6 eV) source in an analysis chamber evacuated to a base pressure of 5 × 10−7 Pa using an ion pump. Peak deconvolution and elemental concentrations were analyzed by the Advantage program. Depth profile of atomic concentration in an initial (pristine) TiAlC film deposited on Si substrate was evaluated with Ar sputtering at 3 keV and 1 µA to a sputter area of 2 mm × 2 mm for 10 min.
## Results
The initial (pristine) TiAlC film deposited on Si substrate was evaluated by depth profile of atomic concentration, as shown in Fig. S1. After removing the native oxide (Al–O, Ti–O, C=O) around 40% oxygen atomic concentration, the ratio of Ti:Al:C:O is around 28:22:40:10. Oxygen also exists inside the TiAlC film around 10%. The oxygen concentration increases to 20% at the interface between TiAlC film and Si substrate surface, that was assigned for the Si–O–C bond.
### Wet etching of TiAlC
In order to develop the etching chemistry for new materials, wet etching of TiAlC was conducted with different liquid mixtures including chlorine-based solutions and non-chlorine solutions. Figure 3a is a cross-sectional SEM images of TiAlC/Si samples before wet etching (35 nm-TiAlC, L0) and after wet etching by chlorine-based solutions and non-chlorine solutions. For chlorine-based solutions, no etching occurred with the mixture of HCl, H2O2, and H2O at a mixture ratio of 1:1:6 (condition L1). A low etch rate of 0.8 nm/min can be obtained with a mixture of HCl and H2O2 (10:1, condition L2).
For non-chlorine solutions, no etching occurred with H2O2 solution (condition L3). A high etch rate of 2.3 nm/min can be obtained by using a NH4OH/H2O2/H2O mixture (2.2:3:52, condition L4).
Wet etching of TiAlC by non-halogen liquid mixtures at different etch time was done by NH4OH/H2O2/H2O mixture. The surface modification of TiAlC film after wet chemical etching in NH4OH/H2O2/H2O mixture at room temperature is analyzed by XPS spectra, as shown Fig. S2. The intensity of Ti 2p and Al 2p was significantly reduced after 5 min and 10 min etching, whereas C–C bond at 284.8 eV (C 1s) was significantly increased. N–H and C–N peaks (around 400 eV) can be found in N 1s spectra after etching. This indicates that the compounds containing C–C and C–N bonds after wet etching of TiAlC surface are not able to dissolve in NH4OH/H2O2/H2O mixture. As a result, Ti and Al could be dissolved in the solution; however, the TiAlC surface was covered with the C-C and C–N bonds.
Based on the XPS results and the studies of Kakihana et al. and Sirijaraensre et al. on dissolution of Ti compounds24,25, the reaction of NH4OH/H2O2/H2O mixture with TiAlC to form hydroxylamine or other compounds can be assumed as follows.
For Ti–Al bond:
$${\text{Ti}}{-}{\text{Al }} + {\text{ H}}_{{2}} {\text{O}}_{{2}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){\text{-OH}} \cdots {\text{H}}_{{2}} {\text{O}}_{{2}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){\text{-OOH}} \cdots {\text{H}}_{{2}} {\text{O,}}$$
(1)
$${\text{Ti}}\left( {{\text{Al-}}} \right){\text{OOH}} \cdots {\text{H}}_{{2}} {\text{O }} + {\text{ NH}}_{{3}} \to {\text{ Ti}}\left( {{\text{Al-}}} \right){\text{OOH}} \cdots {\text{NH}}_{{3}} + {\text{ H}}_{{2}} {\text{O}},$$
(2)
$${\text{Ti}}\left( {{\text{Al}}} \right){\text{-OOH}} \cdots {\text{NH}}_{{3}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){\text{-OH}} \cdots {\text{ONH}}_{{3}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){\text{-OH}} \cdots {\text{NH}}_{{2}} {\text{OH}}.$$
(3)
For Ti–C or Al–C bond:
$${\text{Ti}}\left( {{\text{Al}}} \right){-}{\text{C}} + {\text{ H}}_{{2}} {\text{O}}_{{2}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){-}{\text{C}}{-}{\text{OH}} \cdots {\text{H}}_{{2}} {\text{O}}_{{2}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){-}{\text{C}}( = {\text{O}}){\text{OH}} \cdots {\text{H}}_{{2}} {\text{O}},$$
(4)
$${\text{Ti(Al)C(}}={\text{O)OH}} \ldots {\text{H}}_{2}{\text{O}} + {\text{NH}}_{3} \to {\text{Ti(Al)}}-{\text{C}}(={\text{O}}){\text{OH}} \ldots {\text{NH}}_{2}{\text{OH}}$$
(5)
$${\text{n}}\left[ {{\text{Ti}}\left( {{\text{Al}}} \right) - {\text{C}}} \right] \, + {\text{ nNH}}_{{3}} + {\text{ nH}}_{{2}} {\text{O}}_{{2}} \to {\text{ nTi}}\left( {{\text{Al}}} \right) - {\text{OH}} \cdots {\text{NH}}_{{2}} {\text{OH}} + {\text{ C}}_{{\text{n}}} {\text{H}}_{{{\text{2n}} + {1}}} - {\text{NH}}_{{\text{x}}} .$$
(6)
These hydrogen bonding structures, such as Ti(Al)-OHNH2OH and Ti(Al)-COOHNH2OH, are soluble in water. However, the structures (such as CnC2n+1–NHx), having C–C and C–N, are insoluble in H2O, this forms a barrier to stop wet etching.
Table S1 shows the film thickness of TiAlC film and the surface layer (top layer) etched by the liquid mixture of NH4OH, H2O2, and H2O (condition L4). The film thickness reduces with an increase of etch time. The etch rate decreased from 4.9 to 2.1 nm/min when the etch time was increased from 5 to 15 min. The etch rate reduces owing to a C layer formed on TiAlC surface, this becomes a barrier for the reaction between the liquid and TiAlC surface.
Figure 3b presents the thickness of TiAlC films as a function of etch time in the NH4OH/H2O2/H2O mixture. The etch rate decreases from 4.9 to 2.1 nm/min when the etch time is increased from 5 to 15 min due to a C layer forming on the TiAlC surface. The etch stop at the Si–O–C interface between the TiAlC film and the Si substrate. Therefore, without removal of the C layer, it is difficult to control the etch rate of TiAlC in a NH4OH/H2O2/H2O mixture.
Wet chemical etching of TiAlC film brings potential chemistries for the development of dry etching of TiAlC film. In addition to chlorine-based etching, non-chlorine etching with elements such as H, N, O or their combinations such as NHx, OH, NOx can be candidates for reactive species in plasma etching of TiAlC.
### Dry etching of TiAlC by a remote FW-assisted vapor plasma
With considerations about the volatile products for etching metal compounds, especially for the ternary or more than three-element compounds such as TiAlC, the potential plasma etchants are halogen-based etchants. Fluorine-based plasma forms AlF3 that is a non-volatile product (boiling point (b.p.) more than 1290 °C)26,27. Chlorine, bromide, or iodine-based plasmas can form volatile products, such as TiCl4 (b.p. ~ 136 °C) and AlCl3 (b.p. ~ 183 °C)27,28. In order to obtain high selective removal between Ti compounds, halogen-based plasma is limited to use due to formation of the same volatile product such as TiCl4.
The wet etching of non-halogen mixture of ammonium hydroxide solution, peroxide solution, and deionized water shows promised results with higher etch rate than the halogen-based mixture. This wet chemical etching of TiAlC film brings potential chemistries for the development of non-halogen dry etching of TiAlC film with etchants based on elements such as H, N, and O or their combinations to produce reactive species in plasma etching of TiAlC such as NHx, H, OH, or NOx. In this study, vapors were prepared based on the liquid mixtures that were used in wet etching. These vapors were used for generating reactive species in plasma etching. The remote FW-Ar/vapor plasma can generate various radicals by using different liquid mixtures, this can be used to for selective chemical reactions with TiAlC surface.
Vapor was flowed from upstream region with Ar gas to generate the remote FW-Ar/vapor plasma along the discharge tube. The vapors were prepared without using H2O2. The mixtures of Ar gas and H2O vapor, Ar gas and NH4OH 17% vapor, and Ar gas and NH4OH 28% vapor respectively generate Ar/H2O plasma (condition P1), Ar/NH4OH-17 plasma (condition P2), and Ar/NH4OH-28 plasma (condition P3) (Table 3).
Figure 4 presents the OES of Ar/H2O plasma and Ar/NH4OH plasmas at 100 W and 0.64 kPa. Various plasma colors with different vapor mixtures can be observed from the insets of photograph images. Hβ (486.1 nm) emission is detected in all spectra, whereas no O emission line (777.4 nm) can be seen from Ar/NH4OH-28 plasma. Strong emission lines for OH, NH, and Hα compared with Ar emission are detected. Intensity ratio of OH emission over NH emission can be controlled by different liquid mixtures. In case of using the Ar/NH4OH plasmas, mainly NH3 from NH4OH solution was injected to the chamber due to lower boiling point of NH3 (− 33 °C) compared to H2O (100 °C) at 1 atm29. NH3 can dissociate by energetic electron collisions to form radicals such as NH2 (571 nm), NH (336 nm), N2 (357 nm), and Hβ (486.1 nm), and Hα (656.3 nm)30,31,32,33.
At the power of 100 W, inductive component is dominant (H-mode), and hence, high-density of the Ar/NH4OH plasma can be generated with mainly NH radical (NH*) and H radical (H*) as follows:
$${\text{NH}}_{{3}} + {\text{ e}}^{ - } \to {\text{ NH}}_{{2}} * \, + {\text{ H}}*$$
(7)
$${\text{ and NH}}_{{2}} * + {\text{ e}}^{ - } \to {\text{ NH}}* \, + {\text{H}}*.$$
(8)
The selective generation of reactive species such as OH, O, NH, H radicals can be controlled by using different liquid mixtures. Ar/H2O plasma mainly produces OH, H, and O radicals, whereas Ar/NH4OH plasma mainly produces NH and H radicals. Ar/NH4OH-17 plasma can generate all of species that were listed by both Ar/H2O plasma and Ar/NH4OH-28 plasma such as NH, H, OH, and O radicals. Depending on each application, the generation of selective radicals can be controlled to modify the surface of metal compounds such as oxidation, hydrogenation, nitridation, and methylamination. This modified layer can be removed by heating, ion bombardment, or is exchanged by other ligands for selective removal over other materials.
Film thickness of TiAlC was changed by all of the FW-Ar/vapor plasmas including Ar/H2O plasma (condition P1) and Ar/NH4OH plasmas (condition P2 and P3). Etching effect of TiAlC by FW-Ar/NH4OH plasma was evaluated by ellipsometry. In case of using Ar/H2O plasma for 10 min, the film thickness increases 1.61 nm due to oxidation of metal surface. In case of using Ar/NH4OH plasmas, the film thickness decreases around 1.70 nm after 10 min exposing for both Ar/NH4OH-17 plasma (1.76 nm) and Ar/NH4OH-28 plasma (1.67 nm), proving etching occurred with TiAlC surface when exposing TiAlC to FW-Ar/NH4OH plasma. There is no significant difference in etch rates between Ar/NH4OH-17 and Ar/NH4OH-28 because in addition to forming the volatile products, nitridation also occurred at higher concentration of ammonium hydroxide solution (Fig. 6) at longer treatment time. The experiment in Fig. 6 was done with only 10 min treatment, in which the nitridation was not occurred seriously, so the etch rate in both cases are almost the same. The results prove that radicals (NH, H) from Ar/NH4OH plasma can reacts with TiAlC surface to form of volatile products.
### Surface modification of TiAlC by the remote FW-assisted vapor plasma
The surface modification of TiAlC film before (pristine) and after exposure to Ar/H2O plasma at 100 W and 0.64 kPa was analyzed by XPS spectra, as shown in Fig. 5. In case of using Ar/H2O plasma (Fig. 5b), surface oxidation of TiAlC occurs obviously with the removal of Ti–C, Al–C bonds. Effect of water vapor in water-containing atmospheric-pressure plasma has been studied34,35,36,37,38. The Ar/H2O plasma jet was reported for polymer etching and surface modification at atmospheric pressure, in which OH radical plays a dominant role or is an effective etchant in polymer etching more than H radical and O radical39,40. In this study, the O atomic concentration on TiAlC surface increases from 41% (pristine) to 51% (Ar/H2O plasma). The FW-Ar/H2O plasma can produce very high density of OH and O radicals, and therefore, fully oxidation (may including hydroxylation) of TiAlC surface with only Ti–O(H) and Al–O(H) bonds were detected, and the removal of Ti–C and Al–C bonds were obtained. Only C from TiAlC compound can be etched by Ar/H2O plasma.
The surface modification of TiAlC film after exposure to Ar/NH4OH plasmas at 100 W and 0.64 kPa is shown in Fig. 6. Although etching depths of TiAlC samples after exposing to Ar/NH4OH plasmas for 10 min including Ar/NH4OH-17 plasma and Ar/NH4OH-28 plasma are almost the same, the surface modification of TiAlC for these two samples are quite different. In case of using Ar/NH4OH-17 plasma, NH and H radicals are more dominant compared to OH and O radicals (Fig. 4). A modest amount of N atom on TiAlC surface (less than 2%) can be detected in N 1s spectrum with the formation of N–H and N–O bonds (Fig. 6a). The present of a small amount of OH and O radicals play an important role in hindering nitridation of TiAlC surface. In contrast with the Ar/H2O plasma, the shape of C 1s of sample treated by the Ar/NH4OH plasma shows the same tendency with that of pristine sample, indicating that both Ti–C and Al–C bonds still exist on TiAlC surface. After exposing to Ar/NH4OH-17 plasma, etching occurred with the removal of volatile products, the surface of TiAlC became almost the same with that of pristine sample.
In case of using Ar/NH4OH-28 plasma, nitridation and amination can be detected with a significant change in shape of Ti 2p peak and N 1s peak compared to using Ar/H2O plasma and Ar/NH4OH-17 plasma (Fig. 6b). Main species such as NH and H radicals were detected from OES result (Fig. 4). More N atoms are penetrated into TiAlC with 6.24% (Ar/NH4OH-28 plasma) compared to 1.83% (Ar/NH4OH-17 plasma). In N 1s spectrum, N–H, C–N, Ti–N, and O–Ti–N bonds can be respectively detected at 400.18 eV, 397.99 eV, 396.92 eV, and 396.36 eV. Both of the Ti–C and Al–C bonds of sample treated by the Ar/NH4OH plasmas are still remained (Fig. 7a,b), and Ti(Al)-O bond are removed and replaced by Ti(Al)–N. The remains of C in metal-C bonds are important for developing etching of metal carbides. This C in TiAlC can combine with N(H) and H from the Ar/NH4OH plasma to form Al–CH3, Ti(Al)–N–CH3, and Ti(Al)–O–CnH2n+1 bonds in volatile products, whereas it is impossible to form these bonds on TiN surface by the Ar/NH4OH plasma. The surface of TiAlC is very sensitive with oxidization, and the XPS measurement was conducts after exposure the samples in air, so the atomic concentration of O is over 40% in all cases.
The FW-Ar/vapor plasma has been developed to be used for dry etching of ternary TiAlC material. Surface modification is an indispensable step to approach the atomic layer etching.
## Discussion
### Surface modification and etching of TiAlC using FW-Ar/NH4OH plasma
$${\text{H}}* \, + {\text{ NH}}* \, + {\text{ Ti}}\left( {{\text{Al}}} \right){\text{C }} \to {\text{ Ti}}\left( {{\text{Al}}} \right){\text{C }} + {\text{ N}}_{{{\text{act}}}} + {\text{ H}}_{{{\text{act}}}} + {\text{ H}}_{{2}} \to {\text{ Ti}}\left( {{\text{Al}}} \right) - {\text{N}}{-}{\text{CH}}_{{3}} + {\text{Al}}{-}{\text{CH}}_{{3}} + {\text{ Ti}}\left( {{\text{Al}}} \right){-}{\text{N }} + {\text{ H}}_{{2}} .$$
(9)
Small amount of OH from the plasma can hinder the nitridation by forming NOx gas.
$${\text{H}}* \, + {\text{ NH}}* \, + {\text{ OH}}* \, + {\text{ Ti}}\left( {{\text{Al}}} \right){\text{C }} \to {\text{ Ti}}\left( {{\text{Al}}} \right){\text{C }} + {\text{ N}}_{{{\text{act}}}} + {\text{ H}}_{{{\text{act}}}} + {\text{ O }} + {\text{ H}}_{{2}} \to {\text{ Ti}}\left( {{\text{Al}}} \right){-}{\text{N}}{-}{\text{CH}}_{{3}} + {\text{ Al}}{-}{\text{CH}}_{{3}} + {\text{ N}}{-}{\text{O }} + {\text{ H}}_{{2}} .$$
(10)
Overall, the formation of bonds such as Al–CH3, Ti(Al)–N–CH3 or Ti(Al)–O–CnH2n+1 (in case of pristine sample is TiAlOC) shows a potential of producing volatile products such as Al(R or R′ or R′′)3, and Ti(R or R′ or R′′)4, in which R is –CH3, R′ is –N–CH3, and R′′ is –O–CnH2n+1. This modified layer could be removed by forming volatile products.
### Proposed mechanism of dry etching TiAlC
A plasma etching process for metal carbides MCs such as TiAlC, TiC, and AlC using FW-assisted plasma is demonstrated here (Fig. 8). The surface modification (hydrogenation, and amination) by reactive radicals (NH and H) and the removal of volatile metalorganic products, such as Al(CH3)3, dimer of Al(N(CH3)2)3, and Ti(N(CH3)2)4, are designed for plasma etching of metal carbides. The reactive radicals can be produced by ammonium hydroxide vapor plasma, NH3 plasma, H2 and NH3 mixture (H2/NH3) plasma, N2 and NH3 mixture (N2/NH3) plasma, or N2 and H2 mixture (N2/H2) plasma, or alcohol and ammonium hydroxide vapor mixture (CnH2n+1OH/NH4OH; n = 1–4) plasma. Surface modifications (hydrogenation and amination) of the TiAlC film were controlled by the active radicals produced from FW-assisted non-halogen plasma. The chemical bonds, involving (1) metal and methyl group (Al–CH3), (2) metal and dimethylamine group (Ti(Al)–N(CH3)2), and (3) metal and alkoxy group (Ti(Al)–OCnH2n+1) on TiAlC surface play an important role to form volatile products. Hence, the FW-assisted plasma, that is a rich radical source, is expected to be applied for atomic layer etching of metal and metal compounds in semiconductor device fabrication.
## Conclusions
A dry etching method for a ternary metal carbide TiAlC at atomic level has been developed here by transferring from wet etching to dry plasma etching using FW-assisted non-halogen vapor plasma of ammonium hydroxide. Surface modifications of the TiAlC film were controlled by exposing to the active radicals such as H, NH, and OH radicals, produced from the FW-assisted plasma. Mechanism for removal of metal carbide MC (TiAlC, TiC, AlC) is presented by using NH radicals and H radicals. This FW-assisted plasma technique is expected to be available for highly selective and isotropic atomic layer etching of metal and metal compounds in semiconductor device fabrication. | 2023-03-29 03:47:17 | {"extraction_info": {"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, "math_score": 0.4101608097553253, "perplexity": 6798.612256054099}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00689.warc.gz"} |
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06-29-2008, 07:22 PM #1 (permalink) Occams Razor Join Date: Jul 2007 Location: End of the Earth Posts: 2,470 The Definitve List: Most Overrated Bands\Artists ever Okay; so of course the title is a joke, there is no such thing as a definitive list on any subjective opinion based topic, still this is at the least the most music-banter influenced and among the most well researched compilations on the topic. In addition to scanning the 170 page MB thread and reviewing the results of a Google search, I asked, via email, the 10 people I know whose musical opinions I value most to send me their personal Top Tens. With this thread I will first unveil the list and expose it to scrutiny from the merciless and knowledgeable collective conscious of the site to see how much or how little it resonates with y'all. Following this i will periodically make an individual case against all the these artists based on their commercial success and the type of esteem and regard they are held in by today's music critics and creators. I will essentially try to tell you why they suck, well as least more then they are considered to. This list will not include Led Zeppelin or The Beatles the two, in my opinion, greatest acts ever, despite the fact that they are both often considered highly overrated by their detractors. However I dismiss them as candidates based on the fact that I feel you can't be overrated if your the greatest ever and as I mentioned I believe they are the only two you can make a case for to wear that crown. It also will not include bands like Neutral Milk Hotel, Joy Division or Sonic Youth to cite a few common ones here. Even though their most ardent fans could be accused of overrating them just as ZepHeads or Beatlemaniacs could, on the whole they are largely overlooked by the average music fan and thus can't be overrated to the extent nessacary to make this end all\be all list. This list is a top 20, if it were one notch longer, it would have included Van Halen and I would have argued that never has a collection of such talented musicians written such lousy music ever before or since. And so it starts instead with number 20, right now, with one of my most personally painful and sure to be controversial selections and just gets better from there, without further adieu... 20. Radiohead - Please mark death threat emails Attn: JayJamJah so as not to frighten my children. 19. The Grateful Dead - "What'd mean it's only good if your high?" 18. Snoop Dogg - He has 1 1/2 decent albums and yet seems to have more disposable income and street cred then Richard Branson. 17. Coldplay - How are they even considered good enough to be overrated? 16. The Doors - Turns out were not all poets are we Jim? 15. The Eagles - Together we can annoy millions, but if we split up we can annoy billions. 14. Garth Brooks - That whole Chris Gaines things made me have to hide my "No Fences" CD. Chris Gaines, really, Chris Gaines? 13. John Lennon(post-Beatles) - Whatever, the music sucked. 12. Eminem - Why do good things happen to bad people 11. Kanye West - I'm sure he's upset he's not the winner. 10. The Rolling Stones - They have no right being included in the same sentence as the Beatles, Zeppelin and the Who, combined they never produced half the crap music the Stones have. 9. Nirvana - Not a PC thing to say, but if Cobain doesn't pull the plug they are remembered like Alice in Chains or Soundgarden at best. 8. 2 PAC - Similar to Nirvana, made legendary by the early demise. Not a great emcee nor does he have the track record of many lesser regarded acts. 7. Kiss - Name one person you know who still likes them. 6. Dave Matthews Band - They might have the worst fans ever. 5. White Stripes - If I can create the riffs your best known for on a Casio, I really don't need to hear how great your discography is. 4. Prince - It's so hard to tell when Prince recorded his music, it doesn't sounded dated at all. 3. Bruce Springsteen - I don't care what you say I've meet anyone who liked The Boss that wasn't borderline retarded or an @hole. This guy sucks. 2. U2 - They should be number one, I don't know one person who doesn't loathe them yet they are one of the biggest acts in the world. They top 75% of the online lists, they are constantly mentioned here as over rated. I only keep them at number two because of an incredible argument I heard against my actual title winner, #1!!!!!!!!! Mariah Carey - Just wait until you see the case against. Last edited by Son of JayJamJah; 07-02-2008 at 08:14 AM.
06-29-2008, 07:35 PM #2 (permalink) one big soul Join Date: Feb 2008 Location: Canada Posts: 5,073 I like U2. __________________
06-29-2008, 07:45 PM #3 (permalink) Reformed Jackass Join Date: Sep 2007 Posts: 3,961 I disagree with: Radiohead, are they really overrated or do you just not like them? 2Pac-You niggas made a mistake you shoulda never put my rhymes with Dre them Thug niggas have arrived and it's Judgement Day Hey Homie if ya feel me Tell them tricks that shot me that they missed they ain't killed me I can make a mutha****er shake rattle and roll i'm full of liquor thug nigga quick to jab at them ho's and I can make ya jelous niggas famous **** around with 2Pac and see how good a niggas aim is i'm just a rich mutha****er from tha way If this rappin' bring me money then i'm rappin' till i'm paid i'm getten green like i'm supposed to Nigga, I holla at these ho's and see how many I can go through Look to the star and visualize my debut niggas know me, playa I gotta stay true don't be a dumb mutha****er because it's crazy after dark where the true Thug niggas see ya heart Niggas Can't C Me.... Sure he's not a great MC technically, but he's got an incredible delivery, and all Eyes On Me stands as a landmark of everything Gangsta rap is and was. You could make a case for him being overrated, but really? That much?
06-29-2008, 08:02 PM #4 (permalink) sleepe Join Date: Jun 2008 Location: boston Posts: 1,140 I agree with all except - Doors, Radiohead, and maybe Grateful Dead.
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You're among many. :|
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06-29-2008, 08:25 PM #6 (permalink) Account Disabled Join Date: Dec 2006 Location: Methville Posts: 2,116 As much as I enjoy a few of those artists I can agree that they're all over rated.
06-29-2008, 09:10 PM #7 (permalink) Unrepentant Ass-Mod Join Date: Jun 2008 Location: Pennsylvania Posts: 3,920 Layne Staley DID pull the plug. Where were you? Good list. But where are the Beach Boys? Never before has a boy band been remembered as great. __________________ first.am
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Originally Posted by ProggyMan I disagree with: Radiohead, are they really overrated or do you just not like them? 2Pac-You niggas made a mistake you shoulda never put my rhymes with Dre them Thug niggas have arrived and it's Judgement Day Hey Homie if ya feel me Tell them tricks that shot me that they missed they ain't killed me I can make a mutha****er shake rattle and roll i'm full of liquor thug nigga quick to jab at them ho's and I can make ya jelous niggas famous **** around with 2Pac and see how good a niggas aim is i'm just a rich mutha****er from tha way If this rappin' bring me money then i'm rappin' till i'm paid i'm getten green like i'm supposed to Nigga, I holla at these ho's and see how many I can go through Look to the star and visualize my debut niggas know me, playa I gotta stay true don't be a dumb mutha****er because it's crazy after dark where the true Thug niggas see ya heart Niggas Can't C Me.... Sure he's not a great MC technically, but he's got an incredible delivery, and all Eyes On Me stands as a landmark of everything Gangsta rap is and was. You could make a case for him being overrated, but really? That much?
HAHAHHAHAHAHHAHAHAHAHAHHAHAHAHHAHAHA
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Originally Posted by METALLICA89 Ive seen you on muiltipul forums saying Metallica and slayer are the worst **** you kid go suck your **** while you listen to your ****ing emo **** I bet you do listen to emo music
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Originally Posted by sleepy jack HAHAHHAHAHAHHAHAHAHAHAHHAHAHAHHAHAHA
Ethan has the last laugh!!!
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Originally Posted by loose_lips_sink_ships Ethan gets the last laugh!!!
Whos Ethan? That name sounds familair
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Originally Posted by swim America does folk, hardcore and mathrock better and that's 90% of what I give 2 shits on.
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Originally Posted by chartsengrafs sweet nothing openly flaunts the fact that he is merely the empty shell of an even more unadmirable member. his loneliness and need for attention bleeds through every letter he types. edit: i would just like to add that i'm ashamed that he's from texas. surely you didn't grow up in texas, did you sweet nothing?
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https://dangoldner.wordpress.com/2012/01/02/substitute-then-simplify/ | All year long, the most common “proof” students offer has been to do an example with numbers. On “show that the distance between (a,b) and the origin is $\sqrt{a^2+b^2}$, I’ve been getting: “I chose random numbers for a and b, so I used a=4 and b=7. Then I plugged in and got $\sqrt{16+49}=\sqrt{55}$.” No mention of the Pythagorean Theorem we’d been playing with for the past two days.
To my students—most students?—mathematics is “Substitute, then simplify.” It’s all math has ever been. I have been aware of this, but my response has been inadequate.
• Item: I would put two numerical examples on the board, then a third using variables, crowing that “if you do the same thing with only letters, then you’re doing it for all possible numbers at once! Do it once or twice with numbers, then do the same thing with variables! This is called generalization!” Result: I had one student do a numerical example similar to the problem above, then go on to say “I wanted to solve it with variables, so I made 16=U and 49=V, so the answer is $\sqrt{U+V}$.”
• Item: The idea of a derivation is entirely absent. Later in the year I juxtaposed three student examples of finding a distance to show them that the distance formula is simply the Pythagorean theorem. Students were flabbergasted. I got compliments on my lecture. They had never seen anything like it.
• Item: After developing a meaning for the distance between a point and a line, I asked them, “What is the distance between $(x_1, y_1)$ and y=d?” and gave them some time to work on it. They were totally stumped. “Ok, let’s draw a picture.” Blank stares. “What’s up?” “We don’t know what d is—you have to know what number to use!” Ok. Let’s graph y=2. Fine. How ’bout y=-4. Great. What should we do for y=d? Someone, quiet, tentative: “Draw a flat line?” (indicating horizontal with his hand). Does it matter where I put it? No. What about here on the x-axis? No, don’t do that, that makes it look like zero, and it might not be zero.
In my mind, once you do something with specific numbers, it’s just one more small step to do it again with symbols. After all, it’s the same thing. But that’s the rub: to them, it’s not the same thing!. Torigoe (Thanks, Mylène) has got me thinking about what might be going on.
It never occurred to me that carrying symbols through a problem is more demanding than substituting numbers into as many variables as possible, as soon as possible. By stopping up all those leaky variables with numbers, I can stem the flood of confusion behind a nice high dam of something with just one variable to solve for. The structural relationships among whatever I’m modeling (say, distance between objects on a plane) are eliminated, so I don’t have to think about them anymore.
Anymore? At all! This explains everything! It’s why my precalculus students can’t articulate relationships between abstract quantities. They’ve never had to! So they don’t know how.
At least not yet. The technique of “give one specific problem, then a second, then a general one” and hope they make the leap has not worked. So how can I build the capacity to reason with symbols? | 2017-05-30 01:35:20 | {"extraction_info": {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5912030935287476, "perplexity": 747.856908705494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463613738.67/warc/CC-MAIN-20170530011338-20170530031338-00305.warc.gz"} |
https://cs.stackexchange.com/questions/63284/minor-mistake-in-computability-complexity-and-languages | # Minor Mistake in Computability, Complexity, and Languages?
In the book Computability, Complexity, and Languages (2nd edition), Martin Davis writes in chapter 1 (Preliminaries), section 2 (Functions):
A partial function on a set $S$ is simply a function whose domain is a subset of $S$. An example of a partial function on $\mathbb{N}$ is given by $g(n) = \sqrt n$, where the domain of $g$ is the set of perfect squares.
So far so simple. But he goes ahead and writes a couple lines later at the end of the section:
We will sometimes refer to the idea of closure. If $S$ is a set and $f$ is a partial function on $S$, then $S$ is closed under $f$ if the range of $f$ is a subset of $S$. For example, $\mathbb{N}$ is closed under $f(n) = n^2$, but is not closed under $h(n) = \sqrt n$ (where $h$ is a total function on $\mathbb{N}$).
So in the first quote $\sqrt n$ on $\mathbb{N}$ is an example for a partial function, whereas in the second quote the same function is an example for a total function.
Both examples seem to contradict each other. Or am I missing something in regard to closures here?
• This is pedagogically a bad example of a partial function, because the domain is restricted only by arbitrary definition. A better example of a partial function is f(x) = 1/x, which is undefined when x = 0. – Wildcard Sep 9 '16 at 3:06
• @Wildcard: How is this arbitrary? $\mathbb{N}$ is not an odd choice for the range of $g(n)$.. – MSalters Sep 9 '16 at 8:01
• @MSalters, nethertheless it's a choice as in arbitrary. – Good Night Nerd Pride Sep 9 '16 at 8:06
• @GoodNightNerdPride: If you use that logic, $1/x$ is not a better example, because there are ranges (e.g. projectively extended real line) in which that too is defined. – MSalters Sep 9 '16 at 8:13
• @Wildcard It's not arbitrary at all. $g$ defined as a function from $\mathbb{N}$ to $\mathbb{N}$, which is a completely natural object. The function is undefined in certain places because, e.g., there is no natural number $y$ such that $y^2=2$. That's just the same as your example: it's very natural to have a function from $\mathbb{R}$ to $\mathbb{R}$, and the function you've chosen is undefined in some places because, e.g., there's no real number $y$ such that $y\cdot0=1$. – David Richerby Sep 9 '16 at 8:59
There's no contradiction, here. The first case defines the partial function $g\colon \mathbb{N}\to\mathbb{N}$ given by $$g(n) = \begin{cases} x &\text{if x\in\mathbb{N} and }x^2=n\\ \text{undefined} &\text{if no such x exists.} \end{cases}$$ As the text says, "the domain of $g$ is the set of perfect squares."
The second case defines the total function $h\colon\mathbb{N}\to\mathbb{R}$ given by $$h(n) = x \quad \text{if }x\in\mathbb{R}_{\geq 0} \text{ and } x^2=n\,.$$ The domain of $h$ is the set of all the natural numbers.
You say that these are the same function, but they are not. $g(2)$ is undefined but $h(2)$ is defined (and equal to $\sqrt{2}$). $h$ associates a square root with every natural number but $g$ only associates square roots with natural numbers that are perfect squares.
$\mathbb{N}$ is closed under $g$ because, whenever $g$ is defined, $g(n)\in\mathbb{N}$. $\mathbb{N}$ is not closed under $h$ because, for example, $2\in\mathbb{N}$ but $h(2)\notin\mathbb{N}$.
• +1, though I think the book is in error when it uses "$f$ is a partial function on $S$" to mean "$f$ is a partial function from $S$ to some set". – ruakh Sep 9 '16 at 20:23
• @ruakh What's wrong with that? – David Richerby Sep 9 '16 at 20:32
• In my experience, unless you explicitly add a modifier like "real-valued" (so that the codomain is clear), a function on a set is a function from that set to itself (analogously to how a relation on a set is a relation from that set to itself). But maybe that's not an error, and is just a different terminological tradition? – ruakh Sep 9 '16 at 22:34
In the second example, $h(n)$ is defined for all natural numbers $n$; when $n$ is not a square, $h(n)$ is some irrational quantity, and in particular not a natural number.
In other words, the set of natural numbers is not closed under taking square roots: for example, $\sqrt{2}$ is not a natural number.
• Sorry, I think I wasn't clear enough. Please see the 2nd to last paragraph I added to my question. – Good Night Nerd Pride Sep 8 '16 at 21:10
• @GoodNightNerdPride I think Yuval is exactly right. In the first example, the square root function is a partial function from $\mathbb{N}$ to $\mathbb{N}$; in the second example, it's a total function from $\mathbb{N}$ to $\mathbb{R}$. – David Richerby Sep 8 '16 at 21:23 | 2020-01-20 22:17:06 | {"extraction_info": {"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, "math_score": 0.8996303677558899, "perplexity": 247.8906704620215}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00281.warc.gz"} |
http://nag.com/numeric/fl/nagdoc_fl24/html/C06/c06pff.html | C06 Chapter Contents
C06 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentC06PFF
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
## 1 Purpose
C06PFF computes the discrete Fourier transform of one variable in a multivariate sequence of complex data values.
## 2 Specification
SUBROUTINE C06PFF ( DIRECT, NDIM, L, ND, N, X, WORK, LWORK, IFAIL)
INTEGER NDIM, L, ND(NDIM), N, LWORK, IFAIL COMPLEX (KIND=nag_wp) X(N), WORK(LWORK) CHARACTER(1) DIRECT
## 3 Description
C06PFF computes the discrete Fourier transform of one variable (the $l$th say) in a multivariate sequence of complex data values ${z}_{{j}_{1}{j}_{2}\cdots {j}_{m}}$, where ${j}_{1}=0,1,\dots ,{n}_{1}-1\text{, }{j}_{2}=0,1,\dots ,{n}_{2}-1$, and so on. Thus the individual dimensions are ${n}_{1},{n}_{2},\dots ,{n}_{m}$, and the total number of data values is $n={n}_{1}×{n}_{2}×\cdots ×{n}_{m}$.
The routine computes $n/{n}_{l}$ one-dimensional transforms defined by
$z^ j1 … kl … jm = 1nl ∑ jl=0 nl-1 z j1 … jl … jm × exp ± 2 π i jl kl nl ,$
where ${k}_{l}=0,1,\dots ,{n}_{l}-1$. The plus or minus sign in the argument of the exponential terms in the above definition determine the direction of the transform: a minus sign defines the forward direction and a plus sign defines the backward direction.
(Note the scale factor of $\frac{1}{\sqrt{{n}_{l}}}$ in this definition.)
A call of C06PFF with ${\mathbf{DIRECT}}=\text{'F'}$ followed by a call with ${\mathbf{DIRECT}}=\text{'B'}$ will restore the original data.
The data values must be supplied in a one-dimensional complex array using column-major storage ordering of multidimensional data (i.e., with the first subscript ${j}_{1}$ varying most rapidly).
This routine calls C06PRF to perform one-dimensional discrete Fourier transforms. Hence, the routine uses a variant of the fast Fourier transform (FFT) algorithm (see Brigham (1974)) known as the Stockham self-sorting algorithm, which is described in Temperton (1983).
## 4 References
Brigham E O (1974) The Fast Fourier Transform Prentice–Hall
Temperton C (1983) Self-sorting mixed-radix fast Fourier transforms J. Comput. Phys. 52 1–23
## 5 Parameters
1: DIRECT – CHARACTER(1)Input
On entry: if the forward transform as defined in Section 3 is to be computed, then DIRECT must be set equal to 'F'.
If the backward transform is to be computed then DIRECT must be set equal to 'B'.
Constraint: ${\mathbf{DIRECT}}=\text{'F'}$ or $\text{'B'}$.
2: NDIM – INTEGERInput
On entry: $m$, the number of dimensions (or variables) in the multivariate data.
Constraint: ${\mathbf{NDIM}}\ge 1$.
3: L – INTEGERInput
On entry: $l$, the index of the variable (or dimension) on which the discrete Fourier transform is to be performed.
Constraint: $1\le {\mathbf{L}}\le {\mathbf{NDIM}}$.
4: ND(NDIM) – INTEGER arrayInput
On entry: the elements of ND must contain the dimensions of the NDIM variables; that is, ${\mathbf{ND}}\left(i\right)$ must contain the dimension of the $i$th variable.
Constraints:
• ${\mathbf{ND}}\left(\mathit{i}\right)\ge 1$, for $\mathit{i}=1,2,\dots ,{\mathbf{NDIM}}$;
• ${\mathbf{ND}}\left({\mathbf{L}}\right)$ must have less than $31$ prime factors (counting repetitions).
5: N – INTEGERInput
On entry: $n$, the total number of data values.
Constraint: N must equal the product of the first NDIM elements of the array ND.
6: X(N) – COMPLEX (KIND=nag_wp) arrayInput/Output
On entry: the complex data values. Data values are stored in X using column-major ordering for storing multidimensional arrays; that is, ${z}_{{j}_{1}{j}_{2}\cdots {j}_{m}}$ is stored in ${\mathbf{X}}\left(1+{j}_{1}+{n}_{1}{j}_{2}+{n}_{1}{n}_{2}{j}_{3}+\cdots \right)$.
On exit: the corresponding elements of the computed transform.
7: WORK(LWORK) – COMPLEX (KIND=nag_wp) arrayWorkspace
The workspace requirements as documented for C06PFF may be an overestimate in some implementations.
On exit: the real part of ${\mathbf{WORK}}\left(1\right)$ contains the minimum workspace required for the current value of N with this implementation.
8: LWORK – INTEGERInput
On entry: the dimension of the array WORK as declared in the (sub)program from which C06PFF is called.
Suggested value: ${\mathbf{LWORK}}\ge {\mathbf{N}}+{\mathbf{ND}}\left({\mathbf{L}}\right)+15$
9: IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6 Error Indicators and Warnings
If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
On entry, ${\mathbf{NDIM}}<1$.
${\mathbf{IFAIL}}=2$
On entry, ${\mathbf{L}}<1$ or ${\mathbf{L}}>{\mathbf{NDIM}}$.
${\mathbf{IFAIL}}=3$
On entry, ${\mathbf{DIRECT}}\ne \text{'F'}$ or $\text{'B'}$.
${\mathbf{IFAIL}}=4$
On entry, at least one of the first NDIM elements of ND is less than $1$.
${\mathbf{IFAIL}}=5$
On entry, N does not equal the product of the first NDIM elements of ND.
${\mathbf{IFAIL}}=6$
On entry, LWORK is too small. The minimum amount of workspace required is returned in ${\mathbf{WORK}}\left(1\right)$.
${\mathbf{IFAIL}}=7$
On entry, ${\mathbf{ND}}\left({\mathbf{L}}\right)$ has more than $30$ prime factors.
${\mathbf{IFAIL}}=8$
An unexpected error has occurred in an internal call. Check all subroutine calls and array dimensions. Seek expert help.
## 7 Accuracy
Some indication of accuracy can be obtained by performing a subsequent inverse transform and comparing the results with the original sequence (in exact arithmetic they would be identical).
The time taken is approximately proportional to $n×\mathrm{log}{n}_{l}$, but also depends on the factorization of ${n}_{l}$. C06PFF is faster if the only prime factors of ${n}_{l}$ are $2$, $3$ or $5$; and fastest of all if ${n}_{l}$ is a power of $2$.
## 9 Example
This example reads in a bivariate sequence of complex data values and prints the discrete Fourier transform of the second variable. It then performs an inverse transform and prints the sequence so obtained, which may be compared with the original data values.
### 9.1 Program Text
Program Text (c06pffe.f90)
### 9.2 Program Data
Program Data (c06pffe.d)
### 9.3 Program Results
Program Results (c06pffe.r) | 2016-12-08 09:58:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 63, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9933580756187439, "perplexity": 2107.484926200113}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542520.47/warc/CC-MAIN-20161202170902-00226-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://ncatlab.org/nlab/show/Penrose-Hawking+theorem | # nLab Penrose-Hawking theorem
,
• , ,
• (-)
• -
• , ,
• ,
• , ,
## Spacetimes
vanishing positive
vanishing
positive
• ,
, ,
## Surveys, textbooks and lecture notes
• ,
,
, ,
• , , , ,
• ,
• ,
• ,
• ,
• and
• Axiomatizations
• ,
• -theorem
• Tools
• ,
• ,
• ,
• Structural phenomena
• Types of quantum field thories
• ,
• , ,
• examples
• ,
• ,
• , , , ,
• , ,
# Contents
## Idea
The Penrose-Hawking singularity theorems characterize spacetimes in the theory of Einstein-gravity (general relativity) which have “singularities”, points where the Riemann curvature is undefined (or would be undefined if these points were included in the spacetime manifold) such as appears notably in black hole spacetimes.
Hellman suggest this theorem as an example of noncomputable physics. See Frank for a response.
A related problem is that of the maximal Cauchy development for the Einstein equations. In this case, at least Zorn's lemma can be avoided.
## References
• Stephen Hawking?, Roger Penrose, The Nature of Space and Time Princeton: Princeton University Press. ISBN 0-691-03791-4. (1996)
• Wikipedia, Penrose-Hawking singularity theorem
• Geoffrey Hellman, Mathematical constructivism in spacetime, British Journal for the Philosophy of Science 49 (3):425-450 (1998) PDF
• Matthew Frank, Axioms and aesthetics in constructive mathematics and differential geometry. PhD-thesis, Chicago, 2004.
• Jan Sbierski, On the Existence of a Maximal Cauchy Development for the Einstein Equations - a Dezornification PDF
For further developments see
Last revised on July 24, 2018 at 01:17:14. See the history of this page for a list of all contributions to it. | 2018-09-22 03:36:29 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9693343043327332, "perplexity": 7058.345282071868}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158011.18/warc/CC-MAIN-20180922024918-20180922045318-00123.warc.gz"} |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=solve/function | solve/functions - Maple Programming Help
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : solve : solve/function
solve/functions
for a variable which is used as a function name
Calling Sequence solve(eqn, fcnname)
Parameters
eqn - single equation (as for solve) fcnname - single variable (as for solve)
Description
• When the unknown is a name which is only used as a function in the equation, the solver constructs a function or procedure which solves the equation.
Examples
> $\mathrm{solve}\left(f\left(x\right)-x+2,f\right)$
${x}{→}{x}{-}{2}$ (1)
This is more or less equivalent to:
> $\mathrm{solve}\left(f\left(x\right)-x+2,f\left(x\right)\right)$
${x}{-}{2}$ (2)
> $\mathrm{unapply}\left(,x\right)$
${x}{→}{x}{-}{2}$ (3)
This in turn is more or less equivalent to:
> $\mathrm{solve}\left(y-x+2,y\right)$
${x}{-}{2}$ (4)
> $\mathrm{unapply}\left(,x\right)$
${x}{→}{x}{-}{2}$ (5)
More complicated inputs will generally return answers involving RootOfs
> $\mathrm{solve}\left({f\left(x\right)}^{4}-{f\left(x\right)}^{3}+x+1,f\right)$
${x}{→}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{4}}{-}{{\mathrm{_Z}}}^{{3}}{+}{x}{+}{1}\right)$ (6)
See Also
## Was this information helpful?
Please add your Comment (Optional) E-mail Address (Optional) What is ? This question helps us to combat spam | 2016-06-27 05:53:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9743108153343201, "perplexity": 3007.661249177029}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395621.98/warc/CC-MAIN-20160624154955-00021-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/140341-combinatorics-problem.html | # Math Help - Combinatorics problem
1. ## Combinatorics problem
Hi guys, I'm new to this forum.
So I'm trying to solve this combinatorics/probability problem and I have arrived in a final form of:
[sigma(from i=0 to i=n) of nCi*2^i]/4^n
which is apparently equal to (3/4)^n
Does anyone of you know how sigma(from i=0 to i=n) of nCi*2^i = 3^n?
Thank you for your kind help.
2. Using the binomial theorem we get $\left( {2 + 1} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}2^k }$ | 2015-09-05 08:38:51 | {"extraction_info": {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5926550030708313, "perplexity": 2998.223312580212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645396463.95/warc/CC-MAIN-20150827031636-00111-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?s=d47dad994152758b5feb2b52db018f1a&p=4237287 | ## Thoughts on this Inverse Bijection Proof
Is this sufficient?
Attached Thumbnails
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What are you trying to prove, exactly? These just look like definitions to me, in which case a much simpler description of a one-to-one function would be as follows: A function $f:X \rightarrow Y$ is an injection if, $\forall a,b \in X, \ f(a) = f(b) \implies a = b$.
It is to proof that the inverse is a one-to-one correspondence. I think I get what you are saying though about it looking as a definition rather than a proof. How about this.. Let $f:X\rightarrow Y$ be a one to one correspondence, show $f^{-1}:Y\rightarrow X$ is a one to one correspondence. $\exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2}$ furthermore, $f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})$ (by definition of function $f$ and one to one) kind of stumped from this point on.. I may want to transfer this post over to the hw section though, I did post to just get a confirmation on my thoughts on bijection but it is now turning into something a bit more specific than that
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## Thoughts on this Inverse Bijection Proof
Your proof that $f^{-1}$ is injective is correct. Your proof that it is surjective does not look to me like it actually says anything!
You want to prove that, if $x\in X$ then there exist $y\in Y$ such that $f^{-1}(y)= x$.
Given $x\in X$, let $y= f(x)$. Then it follows that $f^{-1}(y)= f^{-1}(f(x))= x$.
That makes a lot of sense and I am following that thought process, thank you for clearing that up for me. I was just stumped on the direction of the onto.
In the same light, these are my thoughts on my next exercise. If I have this wrong I may need to solidify my idea on the concept a bit more. It reads: Show that if $f:X \rightarrow Y$ is onto $Y$, and $g: Y \rightarrow Z$ is onto $Z$, then $g \circ f:X \rightarrow Z$ is onto $Z$ Prf Given $y \in Y$, let $y = g^{-1}(z)$ and $x = f^{-1}(y)$ $\forall z \in Z$, $f^{-1}(g^{-1}(z)) = f^{-1}(y) = x$
Tags bijection, function, inverse | 2013-05-26 06:58:03 | {"extraction_info": {"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, "math_score": 0.8240703344345093, "perplexity": 333.48959352116066}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706635944/warc/CC-MAIN-20130516121715-00091-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-6-percent-review-exercises-page-469/27 | ## Basic College Mathematics (10th Edition)
$\frac{part}{whole}=\frac{percent}{100}$ (a) 965 out of 1005 have a smoke alarm $\frac{965}{1005}=\frac{p}{100}$ cross multiply to solve for p $1005p=(965)(100)$ $1005p\div1005=96500\div1005$ $p=96.0199$ (b) 461 out of 1005 have a carbon monoxide detector $\frac{461}{1005}=\frac{p}{100}$ cross multiply to solve for p $1005p=(461)(100)$ $1005p\div1005=46100\div1005$ $p=45.8706$ | 2019-10-17 00:07:34 | {"extraction_info": {"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, "math_score": 0.647699773311615, "perplexity": 697.0665853684445}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00463.warc.gz"} |
http://mathhelpforum.com/algebra/74532-inverse-function-print.html | # Inverse of function
• Feb 19th 2009, 02:28 PM
heneri
Inverse of function
y=3x-1/2x
My attempt:
$y=(6x^2-1)/(2x)$
x= $(6y^2-1)/(2y)$
$2yx=6y^2-1$
$2yx-6y^2=-1$
$
2y(x-3y)=-1$
$2y=-1/(x-3y)$
( I get stuck her what do l do with the y in the denominator ? )
• Feb 19th 2009, 02:48 PM
HallsofIvy
Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.
Solve that quadratic equation for x you will get a " $\pm$". This function does not have a true inverse.
• Feb 20th 2009, 01:09 AM
nyasha
Quote:
Originally Posted by HallsofIvy
Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.
Solve that quadratic equation for x you will get a " $\pm$". This function does not have a true inverse.
I solved it thank you very much for your help
• Feb 27th 2009, 05:24 AM
nyasha
Quote:
Originally Posted by HallsofIvy
Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.
Solve that quadratic equation for x you will get a " $\pm$". This function does not have a true inverse.
If my restricted domain is D={x|x<0} does that mean for my inverse function is going to be the one with a minus sign on the quadratic equation. | 2017-01-23 23:14:28 | {"extraction_info": {"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9165202975273132, "perplexity": 1024.6490921110221}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560283301.73/warc/CC-MAIN-20170116095123-00498-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://codeforces.com/blog/entry/46450 | ### DarthKnight's blog
By DarthKnight, history, 3 years ago, ,
Here are the solutions to all problems.
### Div.2 A
Just alternatively print "I hate that" and "I love that", and in the last level change "that" to "it".
Time Complexity:
### Div.2 B
First of all, instead of cycles, imagine we have bamboos (paths). A valid move in the game is now taking a path and deleting an edge from it (to form two new paths). So, every player in his move can delete an edge in the graph (with components equal to paths). So, no matter how they play, winner is always determined by the parity of number of edges (because it decreases by 1 each time). Second player wins if and only if the number of edges is even. At first it's even (0). In a query that adds a cycle (bamboo) with an odd number of vertices, parity and so winner won't change. When a bamboo with even number of vertices (and so odd number of edges) is added, parity and so the winner will change.
Time Complexity:
### A
Consider a queue e for every application and also a queue Q for the notification bar. When an event of the first type happens, increase the number of unread notifications by 1 and push pair (i, x) to Q where i is the index of this event among events of the first type, and also push number i to queue e[x].
When a second type event happens, mark all numbers in queue e[x] and clear this queue (also decreese the number of unread notifications by the number of elements in this queue before clearing).
When a third type query happens, do the following:
while Q is not empty and Q.front().first <= t:
i = Q.front().first
x = Q.front().second
Q.pop()
if mark[i] is false:
mark[i] = true
e[v].pop()
ans = ans - 1 // it always contains the number of unread notifications
But in C++ set works much faster than queue!
Time Complexity:
### B
Reduction to TSP is easy. We need the shortest Hamiltonian path from s to e. Consider the optimal answer. Its graph is a directed path. Consider the induced graph on first i chairs. In this subgraph, there are some components. Each components forms a directed path. Among these paths, there are 3 types of paths:
1. In the future (in chairs in right side of i), we can add vertex to both its beginning and its end.
2. In the future (in chairs in right side of i), we can add vertex to its beginning but not its end (because its end is vertex e).
3. In the future (in chairs in right side of i), we cannot add vertex to its beginning (because its beginning is vertex s) but we can add to its end.
There are at most 1 paths of types 2 and 3 (note that a path with beginning s and ending e can only exist when all chairs are in the subgraph. i.e. induced subgraph on all vertices).
This gives us a dp approach: dp[i][j][k][l] is the answer for when in induced subgraph on the first i vertices there are j components of type 1, k of type 2 and l of type 3. Please note that it contains some informations more than just the answer. For example we count d[i] or - x[i] when we add i to the dp, not j (in the problem statement, when i < j). Updating it requires considering all four ways of incoming and outgoing edges to the last vertex i (4 ways, because each edge has 2 ways, left or right). You may think its code will be hard, but definitely easier than code of B.
Time Complexity:
### C
Build a graph. Assume a vertex for each clause. For every variable that appears twice in the clauses, add an edge between clauses it appears in (variables that appear once are corner cases). Every vertex in this graph has degree at most two. So, every component is either a cycle or a path. We want to solve the problem for a path component. Every edge either appear the same in its endpoints or appears differently. Denote a dp to count the answer. dp[i][j] is the number of ways to value the edges till i - th vertex in the path so that the last clause(i's) value is j so far (j is either 0 or 1). Using the last edge to update dp[i][j] from dp[i - 1] is really easy in theory.
Counting the answer for a cycle is practically the same, just that we also need another dimension in our dp for the value of the first clause (then we convert it into a path). Handling variables that appear once (edges with one endpoint, this endpoint is always an endpoint of a path component) is also hard coding. And finally we need to merge the answers.
Time Complexity:
### D
Assume r < b (if not, just swap the colors). Build a bipartite graph where every vertical line is a vertex in part X and every horizontal line is a vertex in part Y. Now every point(shield) is an edge (between the corresponding vertical and horizontal lines it lies on). We write 1 on an edge if we want to color it in red and 0 if in blue (there may be more than one edge between two vertices). Each constraint says the difference between 0 and 1 edges connected to a certain vertex should be less than or equal to some value. For every vertex, only the constraint with smallest value matters (if there's no constraint on this vertex, we'll add one with di = number of edges connected to i).
Consider vertex i. Assume there are qi edges connected to it and the constraint with smallest d on this vertex has dj = ei. Assume ri will be the number of red (with number 1 written on) edges connected to it at the end. With some algebra, you get that the constraint is fulfilled if and only if . Denote and . So Li ≤ ri ≤ Ri. This gives us a L-R max-flow approach: aside these vertices, add a source S and a sink T. For every vertex v in part X, add an edge with minimum and maximum capacity Lv and Rv from S to v. For every vertex u in part Y, add an edge with minimum and maximum capacity Lu and Ru from u to T. And finally for every edge v - u from X to Y add an edge from v to u with capacity 1 (minimum capacity is 0).
If there's no feasible flow in this network, answer is -1. Otherwise since r ≤ b, we want to maximize the number of red points, that is, maximizing total flow from S to T.
Since the edges in one layer (from X to Y) have unit capacities, Dinic's algorithm works in (Karzanov's theorem) and because and Dinic's algorithm works in .
Time Complexity:
### E
First, we're gonna solve the problem for when the given tree is a bamboo (path). For simplifying, assume vertices are numbered from left to right with 1, 2, .., n (it's an array). There are some events (appearing and vanishing). Sort these events in chronological order. At first (time - ∞) no suit is there. Consider a moment of time t. In time t, consider all available suits sorted in order of their positions. This gives us a vector f(t).
Lemma 1: If i and j are gonna be at the same location (and explode), there's a t such that i and j are both present in f(t) and in f(t) they're neighbours.
This is obvious since if at the moment before they explode there's another suit between them, i or j and that suit will explode (and i and j won't get to the same location).
Lemma 2: If i and j are present in f(t) and in time t, i has position less than j, then there's no time e > t such that in it i has position greater than j.
This hold because they move continuously and the moment they wanna pass by each other they explode.
So this gives us an approach: After sorting the events, process them one by one. consider ans is the best answer we've got so far (earliest explosion, initially ). Consider there's a set se that contains the current available suits at any time, compared by they positions (so comparing function for this set would be a little complicated, because we always want to compare the suits in the current time, i.e. the time when the current event happens). If at any moment of time, time of event to be processed is greater than or equal to ans, we break the loop. When processing events:
First of all, because current event's time is less than current ans, elements in se are still in increasing order of their position due to lemma 2 (because if two elements were gonna switch places, they would explode before this event and ans would be equal to their explosion time). There are two types of events:
1. Suit i appears. After updating the current moment of time (so se's comparing function can use it), we insert i into se. Then we check i with its two neighbours in se to update ans (check when i and its neighbours are gonna share the same position).
2. Suit i vanishes. After updating the current moment of time, we erase i from se and check its two previous neighbours (which are now neighbours to each other) and update ans by their explosion time.
This algorithm will always find the first explosion due to lemma 1 (because the suits that're gonna explode first are gonna be neighbours at some point).
This algorithm only works for bamboos. For the original problem, we'll use heavy-light decompositions. At first, we decompose the path of a suit into heavy-light sub-chains (like l sub-chains) and we replace this suit by l suits, each moving only within a subchain. Now, we solve the problem for each chain (which is a bamboo, and we know how to solve the problem for a bamboo). After replacing each suit, we'll get suits because and we need an extra log for sorting events and using set, so the total time complexity is .
In implementation to avoid double and floating point bugs, we can use a pair of integers (real numbers).
Time Complexity (more precisely):
•
• +164
•
» 3 years ago, # | +28 Thanks for the quick editorial =DFor Div1B/2D, can someone help elaborate a bit more about how to transit from dp[i] to dp[i+1]?
» 3 years ago, # | +51 Tags are so sad! Thanks anyway, nice problem set! :)
• » » 3 years ago, # ^ | -8 why say bye to cf?
• » » 3 years ago, # ^ | +31 They really are..
» 3 years ago, # | ← Rev. 2 → +3 Would some kind gentlemen please explain problem A (Div 1) in simpler terms? Thank you!
• » » 3 years ago, # ^ | +13 It is basically like this : You have an array with all terms initially 0. Operations: op1 : for the x we must a[x]++. op2 : for the x we must a[x]--. op3 : for the first t operations of type 1 i.e for x[1],x[2],....x[t] decrement the values if they are not already updated (ie not already read). if(x[i] != 0) x[i]--; so here brute force will time out . So op1 and op2 are similar . But for 3d one instead of updating all the first t we will update this time , if the previous update time was lesser than current . (I.e we need not read something that has already been read) you can check the update time in code of some submitted solution.Queue is maintained to keep track of last updation time of a particular element . So since each time element is processed at max once O(Q + T). Hope i helped , this is my first explanation on CF so please be kind on me :)
• » » » 3 years ago, # ^ | -15 Oh, lord, I meant problem B (Div 2), but thank you anyway :)
» 3 years ago, # | +23 Why "lastcontest" in Tags? :(
• » » 3 years ago, # ^ | -11 Last contest before IOI
• » » » 3 years ago, # ^ | +41 NOPE! My last contest.
• » » » » 3 years ago, # ^ | ← Rev. 2 → +27 Why? Is there any reason?
• » » » » » 3 years ago, # ^ | -80 Because people usually give up when confronted with some hardship instead of shaking off the dust, standing up and moving forward :)
• » » » » 3 years ago, # ^ | -12 I enjoyed Ur contest amd sir. Finally I am specialist nw. Thanks to u.
• » » » » 3 years ago, # ^ | +31 Come on, don't be so depressed. You have many great rounds before. Looking forward to your next contest!
• » » » » » 3 years ago, # ^ | +30 I don't think he's depressed. He's probably just tired of competitive programming and wants to do something new.
• » » » » » » 3 years ago, # ^ | 0 Oh, maybe you are right.
• » » » » » » 3 years ago, # ^ | +47 Yeah, this is the case. But why do you care anyway? Number of downvotes to the announcement of my last contest says my contest ain't that popular...
• » » » » » » » 3 years ago, # ^ | +43
• » » » » » » » 3 years ago, # ^ | ← Rev. 3 → +16 Why so sad? Sometimes something goes unplanned,this time the tasks were hard but it is not your fault,you can't predict such things,the problem is that a lot of people can't accept fail,most of them just don't know how many time you have to spent to fully prepare a problem.
• » » » » 3 years ago, # ^ | +18 I like your problems and I think your contests were very educational for me , specially your math and query based problems ... I hope you continue your work but if not we will miss your interesting ideas and heros like Duff ... :(Hope you bests in IOI.
» 3 years ago, # | 0 My attempt to Div2 D (Ant Man). is it valid?mark s as visited. start from s, find the next unvisited position with minimum jump cost. Let it be y1.mark y1 as visited. start from 'y1', find the next unvisited position with minimum jump cost. Let it be y2.... doing like this, we get a greedy path: s — y[1] — y[2] — ... — y[n-2] — eFor convenience we can say s = y[0] and e = y[n-1].Let's assume this is not the optimal path. then there exist some i,j such that exchanging y[i] and y[j] decreases the total cost.But exchanging y[i] and y[j] causes following change:from cost[i-1, i] + cost[i, i+1] cost[j-1, j] + cost[j, j+1] to cost[i-1, j] + cost[j, i+1] cost[j-1, i] + cost[i, j+1] But cost[i-1, i] <= cost[i-1, j] for all j cost[i, i+1] <= cost[i, j+1] for all j+1 cost[j-1, j] <= cost[j-1, i] for all i cost[j, j+1] <= cost[j, i+1] for all i+1 There is contradiction. Therefore the greedy path is optimal.
• » » 3 years ago, # ^ | 0 cost[j-1, j] <= cost[j-1, i] for all i.this is wrong. It is for all i that wasn't visited.
» 3 years ago, # | ← Rev. 3 → +4 What is the reason for TLE in Div2C/Div1A on tests ~40-70? Can you look at my code and tell me?Edit: I had a bug: last = x instead of last = max( last, x ). After fixing that, I got AC.
» 3 years ago, # | -7 Nice contest, thank you amd.
» 3 years ago, # | 0 Sorry to talk about this, but I think shouldn't pass problem B (or D in Div.2). Because, I was writing a simple greedy algorithm and it actually pass the system test!! What???
• » » 3 years ago, # ^ | 0 Can you describe it please?
• » » » 3 years ago, # ^ | +39 http://codeforces.com/contest/705/submission/19710317It works as follows: Start with a direct edge S->E. Then keep adding a new node (chair) from 1 to N, excluding S and E. For each chair, find the best location to insert in the path. Thus, we have a very simple greedy solution, though I don't know if it's correct.
• » » » » 3 years ago, # ^ | 0 I think it is not correct(but pass 0_0). For example, now you have 4->1->3 state and trying to add 5. And for example cost of 4->5->1 (if insert between 4 and 1) is less than 1->5->3 (if insert between 1 and 3) — you will have 4->5->1->3 state next and will never get right 4->2->1->6->5->7->3.
• » » » » 3 years ago, # ^ | ← Rev. 3 → +24 I have noticed that it doesn't work if you iterate in a different way. (Not 1 → N)Though the reversed (N → 1) way seems to be fine
• » » » » » 3 years ago, # ^ | 0 So it looks like this algorithm is incorrect，is it？
• » » » » » » 3 years ago, # ^ | 0 How did you come up to this conclusion?
• » » » » » » » 3 years ago, # ^ | 0 I may not understand what you mean, I think if the algorithm is correct, then any order should be correct.Should not mistake it for a sequence.
• » » » » 3 years ago, # ^ | 0 can any red coder give us proof?
• » » 3 years ago, # ^ | +8 Fair player, respect!
» 3 years ago, # | +3 Div2C/Div1A: Has anyone tried Sqrt decomposition?
» 3 years ago, # | ← Rev. 4 → 0 Aren't the time limit for Div2 — C (Python) too strict ?I got TLE on test case 52, which was apparently doing "1 1" x 300000 times. Code for the same:arr.append(). Solution : 19707083 :(
• » » 3 years ago, # ^ | 0 I got TLE on 84, than on 55, and only after removing code of deleting values from list — acc. I think test 52 is doing something interesting in the end, not just "1 1". I think your problem too in "app[app_index].remove(read_pointer)". Try to create pointers for each app (and save them in dict). My accepted (sts[v] — such pointer for app[v]): http://codeforces.com/contest/705/submission/19712655
• » » » 3 years ago, # ^ | 0 https://wiki.python.org/moin/TimeComplexity Delete Item O(n)
» 3 years ago, # | +46 Could anyone please explain the Ant man (DIV2 D/DIV1 B) problem more thoroughly/easily? What property allows us to solve this TSP problem in polynomial time? What is exactly dp[i][j] here? Is this optimal solution for first i vertices, or something like "parts" of optimal solution? How do we update dp? Wouldn't this require iterating over all the components?
• » » 3 years ago, # ^ | ← Rev. 2 → +1 Suppose the following simpler problem: we need to get from chair 1 to chair N and visit each chair exactly once, you can jump from one chair to any other. If you are currently on chair i, it costs li to jump anywhere left and ri to jump anywhere right. Can this problem also be solved in O(N)2 ?Edit: this problem can be reduced to original like the following: assume that landing is free and that jumping off the chair is costs x * 1e8 so that we can neglect the distance between two chairs. And that s = 1, e = N So the same approach should be working for this problem.
» 3 years ago, # | ← Rev. 2 → +15 Please help me with this problem 705A - Hulk, I don't understand why this code (1968850) doesn't write anything in the output of test case 11 (it's very awful for my friend who had passed the pretests). However, when I resubmit this code again after the contest Codeforces Round #366 (Div. 2), it's (19712336) AC!!! MikeMirzayanov, I hope you will look at this comment and take justice back to my friend nghoangphu. Thank you very much !
» 3 years ago, # | ← Rev. 2 → 0 inductive proof for cycles in Div.2 B: f[1] = 0 = 1 - 1 if f[x] = x — 1, then for any c in [1, x], we have: f[x + 1] = f[x + 1 - c] + f[c] + 1 = x + 1 - c - 1 + c - 1 + 1 = x + 1 - 1. It seemed that the conclusion also holds if we can always split an connected graph into two connected graphs until there is only one vertex in each graph.
• » » 3 years ago, # ^ | ← Rev. 2 → 0 Another proof: Suppose there are m vertices in the cycle. In each turn, one cycle will be replaced with two cycles and the number of cycles is increased by 1. After the last turn, the number of cycles will be m (one vertex in each cycle), so the number of turns should always be m - 1.
» 3 years ago, # | +5 For div1B/div2D, how comes that we can solve this problem in O(n^2)? For small values of n I know how to solve the TSP probblem with dp and bitmasks. For me the Antman problem is just like TSP. What actually differentiates them and makes it solvabale in O(n^2)?
• » » 3 years ago, # ^ | +11 In the original Traveling-Salesman Problem, there is no pattern in how your input graph or what costs are there on the edges. For many classic NP-hard problems exist particular restrictions that make them solvable in polynomial time using some observations (ex. Minimum Vertex cover is NP for a general graph, however for bipartite graphs it's reduced to the maximum edge cover using the observations that make up the [König's theorem.](https://en.m.wikipedia.org/wiki/König%27s_theorem_(graph_theory)) . In this particular problem, observations can be made on the fact that the costs on the edges are not random (rather dependent on distance, orientation and the vertexes) and even that the graph is complete.
• » » » 3 years ago, # ^ | 0 This answer my question. Now I'm struggling to understand the solution with DP. The greedy solution seems to work, but I can't prove why. Thank you!
» 3 years ago, # | 0 in Div2D/Div1B I tried so hard , but I couldn't understood the solution can anyone please explain it to me ?
» 3 years ago, # | ← Rev. 5 → 0 ignore, same question above
» 3 years ago, # | 0 Problem B. Noticed that just after coding O(N^2) solution with using Grandi's Theory. So much time wasted for nothing. I really liked this contest, thank you!
» 3 years ago, # | 0 Почему если в задаче С использовать обычный массив и переменные co — количество запросов первого типа и ind — до какого запроса первого типа мы уже читали, и при очередном запросе типа 3 просматривать от ind до t, а потом обновлять ind = t, то получаем тайм лимит на 53 тесте (http://codeforces.com/contest/705/submission/19713056), а если юзать set, то все тесты проходит не больше чем за полсекунды (http://codeforces.com/contest/705/submission/19713545) . Ведь наоборот, должно быть больше времени во втором случае, как минимум вставка нового сообщения происходит за log(количество еще не прочитанных сообщений), для запроса третьего типа, каждое непрочитанное сообщение обрабатывается за O(1), так как мы двигаемся итератором по префиксу сета, а при запросе второго типа, каждое еще непрочитанное сообщение приложения x тоже обрабатывается за log(количество еще не прочитанных сообщений). В первой же программе, каждое из этих действий делается за O(1) и общая сложность O(q). Единственная накладка — это постоянно заполнять и очищать очереди для типов приложений, но ведь это точно также происходит и во второй программе. В общем какая то мистика. Буду благодарен, если кто — нибудь прольет свет на эту тайну)
• » » 3 years ago, # ^ | 0 I have understood. ind = t is not correct. In this case time complicity is O(q^2). I need to write ind = max(ind, t). It is correct and time complicity is O(q) and few faster than the solution with Set.
» 3 years ago, # | ← Rev. 3 → 0 I am having trouble making my solution (19714564) for Problem C pass.Here is my algorithm: int totalUnread = 0 // the answer after every event int prevT = 0 // largest t we have seen so far for eventType=3 int[] byApp = new int[n + 1]; // unread notifications per app Queue notifications = new Queue() // queue of events repeat(q) { int type = readInt() if (type == 1) { int x = readInt() // add new notification from app X totalUnread += 1 byApp[x] += 1 notifications.enqueue(x) } else if (type == 2) { int x = readInt() // mark all notifications from app X as read totalUnread -= byApp(x) byApp[x] = 0 } else if (type == 3) { int t = readInt() // keep removing from queue till we have read t notifications while(t > prevT) { int x = notifications.dequeue() prevT += 1 if (byApp(x) > 0) { totalUnread -= 1 byApp[x] -= 1 } } } println(totalUnread) } But this fails the 3rd test case. Please help.
• » » 3 years ago, # ^ | 0 Yeah, even I have the same issue. "wrong answer 20th numbers differ — expected: '2', found: '1'"
• » » 3 years ago, # ^ | 0 Event 3 is processed incorrectly: the fact that there are some unread application x events (byApp(x) > 0) does not imply that those events are among the first t events. E.g. given 3 4 1 1 2 1 1 1 3 1, your code will delete the second addition event "1 1" whereas it should not have.
• » » 3 years ago, # ^ | 0 byApp(x)>0 does not necessarily mean that the top m messages must contain unread x. unread x may appear in the later message while x in top m messages has already been read.
» 3 years ago, # | 0 Can somebody plz explain that why are we marking the numbers in div2 C.
• » » 3 years ago, # ^ | 0 We mark the notifications so we don't count them twice.
» 3 years ago, # | ← Rev. 2 → 0 Can someone please tell me where I am wrong DIV2 C .My code is http://codeforces.com/contest/705/submission/19705093 Thanks.
• » » 3 years ago, # ^ | 0 It looks like the issue is the same — please see my comment above http://codeforces.com/blog/entry/46450?#comment-309097
• » » 3 years ago, # ^ | ← Rev. 4 → 0 Simply because if the input is : ( same x ) type 1 + x Type 2 + x type 1 + x type 2 + x type 3 + 1 answer is false according to your code , if a [ x ] > 0 then you substract 1 , so that 's why your answer = expected - 1 hope this is helpful
» 3 years ago, # | 0 Some remark and improvement for Div.2 C: You have to write x instead of v in the e[v].pop() We can use vector for e (not queue), delete the e[v].pop() line. "When a second type event happens, check marked and mark all others numbers in vector e[x] and clear this vector "
» 3 years ago, # | +23 I've finally got accepted for problem E. I want to share a bit different geometric view on the same solution as author's, but it may be easier to understand.First, let's apply Heavy-Light decomposition and split each query into parts in such way that each parts belongs to a single path of HLD. Now we can solve problem on a line segment.Let's introduce real-value coordinates on the segment (i. e. points have x-coordinates in range [0, len] where len is the length of a HLD segment). Draw an Oxt plane where t means the time. Each trajectory of a query looks like a segment (possibly, degenerate) on Oxt plane, and what we asked is to check if any two segments share a point, and if so, find the point with smallest t-coordinate. If we had to just check an existence of such point, we could use a well-known sweeping line algorithm that iterates through endpoints of segments in order of their t values and stores segments in some balanced binary search tree checking neighboring pairs of segments for intersection on any insertion/deletion. But this algorithm can be slightly modified in order to find also the point with minimum t-value: all we need is to keep the current known lowest intersection (in sense of t) and stop our algorithm when we pass it.
» 3 years ago, # | ← Rev. 4 → +8 Div2 D , was quite a difficulty jump. Most of the people got AC by using some kind of greedy technique i.e. so they considered a list of vertices in the order of visiting them . which initially contains S,E(path S->E) . then for every i from 1 to N they will try to place i inside the path list in a greedy way (which ever position leads to minimum result, S->i->E).So basically , the question is that this is not DP , surely its not as we are not trying every permutation .So how can we be sure that the solution is right . Though i am not sure , but i think theres actually no need to do a perfect dp . For the moment lets consider this method to be wrong and consider there exists a path in which vertex A should have been placed before placing vertex B where B Xa-Xb + Xb-Xc . so including B before A is always better . can anyone provide more clarity ? or may be explain the given editorial in a better way ?.btw the problem was great .
• » » 3 years ago, # ^ | 0 For example you have 4->2->1->3 and inserting vertex 5 that in the answer is between 6 and 7, at this moment it can happen that insert 5 between 1 and 3 not the best, but answer is 4->2->1->6->5->7->3 http://codeforces.com/blog/entry/46450?#comment-309049
» 3 years ago, # | +36 What is Karzanov's Theorem?If anyone could explain it or just provide a good source, I'd appreciate it :)
» 3 years ago, # | 0 My approach to C (even though I did not participate) was segment trees. We add 1 for unread notification, and 0 for read notification and get the sum to answer.Basically push all notifications into vectors (apps). If type is 1, push the notification and update the value to 1. If type is 2, start from the first notification that was not read during the most recent time where the query gives type 2 and the same app. (This prevents TLE) Then iterate over the vector and update the tree value to 0. If type is 3, we update the "currenttime" value.Then we get the sum from the currrenttime~end. (Because we don't need to look at notifications before "currenttime")This gives a solution in Implementation: http://codeforces.com/contest/704/submission/19718386
» 3 years ago, # | -30 In Div2C, could someone help me to understand how the time complexity is O(q) in the editorial. I wrote the code after reading the editorial, it passed in 233ms only. But i cant understand its complexity. here is the code: #include #define MAX 10000000 #define mpr(x,y) make_pair(x,y) #define pb(x) push_back(x) #define MOD 1000000007ll #define INF 9223372036854775807 #define S(x) scanf("%lld",&x) #define Ss(x) scanf("%s",x) #define Ps(x) printf("%s\n",x) #define P(x) printf("%lld\n",x) #define P_(x) printf("%lld ",x) #define Ps_(x) printf("%s ",x) #define nL printf("\n") using namespace std; typedef long long ll; int main(){ ll n,q,ctr=0; S(n);S(q); deque > Q; vector e[n+1]; bool mark[q+1], markapp[n+1]; memset(mark,0,sizeof(mark)); ll ans = 0; while(q--){ ll x,y; S(x);S(y); if(x==1){ ++ctr; ans++; Q.pb(mpr(y,ctr)); e[y].pb(ctr); } else if(x==2){ for(ll i=0; i >::iterator it = Q.begin(); while(!Q.empty() && (*it).second <= y){ if(!mark[(*it).second]){ mark[(*it).second] = true; ans--; } Q.pop_front(); it++; } } P(ans); } return 0; }
» 3 years ago, # | ← Rev. 2 → +9 In Div1-B, does component here mean "connected component"?PS: And can anyone explain the solution more clearly? I can't imagine how it work.
• » » 3 years ago, # ^ | +5 A pictorial description and sample code of editorial would help.
» 3 years ago, # | 0 In this editorial Div1 B Antman, 'first i chairs' means i visited chairs? or chair[1],chair[2],... and chair[i]?
» 3 years ago, # | ← Rev. 5 → +18 many of you couldn't understand the solution for DIV2 D/Div1 B so here is my solution , hope it will help you suppose you have a directed graph (there is an edge between every i and j ) and you have to find a path from S to E by visiting each vertex once obviously each vertex except S and E should be linked with two different vertices dp[i][j][k] represents minimum cost to reach the destination such that among the last i elements, j of them want to be linked with another vertex and k of them have an outgoing edge which is not connected to anything yet. now we will consider each vertex separately ,we have three cases a vertex with just outgoing edge (if that vertices is S) dp[i][j][k] = min(dp[i+1][j-1][k] + x[i] + c[i], dp[i+1][j][k+1] -x[i] + d[i]) a vertex with just incoming edge (if that vertices is E) dp[i][j][k] = min(dp[i+1][j][k-1] + x[i] + a[i], dp[i+1][j+1][k] -x[i] + b[i]) a vertex with incoming edge and outgoing edge dp[i][j][k] = min( dp[i+1][j-1][k-1] + 2*x[i]+a[i]+c[i] (an edge form vertex x to vertex i (xi)) dp[i+1][j][k] + b[i]+c[i] (an edge form vertex x to vertex i (x>i) and an edge from vertex i to vertex y (yi) and an edge from vertex i to vertex y (y>i)) ) it can be proved that for any (i,j) k will be the same , so we can only memorize dp[i][j] and the complexity will be o(n^2)
• » » 3 years ago, # ^ | ← Rev. 3 → 0 First of all thanks for trying to shed some light on this problem that I can't wrap my head for 12 hours already. represents minimum cost to reach the destination such that between the last i elements, j of them are still wants to be linked with another vertices and k of them have an outgoing edge which not connected to anything yet. What if destination is not in last i vertices? It would be INF? There are j elements without outgoing edges, k elements with outgoing edges that go nowhere, but what if two elements are connected? How can we conclude k then? And of course, why this approach won't work if I take a TSP (say I enumerate all vertices from 1 to N somehow) and apply this DP?
• » » » 3 years ago, # ^ | ← Rev. 2 → +3 j is the number of vertices that doesn't have any edge going from another vertex to it. no it wouldn't be INF , it would be the minimum cost to make a path from S to any vertex by visiting each vertex (from the last i vertices) exactly once and if S in not in last i vertices it would be the minimum cost to make a path from any vertex to any vertex by visiting each vertex (from the last i vertices) exactly once. look at the dp equation, if not either S or E in the last i vertices then (for any (i,j) ) j will be equal to k otherwise the difference between j and k will be at most 1, now you can make it dp[i][j][d] where d represents the difference between j and k, so you can conclude k . it can be proved that for any (i,j) k will be the same . here is my submission for a better understand. to be honest I'm not sure why this approach won't work for TSP but in somehow?? I think it's impossible to enumerate all the vertices, because in this problem dis(i,j) < dis(i,j-1) regardless of jumping and landing and that make this approach work.
• » » 3 years ago, # ^ | 0 I can't understand anything. Can you suggest similar problems which can help?
• » » » 3 years ago, # ^ | +8 I'm sorry but I don't know any similar problems but here is my submission , it will help you to understand the solution
• » » » » 3 years ago, # ^ | ← Rev. 10 → +3 In your submission,solve(i,j,k)means i:=next added point's id(i points are already added) j:=the number of points these are in added i points and want to be connected from a point right than these k:=the number of points these are in added i points and want to connect to a point right than these solve(i,j,k):= with j,k conditions, the sum of cost that is increased by adding point[i],point[i+1],...point[n-1]. And in your memo[i][j], memo[0][0] is finally changed.Because your solve(i,j,k) makes possible transitions and right end condition of the recurrence.In addition,first of all, memo[n][0] is fixed to 0, after that, each memo[i][j] is fixed only when the best is fixed.After adding i points, S and E are not added => j != 0 and k!=0 and j==k and 1<=j<=i S is added and E is not => k==j+1 and 0<=j<=i-1 E is added and S is not => j==k+1 and 0<=k<=i-1 S and E are added => j==k (j==k==0 is possible) and 0<=j<=i-1 Thus,k corresponding to j is only one in all possible right conditions.Is this right?
• » » 3 years ago, # ^ | 0 I really want to know how to simply conclude k，too.
• » » 3 years ago, # ^ | ← Rev. 2 → 0 Finally accepted thanks to you :D submission But there is something I'd like to emphasize, when (j == 0 && 0 == k) and (indexing mode) < i <= (last index) then we stop and indicate that this way won't lead to the intended solution because remaining vertices/chairs can't go to any chair on the left and likewaise, chairs on the left are fully connected to each other without an edge going to the right side, thus we end up having at least two disjoint components.
• » » 3 years ago, # ^ | 0 I've got a problem about the solution. Is there the possibility that the paths form a circur ? Just In dp[i][j][k] = min(dp[i+1][j-1][k-1]+2*x[i]+a[i]+c[i])?
• » » 3 years ago, # ^ | 0 Excuse me,I thought I had got your point about the solution.But my code got wrong answer at test5.I don't know where the problem is.And different from your code,I used simple loop instead of dfs to solve the problem.So I really want a simple loop version of the solution,could u give me an example?
» 3 years ago, # | ← Rev. 2 → 0 Sorry to say, but not a very nice editorial. There are no codes of author too.Edit: Sorry, I didn't see the link before.
• » » 3 years ago, # ^ | +8 There's a link to a gitlab page with all the official solutions. https://gitlab.com/amirmd76/cf-round-366/tree/master
» 3 years ago, # | 0 Hello,For Div.2 problem C, I had an idea during the contest using Segment Tree. We can build the tree on the events sum, then we can query and update our tree, but my solution was not fast enough (for queries of type 2).I am really interested in solving this problem using Segment Tree. Any help?Kindly find here my submission.
• » » 3 years ago, # ^ | 0 Here is my solution with Segment Tree. Actually, solution in editorial is easier.
» 3 years ago, # | 0 المسائل اخرى من وشك يا لوطي
» 3 years ago, # | 0 Can anyone explain why div2D is not np hard problem?I can't understand why it's not np hard.
• » » 3 years ago, # ^ | +8 Consider the middle chairIf you choose one on the right to jump on, you will pay the same amount of money regardless of your choice (which one to choose on the right side), similarly, if you choose one on the left, you will pay another constant value regardless of your choiceSo basically, each chair has 4 states: previous chair and next chair are on the right side previous chair and next chair are on the left side previous chair on right side and next chair on left side previous chair on left side and next chair on right side For S chair: next chair on the right next chair on the left For E chair: previous chair on the right previous chair on the left You can use dynamic programming to solve it keeping in mind that the cost of jumping from a chair and landing on another just depends on the directionYou can see my submission here
• » » » 3 years ago, # ^ | 0 You describe the status by using dp[i][left_go],but there are three arguments:i,left go and sit right,how can you make sure that the status are the same?
» 3 years ago, # | +15 In 1B or 2D, what I overlooked was the constraint "1 ≤ x1 < x2 < ... < xn ≤ 109". After reading the editorial and checking many solutions I still couldn't understand why the applied approach is correct. Reading the question again explained everything. Just posting about this in case someone else is stuck with something similar. BTW, nice question! Thumbs up!
• » » 3 years ago, # ^ | ← Rev. 2 → 0 How does this help? With sufficiently large values for Ai-s, Bi-s, Ci-s, and Di-s, won't Xi-s get abstracted away?
• » » » 2 years ago, # ^ | ← Rev. 3 → 0 Because of the condition for this problem (not just the constraint mentioned above), you can know exactly what is the cost to add a new vertex into existing path only by knowing the unconnected endpoints, and without knowing which vertices this new vertex connected to -> solvable in polynomial time.
» 3 years ago, # | 0 Can anyone please point out what's wrong with my approach for DIV2 D for this contest. I have been trying to find the minimum time from source chair to all the other chairs initially (except end). After iterating over all the chairs I update the source node with the label for the chair which resulted in minimum time.This is my submission
» 3 years ago, # | ← Rev. 2 → 0 can somebody please tell me why the 10 line of judge's answer in test case 4 of problem A (DIV1)is '3' shouldnt it be '4'? | 2019-04-24 09:19:10 | {"extraction_info": {"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, "math_score": 0.4499073922634125, "perplexity": 1735.8484030991658}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578636101.74/warc/CC-MAIN-20190424074540-20190424100540-00329.warc.gz"} |
http://mathhelpforum.com/calculus/217037-f-ind-corresponding-derivative-function-s-graph.html | # Math Help - F ind the corresponding derivative function ’s graph
1. ## F ind the corresponding derivative function ’s graph
See attatcments
I'd appreciate any help!
2. ## Re: F ind the corresponding derivative function ’s graph
Have you attempted these yourself? What do you think the answers might be?
3. ## Re: F ind the corresponding derivative function ’s graph
I thought it was D., G., H., F. but I'm not sure. I looked at the critical numbers. | 2015-01-31 06:32:05 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9945632219314575, "perplexity": 5475.663164958496}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115869320.77/warc/CC-MAIN-20150124161109-00021-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://ericsimmerman.com/blog/2008/06/05/mathml-support-in-openlaszlo/ | # MathML support in OpenLaszlo
I'm happy to announce that my company (Tempest Strings Enterprises) has sponsored my development and release of a MathML library for OpenLaszlo. I'll be working through the OpenLaszlo contributor process, but imagine that it could take a considerable amount of time before you see this work in their repository since I've not yet made any effort to adhere to any coding guidelines. There is very little documentation in the source, no consistent style, and no support for any runtime except SWF. Nevertheless, I've already successfully integrated the library with a larger application and it is functioning well in our beta release.
This library is based upon the Flash Actionscript library that ionel-alexandru (Sourceforge username) of learn-math.info donated to the public domain. I make heavy use of Alex's object tree and MathML rendering logic (thanks Alex!). I've built upon OpenLaszlo's Drawview class from their SVN trunk and added support for rendering stand-alone MathML from a datasource as well as formulas placed inline with other text.
Note that the full set of MathML has not yet been implemented (ex: no support for <mtable>) but here are some examples of supported formulas:
<dataset name="example1">$<mrow><msqrt> <mrow> <msup> <mn>2</mn> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <mrow> <mn>4</mn> <mi>x</mi> </mrow> <mo>+</mo> <mn>4</mn> </mrow> </msup> <mo>+</mo> <mrow> <mn>4</mn> <mi>x</mi> </mrow> <mo>+</mo> <mn>45</mn> </mrow></msqrt><mo>+</mo><mroot linethickness="3"><mrow> <msub> <mi>log</mi> <mn>2</mn> </msub></mrow><mi>5</mi></mroot></mrow>$</dataset>
renders with the library as:
<dataset name="example2">$<mrow> <mo> ( </mo> <mfrac linethickness="0" color="#006699"> <mi fontstyle="bold"> a + b +c </mi> <mi color="#ff0000"> b </mi> </mfrac> <mo> ) </mo> </mrow> <mfrac linethickness="2" fontstyle="italic" > <mrow><mi color="#00ff00">3456</mi><mo>+</mo><mfrac> <mi> a </mi> <mi fontsize="8"> b + c + d</mi> </mfrac> </mrow><mrow><mn>3</mn> <mfrac color="#ffff00"> <mi> c + d </mi> <mi color="#ff0000"> d </mi> </mfrac> </mrow> </mfrac>$</dataset>
makes use of node-level styling to render as
The next one really shows off some of the fancy rendering capabilities:
<dataset name="example3">$<msubsup> <mo><![CDATA[???]]></mo><mn>2</mn><mn>56</mn> </msubsup> <msup> <mi>x</mi> <mn>2</mn> </msup> <mtext fontstyle='italic'>dx</mtext><mo>+</mo> <msubsup> <mo><![CDATA[???]]></mo> <mi>-<![CDATA[&infinity;]]></mi> <mi><![CDATA[&infinity;]]></mi> </msubsup> <msup> <mi>e</mi> <mn>x</mn> </msup> <mtext fontstyle='italic'>dx</mtext><mo>+</mo> <munderover> <mo><![CDATA[???]]></mo> <mrow> <mi>x</mi> <mo>=</mo> <mn>0</mn> </mrow> <mn>56</mn> </munderover> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <munderover> <mo><![CDATA[???]]></mo> <mrow> <mi>x</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi><![CDATA[&infinity;]]></mi> </munderover> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <munderover> <mo><![CDATA[???]]></mo> <mrow><mi>x</mi><mo>=</mo><mn>1</mn></mrow> <mi>n</mi></munderover><mrow> <mi>x</mi></mrow><mo>+</mo> <munder> <mrow> <mi>lim</mi> </mrow> <mrow> <mi>x</mi> <mo><![CDATA[???]]></mo> <mn>0</mn> </mrow></munder><mrow> <mi>x</mi></mrow>$</dataset>
This last example is demonstrating inline rendering with text:
<dataset name="myFormula"><![CDATA[This is an example of pseudo inline MathML:$<mrow> <mi>x</mi> <mo>+</mo> <msup> <mi>y</mi> <mfrac> <mn>2</mn> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> </msup> </mrow>$How do you like them apples?]]></dataset>
All of the formulas above were displayed using the same MathML view using the datasets listed. Here's the view declaration:
<mathmlview font="sans-serif" fontsize="24" fgcolor="#000000" id="math" x="10" y="10" width="800" height="500" datapath="myFormula:/"/>
The current source is released under the MIT Open Source License and available on github. Enjoy! | 2022-12-01 23:50:59 | {"extraction_info": {"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": 1, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.4393734633922577, "perplexity": 5239.4617405214}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00169.warc.gz"} |
https://benjaminwhiteside.com/2014/02/17/topological-invariants/ | Before we can talk about topological invariants we need to know two things: 1) what is a topology, and 2) what is an invariant. Simply put, a topology is a collection of sets, each of which, vaguely speaking, has fuzzy edges. An invariant, on the other hand, is basically something which doesn’t change. In the remainder of this article I will slowly introduce the concept of a topological invariant and then having done so will illustrate what you can do with them.
To properly define a topology let’s start with a non-empty set. We can, of course, give it a name: $X$ and it will consist of one or more objects, or subsets. Suppose further that we collect up a bunch of these subsets in a very particular way (explained in a second) and call it the Greek letter tau, or $\tau$.
If the way in which we pick our subsets adheres to the following rules (or axioms):
1. The empty set $\emptyset$ and the entire set $X$ are in $\tau$
2. Any union (possibly infinite) of any open set is in $\tau$.
3. The intersection of any finite number of the open sets is, again, open.
then the collection of subsets gets a special name, of course, it is called a topology on the set $X$. The objects within this special kind of collection get their own name too, we call them open sets. Open sets are nice sets; we can pick elements out of them without having to worry about accidentally going too far and picking an element that isn’t in it. Formally speaking, an open set does not contain any of its boundary points and this gives us freedom to draw little circles around each and every element inside an open set and know for a fact that the little circle (also called a neighbourhood) is entirely within the set. Taking the set that we began with $X$ and the collection of subsets $\tau$ we form the tuple $(X,\tau)$, and this is called a topological space.
A topological space thus consists of some set along with a collection of subsets that are nice and fuzzy. All in all, this is a pretty big space, almost all of mathematics takes place in some topological space or another. Topological spaces allow you to, obviously define sets and open subsets, which in turn allow you to define neighbourhoods which carry a diameter of sorts, which in turn allows you to define a primitive notion of distance (think of everything being measured in terms of the diameter of the little circles, or $\epsilon$ as it’s denoted). A notion of distance gives rise to the 3 C’s: Continuity, Connectedness and Convergence. And this, my friends, is where calculus lives.
The second concept we need is that of an invariant. An invariant is a class of mathematical objects that don’t change when you move the object out of one environment and in to another (or back to the original one). Technically speaking, moving an object doesn’t really happen, what actually happens is that you move the object’s elements and the way in which that is done is by use of a mapping, usually denoted by $f$. The mapping takes an element, say $x$ of some set $X$, and maps it to another element $y$ in some other set $Y$. Once you do this to each and every element of $X$, if the destination set $Y$ shares a property with $X$ then we say that the original set $X$ is invariant under $f$.
Let’s begin by talking about an invariant of a homeomorphism. In case you don’t recall, a homeomorphism is a mapping (a way of getting from point A to point B), call it $h$, from a topological space $X_1$ to another topological space $X_2$ that preserves the topological properties of the first space. | 2018-05-21 01:13:51 | {"extraction_info": {"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": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8056730628013611, "perplexity": 160.2800862977822}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794863901.24/warc/CC-MAIN-20180521004325-20180521024325-00400.warc.gz"} |
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# Ideal (ring theory)
In ring theory, a branch of abstract algebra, an ideal of a ring is a special subset of its elements. Ideals generalize certain subsets of the integers, such as the even numbers or the multiples of 3. Addition and subtraction of even numbers preserves evenness, and multiplying an even number by any integer (even or odd) results in an even number; these closure and absorption properties are the defining properties of an ideal. An ideal can be used to construct a quotient ring in a way similar to how, in group theory, a normal subgroup can be used to construct a quotient group.
Among the integers, the ideals correspond one-for-one with the non-negative integers: in this ring, every ideal is a principal ideal consisting of the multiples of a single non-negative number. However, in other rings, the ideals may not correspond directly to the ring elements, and certain properties of integers, when generalized to rings, attach more naturally to the ideals than to the elements of the ring. For instance, the prime ideals of a ring are analogous to prime numbers, and the Chinese remainder theorem can be generalized to ideals. There is a version of unique prime factorization for the ideals of a Dedekind domain (a type of ring important in number theory).
The related, but distinct, concept of an ideal in order theory is derived from the notion of ideal in ring theory. A fractional ideal is a generalization of an ideal, and the usual ideals are sometimes called integral ideals for clarity.
## History
Ernst Kummer invented the concept of ideal numbers to serve as the "missing" factors in number rings in which unique factorization fails; here the word "ideal" is in the sense of existing in imagination only, in analogy with "ideal" objects in geometry such as points at infinity.[1] In 1876, Richard Dedekind replaced Kummer's undefined concept by concrete sets of numbers, sets that he called ideals, in the third edition of Dirichlet's book Vorlesungen über Zahlentheorie, to which Dedekind had added many supplements.[1][2][3] Later the notion was extended beyond number rings to the setting of polynomial rings and other commutative rings by David Hilbert and especially Emmy Noether.
## Definitions and motivation
For an arbitrary ring ${\displaystyle (R,+,\cdot )}$, let ${\displaystyle (R,+)}$ be its additive group. A subset ${\displaystyle I}$ is called a left ideal of ${\displaystyle R}$ if it is an additive subgroup of ${\displaystyle R}$ that "absorbs multiplication from the left by elements of ${\displaystyle R}$"; that is, ${\displaystyle I}$ is a left ideal if it satisfies the following two conditions:
1. ${\displaystyle (I,+)}$ is a subgroup of ${\displaystyle (R,+),}$
2. For every ${\displaystyle r\in R}$ and every ${\displaystyle x\in I}$, the product ${\displaystyle rx}$ is in ${\displaystyle I}$.
A right ideal is defined with the condition "r xI" replaced by "x rI". A two-sided ideal is a left ideal that is also a right ideal, and is sometimes simply called an ideal. In the language of modules, the definitions mean that a left (resp. right, two-sided) ideal of R is an R-submodule of R when R is viewed as a left (resp. right, bi-) R-module. When R is a commutative ring, the definitions of left, right, and two-sided ideal coincide, and the term ideal is used alone.
To understand the concept of an ideal, consider how ideals arise in the construction of rings of "elements modulo". For concreteness, let us look at the ring ℤ/nℤ of integers modulo n given integer n ∈ ℤ (note that ℤ is a commutative ring). The key observation here is that we obtain ℤ/nℤ by taking the integer line ℤ and wrapping it around itself so that various integers get identified. In doing so, we must satisfy 2 requirements:
1) n must be identified with 0 since n is congruent to 0 modulo n.
2) the resulting structure must again be a ring.
The second requirement forces us to make additional identifications (i.e., it determines the precise way in which we must wrap ℤ around itself). The notion of an ideal arises when we ask the question:
What is the exact set of integers that we are forced to identify with 0?
The answer is, unsurprisingly, the set nℤ = { nm | m ∈ ℤ } of all integers congruent to 0 modulo n. That is, we must wrap ℤ around itself infinitely many times so that the integers ..., n · (−2), n · (−1), n · (+1), n · (+2), ... will all align with 0. If we look at what properties this set must satisfy in order to ensure that ℤ/nℤ is a ring, then we arrive at the definition of an ideal. Indeed, one can directly verify that nℤ is an ideal of ℤ.
Remark. Identifications with elements other than 0 also need to be made. For example, the elements in 1 + n must be identified with 1, the elements in 2 + n must be identified with 2, and so on. Those, however, are uniquely determined by nℤ since ℤ is an additive group.
We can make a similar construction in any commutative ring R: start with an arbitrary xR, and then identify with 0 all elements of the ideal xR = { x r : rR }. It turns out that the ideal xR is the smallest ideal that contains x, called the ideal generated by x. More generally, we can start with an arbitrary subset SR, and then identify with 0 all the elements in the ideal generated by S: the smallest ideal (S) such that S ⊆ (S). The ring that we obtain after the identification depends only on the ideal (S) and not on the set S that we started with. That is, if (S) = (T), then the resulting rings will be the same.
Therefore, an ideal I of a commutative ring R captures canonically the information needed to obtain the ring of elements of R modulo a given subset SR. The elements of I, by definition, are those that are congruent to zero, that is, identified with zero in the resulting ring. The resulting ring is called the quotient of R by I and is denoted R/I. Intuitively, the definition of an ideal postulates two natural conditions necessary for I to contain all elements designated as "zeros" by R/I:
1. I is an additive subgroup of R: the zero 0 of R is a "zero" 0 ∈ I, and if x1I and x2I are "zeros", then x1x2I is a "zero" too.
2. Any rR multiplied by a "zero" xI is a "zero" rxI.
It turns out that the above conditions are also sufficient for I to contain all the necessary "zeros": no other elements have to be designated as "zero" in order to form R/I. (In fact, no other elements should be designated as "zero" if we want to make the fewest identifications.)
Remark. The above construction still works using two-sided ideals even if R is not necessarily commutative.
## Examples and properties
(For the sake of brevity, some results are stated only for left ideals but are usually also true for right ideals with appropriate notation changes.)
• In a ring R, the set R itself forms a two-sided ideal of R called the unit ideal. It is often also denoted by ${\displaystyle (1)}$ since it is precisely the two-sided ideal generated (see below) by the unity ${\displaystyle 1_{R}}$. Also, the set ${\displaystyle \{0_{R}\}}$ consisting of only the additive identity 0R forms a two-sided ideal called the zero ideal and is denoted by ${\displaystyle (0)}$.[note 1] Every (left, right or two-sided) ideal contains the zero ideal and is contained in the unit ideal.
• An (left, right or two-sided) ideal that is not the unit ideal is called a proper ideal (as it is a proper subset).[4] Note: a left ideal ${\displaystyle {\mathfrak {a}}}$ is proper if and only if it does not contain a unit element, since if ${\displaystyle u\in {\mathfrak {a}}}$ is a unit element, then ${\displaystyle r=(ru^{-1})u\in {\mathfrak {a}}}$ for every ${\displaystyle r\in R}$. Typically there are plenty of proper ideals. In fact, if R is a skew-field, then ${\displaystyle (0),(1)}$ are its only ideals and conversely: that is, a nonzero ring R is a skew-field if ${\displaystyle (0),(1)}$ are the only left (or right) ideals. (Proof: if ${\displaystyle x}$ is a nonzero element, then the principal left ideal ${\displaystyle Rx}$ (see below) is nonzero and thus ${\displaystyle Rx=(1)}$; i.e., ${\displaystyle yx=1}$ for some nonzero ${\displaystyle y}$. Likewise, ${\displaystyle zy=1}$ for some nonzero ${\displaystyle z}$. Then ${\displaystyle z=z(yx)=(zy)x=x}$.)
• The even integers form an ideal in the ring ${\displaystyle \mathbb {Z} }$ of all integers; it is usually denoted by ${\displaystyle 2\mathbb {Z} }$. This is because the sum of any even integers is even, and the product of any integer with an even integer is also even. Similarly, the set of all integers divisible by a fixed integer n is an ideal denoted ${\displaystyle n\mathbb {Z} }$.
• The set of all polynomials with real coefficients which are divisible by the polynomial x2 + 1 is an ideal in the ring of all polynomials.
• The set of all n-by-n matrices whose last row is zero forms a right ideal in the ring of all n-by-n matrices. It is not a left ideal. The set of all n-by-n matrices whose last column is zero forms a left ideal but not a right ideal.
• The ring ${\displaystyle C(\mathbb {R} )}$ of all continuous functions f from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$ under pointwise multiplication contains the ideal of all continuous functions f such that f(1) = 0. Another ideal in ${\displaystyle C(\mathbb {R} )}$ is given by those functions which vanish for large enough arguments, i.e. those continuous functions f for which there exists a number L > 0 such that f(x) = 0 whenever |x| > L.
• A ring is called a simple ring if it is nonzero and has no two-sided ideals other than ${\displaystyle (0),(1)}$. Thus, a skew-field is simple and a simple commutative ring is a field. The matrix ring over a skew-field is a simple ring.
• If ${\displaystyle f:R\to S}$ is a ring homomorphism, then the kernel ${\displaystyle \ker(f)=f^{-1}(0_{S})}$ is a two-sided ideal of ${\displaystyle R}$. By definition, ${\displaystyle f(1_{R})=1_{S}}$, and thus if ${\displaystyle S}$ is not the zero ring (so ${\displaystyle 1_{S}\neq 0_{S}}$), then ${\displaystyle \ker(f)}$ is a proper ideal. More generally, for each left ideal I of S, the pre-image ${\displaystyle f^{-1}(I)}$ is a left ideal. If I is a left ideal of R, then ${\displaystyle f(I)}$ is a left ideal of the subring ${\displaystyle f(R)}$ of S: unless f is surjective, ${\displaystyle f(I)}$ need not be an ideal of S; see also #Extension and contraction of an ideal below.
• Ideal correspondence: Given a surjective ring homomorphism ${\displaystyle f:R\to S}$, there is a bijective order-preserving correspondence between the left (resp. right, two-sided) ideals of ${\displaystyle R}$ containing the kernel of ${\displaystyle f}$ and the left (resp. right, two-sided) ideals of ${\displaystyle S}$: the correspondence is given by ${\displaystyle I\mapsto f(I)}$ and the pre-image ${\displaystyle J\mapsto f^{-1}(J)}$. Moreover, for commutative rings, this bijective correspondence restricts to prime ideals, maximal ideals, and radical ideals (see the Types of ideals section for the definitions of these ideals).
• (For those who know modules) If M is a left R-module and ${\displaystyle S\subset M}$ a subset, then the annihilator ${\displaystyle \operatorname {Ann} _{R}(S)=\{r\in R\mid rs=0,s\in S\}}$ of S is a left ideal. Given ideals ${\displaystyle {\mathfrak {a}},{\mathfrak {b}}}$ of a commutative ring R, the R-annihilator of ${\displaystyle ({\mathfrak {b}}+{\mathfrak {a}})/{\mathfrak {a}}}$ is an ideal of R called the ideal quotient of ${\displaystyle {\mathfrak {a}}}$ by ${\displaystyle {\mathfrak {b}}}$ and is denoted by ${\displaystyle ({\mathfrak {a}}:{\mathfrak {b}})}$; it is an instance of idealizer in commutative algebra.
• Let ${\displaystyle {\mathfrak {a}}_{i},i\in S}$ be an ascending chain of left ideals in a ring R; i.e., ${\displaystyle S}$ is a totally ordered set and ${\displaystyle {\mathfrak {a}}_{i}\subset {\mathfrak {a}}_{j}}$ for each ${\displaystyle i. Then the union ${\displaystyle \textstyle \bigcup _{i\in S}{\mathfrak {a}}_{i}}$ is a left ideal of R. (Note: this fact remains true even if R is without the unity 1.)
• The above fact together with Zorn's lemma proves the following: if ${\displaystyle E\subset R}$ is a possibly empty subset and ${\displaystyle {\mathfrak {a}}_{0}\subset R}$ is a left ideal that is disjoint from E, then there is an ideal that is maximal among the ideals containing ${\displaystyle {\mathfrak {a}}_{0}}$ and disjoint from E. (Again this is still valid if the ring R lacks the unity 1.) When ${\displaystyle R\neq 0}$, taking ${\displaystyle {\mathfrak {a}}_{0}=(0)}$ and ${\displaystyle E=\{1\}}$, in particular, there exists a left ideal that is maximal among proper left ideals (often simply called a maximal left ideal); see Krull's theorem for more.
• An arbitrary union of ideals need not be an ideal, but the following is still true: given a possibly empty subset X of R, there is the smallest left ideal containing X, called the left ideal generated by X and is denoted by ${\displaystyle RX}$. Such an ideal exists since it is the intersection of all left ideals containing X. Equivalently, ${\displaystyle RX}$ is the set of all the (finite) left R-linear combinations of elements of X over R:
${\displaystyle RX=\{r_{1}x_{1}+\dots +r_{n}x_{n}\mid n\in \mathbb {N} ,r_{i}\in R,x_{i}\in X\}.}$
(since such a span is the smallest left ideal containing X.)[note 2] A right (resp. two-sided) ideal generated by X is defined in the similar way. For "two-sided", one has to use linear combinations from both sides; i.e.,
${\displaystyle RXR=\{r_{1}x_{1}s_{1}+\dots +r_{n}x_{n}s_{n}\mid n\in \mathbb {N} ,r_{i}\in R,s_{i}\in R,x_{i}\in X\}.\,}$
• A left (resp. right, two-sided) ideal generated by a single element x is called the principal left (resp. right, two-sided) ideal generated by x and is denoted by ${\displaystyle Rx}$ (resp. ${\displaystyle xR,RxR}$). The principal two-sided ideal ${\displaystyle RxR}$ is often also denoted by ${\displaystyle (x)}$. If ${\displaystyle X=\{x_{1},\dots ,x_{n}\}}$ is a finite set, then ${\displaystyle RXR}$ is also written as ${\displaystyle (x_{1},\dots ,x_{n})}$.
• In the ring ${\displaystyle \mathbb {Z} }$ of integers, every ideal can be generated by a single number (so ${\displaystyle \mathbb {Z} }$ is a principal ideal domain), as a consequence of Euclidean division (or some other way).
• There is a bijective correspondence between ideals and congruence relations (equivalence relations that respect the ring structure) on the ring: Given an ideal I of a ring R, let x ~ y if xyI. Then ~ is a congruence relation on R. Conversely, given a congruence relation ~ on R, let I = { x | x ~ 0 }. Then I is an ideal of R.
## Types of ideals
To simplify the description all rings are assumed to be commutative. The non-commutative case is discussed in detail in the respective articles.
Ideals are important because they appear as kernels of ring homomorphisms and allow one to define factor rings. Different types of ideals are studied because they can be used to construct different types of factor rings.
• Maximal ideal: A proper ideal I is called a maximal ideal if there exists no other proper ideal J with I a proper subset of J. The factor ring of a maximal ideal is a simple ring in general and is a field for commutative rings.[5]
• Minimal ideal: A nonzero ideal is called minimal if it contains no other nonzero ideal.
• Prime ideal: A proper ideal I is called a prime ideal if for any a and b in R, if ab is in I, then at least one of a and b is in I. The factor ring of a prime ideal is a prime ring in general and is an integral domain for commutative rings.
• Radical ideal or semiprime ideal: A proper ideal I is called radical or semiprime if for any a in R, if an is in I for some n, then a is in I. The factor ring of a radical ideal is a semiprime ring for general rings, and is a reduced ring for commutative rings.
• Primary ideal: An ideal I is called a primary ideal if for all a and b in R, if ab is in I, then at least one of a and bn is in I for some natural number n. Every prime ideal is primary, but not conversely. A semiprime primary ideal is prime.
• Principal ideal: An ideal generated by one element.
• Finitely generated ideal: This type of ideal is finitely generated as a module.
• Primitive ideal: A left primitive ideal is the annihilator of a simple left module.
• Irreducible ideal: An ideal is said to be irreducible if it cannot be written as an intersection of ideals which properly contain it.
• Comaximal ideals: Two ideals ${\displaystyle {\mathfrak {i}},{\mathfrak {j}}}$ are said to be comaximal if ${\displaystyle x+y=1}$ for some ${\displaystyle x\in {\mathfrak {i}}}$ and ${\displaystyle y\in {\mathfrak {j}}}$.
• Regular ideal: This term has multiple uses. See the article for a list.
• Nil ideal: An ideal is a nil ideal if each of its elements is nilpotent.
• Nilpotent ideal: Some power of it is zero.
• Parameter ideal: an ideal generated by a system of parameters.
Two other important terms using "ideal" are not always ideals of their ring. See their respective articles for details:
• Fractional ideal: This is usually defined when R is a commutative domain with quotient field K. Despite their names, fractional ideals are R submodules of K with a special property. If the fractional ideal is contained entirely in R, then it is truly an ideal of R.
• Invertible ideal: Usually an invertible ideal A is defined as a fractional ideal for which there is another fractional ideal B such that AB = BA = R. Some authors may also apply "invertible ideal" to ordinary ring ideals A and B with AB = BA = R in rings other than domains.
## Ideal operations
The sum and product of ideals are defined as follows. For ${\displaystyle {\mathfrak {a}}}$ and ${\displaystyle {\mathfrak {b}}}$, left (resp. right) ideals of a ring R, their sum is
${\displaystyle {\mathfrak {a}}+{\mathfrak {b}}:=\{a+b\mid a\in {\mathfrak {a}}{\mbox{ and }}b\in {\mathfrak {b}}\}}$,
which is a left (resp. right) ideal, and, if ${\displaystyle {\mathfrak {a}},{\mathfrak {b}}}$ are two-sided,
${\displaystyle {\mathfrak {a}}{\mathfrak {b}}:=\{a_{1}b_{1}+\dots +a_{n}b_{n}\mid a_{i}\in {\mathfrak {a}}{\mbox{ and }}b_{i}\in {\mathfrak {b}},i=1,2,\dots ,n;{\mbox{ for }}n=1,2,\dots \},}$
i.e. the product is the ideal generated by all products of the form ab with a in ${\displaystyle {\mathfrak {a}}}$ and b in ${\displaystyle {\mathfrak {b}}}$.
Note ${\displaystyle {\mathfrak {a}}+{\mathfrak {b}}}$ is the smallest left (resp. right) ideal containing both ${\displaystyle {\mathfrak {a}}}$ and ${\displaystyle {\mathfrak {b}}}$ (or the union ${\displaystyle {\mathfrak {a}}\cup {\mathfrak {b}}}$), while the product ${\displaystyle {\mathfrak {a}}{\mathfrak {b}}}$ is contained in the intersection of ${\displaystyle {\mathfrak {a}}}$ and ${\displaystyle {\mathfrak {b}}}$.
The distributive law holds for two-sided ideals ${\displaystyle {\mathfrak {a}},{\mathfrak {b}},{\mathfrak {c}}}$,
• ${\displaystyle {\mathfrak {a}}({\mathfrak {b}}+{\mathfrak {c}})={\mathfrak {a}}{\mathfrak {b}}+{\mathfrak {a}}{\mathfrak {c}}}$,
• ${\displaystyle ({\mathfrak {a}}+{\mathfrak {b}}){\mathfrak {c}}={\mathfrak {a}}{\mathfrak {c}}+{\mathfrak {b}}{\mathfrak {c}}}$.
If a product is replaced by an intersection, a partial distributive law holds:
${\displaystyle {\mathfrak {a}}\cap ({\mathfrak {b}}+{\mathfrak {c}})\supset {\mathfrak {a}}\cap {\mathfrak {b}}+{\mathfrak {a}}\cap {\mathfrak {c}}}$
where the equality holds if ${\displaystyle {\mathfrak {a}}}$ contains ${\displaystyle {\mathfrak {b}}}$ or ${\displaystyle {\mathfrak {c}}}$.
Remark: The sum and the intersection of ideals is again an ideal; with these two operations as join and meet, the set of all ideals of a given ring forms a complete modular lattice. The lattice is not, in general, a distributive lattice. The three operations of intersection, sum (or join), and product make the set of ideals of a commutative ring into a quantale.
If ${\displaystyle {\mathfrak {a}},{\mathfrak {b}}}$ are ideals of a commutative ring R, then ${\displaystyle {\mathfrak {a}}\cap {\mathfrak {b}}={\mathfrak {a}}{\mathfrak {b}}}$ in the following two cases (at least)
• ${\displaystyle {\mathfrak {a}}+{\mathfrak {b}}=(1)}$
• ${\displaystyle {\mathfrak {a}}}$ is generated by elements that form a regular sequence modulo ${\displaystyle {\mathfrak {b}}}$.
(More generally, the difference between a product and an intersection of ideals is measured by the Tor functor: ${\displaystyle \operatorname {Tor} _{1}^{R}(R/{\mathfrak {a}},R/{\mathfrak {b}})=({\mathfrak {a}}\cap {\mathfrak {b}})/{\mathfrak {a}}{\mathfrak {b}}.}$[6])
An integral domain is called a Dedekind domain if for each pair of ideals ${\displaystyle {\mathfrak {a}}\subset {\mathfrak {b}}}$, there is an ideal ${\displaystyle {\mathfrak {c}}}$ such that ${\displaystyle {\mathfrak {\mathfrak {a}}}={\mathfrak {b}}{\mathfrak {c}}}$.[7] It can then be shown that every nonzero ideal of a Dedekind domain can be uniquely written as a product of maximal ideals, a generalization of the fundamental theorem of arithmetic.
## Examples of ideal operations
In ${\displaystyle \mathbb {Z} }$ we have
${\displaystyle (n)\cap (m)=\operatorname {lcm} (n,m)\mathbb {Z} }$
since ${\displaystyle (n)\cap (m)}$ is the set of integers which are divisible by both ${\displaystyle n}$ and ${\displaystyle m}$.
Let ${\displaystyle R=\mathbb {C} [x,y,z,w]}$ and let ${\displaystyle {\mathfrak {a}}=(z,w),{\mathfrak {b}}=(x+z,y+w),{\mathfrak {c}}=(x+z,w)}$. Then,
• ${\displaystyle {\mathfrak {a}}+{\mathfrak {b}}=(z,w,x+z,y+w)=(x,y,z,w)}$ and ${\displaystyle {\mathfrak {a}}+{\mathfrak {c}}=(z,w,x+z)}$
• ${\displaystyle {\mathfrak {a}}{\mathfrak {b}}=(z(x+z),z(y+w),w(x+z),w(y+w))=(z^{2}+xz,zy+wz,wx+wz,wy+w^{2})}$
• ${\displaystyle {\mathfrak {a}}{\mathfrak {c}}=(xz+z^{2},zw,xw+zw,w^{2})}$
• ${\displaystyle {\mathfrak {a}}\cap {\mathfrak {b}}={\mathfrak {a}}{\mathfrak {b}}}$ while ${\displaystyle {\mathfrak {a}}\cap {\mathfrak {c}}=(w,xz+z^{2})\neq {\mathfrak {a}}{\mathfrak {c}}}$
In the first computation, we see the general pattern for taking the sum of two finitely generated ideals, it is the ideal generated by the union of their generators. In the last three we observe that products and intersections agree whenever the two ideals intersect in the zero ideal. These computations can be checked using Macaulay2.[8][9][10]
Ideals appear naturally in the study of modules, especially in the form of a radical.
For simplicity, we work with commutative rings but, with some changes, the results are also true for non-commutative rings.
Let R be a commutative ring. By definition, a primitive ideal of R is the annihilator of a (nonzero) simple R-module. The Jacobson radical ${\displaystyle J=\operatorname {Jac} (R)}$ of R is the intersection of all primitive ideals. Equivalently,
${\displaystyle J=\bigcap _{{\mathfrak {m}}{\text{ maximal ideals}}}{\mathfrak {m}}.}$
Indeed, if ${\displaystyle M}$ is a simple module and x is a nonzero element in M, then ${\displaystyle Rx=M}$ and ${\displaystyle R/\operatorname {Ann} (M)=R/\operatorname {Ann} (x)\simeq M}$, meaning ${\displaystyle \operatorname {Ann} (M)}$ is a maximal ideal. Conversely, if ${\displaystyle {\mathfrak {m}}}$ is a maximal ideal, then ${\displaystyle {\mathfrak {m}}}$ is the annihilator of the simple R-module ${\displaystyle R/{\mathfrak {m}}}$. There is also another characterization (the proof is not hard):
${\displaystyle J=\{x\in R\mid 1-yx\,{\text{ is a unit element for every }}y\in R\}.}$
For a not-necessarily-commutative ring, it is a general fact that ${\displaystyle 1-yx}$ is a unit element if and only if ${\displaystyle 1-xy}$ is (see the link) and so this last characterization shows that the radical can be defined both in terms of left and right primitive ideals.
The following simple but important fact (Nakayama's lemma) is built-in to the definition of a Jacobson radical: if M is a module such that ${\displaystyle JM=M}$, then M does not admit a maximal submodule, since if there is a maximal submodule ${\displaystyle L\subsetneq M}$, ${\displaystyle J\cdot (M/L)=0}$ and so ${\displaystyle M=JM\subset L\subsetneq M}$, a contradiction. Since a nonzero finitely generated module admits a maximal submodule, in particular, one has:
If ${\displaystyle JM=M}$ and M is finitely generated, then ${\displaystyle M=0.}$
A maximal ideal is a prime ideal and so one has
${\displaystyle \operatorname {nil} (R)=\bigcap _{{\mathfrak {p}}{\text{ prime ideals }}}{\mathfrak {p}}\subset \operatorname {Jac} (R)}$
where the intersection on the left is called the nilradical of R. As it turns out, ${\displaystyle \operatorname {nil} (R)}$ is also the set of nilpotent elements of R.
If R is an Artinian ring, then ${\displaystyle \operatorname {Jac} (R)}$ is nilpotent and ${\displaystyle \operatorname {nil} (R)=\operatorname {Jac} (R)}$. (Proof: first note the DCC implies ${\displaystyle J^{n}=J^{n+1}}$ for some n. If (DCC) ${\displaystyle {\mathfrak {a}}\supsetneq \operatorname {Ann} (J^{n})}$ is an ideal properly minimal over the latter, then ${\displaystyle J\cdot ({\mathfrak {a}}/\operatorname {Ann} (J^{n}))=0}$. That is, ${\displaystyle J^{n}{\mathfrak {a}}=J^{n+1}{\mathfrak {a}}=0}$, a contradiction.)
## Extension and contraction of an ideal
Let A and B be two commutative rings, and let f : AB be a ring homomorphism. If ${\displaystyle {\mathfrak {a}}}$ is an ideal in A, then ${\displaystyle f({\mathfrak {a}})}$ need not be an ideal in B (e.g. take f to be the inclusion of the ring of integers Z into the field of rationals Q). The extension ${\displaystyle {\mathfrak {a}}^{e}}$ of ${\displaystyle {\mathfrak {a}}}$ in B is defined to be the ideal in B generated by ${\displaystyle f({\mathfrak {a}})}$. Explicitly,
${\displaystyle {\mathfrak {a}}^{e}={\Big \{}\sum y_{i}f(x_{i}):x_{i}\in {\mathfrak {a}},y_{i}\in B{\Big \}}}$
If ${\displaystyle {\mathfrak {b}}}$ is an ideal of B, then ${\displaystyle f^{-1}({\mathfrak {b}})}$ is always an ideal of A, called the contraction ${\displaystyle {\mathfrak {b}}^{c}}$ of ${\displaystyle {\mathfrak {b}}}$ to A.
Assuming f : AB is a ring homomorphism, ${\displaystyle {\mathfrak {a}}}$ is an ideal in A, ${\displaystyle {\mathfrak {b}}}$ is an ideal in B, then:
• ${\displaystyle {\mathfrak {b}}}$ is prime in B ${\displaystyle \Rightarrow }$ ${\displaystyle {\mathfrak {b}}^{c}}$ is prime in A.
• ${\displaystyle {\mathfrak {a}}^{ec}\supseteq {\mathfrak {a}}}$
• ${\displaystyle {\mathfrak {b}}^{ce}\subseteq {\mathfrak {b}}}$
It is false, in general, that ${\displaystyle {\mathfrak {a}}}$ being prime (or maximal) in A implies that ${\displaystyle {\mathfrak {a}}^{e}}$ is prime (or maximal) in B. Many classic examples of this stem from algebraic number theory. For example, embedding ${\displaystyle \mathbb {Z} \to \mathbb {Z} \left\lbrack i\right\rbrack }$. In ${\displaystyle B=\mathbb {Z} \left\lbrack i\right\rbrack }$, the element 2 factors as ${\displaystyle 2=(1+i)(1-i)}$ where (one can show) neither of ${\displaystyle 1+i,1-i}$ are units in B. So ${\displaystyle (2)^{e}}$ is not prime in B (and therefore not maximal, as well). Indeed, ${\displaystyle (1\pm i)^{2}=\pm 2i}$ shows that ${\displaystyle (1+i)=((1-i)-(1-i)^{2})}$, ${\displaystyle (1-i)=((1+i)-(1+i)^{2})}$, and therefore ${\displaystyle (2)^{e}=(1+i)^{2}}$.
On the other hand, if f is surjective and ${\displaystyle {\mathfrak {a}}\supseteq \ker f}$ then:
• ${\displaystyle {\mathfrak {a}}^{ec}={\mathfrak {a}}}$ and ${\displaystyle {\mathfrak {b}}^{ce}={\mathfrak {b}}}$.
• ${\displaystyle {\mathfrak {a}}}$ is a prime ideal in A ${\displaystyle \Leftrightarrow }$ ${\displaystyle {\mathfrak {a}}^{e}}$ is a prime ideal in B.
• ${\displaystyle {\mathfrak {a}}}$ is a maximal ideal in A ${\displaystyle \Leftrightarrow }$ ${\displaystyle {\mathfrak {a}}^{e}}$ is a maximal ideal in B.
Remark: Let K be a field extension of L, and let B and A be the rings of integers of K and L, respectively. Then B is an integral extension of A, and we let f be the inclusion map from A to B. The behaviour of a prime ideal ${\displaystyle {\mathfrak {a}}={\mathfrak {p}}}$ of A under extension is one of the central problems of algebraic number theory.
The following is sometimes useful:[11] a prime ideal ${\displaystyle {\mathfrak {p}}}$ is a contraction of a prime ideal if and only if ${\displaystyle {\mathfrak {p}}={\mathfrak {p}}^{ec}}$. (Proof: Assuming the latter, note ${\displaystyle {\mathfrak {p}}^{e}B_{\mathfrak {p}}=B_{\mathfrak {p}}\Rightarrow {\mathfrak {p}}^{e}}$ intersects ${\displaystyle A-{\mathfrak {p}}}$, a contradiction. Now, the prime ideals of ${\displaystyle B_{\mathfrak {p}}}$ correspond to those in B that are disjoint from ${\displaystyle A-{\mathfrak {p}}}$. Hence, there is a prime ideal ${\displaystyle {\mathfrak {q}}}$ of B, disjoint from ${\displaystyle A-{\mathfrak {p}}}$, such that ${\displaystyle {\mathfrak {q}}B_{\mathfrak {p}}}$ is a maximal ideal containing ${\displaystyle {\mathfrak {p}}^{e}B_{\mathfrak {p}}}$. One then checks that ${\displaystyle {\mathfrak {q}}}$ lies over ${\displaystyle {\mathfrak {p}}}$. The converse is obvious.)
## Generalizations
Ideals can be generalized to any monoid object ${\displaystyle (R,\otimes )}$, where ${\displaystyle R}$ is the object where the monoid structure has been forgotten. A left ideal of ${\displaystyle R}$ is a subobject ${\displaystyle I}$ that "absorbs multiplication from the left by elements of ${\displaystyle R}$"; that is, ${\displaystyle I}$ is a left ideal if it satisfies the following two conditions:
1. ${\displaystyle I}$ is a subobject of ${\displaystyle R}$
2. For every ${\displaystyle r\in (R,\otimes )}$ and every ${\displaystyle x\in (I,\otimes )}$, the product ${\displaystyle r\otimes x}$ is in ${\displaystyle (I,\otimes )}$.
A right ideal is defined with the condition "${\displaystyle r\otimes x\in (I,\otimes )}$" replaced by "'${\displaystyle x\otimes r\in (I,\otimes )}$". A two-sided ideal is a left ideal that is also a right ideal, and is sometimes simply called an ideal. When ${\displaystyle R}$ is a commutative monoid object respectively, the definitions of left, right, and two-sided ideal coincide, and the term ideal is used alone.
An ideal can also be thought of as a specific type of R-module. If we consider ${\displaystyle R}$ as a left ${\displaystyle R}$-module (by left multiplication), then a left ideal ${\displaystyle I}$ is really just a left sub-module of ${\displaystyle R}$. In other words, ${\displaystyle I}$ is a left (right) ideal of ${\displaystyle R}$ if and only if it is a left (right) ${\displaystyle R}$-module which is a subset of ${\displaystyle R}$. ${\displaystyle I}$ is a two-sided ideal if it is a sub-${\displaystyle R}$-bimodule of ${\displaystyle R}$.
Example: If we let ${\displaystyle R=\mathbb {Z} }$, an ideal of ${\displaystyle \mathbb {Z} }$ is an abelian group which is a subset of ${\displaystyle \mathbb {Z} }$, i.e. ${\displaystyle m\mathbb {Z} }$ for some ${\displaystyle m\in \mathbb {Z} }$. So these give all the ideals of ${\displaystyle \mathbb {Z} }$.
## Notes
1. ^ Some authors call the zero and unit ideals of a ring R the trivial ideals of R.
2. ^ If R does not have a unit, then the internal descriptions above must be modified slightly. In addition to the finite sums of products of things in X with things in R, we must allow the addition of n-fold sums of the form x + x + ... + x, and n-fold sums of the form (−x) + (−x) + ... + (−x) for every x in X and every n in the natural numbers. When R has a unit, this extra requirement becomes superfluous.
## References
1. ^ a b John Stillwell (2010). Mathematics and its history. p. 439.
2. ^ Harold M. Edwards (1977). Fermat's last theorem. A genetic introduction to algebraic number theory. p. 76.
3. ^ Everest G., Ward T. (2005). An introduction to number theory. p. 83.
4. ^ Lang 2005, Section III.2
5. ^ Because simple commutative rings are fields. See Lam (2001). A First Course in Noncommutative Rings. p. 39.
6. ^ Eisenbud, Exercise A 3.17
7. ^ Milnor, page 9.
8. ^ "ideals". www.math.uiuc.edu. Archived from the original on 2017-01-16. Retrieved 2017-01-14.
9. ^ "sums, products, and powers of ideals". www.math.uiuc.edu. Archived from the original on 2017-01-16. Retrieved 2017-01-14.
10. ^ "intersection of ideals". www.math.uiuc.edu. Archived from the original on 2017-01-16. Retrieved 2017-01-14.
11. ^ Atiyah–MacDonald, Proposition 3.16.
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation. | 2022-08-11 09:08:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 254, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9433214068412781, "perplexity": 2018.0233334382376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00148.warc.gz"} |
https://www.gamedev.net/forums/topic/174977-text-output-code/ | • ### Announcements
#### Archived
This topic is now archived and is closed to further replies.
# Text output code
## Recommended Posts
Irrelevant 175
void text(char* pstring,float x,float y,DWORD c)
{
if(pstring&&pdev)
{
pdev->SetTexture(0,pfontmap);/*
vertex* pverts = 0;
pvbuf->Lock(0,0,(void**)&pverts,0);
for(int x = 0; x < 2112; x++)
pverts[x].c = c;
pvbuf->Unlock();*/
int col = 0;
int row = 0;
int len = strlen(pstring);
int idx = 0;
int max = 0;
for(int i = 0; i < len; i++)
{
if(pstring[i] > 31 && pstring [i] < 127)
{
idx = (pstring[i] - 32) * 4;
D3DXMatrixTranslation(&viewmat,x+(float)col*5,y-(float)row*5,0.0f);
pdev->SetTransform(D3DTS_VIEW,&viewmat);
col++;
pdev->DrawPrimitive(D3DPT_TRIANGLESTRIP,idx,2);
if(col > max)
max = col;
}
else
if(pstring[i] == ''\t'')
col += 4;
else
if(pstring[i] == ''\n'')
{
col = 0;
row++;
}
}
}
else
{
report("Lost the device. Fix this already.");
func = 0;
}
}
// this is the code i use to create the vertex buffer if anyone is wondering
pdev->CreateVertexBuffer(sizeof(vertex)*2112,D3DUSAGE_DYNAMIC|D3DUSAGE_WRITEONLY,D3DFVF_XYZ|D3DFVF_DIFFUSE|D3DFVF_TEX1,D3DPOOL_DEFAULT,&pvbuf,0)
That code works fun until the locking/unlocking gets uncommented. Why would that mess things up? There are 2112 verts in the buffer, so no overflow or anything in the loop. Can you call Lock on a vb once you''re in BeginScene? this function is called after BeginScene, and when I initialize the verts, that''s before BeginScene so that''s the only thing I can think of.
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Irrelevant 175
And in case it''s not evident from the code, I''m just trying to change the color.
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HaywireGuy 208
Usually if you want text to be drawn on a texture, and then map
it to the polygon, you don''t make each of the characters a
polygon by itself. Try drawin'' text onto your own bitmap (or
whatever) usin'' Win32 DrawText() function, and then lock the
texture and copy that bitmap onto it.
Note that however if you''re usin'' DrawText() to draw on a 32-
bit bitmap, the function is gonna reset the alpha to zero for
those areas your text is coverin'' (this will effectively make
the text area totally transparent).
Hope that helps.
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Hawkeye3 122
In my text drawing function I use:
pdev->SetRenderState(D3DRS_TEXTUREFACTOR,color)
right before drawing the text primitives
and before rendering any text I set the states:
pdev->SetTextureStageState( 0, D3DTSS_COLOROP, D3DTOP_MODULATE );
pdev->SetTextureStageState( 0, D3DTSS_COLORARG1, D3DTA_TEXTURE );
pdev->SetTextureStageState( 0, D3DTSS_COLORARG2, D3DTA_TFACTOR );
Oh, and to speed things up some, you should look at putting all of your characters into one vertex buffer and rendering them in one call. (You seemed to of had the right idea in the commented out code, you just didn''t perform it correctly). This will require that you create an algorithm to create the vertex positions at runtime, but it shouldn''t be too hard. Also, you can lock / unlock buffers while drawing, just don''t do it too much, and make sure they are write only. | 2017-08-20 19:29:28 | {"extraction_info": {"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, "math_score": 0.27431803941726685, "perplexity": 9211.415870472198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106984.52/warc/CC-MAIN-20170820185216-20170820205216-00285.warc.gz"} |
https://direct.mit.edu/view-large/3926283 | In this section, we analyze the impact of the encoder/decoder architecture on the generation quality of considered models. The generation quality experiment of section 5 is repeated on the fMNIST and MNIST data set, where the architecture and hyperparameters are adapted from Dupont (2018). From Table 5 and Figure 9, we see that the overall FID scores and generation quality have improved; however, the relative scores among the models did not change significantly.
Table 5:
FID Scores Computed on Randomly Generated 8000 Images When Trained with Architecture and Hyperparameters.
St-RKMVAE$β$-VAEFactorVAEInfoGAN
MNIST 24.63 (0.22) 36.11 (1.01) 42.81 (2.01) 35.48 (0.07) 45.74 (2.93)
fMNIST 61.44 (1.02) 73.47 (0.73) 75.21 (1.11) 69.73 (1.54) 84.11 (2.58)
St-RKMVAE$β$-VAEFactorVAEInfoGAN
MNIST 24.63 (0.22) 36.11 (1.01) 42.81 (2.01) 35.48 (0.07) 45.74 (2.93)
fMNIST 61.44 (1.02) 73.47 (0.73) 75.21 (1.11) 69.73 (1.54) 84.11 (2.58)
Notes: Lower is better with standard deviations. Adapted from Dupont (2018).
Table 6:
Computing the Diagonalization Scores (see Figure 3).
ModelsdSprites3DShapes3D cars
St-RKM-sl ($σ=10-3$, $U★$0.17 (0.05) 0.23 (0.03) 0.21 (0.04)
St-RKM ($σ=10-3$, $U★$0.26 (0.05) 0.30 (0.10) 0.31 (0.09)
St-RKM ($σ=10-3$, random $U$0.61 (0.02) 0.72 (0.01) 0.69 (0.03)
ModelsdSprites3DShapes3D cars
St-RKM-sl ($σ=10-3$, $U★$0.17 (0.05) 0.23 (0.03) 0.21 (0.04)
St-RKM ($σ=10-3$, $U★$0.26 (0.05) 0.30 (0.10) 0.31 (0.09)
St-RKM ($σ=10-3$, random $U$0.61 (0.02) 0.72 (0.01) 0.69 (0.03)
Notes: Denote $M=1|C|∑i∈CU★⊤∇ψ(yi)∇ψ(yi)⊤U★,withyi=PUϕθ(xi)$ (cf. equation 3.6). Then we compute the score as $M-diag(M)F/MF$, where $diag:Rm×m↦Rm×m$ sets the off-diagonal elements of matrix to zero. The scores are computed for each model over 10 random seeds and show the mean (standard deviation). Lower scores indicate better diagonalization.
Figure 8:
Samples of randomly generated batch of images used to compute FID scores and SWD scores (see Figure 4).
Figure 8:
Samples of randomly generated batch of images used to compute FID scores and SWD scores (see Figure 4).
Close modal
Figure 9:
Samples of randomly generated images used to compute the FID scores. See Table 5.
Figure 9:
Samples of randomly generated images used to compute the FID scores. See Table 5.
Close modal
Figure 10:
(a) Loss evolution ($log$ plot) during the training of equation A.2 over 1000 epochs with $ɛ=10-5$ once with Cayley ADAM optimizer (green curve) and then without (blue curve). (b) Traversals along the principal components when the model was trained with a fixed $U$, that is, with the objective given by equation A.2 and $ɛ=10-5$. There is no clear isolation of a feature along any of the principal components, indicating further that optimizing over $U$ is key to better disentanglement.
Figure 10:
(a) Loss evolution ($log$ plot) during the training of equation A.2 over 1000 epochs with $ɛ=10-5$ once with Cayley ADAM optimizer (green curve) and then without (blue curve). (b) Traversals along the principal components when the model was trained with a fixed $U$, that is, with the objective given by equation A.2 and $ɛ=10-5$. There is no clear isolation of a feature along any of the principal components, indicating further that optimizing over $U$ is key to better disentanglement.
Close modal
Close Modal | 2023-02-08 21:12:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8761239647865295, "perplexity": 2255.3070243337033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00779.warc.gz"} |
https://newproxylists.com/tag/making/ | ## pr.probability – Why does the three points follow by making the two assumptions about the conditioned intensity function?
The intensity function is defined as
$$lambda^*(t)=frac{f(t|H_{t_n})}{1-F(t|H_{t_n})}$$
where $$f$$ is the density function and $$F$$ is the distribution function, and $$H_{t_n}$$ is the history of all the previous points of $$t$$ up to $$t_n$$.
Moreover, there is proven that $$F(t|H_{t_n})$$ is also given as:
$$F(t|H_{t_n})=1-e^{-int_t^{t_n}lambda^*(s)ds}$$.
An assumption is then made, saying that
1. $$lambda^*(t)$$ is non-negative and is integrable on every interval after $$t_n$$.
2. $$int_t^{t_n}lambda^*(s)ds to 1$$ for $$t to infty$$.
It is then said that hence the three points follows:
1. $$0 leq F(t|H_{t_n}) leq 1$$
2. $$F(t|H_{t_n})$$ is a non-decreasing function of $$t$$.
3. $$F(t|H_{t_n}) to 1$$ for $$t to infty$$
Can someone explain to me how these three points follow given the two assumptions above? Thank you.
## making a server management system / control panel
I’m making an open source server management system. I’m starting to make progress on it, so I’d like to share and see if anyone is intereste… | Read the rest of https://www.webhostingtalk.com/showthread.php?t=1835158&goto=newpost
## How can a GM subtly guide characters into making campaign-specific character choices?
Note: the example I’m giving is D&D focussed, but the question is system-agnostic.
We’re in the early days of a new D&D group. If it keeps running, I’m planning to transition the players into Storm King’s Thunder once they reach 5th level. Storm King’s Thunder is an adventure that features a lot of giants.
I allowed the party to defer some character creation choices so we could get playing faster, and one of them has an unused language slot. The party also has a Ranger and the revised version of the class can, at higher levels, choose an epic foe against whom they get combat bonuses.
It would be very sensible if that unused language could turn out to be giantish and that the ranger could take giants as their epic foe. It would also be nice if I didn’t have to lay out to the players that there are giants in store later – they’re all new, and they won’t guess.
Are there any good story-driven ways that a GM can help "guide" players toward making character creation choices that fit the planned campaign?
## algorithm – Making code faster. BITWISE OR operator
Problem:
Given an array A of N integers, you will be asked Q queries. Each query consists of two integers L and R. For each query, find whether bitwise OR of all the array elements between indices L and R is even or odd.
My approach:
Accept the array of N integers. For any query (Qi) scan the array of integers between the given limit. If an odd number is found (x%2==1) then raise a flag and terminate scanning. If flag is found raised tell that the result is odd, else say that it’s even.
On thinking further, I find myself at the dead end. I can’t optimize the code anymore. Only thing I can think of is instead of doing mod 2 I will check the last digit of each number and see if it is one of (0,2,4,6,8). Upon trying that, the time limit still expired (in context to competitive programming (1); Note: the competition has ended a day ago and results declared). Anyways, my question is to find a better method if it exists or optimize the code below.
I assume that the time complexity is O(nQ) where n is the number of elements in the given range.
Assume array is 1-indexed.
Input
First line consists of two space-separated integers N and Q.
Next line consists of N space-separated array elements.
Next Q lines consist of two space-sepaprated integers L and R in each line.
``````#include<stdio.h>
#pragma GCC optimize("O2")
int main()
{
long int c,d,j,N,Q;
int fagvar=0;
scanf("%ld %ldn",&N,&Q);
int a(N);
int i=0;
while(N--)
{
scanf("%ld",&a(i));
i++;
}
while(Q--)
{
scanf("%ld %ld",&c,&d);
for(j=c-1;j<d;j++)
{
if (a(j)%2==1)
{
fagvar=1;
break;
}
}
if (fagvar==1)
{
printf("%dn",0);
fagvar=0;
}
else
{
printf("%dn",1);
}
}
return 0;
}
``````
## modding – Making mods for Game Pass/Microsoft Store Starbound
I want to make mods for the Game Pass/Microsoft Store version of Starbound. I know it is possible to install mods for this version, because I have seen people talk about it (but not write it), and I have seen the mod gear in the bottom right.
But, I have tried everything I could find, and cannot for the life of me find a way to actually get access to these folders. But this probably isn’t the place to go into that. Instead, I would like to know how to unpack the files to write a mod. When I try to use the “unpack_assets.exe” it just says I don’t have permission.
## 9.0 pie – How to build without make clean after making changes in dts with AOSP?
I made few changes in dts and when i gave `make -j8` from my `aosp-root-directory`, I don’t see the changes taking place after i boot up my board. It just quickly builds in a minute or so.
Giving `make clean ` takes about 4 hours. Do i have to give `make clean` everytime i do any changes in my build/dts or can we somehow just clean kernel and build it without cleaning the whole `OUT` directory?
## Simple CPA Siphon RELOADED_Money making Machine | NewProxyLists
Simple CPA Siphon RELOADED_Money making Machine
Hidden Content:
You must reply before you can see the hidden data contained here.
Code:
`http://www.simplecpasiphon.com/`
VirusScan Here:
Code:
`Big File Not Able for Virus Scan`
## macbook pro – How to change USB scheme when making bootable Catalina installer
Background: Upgraded to Big Sur on 2020 MacPro (not the one with M1 chip), hate it, trying to revert to Catalina using a USB bootable installer (followed instructions given at the Apple website How to create a bootable installer for macOS)
When I try to install it using the external drive, I get this screen.
So it seems that my USB has the wrong scheme (should be GUID). I go to Disk Utility and sure enough, it seems like it’s Master Boot Record as shown below.
As per the suggested screen, I try to Erase it, but whenever I try to do that, I get the error shown below.
I’m unsure about how to resolve this and I can’t seem to find someone who experienced the same error online. The USB is a brand-new 32GB SanDisk
## design – How to avoid making User a god object?
Consider typical gym trainings tracker app.
`User` has account related attributes:
``````User {
id
email
fname, lname
isBlocked
}
``````
However, the requirements are that an application’s user manages his trainings, trainings history, achievements, profile, etc. All of those entities should be somehow linked with user account.
How do I link it with an account? What is the common way to do it and its pros/cons?
I can imagine two scenarios:
Possibility 1: Making `User` a large ‘god’ object:
``````User {
id
email
fname, lname
isBlocked
trainings # one to many
training_history # one to one
achievements # one to many
/** possibly many more relations */
}
``````
Possiblity 2: Link `User` with `UserProfile`, and then `UserProfile` holds all the relations.
``````User {
id
email
fname, lname
isBlocked
user_profile # one to one
}
UserProfile {
user_id # one to one
trainings # one to many
training_history # one to one
achievements # one to many
/** possibly many more relations */
}
``````
Is the second option really better than the first one? Can I do better? | 2021-01-20 13:53:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4060554802417755, "perplexity": 1802.5665785840654}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00618.warc.gz"} |