text:
# Write a polyglot that prints the language's name

Your task is to write a program that executes in as many languages as possible, with as few characters as you can. To avoid trivial solutions, the program must print the name of the language it was run in.

## Scoring

Your program has to work in at least 5 languages. A program's score is given as:

$$\frac {\text{(number of languages)}^{1.5}} {\text{(length of program)}}$$

The highest score wins.

• Would using a solution like using print() for Perl, Python, PHP, Processing and R be considered trivial or is it acceptable?
– Rob
Sep 10, 2012 at 23:34
• A can't see a problem with it. The difficulty would be switching print's input per language. Sep 11, 2012 at 2:57
• I should have given a higher weight to the number of languages... it seems choosing to have exactly 5 is the most optimal. I wonder how many languages would have been in the winner entry if the score were n * n / length
– vsz
Sep 11, 2012 at 18:47
• I was going to suggest that too. Based on the answers provided, I don't believe I interpreted your question properly.
– Rob
Sep 11, 2012 at 23:10

For the C and C-like codes I owe a debt to @baby-rabbit. The others are inspired by the recognition that many languages have an eval statement that will grudgingly accept invalid syntax.
Outputs the language name to standard output. Sometimes generates a lot of error messages on standard error, so suppress them with 2>/dev/null as you run them.
#if 0
'PHP<?/*';eval "print\$=?'Perl':'Ruby';echo Bash";exit __END__ #endif #include <stdio.h> main(){puts( #ifdef __OBJC__ "obj-" #endif "C" #ifdef __cplusplus "++" #endif );}//*/?>'  The php solution outputs 'PHP' (including the single quotes), which may be bending the rules a little bit. Last edit: shaved 12 chars from insight that $= is false in Ruby, 60 in Perl, and print$=?... is almost surely an error in Bash. Shaved 7 more from insight the Perl/Ruby/Bash test can now go into a single eval statement. If the rules can tolerate more bending, I present this 8 language, 43 character solution (score 0.5262) print("phperluarscriptrubypythoncatebg13");  for which the output includes the name of the interpreter for php, perl, lua, rscript, ruby, python, cat, and rot13. • TI-Basic supports the last one; just add 6 characters and get print("phperluarscriptrubypythontibasicatebg13"); Dec 31, 2013 at 0:36 • @Timtech no, TI-Basic uses the Disp or Output( or text( tokens, there is no print( command. And including a string as part of an otherwise real-valued expression results in an ERR:INVALID SYNTAX, not the display of the string (not even as a return value). (Also, there is no ;; TI-Basic uses :) Feb 5, 2014 at 21:57 • @AJMansfield Too true, how could I forget :? Feb 5, 2014 at 23:17 • Can you replace echo Bash with ps -ocomm= -p$$? This will output the name of the shell for at least bash, zsh, ksh, csh and tcsh, maybe others. May 30, 2014 at 19:22 # C, C++, BF, BASH and Ruby; 280 chars Score is about 0.040 #include "stdio.h" #define s "C" #ifdef __cplusplus #define s "C++" #endif #ifndef s #"+++++++++[>++++++++++>+++++++++<<-]>>-.<++++.>-.++++++++.<----.>---.<+++++++.>---.++++++++.<<++++++++++.[-]" if [ 1 == 2 ];then puts "Ruby" exit fi echo "BASH" exit end #endif main(){puts(s);}  Note that I am using a Linux system. The code is run or compiled with the following commands (the file's name is test.c) C: gcc test.c  When run with ./a.out, output is C C++: c++ test.c  When run with ./a.out, output is C++ BASH: ./test.c  Outputs: BASH Ruby: ruby test.c  Outputs: Ruby BrainF***: Verified using the following: Outputs: brainfuck Note that if the JS debugger is used, then the first two minus signs need to be removed. They were included to offset the plus signs in the string literal "C++". This was a very fun project, I'm working on adding more languages. Just to add further clarity, here are my interpreter's/compiler's specs: • gcc version 4.6.3 • ruby 1.9.3p194 (2012-04-20 revision 35410) [x86_64-linux] • GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu) SIDE NOTE Using @baby-rabbit's trick I was able to extend my code to be executable in 7 languages (objective-C and objective-c++ being added). This is not my solution since I did copy some, but I thought I would show it off. Update 9.12 Added SmallTalk run with gnu-smalltalk! SmallTalk, C, C++, Objective-C, Objective-C++, BASH, BF, Ruby; 384 chars (Score: 0.059) #if (a) ##(true) ifTrue: ['SmallTalk' printNl] ##(ObjectMemory quit) #"+++++++++++[>++++++++++>+++++++++<<-]>>-.<++++.>-.++++++++.<----.>---.<+++++++.>---.++++++++.<<++++++++++.[-]" if [ 1 == 2 ];then puts 'Ruby' exit fi echo 'BASH' exit end =begin #endif #include "stdio.h" main(){puts( #ifdef __OBJC__ "Objective-" #endif "C" #ifdef __cplusplus "++" #endif );} #ifdef b =end #endif  In the above code you will need to rename the file to produce the langauge's name for objective-c, obj-c++, c and c++. • would it be shorter if you output brainf***? – mob Sep 11, 2012 at 20:15 • Yes it would, I believe. Sep 12, 2012 at 3:42 • You should use if [ 1 = 2 ] instead of if [ 1 == 2 ], because scripts without a hashbang are executed by sh, not bash. sh may or may not be bash, on Debian it's dash. May 1, 2014 at 15:13 ## bash, c, c++, obj-c, obj-c++; 134 chars; score=0.083 #if x echo "bash" exit #endif #include <stdio.h> int main(){puts( #ifdef __OBJC__ "obj-" #endif "c" #ifdef __cplusplus "++" #endif );}  rename file and run/compile as: • sh file.sh • cc file.c • cc file.cpp • cc file.m • cc file.mm (where cc is clang-421.10.42) • +1 since many other upvoted answers were inspired by yours. Jan 1, 2014 at 21:17 ## Lua, Ruby, VimL, Sed, Befunge (129 chars; ~0.087 points) Not sure if this one counts--the sed-part is embedded in the shebang line which is arguably a hack to get around the restriction. This also means that it should be run as an executable and not directly with sed. I was lucky that all the other languages (sans Befunge) automagically ignores the shebang line (though apparently Ruby refuses to run the file if it has a shebang line that doesn't contain the string 'ruby' in it). #!sed sanava;s/.*/sed/;q;ruby --"".to_i#>"egnufeB">:#,_@ if 0 then if IO then puts"Ruby" else print"Lua" end else echo"VimL" end  Sample usage: % lua tmp/glot.poly Lua % ruby tmp/glot.poly Ruby % ./tmp/glot.poly sed % cfunge tmp/glot.poly # Requires Befunge-98 Befunge :source tmp/glot.poly # from vim VimL # displayed as message  • Wow, a solution that does not include C – vsz Sep 11, 2012 at 20:11 ## BF, Bash, Batch, C, vi (163 characters; score ≈ .0686) (<ESC> stands for ASCII code 27.) rem (){ ((0));};true /* rem ;goto(){ rem;} rem ivi<ESC>ZZ'++++++++[>NUL ++++++++<NUL -]>NUL ++.++++.*/;main(){puts("C");}/*' goto a echo Bash exit :a echo Batch rem */  I tested this, as a batch file, with the MS-DOS 6.22 version of COMMAND.COM. By default, that interpreter mixes lines of source code with the output. To prevent that from happening, execute echo off before running the batch file. To execute the vi code, which I have only tested using Vim, use the following command: cat /dev/null > tmpfile && vi -s polyglot.sh tmpfile > /dev/null 2>&1 && cat tmpfile  • You could use touch tmpfile or : > tmpfile instead of cat /dev/null > tmpfile. Dec 30, 2013 at 22:32 # AsciiDots, Befunge-93, C, C++, ><>, Funge-98, GolfScript, lua, MinkoLang, Python 2, Python 3, Ruby, vi (0.170444 points) #/* ["39-egnufeB">:|;ooo"><>" n,n=1,1--[[],0][1]#>52*,@ """""GolfScript" puts"Ruby\n" __END__ :puts{""" print("Python %d"%(3/2*2)); """stoDiicsA"\<ESC>dggivi<ESC>ZZ*/ #include<stdio.h> main(){puts(sizeof'a'-1?"C":"C++");} //>"Minkolang"9[O]." //^}>"89-egnuF"#[v,<]"]]print"lua"--"""  <ESC> stands for the literal escape character 27 I know this is an old post, but it was interesting so I had to give it a shot :P # Every Python Release, 18.37291 points import sys print('Python '+'.'.join(map(str,sys.version_info[:-2])))  Technically valid, if you consider all the python versions to be different languages. There are currently 116 python versions, which I believe gives me a score of around 18.37291. (Also I understand if this isn't considered a valid answer, this was just for fun) • you still get my +1 just for fun. – vsz Aug 5, 2019 at 18:15 • Is printing the complete version number viable? (This golfs down 4 bytes and extends the possible Python versions.) – user85052 Sep 14, 2019 at 12:38 • Also is this okay? tio.run/##K6gsycjPM/7/… – user85052 Sep 14, 2019 at 12:44 # bash, zsh, ksh, csh, tcsh, 15 chars, score 0.745 This one's a bit of a stretch since they're all related languages, but there are differences: ps -ocomm= -p$$  • What about dash? Feb 6, 2016 at 15:37 • @Elliota List of languages that worked on tio.run: Bash, Dash, ksh, OSH, tcsh, Zsh. bosh displayed sh and yash displayed ps. – user100411 Jul 15, 2021 at 9:41 # 16 languages, 348 bytes, score: 0.183908046 #include <stdio.h> #if 0 #replace "C(gcc)" "C(ecpp)" #replace "C++(gcc)" "C++(ecpp)" #endif #ifdef __clang__ #define d 2 #elif __TINYC__ #define d 4 #else #define d 0 #endif // rk:start print: "rk" \ ps -ocomm= -p$$; exit; int x=sizeof('a')%2+d;char*arr[]={"C(gcc)","C++(gcc)","C(clang)","C++(clang)","C(tcc)"};int main(){puts(arr[x]);}  This works in C(gcc), C++(gcc), C(ecpp), C++(ecpp), C(clang), C++(clang), C(tcc), sh, dash, bash, zsh, ksh, csh, tcsh, rk, and SIL. ## Bash, C, Gawk, Perl, vi (145 characters; score ≈ .077) (<BS> stands for ASCII code 8. <ESC> stands for ASCII code 27.) #define echo//<BS><BS><BS>vi<ESC>ZZ #define BEGIN A() #define B\ { echo Bash ;exit; #define C\ } BEGIN {printf(A=="A"?"Perl":"Gawk");exit;} main(){puts("C");}  To execute the vi code, which I have only tested using Vim, use this command: cat /dev/null > tmpfile && vi -s polyglot.sh tmpfile > /dev/null 2>&1 && cat tmpfile  # Java, Lisp, Whitespace, Intercal, PHP, Befunge-98; score =.0189 This was originally an answer to this.  ;\\0"egnufeB">:#,_@SSSTTTSTTTL ;TL ;SSSSSTTSTSSSL ;TL ;SSSSSTTSTSSTL ;TL ;SSSSSTSTSTSSL ;TL ;SSSSSTTSSTSTL ;TL ;SSSSSTSTSSTTL ;TL ;SSSSSTTTSSSSL ;TL ;SSSSSTTSSSSTL ;TL ;SSSSSTTSSSTTL ;TL ;SSSSSTTSSTSTL ;SSL ;L ;L ;L ;public class H{ public static void main(String []a){System.out.println("JAVA");}} ;/* (print "LISP") ;*/ ;//<?php ; echo "PHP" ;//?> ;/* #| DO ,1 <- #8 DO ,1 SUB #1 <- #110 DO ,1 SUB #2 <- #32 DO ,1 SUB #3 <- #72 DO ,1 SUB #4 <- #136 DO ,1 SUB #5 <- #88 DO ,1 SUB #6 <- #136 PLEASE DO ,1 SUB #7 <- #64 DO ,1 SUB #8 <- #80 PLEASE READ OUT ,1 PLEASE NOTE |# ;*/ ;// PLEASE GIVE UP  • This won't work in Java; every line until public class... will cause a compiler problem. May 1, 2014 at 2:53 • Interesting... I distinctly remember testing this in a compiler and it working... or perhaps my memory fails me... May 1, 2014 at 2:55 • @Ypnypn Remember that "T" is a tab,"S" is a space, and "L" is a line feed in white-space, which is why it does not cause a compiler error. May 1, 2014 at 2:57 • Oh; that makes sense. May 1, 2014 at 2:58 ## Befunge, 05AB1E (legacy), Vyxal, Red, Python 1, Vim, Zsh - 71.5 / 114 characters ≈ 0.164 score 'a;# "egnufeB"\\,,,,,,,@"05AB1E",q', 2>2 'a;Vyxal,Q', 2>2#;echo Zsh print 'Red quit;' * 0+'Python', 1>0#<0x1B>}dHiVim  This is a cheap selection of languages, but here goes: ## CoffeeScript, JScript, Mozilla Javascript (≈ JavaScript 1.3), ECMAScript Edition 5, ECMAScript Edition 3, 223 chars, score ≈ 0.0501) a="undefined" x="ECMAScript 3" if(Array.prototype.map) x="ECMAScript 5" if(typeof CoffeeScript!=a) x="CoffeeScript" if(typeof uneval!=a) x="Mozilla JavaScript" if(typeof WScript!=a) WScript.echo("JScript") else alert(x)  • JavaScript, JavaScript, JavaScript (≈ JavaScript), JavaScript, JavaScript, 263 chars, (score ≈ 0). you can also make it shorter a="undefined";if(typeof arguments!=a)...else if(typeof WScript!=a... oh and you can also try golfing it.... Oct 21, 2013 at 12:22 • @tryingToGetProgrammingStraight Putting everything on one line like that is not valid syntax in CoffeeScript (which is a separate language from Javascript). The others are arguable, but since it wasn't specified how different the languages have to be, I think the fact that the different implementations have different specified behaviors allows them to be treated as separate. Oct 21, 2013 at 14:56 • i also cant write a multi-line comment, the only golfing you seem to have used is leaving out {/}/;... the idea is to use a to avoid reusing "undefined", you could also do b=alert and then b("this is also still just javascript"). anyway The golden rule of CoffeeScript is: "It's just JavaScript" (from coffeescript.org), you can also use some jquery to add a sixth language.... Oct 21, 2013 at 15:55 • @tryingToGetProgrammingStraight Adding jQuery would be absurd, since jQuery is a library, not a language. CoffeeScript is definitely not the same language as ECMAScript. The definition of a language is a set of finite strings over a finite alphabet. The set of strings representing valid CoffeeScript programs is not the same as the set of strings representing valid ECMAScript. Oct 22, 2013 at 21:57 • Can you do [].map instead of Array.prototype.map? Feb 6, 2016 at 21:24 ## Windows Batch, TI-Basic, Golf-Basic 84, Quomplex, and GTB 5*2/93 = 0.11 ::¤Quomplex¤:"'*[]'":~"GTB":g;1:d"GOLF-BASIC 84":g1:"TI-BASIC":Lbl 1:End echo Windows Batch  How it works Windows Batch was the easiest, because :: starts a comment. Fairly simple to implement. TI-Basic doesn't support lowercase letters or backticks, causing it to skip the statements d"GOLF-BASIC 84":g1, which Golf-Basic evaulates, Displaying the message and forwarding to Label 1, where it is promptly ended. This is similar for GTB, with its handy display character, ~. By the way, a string with no display will be put in Ans. If there are no Display commands following it, Ans will be outputted (not the case here). Quomplex was snuck in at the beginning because its complex syntax won't allow for much to be skipped. All it does is add "Quomplex" to the output, and then for the mastery of the program... The Mastery of the Programming Syntax :"'*[]'"  Pure genius. Quomplex ignores : and takes "' and '" as strings, leaving it to output the stack and perish in an infinite while loop ([]). Meanwhile, Golf-Basic and TI-Basic take the whole "'*[]'" as a string, because ' is a mathematical operator, not a string operator. • Strictly speaking, : indicates a label in a batch file, not a comment. It's true that it doesn't execute anything on its own, though. – Mark Apr 24, 2014 at 20:19 • @Mark True, but is is :: (still a label though). I'd rather use it than REM, especially since it is required in TI-Basic. Apr 24, 2014 at 22:13 • Right - I was just being pedantic. Also, +1 for using batch files in the first place. :-) – Mark Apr 25, 2014 at 0:26 ## JScript, EcmaScript Edition 3, 5, 6, 2016, Node, Mozilla JavaScript (score ≈ 0.1342) This answer is originally based off Peter Olson's answer, but minus the CoffeeScript (as whitespace-significant languages can be horrible for golfing). I also added Node, ES6 and ES2016 and golfed the code a little, almost tripling the original score. $=this,y="EcmaScript ",x=$.module?"Node":$.uneval?"Mozilla JavaScript":"".padStart?y+2016:$.Map?y+6:[].map?y+5:y+3,($.WScript||$.alert)(x)  # JScript, EcmaScript Edition 3, 5, 6, 2016, 2017, Node, Mozilla JavaScript (score = .174193548387), CoffeeScript (as whitespace-significant languages can be horrible for golfing). $=this,y="EcmaScript ",($.WScript||alert)(x=Object.values?y+2017:$.module?"Node":$.uneval?"Mozilla JavaScript":[].includes?y+2016:$.Map?y+6:[].map?y+5:y+3)


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maximum of the sum of polynomials

Hi

I have $n$ polynomials, each one is positive over a certain range and the maximum value each can attain is 1. Also each polynomial has atmost one peak(maximum). Is there some way that I can show that the sum of these polynomials also has one peak atmost. Each polynomial is something like $x^3(1-x)^2$ where $0\leq x\leq 1$.

is this information enough.

thanks

-
Do you have a particular reason for asking this? –  David Roberts Dec 18 '12 at 4:01
I added some latex (not because I think it good question, just offended my typesetting sensibilities), removed the geometric topology tag, but I'm not happy with those that remain. –  David Roberts Dec 18 '12 at 4:05
I am not sure I understand. On the interval $[0,1]$, take the polynomials $p_1(x)=x^2$ and $p_2(x)=(1-x)^2$. Each has maximum $1$, and exactly $1$ peak on this interval, but their sum has two peaks. For a more extreme example, take $p_1(x)=x$ and $p_2(x)=1-x$. Again, both have maximum $1$ and only $1$ peak on $[0,1]$, but their sum is then the constant function. –  Eric Naslund Dec 18 '12 at 4:08
One fact that might be useful is that on a closed, bounded interval on which the sum is strictly concave, it has exactly one maximum. Of course if all the functions are concave on this interval, and at least one is strictly concave, then the sum is strictly concave. –  Robert Israel Dec 18 '12 at 4:27

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QuantLib Python: why is the NPV different? NPV(0.1,[0,110]) should be 100, not 99.53

I am trying to learn QuantLib for Python. Further to my previous question on the syntax for CashFlows.npv(), now that I understand how the syntax works, I have a question on why the output differs from what I'd expect (which is why I think a separate question is justified).

In my toy example, my cashflows are zero on 15-Jan-2001 and 110 on 15-Jan-2002. If I discount them at 10%, I'd expect the pv to be 100, but I get 99.53211. Why? What am I missing? That would imply that 384 days have gone by, not 365. Those are not leap years. I have tried with act/365 and 30/360: they both give the same result, which is not 100.

import QuantLib as ql
d1 = ql.Date(15,1,2001)
ql.Settings.instance().setEvaluationDate(d1)
cfs = [ql.SimpleCashFlow(0, d1),
ql.SimpleCashFlow(110, d1 + 365)]

calc_date = d1
risk_free_rate = 0.1
curve_act_365 = ql.YieldTermStructureHandle(
ql.FlatForward(calc_date, risk_free_rate, ql.Actual365Fixed()))

pv_act_365 = ql.CashFlows.npv(cfs, curve_act_365, True)

curve_30_360 = ql.YieldTermStructureHandle(
ql.FlatForward(calc_date, risk_free_rate, ql.Thirty360()))

pv_30_360 = ql.CashFlows.npv(cfs, curve_30_360, True)

By default, QuantLib expects a continuously compounded rate in the FlatForward constructor.

So the PV you are getting is basically:

from math import exp
print(110 * exp(-0.1))

If you define your curve with an annually compounded rate like so:

curve_30_360 = ql.YieldTermStructureHandle(
ql.FlatForward(calc_date, risk_free_rate, ql.Thirty360(), ql.Compounded, ql.Annual))

You will get 100.


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# 8.4: Ways to Enhance Memory

Skills to Develop

• Recognize and apply memory-enhancing strategies
• Recognize and apply effective study techniques

Most of us suffer from memory failures of one kind or another, and most of us would like to improve our memories so that we don’t forget where we put the car keys or, more importantly, the material we need to know for an exam. In this section, we’ll look at some ways to help you remember better, and at some strategies for more effective studying.

### Memory-Enhancing Strategies

What are some everyday ways we can improve our memory, including recall? To help make sure information goes from short-term memory to long-term memory, you can use memory-enhancing strategies. One strategy is rehearsal, or the conscious repetition of information to be remembered (Craik & Watkins, 1973). Think about how you learned your multiplication tables as a child. You may recall that $$6 \times 6 = 36$$, $$6 \times 7 = 42$$, and $$6 \times 8 = 48$$. Memorizing these facts is rehearsal.

Another strategy is chunking: you organize information into manageable bits or chunks (Bodie, Powers, & Fitch-Hauser, 2006). Chunking is useful when trying to remember information like dates and phone numbers. Instead of trying to remember $$5205550467$$, you remember the number as $$520-555-0467$$. So, if you met an interesting person at a party and you wanted to remember his phone number, you would naturally chunk it, and you could repeat the number over and over, which is the rehearsal strategy.

You could also enhance memory by using elaborative rehearsal: a technique in which you think about the meaning of the new information and its relation to knowledge already stored in your memory (Tigner, 1999). For example, in this case, you could remember that $$520$$ is an area code for Arizona and the person you met is from Arizona. This would help you better remember the $$520$$ prefix. If the information is retained, it goes into long-term memory.

Mnemonic devices are memory aids that help us organize information for encoding (See figure $$\PageIndex{1}$$). They are especially useful when we want to recall larger bits of information such as steps, stages, phases, and parts of a system (Bellezza, 1981). Brian needs to learn the order of the planets in the solar system, but he’s having a hard time remembering the correct order. His friend Kelly suggests a mnemonic device that can help him remember. Kelly tells Brian to simply remember the name Mr. VEM J. SUN, and he can easily recall the correct order of the planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. You might use a mnemonic device to help you remember someone’s name, a mathematical formula, or the seven levels of Bloom’s taxonomy.

Figure $$\PageIndex{1}$$: This is a knuckle mnemonic to help you remember the number of days in each month. Months with 31 days are represented by the protruding knuckles and shorter months fall in the spots between knuckles. (credit: modification of work by Cory Zanker)

If you have ever watched the television show Modern Family, you might have seen Phil Dunphy explain how he remembers names:

The other day I met this guy named Carl. Now, I might forget that name, but he was wearing a Grateful Dead t-shirt. What’s a band like the Grateful Dead? Phish. Where do fish live? The ocean. What else lives in the ocean? Coral. Hello, Co-arl. (Wrubel & Spiller, 2010)

It seems the more vivid or unusual the mnemonic, the easier it is to remember. The key to using any mnemonic successfully is to find a strategy that works for you.

Some other strategies that are used to improve memory include expressive writing and saying words aloud. Expressive writing helps boost your short-term memory, particularly if you write about a traumatic experience in your life. Masao Yogo and Shuji Fujihara (2008) had participants write for $$20$$-minute intervals several times per month. The participants were instructed to write about a traumatic experience, their best possible future selves, or a trivial topic. The researchers found that this simple writing task increased short-term memory capacity after five weeks, but only for the participants who wrote about traumatic experiences. Psychologists can’t explain why this writing task works, but it does.

What if you want to remember items you need to pick up at the store? Simply say them out loud to yourself. A series of studies (MacLeod, Gopie, Hourihan, Neary, & Ozubko, 2010) found that saying a word out loud improves your memory for the word because it increases the word’s distinctiveness. Feel silly, saying random grocery items aloud? This technique works equally well if you just mouth the words. Using these techniques increased participants’ memory for the words by more than 10%. These techniques can also be used to help you study.

### How to Study Effectively

Based on the information presented in this chapter, here are some strategies and suggestions to help you hone your study techniques (See figure $$\PageIndex{2}$$). The key with any of these strategies is to figure out what works best for you.

Figure $$\PageIndex{2}$$: Memory techniques can be useful when studying for class. (credit: Barry Pousman)

• Use elaborative rehearsal: In a famous article, Craik and Lockhart (1972) discussed their belief that information we process more deeply goes into long-term memory. Their theory is called levels of processing. If we want to remember a piece of information, we should think about it more deeply and link it to other information and memories to make it more meaningful. For example, if we are trying to remember that the hippocampus is involved with memory processing, we might envision a hippopotamus with excellent memory and then we could better remember the hippocampus.
• Apply the self-reference effect: As you go through the process of elaborative rehearsal, it would be even more beneficial to make the material you are trying to memorize personally meaningful to you. In other words, make use of the self-reference effect. Write notes in your own words. Write definitions from the text, and then rewrite them in your own words. Relate the material to something you have already learned for another class, or think how you can apply the concepts to your own life. When you do this, you are building a web of retrieval cues that will help you access the material when you want to remember it.
• Don’t forget the forgetting curve: As you know, the information you learn drops off rapidly with time. Even if you think you know the material, study it again right before test time to increase the likelihood the information will remain in your memory. Overlearning can help prevent storage decay.
• Rehearse, rehearse, rehearse: Review the material over time, in spaced and organized study sessions. Organize and study your notes, and take practice quizzes/exams. Link the new information to other information you already know well.
• Be aware of interference: To reduce the likelihood of interference, study during a quiet time without interruptions or distractions (like television or music).
• Keep moving: Of course you already know that exercise is good for your body, but did you also know it’s also good for your mind? Research suggests that regular aerobic exercise (anything that gets your heart rate elevated) is beneficial for memory (van Praag, 2008). Aerobic exercise promotes neurogenesis: the growth of new brain cells in the hippocampus, an area of the brain known to play a role in memory and learning.
• Get enough sleep: While you are sleeping, your brain is still at work. During sleep the brain organizes and consolidates information to be stored in long-term memory (Abel & Bäuml, 2013).
• Make use of mnemonic devices: As you learned earlier in this chapter, mnemonic devices often help us to remember and recall information. There are different types of mnemonic devices, such as the acronym. An acronym is a word formed by the first letter of each of the words you want to remember. For example, even if you live near one, you might have difficulty recalling the names of all five Great Lakes. What if I told you to think of the word Homes? HOMES is an acronym that represents Huron, Ontario, Michigan, Erie, and Superior: the five Great Lakes. Another type of mnemonic device is an acrostic: you make a phrase of all the first letters of the words. For example, if you are taking a math test and you are having difficulty remembering the order of operations, recalling the following sentence will help you: “Please Excuse My Dear Aunt Sally,” because the order of mathematical operations is Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. There also are jingles, which are rhyming tunes that contain key words related to the concept, such as i before e, except after c.

### Summary

There are many ways to combat the inevitable failures of our memory system. Some common strategies that can be used in everyday situations include mnemonic devices, rehearsal, self-referencing, and adequate sleep. These same strategies also can help you to study more effectively.

### Glossary

chunking
organizing information into manageable bits or chunks
elaborative rehearsal
thinking about the meaning of the new information and its relation to knowledge already stored in your memory
levels of processing
information that is thought of more deeply becomes more meaningful and thus better committed to memory
memory-enhancing strategy
technique to help make sure information goes from short-term memory to long-term memory
mnemonic device
memory aids that help organize information for encoding


text:
## anonymous 5 years ago help me please... does the series converges? if it does , try to find the sum. summation n=1 to infinity n/(2^n) how should i start solving this question ????

1. anonymous

The ratio test shows that this is convergent:$\frac{a_{n+1}}{a_{n}}=\frac{\frac{n+1}{2^n}}{\frac{n}{2^n}}=\frac{n+1}{2n}=\frac{1+\frac{1}{n}}{2}$Taking the limit as n approaches infinity shows that the ratio is 1/2, which is less than 1 (which is what is needed if the ratio test is to work).

2. anonymous

thank you


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# ATLAS Preprints

2021-05-03
15:51
 Performance of the ATLAS Level-1 topological trigger in Run 2 / ATLAS Collaboration During LHC Run 2 (2015-2018) the ATLAS Level-1 topological trigger allowed efficient data-taking by the ATLAS experiment at luminosities up to 2.1x10$^{34}$ cm$^{-2}$s$^{-1}$, which exceeds the design value by a factor of two. [...] arXiv:2105.01416 ; CERN-EP-2021-040. - 2021. - 36 p. Fulltext - Previous draft version - Fulltext

2021-04-27
11:19
 Search for dark matter in events with missing transverse momentum and a Higgs boson decaying into two photons in $pp$ collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector / ATLAS Collaboration A search for dark-matter particles in events with large missing transverse momentum and a Higgs boson candidate decaying into two photons is reported. [...] arXiv:2104.13240 ; CERN-EP-2021-012. - 2021. - 49 p. Fulltext - Previous draft version - Fulltext

2021-04-07
13:04
 A search for the decays of stopped long-lived particles at $\sqrt{s}=13$ TeV with the ATLAS detector / ATLAS Collaboration A search for long-lived particles, which have come to rest within the ATLAS detector, is presented. [...] arXiv:2104.03050 ; CERN-EP-2021-041. - 2021. - 48 p. Fulltext - Previous draft version - Fulltext

2021-04-01
04:35
 Measuring standard model parameters using top-quark cross sections in ATLAS and CMS / Defranchis, Matteo M. (CERN) /ATLAS Collaboration ; CMS Collaboration Theoretical predictions of top-quark cross sections depend on the values of the parameters of the quantum chromodynamics Lagrangian, such as the strong coupling constant and the mass of the top quark, but also on parameters of the electroweak sector of the standard model and parton distribution functions. Comparisons between state-of-the-art calculations and recent measurements at the ATLAS and CMS experiments at the CERN LHC allow for the precise determination of these parameters. [...] arXiv:2103.16384.- 2021-03-30 - 6 p. Fulltext: PDF; In : 13th International Workshop on Top-Quark Physics, Durham, United Kingdom, 14 - 18 Sep 2020

2021-03-26
04:01
 Latest Studies of the SM Higgs Boson Couplings to Fermions at ATLAS / Liu, Yanlin (Michigan U.) This proceeding presents the latest studies on the Yukawa couplings of the Standard Model Higgs boson with 139 fb$^{-1}$ data collected using the ATLAS detector at a center-of-mass energy of 13 TeV. [...] arXiv:2103.12780. - 5 p. Fulltext

2021-03-23
14:43
 Measurements of the inclusive and differential production cross sections of a top-quark-antiquark pair in association with a $Z$ boson at $\sqrt{s} = 13$ TeV with the ATLAS detector Measurements of both the inclusive and differential production cross sections of a top-quark-antiquark pair in association with a $Z$ boson ($t\bar{t}Z$) are presented. [...] CERN-EP-2021-011. - 2021. Fulltext - Previous draft version

2021-03-22
10:03
 Search for supersymmetry in events with four or more charged leptons in 139 fb$^{-1}$ of $\sqrt{s}=13$ TeV $pp$ collisions with the ATLAS detector / ATLAS Collaboration A search for supersymmetry in events with four or more charged leptons (electrons, muons and $\tau$-leptons) is presented [...] arXiv:2103.11684 ; CERN-EP-2021-021. - 2021. - 53 p. Fulltext - Previous draft version - Fulltext

2021-03-18
16:09
 Measurements of $W^+W^-+\ge 1$ jet production cross-sections in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector / ATLAS Collaboration Fiducial and differential measurements of $W^+W^-$ production in events with at least one hadronic jet are presented. [...] arXiv:2103.10319 ; CERN-EP-2021-030. - 2021. - 58 p. Fulltext - Previous draft version - Fulltext

2021-03-18
16:00
 Evidence for Higgs boson decays to a low-mass dilepton system and a photon in pp collisions at $\sqrt{s} =$ 13 TeV with the ATLAS detector / ATLAS Collaboration A search for the Higgs boson decaying into a photon and a pair of electrons or muons with an invariant mass $m_{\ell\ell} < 30$ GeV is presented. [...] arXiv:2103.10322 ; CERN-EP-2021-035. - 2021. - 39 p. Fulltext - Previous draft version - Fulltext

2021-03-16
06:31
 Search for bottom-squark pair production in $pp$ collision events at $\sqrt{s} = 13$ TeV with hadronically decaying $\tau$-leptons, $b$-jets and missing transverse momentum using the ATLAS detector / ATLAS Collaboration A search for pair production of bottom squarks in events with hadronically decaying $\tau$-leptons, $b$-tagged jets and large missing transverse momentum is presented. [...] arXiv:2103.08189 ; CERN-EP-2020-235. - 2021. - 45 p. Fulltext - Previous draft version - Fulltext


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Predictive Hacks

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### Monte Carlo Integration in Python

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# Karl Pearson’s Coefficient Method (Correlation)

## Karl Pearson’s Coefficient Method

Karl Pearson’s method of Coefficient of correlation is also known as Pearsonian coefficient or correlation or product correlation method.

Let X and Y be two random variables, then the correlation of coefficient between the variables X and Y is denoted by r(X, Y) or simply by rXY, and is defined as:${{r}_{XY}}=r\left( X,Y \right)=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}$$Where,Cov\left( X,Y \right)=E\left[ \left\{ X-E\left( X \right) \right\}\left\{ Y-E\left( Y \right) \right\} \right]$$\Rightarrow Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)={{\mu }_{11}}$${{\sigma }_{X}}^{2}=E{{\left\{ X-E\left( X \right) \right\}}^{2}}=\frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}}$${{\sigma }_{Y}}^{2}=E{{\left\{ Y-E\left( Y \right) \right\}}^{2}}=\frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}}$$\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)}{{{\left[ \left\{ \frac{1}{n}{{\sum{\left( {{x}_{i}}-\bar{x} \right)}}^{2}} \right\}\left\{ \frac{1}{n}{{\sum{\left( {{y}_{i}}-\bar{y} \right)}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$Other equivalent forms of Coefficient of correlation formulas are:$Cov\left( X,Y \right)=\frac{1}{n}\sum{\left( {{x}_{i}}-\bar{x} \right)}\left( {{y}_{i}}-\bar{y} \right)$$\Rightarrow Cov\left( X,Y \right)=\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}-\bar{y}\frac{1}{n}\sum{{{x}_{i}}}}-\bar{x}\frac{1}{n}\sum{{{y}_{i}}}+\bar{x}\bar{y}$$\therefore Cov\left( X,Y \right)=\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}}-\bar{x}.\bar{y}$${{\sigma }_{X}}^{2}=\frac{1}{n}\sum{{{x}_{i}}^{2}}-{{\bar{x}}^{2}}$${{\sigma }_{Y}}^{2}=\frac{1}{n}\sum{{{y}_{i}}^{2}}-{{\bar{y}}^{2}}$$\therefore {{r}_{XY}}=\frac{Cov\left( X,Y \right)}{{{\sigma }_{X}}{{\sigma }_{Y}}}=\frac{\frac{1}{n}\sum{{{x}_{i}}{{y}_{i}}}-\bar{x}.\bar{y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{x}_{i}}^{2}}-{{{\bar{x}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{y}_{i}}^{2}}-{{{\bar{y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$

Note:

The values of correlation coefficient cannot exceed unity numerically. It always lies in between -1 and +1. That is -1 ≤ r(X, Y) ≤ 1. If r =+1, the correlation is perfect and positive and if r = -1, correlation is perfect and negative.

1. It is not affected by change of origin or change of scale.
2. It is a relative measure. It does not have any unit attached to it.

 Example 01

Calculate the coefficient of correlation by Karl Pearson’s method based on following values

Solution:

$\bar{X}=\frac{\sum{X}}{5}=\frac{15}{5}=3$$\bar{Y}=\frac{\sum{Y}}{5}=\frac{40}{5}=8$$\frac{1}{n}\sum{{{X}^{2}}=\frac{1}{5}\times 55=11}$$\frac{1}{n}\sum{{{Y}^{2}}=\frac{1}{5}\times 330=66}$$\frac{1}{n}\sum{XY=\frac{1}{5}\times 110=22}$$\therefore {{r}_{XY}}=\frac{\frac{1}{n}\sum{XY}-\bar{X}.\bar{Y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{X}^{2}}}-{{{\bar{X}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{Y}^{2}}}-{{{\bar{Y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$$\therefore {{r}_{XY}}=\frac{22-3\times 8}{{{\left[ \left\{ 11-{{3}^{2}} \right\}\left\{ 66-{{8}^{2}} \right\} \right]}^{\frac{1}{2}}}}=\frac{-2}{2}=-1$

 Example 02

A computer while calculating correlation coefficient between two variables X and Y from 25 pairs of observation obtained the following results: n = 25, ΣX = 125, ΣX2 = 650, ΣY = 100, ΣY2 = 460, ΣXY = 508. It was later noticed that there was some mistake while copied down two pairs as

While the correct values were

Obtain the correct value of the correlation coefficient.

Solution:

Corrected ΣX = 125 – 6 – 8 + 8 + 6 = 125
Corrected ΣY = 100 – 14 – 6 + 12 + 8 = 100
Corrected ΣX2 = 650 – 62 – 82 + 82 + 62 = 650
Corrected ΣY2 = 460 – 142 – 62 + 122 + 82 = 436
Corrected ΣXY = 508 – 6(14) – 8(6) + 8(12) + 6(8) = 520$\bar{X}=\frac{\sum{X}}{25}=\frac{125}{25}=5$$\bar{Y}=\frac{\sum{Y}}{25}=\frac{100}{25}=4$$\therefore \text{Corrected}~~{{r}_{XY}}=\frac{\frac{1}{n}\sum{XY}-\bar{X}.\bar{Y}}{{{\left[ \left\{ \frac{1}{n}\sum{{{X}^{2}}}-{{{\bar{X}}}^{2}} \right\}\left\{ \frac{1}{n}\sum{{{Y}^{2}}}-{{{\bar{Y}}}^{2}} \right\} \right]}^{\frac{1}{2}}}}$$\therefore {{r}_{XY}}=\frac{\frac{1}{25}\times 520-5\times 4}{{{\left[ \left\{ \frac{1}{25}\times 650-{{5}^{2}} \right\}\left\{ \frac{1}{25}\times 436-{{4}^{2}} \right\} \right]}^{\frac{1}{2}}}}$$\therefore {{r}_{XY}}=\frac{\frac{4}{5}}{1\times \frac{6}{5}}=\frac{2}{3}=0.67$

 Example 03

Calculate the correlation coefficient for the following

Solution:

Here,
$\bar{X}=\frac{\sum{X}}{N}=\frac{580}{10}=58$$\bar{Y}=\frac{\sum{Y}}{N}=\frac{140}{10}=14$

$\therefore r(x,y)=\frac{\sum{xy}}{\sqrt{\sum{{{x}^{2}}.\sum{{{y}^{2}}}}}}=\frac{70}{\sqrt{360\times 22}}=0.787$

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# Integration by Parts Help

1. Jan 27, 2009

### fallen186

1. The problem statement, all variables and given/known data
$$\int x*(ln(x))^4dx = 4ln|x|^3-12ln|x|$$

2. Relevant equations

3. The attempt at a solution
I did chart method
u..........dv..........+/-
------------------------
x........ln|x|^4......+
1.....(4ln|x|^3)/x.. -
0.......12ln|x|........+
..........................-

2. Jan 28, 2009

### Staff: Mentor

What's the question?
What's the significance of your chart? Is u = x? And is dv = (ln|x|)^4 *dx? If so, this is not at all a useful substitution.

3. Jan 28, 2009

### Gib Z

The chart is the authors attempt at a Tabular Integration by Parts: http://en.wikipedia.org/wiki/Integration_by_parts#Tabular_integration_by_parts

Hes probably asking us just to check his work.

To fallen186 - When we do Tabular integration, the Column with our chosen "u" is the derivatives column, whilst the "dv" column are for the integrals, not for derivatives again. And as Mark44 said, its probably better if you reconsider your substitutions.

4. Jan 28, 2009

### workerant

I was taught this about picking the correct substitution:

HIGHEST priority for choice of u (i.e. make these things "u")
lnx logx arctanx and things like that LEVEL 1
x^2 (i.e. polynomials) or things like that LEVEL 2
cos x sin x LEVEL 3
e^x LEVEL 4
LOWEST priority of choice of u.

Have you done such?

Also, many textbooks include reduction formulas in the back and sometimes test just to see if you are familiar with such formulas. This may be the case here as well, although using integration by parts is not at all difficult in this problem.

5. Jan 28, 2009

### Gib Z

Yes I have, the algorithim for picking the substitution is often shortened to "ILATE" or "LIATE". I learned ILATE but it doesn't really matter unless you have some nasty product of both an Inverse Trig and a log.

Basically were saying, fallen186, you should pick your log term as your u sub.


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# Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$

Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.

I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$

So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.

The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.

-
So you did all that but you can't find $\tan\left( \dfrac{\pi}{4}\right)$? –  Git Gud Jan 1 '14 at 2:01
Draw an isoceles right-angled triangle. Two of its angles are $\pi/4$. Then $\tan(\pi/4)$ is opposite divided by adjacent. These are equal, so $\tan(\pi/4)=1$. –  André Nicolas Jan 1 '14 at 2:07
You don't have to actually compute $\sin^{-1}(\sqrt{2}/2)$ to solve this problem. Draw a right triangle with opposite side length $\sqrt{2}$ and hypotenuse length $2$. Now use the pythagorean theorem to find the length of the adjacent side; call this length $a$. Then compute $\sqrt{2}/a$. –  Alex Wertheim Jan 1 '14 at 2:12

Hint: $$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}$$
No, because he didn't also misspell $1$. –  John Jan 1 '14 at 2:14
$\hskip2in$
Using the triangle above...& the fact that $$\tan x = \frac{\text{opp}}{\text{adj}}, \space \tan \left(\frac{\pi}{4}\right)=...$$


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# laws of exponents

The so-called laws of exponents are the algebraic rules that arise naturally out of the definition of exponentiation of numbers. These rules are mostly very simple to derive, but there is great value in studying and practicing them until they are committed to memory, as this makes simplifying, combining, and expanding algebraic expressions much quicker and easier.

Recall that to exponentiate a number $$a$$ to the power $$n$$ is to multiply $$a$$ times itself $$n$$ times:

$a^n = \underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{n factors}}$

For reasons that will become clear, we define any number but zero, when raised to the power zero, to be equal to 1:

$a^0 = 1,\,\,\,a \neq 0$

We leave the expression $$0^0$$ undefined.

The complete set of rules is as listed below. Click the show-me-why buttons to show/hide the corresponding derivation and explanation.

$${\Large a^na^m=a^{n+m}}$$

Since $$a^n$$ represents $$n$$ factors of $$a$$ and $$a^m$$ represents $$m$$ factors, their product must be $$n+m$$ factors of $$a$$:

$\begin{eqnarray*}a^n a^m & = & \underbrace{\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{n factors}}\times\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{m factors}}\,\,}_{\text{n+m factors}} \\ & = & a^{n+m}\end{eqnarray*}$

$${\Large \left(a^n\right)^m=a^{nm}}$$

Since $$a^n$$ represents $$n$$ factors of $$a$$, raising this to the $$m$$th power must represent $$n\times m$$ factors of $$a$$:

$\begin{eqnarray*}\left(a^n\right)^m & = & \underbrace{\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{n factors}}\times\cdots\times\underbrace{\,a \times a \times \cdots \times a\,\,}_{\text{n factors}}\,\,\,}_{\text{m times}} \\ & = & a^{nm}\end{eqnarray*}$

$${\Large a^{-n}=\displaystyle\frac{1}{a^n}}$$

This is by definition, and is so defined because it is needed to make the other rules work properly. For instance, according to the rules already defined, $$a^na^{-n}=a^{n-n}=a^0=1$$, but also $$a^na^{-n}=\displaystyle\frac{a^n}{a^n}=1$$.

$${\Large (ab)^n=a^nb^n}$$

This is by commutivity of multiplication:

$\begin{eqnarray*}(ab)^n & = & \underbrace{\,ab \times ab \times \cdots \times ab\,\,\,}_{\text{n factors}} \\ & = & \underbrace{\,a \times a \times \cdots \times a\,\,\,}_{\text{n factors}}\,\times\underbrace{\,b \times b \times \cdots \times b\,\,}_{\text{n factors}} \\ & = & a^nb^n\end{eqnarray*}$

$${\Large \displaystyle\frac{a^n}{a^m}=a^{n-m}}$$

This follows immediately from the previous rules:

$\begin{eqnarray*}\displaystyle\frac{a^n}{a^m} & = & a^n\displaystyle\frac{1}{a^m} \\ & & \\ & = & a^na^{-m} \\ & & \\ & = & a^{n-m}\end{eqnarray*}$

$${\Large \left(\displaystyle\frac{a}{b}\right)^n=\displaystyle\frac{a^n}{b^n}}$$

By the previous rules:

$\begin{eqnarray*}\left(\displaystyle\frac{a}{b}\right)^n & = & \left(a \times b^{-1}\right)^n \\ & &\\ & = & a^n \times \left(b^{-1}\right)^n \\ & &\\ & = & a^n \times b^{-n} \\ & & \\ & = & \displaystyle\frac{a^n}{b^n}\end{eqnarray*}$

$${\Large a^{\frac{p}{q}}=\sqrt[q]{a^p}=\left(\sqrt[q]{a}\right)^p}$$

We observe that since we may write 1 as $$\frac{n}{n}$$ for any non-zero $$n$$, we have that

$a = a^1 = a^{\frac{n}{n}} = \left(a^{\frac{1}{n}}\right)^n$

and consequently $$a^{\frac{1}{n}}$$ must represent the $$n$$th root of $$a$$. Since also

$a^{\frac{p}{q}} = \left(a^{\frac{1}{q}}\right)^p = \left(a^p\right)^{\frac{1}{q}}$

we have that $$a^{\frac{p}{q}}$$ is the $$q$$th root of $$a^p$$, or equivalently the $$p$$th power of $$a^{\frac{1}{q}}$$.

Since any exponential expression can be written as an equivalent logarithmic expression, these rules have their counterparts in the laws of logarithms, which any good student of mathematics should also seek to master.


text:
# Suppose you purchase one California SuperLotto Plus lottery ticket. Here's how to play: Pick five numbers from 1 to 47 (no repeats allowed) and then pick one Mega number from 1 to 27. a. Find the probability you get the Mega number, but none of your five numbers match the winning numbers. b. Find the probability you get the Mega number and exactly two of your five numbers match the winning numbers. c. Find the probability you get the Mega number and all give of your five numbers match the winning numbers.

Suppose you purchase one California SuperLotto Plus lottery ticket. Here's how to play: Pick five numbers from 1 to 47 (no repeats allowed) and then pick one Mega number from 1 to 27.
a. Find the probability you get the Mega number, but none of your five numbers match the winning numbers.
b. Find the probability you get the Mega number and exactly two of your five numbers match the winning numbers.
c. Find the probability you get the Mega number and all give of your five numbers match the winning numbers.
You can still ask an expert for help

## Want to know more about Probability?

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a)Required probability= \frac{1}{27} \times (\frac{42}{47} \times \frac{41}{46} \times \frac{40}{45} \times \frac{39}{44}\times \frac{38}{43})
=0.02054
b)Required probability= \frac{1}{27} \times 5/2 (\frac{5}{47} \times \frac{4}{46} \times \frac{42}{45} \times \frac{41}{44}\times \frac{40}{43})
=0.00277
a)Required probability= \frac{1}{27} \times \frac{5 \times 4 \times 3 \times 2 \times 1}{47 \times 46 \times 45 \times 44 \times 43}
=0.000000024


text:
# Without regard to the transactions or events described below, Cher Holder, who is a single taxpay...

Without regard to the transactions or events described below, Cher Holder, who is a single taxpayer, has gross ordinary income of $60,000, Â§62 and Â§63 deductions of$20,000, and taxable income of $40,000 in the current year. Consider together the following further facts and then answer questions (a) through (d) which follow: (1) Cher owns a$5,000 â€œnoteâ€ of Flibinite Corporation which she got from the Corporation for a loan of that amount and which is supposed to pay five percent interest each year. Flibinite goes bankrupt and Cherâ€™s â€œnoteâ€ is worthless. The Commissioner successfully asserts that Cherâ€™s â€œnoteâ€ represents an equity contribution to Flibinite. Cher acquired the â€œnoteâ€ two years ago.

(2) Cher owns common stock in Flibinite which also becomes worthless in the current year. She paid $3,000 for the stock three years ago (3) Two years ago Cher loaned her friend Mooney$2,600. That loan becomes worthless in the current year. (What factors would be considered in determining if the loan created a bona fide debt?) Assume the debt was bona fide.

(4) Cher owned some tax exempt state bonds which she purchased for $8,000 four years ago. When they were worth$12,000 they were stolen and Cher received $12,000 in insurance proceeds in the current year. A) To what extent will the above transactions reduce Cherâ€™s taxable income for the year B) What, if any, is Cherâ€™s capital loss carryover to the succeeding year? C) If, in addition, Cher had sold some stock for$20,000 which she had purchased three years earlier for $9,400, what is her taxable income for the current year? (Assume her other Â§62 and Â§63 deductions are still$20,000)

D) Assuming the facts of all parts including part (c) in the above problem, must Cher report any income (and if so what character of income) if in the following year Mooney inherits some money and repays her \$2,600 obligation to Cher?


text:
# Hydrology - Depression cone VS transmissivity

My question concerns the depression cone of a pumping well in a confined aquifer: Say the transmissivity of the aquifer is 1200 m^2/day how will the cone change if the transmissivity is 600 m^2/day? All the aquifers parameters stay constant in both scenarios and so is the pumping rate (say- 300 cm/hr).

The main interest is what happens to the radius of influence (where there will be no change in head from the aquifers regional head)? and also what happens to the head at the well it self?

Thanks, Asher

• If you are interested in the head at the well, then you need to consider well-loss in addition to the aquifer response. – haresfur Aug 3 '17 at 0:32

$$s = \frac{Q}{4T\pi}\int_u^\infty \frac{e^{-y}}y dy$$ and $$u = \frac{r^{2}s}{4Tt}$$ Where s is drawdown, r is radius from well, t is time, T is transmissivity, Q is pumping rate, and u is the well function.


text:
# AUSTRALIA TELESCOPE OBSERVATIONSOF THE CTB 33 COMPLEX

A. P. SARMA
Physics & Astronomy Department, University of Kentucky, Lexington, KY 40506
sarma@pa.uky.edu

W. M. GOSS
National Radio Astronomy Observatory, P. O. Box 0, Socorro, NM 87801
mgoss@nrao.edu

A. J. GREEN
School of Physics, University of Sydney, NSW 2006, Australia
agreen@Physics.usyd.edu.au

D. A. FRAIL
National Radio Astronomy Observatory, P. O. Box 0, Socorro, NM 87801
dfrail@nrao.edu

(accepted Feb 21 1997, scheduled for publication, July 1997, ApJ, 483)

We present high resolution radio continuum, \ion{H}{1} absorption and H90$\alpha$ recombination line observations of the CTB 33 complex and nearby sources (l=337$\arcdeg$, b=0$\arcdeg$) taken with the Australia Telescope Compact Array. We estimate distances to several discrete sources based on \ion{H}{1} absorption. Radio recombination lines are detected for two sources in the field. The compact source G337.0$-$0.1 in the CTB 33 complex, whose nature was previously in doubt, is identified as a supernova remnant with a diameter of $\sim$ 5 pc. This classification is based on several contributing features, a non-thermal spectral index ($-$0.6), a shell structure and the absence of detectable recombination lines. In addition a 1720 MHz OH maser is probably associated with G337.0$-$0.1. We conclude that the CTB 33 region is a SNR-\ion{H}{2} complex at a distance of $\sim$ 11 kpc.


text:
# Tag Info

24

There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove $a = c$ and $b = c$ ...

5

short answer: equality is symmetric, implication is not (both are however transitive) longer answer: You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = C (because "=" is transitive and symmetric) Your teacher is right: if you prove that something true follows from A = B, i.e. A = ...

5

It is enough.. Consider this example: To prove: $a=b$ Proof: $$a=c$$ $$b=c$$ Since $a$ and $b$ are equal to the same thing, $a=b$. That is the exact technique you are using and it sure can be used.

5

First observe that $(x-y)^2 + (x-z)^2 + (y-z)^2 \geq 0$. Next, expand the LHS to obtain: $2x^2 + 2y^2 + 2z^2 - 2xy - 2xz - 2yz \geq 0$ Now you simply divide by two and add $xy + xz + yz$ to both sides.

5

The easiest here is to prove the contrapositive: if $A=\emptyset$ and $B=\emptyset$, then clearly $A\cup B=\emptyset$ and we're done.

4

To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of ...

4

Assume by contradiction that $A$ is unbounded. Let $a \in A$. Then for each $n$ there exists some $x_n$ such that $$d(x_n,a) >n$$ Now, $x_n$ has a converging subsequence $x_{k_n} \to b$. Now, for $\epsilon=1$, since $x_{k_n} \to b$ there exists some $N$ so that for all $n >N$ we have $$d(x_{k_n}, b) <1$$ Then, for all $n >N$ we have $$d(a,b) ... 4 Short answer: It's the wrong kind of proof Longer answer: In your proof, you cannot assume the conlcusion. If I assume that all squirrels have two tails, then it follows from my assumptions that all squirrels have two tails. This does not mean I proved that they do, of course. 3 My favorite : let k be your ground field, and let A = k[X_{ij}]_{1\leqslant i,j\leqslant n} be the ring of polynomials in n^2 indeterminates over k, and K = Frac(A). Then put M = (X_{ij})_{ij}\in M_n(A) the "generic matrix". For any N=(a_{ij})_{ij}\in M_n(k), there is a unique k-algebra morphism \varphi_N:A\to k defined by ... 3 You have the correct approach. Just clean up a few details. You are trying to show that given any \epsilon > 0, there exists a partition P_\epsilon such that U(P_\epsilon,f) - L(P_\epsilon,f) < \epsilon. First define M,$$M := \sup_{x \in [a,b]} |f(x)|.$$Then it follows that for i = 1, 2, \ldots, n we have$$M_i - m_i \leqslant 2M,... 3 There are two common errors that these methods are preventing: \begin{align} 1 &< 2 &&\text{True.} \\ 0 \cdot 1 &< 0 \cdot 2 &&\text{Um...} \\ 0 &< 0 &&\text{False.} \end{align} \begin{align} 1 &= 2 &&\text{False.} \\ 0 \cdot 1 &= 0 \cdot 2 &&\text{Um...} \\ ... 3 I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like ... 3 An alternative solution \begin{align} E(X^r) & = \int_0^{\infty} r x^{r-1} P(X > x) \:dx \\ &= \int_0^{\infty} r x^{r-1} \left[ \int_{y=x}^{\infty} f(y) \:dy \right] \:dx \\ &= \int_{y=0}^{\infty} f(y) \left[ \int_{x=0}^{y}r x^{r-1} \,dx \right]\:dy \\ &= \int_0^{\infty} y^r f(y)\,dy \end{align} $$2 Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form "A = B" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. Not necessary Let us first deal with the false ... 2 In finite dimensions (which is the case for you), if you know one basis \mathcal B, then proving another set \mathcal L is a basis is easiest if you prove that \mathcal L and \mathcal B have the same number of elements every element from \mathcal B can be expressed as a linear combination of elements from \mathbb L. 2 Let L be the line y=-mx+c, m>0. The line shifted right by a has the equation$$\begin{align*} y &= -m(x-a)+c\\ y &= -mx +c+ma\\ y-ma &= -mx+c \end{align*}$$Since m>0 and a>0, ma>0, so the graph is equivalent to L shifted up by ma. 2$$2r-2s-1=2r-2s-1-1+1=2r-2s-2+1=2(r-s-1)+1$$You can subtract 1 and plus 1 to balance things up. Something like adding zero, hence you preserve the equality. 2 Take$$S_n=\sum_{k=1}^n\left|f_k\right|$$and apply the monotone convergence theorem for \left\{S_n\right\}_{n=1}^\infty. 2 It's just a matter of cardinalities. A countable union of countable unions of closed sets is still a countable union of closed sets. 1 Here is a neat proof from Qiaochu Yuan's answer to this question: If L is diagonalizable with eigenvalues \lambda_1, \dots \lambda_n, then it's clear that (L - \lambda_1) \dots (L - \lambda_n) = 0, which is the Cayley-Hamilton theorem for L. But the Cayley-Hamilton theorem is a "continuous" fact: for an n \times n matrix it asserts that n^2 ... 1 I'm not too sure what the etiquette is on Math SE regarding answering old questions but I stumbled across this question and felt it would be good to drop an answer for future users that happen to ask the same question in the future. Anyways, I'm going to do the particular case when \delta(G) = 3 and show that there exists a path of at least length 3. This ... 1 If we denote by e_1,\dotsc,e_n the standard basis of K^n (assuming A\in {\rm GL}_n(K)), then we have Ae_1=0 and Ae_{i+1}=i\cdot e_i for i=1,\dotsc,n-1. Hence A^{i}e_{i}=0 for i=1,\dotsc,n. Then A^n e_i = 0 for all i=1,\dotsc,n, i. e. A^n=0 since e_1,\dotsc,e_n is a basis. Note, that we have A^{n-1}e_n = (n-1)!\cdot e_1, which is ... 1 Hint: (x-y)^2>0.Can you take it from here? 1 After some guidance, I think I might have the answer to this: Let f,g be continuous on [a,b] and f(a)>g(a) but g(b)>f(b). Prove that \exists c \in[a,b]:f(c)=g(c). Proof: Define h(x) as follows:$$h(x) = g(x) - f(x)$$Since h(x) is a composition of two continuous functions, it shows that h(x) is also continuous. Also, since ... 1 Since the set you're given has n+1 elements, all you need to show is that it is linearly independent, because \dim P_n=n+1. If you consider$$ a_0\cdot 1+a_1(x-1)+a_2(x^2-x)+\dots+a_n(x^n-x^{n-1})=0 $$you get the matrix$$ \begin{bmatrix} 1 & -1 & 0 & 0 & \dots & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & \dots & 0 ...

1

The first term should be $[x^r(1-F(x)]_0^{\infty}=0-0$.

1

You need to prove a little lemma: (1) Sum of evens is even. (2) Sum of odd number of odds is odd. Prove (1) by factoring out a $2$, and prove (2) by induction on the number of terms. Then we can prove what you want. $E(G)=\{v\in V(G):d(v)\text{ is even}\}$ $O(G)=\{v\in V(G):d(v)\text{ is odd}\}$ $\sum_{v\in V(G)} d(v) =\sum _{v\in E(G)}d(v)+ ... 1 Statement:$A \cap (B' \cap C') = (A \cap B') \cap (A \cap C')A \cap (B' \cap C') \subset (A \cap B') \cap (A \cap C')x \in A \cap (B' \cap C') \implies x \in A$and$x \in B' \cap C' \implies x \in A, x \in B', x \in C' \implies x \in A \cap B'$and$x \in A \cap C' \implies x \in (A \cap B') \cap (A \cap C')(A \cap B') \cap (A \cap C') ...

1

By definition, the elements of $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$ are the units of the ring $R = \mathbb{Z}/n\mathbb{Z}$. To say that $a\in R$ is a unit is to say that there exists $x\in R$ such that $ax = 1$, but by definition of $R$ that just means that $ax\equiv 1\pmod{n}$. At this point you can write everything out in prime factors if you ...

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# The flux of the vector field F ( x , y , z ) = 3 z k across the closed rectangular box with opposing vertices at ( 0 , 0 , 0 ) and ( 6 , 5 , 7 ) is:

Using Surface Integrals, not divergence theorem, find The flux of the vector field F ( x , y , z ) = 3 z k across the closed rectangular box with opposing vertices at ( 0 , 0 , 0 ) and ( 6 , 5 , 7 ) .

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Mathematical Omnibus: Thirty Lectures on Classic Mathematics

Dmitry Fuchs University of California, Davis, CA
Serge Tabachnikov Pennsylvania State University, University Park, PA
Available Formats:
Hardcover ISBN: 978-0-8218-4316-1
Product Code: MBK/46
List Price: $69.00 MAA Member Price:$62.10
AMS Member Price: $55.20 Electronic ISBN: 978-1-4704-1812-0 Product Code: MBK/46.E List Price:$65.00
MAA Member Price: $58.50 AMS Member Price:$52.00
Bundle Print and Electronic Formats and Save!
This product is available for purchase as a bundle. Purchasing as a bundle enables you to save on the electronic version.
List Price: $103.50 MAA Member Price:$93.15
AMS Member Price: $82.80 Click above image for expanded view Mathematical Omnibus: Thirty Lectures on Classic Mathematics Dmitry Fuchs University of California, Davis, CA Serge Tabachnikov Pennsylvania State University, University Park, PA Available Formats:  Hardcover ISBN: 978-0-8218-4316-1 Product Code: MBK/46  List Price:$69.00 MAA Member Price: $62.10 AMS Member Price:$55.20
 Electronic ISBN: 978-1-4704-1812-0 Product Code: MBK/46.E
 List Price: $65.00 MAA Member Price:$58.50 AMS Member Price: $52.00 Bundle Print and Electronic Formats and Save! This product is available for purchase as a bundle. Purchasing as a bundle enables you to save on the electronic version.  List Price:$103.50 MAA Member Price: $93.15 AMS Member Price:$82.80
• Book Details

2007; 463 pp
MSC: Primary 00;

The book consists of thirty lectures on diverse topics, covering much of the mathematical landscape rather than focusing on one area. The reader will learn numerous results that often belong to neither the standard undergraduate nor graduate curriculum and will discover connections between classical and contemporary ideas in algebra, combinatorics, geometry, and topology. The reader's effort will be rewarded in seeing the harmony of each subject. The common thread in the selected subjects is their illustration of the unity and beauty of mathematics. Most lectures contain exercises, and solutions or answers are given to selected exercises. A special feature of the book is an abundance of drawings (more than four hundred), artwork by an award-winning artist, and about a hundred portraits of mathematicians. Almost every lecture contains surprises for even the seasoned researcher.

• Algebra and arithmetics
• Chapter 1. Arithmetic and combinatorics
• Lecture 1. Can a number be approximately rational?
• Lecture 2. Arithmetical properties of binomial coefficients
• Lecture 3. On collecting like terms, on Euler, Gauss, and MacDonald, and on missed opportunities
• Chapter 2. Equations
• Lecture 4. Equations of degree three and four
• Lecture 5. Equations of degree five
• Lecture 6. How many roots does a polynomial have?
• Lecture 7. Chebyshev polynomials
• Lecture 8. Geometry of equations
• Geometry and topology
• Chapter 3. Envelopes and singularities
• Lecture 9. Cusps
• Lecture 10. Around four vertices
• Lecture 11. Segments of equal areas
• Lecture 12. On plane curves
• Chapter 4. Developable surfaces
• Lecture 13. Paper sheet geometry
• Lecture 14. Paper Möbius band
• Lecture 15. More on paper folding
• Chapter 5. Straight lines
• Lecture 16. Straight lines on curved surfaces
• Lecture 17. Twenty-seven lines
• Lecture 18. Web geometry
• Lecture 19. The Crofton formula
• Chapter 6. Polyhedra
• Lecture 20. Curvature and polyhedra
• Lecture 21. Non-inscribable polyhedra
• Lecture 22. Can one make a tetrahedron out of a cube?
• Lecture 23. Impossible tilings
• Lecture 24. Rigidity of polyhedra
• Lecture 25. Flexible polyhedra
• Chapter 7. Two surprising topological constructions
• Lecture 26. Alexander’s horned sphere
• Lecture 27. Cone eversion
• Chapter 8. On ellipses and ellipsoids
• Lecture 28. Billiards in ellipses and geodesics on ellipsoids
• Lecture 29. The Poncelet porism and other closure theorems
• Lecture 30. Gravitational attraction of ellipsoids
• Lecture 31. Solutions to selected exercises

• Reviews

• The authors manage to breathe new life into topics that at first glance appear to be old hat.

• This is an enjoyable book with suggested uses ranging from a text for a undergraduate Honors Mathematics Seminar to a coffee table book. It is appropriate for either It could also be used as a starting point for undergraduate research topics or a place to find a short undergraduate seminar talk. This is a wonderful book that is not only fun to read, but gives the reader new ideas to think about.

MAA Reviews
• Dmitry Fuchs and Serge Tabachnikov display impeccable taste in their choice of the material, level of exposition, and the balance between concrete and more conceptual mathematical themes. Each of the thirty lectures tells a unique mathematical story, each with a display of mathematical narrative art, with great care for the details, revealing masters of their craft at work. Both novice and more experienced readers will find many pleasant surprises at all levels of exposition. ...[A] book suitable for such a noble and demanding goal to serve as an introduction to the world of 'serious mathematics' for new generations of mathematicians. ...[E]very page has one or more diagrams, graphs, and pictures illustrating the material. ...[T]he special artistic spirit and atmosphere the book owes to numerous, witty, humorous, and mysterious illustrations of the artist Sergey Ivanov. Without much exaggeration, one may say that these provocative, yet mathematically correct drawings can alone serve as a layman's guide to the beauty and mystery of mathematics. Summarizing, we can say that Mathematical Omnibus is a 'desert island book,' a 'coffee table book,' a book to share with friends, colleagues, and students, a gift for a beginner and an expert alike. In short, it is a wonderful addition to our personal, school, and university libraries.

Rade T. Zivaljevic, The American Mathematical Monthly
• Requests

Review Copy – for reviewers who would like to review an AMS book
Permission – for use of book, eBook, or Journal content
Accessibility – to request an alternate format of an AMS title
2007; 463 pp
MSC: Primary 00;

The book consists of thirty lectures on diverse topics, covering much of the mathematical landscape rather than focusing on one area. The reader will learn numerous results that often belong to neither the standard undergraduate nor graduate curriculum and will discover connections between classical and contemporary ideas in algebra, combinatorics, geometry, and topology. The reader's effort will be rewarded in seeing the harmony of each subject. The common thread in the selected subjects is their illustration of the unity and beauty of mathematics. Most lectures contain exercises, and solutions or answers are given to selected exercises. A special feature of the book is an abundance of drawings (more than four hundred), artwork by an award-winning artist, and about a hundred portraits of mathematicians. Almost every lecture contains surprises for even the seasoned researcher.

• Algebra and arithmetics
• Chapter 1. Arithmetic and combinatorics
• Lecture 1. Can a number be approximately rational?
• Lecture 2. Arithmetical properties of binomial coefficients
• Lecture 3. On collecting like terms, on Euler, Gauss, and MacDonald, and on missed opportunities
• Chapter 2. Equations
• Lecture 4. Equations of degree three and four
• Lecture 5. Equations of degree five
• Lecture 6. How many roots does a polynomial have?
• Lecture 7. Chebyshev polynomials
• Lecture 8. Geometry of equations
• Geometry and topology
• Chapter 3. Envelopes and singularities
• Lecture 9. Cusps
• Lecture 10. Around four vertices
• Lecture 11. Segments of equal areas
• Lecture 12. On plane curves
• Chapter 4. Developable surfaces
• Lecture 13. Paper sheet geometry
• Lecture 14. Paper Möbius band
• Lecture 15. More on paper folding
• Chapter 5. Straight lines
• Lecture 16. Straight lines on curved surfaces
• Lecture 17. Twenty-seven lines
• Lecture 18. Web geometry
• Lecture 19. The Crofton formula
• Chapter 6. Polyhedra
• Lecture 20. Curvature and polyhedra
• Lecture 21. Non-inscribable polyhedra
• Lecture 22. Can one make a tetrahedron out of a cube?
• Lecture 23. Impossible tilings
• Lecture 24. Rigidity of polyhedra
• Lecture 25. Flexible polyhedra
• Chapter 7. Two surprising topological constructions
• Lecture 26. Alexander’s horned sphere
• Lecture 27. Cone eversion
• Chapter 8. On ellipses and ellipsoids
• Lecture 28. Billiards in ellipses and geodesics on ellipsoids
• Lecture 29. The Poncelet porism and other closure theorems
• Lecture 30. Gravitational attraction of ellipsoids
• Lecture 31. Solutions to selected exercises
• The authors manage to breathe new life into topics that at first glance appear to be old hat.


text: | 5 | 6 | 7  | var now = new Date(); google_ad_client = "pub-0863636157410944"; 2". //--> is not equal to and any corresponding bookmarks? page, Finding Mathematically, this means that. Now to verify that this is indeed the inverse. The calculator will find the inverse of the given function, with steps shown. 2" and the Previous How to Find the Inverse of a Function 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. I n an equation, the domain is represented by the x variable and the range by the y variable. If it isn't, restrict the domain to pass the horizontal line test. (fourdigityear(now.getYear())); is not equal to in Order  |  Print-friendly This analysis method can be used two different ways, when conducting the overall market analysis using trends and the larger time frames, and also at the point of trade entry to increase overall trading accuracy. In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, and vice versa, i.e., f(x) = y if and only if g(y) = x. Do both of these have same volatility and if so how would you prove it? the Inverse of a Function (page Find the Inverse Function f(x) = log of x. and the inverse is also a function, with domain of all x the vertex is at (x, 1jaiz4 and 19 more users found this answer helpful. so this graph is the left-hand "half" of the parabola. $inverse\:y=\frac {x} {x^2-6x+8}$. Let St = Fx and St(Inverse) = 1/Fx. The inverse function for f( x), labeled f −1 ( x) (which is read “ f inverse of x”), contains the same domain and range elements as the original function, f( x).However, the sets are switched. 1jaiz4 and 19 more users found this answer helpful. To find the inverse of any function, first, replace the function variable with the other variable and then solve for the other variable by replacing each other. bookmarked pages associated with this title. asymptotes) that Because these functions have range elements that correspond to only one domain element each, there's no danger that their inverses will not be functions. In this short tutorial we will learn how you can easily find the inverse of a matrix using a Casio fx-991ES plus.  Return to Index  Next 'June','July','August','September','October', , Copyright © 2020  Elizabeth Stapel   |   About   |   Terms of Use   |   Linking   |   Site Licensing, Return to the So the for non-inverse function y=4. This half of the parabola [Date] [Month] 2016, The "Homework here because I need to know the domain and range of the inverse. the algebra: Since the inverse f(x): took an element from the domain and added 1 … "0" : "")+ now.getDate(); Then the inverse is y = (–2x – 2) / (x – 1), and the inverse is also a function, with domain of all x not equal to 1 and range of all y not equal to –2. 2. $inverse\:f\left (x\right)=\ln\left (x-5\right)$. The slope-intercept form gives you the y-intercept at (0, –2). 99. profile. Forex analysis with parallel and inverse pairs can be learned in a very short period of time, perhaps in just a few weeks. Since functions and inverse functions contain the same numbers in their ordered pair, just in reverse order, their graphs will be reflections of one another across the line y = x, as shown in Figure 1. | 2 | 3 | 4 The inverse of f(x) is f-1 (y) We can find an inverse by reversing the "flow diagram" Or we can find an inverse by using Algebra: Put "y" for "f(x)", and ; Solve for x; We may need to restrict the domain for the function to have an inverse So, here the sum of the two limited integral referred in the question is- The Area of the big rectangle – the Area of the smaller one = 10×5–2×1 =50−2 = 48 google_ad_height = 600; Here's 5. is a rational function. from the picture (and recalling the concept of horizontal Since , is the inverse of . 'January','February','March','April','May', not equal The key steps involved include isolating the log expression and then rewriting the … Inverse of Logarithmic Function Read More » But how to solve for the inverse? will be "y You will realize later after seeing some examples that most of the work boils down to solving an equation. An example is provided below for better understanding. To find the inverse function for a one‐to‐one function, follow these steps: 1. Removing #book# You should use that portion of the graph because it is the reflection of f( x) across the line y = x, unlike the portion on x < 0. Find a local math tutor,


text:
Electronic Communications in Probability

Extension of time for decomposition of stochastic flows in spaces with complementary foliations

Abstract

Let $M$ be a manifold equipped (locally) with a pair of complementary foliations. In Catuogno, da Silva and Ruffino (Stoch. Dyn. 2013), it is shown that, up to a stopping time $\tau$, a stochastic flow of local diffeomorphisms $\varphi_t$ in $M$ can be decomposed in diffeomorphisms that preserves this foliations. In this article we present techniques which allows us to extend the time of this decomposition. For this extension, we use two techniques: In the first one, assuming that the vector fields of the system commute with each other, we apply Marcus equation to jump nondecomposable diffeomorphisms. The second approach deals with the general case: we introduce a "stop and go" technique that allows us to construct a process that follows the original flow in the "good zones" for the decomposition, and remains paused in "bad zones". Among other applications, our results open the possibility of studying the asymptotic behaviour of each component.

Article information

Source
Electron. Commun. Probab., Volume 20 (2015), paper no. 38, 9 pp.

Dates
Accepted: 17 May 2015
First available in Project Euclid: 7 June 2016

https://projecteuclid.org/euclid.ecp/1465320965

Digital Object Identifier
doi:10.1214/ECP.v20-3762

Mathematical Reviews number (MathSciNet)
MR3352333

Zentralblatt MATH identifier
1332.60086

Rights

Citation

Morgado, Leandro; Ruffino, Paulo. Extension of time for decomposition of stochastic flows in spaces with complementary foliations. Electron. Commun. Probab. 20 (2015), paper no. 38, 9 pp. doi:10.1214/ECP.v20-3762. https://projecteuclid.org/euclid.ecp/1465320965

References

• Applebaum, David. Lévy processes and stochastic calculus. Cambridge Studies in Advanced Mathematics, 93. Cambridge University Press, Cambridge, 2004. xxiv+384 pp. ISBN: 0-521-83263-2
• Bismut, Jean-Michel. Mécanique aléatoire. (French) [Random mechanics] With an English summary. Lecture Notes in Mathematics, 866. Springer-Verlag, Berlin-New York, 1981. xvi+563 pp. ISBN: 3-540-10840-8
• Catuogno, Pedro J.; da Silva, Fabiano B.; Ruffino, Paulo R. Decomposition of stochastic flows with automorphism of subbundles component. Stoch. Dyn. 12 (2012), no. 2, 1150013, 15 pp.
• Catuogno, Pedro J.; da Silva, Fabiano B.; Ruffino, Paulo R. Decomposition of stochastic flows in manifolds with complementary distributions. Stoch. Dyn. 13 (2013), no. 4, 1350009, 12 pp.
• Colonius, F.; Ruffino, P. R. C. Nonlinear Iwasawa decomposition of control flows. Discrete Contin. Dyn. Syst. 18 (2007), no. 2-3, 339–354.
• Fujiwara, Tsukasa; Kunita, Hiroshi. Canonical SDE's based on semimartingales with spatial parameters. I. Stochastic flows of diffeomorphisms. Kyushu J. Math. 53 (1999), no. 2, 265–300.
• Kunita, H. Stochastic differential equations and stochastic flows of diffeomorphisms. Ã‰cole d'été de probabilités de Saint-Flour, XIIâ€”1982, 143–303, Lecture Notes in Math., 1097, Springer, Berlin, 1984.
• Kunita, Hiroshi. Stochastic flows and stochastic differential equations. Reprint of the 1990 original. Cambridge Studies in Advanced Mathematics, 24. Cambridge University Press, Cambridge, 1997. xiv+346 pp. ISBN: 0-521-35050-6; 0-521-59925-3
• Kurtz, Thomas G.; Pardoux, Ã‰tienne; Protter, Philip. Stratonovich stochastic differential equations driven by general semimartingales. Ann. Inst. H. Poincaré Probab. Statist. 31 (1995), no. 2, 351–377.
• Liao, Ming. Decomposition of stochastic flows and Lyapunov exponents. Probab. Theory Related Fields 117 (2000), no. 4, 589–607.
• Milnor, J. Remarks on infinite-dimensional Lie groups. Relativity, groups and topology, II (Les Houches, 1983), 1007–1057, North-Holland, Amsterdam, 1984.
• K.-H. Neeb – Infinite Dimensional Lie Groups. Monastir Summer Schoool, 2009. Available from hal.archives-ouvertes.fr/docs/00/39/.../CoursKarl-HermannNeeb.pdf
• Nualart, David. The Malliavin calculus and related topics. Probability and its Applications (New York). Springer-Verlag, New York, 1995. xii+266 pp. ISBN: 0-387-94432-X
• Omori, Hideki. Infinite-dimensional Lie groups. Translated from the 1979 Japanese original and revised by the author. Translations of Mathematical Monographs, 158. American Mathematical Society, Providence, RI, 1997. xii+415 pp. ISBN: 0-8218-4575-6
• Ruffino, Paulo R. C. Decomposition of stochastic flows and rotation matrix. Stoch. Dyn. 2 (2002), no. 1, 93–107.


text:
## anonymous 4 years ago The radioactive substance krypton-91 has a very short half-life of about 10 seconds. What is the decay constant for this substance?

• This Question is Open
1. Kainui

$\frac{ \ln2 }{ t _{1/2} }=k$ I can derive this for you if you'd like.

2. anonymous

half life= 0.693/decay constant


text:
Hardcover | $21.95 Trade | £14.95 | ISBN: 9780262134729 | 117 pp. | 5.375 x 8 in | 30 illus.| July 2006 ebook |$15.95 Trade | ISBN: 9780262252881 | 117 pp. | 5.375 x 8 in | 30 illus.| July 2006

# The Laws of Simplicity

## Overview

John Maeda joined Kleiner Perkins Caufield & Byers in 2013 as Design Partner, where he works within KPCB’s ProductWorks program helping entrepreneurs and portfolio companies to build design into their company cultures. From 2008 to 2013 he was President of Rhode Island School of Design and before that was Associate Director of the MIT Media Lab. In 2008 Esquire magazine named Maeda one of the 75 most influential people of the twenty-first century. He is the author of The Laws of Simplicity (MIT Press, 2006) and other books.

## Reviews

Abstract recommends this book particularly to marketing people, product designers and technical writers.”—GetAbstract"—

"FINALLY, a book about simplicity that is not too academic to read.... At the book's heart is the Shinto belief in animism, the spirit in all objects. Nicholas Negroponte, one of Maeda's mentors, once told him to become a lightbulb, not a laser beam. This he has done; all this and more." —Susan Salter Reynolds, The LA Times"—

"John Maeda's new book, The Laws of Simplicity, is simply terrific. It's exactly 100 pages, the illustrations are brilliant and the 10 Laws of Simplicity (plus Three Keys) are a canon to design one's entire life, much less specific products, services or business models. The subtitle is: Design, Technology, Business, Life." —Bruce Nussbaum, BusinessWeek's blog "NussbaumOnDesign"—

"Keep it simple, Stupid" is an old piece of advice, so much so that it's often abbreviated as the "KISS principle." But it's advice that's often ignored, and MIT Professor John Maeda aims to change that. . . . Designers and marketers will find Maeda's book both interesting and useful...." —New York Post"—

"Maeda's Laws and Keys have an obvious practical application in everyday running of a busy life (and desktop); they also have the potential to translate into a productive methodology for any craft or design practice.... A very humble, enlightened and caring human, John's written a little bible."—Liz Farrelly, Crafts Magazine"—

"Maeda's upbeat explanations usefully break down the power of less-fewer features, fewer buttons and fewer distractions-while providing practical strategies for harnessing that power. . . . Emphasizing the delicate balance-work involved in simplifying the complex, Maeda admits the process isn't easy, and that his ten laws don't necessarily provide all the answers-in numerous places, he directs readers to the web site where his theories continue to develop. Despite that, this slim book feels complete in itself; not only will it stimulate ideas, it will keep readers thumbing back for a second and third look at Maeda's deceptively simple advice." —Publishers Weekly"—

"Technology and life seem to be getting more complicated, yet two great success stories, Google and the iPod, both provide the antidote of simplicity. In The Laws of Simplicity, John Maeda uses humble prose to provide an accessible guide, business and life, observing the principle: 'Simplicity equals sanity.'" —David Smith, The Observer"—

## Endorsements

"If brevity is the soul of wit, simplicity is the soul of design. John Maeda uses the concept of simplicity to get at the nature of human thought and perception while drawing out tangible applications for business, technology, and life in general. *The Laws of Simplicity* is thoroughly optimistic, entertaining, and erudite, just as you would expect from Maeda. It is also the most compelling one hundred pages of design writing I have read this year."—Rob Forbes, Founder, Design Within Reach"—

"I planned to skim/sample John Maeda's book, then decide to endorse it—or not. I quickly found myself mesmerized—and thence the only issue was deciding what were the strongest words I could muster in support of The Laws of Simplicity. The book is important; and Maeda has made an absurdly complex subject—simplicity—approachable and usable. Bravo! I hope the people who design the products I'll acquire in the next ten years take this book to heart." —Tom Peters"—

"I planned to skim/sample John Maeda's book, then decide to endorse it -- or not. I quickly found myself mesmerized -- and thence the only issue was deciding what were the strongest words I could muster in support of *The Laws of Simplicity*. The book is important; and Maeda has made an absurdly complex subject -- simplicity -- approachable and usable. Bravo! I hope the people who design the products I'll acquire in the next ten years take this book to heart."—Tom Peters"—


text:
# Which die should I choose?

Some days ago I read this problem, and I haven't arrived yet at a solution: Imagine you have 3 kinds of regular number die:

d1 has 12 faces (from 1 to 12), d2 has 6 faces (from 1 to 6), d3 has 4 faces (from 1 to 4).

Now, you can choose between using one d1, two d2's, or three d3's. Which configuration of dice would you choose if the one who gets the higher number wins?

I would choose the first one, because the probability of winning (getting the highest number) is 1/12. Instead, throwing two d2's the probability is (1/6)^2, and for three d3's it's (1/4)^3. But I'm not sure.

• What do you think? Also, what numbers are in the dice? – Arthur Oct 7 '17 at 16:49
• There are some subtleties. With one D1, you will have nonzero chance to roll a 1. But you will always roll something higher than 1 if you roll two D2 or three D3. This kind of consideration counters your observation about what it takes to roll a 12. And ultimately, these considerations outweigh your observation, as @Austin's answer shows – alex.jordan Oct 7 '17 at 16:56

Simply calculate the expected value of the sum of these dice:

$$\Bbb E[d_1]=\sum^{12}_{n=1}\frac1{12}n=\frac{13}2=6.5$$ $$\Bbb E[2d_2]=2\left(\sum^6_{n=1}\frac16n\right) = 2\cdot\frac72=7$$ $$\Bbb E[3d_3]=3\left(\sum^4_{n=1}\frac14n\right) = 3\cdot\frac52=7.5$$

Therefore, $3d_3$ is to be the best choice.

EDIT:

I might have misinterpreted this question. My answer is correct if the number referred to in "highest number" is the sum. Alternatively, with two players, this could mean the highest number rolled on any die. I will answer this now:

Expected value of the highest die rolled is relatively simple:

$$\Bbb E[d_1]=\sum^{12}_{n=1}\frac1{12}n=\frac{13}2=6.5$$ $$\Bbb E[2d_2]=\sum^6_{n=1}\frac{n^2-(n-1)^2}{6^2}n = 4.47\bar2$$ $$\Bbb E[3d_3]=\sum^4_{n=1}\frac{n^3-(n-1)^3}{4^3}n = 3.4375$$

So the answer in the case of the maximum die winning, you should choose the $12$-sided die.

• The expected values can also be calculated thusly with no summation. D1's distribution is symmetric, ranging from 1 to 12. So $\frac{1}{2}(1+12)$ is its expected value. And D2+D2's distribution is symmetric, ranging from 2 to 12. So $\frac{1}{2}(2+12)$ is its expected value. And similarly for three D3s. – alex.jordan Oct 7 '17 at 17:01
• The question cannot be answered simply by expectation values. As far as I understand, the OP is considering a game in which two players will each roll (with their chosen set of dice) and the higher score wins (quote: "the one who gets the higher number wins"). One should really consider the probability of getting a higher score than an opponent, not simply as high as possible. This might depend on the opponent's choice. I admit that here the expected values are sufficiently different to make this unlikely, but I can certainly devise a set-up in which the highest E is not the optimal choice. – Nick Pavlov Oct 7 '17 at 18:54
• @Nick check out the edit. I have answered the question both ways. – Austin Weaver Oct 7 '17 at 19:17
• @AustinWeaver Actually, that is not what I meant. I meant that the aim might not be to maximize an expected value, but to maximize probability of getting more than the result of another random event (the opponent's roll). As in: player A rolls one of the options (12-sided, or two 6-sided, or three 4-sided) and gets a sum of $x$ (a random variable with a certain distribution which depends on what A has chosen to roll). What choice of dice for player B gives him a better ratio of getting more than $x$ (winning) vs. less than $x$ (losing)? – Nick Pavlov Oct 7 '17 at 19:57
• @AustinWeaver In this situation (and I suspect whenever the distributions are symmetric), the distribution with the higher expected value has an advantage. But it is conceivable that it is not always the case (in fact I have a counter-example with non-symmetric distributions) – Nick Pavlov Oct 7 '17 at 20:00


text:
Underscript, subscript and superscript add the same time possibly via \mathoperator* without trailing space

I have a symbol, which I want to give a underscipt, subscript and superscript.

I tried:

\DeclareMathOperator*{\wmat}{w}
\wmat_2_i^j

But now I get a space between my symbol and the sub and superscript. Can this space be removed?

Or if someone knows a different method without any use of \DeclareMathOperator: you are more than welcome.

N.B. I tried many other things, with worse result.

Here's my proposal:

\documentclass{article}
\usepackage{amsmath}

\makeatletter % access "private" commands
\newcommand{\wmat}[1][]{%
\mathop{\wmat@false\mathpalette\wmat@{#1}}%
\!%
\mathop{\wmat@true\mathpalette\wmat@{#1}}%
\nolimits
}
\newif\ifwmat@
\newcommand{\wmat@}[2]{%
\begingroup
\setbox\z@=\vtop{\offinterlineskip
\ialign{%
\hfil$\m@th##$\hfil\cr
#1\mathrm{w}\cr
\noalign{\kern1pt}
\scriptscriptstyle#2\cr
}%
}%
\ifwmat@
\sbox\tw@{$#1\mathrm{w}$}%
\dp\z@=\dp\tw@
\box\z@
\else
\vrule height \ht\z@ depth \dp\z@ width \z@
\fi
\endgroup
}

\begin{document}
\begin{center}
\fboxsep=0pt\fboxrule=0.1pt
$\wmat[2]_i^j$ $\scriptstyle\wmat[2]_i^j$ $\scriptscriptstyle\wmat[2]_i^j$
\fbox{$\wmat[2]_i^j$}
\end{center}
$\wmat[2]_i^j$

\end{document}

The centered subscript is always in \scriptscriptstyle so as to avoid it making for bad spacing.

The strategy is to define a math operator that accepts an optional argument for the centered subscript. However, this subscript should not influence the vertical size of the ‘w’, in order that the standard subscript and superscript are set to the letter.

So I typeset ‘w with the subscript’ twice, one with \wmat@false and one with \wmat@true. The first instance is used to make a phantom that has the same vertical size of ‘w with the subscript’, placed in a \mathop atom of its own. The second instance is also typeset in a \mathop atom; a \! between them removes the space that TeX adds between consecutive \mathop atoms.

I build a box containing ‘w with the subscript’ (details later); in the \wmat@false case this is used to make a vertical rule as high and deep as the box, with zero width. This will reserve the needed vertical space but produce no visible output. In the \wmat@true case the box is resized to have the same vertical dimensions as a plain ‘w’. This will be the last item that TeX will see and to which it will attach the _j^i items that follow.

Now how's the box built? I found that using \operatorname*{w}_2 places the subscript too far down below the ‘w’, so a different method is needed. I use the primitive TeX method for alignments. I open a \vtop, that is a box containing vertical material, with the reference point given by the top item inside it (it will be \mathrm{w}).

Inside the \vtop I build an \ialign (a wrapper around \halign with some initialization). This does a one column table, with no vertical space between rows (by \offinterlineskip) except the explicitly added \kern1pt. The top row contains, as said, \mathrm{w} in the needed size (via \mathpalette); the bottom row contains the subscript, always in \scriptscriptstyle.

The \vtop is assigned to a box register, so it is possible to access its dimensions (for typesetting the zero width rule in the \wmat@false case) or change them (for making it appear the same as ‘w’). Then the box is used in the \wmat@true case).

Information about \halign is found in the TeXbook or in TeX by Topic.

• @JensWagemaker xparse is a relic of a previous attempt and is not needed here. I'll add some explanations. For the @, it is customary to use it in “private” command names, so as not to conflict with user level commands. – egreg Jan 30 at 11:15
• @JensWagemaker I added some explanations. – egreg Jan 30 at 11:38
• Can you explain what \wmat@false is? Because it looks like a control sequence, but the only control sequence that is defined until the first use of \wmat@false is \wmat. – Jens Wagemaker Feb 1 at 11:43
• @JensWagemaker \wmat@false is defined by \newif\ifwmat@ and makes the conditional return false when used. Similarly for \wmat@true. – egreg Feb 1 at 11:47

You code generates a double subscript error. Perhaps what you want is this:

\documentclass{article}
\usepackage{mathtools}
\DeclareMathOperator*{\wmat}{w}
\begin{document}
$\smash{\wmat_2}\mathllap{\phantom{\wmat}}_i^j$
\end{document}

You might want to define a macro if you want to use this more than once.


text:
Fitting one Polygon in another

I have two Polygons A and B and I want to find the position, rotation and scale of B, so it fits into A and has the maximum Area possible. Also both can be concave.

I did some research but couldn't find anything fitting. Maybe someone here can help or give me something that could be helpful.

• When you say "I have", do you mean that, in a computer, you have a list of the vertices of $A$ and a list of the vertices of $B$, in order, and when you say "I want to find", do you mean that you want an algorithm that will find those numbers in finite time? – Ben McKay Jun 27 '18 at 16:17
• These questions are related and their answers give references that could be a good place to start mathoverflow.net/questions/93704/… and mathoverflow.net/questions/282158/… – j.c. Jun 27 '18 at 17:27

Here are two references. The first, a 1981 paper, provides an algorithm to solve the problem allowing translations and rotations, but not scaling:

Chazelle, Bernard. "The polygon containment problem." Carnegie-Mellon University, Department of Computer Science, 1981.

This was followed in 1994 by an algorithm that includes scaling. The algorithm uses parametric search:

Sharir, Micha, and Sivan Toledo. "Extremal polygon containment problems." Computational Geometry-Theory and Application 4, no. 2 (1994): 99. (Elsevier link.)

All algorithms are polynomial in the number of vertices of the two polygons. In the latter paper, the time complexity is $O(k^2 n \lambda_6(kn) \log^3(kn) \log\log(kn))$, where the containing polygon has $n$ vertices and the contained has $k$ vertices. For $k=n$, this is a bit beyond $O(n^5)$. Various speed-ups can be achieved if one or the other polygon is known to be convex, or if one restricts the rigid motions allowed.

Google Scholar shows about 50 papers that subsequently cite Sharir-Toledo. Packing applications have led some researchers to more practical approximation algorithms.

Here are some observations which might help:

• The space of translations, rotations, and scales you need to consider for fixed A and B is compact, and can be parameterized by a bounded subset of $\mathbb{R}^4$, with an upper bound on the radius of this subset computable from the list of vertices of $A$ and $B$.

• The area of $A \cap B$ is uniformly continuous with respect to (and can be computed from) the position, scale, and rotation of B, with some modulus of continuity computable from the list of vertices for $A$ and $B$.

• The basic technique for doing this is to find triangulations of A and (the rotated/scaled/translated copy of) B. The area of the intersection is just the sum of the areas of the intersections of triangles, so the problem reduces to computing the area of the intersection of two triangles whose vertices you know.
• The distance between $B$ and $A^c$ is uniformly continuous with respect to (and can be computed from) the position, scale, and rotation of B, with some modulus of continuity computable from a list of vertices for $A$ and $B$.

• Similar idea as a above using triangulations. You can use this to figure out whether, given some rotation/scaling/translation, B is strictly contained entirely in A: this is so if and only if this distance is strictly positive.

What these tell us is that the maximum area is computable from a list of vertices of A and B. To find an approximation to the maximum area, just check an $\epsilon$-net of points in the parameter space. Your approximation is just the maximal area obtained by those of these finitely many possibilities which have distance between $B$ and $A^c$ at least $\epsilon$. You can use the modulii of continuity computed above to figure out how close your approximation to the maximal area is to the actual maximal area. You can prove as $\epsilon \rightarrow 0$ this approximation converges to the correct area, and you can even compute a decent a priori bound on the rate of convergence.

For actually computing the rotation, translation, and scale that yield this maximal area: it is computable if it is unique.