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http://peterashmathedblog.blogspot.com/2015/05/ | 1,508,469,913,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823630.63/warc/CC-MAIN-20171020025810-20171020045810-00671.warc.gz | 264,772,781 | 11,050 | ## Math Tutoring Service
See my Mathematics Tutoring Service on Thumbtack
### Embarrasing mistake
I just realized that my last post was far off the mark. There is a super-obvious example that shows that f(n) >= n/2, and so in fact f(n) = n/2. The example is the subset {n/2 + 1, n/2 + 2, ... , n} which contains n/2 elements and clearly has the non-divisible property. The problem has very little to do with prime numbers. The previously-published result was sharp.
### A simple (?) number theory conjecture
Call a set S of positive integers non-divisible if, whenever a and b belong to S, it is not true that (a|b or b|a). For example, the set of prime numbers is non-divisible. Let S(n) be the set of the first n positive integers, and let f(n) be the cardinality of the largest non-divisible subset of S(n). Then clearly f(n) >= pi(n), where pi(n) denotes the number of prime numbers <= n. Furthermore, I know a pretty proof (published, but not by me) that shows f(n) <= n/2 (n even). That is, in any subset of n/2 + 1 positive integers all <= n, at least one integer divides another integer in the set. This proof involves the pigeonhole principle. So, pi(n) <= f(n) <= n/2. (n even) My conjecture is that, in fact, f(n) = pi(n) for all n > 1. Does anyone have a counterexample or a proof? It seems like it should be very simple. | 363 | 1,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-43 | longest | en | 0.902427 |
http://forum.arduino.cc/index.php?action=profile;u=14246;sa=showPosts | 1,406,271,672,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997893881.91/warc/CC-MAIN-20140722025813-00172-ip-10-33-131-23.ec2.internal.warc.gz | 143,306,547 | 11,011 | | Arduino Forum :: Members :: frostin
Show Posts Pages: [1] 2 3 ... 28
1 Using Arduino / Programming Questions / Re: Dht22 on: August 31, 2011, 03:07:50 pm I was meaning by reading the value of c (hence it's in celcius when on or off). Couldn't I use #define 120 in f not in c by using some math or am I just going to have to calculate it only for displaying ane figure the temp I'd have to set at in c for the relay?
2 Using Arduino / Programming Questions / Re: Dht22 on: August 28, 2011, 06:02:33 pm How so? I believe standard the library is in Celcius not fahrenheit which I had to calculate to display F Edit:Basically the relay which is connected to say pin 4 its HIGH (relay closed) crock pot will be on till about 120ish degrees F. If it goes higher than 120 then the relay will be LOW (relay open).
3 Using Arduino / Programming Questions / Re: Dht22 on: August 28, 2011, 05:38:54 pm Didn't compile correctly... But this does:Code:#include "DHT.h"#define DHTTYPE DHT22#define DHTPIN 2DHT dht(DHTPIN, DHTTYPE);#define CROCKPOTPIN 4#define TEMP_HIGH 22#define TEMP_LOW 20#define HUMID_HIGH 60#define HUMID_LOW 20float minTemp = 100;float maxTemp = 0;float minHumid = 100;float maxHumid = 0;int crockValue = 0;void setup(){ Serial.begin(115200); Serial.println("TEST PROGRAM "); pinMode(CROCKPOTPIN, OUTPUT);}void loop(){ // READ DATAint t = dht.readTemperature();int temperatureF = (t * 9 / 5) +32.5; // DO THE MIN/MAX minTemp = min(minTemp, t); maxTemp = max(maxTemp , t); // DISPLAT DATA Serial.println(temperatureF); // CROCKPOT if (t > TEMP_HIGH) crockValue = 1; else if (t < TEMP_LOW) crockValue = 0; digitalWrite(CROCKPOTPIN, crockValue); delay(2000); // DHT's need a delay between readings ...}
4 Using Arduino / Programming Questions / Dht22 on: August 28, 2011, 03:07:34 pm I have the dht22 wired sensor from adafruit conected to digital pin 2. I haven't played with arduino in over a year so any help or guidance from a couple things:1:I'm using this sensor so I can activate a relay going to a crock pot. This is where it'll turn on or off at a certain temperature.2: I'm using the humidty part to know when I need to open a valve controlled by a servo to control the humidty at a certain percentage.3: I have got it working on a LCD 16x2 display temp (f) and humidty. This will be just an extra add on and later I made add a min/max feature.Also out of question (may need to be in hardware) has any interfaced this small relay off eBay that's already on a pcb? Can hardly see the pinout on silkscreen (I bought one).
5 Community / Bar Sport / Getting rid of goodies on: February 28, 2011, 11:46:39 pm 1x arduino duemilanove1x sparkfun 5v ftdi adapter1x adafruit breadboard power supply1x wii nunchuck1x wii nunchuck adapter1x max7219 ic (dip)3x 74hc595 ics (dip)1x glcd 1x usbtinyisp avr programmer2 USB cables (one for arduino, one for ftdi)Want to get rid of in one batch.Pm me.
6 Forum 2005-2010 (read only) / Syntax & Programs / Re: GLCD Wii Tilt Code on: August 10, 2010, 08:59:42 pm @memNo more flickering The DrawLine isn't even at at the center of the Horizontal line.
7 Forum 2005-2010 (read only) / Syntax & Programs / Re: GLCD Wii Tilt Code on: August 08, 2010, 09:42:46 pm I tried and I dont get anything from the display. Nothing in serial monitor. I took out the loop_cnt. Could be why it isn't displaying anything. Maybe I've got it coded wrong... I tried this:Code:#include #include "fonts/Arial14.h" // proportional font#include "fonts/SystemFont5x7.h" // system font#include // initialize wire#include "nunchuck_funcs.h"byte accx;void setup(){ Serial.begin(19200); nunchuck_setpowerpins(); nunchuck_init(); GLCD.Init(NON_INVERTED); GLCD.SelectFont(System5x7);}void loop(){ if( accx >= 75 && accx <= 185) { nunchuck_get_data(); accx = nunchuck_accelx(); Serial.println((byte)accx,DEC); byte y = map(accx, 75, 185, 0, 63); // map returns a value from 0 to 63 for values from 75 to 185 GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 63-y, 127, y); // draw the line } delay(100); // the time in milliseconds between redraws}
8 Forum 2005-2010 (read only) / Syntax & Programs / Re: GLCD Wii Tilt Code on: August 07, 2010, 10:46:59 pm This works surprisingly I have a tiny bit of flicker from 64,32 all the way to 127, 0 its not bad.Code:#include #include "fonts/Arial14.h" // proportional font#include "fonts/SystemFont5x7.h" // system font#include // initialize wire#include "nunchuck_funcs.h"int loop_cnt=0;byte accx;void setup(){ Serial.begin(19200); nunchuck_setpowerpins(); nunchuck_init(); GLCD.Init(NON_INVERTED); GLCD.ClearScreen(); GLCD.SelectFont(System5x7); Serial.print("Wii nunchuck ready"); GLCD.DrawHLine(0, 32, 127);}void loop(){ if( loop_cnt > 100 ) loop_cnt = 0; nunchuck_get_data(); accx = nunchuck_accelx(); Serial.println((byte)accx,DEC); if ((accx <= 185) && (accx > 180)){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 0, 127, 63);}else if(( accx < 180) && (accx >= 175)){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0 , 3, 127, 60); }else if ((accx <= 175) && (accx > 170)){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127);GLCD.DrawLine(0, 6, 127, 57);}else if((accx < 170) && (accx >= 165)){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 9, 127, 54); } else if((accx <= 165) && (accx > 160)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 12, 127, 51); } else if((accx < 160) && (accx >= 155)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 15, 127, 48); } else if((accx <= 155) && (accx > 150)) {GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 18, 127, 45); } else if((accx < 150) && (accx >= 145)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 21, 127, 42); } else if((accx <= 145) && (accx > 140)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 24, 127, 39); } else if((accx < 140) && (accx >= 135)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 27, 127, 36); } else if((accx <= 135) && (accx > 130)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 30, 127, 33); } else if((accx < 130) && (accx >= 125)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 32, 127, 32); } else if((accx <= 125) && (accx > 120)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 34, 127, 31); } else if((accx < 120) && (accx >= 115)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 37, 127, 28); } else if((accx <= 115) && (accx > 110)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 40, 127, 25); } else if((accx < 110) && (accx >= 105)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 43, 127, 22); } else if((accx <= 105) && (accx > 100)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 46, 127, 19); } else if((accx < 100) && (accx >= 95)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 49, 127, 16); } else if((accx <= 95) && (accx > 90)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 52, 127, 13); } else if((accx < 90) && (accx >= 85)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 55, 127, 10); } else if((accx <= 85) && (accx > 80)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 58, 127, 7); } else if ((accx < 80) && (accx >= 75)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 63, 127, 0); } loop_cnt++; delay(1);} I still left the draw line and clear screen in each else if.
9 Forum 2005-2010 (read only) / Syntax & Programs / Re: GLCD Wii Tilt Code on: August 06, 2010, 02:21:58 pm Yes it does compile. Thanks guys for noticing. I was trying to get to if a value is greater than x but less than x do a glcd print. I'll rework it tonight and see what I get.@billI'll look for the example. I noticed in the examples you guys call a void () glcd clear.
10 Forum 2005-2010 (read only) / Syntax & Programs / Re: GLCD Wii Tilt Code on: August 03, 2010, 06:23:24 pm Code:#include #include "fonts/Arial14.h" // proportional font#include "fonts/SystemFont5x7.h" // system font#include // initialize wire#include "nunchuck_funcs.h"int loop_cnt=0;byte accx;void setup(){ Serial.begin(19200); nunchuck_setpowerpins(); nunchuck_init(); GLCD.Init(NON_INVERTED); GLCD.ClearScreen(); GLCD.SelectFont(System5x7); Serial.print("Wii nunchuck ready"); GLCD.DrawHLine(0, 32, 127);}void loop(){ if( loop_cnt > 100 ) { loop_cnt = 0; nunchuck_get_data(); accx = nunchuck_accelx(); Serial.println((byte)accx,DEC); if ((accx > 180) && (accx > 185)){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 0, 127, 63);}else if(( accx < 180) && (accx > 175)){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0 , 3, 127, 60); }else if ((accx < 175) && (accx > 170)){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127);GLCD.DrawLine(0, 6, 127, 57);}else if((accx < 170) && (accx > 165)){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 9, 127, 54); } else if((accx < 165) && (accx > 160)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 12, 127, 51); } else if((accx < 160) && (accx > 155)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 15, 127, 48); } else if((accx < 155) && (accx > 150)) {GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 18, 127, 45); } else if((accx < 150) && (accx > 145)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 21, 127, 42); } else if((accx < 145) && (accx > 140)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 24, 127, 39); } else if((accx < 140) && (accx > 135)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 27, 127, 36); } else if((accx < 130) && (accx > 125)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 32, 127, 32); } else if((accx < 125) && (accx > 120)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 37, 127, 28); } else if((accx < 120) && (accx > 115)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 40, 127, 25); } else if((accx < 115) && (accx > 110)) GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 43, 127, 22); } else if((accx < 110) && (accx > 105)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 46, 127, 19); } else if((accx < 105) && (accx > 100)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 49, 127, 16); } else if((accx < 100) && (accx > 95)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 52, 127, 13); } else if((accx < 95) && (accx > 90)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 55, 127, 10); } else if((accx < 90) && (accx > 85)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 58, 127, 7); } else if((accx < 85) && (accx > 80)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 61, 127, 4); } else if ((accx < 80) && (accx > 75)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 63, 127, 0); } loop_cnt++; delay(1);}Works until tilted to the left and some garbled lines and the serial monitor freezes.Looks like the data 75 to 125 hangs up. I took out 125 to 185 and does the same thing... :oShot a video:
11 Forum 2005-2010 (read only) / Syntax & Programs / GLCD Wii Tilt Code on: August 03, 2010, 04:38:05 pm I started a new thread from this post:http://www.arduino.cc/cgi-bin/yabb2/YaBB.pl?num=1279128237/45#45Code:#include #include "fonts/Arial14.h" // proportional font#include "fonts/SystemFont5x7.h" // system font#include // initialize wire#include "nunchuck_funcs.h"int loop_cnt=0;byte accx;void setup(){ Serial.begin(19200); nunchuck_setpowerpins(); nunchuck_init(); GLCD.Init(NON_INVERTED); GLCD.ClearScreen(); GLCD.SelectFont(System5x7); Serial.print("Wii nunchuck ready"); GLCD.DrawHLine(0, 32, 127);}void loop(){ if( loop_cnt > 100 ) { loop_cnt = 0; nunchuck_get_data(); accx = nunchuck_accelx(); Serial.println((byte)accx,DEC); if ( accx > 180){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 63, 127, 0);}else if ( accx > 130){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0 , 32, 64, 0); }else if ( accx < 130){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127);GLCD.DrawLine(0, 0, 127, 0); } } loop_cnt++; delay(1);}It's working but I need to get it fine tuned.I prefer the math method as its easiest for me to use.Based on tilt detection from a value of 73 to 185 I'm trying to make a tilt gauge. I could use alot of the else if statements but thats going to be alot of code .Whenever read 73 to 80 id implement a GLCD.DrawLine(0, 63, 127, 0)from 80 to 85 id do a GLCD.DrawLine(0, 58, 127, 5) all the way till I get past a value of 130 and it'd go the oppposite way.Edit: New code working goodCode:#include #include "fonts/Arial14.h" // proportional font#include "fonts/SystemFont5x7.h" // system font#include // initialize wire#include "nunchuck_funcs.h"int loop_cnt=0;byte accx;void setup(){ Serial.begin(19200); nunchuck_setpowerpins(); nunchuck_init(); GLCD.Init(NON_INVERTED); GLCD.ClearScreen(); GLCD.SelectFont(System5x7); Serial.print("Wii nunchuck ready"); GLCD.DrawHLine(0, 32, 127);}void loop(){ if( loop_cnt > 100 ) { loop_cnt = 0; nunchuck_get_data(); accx = nunchuck_accelx(); Serial.println((byte)accx,DEC); if ( accx > 180){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 63, 127, 0);}else if(( accx < 179) && (accx > 170)){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0 , 59, 127, 5); }else if ((accx < 169) && (accx > 160)){GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127);GLCD.DrawLine(0, 54, 127, 10);}else if((accx < 159) && (accx > 150)){ GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 49, 127, 15); } else if((accx < 149) && (accx > 140)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 45, 127, 20); } else if((accx < 139) && (accx > 130)) { GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 41, 127, 25); } else if((accx < 129) && (accx > 120)) {GLCD.ClearScreen(); GLCD.DrawHLine(0, 32, 127); GLCD.DrawLine(0, 32, 127, 32); } } loop_cnt++; delay(1);}
12 Forum 2005-2010 (read only) / Syntax & Programs / Re: Trouble with code on: March 09, 2010, 04:37:18 pm I was missing a }
13 Forum 2005-2010 (read only) / Syntax & Programs / Re: Trouble with code on: March 09, 2010, 04:26:02 pm Yes it uploads fine... Problem is when hits button 4 it wont physically do whats in the code...:Code:if (lightMode == 4) { analogWrite(redPin, 255); delay(500); analogWrite(greenPin, 255); delay(500); analogWrite(redPin, 100); analogWrite(greenPin, 180); delay(500); analogWrite(redPin, 0); analogWrite(greenPin, 255); analogWrite(bluePin, 255); delay(500); analogWrite(redPin, 0); analogWrite(bluePin, 100); analogWrite(greenPin, 100); delay(500); analogWrite(redPin, 255); analogWrite(greenPin, 255); analogWrite(bluePin, 0); delay(500); analogWrite(redPin, 0); analogWrite(greenPin, 170); analogWrite(bluePin, 80);
14 Forum 2005-2010 (read only) / Syntax & Programs / Trouble with code on: March 09, 2010, 03:09:37 pm Code:// RGB Lightint switchPin = 2; int redPin = 5;int greenPin = 6;int bluePin = 3;int val; int val2; int buttonState; int lightMode = 0; void setup() { pinMode(switchPin, INPUT); pinMode(redPin, OUTPUT); pinMode(greenPin, OUTPUT); pinMode(bluePin, OUTPUT); Serial.begin(9600); buttonState = digitalRead(switchPin); }void loop(){ val = digitalRead(switchPin); delay(10); val2 = digitalRead(switchPin); if (val == val2) { if (val != buttonState) { if (val == LOW) { if (lightMode == 0) { lightMode = 1; } else { if (lightMode == 1) { lightMode = 2; } else { if (lightMode == 2) { lightMode = 3; } else { if (lightMode == 3) { lightMode = 4; } else { lightMode = 0; } } } } } } buttonState = val; } // if (lightMode == 0) {analogWrite(redPin, 0);analogWrite(greenPin, 0);analogWrite(bluePin, 0); } if (lightMode == 1) { analogWrite(redPin, 255); analogWrite(bluePin, 255); analogWrite(greenPin, 0); } if (lightMode == 2) { analogWrite(redPin, 255); analogWrite(bluePin, 0); analogWrite(greenPin, 255); } if (lightMode == 3) { analogWrite(redPin, 0); analogWrite(greenPin, 255); analogWrite(bluePin, 255); if (lightMode == 4) { analogWrite(redPin, 255); delay(500); analogWrite(greenPin, 255); delay(500); analogWrite(redPin, 100); analogWrite(greenPin, 180); delay(500); analogWrite(redPin, 0); analogWrite(greenPin, 255); analogWrite(bluePin, 255); delay(500); analogWrite(redPin, 0); analogWrite(bluePin, 100); analogWrite(greenPin, 100); delay(500); analogWrite(redPin, 255); analogWrite(greenPin, 255); analogWrite(bluePin, 0); delay(500); analogWrite(redPin, 0); analogWrite(greenPin, 170); analogWrite(bluePin, 80); } }} Works fine till I press the button the 4th time to the delayed analogWrite.I adapted from lady ada's code.
15 Forum 2005-2010 (read only) / Syntax & Programs / Re: Map() Function - am i missing something? on: February 28, 2010, 10:52:01 pm Quotedo you think changing the D9 to D1 would help?give it a try see what happens
Pages: [1] 2 3 ... 28 | 7,241 | 18,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-23 | longest | en | 0.772908 |
https://www.livecasinodirect.com/martingale-betting-system-how-to-use-it-when-playing-at-online-casinos | 1,660,635,459,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572221.38/warc/CC-MAIN-20220816060335-20220816090335-00015.warc.gz | 736,338,467 | 13,987 | # Martingale betting system: how to use it when playing at online casinos
The Martingale method is one of the most well-liked wagering systems in both live and virtual casinos. The Martingale betting technique has several glaring drawbacks and hazards, even though many best casino bonus Canada fans swear by its effectiveness. To avoid putting any cash in danger, it is essential to be aware of all the benefits and drawbacks of this technique and the most acceptable advice for utilising it effectively. This manual will assist you in fully comprehending the Martingale system. Additionally, it will provide a response to the crucial query, “Can this tactic help you win?” Discover more by reading on.
## The Martingale System
A wagering method with a very easy-to-understand notion is the Martingale system. The fundamental idea is to boost your wager each time you lose. This is done in order to receive your cashback as fast as you win, regardless of how many consecutive deficits you have. You can begin with the starting wager size as you did previously after only one win to replenish your money. This appears to be utterly rational on the surface because it is simple to grasp and doesn’t require a lot of focus to follow. Additionally, you don’t have to be an expert in mathematics to calculate and determine how much to wager in each round.
## What the System Entails
Every wager you make must result in a 50/50 conclusion for the Martingale strategy to be successful. Doing this is the only way to guarantee that you can make up your deficits after a win. Nevertheless, the first problem with the Martingale wagering strategy manifests itself here. No wager, including those that are marketed as even cash, is truly 50-50 since the facility has a house advantage in every game you play. There is always a tiny possibility that you may lose everything. For instance, there is always a probability that the ball will land on zero if you play roulette and only put even cash wagers like odds/evens, red/black, or high/low. There is no way of winning except if you wager explicitly on zero, as zero is not a component of either of these even cash bets.
This is one of the rule factors why the Martingale approach is a fantastic one for quick gains but a bad one for long-term success. You might readily argue that this is not a wise approach to wager over the long run because you’re wagering big to gain small, and you risk losing a great deal of income. These factors make the Martingale system one of the highest-risk wagering strategies you can employ. It’s another reason why a lot of seasoned bettors never use it.
## How the Martingale System Operates
Now that we have a theoretical understanding of how the Martingale wagering system functions, let’s move on to some real-world instances. As we’ve already said, the Martingale gambling method is most effectively applied when just even cash wagers are made. The best illustration of it is roulette, which gives you a number of even cash wagers. It follows a relatively straightforward trajectory that never changes.
### 1. Grand Martingale System of Gambling
The gambling restrictions on your money will determine how high you may scale this series. You can see how the bets might go along fast by concentrating on the thirteen digits we specified before. Due to this, we usually advise you to place bets with a minimal sum. This should ideally be the lowest wager for the gambling game you are enjoying. Therefore, if you place a \$1 chance at the beginning, you should continue to do so as long as you are winning. You increase your wager to \$2 after you lose. You keep going through the steps upwards and raising your stakes till you win if you keep failing. When you win, go back to your original wager and begin again. This will enable you to continually gain back whatever cash you lose while increasing your overall winnings.
### 2. The Mini Martingale
The system restricts how frequently you double your wagers, thereby capping your winnings while also assisting you in preventing significant losses. However, the Grand Martingale wagering strategy is virtually identical to the traditional Martingale strategy. The only distinction is that you add a new stake unit each time you fail. This implies that while this variety is costlier than the first, it also recovers deficits faster.
## The Anti-Martingale: What Is It?
Some gamers attempt to go against the trend and create a wagering technique that is entirely contrary as soon as a gaming method finds popularity in the wagering market. This is the Reverse Martingale wagering strategy, also known as the anti-Martingale gambling strategy. This approach is straightforward because it essentially promotes extending hot streaks and using a stop-loss strategy to avert significant losses during protracted losing streaks. This makes it far less dangerous than the standard Martingale wagering strategy. As the name suggests, the anti-Martingale method involves half your wager whenever you lose a stake and increasing it if you win one.
As an illustration, if you spend \$10 on an even roulette play and succeed, you would stake \$20 on the next spin. You might wager \$40 if you won again. If the next spin is a loss, you would go back one step and place another \$20 wager. There are a few more significant iterations of this strategy in addition to the anti-Martingale gambling strategy. The Mini Martingale and the Grand Martingale are the top renowned well variations.
## The Martingale Betting System’s Drawbacks
The dangerous nature of the Martingale betting strategy has already been addressed. We should go through the drawbacks of this betting strategy in greater detail to better appreciate how risky it might be for your account. Of course, the biggest problem with the Martingale technique is how rapidly it may deplete your budget. Before you know what’s going on, you might squander your real money if you get into a rut. This applies particularly if you don’t have a big budget and must exercise caution while placing bets. It’s advised to steer clear of using the Martingale wagering strategy in this situation.
To put this into perspective, imagine that you have a \$1,000 gambling budget and wish to use the Martingale strategy to place bets while enjoying roulette or blackjack. In this case, you have a little longer winless skid where you lose consecutive spins. The Martingale System’s Merits and Drawbacks
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### 1-3: Subtracting Rational Numbers
#### Key Concept
You subtract a fraction or a decimal the same way you subtract an integer.
For any rational numbers a and b:
ab = a + (− b)
a − (− b) = a + b
table with 4 rows and 3 columns , row1 column 1 , cap writesubtractionasaddingtheopposite. , column 2 minus , 3 fifths , minus , 4 fifths , column 3 equals negative , 3 fifths , plus . open , negative , 4 fifths , close , row2 column 1 , cap addnumberswiththesamesign. , column 2 , column 3 equals negative , 7 fifths , row3 column 1 , cap writesubtractionasaddingtheopposite. , column 2 negative 9.1 minus . open , negative 7.6 , close , column 3 equals negative 9.1 plus 7.6 , row4 column 1 , cap addnumberswithdifferentsigns. , column 2 , column 3 equals negative 1.5 , end table
#### Part 1
Example Predicting The Sign of Differences of Rational Numbers
Determine if each difference is negative, zero, or positive.
• a. 1 half , minus . open , negative , 1 fourth , close
• b. negative 5 minus . open . negative 3 , and 3 fourths . close
• c. negative 4.7 minus . open , negative 4.7 , close
• d. 5.7 minus 8 , and 1 half
• e. − 2.6 − (− 3.1)
• f. negative 3 , and 1 half , minus . open . negative 3 , and 1 half . close
• g. negative , 4 thirds , minus . open , negative , 2 thirds , close
• h. negative 0.8 minus 0.8
Solution
Rewrite the subtraction as adding the opposite. Then think about the relationship of the addends.
• a.
• b.
• c.
• d.
• e.
End ofPage 17 | 480 | 1,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-45 | longest | en | 0.481862 |
https://www.physicsforums.com/threads/simple-question.161263/ | 1,477,010,524,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717959.91/warc/CC-MAIN-20161020183837-00048-ip-10-142-188-19.ec2.internal.warc.gz | 825,703,263 | 15,013 | # Simple question
1. Mar 17, 2007
### ak416
does a set with empty interior have measure zero? I think it does...
2. Mar 17, 2007
Doesn't the Cantor set have empty interior?
3. Mar 18, 2007
### ak416
ya and I read somewhere that it has measure zero. So this wouldnt be a counterexample.
4. Mar 18, 2007
### ak416
ok i may be wrong on this...
5. Mar 18, 2007
### JasonRox
The Cantor set does have an empty interior and also has measure zero.
I never worked with measurable spaces yet, but if I had to guess, I would say a set with empty interior does in fact have measure zero.
6. Mar 18, 2007
### ak416
no actually it says in my book that the standard middle thirds cantor set has measure zero, but not any cantor set. So I guess its incorrect.
7. Mar 18, 2007
### mathwonk
come on guys, look at the set of irrational numbers betwen 0 and 1. what is the interior? what is the measure? | 261 | 904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-44 | longest | en | 0.957592 |
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#### Resources tagged with Interactivities similar to Mystery Matrix:
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### Combining Cuisenaire
##### Age 7 to 11 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
##### Age 5 to 11 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### A Square of Numbers
##### Age 7 to 11 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Difference
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### One to Fifteen
##### Age 7 to 11 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
##### Age 7 to 11 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
##### Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### More Transformations on a Pegboard
##### Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Coded Hundred Square
##### Age 7 to 11 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Which Symbol?
##### Age 7 to 11 Challenge Level:
Choose a symbol to put into the number sentence.
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Winning the Lottery
##### Age 7 to 11 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
### Junior Frogs
##### Age 5 to 11 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Times Tables Shifts
##### Age 7 to 11 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### First Connect Three for Two
##### Age 7 to 11 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Tetrafit
##### Age 7 to 11 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Multiplication Square Jigsaw
##### Age 7 to 11 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### Teddy Town
##### Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Colour Wheels
##### Age 7 to 11 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Triangles All Around
##### Age 7 to 11 Challenge Level:
Can you find all the different triangles on these peg boards, and find their angles?
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Cuisenaire Environment
##### Age 5 to 11 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Countdown
##### Age 7 to 14 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Four Triangles Puzzle
##### Age 5 to 11 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Fault-free Rectangles
##### Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Have You Got It?
##### Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Venn Diagrams
##### Age 5 to 11 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Sorting Symmetries
##### Age 7 to 11 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Fractions and Coins Game
##### Age 7 to 11 Challenge Level:
Work out the fractions to match the cards with the same amount of money.
### Train for Two
##### Age 7 to 11 Challenge Level:
Train game for an adult and child. Who will be the first to make the train?
### Triangle Pin-down
##### Age 7 to 11 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Rod Ratios
##### Age 7 to 11 Challenge Level:
Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6?
### Spot Thirteen
##### Age 7 to 11 Challenge Level:
Choose 13 spots on the grid. Can you work out the scoring system? What is the maximum possible score?
### Coordinate Tan
##### Age 7 to 11 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### Light the Lights Again
##### Age 7 to 11 Challenge Level:
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
### Take the Right Angle
##### Age 7 to 11 Challenge Level:
How many times in twelve hours do the hands of a clock form a right angle? Use the interactivity to check your answers.
### 100 Percent
##### Age 7 to 11 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Noughts and Crosses
##### Age 7 to 11 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
### Building Stars
##### Age 7 to 11 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
### Calculator Bingo
##### Age 7 to 11 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Dominoes Environment
##### Age 5 to 11 Challenge Level:
These interactive dominoes can be dragged around the screen.
### Shapely Tiling
##### Age 7 to 11 Challenge Level:
Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle? | 2,132 | 9,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-39 | latest | en | 0.874361 |
https://www.dummies.com/article/academics-the-arts/math/pre-calculus/trigonometry-proofs-and-pythagorean-identities-167821/ | 1,716,778,056,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00544.warc.gz | 639,706,091 | 14,691 | ##### Trigonometry Workbook For Dummies
The Pythagorean identities pop up frequently in trig proofs. Pay attention and look for trig functions being squared. Try changing them to a Pythagorean identity and see whether anything interesting happens.
The three Pythagorean identities are
After you change all trig terms in the expression to sines and cosines, the proof simplifies and makes your job that much easier. For example, follow these steps to prove
1. Convert all the functions in the equality to sines and cosines.
2. Use the properties of fractions to simplify.
Dividing by a fraction is the same as multiplying by its reciprocal, so
3. Identify the Pythagorean identity on the left side of the equality.
Because sin2 x + cos2 x = 1, you can say that 1 = 1. | 173 | 774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.898052 |
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Driving 1.5 times slower, Bill was late for school today.
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Driving 1.5 times slower, Bill was late for school today. [#permalink]
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18 Sep 2008, 20:53
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Driving 1.5 times slower, Bill was late for school today. What is the usual time it takes Bill to drive to school? (Assume that each day Bill takes the same route).
(1) It took Bill 15 more minutes to drive to school today than usually
(2) The distance between home and school is 15 miles
This question may be the end of me, but if you go 1.5 times slower aren't you going a negative speed? For example, if you are going 10 mph, and you drive 1.5 times slower (10 * 1.5 = 15) ---> 10 - 15 = -5, doesn't this mean you are going -5 mph?
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19 Sep 2008, 00:02
Hi rate cannot be negative. it can be direction. Let's say it 10 mph then it's 1.5 times down is 10 /1.5 .
I think if we combine both stats then u have 1.5 x 15/x = 15/x + 15/60 where x is normal speed mph.
so u get x and then usual time for 15 miles
so Ans is C
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19 Sep 2008, 00:19
1.5 times is supposed to be multiplied and not divided right ? How can something be 1.5 times slow ? do we have to multiply by -1.5 ? What am I missing here ?
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19 Sep 2008, 01:19
1.5 times low doesnt mean multiply by -1.5 times.
if rate is 1.5 less means (x * 1.5) = 10 so x = 10 /1.5
Hope it clarifies..
cheers
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Joined: 19 Sep 2008
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19 Sep 2008, 01:42
The rate is absolutely possible, it just meens that the speed was 1.5 time less than Bill's everyday speed. Find below my logic:
Actually we don't need to know the distance to find how much time it takes to Bill to get to the school, so we can assume that distance is "1", then we get the following equation: 1/x=1/x/1.5-15/60, where x - his normal speed, 1/x - usual time he spends everyday to get to the school, 1/x/1.5 - this particular day time, 15/60 - minutes in hours, thus we have:
1/x=3/2x-0.25 => x = 2
The second statement doesn't give us anything, so I think the asnwer is "A".
PS: Sorry for my English
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19 Sep 2008, 08:56
rakeshmrana wrote:
1.5 times low doesnt mean multiply by -1.5 times.
if rate is 1.5 less means (x * 1.5) = 10 so x = 10 /1.5
Hope it clarifies..
cheers
But according to your formula, which is (old rate)/(factor) = new rate ---> new rate = 10/1.5, if the question were to state that "if you are going 10 mph, and you drive 50% slower" then 10/0.5 = 20 --> you would be going faster not slower. How does this make sense?
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19 Sep 2008, 09:17
Basically the distance is the same, so:
Distance = rate x time
rate1 x time1 = rate1 x 1.5 x time2
We have an equation with three variables, but rate1 can be cancelled out. Using Statement 1 you can get an equation with only one variable. Answer should be A.
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19 Sep 2008, 09:35
Nerdboy wrote:
Basically the distance is the same, so:
Distance = rate x time
rate1 x time1 = rate1 x 1.5 x time2
We have an equation with three variables, but rate1 can be cancelled out. Using Statement 1 you can get an equation with only one variable. Answer should be A.
Ahh, that makes sense. So, in a way, the new time is 1.5 times as long. This is easier to think of then trying to manipulate the rate. Thanks.
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19 Sep 2008, 10:28
Nerdboy wrote:
Basically the distance is the same, so:
Distance = rate x time
rate1 x time1 = rate1 x 1.5 x time2
We have an equation with three variables, but rate1 can be cancelled out. Using Statement 1 you can get an equation with only one variable. Answer should be A.
Get it now !!!!
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05 Nov 2008, 01:27
Nerdboy - you still have 3 variables left as t1 and t2 are not the same? Would you please explain by solving the equation completely, considering it a problem solving question. Thanks
Re: possibly impossible rate [#permalink] 05 Nov 2008, 01:27
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Driving 1.5 times slower, Bill was late for school today.
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# Pilot Operated Safety Valves in Liquid Systems
## Calculate pilot operated relief valves in liquid systems.
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The minimum discharge area of a pilot operated relief safety valve in a liquid system can be calculated as
A = q SG1/2 / (36.81 Kvisc dp1/2) (1)
where
A = discharge area (Square Inches)
q = relieving capacity (Gallons per Minute)
SG = Specific Gravity of the fluid
Kvisc = correction factor due to velocity - 1.0 for most water systems
dp = differential pressure - set pressure (psig) + over pressure (psig) - back pressure (psig)
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. | 802 | 3,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-50 | longest | en | 0.781342 |
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# Daily Math Review KINDERGARTEN Quarter 3
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Daily Math Review for Kindergarten Quarter 3 includes 5 questions each day. Also included is a Problem of the Week to use in a math journal or on its own.
This is a great way to get little minds warmed up for math!
Reviews Common Core Math Standards including:
K.OA.1 Represent addition and subtraction with objects, fingers, mental images, and drawings.
K.OA.2 Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
K.OA.3 Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation.
K.OA.4 For any number from 1 to 9, find the number that makes 10 when added to the given number.
K.OA.5 Fluently add and subtract within 5.
9 weeks are included and each week also has a problem of the week and encourages students to draw a picture and write a complete sentence to solve the problem.
All 4 Quarters are bundled! Save 25% when you buy all 4!
Daily Math Review KINDERGARTEN BUNDLE
I have a full year of 1st grade also available. Tell your 1st grade friends!
Daily Math Review 1st Grade Quarter 1
Daily Math Review 1st Grade Quarter 2
Daily Math Review 1st Grade Quarter 3
Daily Math Review 1st Grade Quarter 4
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Sign up | 567 | 2,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-05 | latest | en | 0.87421 |
http://english.links123.com/tingli/meiyin/2717.html | 1,527,127,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00284.warc.gz | 96,241,215 | 7,396 | How To Play Chess
Question
What is the most important chess piece?
Chess is played on a square board which is divided into 64 squares of alternating colours. By convention these are referred to as light and dark squares.
A chess set consists of two identical groups of pieces. One is white and one black. One player plays with the white pieces, the other with the black. Each player has: one king, one queen, two bishops, two knights, two rooks, and eight pawns. White always makes the first move in any game.
The object of chess is to capture your opponent's king. In practice, the king is never actually captured, but trapped so that he cannot move without being taken. This is known as checkmate, and is the end of the game.
The chess pieces are lined up at either end of the board in predetermined places. They are configured as follows:The queen is always placed on her own colour, nearest the middle of the row. Next to her, in the middle, is the king. The king and queen are flanked by their bishops. Next come the knights and finally the rooks. The pawns are lined up in front of the other pieces, on the second row in.
KingThe king is the most important piece in the game. He can move one square in any direction.QueenThe queen is the most powerful piece in the game. She can move any number of spaces in any direction - horizontally, vertically, or diagonally. Bishops Bishops can only move diagonally, but they can move any number of spaces in one move.KnightsThe knight is the only piece that can 'jump' over other pieces. He moves one space horizontally and two spaces vertically, or two spaces horizontally and one space vertically.Rooks Rooks can move any number of spaces in one move, but they can only move horizontally or vertically.When any of these pieces land on a square occupied by an opponent's piece, the player "takes" that piece, removing it from the board and out of play.Pawns Pawns have three different standard moves. The first time they are moved, they can move either one or two spaces forward. For the rest of the game they can only move one space forward at a time. To take an opponent's piece, the pawn must move one space diagonally forwards. They cannot take by moving vertically forwards, nor can they move backwards.
PromotionIf a pawn reaches the other side of the board, it can be promoted to the rank of queen, bishop, knight or rook. It will then be able to move in the same way as that piece.En PassantIn special circumstances, a pawn can also take en passant - by passing another pawn. The first time a player moves their pawn they may move it two squares vertically. If, when he does this, there is an opposing pawn on an adjacent square, the opponent can take the first pawn by moving diagonally into the square passed over by the first pawn's double-step. If a player decides to take En passant, he must do so immediately following the double step move.CastlingUnder certain conditions, a king and a rook may move simultaneously in a move known as castling. This is a move to a strong defensive position, and is worth making if you can. In order to castle, the following conditions must be met: Neither the king nor the rook have yet moved in the game; there are no pieces in between the rook and the king; the king is not moving into check; none of the opponent's pieces are able to attack any of the squares in between the king and the rook. Provided these conditions are met, the castling move may take place. The king moves two spaces towards the rook, and the rook moves over the king to the next square.
Touching pieces[/steph | 787 | 3,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-22 | latest | en | 0.965769 |
https://arc.nesa.nsw.edu.au/go/stage-2/maths/activities/classroom-map | 1,686,307,534,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00429.warc.gz | 123,963,036 | 6,046 | NESA is regularly updating its advice as the coronavirus outbreak unfolds. Get our latest COVID-19 advice
This webpage has been archived to prepare for transfer to the new NESA website. Reference to syllabus outcomes and content on this webpage may not be current. Teachers are encouraged to visit the Key Learning Area page for recent student work samples on the NESA website.
Assessment Resource Centre (ARC)
Home
1. Stage 2
2. Mathematics
3. Activities
4. Classroom map
## Classroom map
Midway through Stage 2 (end of Year 3)
New Work Samples
Tai
Cameron
Morgan
Casey
Kendall
Marley
Jules
Mackenzie
Indra
## Foundation Statement strands
The following strands are covered in this activity:
## Description of activity
Students draw a simple map of the classroom on grid paper or create their own grid for the map. They plot coordinates on the map and include a key. Students describe the location of items in the classroom using coordinates. They describe how to get from one point to another on the map.
## Suggested materials
Paper, pencils, grid paper
## Prior Learning
Students have represented the position of objects using models and drawings. They have described the position of objects using everyday language such as ‘left’ and ‘right’. Students have used simple maps and grids to represent position and follow routes. They have described the location of an object on a simple map using coordinates or directions.
Board of Studies NSW, Mathematics K–6 Sample Units of Work, p 119
## Outcomes
Position (SGS2.3)
Uses simple maps and grids to represent position and follow routes.
Communicating (WMS2.3)
Uses appropriate terminology to describe, and symbols to represent, mathematical ideas.
## Criteria for assessing learning
Students will be assessed on their ability to:
• construct a simple map
• describe the position of items using coordinates
• describe how to get from one point to another on a map. | 414 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.881829 |
https://stats.libretexts.org/Bookshelves/Introductory_Statistics/OpenIntro_Statistics_(Diez_et_al)./02%3A_Probability | 1,726,856,983,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00109.warc.gz | 487,468,018 | 30,063 | # 2: Probability
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
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$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
Probability forms a foundation for statistics. You might already be familiar with many aspects of probability, however, formalization of the concepts is new for most. This chapter aims to introduce probability on familiar terms using processes most people have seen before.
This page titled 2: Probability is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by David Diez, Christopher Barr, & Mine Çetinkaya-Rundel via source content that was edited to the style and standards of the LibreTexts platform. | 1,871 | 4,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.195464 |
https://gis.stackexchange.com/questions/155629/nnjoin-mmqgis-and-nearest-neighbor-qgis-failed-to-measure-shortest-distance | 1,722,897,786,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00615.warc.gz | 226,811,372 | 40,866 | # NNJoin, MMQGIS and Nearest Neighbor QGIS failed to measure shortest distance
I have two layers, one point and one line. I need the distance to the closest line from the point layer. Not NNJoin, MMQGIS nor the distance hub in QGIS provided accurate measures. Any ideas?
• Just to get it right: You want for each point the shortest distance to some line? You are not looking for the distance to vertices but to some position along the lines? Commented Jul 24, 2015 at 6:17
• Yes. I ended up using the measure tool for each point to the nearest line. We wanted to know what stream was the closest to a camera trap site for a wildlife study. Commented Jul 25, 2015 at 15:56
If you have to measure the distances once again, the following Python code may save you some time.
Have your points and lines loaded as map layers. Change the names 'point' and 'line' in line 4 and 5 resp. to the names of your layers. Copy the code into the Python console.
The function iterates over all points and finds the closest segment of all lines, and writes the closest position along the segment to a list. Then takes the position with the minimum distance to a points and builds a connecting line. The length of the line goes to the field distance.
``````d_lyr = QgsVectorLayer('LineString?field=distance:float', 'dist', 'memory')
p_lyr = QgsMapLayerRegistry.instance().mapLayersByName('point')[0]
l_lyr = QgsMapLayerRegistry.instance().mapLayersByName('line')[0]
d_lyr = QgsMapLayerRegistry.instance().mapLayersByName('dist')[0]
prov = d_lyr.dataProvider()
feats = []
for p in p_lyr.getFeatures():
minDistPoint = min([l.geometry().closestSegmentWithContext(QgsPoint(p.geometry().asPoint())) for l in l_lyr.getFeatures()])[1]
feat = QgsFeature()
feat.setGeometry(QgsGeometry.fromPolyline([QgsPoint(p.geometry().asPoint()), QgsPoint(minDistPoint[0], minDistPoint[1])]))
feat.setAttributes([feat.geometry().length()])
feats.append(feat)
d_lyr.updateExtents()
d_lyr.triggerRepaint()
``````
To test the code initiate 2 memory layers, a point layer named 'point' and a line layer named 'line', digitize some features, and run the code.
An example:
• I tried out your answer and it works perfectly. +1 Commented Nov 8, 2015 at 11:30
The NNJoin plugin uses QgsGeometry.distance, which in turn uses GEOS. It should return the minimum distance:
Returns the minimum distanace between this geometry and another geometry, using GEOS.
So I am a bit surprised that NNJoin does not work as expected (as long as you do not choose to use approximate geometries).
Are you using a non-projected coordinate system? Geos does not support that:
GEOS distance calculations are linear – in other words, GEOS does not perform a spherical calculation even if the SRID specifies a geographic coordinate system.
If you are not using a non-projected coordinate system, it would be interesting to see example data sets. You could also open an issue on the NNJoin issue tracker.
• I went back and re-ran the NNJoin and it worked. It gave measurement close to the hand calculated ones. Not sure what I did wrong. The data is projected, and I did not use approximate geometries, today. Maybe that was the problem. Thank you. Commented Aug 5, 2015 at 21:27 | 780 | 3,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.789669 |
http://mizar.uwb.edu.pl/version/current/html/proofs/flang_3/19 | 1,568,806,229,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00341.warc.gz | 124,494,256 | 2,120 | let E be set ; :: thesis: for A being Subset of ()
for m, n being Nat st n > 0 holds
(A |^.. m) |^ n = A |^.. (m * n)
let A be Subset of (); :: thesis: for m, n being Nat st n > 0 holds
(A |^.. m) |^ n = A |^.. (m * n)
let m, n be Nat; :: thesis: ( n > 0 implies (A |^.. m) |^ n = A |^.. (m * n) )
defpred S1[ Nat] means ( \$1 > 0 implies (A |^.. m) |^ \$1 = A |^.. (m * \$1) );
A1: now :: thesis: for n being Nat st S1[n] holds
S1[n + 1]
let n be Nat; :: thesis: ( S1[n] implies S1[n + 1] )
assume A2: S1[n] ; :: thesis: S1[n + 1]
now :: thesis: ( n + 1 > 0 implies (A |^.. m) |^ (n + 1) = A |^.. (m * (n + 1)) )
assume n + 1 > 0 ; :: thesis: (A |^.. m) |^ (n + 1) = A |^.. (m * (n + 1))
per cases ( n = 0 or n > 0 ) ;
suppose n = 0 ; :: thesis: (A |^.. m) |^ (n + 1) = A |^.. (m * (n + 1))
hence (A |^.. m) |^ (n + 1) = A |^.. (m * (n + 1)) by FLANG_1:25; :: thesis: verum
end;
suppose n > 0 ; :: thesis: (A |^.. m) |^ (n + 1) = A |^.. (m * (n + 1))
hence (A |^.. m) |^ (n + 1) = (A |^.. (m * n)) ^^ (A |^.. m) by
.= A |^.. ((m * n) + m) by Th18
.= A |^.. (m * (n + 1)) ;
:: thesis: verum
end;
end;
end;
hence S1[n + 1] ; :: thesis: verum
end;
A3: S1[ 0 ] ;
for n being Nat holds S1[n] from NAT_1:sch 2(A3, A1);
hence ( n > 0 implies (A |^.. m) |^ n = A |^.. (m * n) ) ; :: thesis: verum | 620 | 1,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-39 | latest | en | 0.635078 |
https://www.physicsforums.com/threads/special-relativity-and-force.428186/ | 1,544,882,789,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.91/warc/CC-MAIN-20181215131038-20181215153038-00141.warc.gz | 998,403,850 | 14,611 | # Homework Help: Special Relativity and Force
1. Sep 11, 2010
### StephenD420
Hello everyone
I was wondering why in special relativity that the force when the velocity is parallel to the force the force = dP/dt= $$\gamma$$^3 ma, whereas when the velocity is perpendicular to the force the force = dP/dt= $$\gamma$$ ma. Why is this? Is it because when the velocity is parallel to the force both the velocity of the gamma and v in the momentum, $$\gamma$$mv, changes, thus the product rule applies, whereas when the velocity is perpendicular only the velocity in the momentum changes not the velocity in the gamma, thus the product rule does not apply?
Thanks for the help in explain this.
Stephen
2. Sep 12, 2010
### StephenD420
bump...
3. Sep 12, 2010
### hikaru1221
Well you're right When force is perpendicular to velocity or momentum, it doesn't change the MAGNITUDE of momentum and thus speed. Because the gamma coefficient only depends on speed (or magnitude of velocity), in that case, gamma doesn't change. But in the case where force is parallel to velocity, it changes both direction and magnitude of velocity, so gamma also changes.
4. Sep 12, 2010
### collinsmark
Hello Stephen
First, I plead with you to be careful with your terminology. Three momentum is
$$\vec p = \gamma m \vec v$$
Gamma is not multiplied by the momentum, gamma actually is part of the momentum! In special relativity, that's a pretty important point.
Semantics/terminology aside, I believe you are asking whether you can simply treat gamma as a constant and simply bring it out of the time derivative. Here is my answer: I wouldn't do that unless you are prepared to back it up with some math to show that it is a constant.* It's not intuitively obvious that gamma remains constant when the 3-force is perpendicular to the 3-velocity. Generally speaking, The v in γmv is the same v as the v within γ itself (unless you work with components individually, but even then, the component in question is still within γ).
But there is an easier way to show that f = γma if the force is perpendicular to the velocity. This is the route I suggest taking if you are ever quizzed on this. It's the easy way. Note that,
$$\vec v = (v_x, v_y, v_z)$$
Let's assume that the object is moving along the x-axis only (vy and vz are zero), and the 3-force is applied in the y direction. Now evaluate
$$F_y = m \frac{d}{dt}(\gamma v_y) \left|_{v_y = 0, v_z = 0}$$
But be careful, make sure you do the evaluation after you take the derivative. This is very important. Just like in Newtonian mechanics, just because the component's velocity happens to be zero doesn't mean the component's acceleration happens to be zero! However, once you come across a vy that is outside any derivative, then it is okay to to zero that whole term. Go ahead and do this (using the product rule), and you'll immediately see why F = γma if the force is perpendicular to the velocity. That's the easy way.
*Now, as hikaru1221 points out, gamma is only dependent on speed, so a 3-force perpendicular to the 3-velocity should not change gamma. But you should be prepared to back that up mathematically if you ever use that (or risked getting "dinged" by your instructor, who may not have gone through the math himself/herself). But it can be done (I proved it to myself last night on a paper napkin). I can't give you the full thing but here is a hint on how to do it.
If we use the units c = 1, then gamma in longhand form is
$$\frac{1}{\sqrt{1 - (v_x^2 + v_y^2 + v_z^2)}}$$
Now solve for $d(\gamma)/dt [/tex], with the nonzero velocity only on the x-axis (but again, be careful -- don't do any evaluations until after taking the derivative!), and noting that the dot product between perpendicular vectors is zero. Then point out that if [itex] \dot \gamma$ is zero, it implies that $\gamma$ is a constant.
Last edited: Sep 12, 2010
5. Sep 12, 2010
### hikaru1221
Hurray to collinsmark for a very detailed, helpful post
@to the OP: You can learn a lot besides your particular problem from collinsmark's post | 1,022 | 4,076 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-51 | latest | en | 0.919229 |
https://www.askiitians.com/iit-jee-coordinate-geometry/propositions-of-an-ellipse.aspx | 1,725,752,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00559.warc.gz | 641,597,956 | 45,063 | # Propositions on Ellipse
### Table of Content
The standard equation of an ellipse referred to its principal axes along the coordiante axes is x2/a2 + y2/b2 =1, where
a > b and b2 = a2(1-e2), where e = eccentricity lies between 0 and 1, i.e. 0 < e < 1.
It has two focii S(ae, 0),and S’(-ae, 0).
## Position of a point with respect to Ellipse
The point P(x1, y1) lies outside, inside or on the ellipse according as x12/a2 + y12/b2 – 1 >, < or equal to 0.
## Auxiliary circle of an Ellipse
Auxiliary circle of an ellipse which is a circle described on the major axis of an ellipse as its diameter.
Let the ellipse be
x2/a2 + y2/b2 =1 ...... (1)
Then the equation of its auxiliary circle is x2 + y2 = a2 ...... (2)
Take a point P(x1, y1) on (1).
Through P, draw a line perpendicular to major axis intersecting major axis in N and auxiliary circle in P'.
The points P and P’ are called as correspoding points on the ellipse and auxiliary circle respectively.
This angle is known as the eccentric angle of the poitn P on the ellipse and auxialiary circle respectively.
## Parametric Representation
The coordinates x = a cos θ and y = b sin θ satisfy the equation x2/a2 + y2/b2 = 1, for all real values of θ.
Thus x = a cos θ, y = b sin θ are the parametric equations of the ellipse x2/a2 + y2/b2 = 1, where 0 θ ≤ 2π.
Therefore the coordinates of any point on the ellipse x2/a2 + y2/b2 = 1 may be taken as (a cos θ, b sin θ). The angle θ is called the eccentric angle of the point (a cos θ, b sin θ) on the ellipse.
## Tangents to the Ellipse
The condition for the line y = mx + c to be a tangent to the ellipse x2/a2 + y2/b2 = 1 is that c2 = a2m2 + band the coordinates of the points of contact are ± (a2m/√(a2m2 + b2), (- b2/√(a2m2 + b2).
Also, x cos a + y sin a = p is a tangent if p2 = a2 cos2α + b2 sin2α
Moreover, lx + my + n = 0 is a tangent if n2 = a2l2 + b2m2.
Fro more detalis on equations of tangent in different forms, stduents may refer the following sections.
## Chord of Contact
The equation of chord of contact of atngent drawn from a point P(x1, y1) to the ellipse x2/a2 + y2/b2 = 1 is T = 0, where T = xx1/a2 + yy1/b2 -1.
## Diameter of an Ellipse
The locus of the middle points of a system of parallel chords of an ellipse is called the diameter of the ellipse.
Let y = mx + c ...... (1)
be the equation of a system of parallel chords of the ellipse x2/a2 + y2/b2 = 1 ...... (2)
In (1) m is constant and c varies from chord to chord. Let K (x1, y1) be the midpoint of a chord PQ of this system.
Eliminating y between (1) and (2) we get
(a2m2 + b2)x2 + 2a2mcx + a2(c2 - b2) = 0 ...... (3)
⇒ x2 + x3 = (-2a2 mc)/(a2 m2+b2 )
But x1 = (x2+x3)/2 =(-a2 mc)/(a2 m2+b2)
Or c = (-x(a2 m2+b2))/(a2 m2)
Also K (x1, y1) is a point on (1) so
y1 = mx1 + c ⇒ y1= -b2x1/a2m
.·. The locus of K (x1, y1) is y = -b2x/a2m which is a diameter of the ellipse x2/a2 + y2/b2 = 1.
## Conjugate Diameters
Two diameters of an ellipse which bisects chords parallel to each other are called conjugated diameters. Therefore the diameters y = mx and y = m1x of the ellipse x2/a2 +y2/b2 = 1 are conjugate if mm1 = -b2/a2.
Key Points:
• In an ellipse, the major axes bisects all chords parallel to the minor axes and vice-versa, therefore major axes and minor axes of an ellipse are conjugate diameters but they do not satisfy the condition mm1 = -b2/a2 and are the only perpendicular conjugate diameters.
• The eccentric angles of the ends of a pair of conjugate diameters differ by a right angle.
• Equi-conjugate diameter: If the length of two conjugate diameters of ellipse be equal then they are called equi conjugate diameters.
• The equation of equi conjugate diameters are x2/a2 ± y2/b2 = 1.
• The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by a right angle i.e., if one end of a diameter (PQ) is P(a sin Φ, b cos Φ).
• The sum of the squares of any two conjugate semi-diameters of an ellipse is constant and equal to the sum of the squares of the semi-axis of the ellipse i.e. OP2 + OP2 = a2 + b2.
• The produce of the focal distances of a point on an ellipse is equal to the square of the semi-diameters, which is conjugate to the diameter through the point.
• The tangents at the ends of a pair of conjugate diameters of an ellipse form a parallelogram and the area of the parallelogram is constant and is equal to the product of the axis i.e. equal to 4ab.
## Director circle of an Ellipse
The director circle is the locus of the point of intersection of pair of perpendicular tangents to an ellipse.
Two perpendicular tangents of ellipse x2/a2 + y2/b2 = 1 are
y - mx = √(a2m2+b2) ...... (1)
and my + x = √(a2+b2 m2) ...... (2)
To obtain the locus of the point of intersection (1) and (2) we have to eliminate m squaring and adding (1) and (2), we get
(y - mx)2 + (my + x)2 = (a2m2 + b2) + (a2 + b2m)
⇒ x2 + y2 = a2 + b2, which is the equation of the director circle.
## How to find equation of a chord of an Ellipse whose mid point is (x1, y1)?
Let the ellipse be x2/a2 + y2/b2 = 1 ...... (1)
Any line through (x1, y1) is y - y1 = m(x - x1) ...... (2)
Eliminating y between (1) and (2) we get
x2 (b2 + a2m2) + 2ma2 (y1 - mx1) x + a2 [(y1 - mx1)2 - b2] = 0
If x2 and x3 are the roots of this equation then
x2 + x3 = -2ma2 (y1 - mx1)/(b2 + a2m2)
But (x1, y1) the mid point of the chord.
.·. x1 = (x+ x3)/2 = (-ma2 (y- mx1 ))/(b+ a2m2)
⇒ m = (-b2x1)/(a2y1)
.·. From (2) and (3) the equation of the chord where mid point is (x1, y1) is
y - y1 = ((-b2x1)/(a2y1)) (x - x1)
⇒ xx1/a2 + yy1/b2 = (x12)/a2 +(y12)/b2
Or T = S1 where T = xx1/a2 + yy1/b2 -1
and S1 = (x12)/a2 + (y12)/b2 -1
## Pole and polar of an Ellipse
The locus of the points of intersection of tangents drawn at the point extremities of the chords passing through a fixed point is called the polar of that fixed point and the fixed point is called the pole.
Equation of the polar of P(x1, y1) with respect to the ellipse
x2/a2 + y2/b2 = 1 is
xx1/a2 + yy1/b2 = 1
### Key Points:
• Polar of the focus is the directirx.
• Any tangent is the polar of its points of contact.
• If the polar of P(x1, y1) passes through Q(x2, y2) then the polar of Q will pass through P and such points are siad to be conjugate points.
• The point of intersection of any two lines is the pole of the line joining the pole of the two lines.
• If the pole of the line lx + my + n = 0 lies on another line l’x + m’y + n’ = 0, then the pole of the second line will lie on the first and such lines are said to be conjugate lines.
### Illustration:
How to find out pole of the line lx + my + n = 0 w.r.t. the ellipse x2/a2+y2/b2 = 1.
### Solution:
Let (x1, y1) be the pole of line lx + my + n = 0. ...... (1)
w.r.t. the ellipse x2/a2 + y2/b2 = 1 ...... (2)
Now the polar of (x1, y1) w.r.t. (2) is xx1/a2 + yy1/b2 = 1 ...... (3)
Since (1) and (3) represents the same polar, so comparing them we have
(x1/a2)/l + (y1/b2)/b2 = (-1)/n
or x1 = (-a2 l)/n y1 = (-b2m)/n
.·. The required pole is ((-a2 l)/n,(-b2 m)/n)
### Illustration:
The number of values of c such that the straight line y = 4x + c touches the curve x2/4 + y2 = 1 is ? (IIT JEE 1998)
### Solution:
For an ellipse, we know that the condition for tangency is c2 = a2m2 + b2
The given line is y = 4x + c and the curve x2/4 + y2 = 1
hence, c2 = 4.42 + 1 = 65
Hence, c = √65.
## Related Resources
To read more, Buy study materials of Ellipse comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
### Course Features
• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
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This Sudoku, based on differences. Using the one clue number can you find the solution? | 2,110 | 9,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-43 | latest | en | 0.880245 |
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### What is your watthours per mile?
Posted: Aug 06 2013 6:00pm
I'm trying to figure out whether my bike is as efficient like I'm expecting, so I'm doing testing. I'm wondering if my squeaking wheel is affecting efficiency and whether that warrants servicing it?
I was doing some testing on my bike recently, it's a Bike-E recumbent like this.
UNASSISTED:
Normal riding position
Voltage: 24v
avg-speed: 13.5 mph
Rear tire: 40psi Knobby
Wh/mi: 15.5
Normal riding position
Voltage: 48v
avg speed: 12.5 mph
Rear tire: 100psi comet primo low rolling resistance tire
Wh/mi: 11.7 wh/mi
Draw legs close to the body, bring in head. (Curl up into a ball) [To minimize frontal area]
Voltage: 48v
avg speed: 12.5 mph
Rear tire: 100psi comet primo low rolling resistance tire
Wh/mi: 10.5 wh/mi
One time, I was trying to maximize my range because I had no idea how far my bike could go and my derailleur was busted so I couldn't pedal, so I went 10 mph and coasted wherever possible and minimized braking (To conserve energy).
Normal riding position
Voltage: 48v
avg speed: 10 mph
Rear tire: 100psi comet primo low rolling resistance tire
Wh/mi: 9.2 wh/mi
A neat way to figure out your watthours per mile, without relying on a CA to do it for you, is by dividing your power output by your speed.
So, this one time I was going 26 mph and my motor was consuming 450 watts, so my wh/mi at 26 mph was (450/26)=17.3 wh/mi. (I had my 40psi knobbies on at the time)
On my older upright bike like this:
It was consuming 770 watts to sustain 25.5 mph, so my wh/mi. was (770/25.5) = 30wh/mi which is a figure I remembered seeing on my CA.
### Re: What is your watthours per mile?
Posted: Aug 06 2013 6:29pm
that is a clown bike
i will measure my consumption tomorrow
i do about 70km and consume about 20ah
we're talking a cargo bike though- 310lbs with me on it...and going 20-25mph
tire pressure does make a super difference....
i am still working on improving my efficiency- im sure i can upgrage bearings and do a few things
Edit- my voltage is 48V - so about 52v nom -
Edit2- generally on an efficient cycle going @ 20-21mph 13 to 15wh/km is normal
10 is also possible if your rig is really efficient possibly with skinny tires/small weight etc....
mine is probably 20wh/km or close as my rig is heavy and i go fast
not bad looks like im using 15wh per KM
70KM * 15WH/KM / 52 = 20ah
Edit: new record did 140km trip used up 24AH
this is on a 60v battery (added 2 cells my my 16s to make it 18s) and i wasnt going fast this time
### Re: What is your watthours per mile?
Posted: Aug 06 2013 6:32pm
davec wrote:that is a clown bike
i will measure my consumption tomorrow
i do about 70km and consume about 20ah
we're talking a cargo bike though- 310lbs with me on it...and going 20-25mph
tire pressure does make a super difference....
A super comfortable, highly efficient and fast clown bike at that! (Also, just about the only recumbent with a suspension for under \$500. )
Essentially, perfect for long-distance trips over 60 miles. Doesn't have the usual burning urethra sensation that follows riding on a lycra's bike for more than 10 miles.
Do you have the voltage? With that info, I could calculate the wh/km.
### Re: What is your watthours per mile?
Posted: Aug 06 2013 6:45pm
On my monster bike, running 20lbs in some extra fat knobby tires, I get about 10wh/mile at 10mph, and thats with the aerodynamics of a brick wall wearing a parachute.
My Kona gets sbout the same with it's Clyte. I think my last bike with a 9C got slightly better. 9-point-something at 10mph
If youe wheel is making noise, it's burning watts. 30db measured at 1 meter is 1 watt. And it doubles for every 3db increase. 33db is 2 watts. 36 is 4 watts.
### Re: What is your watthours per mile?
Posted: Aug 06 2013 6:46pm
Typically between 90 and 110 wh/mile on the race bike commuting. At laguna seca we were averaging about 280 wh/mi.
### Re: What is your watthours per mile?
Posted: Aug 06 2013 7:02pm
Farfle wrote:Typically between 90 and 110 wh/mile on the race bike commuting. At laguna seca we were averaging about 280 wh/mi.
wh/mile like that is some FUN riding.
### Re: What is your watthours per mile?
Posted: Aug 06 2013 7:12pm
On my Catrike 700 with Bionx @36v (41v hoc) Tadpole Trike I'm using 7-9wh per km speeds between 35-40kph.
And I'm pedaling for a workout as well.
Tommy L sends.....
### Re: What is your watthours per mile?
Posted: Aug 06 2013 7:14pm
Drunkskunk wrote:If youe wheel is making noise, it's burning watts. 30db measured at 1 meter is 1 watt. And it doubles for every 3db increase. 33db is 2 watts. 36 is 4 watts.
Oh interesting. Didn't know one could directly infer power levels from volume, but makes sense.
It seems that wikipedia thinks that 90 dB is .001W. Wiki's Sound power
Chain saw 0.1 W 110 dB
Helicopter 0.01 W 100 dB
Loud speech 0.001 W 90 dB
But it's not necessarily the energy wasted on sound that I'm worried about, it's the energy wasting on the rubbing in the form of heat, the squeak just lets me know something's rubbing.
It seems that the wheel started squeaking after I changed out my 40psi knobby tire for a 110psi comet primo tire and pumped it up to 100-110 psi. So, I'm not sure if the high air pressure is causing the inner tube's rubber to rub against the casing or against the Mr.Tuffy Tire Liner or what, or if some problem with the hub bearings suddenly arose after changing out the tire. (Seems unlikely, but possible, I suppose.)
If it's the inner tube, like I suspect, not entirely sure what I can do about that that doesn't involve deflating the tire... (Deflating the tire would presumably cause me to lose precious watts)
### Re: What is your watthours per mile?
Posted: Aug 06 2013 7:29pm
Switching to a 100 psi front tire passes the bump shock to the wheel bearing harder than with a 40psi tire.
If you can, replace with a high quality ceramic bearing
I will help reduce wh too
Tommy L sends......
### Re: What is your watthours per mile?
Posted: Aug 06 2013 7:50pm
Farfle wrote:... At laguna seca we were averaging about 280 wh/mi.
What were your average speed?
50mph only requires about 3,500W. At 50% efficiency, that's still only 7,000W or 140Wh/mi.
Using the same calculation, 62mph only requires 12,400W OR 200Wh/mi.
4,000+ lb electric cars consume less than 280Wh/mi on flat at 60mph.
### Re: What is your watthours per mile?
Posted: Aug 06 2013 11:27pm
Our fastest lap was 2:09 , and iirc the track is 2.3 miles long. So somewhere around 65mph average speed. We were hitting the rev limiter often, and that happems around 88mph.
### Re: What is your watthours per mile?
Posted: Aug 07 2013 1:30am
swbluto wrote:just about the only recumbent with a suspension for under \$500. )
actually it's more like a heavy-duty glorified thudbuster than any real sort of suspension.
is this like the 6th or 7th energy per distance thread?
half of them started by swb it seems.
there's a sticky & everything for this.
why didn't you use it?
i sense moderation in your future.
### Re: What is your watthours per mile?
Posted: Aug 07 2013 1:40am
Toorbough ULL-Zeveigh wrote:
swbluto wrote:just about the only recumbent with a suspension for under \$500. )
actually it's more like a heavy-duty glorified thudbuster than any real sort of suspension.
If it's a glorified thud-buster, thud busters must be really NIIIIICCCE...
is this like the 6th or 7th energy per distance thread?
half of them started by swb it seems.
there's a sticky & everything for this.
why didn't you use it?
i sense moderation in your future.
Please link to the sticky. I didn't spot it. Even googled and everything, and got anything between flamewars from 2007 and craptastic "wh/mi/lb" threads, but didn't see much in the way of wh/mi.
It's be nice if you linked to my "energy per distance" threads as I have no idea what you're talking about. I like energy efficiency concepts, but started very few to no threads around the "Wh per mile" concept, specifically, AFAIK.
### Re: What is your watthours per mile?
Posted: Aug 07 2013 2:05am
Toorbough ULL-Zeveigh wrote:
swbluto wrote:just about the only recumbent with a suspension for under \$500. )
actually it's more like a heavy-duty glorified thudbuster than any real sort of suspension.
Oh interesting, it appears Cane-Creek made the air-suspension to my recumbent and it also makes the thud-buster. Maybe it IS the exact same technology. (Actually, just checked; it uses elastomers instead of air for cushioning.)
### Re: What is your watthours per mile?
Posted: Aug 07 2013 9:57am
A 25mph commute drains about 45 wh/mi from my upright off-roader bike. Actual off-roading drains quite a bit more energy.
On my recumbent tadpole I can drain as much as 70 wh/mi when I go on little 3500 watt joy rides and as little as 7wh/mi when I go on energy save runs. A realistic good clip commuting to work puts me in the 25-30 wh/mi range amperage set to 2000watts maxout.
### Re: What is your watthours per mile?
Posted: Aug 08 2013 4:00am
My commute around 21mph drains a 20Ah 6s Lipo in about 20miles. I make that 23Wh/mile. It's the world's cheapest old mountain bike with the knobblies replaced by "city tyres" (that was worth about 10% more range thankyou very much). goodish roads, no big hills.
### Re: What is your watthours per mile?
Posted: Aug 08 2013 7:19am
Wh/mi will vary wildly with speed, and if you pedal or not at the slower speeds.
My super un aerodynamic longtail can get 12 wh/mi, at about 12 mph. But at 30 mph, 40 is more like it. I aim for about 25 wh/mi as a compromise between speed and efficiency. That's about 20 mph max.
On a slightly more aero commuter bike, I'd get closer to 20 wh/mi at 20 mph.
Dirt riding could easily get as poor as 60 wh/mi. Without going fast too. Just lots of brake, corner, throttle, brake corner throttle. Just like stopping every 50 feet or so would be.
### Re: What is your watthours per mile?
Posted: Aug 08 2013 8:30am
33 Wh/mi city commute averaging 17MPH running 26" wheel 9C, 9FET Lyen, 30A, 60V hot off charger.
17MPH is mostly due to frquent stop lights. My average traffic speed is around 20-25MPH.
### Re: What is your watthours per mile?
Posted: Aug 08 2013 3:52pm
Now, talking about with assistance, I can go 13 mph no matter how windy it is, with somewhat hilly terrain, and get 3wh/mi. comfortably.
But, it seems silly to compare "with assistance" stats, as how much assistance varies from rider to rider and ride to ride.
### Re: What is your watthours per mile?
Posted: Aug 08 2013 8:19pm
I have tried to think up a way for everyone to realistically and honestly compare wh/mi numbers. So many variables to try and keep consistent.
The best thing I have come up with is probably most people on this forum (at least in the states & Canada) live somewhat close to a football/soccer field that has a track around it for track events. These are consistent in length, always dead level and are very common. Usually they are open for people that like to walk them for exercise too.
1.jpeg (44.08 KiB) Viewed 2875 times
I think a comparison of like five laps on the outside track at as consistent speed as you can maintain (probably around 15mph because of the turns) would give a nice real world comparison of the figures.
I have a track at the end of my block. If there's any interest in doing something like this I'll put up my numbers.
### Re: What is your watthours per mile?
Posted: Aug 08 2013 9:03pm
StudEbiker wrote:I have tried to think up a way for everyone to realistically and honestly compare wh/mi numbers. So many variables to try and keep consistent.
The best thing I have come up with is probably most people on this forum (at least in the states & Canada) live somewhat close to a football/soccer field that has a track around it for track events. These are consistent in length, always dead level and are very common. Usually they are open for people that like to walk them for exercise too.
1.jpeg
I think a comparison of like five laps on the outside track at as consistent speed as you can maintain (probably around 15mph because of the turns) would give a nice real world comparison of the figures.
I have a track at the end of my block. If there's any interest in doing something like this I'll put up my numbers.
I used to have a track like that, but they tore it down for the summer.
You can't imagine what affect that had on my budding running exercise plans. (And, wh/mi testing)
There's another one around here, but it feels all sorts of weird since it's hidden from view (Recessed below ground level) and there's only one way in and out.
I used, instead, a relatively flat middle school parking lot that has a half-mile loop and no stops. Does the trick.
### Re: What is your watthours per mile?
Posted: Aug 09 2013 2:25am
Normally 35-50Wh/mile on my fast DD mountain ebike
On my hybrid with heavy pedal assist for highest range: 9.8wh/mile otherwise normally @ 35wh/mile
### Re: What is your watthours per mile?
Posted: Aug 11 2013 7:25pm
Just did some trip where I was trying to ride at 20mph along those busy 25mph roads and found that it took about 150 watts on average to sustain 20mph with my weak, untrained muscles (My bikes have been too broke this summer to train. ). That implies the wh/mi is (150/20) = 7.5 wh/mi at 20mph with assistance on my recumbent.
With the addition of 100 watts of solar panels on a rear rack, that implies that I could potentially cut my power usage of the batteries from 150 watts to 50 watts, thus tripling my effective range of 133 miles@20mph with 1kwh of batts to around 400 miles. In practice, I'd probably get around 200-300 miles in a day and the battery would never fully deplete.
### Re: What is your watthours per mile?
Posted: Aug 11 2013 7:44pm
swbluto wrote:Just did some trip where I was trying to ride at 20mph along those busy 25mph roads and found that it took about 150 watts on average to sustain 20mph with my weak, untrained muscles (My bikes have been too broke this summer to train. ). That implies the wh/mi is (150/20) = 7.5 wh/mi at 20mph with assistance on my recumbent.
With the addition of 100 watts of solar panels on a rear rack, that implies that I could potentially cut my power usage of the batteries from 150 watts to 50 watts, thus tripling my effective range of 133 miles@20mph with 1kwh of batts to around 400 miles. In practice, I'd probably get around 200-300 miles in a day and the battery would never fully deplete.
Only one way to tell! Try it out! I suspect you won't get 100W out of the solar panels. I mean there will be clouds. Also carrying solar panels will increase weights and probably kills your aero. Then the hills and wind will cut down your range significantly. Finally keep in mind you only get x number of sunlight. 200/20mph = 10 hours. You will have to take breaks. Overall, I think when you do take the long trip. You find you will get around 150-200 max per day even with heavy assist. Unless you go and charge and go and charge by doing opportunity charging with high power charger.
### Re: What is your watthours per mile?
Posted: Aug 11 2013 7:54pm
mvly wrote:
swbluto wrote:Just did some trip where I was trying to ride at 20mph along those busy 25mph roads and found that it took about 150 watts on average to sustain 20mph with my weak, untrained muscles (My bikes have been too broke this summer to train. ). That implies the wh/mi is (150/20) = 7.5 wh/mi at 20mph with assistance on my recumbent.
With the addition of 100 watts of solar panels on a rear rack, that implies that I could potentially cut my power usage of the batteries from 150 watts to 50 watts, thus tripling my effective range of 133 miles@20mph with 1kwh of batts to around 400 miles. In practice, I'd probably get around 200-300 miles in a day and the battery would never fully deplete.
Only one way to tell! Try it out! I suspect you won't get 100W out of the solar panels. I mean there will be clouds. Also carrying solar panels will increase weights and probably kills your aero. Then the hills and wind will cut down your range significantly. Finally keep in mind you only get x number of sunlight. 200/20mph = 10 hours. You will have to take breaks. Overall, I think when you do take the long trip. You find you will get around 150-200 max per day even with heavy assist. Unless you go and charge and go and charge by doing opportunity charging with high power charger.
I'm assuming I'd be using "150 watts worth of solar panels" to get 100w average and for a longdistance trip, I'd probably be willing to sink in the \$3/watt for the ultra-lightweight and high efficiency panels on ebay (1.5 pounds per 50 watts, so 150 watts would be 5 pounds). Since you can design it so that the panels lay flat on the back of the bike on a custom rear rack, the affect on aero should be negligible (Since flat panels have a very low drag coefficient and obviously very low frontal area). Would probably become significant at 50+mph, sure, but I wouldn't suspect it'd be significant at 20mph.
I'm assuming one would take such a hypothetical long-distance trip during the summer time when cloud cover is minimal and the days are long. Even in the case of cloud cover, the data shows that power output is usually cut in half, so it's still producing a useful amount of energy -- i.e., might end up merely doubling your range instead of tripling. There's one welcome affect in any case, however - it could negate the need to stop somewhere to charge.
This is all hypothetical, however. There's no place in the United States, AFAIK, I'd be willing to bicycle for 1000s of miles -- highways are too dangerous with people whizzing by at 80 mph. Maybe when they get that eastern coastal interstate bikepath done, there'll be a reason to try it out. | 4,821 | 17,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-34 | latest | en | 0.95279 |
https://www.scribd.com/document/78606410/Account-Formule | 1,524,146,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936969.10/warc/CC-MAIN-20180419130550-20180419150550-00059.warc.gz | 872,116,424 | 25,123 | # Ratios and Formulas in Customer Financial Analysis
Financial statement analysis is a judgmental process. One of the primary objectives is identification of major changes in trends, and relationships and the investigation of the reasons underlying those changes. The judgment process can be improved by experience and the use of analytical tools. Probably the most widely used financial analysis technique is ratio analysis, the analysis of relationships between two or more line items on the financial statement. Financial ratios are usually expressed in percentage or times. Generally, financial ratios are calculated for the purpose of evaluating aspects of a company's operations and fall into the following categories:
y y y
y
liquidity ratios measure a firm's ability to meet its current obligations. profitability ratios measure management's ability to control expenses and to earn a return on the resources committed to the business. leverage ratios measure the degree of protection of suppliers of long-term funds and can also aid in judging a firm's ability to raise additional debt and its capacity to pay its liabilities on time. efficiency, activity or turnover ratios provide information about management's ability to control expenses and to earn a return on the resources committed to the business.
A ratio can be computed from any pair of numbers. Given the large quantity of variables included in financial statements, a very long list of meaningful ratios can be derived. A standard list of ratios or standard computation of them does not exist. The following ratio presentation includes ratios that are most often used when evaluating the credit worthiness of a customer. Ratio analysis becomes a very personal or company driven procedure. Analysts are drawn to and use the ones they are comfortable with and understand.
Liquidity Ratios
Working Capital Working capital compares current assets to current liabilities, and serves as the liquid reserve available to satisfy contingencies and uncertainties. A high working capital balance is mandated if the entity is unable to borrow on short notice. The ratio indicates the short-term solvency of a business and in determining if a firm can pay its current liabilities when due. y Formula Current Assets - Current Liabilities Acid Test or Quick Ratio A measurement of the liquidity position of the business. The quick ratio compares the cash plus cash equivalents and accounts receivable to the current liabilities. The primary difference between the current ratio and the quick ratio is the quick ratio does not include inventory and prepaid expenses in the calculation. Consequently, a business's quick ratio will be lower than its current ratio. It is a stringent test of liquidity.
and inventories. Return on Assets Measures the company's ability to utilize its assets to create profits. However. the normal current ratio fluctuates from industry to industry. y Formula Net Income * Net Sales * Refinements to the net income figure can make it more accurate than this ratio computation. a current ratio significantly lower than the industry average could indicate a lack of liquidity. y Formula Current Assets Current Liabilities Cash Ratio Indicates a conservative view of liquidity such as when a company has pledged its receivables and its inventory. y Formula Cash Equivalents + Marketable Securities Current Liabilities Profitability Ratios Net Profit Margin (Return on Sales) A measure of net income dollars generated by each dollar of sales. They could include removal of equity earnings from investments. A current ratio significantly higher than the industry average could indicate the existence of redundant assets. Current liabilities include accounts payable.y Formula Cash + Marketable Securities + Accounts Receivable Current Liabilities Current Ratio Provides an indication of the liquidity of the business by comparing the amount of current assets to current liabilities. A business's current assets generally consist of cash. current maturities of long-term debt. businesses prefer to have at least one dollar of current assets for every dollar of current liabilities. In general. marketable securities. Conversely. "other income" and "other expense" items as well as minority share of earnings and nonrecuring items. and other accrued expenses that are due within one year. y Formula . or the analyst suspects severe liquidity problems with inventory and receivables. accrued income taxes. accounts receivable.
. The benefit of the method is that it provides an understanding of how the company generates its return.Net Income * (Beginning + Ending Total Assets) / 2 Operating Income Margin A measure of the operating income generated by each dollar of sales. y Formula Net Income * Long-term Liabilities + Equity Return on Equity Measures the income earned on the shareholder's investment in the business. This ratio should be compared with industry data as it may indicate insufficient volume and excessive purchasing or labor costs. y Formula Net Income * Equity Du Pont Return on Assets A combination of financial ratios in a series to evaluate investment return. y Formula Operating Income Net Sales Return on Investment Measures the income earned on the invested capital. y Formula Gross Profit Net Sales Financial Leverage Ratio Total Debts to Assets Provides information about the company's ability to absorb asset reductions arising from losses without jeopardizing the interest of creditors. y Formula Net Income * Sales Assets x x Sales Assets Equity Gross Profit Margin Indicates the relationship between net sales revenue and the cost of goods sold.
y Formula Net Sales Cash Sales to Working Capital (Net Working Capital Turnover) Indicates the turnover in working capital per year.y Formula Total Liabilities Total Assets Capitalization Ratio Indicates long-term debt usage. y Formula Total Debt Total Equity Interest Coverage Ratio (Times Interest Earned) Indicates a company's capacity to meet interest payments. y Formula Long-Term Debt Long-Term Debt + Owners' Equity Debt to Equity Indicates how well creditors are protected in case of the company's insolvency. . A low ratio indicates inefficiency. while a high level implies that the company's working capital is working too hard. y Formula Long-term Debt Current Assets . Uses EBIT (Earnings Before Interest and Taxes) y Formula EBIT Interest Expense Long-term Debt to Net Working Capital Provides insight into the ability to pay long term debt from current assets after paying current liabilities.Current Liabilities Efficiency Ratios Cash Turnover Measures how effective a company is utilizing its cash.
or to another factor such as a change in selling terms. An analyst might compare the days' sales in receivables with the company's credit terms as an indication of how efficiently the company manages its receivables. y Formula Gross Receivables Annual Net Sales / 365 Accounts Receivable Turnover Indicates the liquidity of the company's receivables. . y Formula Average Gross Receivables Annual Net Sales / 365 Days' Sales in Inventory Indicates the length of time that it will take to use up the inventory through sales. y Formula Net Sales Net Fixed Assets Days' Sales in Receivables Indicates the average time in days. y Formula Net Sales Average Total Assets Fixed Asset Turnover Measures the capacity utilization and the quality of fixed assets.y Formula Net Sales Average Working Capital Total Asset Turnover Measures the activity of the assets and the ability of the business to generate sales through the use of the assets. that receivables are outstanding (DSO). It helps determine if a change in receivables is due to a change in sales. y Formula Net Sales Average Gross Receivables Accounts Receivable Turnover in Days Indicates the liquidity of the company's receivables in days.
For most companies the operating cycle is less than one year. but in some industries it is longer. y Formula Cost of Goods Sold Average Inventory Inventory Turnover in Days Indicates the liquidity of the inventory in days. y Formula Purchases Average Accounts Payable Payables Turnover in Days Indicates the liquidity of the firm's payables in days. y Formula Average Inventory Cost of Goods Sold / 365 Operating Cycle Indicates the time between the acquisition of inventory and the realization of cash from sales of inventory. y Formula Ending Accounts Payable Purchases / 365 Payables Turnover Indicates the liquidity of the firm's payables. y Formula .y Formula Ending Inventory Cost of Goods Sold / 365 Inventory Turnover Indicates the liquidity of the inventory. y Formula Accounts Receivable Turnover in Days + Inventory Turnover in Day Days' Payables Outstanding Indicates how the firm handles obligations of its suppliers. | 1,614 | 8,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-17 | latest | en | 0.945508 |
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### AVL trees are not weight-balanced?
In a previous question there was a definition of weight balanced trees and a question regarding red-black trees. This question is to ask the same question, but for AVL trees. The question is, ...
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### A median of an AVL. How to take advantage of the AVL?
Here is the source of my question. Given a self-balancing tree (AVL), code a method that returns the median. (Median: the numerical value separating the higher half of a data sample from ...
2k views
### Balance factor changes after local rotations in AVL tree
I try to understand balance factors change after local rotations in AVL trees. Given the rotate_left operation: ...
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### Compute height of AVL tree as efficiently as possible
Given an AVL tree, I want to compute its height as efficiently as possible. $\newcommand{\bf}{\text{bf}}\newcommand{\height}{\text{height}}$ Each node of an AVL tree stores its balance factor ($\bf$),...
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### Binary tree To Red-Black tree
I have a question regarding the solution provided by Karolis Juodelė. Given in this question; Colour a binary tree to be a red-black tree Black = black nodes, white = red nodes So for this tree when ...
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### Traversals from the root in AVL trees and Red Black Trees
We all know that for insertion() operation in AVL tree following can happen: We traverse down the tree from root to appropriate node and there insert the key and then for maintaining height balance ...
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### Why do we need double-rotations to rebalance AVL trees?
I was reading about AVL tree rebalancing from Behrouz Forouzan's book. The book first defines Left High and Right High tree: Left High (LH) tree is a tree tree with the height of the left ...
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### Prove that given a number we can find whether there're 2 elements in a red/black tree that their sum equals that number in $\Theta(n)$ time
Prove that given a number we can find whether there're 2 elements in a red/black tree that their sum equals that number in $\Theta(n)$ time and constant space. The original problem appears here, ...
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### Every AVL tree may be red black tree
I proved by induction that every AVL tree may be colored such that it will be red black tree. The problem is that I can't see an error in my proof. Look at my proof. Induction for height. Let's ...
558 views
### Reb-black tree amortized cost of the rebalancing
I've read in different sources that the amortized cost of a red-black tree rebalancing is constant (at least during the tree creation using only insertions). How can it be proved?
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### Can a Red Black tree be constructed of only black nodes using RB insert only?
I am trying to construct a red black tree out of only black nodes. I know it is possible getting it after some deletions but I am trying to construct one only via insertion orders. Is it possible? I ...
353 views
### Left-Right-Rotation of AVL-Tree
For AVL-Tree there exists the following Rotations for Balancing: Left Rotation Right Rotation Left-Right Rotation Right-Left Rotation My Question is about the Naming for the Double-Rotations. ...
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### Does the rebalancing propagate upwards only to update the height of the nodes in an AVL tree?
I was studying AVL trees and was wondering if the only reason one propagates upwards to the node in an insert is to change the height. It seems to me that rebalancing does not recursively propagate ...
### Why not use large $k$ in a $k$-ary tree?
Obviously binary trees are great because of $O(\log_2 n)$ search, inserts, and deletes in best case. To "maximize" occurrence of best case, we can use self-balancing trees like red-black trees, AVLs, ...
The worst case height of AVL tree is $1.44 \log n$. How do we prove that? I read somewhere about Fibonacci quicks but did not understand it. | 912 | 3,967 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-35 | latest | en | 0.898307 |
https://es.scribd.com/document/45174203/Econ-Assignment-1 | 1,508,517,708,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00703.warc.gz | 688,040,019 | 31,892 | # A1-1 False The opportunity cost of going skiing during reading week must include the value of the time
In this case, the opportunity cost:
Cell phone Computer Japan 1/3 computer 3 cell phones Finland 1/4 computer 4 cell phones Therefore, Finland has a comparative advantage on cell phone production compare to Japan.
A1-9 the table below lists points on a production possibility frontier or PPF (sometimes called the production possibility boundary or PPB). Good x Good y a) 0 55 1 54 2 52 3 49 4 45 5 40 6 34 7 27 8 19 9 10 10 0 b) Opportunity cost of the nth unit of x in terms of unit of y 1 55 – 54 1 2 54 – 52 2 3 52 – 49 3 4 49 − 45 4 5 45 − 40 5 6 40 – 34 6 7 34 – 27 7 8 27 – 19 8 9 19 – 10 9 10 10 .0 10 The opportunity cost of x increases as more of it is produce is because the each factor of production is not equally useful in producing both x and y. n= .
Plug in the number. Plug in the number. we must shift more resources that are quite suitable for making y and maybe less suitable for making x. c) The opportunity cost per unit of x is given by.86 units of y. and gradually shift more resources toward the production of x. Therefore. the opportunity cost of x given this production point is 5 units of y. This shift of resources will therefore lead to a small reduction in y. d) The relative price per unit of x is given by. Therefore. . the relative price of x given this production point is 3. Then. Therefore. However. Some resources that may not be very useful for making y. is very useful for making x. but a substantial increase in x. the opportunity cost of producing x rises as more x is produced. as we produce more and more x. the amount of y that must be forgone to produce one extra unit of x rises. as we produce more x.Let’s start from the point when production of x is zero.
f) We do not expect the economy to remain at the production point in part d. By being opened to trade. the economy will be able to provide units of x and y that was previously unattainable as shown in the graph. the consumption possibility curve can lie beyond the production possibility curve. The point at which it should be produced should shift towards the left on the table given. . Yes. the economy will be better able to provide x and y to its members by being opened to trade.e) With trade.
a point outside of the PPF-without-trade will be possible. if we choose to trade 21 y of our production of 7 x with the rest of the world. This combination is unattainable without trade. . if we choose to produce 0 x and 55 y as shown in the PPF.For example. a combination of 7 x and (55-21 = 34) y will be achieved. For example.
50.00 To find the equilibrium quantity.00. QD = 0 100-10P=0 P=10 If consumption falls to 0. then the price would be \$10. Qs’ = -60+30P 0 = -60+30P P=2 .50 to \$2.5 If production falls to 0.00 By setting P=0. we find that the equilibrium quantity is 50. b) To find the equilibrium price.A1-10 a) QS =0 -50+20P=0 20P=50 P=2. then the price would be \$2. subbing in P=5 into the equation for QS. the maximum amount that would ever be consumed would be 100. -50+20P = 100-10P 20P+10P = 50+100 30P = 150 P=5 The equilibrium price is \$5. c) The price at which there will be no production has changed from \$2.
An increase in supply creates a surplus at the initial equilibrium price. e) QS’ = QD -60+30P = 100-10P 40P = 40 P=1 The consumer is better off because the equilibrium prices decreased from \$5. they made some profit which is better off than not making any money at all before they were in the market.5. At P = 2.00 to \$1. they were not making any profit at all. The old producer is worse off because a increase in the supply causes a decrease in the equilibrium price which means that they’re selling their product at a lower price and therefore are losing money. The new producer is better off because before they joined the market.00. and the unsuccessful supplier force the price down. This drop in price increases the quantity demanded and the new equilibrium is at a lower price and a higher quantity exchanged. there is now excess supply. .d) The price we calculated is no longer the equilibrium price because an increase in supply puts a downward pressure on the price and causes a decrease in the equilibrium price. After joining the market. | 1,132 | 4,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-43 | latest | en | 0.918499 |
https://virtualpsychcentre.com/characteristics-of-polynomials/ | 1,716,098,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057684.4/warc/CC-MAIN-20240519035827-20240519065827-00780.warc.gz | 533,247,987 | 11,620 | ## What is the facts characteristics of polynomial equation?
The equations formed with variables, exponents and coefficients are called as polynomial equations. It can have different exponents, where the higher one is called the degree of the equation. We can solve polynomials by factoring them in terms of degree and variables present in the equation.
## What are the characteristics of a polynomial graph?
Recognizing Characteristics of Graphs of Polynomial Functions. Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous.
## What are the 4 types of polynomials?
They are monomial, binomial, trinomial. Based on the degree of a polynomial, it can be classified into 4 types. They are zero polynomial, linear polynomial, quadratic polynomial, cubic polynomial.
## How do you identify the characteristics of a polynomial function?
Identify the end behavior of a function based on the degree and coefficient. Understand how the multiplicity of a factor affects the behavior of the graph of a polynomial at its intercept. Determine the intervals where a polynomial is positive and negative.
## What are polynomial functions?
A polynomial function is a function that involves only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc. For example, 2x+5 is a polynomial that has exponent equal to 1.
## What is a polynomial simple definition?
Definition of polynomial
(Entry 1 of 2) : a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a nonnegative integral power (such as a + bx + cx2)
## How many terms are in polynomial?
Polynomials are classified according to their number of terms. 4x3 +3y + 3x2 has three terms, -12zy has 1 term, and 15 – x2 has two terms. As already mentioned, a polynomial with 1 term is a monomial. A polynomial with two terms is a binomial, and a polynomial with three terms is a trinomial.
## What are the characteristics of rational functions?
A rational function is defined as the quotient of polynomials in which the denominator has a degree of at least 1 . In other words, there must be a variable in the denominator. The general form of a rational function is p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .
## What are polynomials class 9th?
Polynomial Definition. Polynomials are expressions with one or more terms with a non-zero coefficient. A polynomial can have more than one term. An algebraic expression p(x) = a0xn + a1xn1 + a2xn2 + … an is a polynomial where a0, a1, ………. an are real numbers and n is non-negative integer.
## Is polynomial an equation?
A polynomial equation is an equation that has varied terms and generally includes variables coefficient and exponent. Polynomials can retain various exponents. The degree of a polynomial is considered as the greatest exponent.
## What is a constant term of a polynomial?
A constant term is a term that contains only a number. In other words, there is no variable in a constant term. Examples of constant terms are 4, 100, and -5. Standard Form of a Polynomial.
## What is the range of polynomial function?
The domain of any polynomial function (including quadratic functions) is x∈(−∞,∞). Functions of even degree will have a bounded range (from below if the leading coefficient is positive, from above if it’s negative), and functions of odd degree will have range y∈(−∞,∞).
## What is standard form of a polynomial?
The standard form of a polynomial is a way of writing a polynomial such that the term with the highest power of the variables comes first followed by the other terms in decreasing order of the power of the variable.
## What is a 4 term polynomial called?
The term “quadrinomial” is occasionally used for a four-term polynomial.
## Which one is not a polynomial?
Terms containing fractional exponents (such as 3x+2y1/2-1) are not considered polynomials. Polynomials cannot contain radicals. For example, 2y2 +√3x + 4 is not a polynomial. | 958 | 4,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-22 | latest | en | 0.942819 |
https://aaronice.gitbook.io/lintcode/string/reverse-words-in-a-string-ii | 1,716,859,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00860.warc.gz | 60,746,067 | 48,414 | # Reverse Words in a String II
Given an input string, reverse the string word by word.
Example:
``````Input:
["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output:
["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]``````
Note:
• A word is defined as a sequence of non-space characters.
• The input string does not contain leading or trailing spaces.
• The words are always separated by a single space.
Follow up: Could you do itin-placewithout allocating extra space?
Topics: Two Pointers, String
## Analysis
`void reverseEachWords(char[] str)`中，快指针找到word结尾后的index，满指针指向word的开始第一个index。
## Solution
Two-Pointer + Multi-reversal - O(n) time, O(1) space (In-place) - (5ms AC)
``````class Solution {
public void reverseWords(char[] str) {
// reverse the whole input
reverse(str, 0, str.length - 1);
// reverse each individual word
reverseEachWords(str);
}
void reverse(char[] str, int s, int t) {
while (s < t) {
char tmp = str[s];
str[s] = str[t];
str[t] = tmp;
s++;
t--;
}
}
void reverseEachWords(char[] str) {
int i = 0, j = 0;
int n = str.length;
while (i < n && j < n) {
while (i < n && str[i] == ' ') {
i++;
}
j = i;
while (j < n && str[j] != ' ') {
j++;
}
reverse(str, i, j - 1);
i = j;
}
}
}``````
This solution (by @siyang3) takes advantage of the conditions in the problem: 1. The input string does not contain leading or trailing spaces. 2. The words are always separated by a single space.
``````public void reverseWords(char[] s){
reverseWords(s,0,s.length-1);
for(int i = 0, j = 0;i <= s.length;i++){
if(i==s.length || s[i] == ' '){
reverseWords(s,j,i-1);
j = i+1;
}
}
}
private void reverseWords(char[] s, int begin, int end){
while(begin < end){
char c = s[begin];
s[begin] = s[end];
s[end] = c;
begin++;
end--;
}
}``````
Last updated | 565 | 1,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.550902 |
https://us.metamath.org/nfeuni/mp3anr1.html | 1,713,483,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00889.warc.gz | 522,485,666 | 2,849 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > mp3anr1 GIF version
Theorem mp3anr1 1274
Description: An inference based on modus ponens. (Contributed by NM, 4-Nov-2006.)
Hypotheses
Ref Expression
mp3anr1.1 ψ
mp3anr1.2 ((φ (ψ χ θ)) → τ)
Assertion
Ref Expression
mp3anr1 ((φ (χ θ)) → τ)
Proof of Theorem mp3anr1
StepHypRef Expression
1 mp3anr1.1 . . 3 ψ
2 mp3anr1.2 . . . 4 ((φ (ψ χ θ)) → τ)
32ancoms 439 . . 3 (((ψ χ θ) φ) → τ)
41, 3mp3anl1 1271 . 2 (((χ θ) φ) → τ)
54ancoms 439 1 ((φ (χ θ)) → τ)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 358 ∧ w3a 934 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 177 df-an 360 df-3an 936 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 358 | 869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.342726 |
https://studydaddy.com/question/question-how-many-boxes-to-place-on-a-full-pallet-2-how-many-full-pallets-to-shi | 1,571,709,408,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00259.warc.gz | 717,531,890 | 7,826 | Answered You can hire a professional tutor to get the answer.
QUESTION
# Question: How many boxes to place on a full pallet? 2.How many full pallets to ship? 3.Do you need a partial pallet? How many boxes will it have?
3.Do you need a partial pallet? How many boxes will it have?
Details about the bricks:
High-Temperature Industrial Ceramic Brick each are
11″ X 8″ X 4″ (length x width x height) and weight 3 pounds each
The customer has ordered 5632 bricks
Each Box -add 1 pound for packing materials and 1" of height
Note: boxes must be laid flat on the pallet-
Pallets:
Maximum pallet weight is 1000 lbs.
Each pallet weighs 55 lbs.
Pallets are 44" x 48" x 4"
Maximum pallet height allowed is 72 inches
All pallets must be built with full rows to be uniform
For example: if each pallet has 6 boxes per row and we could place 20 boxes on a pallet using weight only requirements, we would build our full pallets with three rows of six boxes each for 18 boxes per pallet. The two remaining boxes would be a partial pallet. | 264 | 1,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-43 | latest | en | 0.903196 |
https://www.mathsisfun.com/definitions/integral.html | 1,643,275,620,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00131.warc.gz | 913,981,513 | 2,400 | Definition of
# Integral
Two definitions:
• being an integer (a number with no fractional part)
Example: "there are only integral changes" means any change won't have a fractional part.
• the result of integration.
Integration is a way of adding slices to find the whole.
It can be used to find areas, volumes, central points and many useful things.
The slices head towards zero in width ("dx"), and in many cases we can find a formula that gives us an exact answer. | 102 | 473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.904939 |
http://math.stackexchange.com/questions/249927/why-do-we-say-that-two-vectors-in-r3-cannot-span-r2 | 1,466,943,716,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395166.84/warc/CC-MAIN-20160624154955-00185-ip-10-164-35-72.ec2.internal.warc.gz | 191,941,958 | 19,765 | # Why do we say that two vectors in $R^3$ cannot span $R^2$?
If we have two vectors in $R^3$, $v=(1,2,0)$ and $u=(5,3,0)$, and if we draw these vectors in $R^3$ they will be in the $xy$-plane or $R^2$ and also these two vectors are not multiples of each other then why can’t we say that these vectors span $R^2$?
-
Who says that we "can’t say that"? – draks ... Dec 3 '12 at 12:43
Probably his linear algebra professor :) – Neal Dec 3 '12 at 12:45
His linear algebra prof....or any other mathematician, of course. – DonAntonio Dec 3 '12 at 12:56
Because $\,\Bbb R^3\rlap{\;\;/}\subset \Bbb R^2\,$ , so vectors in the former are not even vectors in the latter.
Note that you cannot draw the given vectors in the plane $\,\Bbb R^2\,$: what you can do is draw their projections on some plane in $\,\Bbb R^3\,$ and identify this plane with $\,\Bbb R^2\,$, but this can be done in an infinite number of different ways.
-
But these vectors span a $2$-dimensional subset (=a plane) of $\mathbb R^3$: Assume that $av + bu = 0$. Then $a + 5b = 0$ and $2a + 3b = 0$. Then $a = -5b$ and hence $2a +3b = -10b + 3b = -7b = 0$ so that $b = 0$ and hence $a=0$. Hence $u$ and $v$ are linearly independent.
-
Those vectors span the image of the most obvious embedding of $\def\R{\mathbf R}\R^2$ in $\R^3$, but since this embedding does not have many properties to make it stand out among other linear embeddings, it is unusual to identify $\R^2$ with a part of $\R^3$ in this way, in the same manner as one identifies for instance $\R$ with a part of $\mathbf C$.
-
They will span a subspace of $\Bbb R^3$ which will look like $\Bbb R^2$. Your intuition is that
$$f:\Bbb R^3 \to \Bbb R^2/f(x,y,z)=(x,y)$$
will be onto, which is the case. But note we don't get an isomorphism for this isn't an injection. You're identifying a plane in $\Bbb R^3$ with $\Bbb R^2$ by a projection, in some sense.
-
Shortly speaking, $\mathbb R^3\cap \mathbb R^2=\emptyset$. – yo' Dec 4 '12 at 8:47
@tohecz What is your point? – Pedro Tamaroff Dec 4 '12 at 14:04 | 681 | 2,037 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-26 | latest | en | 0.89067 |
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# 180_pdfsam_math 54 differential equation solutions odd -...
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Chapter 4 Therefore, the auxiliary equation has a double root 2 and a root 2. The functions e 2 t , te 2 t ,and e 2 t form a linearly independent solution set. Therefore, a general solution in this problem is z ( t )= c 1 e 2 t + c 2 te 2 t + c 3 e 2 t . 45. By inspection, we see that r = 2 is a root of the auxiliary equation, r 3 +3 r 2 4 r 12 = 0. Dividing the polynomial r 3 r 2 4 r 12 by r 2 yields r 3 r 2 4 r 12 = ( r 2) ( r 2 +5 r +6 ) =( r 2)( r +2)( r +3) . Hence, two other roots of the auxiliary equation are r = 2and r = 3. The functions e 3 t , e 2 t e 2 t are three linearly independent solutions to the given equation, and a general solution is given by y ( t c 1 e 3 t + c 2 e 2 t + c 3 e 2 t . 47. First we ±nd a general solution to the equation y 0 y 0 = 0. Its characteristic equation, r 3 r = 0, has roots
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Ask a homework question - tutors are online | 391 | 1,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-09 | latest | en | 0.820825 |
https://economyincrisis.org/quebec/annual-average-daily-traffic-calculation-example.php | 1,642,730,585,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00411.warc.gz | 288,069,630 | 12,238 | # Quebec Annual Average Daily Traffic Calculation Example
## Estimation of Annual Average Daily Traffic in a Florida
### NSW Roads Traffic Volume Counts API Transport for NSW
ESTIMATES OF AADT Q UANTIFYING THE UNCER TAINTY ECJ 6.9. > Publications > Ontario Provincial Highways Traffic Volumes report on both that segment's distance in kilometers and the annual average daily traffic, > Publications > Ontario Provincial Highways Traffic Volumes report on both that segment's distance in kilometers and the annual average daily traffic.
### Roadway Design Manual Traffic Characteristics
Improved Annual Average Daily Traffic (AADT) Estimation. Section 2: Traffic Characteristics Traffic volumes may be expressed in terms of average daily traffic or design hourly volumes. For example,, Annual average daily traffic is abbreviated as AADT, is an assessment measure used for planning of transportation and transportation engineering. The below given is.
Example Export File; AADT Calculation – AASHTO Method. AADT = Annual Average Daily Traffic. Calculation: AADT = VOL x SF x AF > Publications > Ontario Provincial Highways Traffic Volumes report on both that segment's distance in kilometers and the annual average daily traffic
An example of the complete calculation for a 10-year average monthly Take daily rainfall readings at "How to Calculate Average Monthly Rainfall Annual Average Daily Traffic (AADT) For example, revised accident rates and costs Volume 5 Section 1 Part 3 TA 46/97
Annual Average Daily Traffic (AADT) These are defined as discrete sections of road where the actual values of traffic volumes anywhere along that section are The Annual Traffic Report contains Annual Average Daily Traffic Data from these are used in the calculation of seasonal traffic variations For example: 155
Calculations convert current and forecast average daily traffic, For example, if the design year "How to Calculate ESAL From ADT." Career Trend, Evaluating Annual Average Daily Traffic Calculation Methods with Continuous Truck Traffic Data. Traffic volume, often measured in relation to annual average daily
Southern Windsor County Annual Traffic Count Program Understanding Traffic Terminology AADT - Annual Average Daily Traffic is the term used to show the average Traffic Reports User Documentation (Annual Average Daily Traffic): Fridays are excluded from the calculation as they tend to have traffic patterns that do not
The ideal scenario for estimation of Annual Average Daily Traffic (AADT) is count the number of vehicles passing over the project road throughout 365 days for 24 hours. Design Controls and Criteria. For example, given a traffic count of Another commonly used measure of traffic volume is the annual average daily traffic
Design Controls and Criteria. For example, given a traffic count of Another commonly used measure of traffic volume is the annual average daily traffic 19/07/2017В В· How to Calculate Compounded Annual Growth Rate. indicates the average annual rate Calculating Compound Annual Growth Rate Calculating CAGR in Excel
Annual Average Daily Traffic (AADT) is the estimated mean daily traffic volume. For continuous sites, calculated by summing the Annual Average Days of the Week and Annual average daily traffic is abbreviated as AADT, is an assessment measure used for planning of transportation and transportation engineering. The below given is
How to calculate ADT based on a 1 hour traffic survey? How to calculate ADT More for example here - http://en.wikipedia.org/wiki/Annual_average_daily_traffic. Average Annual Daily Potential Solar Energy The average daily potential watt hours of solar energy were A The calculation of average annual daily potential is
Michigan Department of Transportation - The 2003 Average Daily Traffic (ADT) Map Example Export File; AADT Calculation – AASHTO Method. AADT = Annual Average Daily Traffic. Calculation: AADT = VOL x SF x AF
The precision of average annual daily traffic volume estimates from seasonal traffic counts: Alberta example Satish C. Sharma, Peter Kilburn, and Yongquiang Wu Improved Annual Average Daily Traffic For example, AADTs are used by on the interested roads are then used to calculate Average Daily Traffic
How to calculate ADT based on a 1 hour traffic survey? How to calculate ADT More for example here - http://en.wikipedia.org/wiki/Annual_average_daily_traffic. Estimating Annual Average Daily Traffic Using Daily Adjustment Factor while maintaining an appropriate level of traffic service. To calculate accurate AADT,
ABSTRACT This thesis is aimed at estimation of Annual Average Daily Traffic (AADT) for Indian highways. Three basic topics regarding AADT estimation problem has been Traffic Census Program. TRAFFIC COUNTS See examples below. (PM) peak periods are expressed as a percentage of Annual Average Daily Traffic (AADT).
Request PDF Evaluating Annual Average Daily Traffic Calculation Methods with Continuous Truck Traffic Data Traffic volume, often measured in relation to annual calculation? each weekday using Annual Average daily traffic provides an example of one possible method to estimate hourly traffic volumes
### Traffic Census Program Caltrans - California Department
Understanding Traffic Terminology AADT ADT. Estimating Annual Average Daily Traffic Using Daily Adjustment Factor while maintaining an appropriate level of traffic service. To calculate accurate AADT,, Annual Average Daily Traffic (AADT) is the estimated mean daily traffic volume. For continuous sites, calculated by summing the Annual Average Days of the Week and.
The precision of average annual daily traffic volume. About this web site Indices definition hourly value / maximum hourly value or adjusted daily average in Common annual air quality index calculation, Average Annual Daily Traffic Volumes . Planning & Engineering Tel: (306) 786-1730 Fax: (306) 786-6880 e-mail: Average Annual Daily Traffic (AADT) is the average.
### National Performance Measures for Congestion Reliability
Project Appraisal Guidelines Railway Procurement Agency. There are three main data sources which are used to calculate annual traffic estimates. instead a representative sample of . 1 Annual average daily flow TRB Paper 01-3440 Estimation of Annual Average Daily Traffic in a Florida County Using GIS and Regression Fang Zhao, Lehman Center for Transportation Research.
• Assignment of estimated average annual daily traffic on
• The precision of average annual daily traffic volume
• 2004 Annual Average Daily Traffic Ontario
• Improved Annual Average Daily Traffic For example, AADTs are used by on the interested roads are then used to calculate Average Daily Traffic NSW Roads Traffic Volume Counts API Annual Average Traffic Count Summary - This station post 2006 at a daily level. Sample Hourly Traffic Counts - This table
The ideal scenario for estimation of Annual Average Daily Traffic (AADT) is count the number of vehicles passing over the project road throughout 365 days for 24 hours. Derivation of AADT and AAWT Factors Factors are required to convert the peak hour and inter peak flows to Average Annual Daily Traffic Calculation of
Southern Windsor County Annual Traffic Count Program Understanding Traffic Terminology AADT - Annual Average Daily Traffic is the term used to show the average Annual Average Daily Vehicle Trips (AADT) Of Selected Land Uses The following table was developed to provide general guidance for determining the Annual Average Daily
Steps on How to Calculate Noise Levels Identify the Annual Average Daily Traffic Environmental Officer for an example of a complete DNL Calculation with Derivation of AADT and AAWT Factors Factors are required to convert the peak hour and inter peak flows to Average Annual Daily Traffic Calculation of
Examples ; Tutorials ; Answers The given below is the annual average daily traffic AADT formula which shows you how to calculate the annual average daily traffic Design Controls and Criteria. For example, given a traffic count of Another commonly used measure of traffic volume is the annual average daily traffic
Request PDF Evaluating Annual Average Daily Traffic Calculation Methods with Continuous Truck Traffic Data Traffic volume, often measured in relation to annual Traffic Calculation No. of Years to Project Traffic (yrs): Determine Past and Future ESALs Two-Way Average Daily Traffic (ADT): Directional Distribution Factor (%):
Analysis Techniques: Annual Analysis. It is important to calculate the mean annual flow for each year in the period of (=average( ) )can be used. For example, Traffic Reports User Documentation (Annual Average Daily Traffic): Fridays are excluded from the calculation as they tend to have traffic patterns that do not
ESTIMATION OF ANNUAL AVERAGE DAILY TRAFFIC ON LOCAL ROADS IN Annual average daily traffic For instance, AADT assists in the calculation Traffic Census Program. TRAFFIC COUNTS See examples below. (PM) peak periods are expressed as a percentage of Annual Average Daily Traffic (AADT).
## Project Appraisal Guidelines Railway Procurement Agency
Traffic Data and Analysis Manual Texas Department of. Traffic volume, often measured in relation to annual average daily traffic (AADT), is a fundamental output of traffic monitoring programs. At continuous count sites, How can I calculate the equivalent single axle load (ESAL) How do I calculate Annual Average Daily Traffic How will I start the calculation of load for.
### How to calculate ADT based on a 1 hour traffic survey?
Estimation of Annual Average Daily Traffic in a Florida. calculation? each weekday using Annual Average daily traffic provides an example of one possible method to estimate hourly traffic volumes, Analysis Techniques: Annual Analysis. It is important to calculate the mean annual flow for each year in the period of (=average( ) )can be used. For example,.
Most Travelled Urban Highways Average Annual Daily Traffic (AADT) > 250,000 Steps on How to Calculate Noise Levels Identify the Annual Average Daily Traffic Environmental Officer for an example of a complete DNL Calculation with
The Annual Traffic Report contains Annual Average Daily Traffic Data from these are used in the calculation of seasonal traffic variations For example: 155 There are three main data sources which are used to calculate annual traffic estimates. instead a representative sample of . 1 Annual average daily flow
Estimating Annual Average Daily Traffic Using Daily Adjustment Factor while maintaining an appropriate level of traffic service. To calculate accurate AADT, An example of the complete calculation for a 10-year average monthly Take daily rainfall readings at "How to Calculate Average Monthly Rainfall
Annual Average Daily Traffic Estimation from Seasonal Traffic Counts Estimation of Annual Average Daily truck Traffic volume. Example Export File; AADT Calculation – AASHTO Method. AADT = Annual Average Daily Traffic. Calculation: AADT = VOL x SF x AF
Daily Vehicle Miles Traveled (DVMT) is a simple mechanism to measure how much traffic is flowing along a roadway during an average 24 hour period. Annual Average Daily Traffic (AADT) is an estimate of the average daily traffic along a defined segment of roadway. This value is calculated from short term counts
Annual Average Daily Traffic (AADT): Beginning 1977 - Annual Average Daily Traffic (AADT) is an estimate of the average daily traffic along a defined segment of roadway. The precision of average annual daily traffic volume estimates from seasonal traffic counts: Alberta example Satish C. Sharma, Peter Kilburn, and Yongquiang Wu
19/07/2017В В· How to Calculate Compounded Annual Growth Rate. indicates the average annual rate Calculating Compound Annual Growth Rate Calculating CAGR in Excel Section 2: Traffic Characteristics Traffic volumes may be expressed in terms of average daily traffic or design hourly volumes. For example,
Annual Average Daily Vehicle Trips (AADT) Of Selected Land Uses The following table was developed to provide general guidance for determining the Annual Average Daily Annual average daily traffic is abbreviated as AADT, is an assessment measure used for planning of transportation and transportation engineering. The below given is
Traffic Calculation No. of Years to Project Traffic (yrs): Determine Past and Future ESALs Two-Way Average Daily Traffic (ADT): Directional Distribution Factor (%): Definition, Interpretation, and Calculation of Annual average daily traffic, tools with the intent of obtaining a representative sample of the range of
Southern Windsor County Annual Traffic Count Program Understanding Traffic Terminology AADT - Annual Average Daily Traffic is the term used to show the average How to calculate ADT based on a 1 hour traffic survey? How to calculate ADT More for example here - http://en.wikipedia.org/wiki/Annual_average_daily_traffic.
Calculations convert current and forecast average daily traffic, For example, if the design year "How to Calculate ESAL From ADT." Career Trend, Annual average daily traffic is abbreviated as AADT, is an assessment measure used for planning of transportation and transportation engineering. The below given is
AADT Calculation Short Tutorials. Annual Average Daily Vehicle Trips (AADT) Of Selected Land Uses The following table was developed to provide general guidance for determining the Annual Average Daily, How can I calculate the equivalent single axle load (ESAL) How do I calculate Annual Average Daily Traffic How will I start the calculation of load for.
### Traffic Data Program British Columbia
GIS TOOLS TO ESTIMATE AVERAGE ANNUAL DAILY TRAFFIC. (ACR) axle counts to develop average annual daily traffic (AADT) figures. Statewide Traffic Data Traffic Data and Analysis Manual 1-8 TxDOT 9/2001 Section 4, About this web site Indices definition hourly value / maximum hourly value or adjusted daily average in Common annual air quality index calculation.
Annual average daily traffic Wikipedia. Improved Annual Average Daily Traffic For example, AADTs are used by on the interested roads are then used to calculate Average Daily Traffic, Design Controls and Criteria. For example, given a traffic count of Another commonly used measure of traffic volume is the annual average daily traffic.
### MDOT Annual Average Daily Traffic (AADT) Maps
Estimation of Annual Average Daily Traffic (AADT) for. GIS TOOLS TO ESTIMATE AVERAGE ANNUAL DAILY TRAFFIC GIS Tools to Estimate Average Annual Daily Traffic For example, Figure ES-1 shows where traffic engineers have Average Annual Daily Potential Solar Energy The average daily potential watt hours of solar energy were A The calculation of average annual daily potential is.
The ideal scenario for estimation of Annual Average Daily Traffic (AADT) is count the number of vehicles passing over the project road throughout 365 days for 24 hours. Evaluating Annual Average Daily Traffic Calculation Methods with Continuous Truck Traffic Data. Traffic volume, often measured in relation to annual average daily
COMMONWEALTH of VIRGINIA DEPARTMENT of TRANSPORTATION. Examples of such roadways are the interstate system and routes with Annual Average Daily Traffic. Derivation of AADT and AAWT Factors Factors are required to convert the peak hour and inter peak flows to Average Annual Daily Traffic Calculation of
AADT Calculation. How do you calculate Average Annual Daily Traffic (aadt)? How to Calculate Average Annual Daily Traffic? AADT is the abbreviation of Annual average NSW Roads Traffic Volume Counts API Annual Average Traffic Count Summary - This station post 2006 at a daily level. Sample Hourly Traffic Counts - This table
The Annual Traffic Report contains Annual Average Daily Traffic Data from these are used in the calculation of seasonal traffic variations For example: 155 Derivation of AADT and AAWT Factors Factors are required to convert the peak hour and inter peak flows to Average Annual Daily Traffic Calculation of
AADT Calculation. How do you calculate Average Annual Daily Traffic (aadt)? How to Calculate Average Annual Daily Traffic? AADT is the abbreviation of Annual average Annual Average Daily Traffic Estimation from Seasonal Traffic Counts Estimation of Annual Average Daily truck Traffic volume.
Calculations convert current and forecast average daily traffic, For example, if the design year "How to Calculate ESAL From ADT." Career Trend, GIS TOOLS TO ESTIMATE AVERAGE ANNUAL DAILY TRAFFIC GIS Tools to Estimate Average Annual Daily Traffic For example, Figure ES-1 shows where traffic engineers have
Annual Average Daily Traffic (AADT) For example, revised accident rates and costs Volume 5 Section 1 Part 3 TA 46/97 Annual Average Daily Traffic (AADT) For example, revised accident rates and costs Volume 5 Section 1 Part 3 TA 46/97
NRA Project Appraisal Guidelines Unit 16.1: Annual Average Daily Traffic AADT is best calculated by using a long sample set of data, > Publications > Ontario Provincial Highways Traffic Volumes report on both that segment's distance in kilometers and the annual average daily traffic
View all posts in Quebec category | 3,354 | 17,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.867256 |
https://tutorme.com/blog/tag/algebra/page2/ | 1,642,360,727,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00679.warc.gz | 656,932,053 | 41,600 | Enable contrast version
# TutorMe Blog
## Is a Rectangle a Parallelogram? Or Is a Parallelogram a Rectangle?
Inactive
Andrew Lee
June 24, 2021
Both rectangles and parallelograms are quadrilaterals, which means that they are polygons with four sides. But is a rectangle a parallelogram? Or is a parallelogram a rectangle? Here's a review of the different properties of both parallelograms and rectangles.
A parallelogram is a quadrilateral that has:
• Two pairs of parallel sides
• Two pairs of opposite sides that are congruent
A rectangle is a quadrilateral that has:
• Four right angles
• Two pairs of opposite sides that are congruent
A rhombus is a quadrilateral that has:
• Four sides of equal length
• Two pairs of equal opposite angles
## What's the Difference Between Expressions and Equations?
Inactive
Jana Russick
June 23, 2021
Expressions and equations are terms you've probably heard in real-life, most often in a high school math class. But do you know the real difference? They both can have numbers and variables, but there’s one key difference. Let's explore what this is and show you how to simplify and evaluate expressions and equations.
## How to Work With Absolute Value Inequalities, Step by Step
Inactive
Andrew Lee
June 23, 2021
You might already be familiar with solving inequalities, but what happens when you add an absolute value expression into the mix? Let’s go through the steps it takes to solve absolute value inequalities.
The absolute value symbol looks like this: |a|, where a is any real number. The absolute value sign means that the expression is no longer just a, but the distance a is from 0 on a number line.
A quick example: |2| = 2, because positive 2 is a distance of 2 from 0 on a number line.
Similarly, |-2| = 2, because negative 2 is also a distance of 2 from 0 on a number line.
## What Are Like Terms, and How Do You Combine Them?
Inactive
Jana Russick
June 13, 2021
What are like terms? They are terms whose variables and exponents have the same value. If two terms have different variables or different exponents, they are called unlike terms. Here are some examples of like and unlike terms:
Like terms:
Unlike terms:
Let's break down the defining features of like terms and show you how to combine them.
## How to Write a Fraction as a Decimal: 2 Simple Methods
Inactive
Jana Russick
June 10, 2021
If you’re looking for the easiest solution for how to write a fraction as a decimal, we’re here to help. Here, we’ll guide you through three simple ways to convert fractions into decimals. | 603 | 2,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-05 | latest | en | 0.912287 |
https://www.numbersaplenty.com/7718269 | 1,624,404,647,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00154.warc.gz | 824,728,598 | 3,426 | Cookie Consent by FreePrivacyPolicy.com
Search a number
7718269 = 13619733
BaseRepresentation
bin11101011100010101111101
3112112010110211
4131130111331
53433441034
6433232421
7122414146
oct35342575
915463424
107718269
1143a1939
122702711
1317a3130
14104cacd
15a26d64
hex75c57d
7718269 has 8 divisors (see below), whose sum is σ = 8449112. Its totient is φ = 7007040.
The previous prime is 7718257. The next prime is 7718279. The reversal of 7718269 is 9628177.
It can be written as a sum of positive squares in 4 ways, for example, as 84100 + 7634169 = 290^2 + 2763^2 .
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 7718269 - 215 = 7685501 is a prime.
It is a super-2 number, since 2×77182692 = 119143352712722, which contains 22 as substring.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (7718219) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 4074 + ... + 5659.
It is an arithmetic number, because the mean of its divisors is an integer number (1056139).
Almost surely, 27718269 is an apocalyptic number.
It is an amenable number.
7718269 is a deficient number, since it is larger than the sum of its proper divisors (730843).
7718269 is a wasteful number, since it uses less digits than its factorization.
7718269 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 9807.
The product of its digits is 42336, while the sum is 40.
The square root of 7718269 is about 2778.1772801605. The cubic root of 7718269 is about 197.6241296267.
The spelling of 7718269 in words is "seven million, seven hundred eighteen thousand, two hundred sixty-nine".
Divisors: 1 13 61 793 9733 126529 593713 7718269 | 590 | 1,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-25 | latest | en | 0.866603 |
https://stonespounds.com/16-2-pounds-in-stones-and-pounds | 1,637,989,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00142.warc.gz | 641,840,063 | 4,885 | # 16.2 pounds in stones and pounds
## Result
16.2 pounds equals 1 stones and 2.2 pounds
You can also convert 16.2 pounds to stones.
## Converter
Sixteen point two pounds is equal to one stones and two point two pounds (16.2lbs = 1st 2.2lb). | 75 | 245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-49 | latest | en | 0.8695 |
https://lnipe.net/questions/1207790-a-050-kilogram-cart-is-rolling-at-a-speed-of-040-meter-per-second-if-the-speed-of-the-cart-is-d.html | 1,620,934,518,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00413.warc.gz | 395,474,554 | 10,408 | Scotty383
2021-03-21 16:20:27
A 0.50-kilogram cart is rolling at a speed of 0.40 meter per second. If the speed of the cart is doubled, the inertia of the cart is (1) halved (3) quadrupled (2) doubled (4) unchanged | 75 | 214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-21 | latest | en | 0.765886 |
https://subscription.packtpub.com/book/data/9781789956399/5/ch05lvl1sec24/comparison-of-signatures | 1,725,754,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00641.warc.gz | 533,697,061 | 28,364 | #### Overview of this book
Starting with the basics, Applied Unsupervised Learning with R explains clustering methods, distribution analysis, data encoders, and features of R that enable you to understand your data better and get answers to your most pressing business questions. This book begins with the most important and commonly used method for unsupervised learning - clustering - and explains the three main clustering algorithms - k-means, divisive, and agglomerative. Following this, you'll study market basket analysis, kernel density estimation, principal component analysis, and anomaly detection. You'll be introduced to these methods using code written in R, with further instructions on how to work with, edit, and improve R code. To help you gain a practical understanding, the book also features useful tips on applying these methods to real business problems, including market segmentation and fraud detection. By working through interesting activities, you'll explore data encoders and latent variable models. By the end of this book, you will have a better understanding of different anomaly detection methods, such as outlier detection, Mahalanobis distances, and contextual and collective anomaly detection.
Applied Unsupervised Learning with R
Preface
Free Chapter
Introduction to Clustering Methods
Probability Distributions
Dimension Reduction
Data Comparison Methods
Anomaly Detection
## Comparison of Signatures
Next, we can compare these two signatures, to see whether they have mapped our different images to different signature values.
You can compare the signatures with one simple line of R code as follows:
`comparison<-mean(abs(borges_signature-building_signature))`
This comparison takes the absolute value of the difference between each element of the two signatures, and then calculates the mean of those values. If two signatures are identical, then this difference will be 0. The larger the value of comparison, the more different the two images are.
In this case, the value of comparison is 0.644, indicating that on average, the corresponding signature entries are about 0.644 apart. This difference is substantial for a dataset where the values only range between 1 and -1. So we see that our method for creating signatures has created very different signatures for very different images, as we would expect.
Now, we can calculate a signature for an image that is very similar... | 460 | 2,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-38 | latest | en | 0.894085 |
http://mathhelpforum.com/algebra/8110-please-help-step-step.html | 1,526,866,704,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863901.24/warc/CC-MAIN-20180521004325-20180521024325-00206.warc.gz | 189,273,919 | 9,579 | 8^-1 /y +3^-1 =-3
2. Originally Posted by jerryramos
8^-1 /y +3^-1 =-3
$\displaystyle \frac{8^{-1}}{y} + 3^{-1} = -3$
$\displaystyle \frac{1}{8y} + \frac{1}{3} = -3$
I'm going to clear the fractions: The least common multiple of 8y and 3 is 24y.
$\displaystyle 24y \left ( \frac{1}{8y} + \frac{1}{3} \right ) = 24y \cdot (-3)$
$\displaystyle 3 + 8y = -72y$
$\displaystyle 3 + 8y - 8y = -72y - 8y$
$\displaystyle 3 = -80y$
$\displaystyle y = - \frac{3}{80}$
-Dan | 207 | 470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.540437 |
http://www.justanswer.com/homework/5pdf3-accounting.html | 1,464,578,693,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049288709.26/warc/CC-MAIN-20160524002128-00121-ip-10-185-217-139.ec2.internal.warc.gz | 605,847,840 | 26,132 | • 100% Satisfaction Guarantee
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# Accounting
### Customer Question
1) Anthony Company uses a perpetual inventory system. It entered into the following purchases and sales transactions for March.
Date Activities Units Acquired at Cost Units Sold at Retail Mar. 1 Beginning inventory 160 units @ \$52.20/unit Mar. 5 Purchase 255 units @ \$57.20/unit Mar. 9 Sales 320 units @ \$87.20/unit Mar. 18 Purchase 115 units @ \$62.20/unit Mar. 25 Purchase 210 units @ \$64.20/unit Mar. 29 Sales 190 units @ \$97.20/unit Totals 740 units 510 units
10 points
3 Compute the cost assigned to ending inventory using (a) FIFO, (b) LIFO, (c) weighted average, and (d) specific identification. For specific identification, the March 9 sale consisted of 95 units from beginning inventory and 225 units from the March 5 purchase; the March 29 sale consisted of 75 units from the March 18 purchase and 115 units from the March 25 purchase. (Due to rounding, the sum of Cost of Goods Sold and Ending inventory may not equal the Cost of Good available for sales. Round your per unit costs to 3 decimal places and inventory balances to the nearest dollar amount. Omit the "\$" sign in your response.)
2) The inventory turnover ratio:
Calculation depends on the company's inventory valuation method. Reveals how many times a company turns over (sells) its merchandise inventory. Is used to measure solvency. Validates the acid-test ratio. Is used to analyze profitability.
3) On December 31 of the current year, Hewett Company reported an ending inventory balance of \$213,000. The following additional information is also available:
• Hewett sold goods costing \$37,600 to Trump Enterprises on December 28 and shipped the goods on that date with shipping terms of FOB shipping point. The goods were not included in the ending inventory amount of \$213,000 because they were not in Hewett's warehouse. • Hewett purchased goods costing \$43,600 on December 29. The goods were shipped FOB destination and were received by Hewett on January 2 of the following year. The shipment was a rush order that was supposed to arrive by December 31. These goods were included in the ending inventory balance of \$213,000. • Hewett's ending inventory balance of \$213,000 included \$14,600 of goods being held on consignment from Rumsfeld Company. (Hewett Company is the consignee.) • Hewett's ending inventory balance of \$213,000 did not include goods costing \$94,600 that were shipped to Hewett on December 27 with shipping terms of FOB destination and were still in transit at year-end.
Based on the above information, the correct balance for ending inventory on December 31 is:
\$198,400 \$169,400 \$192,400 \$154,800 \$207,000
4) The amount recorded for merchandise inventory includes all of the following except:
Freight costs paid by the buyer.
Purchase discounts.
Returns and allowances.
Freight costs paid by the seller.
5)
Aug. 1 Purchased merchandise from Abilene Company for \$6,000 under credit terms of 1/10, n/30, FOB destination, invoice dated August 1. 4 At Abilene's request, Stone paid \$100 cash for freight charges on the August 1 purchase, reducing the amount owed to Abilene. 5 Sold merchandise to Lux Corp. for \$4,200 under credit terms of 2/10, n/60, FOB destination, invoice dated August 5. The merchandise had cost \$3,000. 8 Purchased merchandise from Welch Corporation for \$5,300 under credit terms of 1/10, n/45, FOB shipping point, invoice dated August 8. The invoice showed that at Stone’s request, Welch paid the \$240 shipping charges and added that amount to the bill. (Hint: Discounts are not applied to freight and shipping charges.) 9 Paid \$120 cash for shipping charges related to the August 5 sale to Lux Corp. 10 Lux returned merchandise from the August 5 sale that had cost Stone \$500 and been sold for \$700. The merchandise was restored to inventory. 12 After negotiations with Welch Corporation concerning problems with the merchandise purchased on August 8, Stone received a credit memorandum from Welch granting a price reduction of \$800. 15 Received balance due from Lux Corp. for the August 5 sale less the return on August 10. 18 Paid the amount due to Welch Corporation for the August 8 purchase less the price reduction granted. 19 Sold merchandise to Trax Co. for \$3,600 under credit terms of 1/10, n/30, FOB shipping point, invoice dated August 19. The merchandise had cost \$2,500. 22 Trax requested a price reduction on the August 19 sale because the merchandise did not meet specifications. Stone sent Trax a \$600 credit memorandum to resolve the issue. 29 Received Trax's cash payment for the amount due from the August 19 sale. 30 Paid Abilene Company the amount due from the August 1 purchase.
Prepare journal entries to record the above merchandising transactions of Stone Company, which applies the perpetual inventory system. (Identify each receivable and payable; for example, record the purchase on August 1 in Accounts Payable—Abilene.) (Omit the "\$" sign in your response.)
Date General Journal Debit Credit Aug. 1 4 5 8 9 10 12 15 18 19 22 29 30
6)
A company has inventory of 19 units at a cost of \$21 each on June 1. On June 3, it purchased 31 units at \$23 each. 22 units are sold on June 5. Using the FIFO periodic inventory method, what is the cost of the 22 units that were sold?
\$472. \$462. \$483. \$468. \$493.
A company uses the perpetual inventory system and recorded the following entry
This entry reflects a:
Payment of the account payable and recognition of a 2% cash discount taken. Payment of the account payable and recognition of a 1% cash discount taken. Purchase of merchandise on credit. Sale of merchandise on credit. Return of merchandise.
7) On October 1, Robinson Company sold merchandise in the amount of \$7,200 to Rosser, with credit terms of 1/10, n/30. The cost of the items sold is \$5,400. Robinson uses the perpetual inventory system. The journal entry or entries that Robinson will make on October 1 is:
8) A company purchased \$2,600 of merchandise on December 5. On December 7, it returned \$600 worth of merchandise. On December 8, it paid the balance in full, taking a 3% discount. The amount of the cash paid on December 8 equals:
\$2,600. \$600. \$1,940. \$2,522. \$2,000.
Bottom of Form
9)
A company had inventory on November 1 of 5 units at a cost of \$11 each. On November 2, they purchased 18 units at \$13 each. On November 6 they purchased 14 units at \$16 each. On November 8, 16 units were sold for \$46 each. Using the LIFO perpetual inventory method, what was the value of the inventory on November 8 after the sale?
\$263 \$258 \$242 \$274 \$315
Submitted: 4 years ago.
Category: Homework
Expert: F. Naz replied 4 years ago.
Data is so confusing, please arrange and send it properly. You may upload the word format file at www.mediafire.com and give me the sharing key, thanks.
Customer: replied 4 years ago.
There were some other questions earlier. I really don't understand the perpetual inventory system.
Customer: replied 4 years ago.
http://www.mediafire.com/?p77bocmsn6b94d5
Customer: replied 4 years ago.
Just curious as to whether or not the data was received and how long a response will take?
Expert: F. Naz replied 4 years ago.
I have recieved but they are too many, You will receive the email for fair price soon. thanks.
Customer: replied 4 years ago.
I no longer need an answer, this process was too long and drawn out and it took too long for a response.
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3,000+ satisfied customers, all topics, A+ work | 2,524 | 10,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2016-22 | longest | en | 0.888229 |
https://www.coursehero.com/file/58825/Physics-Lab-report-5-Force-Mass-and-Acceleration/ | 1,495,754,017,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00614.warc.gz | 864,168,153 | 24,811 | Physics Lab report 5, Force, Mass and Acceleration
# Physics Lab report 5, Force, Mass and Acceleration -...
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Unformatted text preview: Alexander Hyatt Physics Lab 124-021 February 21, 2007 Abstract This experiment demonstrates the relationship between force, mass and acceleration. The lab was set up so that a glider was set on an air track, and a constant force would push the glider along the track. The first experiment involved an air track raised at different heights only on one side. The higher the track was raised, the greater the acceleration of the glider. The other experiments, differentiated only by differing glider masses, both used the air track at a level position, and used a hanging mass to pull the glider along the track. The acceleration was found experimentally. For the first experiment, the obtained data was used to find an experimental value for g, and then the % error was found for the difference between the experimental and actual values of g. The same was done for the other experiments except with respect to m1 instead of g. Experiment Table 1 Data Table Exercise 1 D (cm) =100 H (cm) sin(theta ) a (m/s^2) 1.29 0.0129 0.132 2.59 0.0259 0.263 3.89 0.0389 0.384 5.18 0.0518 0.516 Calculated G 9.8148 Percent Error 0.05% Table 1 deals with the data used to find the experimental value for gravity, or g. In this exercise, the air track was raised up a certain height, and then the acceleration was recorded at each height. Also, the angle of the incline was found by take arcsin of the Height (H) divided by the Distance (D). The g value was found by making a graph and making a line whos slope is: a/sin(theta). The experimental g value was found to be 9.8148m/s^2, while the actual value is 9.81m/s^2. The percent error between the two values is a negligible .05%. Alexander Hyatt Physics Lab 124-021 February 21, 2007 Table 2 Data Table Exercise 2 m1= .3308 m2 1/m2 a g/a 0.070 14.286 1.661 5.906 0.090 11.111 2.035 4.821 0.110 9.091 2.395 4.096 0.130 7.692 2.669 3.676 0.150 6.667 3.013 3.256 0.170 5.882 3.295 2.977 0.190 5.263 3.416 2.872 0.210 4.762 3.621 2.709 Calculated m1 0.3393 Percent Difference 2.54% Calculated y- intercept 1.043 Percent Difference 4.21% Table 2 is the data collected from exercise 2. Table 2 is the data collected from exercise 2....
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## This note was uploaded on 04/03/2008 for the course PHYS 124 taught by Professor Jay during the Spring '07 term at Clemson.
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Ask a homework question - tutors are online | 822 | 2,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-22 | longest | en | 0.932591 |
http://yetanothermathprogrammingconsultant.blogspot.com/2015/07/monotonic-transformations.html | 1,531,889,881,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00535.warc.gz | 569,048,039 | 12,854 | ## Tuesday, July 14, 2015
### monotonic transformations
From: lp_solve@yahoogroups.com
To: lp_solve@yahoogroups.com
Date: Thu, 9 Jul 2015 00:32:47 -0700
Subject: [lp_solve] C# to set the object with the form of square
Could LP set the object in a form of square? Just like this :
min: Q^2
If we have Q≥0 then just use MIN Q. We can apply a monotonic function. I have used this trick a lot in the context of minimizing distances. Instead of
we can keep things quadratic by using
This got rid of a nasty square root and this allowed us use quadratic solvers.
Back to the original problem where we want to remove a square. What if we don’t know in advance that Q≥0? Then we can solve two problems:
and pick the best one.
The suggested solution in the thread: “Have a look into piecewise linear approximations” is at least much overkill. | 224 | 842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-30 | latest | en | 0.894781 |
https://www.kimdutoit.com/2018/02/19/just-to-mess-with-ya/ | 1,709,299,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00201.warc.gz | 832,193,401 | 12,861 | # Just To Mess With Ya
Here’s an interesting math situation, wherein I prove that 2=1:
1.) Suppose you have quantities A and B, and suppose they are equal. That is,
A = B
2.) Multiply both sides by A:
A^2 = AB
3.) Now subtract B^2 from both sides:
A^2-B^2 = AB-B^2
4.) Factor both sides:
(A+B)(A-B) = B(A-B)
5.) Divide both sides by the common factor (A-B):
A+B = B
6.) Now, remembering that A=B, we have
B+B=B, or 2B=B
7.) Divide both sides by B:
2=1
And now, children, you will understand how Congress creates the national budget.
/Lewis Carroll
1. fastrichard says:
I had to look at that for a little while to figure out where it goes wrong. I’m not convinced that congress uses even that level of rational logic to arrive at what they do.
2. Weetabix says:
You just caused a discontinuity in space-time by doing the impermissible. đ
To preserve the joy of the next readers, I won’t spoil it by saying what.
1. Mike M. says:
Yup, this is an old one….
3. Roy says:
I won’t give it away exactly, but I will point out that if A=B then A-B=?. You do the math.
1. Kim du Toit says:
No fair using calculus…
;=) | 328 | 1,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-10 | latest | en | 0.909345 |
https://www.numbersaplenty.com/8578091 | 1,624,338,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488507640.82/warc/CC-MAIN-20210622033023-20210622063023-00502.warc.gz | 814,928,315 | 3,238 | Search a number
8578091 is a prime number
BaseRepresentation
bin100000101110…
…010000101011
3121010210221002
4200232100223
54143444331
6503505215
7132625004
oct40562053
917123832
108578091
114929935
122a5820b
131a145c2
1411d41ab
15b469cb
hex82e42b
8578091 has 2 divisors, whose sum is σ = 8578092. Its totient is φ = 8578090.
The previous prime is 8578067. The next prime is 8578093. The reversal of 8578091 is 1908758.
8578091 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-8578091 is a prime.
Together with 8578093, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (8578093) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4289045 + 4289046.
It is an arithmetic number, because the mean of its divisors is an integer number (4289046).
Almost surely, 28578091 is an apocalyptic number.
8578091 is a deficient number, since it is larger than the sum of its proper divisors (1).
8578091 is an equidigital number, since it uses as much as digits as its factorization.
8578091 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20160, while the sum is 38.
The square root of 8578091 is about 2928.8378241207. The cubic root of 8578091 is about 204.7058322955.
The spelling of 8578091 in words is "eight million, five hundred seventy-eight thousand, ninety-one". | 485 | 1,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-25 | latest | en | 0.868586 |
https://betterlesson.com/lesson/473288/spread-out?from=breadcrumb_lesson | 1,628,196,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00178.warc.gz | 147,256,898 | 24,739 | 2 teachers like this lesson
Print Lesson
## Objective
SWBAT rank data sets from least to most spread out based on their own reasoning. SWBAT examine and compare three different methods for ranking data. SWBAT understand mean absolute deviation.
#### Big Idea
How would you determine how spread out a data set is? Students put data sets in order from least to most spread out based on their own intuition and reasoning.
## Opening
10 minutes
The two activities in today's lesson lay the foundation for students to understand standard deviation. The first activity is called "Data Spread" and can be found on page 323 of IMP's Year 1 textbook (2009). The activity asks students to rank 4 sets of data that have the same mean and median from the least spread out to the most spread out and asks them explain their reasoning. I can begin class by listing the 4 sets of data on the board and asking students to rank them on their own from least to most spread out. The data are named as groups from A to D. I then ask students to share their rankings and their reasons why they ranked them in that order. We will refer back to these lists later, so I save them on my Smartboard (or take a picture or record them in some other way).
Here is a screen capture from our rankings sharing and discussion this year. I tried to keep track of some of the ways students chose to rank their data from least to most spread out.
## Investigation
30 minutes
Students now look at three different ways to arrange the data. These three different methods ask them to consider what spread out means in different ways.
• The first method is in the same activity and asks students to rank the data sets by range. The second and third methods are in the following activity "Kai and Mai Spread Data" on page 324-325 of IMP's Year 1 textbook (2009).
• The second method, Kai's method, is very close to mean absolute deviation. Kai's method takes the distance from the mean for each data item and adds those distances together to "score" the data sets.
• The third method asks students to remove the highest and lowest values and then find the data item that is farthest from the original mean. This number is then assigned as the score to that data set and the sets are again ranked.
Students work to arrange the data from least to most spread out for each of the three methods. Here is an example of some student work from this activity. We can see from this piece of work how the student "scored" each set of data according to the different methods that we assigned.
As students work, they may find that they like one of the methods better than the others. I try to get them to capture this thinking by writing it down and recording their thoughts. I also want to encourage them to think about why this method is a better way of measuring spread. These thoughts and comparisons can often lead to rich discussion.
Once they've sorted the data using the three different methods, students compare the new spreads with their original ranking of the data to see which method is closest to their own (it is likely that some students will have used some of the exact methods, especially using the range).
## Discussion + Closing
20 minutes
Next, I bring the class back together as a whole group and ask if anyone would like to change their original ranking. I also ask students what method they liked best and why and how similar their own rankings were to the three methods.
I bring up the concept of outliers in the context of the third method, though only one of the sets of data really has true outliers.
We spend some time talking about the second method, which is very close to mean absolute deviation. I bring up absolute value in the context of the data items. I point out to students that sometimes they are subtracting the mean value from a data item that is larger than the mean and sometimes they are subtracting a smaller value than the mean from the mean. I ask them how they could standardize this process and try to elicit the idea of absolute value. I introduce the symbol "x bar" to students and ask them to generalize this difference from the mean.
I also show students here that mean absolute deviation is just one more step from this point. Once they "score" the data set based on the sum of the differences from the mean, they only need to divide by the number of data items to get mean absolute deviation. I let them know that this is a simpler form of something called standard deviation that they will be learning about next. Teaching mean absolute deviation here helps students to have a more intuitive understanding of standard deviation later so I try to make sure students have a solid understanding of the concept.
If students are familiar with summation notation, I might show them how mean absolute deviation is written here. I find students like to be able to interpret what looks like complicated symbols and making sense of them gives them a sense of accomplishment.
To close class today, I give students the opportunity to reflect on these three different methods. I ask students to complete an exit ticket in response to the following prompt:
Which method did you like best to determine the ranking of least spread out to most spread out data sets? Why? How was this method similar or different from your original way of sorting the sets?
## Citations
1. This material is adapted from the IMP Teacher’s Guide, © 2010 Interactive Mathematics Program. Some rights reserved. | 1,145 | 5,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-31 | latest | en | 0.953246 |
https://mathematica.stackexchange.com/questions/116457/how-to-use-abs-within-event-with-a-function-of-several-variables | 1,716,513,160,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00255.warc.gz | 316,498,778 | 40,243 | # How to use Abs within Event with a function of several variables?
I would like to use EventLocator to stop NDSolve with the Event command
"Event" -> Abs[D[f[x,y,t],t]] < threshold
That is, I need to check the time derivative of function f[x,y,t] over a spatial domain a >= 0 && a <= L, b >= 0 && b <= L at each integration step. Beside, it is clear that at that moment the temporal boundary is t. When I run it, NDSolve gives:
"The function value \ Abs[InterpolatingFunction[...]+<<1>><<<1>> is not True or False \ when the arguments are {3.187144811900335*^-8,<<4>>}"
I am not familiar with Event, but I understand that one should use a and b instead of x and y in the Abs[...], because x and y are NDSolve variables and it is better to avoid inserting them in the numerical functions.
Actually, what I need is something like this:
"Event" -> Abs[D[f[a,b,t],t], a >= 0 && a <= L, b >= 0 && b <= L] < threshold
Obviously, it is not correct in the syntax of Abs, how can I implement the 'Event' properly to this end. Thank you!
Here is an example for a 1D PDE modified from the documentation, where the integration will stop as long as Derivative[1, 0][u][t, x] > 1:
eqn = {
D[u[t, x], t, t] == D[u[t, x], x, x] - Sin[u[t, x]],
u[0, x] == E^(-(x - 5)^2) + E^(-(x + 5)^2/2), Derivative[1, 0][u][0, x] == 0,
u[t, -50] == u[t, 50]
}
NDSolve[eqn, u, {t, 0, 100}, {x, -50, 50},
Method -> {
"MethodOfLines", "DiscretizedMonitorVariables" -> True,
"SpatialDiscretization" -> {"TensorProductGrid", "DifferenceOrder" -> "Pseudospectral"},
Method -> {
"EventLocator",
"Event" :>
If[And @@ (# < 1 & /@ Abs[Derivative[1, 0][u][t, x]]), 0, 1],
"EventLocationMethod" -> "StepBegin"
}
}
]
• Perhaps you could add a link to the relevant documentation page as well. Jun 1, 2016 at 17:15
• @MarcoB Thanks for the editing. I have added the link. Jun 1, 2016 at 17:51
• @can Derivative[1, 0][u][t, x] represents a series of values corresponding to different x in the solving region at the current time t. For example, {u[t,x1],u[t,x2],...}.I'm trying to check whether there is at least one of those values breaks our condition, so we will stop the integration. Jun 2, 2016 at 1:59
• @ xslittlegrass, thanks for your answer. In the condition of the If I understand that you Map the pure function #<1 & onto Abs to give 0 or 1 when the condition is true or false, respectively. But I don't understand that why you Apply the And to (...)? What is the intention that And only Apply only one argument? Jun 2, 2016 at 2:00
• @xslittlegrass, in my real problem, when I run the NDSolve` it gives >"The function value If[<<1>>,0,1] is not True or False when the arguments are {...}." After three warning like this, and this line: >Further output of NDSolve::nbnum1 will be suppressed during this calculation. >>. It runs normally so far, however I will accept it if the solution is numerical reasonably. Thank you! Jun 4, 2016 at 3:13 | 893 | 2,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.810597 |
http://forums.wolfram.com/mathgroup/archive/2001/Nov/msg00156.html | 1,579,700,711,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607118.51/warc/CC-MAIN-20200122131612-20200122160612-00350.warc.gz | 71,412,128 | 8,753 | Re: Limit question
• To: mathgroup at smc.vnet.net
• Subject: [mg31561] Re: Limit question
• From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
• Date: Sun, 11 Nov 2001 00:34:53 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```Allan
I think this argument has moved somewhat beyond Mathematica so I will
this aspect first. I don't think on this point there is any real
disagreement involved. I wrote in my message: "Of course one could
formulate it in different but equivalent terms...". Thus the
epsilon-delta approach is less general than talking about topology since
it uses metric space structures, but apart from that is equivalent. It
is not important whether one starts with the notion of continuity of a
function or convergence, but topology is in any case used, even if only
implicitly, since it underlies any metric space structure. One need not
to consider infinities to be objects which you "add" , but even if one
does not one ends up describing with mathematically equivalent concept
of "converging to infinity", expressed in a more metaphorical and, in
my opinion less intuitive, language.
However, let's come back to Mathematica. This thread started with the
following observation:
In[1]:=
Limit[Tan[x],x->Pi/2]
Out[1]=
-Infinity
I agree with the original poster that this is clearly wrong. It just
does not make sense to give this answer from any mathematical view
point. To explain how Mathematica arrives at this sort of thing (by
making clear out it's hidden defaults) does not, in my opinion, address
the original question: what is the mathematical justification for this
answer? I know there are a number of situations (particularly involving
integration) when Mathematica faced with an input which which from its
point of view is not sufficiently informative, will default to some
particular defaults. I think this is to some extent unavoidable, because
of the way Mathematica is constructed. Unlike more specialized programs
(like, for example, Macaulay) you do not start a Mathematica computation
by specifying rigorously the mathematical setting (e.g. domain and range
of your functions.). This results in wider applicability of Mathematica
constructions and much easier programming language, but at a cost of a
certain vagueness. Personally I am not bothered by these matters,
because in almost all all cases I can think of the "confusion" is not
really significant and one can adjust things to fit ones own intentions.
But still I think whenever possible Mathematica ought to give
mathematically sensible answers, and -Infinity is not really a sensible
With best regards
Andrzej
On Sunday, November 11, 2001, at 02:43 AM, Allan Hayes wrote:
> I agree that if we want Infinity etc. to be objects, rather than just
> words
> in a phrase, then we need to "add them" to the original spaces (or
> embed the
> original space suitably). But the language seems to me to come first,
> and to
> be more directly related to the epsilon-delta definitions and more
> amenable
> to computational treatment. Also it provides the motivation for the
> extension and an opportunity to illustrate the fascinating idea of
> creating
> structures to our needs --- Cauchy sequences ---> completions,
> ultrafilters --> compactifications, prime ideals --> .... and
> consistency --> model.
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
```
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• Next by thread: Re: Re: Re: Limit question | 869 | 3,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-05 | latest | en | 0.924543 |
https://onemathematicalcat.org/algebra_book/online_problems/solve_abs_val_ineq_gt.htm | 1,686,035,121,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00742.warc.gz | 474,321,287 | 10,976 | # Solving Absolute Value Inequalities Involving ‘Greater Than’
Need some simpler practice with absolute value and related concepts first?
This section should feel remarkably similar to the previous ones.
This section presents the tools needed to solve absolute value inequalities involving ‘greater than’, like these:
$$\begin{gather} \cssId{s6}{|x|\gt 5} \\ \\ \cssId{s7}{|x + 1|\ge 3} \\ \\ \cssId{s8}{|2 - 3x| \gt 7} \end{gather}$$
Each of these inequalities has only a single set of absolute value symbols which is by itself on the left-hand side of the sentence, and has a variable inside the absolute value. The verb is either ‘$\,\gt\,$’ (greater than) or ‘$\,\ge\,$’ (greater than or equal to).
As in the previous sections, solving sentences like these is easy, if you remember the critical fact that absolute value gives distance from $\,0\,.$
Keep this in mind as you read the following theorem:
THEOREM solving absolute value inequalities involving ‘greater than’
Let $\,x\in\mathbb{R}\,,$ and let $\,k\ge 0\,.$ Then:
$$\begin{gather} |x| \gt k\ \ \ \text{ is equivalent to }\ \ x\lt -k\ \ \text{ or }\ \ x\gt k \\ \\ |x| \ge k\ \ \ \text{ is equivalent to }\ \ x\le -k\ \ \text{ or }\ \ x\ge k \\ \end{gather}$$
## Translating the Theorem
Recall first that normal mathematical conventions dictate that ‘$\,|x| \gt k\$’ represents an entire class of sentences, including the members:
$$\begin{gather} \cssId{s22}{|x| \gt 2} \cr\cr \cssId{s23}{|x| \gt 5.7} \cr\cr \cssId{s24}{|x| \gt \frac{1}{3}} \end{gather}$$
The variable $\,k\,$ changes from sentence to sentence, but is constant within a given sentence.
Recall that ‘$\,x\lt -k\ \ \text{ or }\ \ x\gt k\$’ is an ‘or’ sentence, where that's the mathematical word ‘or’. An ‘or’ sentence is true when at least one of the subsentences is true. Thus, the sentence ‘$\,x\lt -k\,$ or $\,x\gt k\,$’ is true for all the numbers to the left of $\,-k\,,$ put together with all the numbers to the right of $\,k\,$:
The values of $\,x\,$ that make the sentence ‘$\,x\lt -k\ \text{ or }\ x\gt k\$’ true (for $\,k \gt 0$)
When you see a sentence of the form $\,|x| \gt k\,,$ here's what you should do:
• Check that $\,k\,$ is a nonnegative number (greater than or equal to zero).
• The symbol $\,|x|\,$ represents the distance between $\,x\,$ and $\,0\,.$
• Thus, you want the numbers $\,x\,,$ whose distance from $\,0\,$ is greater than $\,k\,$:
• You can walk from $\,0\,$ in two directions: to the left, or to the right. Walk more than $\,k\,$ units to the left, and you get all the numbers to the left of $\,-k\,.$ Walk more than $\,k\,$ units to the right, and you get all the numbers to the right of $\,k\,.$ Throw these two sets together!
• Thus, $\,|x| \gt k\$ is equivalent to $\,x\lt -k\,$ or $\,x\gt k\,.$
• Equivalent sentences are completely interchangeable, and you can use whichever is easiest to work with. In this case, you're getting rid of the troublesome absolute value in exchange for a less-troublesome compound inequality.
Recall that ‘$\iff$’ is a symbol for ‘is equivalent to’.
The power of the sentence-transforming tool
$$\cssId{s46}{|x| \gt k \ \ \iff\ \ x\lt -k\ \ \text{ or }\ \ x\gt k}$$
goes far beyond solving simple sentences like $\,|x| \gt 5\,$!
Since $\,x\,$ can be any real number, you should think of $\,x\,$ as merely representing the stuff inside the absolute value symbols. Thus, you could think of rewriting the tool as: $$\cssId{s50}{|\text{stuff}| \gt k \ \ \iff\ \ \text{stuff}\lt -k\ \ \text{ or }\ \ \text{stuff}\gt k}$$
See how this idea is used in the following examples:
## Example
Solve: $|2 - 3x| \gt 7$
Solution: Write a nice, clean list of equivalent sentences:
$|2 - 3x| \gt 7$ original sentence $2-3x\lt -7$ or $\,2-3x \gt 7$ check that $\,k\ge 0\,$; use the theorem $-3x\lt -9$ or $\,-3x \gt 5$ subtract $\,2\,$ from both sides of both subsentences $\displaystyle x\gt 3\ \ \text{or}\ \ x\lt -\frac{5}{3}$ divide by $\,-3\,$; change direction of inequality symbols $\displaystyle x\lt -\frac{5}{3}\ \ \text{or}\ \ x\gt 3$ in the web exercise, the ‘less than’ part is always reported first
It's a good idea to check the ‘boundaries’ of the solution set:
$|2 - 3(-\frac{5}{3})| = |2 + 5| = 7$
Check!
$|2 - 3(3)| = |2 - 9| = |-7| = 7$
Check!
## Example
Solve: $3|-6x + 7| \ge 9$
Solution: To use the theorem, you must have the absolute value all by itself on one side of the equation. Thus, your first job is to isolate the absolute value:
$3|-6x + 7| \ge 9$ original sentence $|-6x + 7| \ge 3$ divide both sides by $\,3$ $-6x + 7 \le -3$ or $\,-6x + 7\ge 3$ check that $\,k \ge 0\,$; use the theorem $-6x\le -10$ or $\,-6x\ge -4$ subtract $\,7\,$ from both sides of both subsentences $\displaystyle x\ge\frac{10}{6}\ \ \text{or}\ \ x\le \frac{4}{6}$ divide by $\,-6\,$; change direction of inequality symbols $\displaystyle x\ge\frac{5}{3}\ \ \text{or}\ \ x\le \frac{2}{3}$ simplify fractions $\displaystyle x\le \frac{2}{3}\ \ \text{or}\ \ x\ge\frac{5}{3}$ in the web exercise, the ‘less than’ part is always reported first
Check the ‘boundaries’ of the solution set:
$3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$
Check!
$3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$
Check!
## Example
Solve: $|5 - 2x| \gt -3$
Solution: The theorem can't be used here, since $\,k\,$ is negative. In this case, you need to stop and think.
No matter what number you substitute for $\,x\,,$ the left-hand side of the inequality will always be a number that is greater than or equal to zero, so it will always be greater than $\,-3\,.$ Therefore, this sentence has all real numbers as solutions. It is always true.
## Concept Practice
Solve the given absolute value inequality. Write the result in the most conventional way.
For more advanced students, a graph is available. For example, the inequality $\,|2 - 3x| \gt 7\,$ is optionally accompanied by the graph of $\,y = |2 - 3x|\,$ (the left side of the inequality, dashed green) and the graph of $\,y = 7\,$ (the right side of the inequality, solid purple). In this example, you are finding the values of $\,x\,$ where the green graph lies above the purple graph.
Click the ‘Show/Hide Graph’ button to toggle the graph.
Solve: | 2,023 | 6,258 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-23 | latest | en | 0.772159 |
https://www.sabanciuniv.edu/syllabus/courses.cfm?year=2020&term=01&subject=ENS&code=202&lan=eng | 1,601,478,414,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00439.warc.gz | 1,004,789,521 | 8,459 | ### Thermodynamics (ENS 202)
2020 Fall
Faculty of Engineering and Natural Sciences
Engineering Sciences(ENS)
3
6.00 / 6.00 ECTS (for students admitted in the 2013-14 Academic Year or following years)
Gözde İnce gozdeince@sabanciuniv.edu,
English
NS102
Formal lecture,Recitation
Interactive,Learner centered,Communicative,Discussion based learning,Task based learning
### CONTENT
Fundamental concepts and mathematical tools ; thermal equilibrium; Zeroth Law and definition of temperature; equations of state; First and Second Laws; thermodynamic potentials (enthalpy, Helmholtz, Gibbs) and the Maxwell relations; first order phase transitions; critical phenomena; Third Law, negative temperatures; introduction to statistical mechanics. Also part of the "core course" pools for the BIO, MAT, ME,TE degree programs.
### OBJECTIVE
To equip the students with a basic understanding of equilibrium, both from a macroscopic and a microscopic viewpoint, so that they can (i) perform the energy balance for a system and analyze the energy transfer processes in the system; (ii) interrelate various thermodynamic functions so that hard-to-measure properties may be determined through the measurable ones; (iii) develop a basic understanding of phase behavior.
### LEARNING OUTCOME
State and explain general concepts used in thermodynamics including the system and its surroundings, mechanisms of energy transfer; state versus path function.
Interpret the basic assumptions of the ideal gas law and illustrate how the van der Waals equation of state rectifies these assumptions to lead to a gas <-> liquid phase transition behavior and the critical point.
Using published data, such as heat capacity, calculate the internal energy, enthalpy changes of a system with respect to a reference state.
Apply the first law of thermodynamics by performing a detailed balance of energy transfer for a variety of real systems involving thermal energy, calculate efficiency in energy conversion
Define second law of thermodynamics and using published data calculate the entropy change of a system and surroundings
Write the entropy rate balance for control values and calculate the entropy production
Define and calculate the Gibbs and Helmholtz free energy changes in various systems using Maxwell's relations, write the differential forms of state functions
Define chemical potential and relate it to change in Gibbs energy and identify reversibility and spontaneity in changes towards equilibrium.
Describe the physical, structural, and thermodynamic properties of equilibrium phases and phase transformations in single and two-component systems
Determine the changes in thermodynamic properties in ideal, non-ideal, dilute, and in regular solutions
Draw P-V and T-V diagram of pure substances, determine the phase of a substance at different conditions
Calculate the activities and activity coefficients for real solutions
Apply the Lever Rule to determine the phase composition in a multi-phase field;
Define the ideal thermodynamic cycles for gas and gas-vapor systems and calculate the thermal efficiency
### PROGRAMME OUTCOMES
1. Understand the world, their country, their society, as well as themselves and have awareness of ethical problems, social rights, values and responsibility to the self and to others. 2
2. Understand different disciplines from natural and social sciences to mathematics and art, and develop interdisciplinary approaches in thinking and practice. 3
3. Think critically, follow innovations and developments in science and technology, demonstrate personal and organizational entrepreneurship and engage in life-long learning in various subjects. 4
4. Communicate effectively in Turkish and English by oral, written, graphical and technological means. 4
5. Take individual and team responsibility, function effectively and respectively as an individual and a member or a leader of a team; and have the skills to work effectively in multi-disciplinary teams. 4
1. Possess sufficient knowledge of mathematics, science and program-specific engineering topics; use theoretical and applied knowledge of these areas in complex engineering problems. 5
2. Identify, define, formulate and solve complex engineering problems; choose and apply suitable analysis and modeling methods for this purpose. 5
3. Develop, choose and use modern techniques and tools that are needed for analysis and solution of complex problems faced in engineering applications; possess knowledge of standards used in engineering applications; use information technologies effectively. 5
4. Ability to design a complex system, process, instrument or a product under realistic constraints and conditions, with the goal of fulfilling specified needs; apply modern design techniques for this purpose. 3
5. Design and conduct experiments, collect data, analyze and interpret the results to investigate complex engineering problems or program-specific research areas. 3
6. Knowledge of business practices such as project management, risk management and change management; awareness on innovation; knowledge of sustainable development. 2
7. Knowledge of impact of engineering solutions in a global, economic, environmental, health and societal context; knowledge of contemporary issues; awareness on legal outcomes of engineering solutions; understanding of professional and ethical responsibility. 4
1. Applying fundamental and advanced knowledge of natural sciences as well as engineering principles to develop and design new materials and establish the relation between internal structure and physical properties using experimental, computational and theoretical tools. 5
2. Merging the existing knowledge on physical properties, design limits and fabrication methods in materials selection for a particular application or to resolve material performance related problems. 4
3. Predicting and understanding the behavior of a material under use in a specific environment knowing the internal structure or vice versa. 5
1. Familiarity with concepts in statistics and optimization, knowledge in basic differential and integral calculus, linear algebra, differential equations, complex variables, multi-variable calculus, as well as physics and computer science, and ability to use this knowledge in modeling, design and analysis of complex dynamical systems containing hardware and software components. 5
2. Ability to work in design, implementation and integration of engineering applications, such as electronic, mechanical, electromechanical, control and computer systems that contain software and hardware components, including sensors, actuators and controllers. 3
1. Comprehend key concepts in biology and physiology, with emphasis on molecular genetics, biochemistry and molecular and cell biology as well as advanced mathematics and statistics. 4
2. Develop conceptual background for interfacing of biology with engineering for a professional awareness of contemporary biological research questions and the experimental and theoretical methods used to address them. 5
1. Formulate and analyze problems in complex manufacturing and service systems by comprehending and applying the basic tools of industrial engineering such as modeling and optimization, stochastics, statistics. 3
2. Design and develop appropriate analytical solution strategies for problems in integrated production and service systems involving human capital, materials, information, equipment, and energy. 4
3. Implement solution strategies on a computer platform for decision-support purposes by employing effective computational and experimental tools. 2
1. Use mathematics (including derivative and integral calculations, probability and statistics, differential equations, linear algebra, complex variables and discrete mathematics), basic sciences, computer and programming, and electronics engineering knowledge to (a) Design and analyze complex electronic circuits, instruments, software and electronics systems with hardware/software. or (b) Design and analyze communication networks and systems, signal processing algorithms or software
### ASSESSMENT METHODS and CRITERIA
Percentage (%) Final 35 Midterm 50 Participation 5 Homework 10 | 1,504 | 8,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-40 | latest | en | 0.845037 |
https://www.wyzant.com/resources/answers/21783/word_problem | 1,524,373,836,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945493.69/warc/CC-MAIN-20180422041610-20180422061610-00464.warc.gz | 920,973,833 | 18,191 | 0
Word Problem
The cost to produce one laptop computer is 890.00 plus a one time fixed cost of 100,000.00 for research and development. Each laptop computer will be sold for 1520.00.
A. Write a formula that gives the cost C of producing x laptops
B. Write a formula that gives the revenue R from selling x laptop computers
C. Profit equally revenue minus cost. Write formula that calculates the profit P from selling X laptop computers.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
2
a)
C = 890 X + 100,000
b)
R = 1520 X
c) P= 1520X - 890X - 100,000
d> 1520 X -890X - 100,000 >0
630 X > 100,000
X > 100,000/630 = 158.7
Has to sell at least 159 laptop to make a profit.
159 sold is a breakeven point.
William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
1
Dear Jason,
(A.) The cost, C, of producing x laptops would be
C = (\$890)*x + (\$100,000)/x
(B.) Revenue, R, from selling x laptops would be
R = (\$1520)*x
(C.) The profit, P, from selling x laptops is
P = (\$1520)*x - [(\$890)*x + (\$100,000/x)] or (\$630)*x - (\$100,000/x)
(D.) Let P = 0
Then
(\$630)*x - (\$100,000/x) = 0
(\$630)*x = (\$100,000/x)
Multiply both sides by x
(\$630)*x2 = \$100,000
Divide both sides by \$630
x2 = 158.73
x = 12.5988
In this case x is the minimum number of laptops that can be made and sold to start realizing a profit. However, since no one is going to make, let alone sell, 0.5988 of a computer, the company will have to make at least 13 laptops to start realizing a profit.
The R&D cost is incurred no matter how many computers are made.
Sorry, Jason, I was trying to do things in such a way as to break up that R&D cost on a "per unit" basis and obviously got it wrong. Chaitali and Parviz got it right.
100,000 is going to be distributed among the entire numbers of laptop sold, therefore:
C = 630 X + 100,000 ( is a total costs of producing X numbers of laptop)
C = \$890X + 100,000
\$890 X = cost of production of X unit
Total cost , should be calculated by adding fixed cost + variable cost (\$890X)
However you have added per unit cost of fixed cost 100,000/x.
100, 000 is fixed and is not depended on number of production.
You see that in next question asks how many unit needed to be produced to incur profit.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
0
A. C = 100000 + 890 x
B. R = 1520 x
C. Profit = Revenue - Cost; P = R - C = 1520 x - 890 x - 100000 = 630 x - 100000
D. P > 0 => 630 x - 100000 > 0 => x > 100000/630 ≈ 158.73015873015873,
but since x has to be an integer, x > 158 for a positive profit. If x <= 158 there will be a loss.
Chaitali M. | An Experienced Tutor in Elementary to College level Maths/Science.An Experienced Tutor in Elementary to Co...
4.7 4.7 (110 lesson ratings) (110)
0
A) C = 890x + 100,000
B) R = 1520x
C) P = 1520x - (890x + 100,000)
D) P = 1520x - (890x + 100,000)
= 1520x - 890x - 100,000
P = 630x - 100,000
P + 100,000 = 630x
x = (P + 100,000)/630 | 1,034 | 3,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-17 | latest | en | 0.89428 |
https://www.hackmath.net/en/math-problem/74684 | 1,701,636,790,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00322.warc.gz | 899,476,983 | 9,011 | # A rectangle 12
A rectangle has a width of 7.2 cm shorter than its length. The perimeter of this rectangle is 22.8 cm. Write the width-to-length ratio of this rectangle in the simplest form and show how you solve the problem.
r = 7:31
### Step-by-step explanation:
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#### Grade of the word problem:
We encourage you to watch this tutorial video on this math problem: | 153 | 669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.890653 |
http://girlscantcode.blogspot.com/2013/09/ | 1,531,701,943,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589029.26/warc/CC-MAIN-20180716002413-20180716022413-00088.warc.gz | 150,357,218 | 20,526 | ## Monday, September 30, 2013
### Lesson 24 - Accidental Arithmetics
Tonight she wanted to have a lesson just before bedtime. It's usually a bad idea, but I came around because she was so eager and didn't seem too tired.
The plan was to complete the Turtle Roy drawing from the last time, until an accident happened.
We started with turtle graphics and this time, I think, was the first time she remembered without any help that you use "fd 100" to move forward 100 pixels. She also mastered the clock trick from last time to determine whether we should turn right or left.
Then she accidentally typed just
1
1
Then teased her to try if the computer can do 1 + 1. And then we had a lot of fun doing arithmetics. She was really excited, for example, what 77 + 77 is.
## Sunday, September 29, 2013
### Lesson 23 - Clock
Today we did a lot of drawing with Turtle Roy. The key thing that was learnt was that turning right is turning clockwise and vice versa. The directions left and right were pretty well learnt already but when the turtle is facing downwards, it's not trivial to say whether you should turn left or right to get to the direction you want to go. (not even for me)
So we placed a pen on the computer screen and turned it to the direction we wanted to go. Then, if it appeared to be the direction the clock always turns, it's Right.
What we actually tried to draw with Turtle Roy this time is her name. We go this far. We'll continue with that on the next lesson. It's amazing that she doesn't really care whether we should write from left to right or the other way. She chose to write the arabic way because there was "more room to the left".
She also knew exactly when we should lift the pen up and when to put it down. You have to lift the pen up to move it without drawing on the screen, you know.
This lesson was a fun and rewarding one, thanks to not having it just before bedtime. She remembered the space between the function name and the argument very well. That's because I've given her a cookie every time she remembers it.
Bribery works. We've run out of cookies now, but we don't need them anymore either.
## Thursday, September 26, 2013
### Lesson 22 - Reverse
This time we started too late. She was too tired to learn. We should never do this just before bedtime...
Anyway, it was Haskell time. The idea was to teach the computer who the girls are and then turn them upside down.
So we opened ghci and started teaching. Like last time with Turtle Roy, we used let to define a new thing. I suggested the word "tytöt" (girls) but we chose to use the name "t" because it's shorter and the computer doesn't really care. So,
> let t = ["mila", "elena"]
Then we discussed head and tail a bit and tried them
"mila"
> tail t
["elena"]
I took a while to make a distinction between a character string and a list and that tail returns a list of strings in this case. I'm quite sure she ignored my talk. But let's see. Maybe she's a natural with types.
Next we tried reverse:
> reverse t
["elena", "mila"]
And finally the funniest thing. I introduced a new function map which takes a list and another function (POW!) and performs the other function for all things on the list. So we applied reverse to all the girls:
> map reverse t
["alim", "anele"]
That was fun. Then we reversed daddy too.
## Tuesday, September 24, 2013
### Lesson 21 - Characters
Last time we talked about getting to know all the characters that one can type using the keyboard. Today she wanted to get on with that, so we skipped the turtles and went for vim.
She typed vim merkit by herself. In vim she instantly knew she has to press i to start writing text. Then she wrote something like
and so on. We went through the numbers, then numbers with the shift key, then with alt and finally with both shift and alt pressed. We discovered a host of fun characters that I had no idea were hidden behind the number keys.
Then the same excercise for all the letter keys. There was a lot of discovery and she managed to save the document after each line of text to make sure we won't lose our work if the battery goes out. That's an important lesson for you too :)
So, even though we hadn't used vim for a week, she mastered it. She also remembered that cat is used to print file contents and, with some assistance, that we can pipe it to say to have the computer speak.
All in all, I noticed she had improved writing skills and general confidence in her own abilities. And it seems that my fears about her forgetting all the learnt skills in a day were without foundation. Even though it sometimes feels that nothing seems to "stick", a lot does indeed.
## Saturday, September 21, 2013
### Lesson 20 - Spider
After yesterday's repetitive practise on Turle Roy basic movements, we were quite well prepared for today's lesson. We started teaching the computer immediately and wrote (again)
let askel = s [fd 100, lt 90]
Then
let neliö = r 4 askel
Then (the excitement thickens) we ran our neliö function and had the turtle draw a perfect square. And again. And again.
Then we set to implementing the "spiderweb" thing that I showed her earlierly. The idea is to draw a square, then turn right, say, 5 degrees, then rinse and repeat. So first we taught the computer to do the "square and then turn" thing, the naming of which was left to her. She chose to name it xz. And then we wrote
let xz = s [neliö, lt 1]
She chose the 1-degree turn against my advice. And finally
let spider = r 360 xz
And we got a very fancy picture. Not the spiderweb thing though, because of the 1-degree turn, but a cool thing anyway. Then we modified the program a bit to draw the actual spiderweb. We saved our work, so that you can try it out yourself.
This was a fun lesson! She would have wanted to do some keyboard practise in vim too, but we ran out of time.
### Lesson 19 - Left & Right
Yesterday she announced that she wants to learn to code. So we went back to Turtle Roy to do some programming.
We started, once again, with some basic turtle graphics stuff like
fd 100
lt 90
fd 100
.. and so on. I was a overly optimistic to presume she'd be able to memorize how to turn to left and right and how to move forward in a blink of the eye. In fact, it proved quite hard this time. Maybe the problem is that left and right themselves are quite tough concepts for a 4-year old. When you add that we discuss in Finnish and code in English, and that English is not written the same way it's pronounced, I can understand that it's not easy.
So this time we spent most of our time learning turtle graphics and trying to memorize left and right. On the other hand she easily remembered how to construct lists using square brackets around them and commas between the list elements...
All in all, this lesson felt like a step backwards. Or maybe more like a reminder of the fact that she's just 4. So let's take it easy and have fun.
It is said that Logo and turtle graphics are a great way for kids to learn programming. However, I realised yesterday that teaching is usually started at age 12 or 13. Finnish children of that age can
• speak / read / write English
• understand numbers up to 1000 (?)
• understand some geometry, including angles
Children at four, on the other hand, can do none of these things. So the learning curve is in fact quite steep.
But we won't give up!
I'm open to new ideas too. What should we do instead, or in addition to Turtle Roy?
## Tuesday, September 17, 2013
### Lesson 18 - IO Monad
I'm slipping. I forgot to post yesterday even though we had a lesson.
Yesterday we visited Haskell for the first time.
That was just because the "say" function in Turtle Roy didn't produce any speech though. Fortunately I had setup her GHCI so that it includes my special helper module Napero which provides easy single-letter shorthands to some common functions. For those not in the know, the Roy language used in Turtle Roy is syntactically very similar to Haskell, so we can switch between the two without struggling with different syntaxes.
The main thing of lesson 18 was the idea that you can define a list of actions and then sequence them into a single action that performs all of the actions on the list. So, in Haskell we did
s [say "mila", say "ella"]
Where s is a shorthand for sequence and say is a wrapper for the command-line say command. Composition is a powerful tool! This is coding!
Next we started "teaching the computer". The idea was to teach the Turtle Roy to draw a rectangle, which consists of four equal parts. Each of the 4 parts proceeds 100 pixels and then turns right. We called this operation "askel" and taught the computer to do "askel" like this
let askel = s [fd 100, rt 90]
When we had taught the computer, i.e. defined a function, we called askel 4 times and voilá, got the rectangle together!
Next up: define the rectangle function.
The list syntax is something we'll be revisiting time after time until it feels natural to her.
## Sunday, September 15, 2013
### Lesson 17 - Song Finished
Today we completed the "head shoulders knees and toes" song by adding a new line "eyes and ears and mouth and nose" and yank-pasting a couple of lines to the end. It took about 15 minutes most of it we spent typing the 5-word line.
She remembered the space between words. She had more confidence with the letters G and B which we practiced quite a bit when moving around the document.
Next up: programming. I'be prepared a bunch of Haskell helpers already. Not sure whether to start Haskell next time or stick with Turtle Roy...
She taught Mom some more vim after our lesson. Mom proved to have mad unix skillz.
## Thursday, September 12, 2013
### Lesson 16 - Mom Learns Vim
This time we practised vim movements quite a bit
• 0/\$ - start/end of line
• 1G - first line
• G - last line
• 2G -second line
• w - forward one word
• b - back one word
• j/k - down/up
And that's quite enough for most of your movement needs. We'll practice those more later.
Then we did some editing, using dw to "delete word" and then i to insert new text. She remembered to exit insert mode, of course.
Then we yanked a line with yy and pasted with p. Those she remembered too. And deleted one line with dd, so that our little song goes now like
knees and toes
knees and toes
Then we made the computer speak this out load again with cat pää|say. She taught her mother how to do that.
Then she insisted on teach mom some vim too. And so also mom is capable of entering and exiting vim! She'll get another lesson tomorrow.
## Tuesday, September 10, 2013
### Lesson 15 - Gravity
Today we implemented the GOF Patterns starting from Singleton and Visitor.
Not.
We practised the most popular method of code re-use: copy-paste. In vim it's called yank-paste. First we added a new line to our song. It was easy. Now the song goes as in
head toes and knees and toes
knees and toes
Then I teached her how to move up and down in vim, using the j and k keys. We had an exercise where I shouted UP and DOWN randomly and she had to follow the orders with the cursor. First she tried to use the mouse but we soon established that vim doesn't obey the mouse. Soon she mastered the movevents and we were ready for copy-pasting. So, we moved to line 1 and yanked it using "yy". Then went to the end and pasted it using "p". This for both lines. She was happy to achieve so much with so little effort!
Then she told me the lyrics are all wrong. They should go like "head shoulders knees and toes" instead. Damn! We decided to fix that next time and had fun with "cat pää|say" instead. The computer was a lousy singer but we listened to it many times nevertheless. This was a fun lesson!
After the lesson I questioned her and she still remembered the movements, yanking and pasting. The hardest thing is still remembering to enter a space between words. In fact, I promised her a cookie for each space she remembers and she actually remembered on this time. The offer still stands, so let's see if she remembers better next time.
Oh, and yesterday we discussed whether people on the other side of Earth stand head-down. We discussed gravity and she drew a very nice diagram that show the force vectors for two girls on the opposite sides of the world:
## Friday, September 6, 2013
### Lesson 14 - Back And Forth
We started by moving the "pää" file back to the home directory. It took some time to establish that the snake-like tilde character represents the home directory and she decided to call it the "waves of the ocean" character. I think it's good to have a memorable name for that.
Then it was vim time. She knew that to edit the file, she has to use vim, and to open the file in vim she has to type "vim pää". She also types it correctly, even including the space between the words.
In vim, we practiced movements. Use 0/\$ to go to the beginning/end of the line. Then we used w to go to the end of the current/next word. We moved around quite a bit, then appended the missing "the" word in the file that will eventually contain the lyrics of the "head toes knees and toes" song. She didn't remember the a/append command, but we memorized that append is a lot like "Abhem", who is one of her friends. Maybe remember next time?
We enjoyed the power of vim by repeating the previous command with "." and even tried "10." to repeat it ten times. Then u for undo. At last we had the first line of the lyrics ready!
And that was enough for here this time. Too tired to go on. Better quit while winning, right?
## Tuesday, September 3, 2013
### Lesson 13 - Desktop
The main challenge this time was to keep the smaller girl entertained while practising mad skills with the other. Fortunately there was a certain "makkarapeli", i.e. a monster-feeding game on the iPad...
She remembered the file name from the last time and that we can use vim to edit it. She also typed both correctly. The challenge is still to remember to type a space between the program and the file name. Anyway, we added a new word to our pää file and had the computer recite the file again. All this went quite smoothly and quickly this time.
Then we discussed directories for a while and moved the pää file to a directory called "Desktop". She remembered the word move from the last time and that the program is actualy spelled mv. It was fun to discover that the pää file could now be found on the computer's desktop. We even opened the file in Textedit, but she was not very enthusiastic about doing any editing to the file this time. So we decided that it was enough.
But it wasn't. With some help, she googled for pictures of pricesses, then saved a photo of a cake on the Desktop. We used "ls Desktop" to list the files on the desktop on the command-line. Then discussed the hard disc and directories again.
Then to Turtle Roy, where she demanded that we must make a circle this time. While I was making oatmeal and pampering the hysterically crying smaller girl, my vim-girl learned how to type square brackets and managed to produce the following program (I practically dictated this one):
s [lt 10, fd 10]
Here s is a function that runs the listed actions sequentially, i.e. turns left 10 degrees and then proceeds 10 pixels. Then she repeated this a few times and declared that she can see the angles in the output, hence it won't form a circle. So I taught her how to edit the command line so that it will proceed just 1 pixel and turn 1 degree. Then she repeated that one for some time and was convinced that a circle would appear in the long run.
The little one was now playing makkarapeli and not yelling or bashing at us or the computer, so we were now able to discuss programming a bit. The thing was that repeating the steps manually would be tiresome and we should use a function repeat to repeat the step 360 times for a full circle. I demonstrated this and we were both happy with the result. Next time she'll do it herself.
Afterwards, while eating oatmeal, she proclaimed cheerfully:
"I can code!"
Then with a smaller voice: "Not as well as daddy though". Practice makes perfect.
## Monday, September 2, 2013
### Lesson 12 - Head and Toes
This time we started with a new command mv that we used for renaming the file "ihmisentarvikkeet" to "pää". I think she had great insight when she explained to me why you cannot use tab-completion when typing the new file (the file doesn't exist yet so the computer cannot guess it).
We used ls to list all files and find the new one in the listing. Then we piped the output of ls to say to have the computer speak it out loud. I promised to install a better voice to the computer so that it would be better at Finnish. She made me promise that we install it together when I've figured out how to do it.
Then it was Vim time again. We learnt to navigate to the end of the line (\$) and to the start of the line (0). Then we added another word to our song (with a=append), which now goes like | 4,016 | 17,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-30 | longest | en | 0.980695 |
https://au.mathworks.com/matlabcentral/answers/1983399-how-do-i-compute-the-p-value-of-the-highest-r-value-from-the-function-c-lags-xcorr-a-b | 1,723,677,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00009.warc.gz | 80,730,736 | 25,058 | # How do I compute the p-value of the highest r-value from the function [c,lags]=xcorr(A,B)?
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Sammy on 15 Jun 2023
Commented: Sammy on 16 Jun 2023
When using the [c,lags]=xcorr(A,B) and [r,p]=corrcoef(A,B), the c and r values are different so I can't use both. What I would like to know is the p-value from the max value of c from xcorr. But I need to use xcorr for its lag and maxlag features. Thanks.
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### Answers (1)
Anush on 15 Jun 2023
Hello Sammy,
To compute the p-value of the highest correlation value obtained from the xcorr() function, you can do the following:
[c, lags] = xcorr(A, B);
maxIndex = find(c == max(c));
maxLag = lags(maxIndex);
% Shift one of the signals by the maxLag
shiftedB = circshift(B, maxLag);
[r, p] = corrcoef(A, shiftedB);
This way you obtain the values of lag, maxlag and p-value from the max value of c from xcorr,
##### 1 CommentShow -1 older commentsHide -1 older comments
Sammy on 16 Jun 2023
Hello, I still couldn't get the max c and r values to match so I can use its corresponding p-value. I think the code could also improve if you place an absolute value of c because there are times when it's value is nearest to negative 1. So...
maxIndex = find(c == max(c));
Thanks for the info by the way. (:
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Start Hunting! | 429 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.853458 |
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A familiar footstep outside my office, “C’mon in, Vinnie, the door’s open.”
“Hi, Sy, how ya doin’?”
“Can’t complain. Yourself?”
“Fine, fine. Hey, I been thinking about something you said while Al and us were talking about rockets and orbits and such. You remember that?”
“We’ve done that in quantity. What statement in particular?”
“It was about when you’re in the ISS, you still see like 88% of Earth’s gravity. But I seen video of those astronauts just floating around in the station. Seems to me those two don’t add up.”
“Hah! We’re talking physics of motion here. What’s the magic word?”
“You’re saying it’s frames? I thought black holes did that.”
“Black holes are an extreme example, but frame‑thinking is an essential tool in analyzing any kind of relative motion. Einstein’s famous ‘happy thought‘ about a man in a free‑falling elevator—”
“Whoa, why is that a happy thought? I been nervous about elevators ever since that time we got stuck in one.”
“At least it wasn’t falling, right? Point is, the elevator and whoever’s in it agree that Newton’s First Law of Motion is valid for everything they see in there.”
“Wait, which Law is that?”
“‘Things either don’t move or else they move at a steady pace along a straight line.’ Suppose you’re that guy—”
“I’d rather not.”
“… and the elevator is in a zero‑gravity field. You take something out of your pocket, put it the air in front of you and it stays there. You give it a tap and it floats away in a straight line. Any different behavior means that your entire frame — you, the elevator and anything else in there — is being accelerated by some force. Let’s take two possibilities. Case one, you and the elevator are resting on terra firma, tightly held by the force of gravity.”
“I like that one.”
“Case two, you and the elevator are way out in space, zero‑gravity again, but you’re in a rocket under 1-g acceleration. Einstein got happy because he realized that you’d feel the same either way. You’d have no mechanical way to distinguish between the two cases.”
“What’s that mean, mechanical?”
“It excludes sneaky ways of outside influence by magnetic fields and such. Anyhow, Einstein’s insight was key to extending Newton’s First Law to figuring acceleration for an entire frame. Like, for instance, an orbiting ISS.”
“Ah, you’re saying that floating astronauts in an 88% Earth-gravity field is fine because the ISS and the guys share the frame feeling that 88% but the guys are floating relative to that frame. But down here if we could look in there we’d see how both kinds of motion literally add up.”
“Exactly. It’s just much easier to think about only one kind at a time.”
“Wait. You said the ISS is being accelerated. I thought it’s going a steady 17500 miles an hour which it’s got to do to stay 250 miles up.”
“Is it going in a straight line?”
“Well, no, it’s going in a circle, mostly, except when it has to dodge some space junk.”
“So the First Law doesn’t apply. Acceleration is change in momentum, and the ISS momentum is constantly changing.”
“But not in a straight line. Momentum is a vector that points in a specific direction. Change the direction, you change the momentum. Newton’s Second Law links momentum change with force and acceleration. Any orbiting object undergoes angular acceleration.”
“Angular acceleration, that’s a new one. It’s degrees per second per second?”
“Yup, or radians. There’s two kinds, though — orbiting and spinning. The ISS doesn’t spin because it has to keep its solar panels facing the Sun.”
“But I’ve seen sci-fi movies set in something that spins to create artificial gravity. Like that 2001 Space Odyssey where the guy does his running exercise inside the ship.”
“Sure, and people have designed space stations that spin for the same reason. You’d have a cascade of frames — the station orbiting some planet, the station spinning, maybe even a ballerina inside doing pirouettes.”
“How do you calculate all that?”
“You don’t. You work with whichever frame is useful for what you’re trying to accomplish.”
~~ Rich Olcott
# Turn This Way to Turn That Way
“I don’t understand, Sy. I get that James Webb Space Telescope uses its reaction wheels like a ship uses a rudder to change direction by pushing against something outside. Except the rudder pushes against water but the reaction wheels push against … what, the Universe?”
“Maybe probably, Al. We simply don’t know how inertia works. Newton just took inertia as a given. His Laws of Motion say that things remain at rest or persist in linear motion unless acted upon by some force. He didn’t say why. Einstein’s General Relativity starts from his Equivalence Principle — gravitational inertia is identical to mechanical inertia. That’s held up to painstaking experimental tests, but why it works is still an open question. Einstein liked Mach’s explanation, that we experience these inertias because matter interacts somehow with the rest of the Universe. He didn’t speculate how that interaction works because he didn’t like Action At A Distance. The quantum field theory people say that everything’s part of the universal field structure, which sounds cool but doesn’t help much. String theory … ’nuff said.”
“Hey, Moire, what’s all that got to do with the reaction wheel thing? JWST can push against one all it wants but it won’t go anywhere ’cause the wheel’s inside it. What’s magic about the wheels?”
JWST doesn’t want to go anywhere else, Mr Feder. We’re happy with it being in its proper orbit, but it needs to be able to point to different angles. Reaction wheels and gyroscopes are all about angular momentum, not about the linear kind that’s involved with moving from place to place.”
“HAH! JWST is moving place to place, in that orbit! Ain’t it got linear momentum then?”
“In a limited way, pun intended. Angular momentum is linear momentum plus a radial constraint. This goes back to Newton and his Principia book. I’ve got a copy of his basic arc‑splitting diagram here in Old Reliable. The ABCDEF line is a section of some curve around point S. He treated it as a succession of short line segments ABc, BCd, CDe and so on. If JWST is at point B, for instance, Newton would say that it’s traveling with a certain linear momentum along the BCd line. However, it’s constrained to move along the arc so it winds up at D instead d. To account for the constraint Newton invented centripetal force to pull along the Sd line. He then mentally made the steps smaller and smaller until the sequence of short lines matched the curve. At the limit, a sequence of little bits of linear momentum becomes angular momentum. By the way, this step‑reduction process is at the heart of calculus. Anyway, JWST uses its reaction wheels to swing itself around, not to propel itself.”
“And we’re back to my original question, Sy. What makes that swinging happen?”
“Oh, you mean the mechanical reality. Easy, Al. Like I said, three pairs of motorized wheels are mounted on JWST‘s frame near the center of mass. Their axles are at mutual right angles. Change a wheel’s angular momentum, you get an equal opposing change to the satellite’s. Suppose the Attitude Control System wants the satellite to swing to starboard. That’d be clockwise viewed from the cold side. ACS must tell a port/starboard motor to spin its wheel faster counterclockwise. If it’s already spinning clockwise, the command would be to put on the brakes, right? Either way, JWST swings clockwise. With the forward/aft motors and the hot‑side/cold‑side motors, the ACS is equipped to get to any orientation. See how that works?”
“Hang on.” <handwaving ensues> “Yeah, I guess so.”
“Hey, Moire. What if the wheel’s already spinning at top speed in the direction the ACS wants more of?”
“Ah, that calls for a momentum dump. JWST‘s equipped with eight small rocket engines called thrusters. They convert angular momentum back to linear momentum in rocket exhaust. Suppose we need a further turn to starboard but a port/starboard wheel is nearing threshold spin rate. ACS puts the brakes on that wheel, which by itself would turn the satellite to port. However, ACS simultaneously activates selected thrusters to oppose the portward slew. Cute, huh?”
~~ Rich Olcott
Mr Feder has a snarky grin on his face and a far‑away look in his eye. “Got another one. James Webb Space Telescope flies in this big circle crosswise to the Sun‑Earth line, right? But the Earth doesn’t stand still, it goes around the Sun, right? The circle keeps JWST the same distance from the Sun in maybe January, but it’ll fly towards the Sun three months later and get flung out of position.” <grabs a paper napkin> “Lemme show you. Like this and … like this.”
“Sorry, Mr Feder, that’s not how either JWST or L2 works. The satellite’s on a 6-month orbit around L2 — spiraling, not flinging. Your thinking would be correct for a solid gyroscope but it doesn’t apply to how JWST keeps station around L2. Show him, Sy.”
“Gimme a sec with Old Reliable, Cathleen.” <tapping> “OK, here’s an animation over a few months. What happens to JWST goes back to why L2 is a special point. The five Lagrange points are all about balance. Near L2 JWST will feel gravitational pulls towards the Sun and the Earth, but their combined attraction is opposed by the centrifugal force acting to move the satellite further out. L2 is where the three balance out radially. But JWST and anything else near the extended Sun‑Earth line are affected by an additional blended force pointing toward the line itself. If you’re close to it, sideways gravitational forces from the Sun and the Earth combine to attract you back towards the line where the sideways forces balance out. Doesn’t matter whether you’re north or south, spinward or widdershins, you’ll be drawn back to the line.”
Al’s on refill patrol, eavesdropping a little of course. He gets to our table, puts down the coffee pot and pulls up a chair. “You’re talking about the JWST. Can someone answer a question for me?”
“We can try.”
”What’s the question?”
Mr Feder, not being the guy asking the question, pooches out his lower lip.
“OK, how do they get it to point in the right direction and stay there? My little backyard telescope gives me fits just centering on some star. That’s while the tripod’s standing on good, solid Earth. JWST‘s out there standing on nothing.”
JWST‘s Attitude Control System has a whole set of functions to do that. It monitors JWST‘s current orientation. It accepts targeting orders for where to point the scope. It computes scope and satellite rotations to get from here to there. Then it revises as necessary in case the first‑draft rotations would swing JWST‘s cold side into the sunlight. It picks a convenient guide star from its million‑star catalog. Finally, ACS commands its attitude control motors to swing everything into the new position. Every few milliseconds it checks the guide star’s image in a separate sensor and issues tweak commands to keep the scope in proper orientation.”
“I get the sequence, Sy, but it doesn’t answer the how. They can’t use rockets for all that maneuvering or they’d run out of fuel real fast.”
“Not to mention cluttering up the view field with exhaust gases.”
“Good point, Cathleen. You’re right, Al, they don’t use rockets, they use reaction wheels, mostly.”
“Uh-oh, didn’t broken reaction wheels kill Kepler and a few other missions?”
“That sounds familiar, Mr Feder. What’s a reaction wheel, Sy, and don’t they put JWST in jeopardy?”
“A reaction wheel is a massive doughnut that can spin at high speed, like a classical gyroscope but not on gimbals.”
“Hey, Moire, what’s a gimbal?”
“It’s a rotating frame with two pivots for something else that rotates. Two or three gimbals at mutual right angles let what’s inside orient independent of what’s outside. The difference between a classical gyroscope and a reaction wheel is that the gyroscope’s pivots rotate freely but the reaction wheel’s axis is fixed to a structure. Operationally, the difference is that you use a gyroscope’s angular inertia to detect change of orientation but you push against a reaction wheel’s angular inertia to create a change of orientation.”
Kepler‘s failing wheels used metal bearings. JWST‘s are hardened ceramic.”
<whew>
~~ Rich Olcott
# Rotation, Revolution and The Answer
“Sy, I’m startin’ to think you got nothin’. Al and me, we ask what’s pushing the Moon away from us and you give us angular momentum and energy transfers. C’mon, stop dancin’ around and tell us the answer.”
“Yeah, Sy, gravity pulls things together, right, so how come the Moon doesn’t fall right onto us?”
“Not dancing, Vinnie, just laying some groundwork for you. Newton answered Al’s question — the Moon is falling towards us, but it’s going so fast it overshoots. That’s where momentum comes in, Vinnie. Newton showed that a ball shot from a cannon files further depending on how much momentum it gets from the initial kick. If you give it enough momentum, and set your cannon high enough that the ball doesn’t hit trees or mountains, the ball falls beyond the planet and keeps on falling forever in an elliptical orbit.”
“Forever until it hits the cannon.”
“hahaha, Al. Anyway, the ball achieves orbit by converting its linear momentum to angular momentum with the help of gravity. The angular momentum pretty much defines the orbit. In Newton’s gravity‑determined universe, momentum and position together let you predict everything.”
“Linear and angular momentum work the same way?”
“Mostly. There’s only one kind of linear momentum — straight ahead — but there are two kinds of angular momentum — rotation and revolution.”
“Aw geez, there’s another pair of words I can never keep straight.”
“You and lots of people, Vinnie. They’re synonyms unless you’re talking technicalese. In Physics and Astronomy, rotation with the O gyrates around an object’s own center, like a top or a planet rotating on its axis. Revolution with the E gyrates around some external location, like the planet revolving around its sun. Does that help?”
“Cool, that may come in handy. So Newton’s cannon ball got its umm, revolution angular momentum from linear momentum so where does rotation angular momentum come from?”
“Subtle question, Vinnie, but they’re actually all just momentum. Fair warning, I’m going to avoid a few issues that’d get us too far into the relativity weeds. Let’s just say that momentum is one of those conserved quantities. You can transfer momentum from one object to another and convert between forms of momentum, but you can’t create momentum in an isolated system.”
“That sounds a lot like energy, Sy.”
“You’re right, Al, the two are closely related. Newton thought that momentum was THE conserved quantity and all motion depended on it. His arch‑enemy Leibniz said THE conserved quantity was kinetic energy, which he called vis viva. That disagreement was just one battle in the Newton‑Leibniz war. It took science 200 years to understand the momentum/kinetic energy/potential energy triad.”
“Wait, Sy, I’ve seen NASA steer a rocketship and give it a whole different momentum. I don’t see no conservation.”
“You missed an important word, Vinnie — isolated. Momentum calculations apply to mechanical systems — no inputs of mass or non‑mechanical energy. Chemical or nuclear fuels break that rule and get you into a different game.”
“Ah-hahh, so if the Earth and Moon are isolated…”
“Exactly, and you’re way ahead of me. Like we said, no significant net forces coming from the Sun or Jupiter, so no change to our angular momentum.”
“Hey, wait, guys. Solar power. I know we’ve got a ton of sunlight coming in every day.”
“Not relevant, Al. Even though sunlight heats the Earth, mass and momentum aren’t affected by temperature. Anyhow, we’re finally at the point where I can answer your question.”
“Hush. OK, here’s the chain. Earth rotates beneath the Moon and gets its insides stirred up by the Moon’s gravity. The stirring is kinetic energy extracted from the energy of the Earth‑Moon system. The Moon’s revolution or the Earth’s rotation or both must slow down. Remember the M=m·r·c/t equation for angular momentum? The Earth‑Moon system is isolated so the angular momentum M can’t change but the angular velocity c/t goes down. Something’s got to compensate. The system’s mass m doesn’t change. The only thing that can increase is distance r. There’s your answer, guys — conservation of angular momentum forces the Moon to drift outward.”
“To the Moon and back.”
~~ Rich Olcott
# Here’s a Different Angle
“OK, Sy, so there’s a bulge on the Moon’s side of the Earth and the Earth rotates but the bulge doesn’t and that makes the Moon’s orbit just a little bigger and you’ve figured out that the energy it took to lift the Moon raised Earth’s temperature by a gazillionth of a degree, I got all that, but you still haven’t told Al and me how the lifting works.”
“You wouldn’t accept it if I just said, ‘The Moon lifts itself by its bootstraps,’ would you?”
“Not for a minute.”
“And you don’t like equations. <sigh> OK, Al, pass over some of those paper napkins.”
“Aw geez, Sy.”
“You guys asked the question and this’ll take diagrams, Al. Ante up. … Thanks. OK, remember the time Cathleen and I caught Vinnie here at Al’s shop playing with a top?”
“Yeah, and he was spraying paper wads all over the place.”
“I wasn’t either, Al, it was the top sending them out with centri–…, some force I can never remember whether it’s centrifugal or centripetal.”
“Centrifugal, Vinnie, –fugal– like fugitive, outward‑escaping force. It’s one of those ‘depends on how you look at itfictitious forces. From where you were sitting, the wads looked like they were flying outward perpendicular to the top’s circle. From a wad’s point of view, it flew in a straight line tangent to the circle. It’s like we have two languages, Room and Rotor. They describe the same phenomena but from different perspectives.”
“Hey, it’s frames again, ain’t it?”
“Newton’s inertial frames? Sort‑of but not quite. Newton’s First Law only holds in the Room frame — no acceleration, motion is measured by distance, objects at rest stay put. Any other object moves in a straight line unless its momentum is changed by a force. You can tackle a problem by considering momentum and force components along separate X and Y axes. Both X and Y components work the same way — push twice as hard in either direction, get twice the acceleration in that direction. Nice rules that the Rotor frame doesn’t play by.”
“I guess not. The middle’s the only place an object can stay put, right?”
“Exactly, Al. Everything else looks like it’s affected by weird, constantly‑varying forces that’re hard to describe in X‑Y terms.”
“So that breaks Newton’s physics?”
“Of course not. We just have to adapt his F=m·a equation (sorry, Vinnie!) to Rotor conditions. For small movements we wind up with two equations. In the strict radial direction it’s still F=m·a where m is mass like we know it, a is acceleration outward or inward, and F is centrifugal or centripetal, depending. Easy. Perpendicular to ‘radial‘ we’ve got ‘angular.’ Things look different there because in that direction motion’s measured by angle but Newton’s Laws are all about distances — speed is distance per time, acceleration is speed change per time and so forth.”
“So what do you do?”
“Use arc length. Distance along an arc is proportional to the angle, and it’s also proportional to the radius of the arc, so just multiply them together.”
“What, like a 45° bend around a 2-foot radius takes 90 feet? That’s just wrong!”
“No question, Al. You have to measure the angle in the right units. Remember the formula for a circle’s circumference?”
“Sure, it’s 2πr.”
“Which tells you that a full turn’s length is times the radius. We can bridge from angle to arc length using rotational units so that a full turn, 360°, is units. We’ll call that unit a radian. Half a circle is π radians. Your 45° angle in radians is π/4 or about ¾ of a radian. You’d need about (¾)×(2) or 1½ feet of whatever to get 45° along that 2-foot arc. Make sense?”
“Gimme a sec … OK, I’m with you.”
“Great. So if angular distance is radius times angle, then angular momentum which is mass times distance per time becomes mass times radius times angle per time.”
“”Hold on, Sy … so if I double the mass I double the momentum just like always, but if something’s spinning I could also double the angular momentum by doubling the radius or spinning it twice as fast?”
“Couldn’t have put it better myself, Vinnie.”
~~ Rich Olcott
# Two Against One, And It’s Not Even Close
On a brisk walk across campus when I hear Vinnie yell from Al’s coffee shop. “Hey! Sy! Me and Al got this argument going you gotta settle.”
“Happy to be a peacemaker, but it’ll cost you a mug of Al’s coffee and a strawberry scone.”
“Coffee’s no charge, Sy, but the scone goes on Vinnie’s tab. What’s your pleasure?”
“It’s morning, Al, time for black mud. What’s the argument, Vinnie?”
“Al read in one of his astronomy magazines that the Moon’s drifting away from us. Is that true, and if it is, how’s it happen? Al thinks Jupiter’s gravity’s lifting it but I think it’s because of Solar winds pushing it. So which is it?”
“Here you go, Sy, straight from the bottom of the pot.”
“Perfect, Al, thanks. Yes, it’s true. The drift rate is about 1¼ nanometers per second, 1½ inches per year. As to your argument, you’re both wrong.”
“Huh?”
”Aw, c’mon!”
“Al, let’s put some numbers to your hypothesis. <pulling out Old Reliable and screen‑tapping> I’m going to compare Jupiter’s pull on the Moon to Earth’s when the two planets are closest together. OK?”
“I suppose.”
“Alright. Newton’s Law tells us the pull is proportional to the mass. Jupiter’s mass is about 320 times Earth, which is pretty impressive, right? But the attraction drops with the square of the distance. The Moon is 1¼ lightseconds from Earth. At closest approach, Jupiter is almost 2100 lightseconds away, 1680 times further than the Moon. We need to divide the 320 mass factor by a 1680‑squared distance factor and that makes <key taps> Jupiter’s pull on the Moon is only 0.011 percent of Earth’s. It’ll be <taps> half that when Jupiter’s on the other side of the Sun. Not much competition, eh?”
“Yeah, but a little bit at a time, it adds up.”
“We’re not done yet. The Moon feels the big guy’s pull on both sides of its orbit around Earth. On the side where the Moon’s moving away from Jupiter, you’re right, Jupiter’s gravity slows the Moon down, a little. But on the moving-toward-Jupiter side, the motion’s sped up. Put it all together, Jupiter’s teeny pull cancels itself out over every month’s orbiting.”
“Gotcha, Al. So what about my theory, Sy?”
“Basically the same logic, Vinnie. The Solar wind varies, thanks to the Sun’s variable activity, but satellite measurements put its pressure somewhere around a nanopascal, a nanonewton per square meter. Multiply that by the Moon’s cross‑sectional area and we get <tap, tap> a bit less than ten thousand newtons of force on the Moon. Meanwhile, Newton’s Law says the Earth’s pull on the Moon comes to <tapping>
G×(Earth’s mass)×(Moon’s mass)/(Earth-Moon distance)²
and that comes to 2×1011 newtons. Earth wins by a 107‑fold landslide. Anyway, the pressure slows the Moon for only half of each month and speeds it up the other half so we’ve got another cancellation going on.”
“So what is it then?”
”So what is it then?”
“Tides. Not just ocean tides, rock tides in Earth’s fluid outer mantle. Earth bulges, just a bit, toward the Moon. But Earth also rotates, so the bulge circles the planet every day.”
“Reminds me of the wave in the Interstellar movie, but why don’t we see it?”
“The movie’s wave was hundreds of times higher than ours, Al. It was water, not rock, and the wave‑raiser was a huge black hole close by the planet. The Moon’s tidal pull on Earth produces only a one‑meter variation on a 6,400,000‑meter radius. Not a big deal to us. Of course, it makes a lot of difference to the material that’s being kneaded up and down. There’s a lot of friction in those layers.”
“Friction makes heat, Sy. Rock tides oughta heat up the planet, right?”
“Sure, Vinnie, the process does generate heat. Force times distance equals energy. Raising the Moon by 1¼ nanometers per second against a force of 2×1021 newtons gives us <taping furiously> an energy transfer rate of 4×10‑23 joules per second per kilogram of Earth’s 6×1024‑kilogram mass. It takes about a thousand joules to heat a kilogram of rock by one kelvin so we’re looking at a temperature rise near 10‑27 kelvins per second. Not significant.”
“No blaming climate change on the Moon, huh?”
~~ Rich Olcott
# The Top Choice
Al grabs me as I step into his coffee shop. “Sy, ya gotta stop Vinnie, he’s using up paper napkins again, and he’s making a mess!”
Sure enough, there’s Vinnie at his usual table by the door. He’s got a kid’s top, a big one, spinning on a little stand. He’s methodically dropping crumpled-up paper wads onto it and watching them fly off onto the floor. “Hey, Vinnie, what’s the project?”
“Hi, Sy. I’m trying to figure how come these paper balls are doing a circle but when they fly off they always go in a straight line, at least at first. They got going-around momentum, right, so how come they don’t make a spiral like stars in a galaxy?”
Astronomy professor Cathleen’s standing in the scone line. She never misses an opportunity to correct a misconception. “Galaxy stars don’t spray out of the center in a spiral, Vinnie. Like planets going around a star, stars generally follow elliptical orbits around the galactic center. A star that’s between spiral arms now could be buried in one ten million years from now. The spiral arms appear because of how the orbits work. One theory is that the innermost star orbits rotate their ellipse axes more quickly than the outer ones and the spirals form where the ellipses pile up. Other theories have to do with increased star formation or increased gravitational attraction within the pile-up regions. Probably all three contribute to the structures. Anyhow, spirals don’t form from the center outward.”
My cue for some physics. “What happens in a galaxy is controlled by gravity, Vinnie, and gravity doesn’t enter into what you’re doing. Except for all that paper falling onto Al’s floor. There’s no in-plane gravitational or electromagnetic attraction in play when your paper wads leave the toy. Newton would say there’s no force acting to make them follow anything other than straight lines once they break free.”
“What about momentum? They’ve got going-around momentum, right, shouldn’t that keep them moving spirally?”
I haul out Old Reliable for a diagram. “Thing is, your ‘going-around momentum,’ also known as ‘angular momentum,’ doesn’t exist. Calm down, Vinnie, I mean it’s a ‘fictitious force‘ that depends on how you look at it.”
“Is this gonna be frames again?”
“Yup. Frames are one of our most important analytical tools in Physics. Here’s your toy and just for grins I’ve got it going around counterclockwise. That little white circle is one of your paper wads. In the room’s frame that wad in its path is constantly converting linear momentum between the x-direction and the y-direction, right?”
“East-West to North-South and back, yeah, I get that.”
“Such a mess to calculate. Let’s make it easier. Switch to the perspective of a frame locked to the toy. In that frame the wad can move in two directions. It can fly away along the radial direction I’ve called r, or it can ride along sideways in the s-direction.”
“So why hasn’t it flown away?”
“Because you put some spit on it to make it stick — don’t deny it, I saw you. While it’s stuck, does it travel in the r direction?”
“Nope, only in the s direction. Which should make it spiral like I said.”
“I’m not done yet. One of Newton’s major innovations was the idea of infinitesimal changes, also known as little-bits. The s-direction is straight, not curved, but it shifts around little-bit by little-bit as the top rotates. Newton’s Laws say force is required to alter momentum. What force influences the wad’s s-momentum?”
“Umm … that line you’ve marked c.”
“Which is the your spit’s adhesive force between the paper and the top. The wad stays stuck until the spit dries out and no more adhesion so no more c-force. Then what happens?”
“It flies off.”
“In which direction?”
“Huh! In the r-direction.”
“And in a straight line, just like Newton said. What you called ‘going-around momentum’ becomes ‘radial momentum’ and there’s no spiraling, right?”
“I guess you’re right, but I miss spirals.”
Al comes over with a broom. “Now that’s settled, Vinnie, clean up!”
~~ Rich Olcott
• Thanks for the question, Jen Keeler. Stay tuned.
# Conversation of Momentum
Teena bounces out of the sandbox, races over to the playground’s little merry-go-round and shoves it into motion. “Come help turn this, Uncle Sy, I wanna go fast!” She leaps onto the moving wheel and of course she promptly falls off. The good news is that she rolls with the fall like I taught her to do.
“Why can’t I stay on, Uncle Sy?”
“What’s your new favorite word again?”
“Mmmo-MMENN-tumm. But that had to do with swings.”
“Swings and lots of other stuff, including merry-go-rounds and even why you should roll with the fall. Which, by the way, you did very well and I’m glad about that because we don’t want you getting hurt on the playground.”
“Well, it does hurt a little on my elbow, see?”
“Let me look … ah, no bleeding, things only bend where they’re supposed to … I think no damage done but you can ask your Mommie to kiss it if it still hurts when we get home. But you wanted to know why you fell off so let’s go back to the sandbox to figure that out.”
<scamper!> “I beat you here!”
“Of course you did. OK, let’s draw a big arc and pretend that’s looking down on part of the merry-go-round. I’ll add some lines for the spokes and handles. Now I’ll add some dots and arrows to show what I saw from over here. See, the merry-go-round is turning like this curvy arrow shows. You started at this dot and jumped onto this dot which moved along and then you fell off over here. Poor Teena. So you and your momentum mostly went left-to-right.”
“But that’s not what happened, Uncle Sy. Here, I’ll draw it. I jumped on but something tried to push me off and then I did fall off and then I rolled. Poor me. Hey, my arm doesn’t hurt any more!”
“How about that? I’ve often found that thinking about something else makes hurts go away. So what do you think was trying to push you off? I’ll give you a hint with these extra arrows on the arc.”
“That looks like Mr Newton’s new directions, the in-and-out direction and the going-around one. Oh! I fell off along the in-and-out direction! Like I was a planet and the Sun wasn’t holding me in my orbit! Is that what happened, I had out-momentum?”
“Good thinking, Teena. Mr Newton would say that you got that momentum from a force in the out-direction. He’d also say that if you want to stand steady you need all the forces around you to balance each other. What does that tell you about what you need to do to stay on the merry-go-round?”
“I need an in-direction force … Hah, that’s what I did wrong! I jumped on but I didn’t grab the handles.”
“Lesson learned. Good.”
“Well, in general when you fall it’s nearly always good to roll the way your body’s spinning and only try to slow it down. People who put out an arm or leg to stop a fall often stress it and and maybe even tear or break something.”
“That’s what you’ve told me. But what made me spin?”
“One of Mr Newton’s basic principles was a rule called ‘Conservation of Momentum.’ It says that you can transfer momentum from one thing to another but you can’t create it or destroy it. There are some important exceptions but it’s a pretty good rule for the cases he studied. Your adventure was one of them. Look back at the picture I drew. You’d built up a lot of going-around momentum from pushing the merry-go-round to get it started. You still had momentum in that direction when you fell off. Sure enough, that’s the direction you rolled.”
“Is that the ‘Conversation of Energy’ thing that you and Mommie were talking about?”
“Conservation. It’s not the same but it’s closely related.”
“Why does it even work?”
“Ah, that’s such a deep question that most physicists don’t even think about it. Like gravity, Mr Newton described what inertia and momentum do, but not how they work. Einstein explained gravity, but I’m not convinced that we understand mass yet.”
~~ Rich Olcott
# A Momentous Occasion
<creak> Teena’s enjoying her new-found power in the swings. “Hey, Uncle Sy? <creak> Why doesn’t the Earth fall into the Sun?”
“What in the world got you thinking about that on such a lovely day?”
“The Sun gets in my eyes when I swing forward <creak> and that reminded me of the time we saw the eclipse <creak> and that reminded of how the planets and moons are all floating in space <creak> and the Sun’s gravity’s holding them together but if <creak> the Sun’s pulling on us why don’t we just fall in?” <creak>
“An excellent question, young lady. Isaac Newton thought about it long and hard back when he was inventing Physics.”
“Isaac Newton? Is he the one with all the hair and a long, skinny nose and William Tell shot an arrow off his head?”
“Well, you’ve described his picture, but you’ve mixed up two different stories. William Tell’s apple story was hundreds of years before Newton. Isaac’s apple story had the fruit falling onto his head, not being shot off of it. That apple got him thinking about gravity and how Earth’s gravity pulling on the apple was like the Sun’s gravity pulling on the planets. When he was done explaining planet orbits, he’d also explained how your swing works.”
“My swing works like a planet? No, my swing goes back and forth, but planets go round and round.”
“Jump down and we can draw pictures over there in the sandbox.”
<thump!! scamper!> “I beat you here!”
“Of course you did. OK, what’s your new M-word?”
“Mmmo-MMENN-tummm!”
“Right. Mr Newton’s Law of Inertia is about momentum. It says that things go in a straight line unless something interferes. It’s momentum that keeps your swing going.”
“B-u-u-t, I wasn’t going in a straight line, I was going in part of a circle.”
“Good observing, Teena, that’s exactly right. Mr Newton’s trick was that a really small piece of a circle looks like a straight line. Look here. I’ll draw a circle … and inside it I’ll put a triangle… and between them I’ll put a hexagon — see how it has an extra point halfway between each of the triangle’s points? — and up top I’ll put the top part of whatever has 12 sides. See how the 12-thing’s sides are almost on the circle?”
“Ooo, that’s pretty! Can we do that with a square, too?”
“Sure. Here’s the circle … and the square … and an octagon … and a 16-thing. See, that’s even closer to being a circle.”
“Ha-ha — ‘octagon’ — that’s like ‘octopus’.”
“For good reason. An octopus has eight arms and an octagon has eight sides. ‘Octo-‘ means ‘eight.’ So anyway, Mr Newton realized that his momentum law would apply to something moving along that tiny straight line on a circle. But then he had another idea — you can move in two directions at once so you can have momentum in two directions at once.”
“That’s silly, Uncle Sy. There’s only one of me so I can’t move in two directions at once.”
“Can you move North?”
“Uh-huh.”
“Can you move East?”
“Sure.”
“Can you move Northeast?”
“Oh … does that count as two?”
“It can for some situations, like planets in orbit or you swinging on a swing. You move side-to-side and up-and-down at the same time, right?”
“Uh-huh.”
“When you’re at either end of the trip and as far up as you can get, you stop for that little moment and you have no momentum. When you’re at the bottom, you’ve got a lot of side-to-side momentum across the ground. Anywhere in between, you’ve got up-down momentum and side-to-side momentum. One kind turns into the other and back again.”
“So complicated.”
“Well, it is. Newton simplified things with revised directions — one’s in-or-out from the center, the other’s the going-around angle. Each has its own momentum. The swing’s ropes don’t change length so your in-out momentum is always zero. Your angle-momentum is what keeps you going past your swing’s bottom point. Planets don’t have much in-out momentum, either — they stay about their favorite distance from the Sun.”
“Earth’s angle-momentum is why we don’t fall in?”
“Yep, we’ve got so much that we’re always falling past the Sun.”
~~ Rich Olcott
# Gettin’ kinky in space
Things were simpler in the pre-Enlightenment days when we only five planets to keep track of. But Haley realized that comets could have orbits, Herschel discovered Uranus, and Galle (with Le Verrier’s guidance) found Neptune. Then a host of other astronomers detected Ceres and a host of other asteroids, and Tombaugh observed Pluto in 1930.
Astronomers relished the proliferation — every new-found object up there was a new test case for challenging one or another competing theory.
Here’s the currently accepted narrative… Long ago but quite close-by, there was a cloud of dust in the Milky Way galaxy. Random motion within it produced a swirl that grew into a vortex dozens of lightyears long.
Consider one dust particle (we’ll call it Isaac) afloat in a slice perpendicular to the vortex. Assume for the moment that the vortex is perfectly straight, the dust is evenly spread across it, and all particles have the same mass. Isaac is subject to two influences — gravitational and rotational.
Gravity pulls Isaac towards towards every other particle in the slice. Except for very near the slice’s center there are generally more particles (and thus more mass) toward and beyond the center than back toward the edge behind him. Furthermore, there will generally be as many particles to Isaac’s left as to his right. Gravity’s net effect is to pull Isaac toward the vortex center.
But the vortex spins. Isaac and his cohorts have angular momentum, which is like straight-line momentum except you’re rotating about a center. Both of them are conserved quantities — you can only get rid of either kind of momentum by passing it along to something else. Angular momentum keeps Isaac rotating within the plane of his slice.
An object’s angular momentum is its linear momentum multiplied by its distance from the center. If Isaac drifts towards the slice’s center (radial distance decreases), either he speeds up to compensate or he transfers angular momentum to other particles by colliding with them.
But vortices are rarely perfectly straight. Moreover, the galactic-cloud kind are generally lumpy and composed of different-sized particles. Suppose our vortex gets kinked by passing a star or a magnetic field or even another vortex. Between-slice gravity near the kink shifts mass kinkward and unbalances the slices to form a lump (see the diagram). The lump’s concentrated mass in turn attracts particles from adjacent slices in a viscous cycle (pun intended).
After a while the lumpward drift depletes the whole neighborhood near the kink. The vortex becomes host to a solar nebula, a concentrated disk of dust whirling about its center because even when you come in from a different slice, you’ve still got your angular momentum. When gravity smacks together Isaac and a few billion other particles, the whole ball of whacks inherits the angular momentum that each of its stuck-together components had. Any particle or planetoid that tries to make a break for it up- or down-vortex gets pulled back into the disk by gravity.
That theory does a pretty good job on the conventional Solar System — four rocky Inner Planets, four gas giant Outer Planets, plus that host of asteroids and such, all tightly held in the Plane of The Ecliptic.
How then to explain out-of-plane objects like Pluto and Eris, not to mention long-period comets with orbits at all angles?
We now know that the Solar System holds more than we used to believe. Who’s in is still “objects whose motion is dominated by the Sun’s gravitational field,” but the Sun’s net spreads far further than we’d thought. Astronomers now hypothesize that after its creation in the vortex, the Sun accumulated an Oort cloud — a 100-billion-mile spherical shell containing a trillion objects, pebbles to planet-sized.
At the shell’s average distance from the Sun (see how tiny Neptune’s path is in the diagram) Solar gravity is a millionth of its strength at Earth’s orbit. The gravity of a passing star or even a conjunction of our own gas giants is enough to start an Oort-cloud object on an inward journey.
These trans-Neptunian objects are small and hard to see, but they’re revolutionizing planetary astronomy.
~~ Rich Olcott | 9,428 | 40,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.954342 |
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## What’s The Meaning Of “N, I, T, P, E” In The Performance Chart?
A motor performance curve chart is a graphical representation of how a motor performs under various conditions. The most common type of motor performance curve chart is a torque vs. speed curve, which shows how much torque a motor can produce at various speeds.
Other types of motor performance curve charts can show power vs. speed, efficiency vs. speed, or current vs. speed. Motor performance curve charts are useful for determining the best operating point for a given application, and for comparing the performance of different motors.
N: It indicates the number of revolution per minute (rpm) of the rotor. Speed would be varied if either voltage (V) supplied or load applied to the motor is varying.
I: It indicates the amount of electrons (current) which is discharged when voltage is applied between the terminals of the motor. The load applied to the motor is directly proportional to the current drawn by the motor.
T: This is the twisting force that causing rotation of the rotor. The value of torque is depended on the load applied to and the current drawn by the motor.
P: This is the work done by the motor which is the product of N and t The maximum output (Pmax) would be occurred at a point of Ts/2 on the P curve.
E: This is the percentage of the work done by the motor to the energy supplied to the motor. That is, the ratio of the mechanical output to the electrical input(E=P/VI x 100%).The maximum efficiency (Emax) would be found at a point in the torque range less than TS/2.
## Using NFP-Motor Instructions
10 Points Should Be Awared When Using Small DC Motors 1. You should be aware of using the adhesive while mounting the brush DC motor, the residue may cause bearing damaged and motor run incorrectly.2. You should check the loading is followed the motor specification,...
## The Working Principle Of Motor Armature In Motor
What is a motor armature? It is a key and pivotal component in the process of mutual conversion between mechanical energy and electric energy of the motor. For a generator, it is a component that generates electromotive force, such as the rotor in a DC generator, and...
## How To Use The Coupon In NFP-Shop
First of all, the most important thing is that the total amount of products you have purchased must be greater than USD \$100, such as the folloiwng cart: Then enter "FREE-SHIPPING" to the coupon code box as follow. Then you can click proceed to checkout or Paypal...
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# How reliable are answers to Gmat + RC material?
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How reliable are answers to Gmat + RC material? [#permalink]
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21 Jan 2004, 11:57
I was going through RC GMAT PLUS material, more specifically, Gmat plus 4, RC passages 7 and 8 and found so many anwers that were totally different from mine. I would like to know how reliable are those answers.
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25 Jan 2004, 23:38
GMAT+ answers are pretty reliable from my experience. If you have trouble or are not ok with the answers, the best thing would be to post the original question in the verbal forum and get feedback. It would be time consuming if you only wrote the question no etc. GMAT+ being a free source, you can put the questions as is.
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# How reliable are answers to Gmat + RC material?
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 650 | 2,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-39 | latest | en | 0.916223 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/zinc-blank-cyanide | 1,709,319,080,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00682.warc.gz | 653,368,862 | 7,953 | # Weight of Zinc cyanide
## zinc cyanide: convert volume to weight
### Weight of 1 cubic centimeter of Zinc cyanide
carat 9.26 ounce 0.07 gram 1.85 pound 0 kilogram 0 tonne 1.85 × 10-6 milligram 1 852
#### How many moles in 1 cubic centimeter of Zinc cyanide?
There are 15.77 millimoles in 1 cubic centimeter of Zinc cyanide
### The entered volume of Zinc cyanide in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
SWEETENED ICED TEA, UPC: 011225108921 contain(s) 25 calories per 100 grams (≈3.53 ounces) [ price ]
1866 foods that contain Fiber, soluble. List of these foods starting with the highest contents of Fiber, soluble and the lowest contents of Fiber, soluble
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CaribSea, Freshwater, Instant Aquarium, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
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Volume to weight and weight to volume conversions using density | 665 | 2,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.672099 |
https://stats.stackexchange.com/questions/80342/casting-a-multivariate-linear-model-as-a-multiple-regression | 1,721,687,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00203.warc.gz | 484,287,808 | 42,694 | # Casting a multivariate linear model as a multiple regression
Is recasting a multivariate linear regression model as a multiple linear regression entirely equivalent? I'm not referring to simply running $t$ separate regressions.
I have read this in a few places (Bayesian Data Analysis -- Gelman et al., and Multivariate Old School -- Marden) that a multivariate linear model can easily be reparameterized as multiple regression. However, neither source elaborates on this at all. They essentially just mention it, then continue using the multivariate model. Mathematically, I'll write the multivariate version first,
$$\underset{n \times t}{\mathbf{Y}} = \underset{n \times k}{\mathbf{X}} \hspace{2mm}\underset{k \times t}{\mathbf{B}} + \underset{n \times t}{\mathbf{R}},$$ where the bold variables are matrices with their sizes below them. As usual, $\mathbf{Y}$ is data, $\mathbf{X}$ is the design matrix, $\mathbf{R}$ are normally distributed residuals, and $\mathbf{B}$ is what we are interested in making inferences with.
To reparameterize this as the familiar multiple linear regression, one simply rewrites the variables as:
$$\underset{nt \times 1}{\mathbf{y}} = \underset{nt \times nk}{\mathbf{D}} \hspace{2mm} \underset{nk \times 1}{\boldsymbol{\beta}} + \underset{nt \times 1}{\mathbf{r}},$$
where the reparameterizations used are $\mathbf{y} = row(\mathbf{Y})$, $\boldsymbol\beta = row(\mathbf{B})$, and $\mathbf{D} = \mathbf{X} \otimes \mathbf{I}_{n}$. $row()$ means that the rows of the matrix are arranged end to end into a long vector, and $\otimes$ is the kronecker, or outer, product.
So, if this is so easy, why bother writing books on multivariate models, test statistics for them etc.? It is most effective to just transform the variables first and use common univariate techniques. I'm sure there is a good reason, I just am having a hard time thinking of one, at least in the case of a linear model. Are there situations with the multivariate linear model and normally distributed random errors where this reparameterization does not apply, or limits the possibilities of the analysis you can undertake?
Sources I have seen this: Marden - Multivariate Statistics: Old School. See sections 5.3 - 5.5. The book is available free from: http://istics.net/stat/
Gelman et al. - Bayesian Data Analysis. I have the second edition, and in this version there is a small paragraph in Ch. 19 'Multivariate Regression Models' titled: "The equivalent univariate regression model"
Basically, can you do everything with the equivalent linear univariate regression model that you could with the multivariate model? If so, why develop methods for multivariate linear models at all?
• It is good question. May be you could ask for more in terms of foundations rather than a structure.
– user10619
Commented Dec 22, 2013 at 10:37
• What do you mean by foundations rather than structure? Could you elaborate? Commented Dec 22, 2013 at 10:51
• May note that I learnt only two papers as part of my first and postgraduate degree long back, I do not have grooming in technical descriptions. I understand that Multivariate analysis has different assumptions when compared with a multiple linear regression or simply linear regression model. The assumptions for Multivariate analysis are different i.e. mathematical expectation prevails upon. multiple linear regression makes certain other assumptions that result in heteroscedatisticity. The structure here I mean refers to your equations.
– user10619
Commented Dec 22, 2013 at 13:35
• You should say it clearly in the title or the beginning whether you are speaking of multivariate (general) linear model or about bayesian multivariate regression. Commented Dec 22, 2013 at 17:11
• Ok, so.. its not my approach, I pointed out two places I have seen this. The approach is the crux of the issue. What is the difference between the multivariate version and the reparameterized univariate version? Commented Dec 22, 2013 at 20:03
Basically, can you do everything with the equivalent linear univariate regression model that you could with the multivariate model?
I believe the answer is no.
If your goal is simply either to estimate the effects (parameters in $\mathbf{B}$) or to further make predictions based on the model, then yes it does not matter to adopt which model formulation between the two.
However, to make statistical inferences especially to perform the classical significance testing, the multivariate formulation seems practically irreplaceable. More specifically let me use the typical data analysis in psychology as an example. The data from $n$ subjects are expressed as
$$\underset{n \times t}{\mathbf{Y}} = \underset{n \times k}{\mathbf{X}} \hspace{2mm}\underset{k \times t}{\mathbf{B}} + \underset{n \times t}{\mathbf{R}},$$
where the $k-1$ between-subjects explanatory variables (factor or/and quantitative covariates) are coded as the columns in $\mathbf{X}$ while the $t$ repeated-measures (or within-subject) factor levels are represented as simultaneous variables or the columns in $\mathbf{Y}$.
With the above formulation, any general linear hypothesis can be easily expressed as
$$\mathbf{L} \mathbf{B} \mathbf{M} = \mathbf{C},$$
where $\mathbf{L}$ is composed of the weights among the between-subjects explanatory variables while $\mathbf{L}$ contains the weights among levels of the repeated-measures factors, and $\mathbf{C}$ is a constant matrix, usually $\mathbf{0}$.
The beauty of the multivariate system lies in its separation between the two types of variables, between- and within-subject. It is this separation that allows for the easy formulation for three types of significance testing under the multivariate framework: the classical multivariate testing, repeated-measures multivariate testing, and repeated-measures univariate testing. Furthermore, Mauchly testing for sphericity violation and the corresponding correction methods (Greenhouse-Geisser and Huynh-Feldt) also become natural for univariate testing in the multivariate system. This is exactly how the statistical packages implemented those tests such as car in R, GLM in IBM SPSS Statistics, and REPEATED statement in PROC GLM of SAS.
I'm not so sure whether the formulation matters in Bayesian data analysis, but I doubt the above testing capability could be formulated and implemented under the univariate platform.
• I see, this makes sense. Thank you for the great answer. I'd love to hear a Bayesian perspective too. Commented Dec 25, 2013 at 5:37
• @PeterRabbit If you like the answer, please express your gratitude to bluepole by accepting his answer. He'll get points. Commented Dec 26, 2013 at 22:58
• I will, I was just holding out a bit to see if anyone would offer a Bayesians perspective though. Commented Dec 27, 2013 at 19:22
Both models are equivalent if you fit appropriate variance-covariance structure. In transformed linear model we need to fit variance-covariance matrix of error component with kronecker product which has limited availability in available computing softwares. Linear Model Theory-Univariate, Multivariate, and Mixed Models is excellent reference for this topic.
Edited
Here is another nice reference freely available.
• Oh ok, so in a normal univariate model, there is no type of covariance structure "within" the DVs. Therefore hypothesis tests concerned with that don't exist. Thank you! I'll see if I can pick up that book. Commented Dec 25, 2013 at 8:14 | 1,779 | 7,475 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.820161 |
https://math.answers.com/math-and-arithmetic/How_many_number_combinations_are_there_from_rolling_two_number_cubes | 1,726,644,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00377.warc.gz | 338,555,433 | 48,879 | 0
# How many number combinations are there from rolling two number cubes?
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### How many outcomes are possible when rolling 5 number cubes at one time?
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### How many possible combinations of 6 cubes are there?
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### When rolling a cube what is the probability of rolling a 1 and a 6 at the same time?
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There are 5,461,512 such combinations. | 488 | 2,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.938057 |
https://drawabox.com/community/sketchbook/DarkOsyris/replies | 1,726,851,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00324.warc.gz | 185,581,056 | 12,873 | The Autumn Promptathon is Coming
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# DarkOsyris
## darkosyris's Sketchbook
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##### 3:44 PM, Monday November 9th 2020
Hello Restlegless! Let's get right into your 250 boxes exercise
I see that your boxes is really good! they have a very confident lines, great lineworks, and great variety, and your boxes are improving over time! looks like you understand the concept of 3 points perspective and i have nothing to say anymore other than great job! congratulations on finishing this exercise
Next Steps:
Continue to lesson 2
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##### 3:08 PM, Monday November 9th 2020
Hello Devvy! let's get right into your 250 boxes exercise~
So i noticed that your boxes are inconsistently converging into the vp's like you improved some, but overtime your boxes are converging into the infinity again like there are little convergence to your boxes, but from what i see is that your lineworks is pretty confident and smooth so good job on that! but i feel like you still don't understand where the lines needs to be converging so for that i recommend you to check this video again for understanding https://www.youtube.com/watch?v=mteUPdCHn4s where you can see that each lines are converging into 1 vps instead of going infinity, but all in all congratulations on finishing this exercise! try to ask more on discord group so you can get a better understanding on the subject that you got stuck.
Next Steps:
Lesson 2
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##### 4:54 AM, Sunday November 8th 2020
Hello Hamoginabog! let's jump right into your exercise!
Super Imposed Lines
good job with this exercise! i see that some of your lines are wobbly but still this is just an exercise so keep doing these as an warmup
Ghosted Planes
Good job with this exercise! some of your planes have a wobbly lines, and again just keep doing these as an warmup
Ghosted Lines
Good job with this exercise! i can't say much about this exercise cause you've done mostly a good job with these ghosted lines
Table of Ellipses
Good job with this exercise! can't say much with these exercise other than a good job
Ellipses in Planes
Good job! you managed to fit the ellipses inside your planes!
Ellipses in Funnels
Great job! your funnels exercise is really well done!
Plotted Perspective
Another great job! you seem to understand or grasp the concept of 2 point perspective in this exercise
Rough Perspective
Great job with this exercise! this exercise is really hard cause you eyeball your perspective, but from what i see is that you understand the concept of 1 point perspective
Rotated Boxes
Good job with these exercise! your boxes are all rotating! while some of them are rotating in the wrong perspective, you still done a good job!
Organic Perspective
Good job with this exercise! the boxes are very few lol, and some of the boxes are in the same size, but mostly you're boxes is seems to be rotating in 3D space
Next Steps:
250 boxes exercise
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##### 4:19 AM, Sunday November 8th 2020
Hello D2GHER let's get right into your exercise
Organic Arrows
Good job with your arrows! your arrows is vert flowy and they're feel like moving in 3D space
Organic Contour Lines
Good job with this exercise! i see that some of your inner contour lines are varied and some are not, but you seem to understand the concept of this exercise!
Texture Analysis
Great Job with this exercise! you seem to include alot of details and you've done a bang up job with the lizard eye
Dissection
Another great job! i can't say much about this exercise other than you doing this exercise with such a great results!
Form Intersection
Again, great job with this exercise! all of you're forms are intersecting into one another!
Organic Intersection
Good job with this exercise! i see that some of your organic sausage are very stiff, other than that good job!
Next Steps:
Continue to lessons 3 constructing the plants
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##### 2:58 PM, Saturday November 7th 2020
Hello Kamiyasora!
Let's jump right into your exercise
Organic Arrows
So the arrows exercise is very well done! they feel like they're moving in a 3D space and they're very flowy!
Contour Ellipses
Good job on your contour ellipses! i can see that you're a bit struggling with this exercise, i see some of the sausages are very stiff and elongated, but still you've try your best with this exercise so keep it up!
Texture Analysis
Good job with the textures analysis, you've fill your textures with solid shapes before filling your shadows
Dissection Exercise
I can't say much about this exercise other than a good job! i know this exercise is the most difficult and boring, but you keep going and the results are looking great!
Forms Intersection
Great job on this exercise! all of your forms are intersecting into one another
Organic Intersections
Good job with this exercise! i can see that some of your sausage are a bit stiff and wonky, but what's important is that you understand the concept of putting stuff on top of each other!
Well, congrats Kamisayora! after you finish exercise 2, go right ahead into exercise 3! best of luck kamisa :D
Next Steps:
lesson 3
This community member feels the lesson should be marked as complete, and 5 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
##### 5:07 AM, Sunday August 23rd 2020
What i meant about the organic perspective is that some of your boxes are very long, and again, i think it's fine cause your organic perspective exercise looks very convincing and good :)
2 users agree
##### 1:29 PM, Saturday August 22nd 2020
Hi Jamb! let's just get right into your exercise
First thing first, congratulations on finishing your exercise!, it must have been very long, tedious, and tiring, and i can see that you put more effort or doing it seriously on this exercise, cause it shows! for me personally i don't find a major fault or something seriously wrong with all of your boxes cause all of them are improvements, so good job on doing it and keep it up!
Next Steps:
Started doing Lessons 2
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2 users agree
##### 1:14 PM, Saturday August 22nd 2020
Hi Chironex! let's just get into your exercise.
For starters, congratulations on finishing this exercise! it's really a long and tedious exercise, but i can see the improvements on your convergence lines, but i do have some critique on your works
1. Your hatching lines is really random, by that i mean you've done most of the boxes with hatching lines, then the 215 - 230rd boxes you didn't hatch your boxes
2. Your lines are somewhat still wobbly, i recommend to do ghosted lines, sil, and the ghosted planes exercise everyday
Other than that, congratulations again for finishing this exercise! feel free to move on to lessons 2 if you wanted to
Next Steps:
Lessons 2
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##### 12:39 PM, Saturday August 22nd 2020
Hi Hall! let's just get right into all of your exercise
1. Lines for most the part all of your lines are great! even though i noticed that some of your lines are wobbly in your SIL exercise and it's okay cause you've done a great job on all of your lines exercise!
2. Ellipses good job on all of your ellipses exercise too! it is really difficult to draw ellipses ( but don't worry, just remember that this is all just an exercise or a fundamentals, not a final artwork :D )
3. Boxes i don't have any comments on your plotted perspective and rotating boxes exercise other than a great job! as for your plotted perspective, again you've done a great job on it! it is hard though because you are drawing it without a ruler, but atleast you understand the concept of doing rough 1 point perspective, and as for your organic perspective, i see that some of your boxes are elongated, they're supposed to be on the same size. Nonetheless congratulations on finishing Lessons 1 :D
Next Steps:
i request thou to do the 250 boxes exercise
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
2 users agree
##### 3:24 PM, Friday August 21st 2020
Hi Salty_Is_Salty! Let's just get to your exercise :
1. Lines : Overall, great job on all of your lines exercise! however, i see that some of your lines is still wobbly ( but it's okay because for the most part everything looks great! ) and as for the SIL exercise, you did a good job on it too!
2. Ellipses : Again, great job on your ellipses exercise! however, i see that you draw over your ellipses for more over 3 times, just remember that when you drawing your ellipses again, the base maximum is 2 - 3 times :)
3. Boxes : Overall, great job on finishing all of the boxes exercise! it can be difficult and tedious at first, but when you do the 250 boxes challenge hopefully you became much more accustomed with drawing them boxes :)
Next Steps:
Go forth, do the 250 boxes challenge
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
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### Rapid Viz
Rapid Viz is a book after mine own heart, and exists very much in the same spirit of the concepts that inspired Drawabox. It's all about getting your ideas down on the page, doing so quickly and clearly, so as to communicate them to others. These skills are not only critical in design, but also in the myriad of technical and STEM fields that can really benefit from having someone who can facilitate getting one person's idea across to another.
Where Drawabox focuses on developing underlying spatial thinking skills to help facilitate that kind of communication, Rapid Viz's quick and dirty approach can help students loosen up and really move past the irrelevant matters of being "perfect" or "correct", and focus instead on getting your ideas from your brain, onto the page, and into someone else's brain as efficiently as possible. | 2,651 | 11,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.939786 |
https://brilliant.org/problems/harry-potter-is-late/ | 1,477,210,029,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719192.24/warc/CC-MAIN-20161020183839-00333-ip-10-171-6-4.ec2.internal.warc.gz | 818,951,781 | 15,547 | # Well, well, well, we are in trouble
Logic Level 1
Harry Potter and Ron Weasley need to get to Paddington station as quickly as possible since the Hogwarts Express leaves from platform $$9 \frac{3}{4}$$ in 19 minutes. If they were to both run, they would get to the station in only 25 minutes.
However, by a stroke of luck, one of them has a broom, which, alas, can only carry one person at a time. The broom can fly twice as fast as one of the boys can run. But this is an old, cheap broom: it cannot fly by itself. Either Harry or Ron must sit on the broom, and the other must run.
Can they catch the train?
Details and Assumptions:
• They are only in their second year, so they can't Apparate.
• Keep to the constraints of the problem. The broom can't be snapped in half, kids.
• Bonus points for spotting a Harry Potter mistake in the question.
× | 218 | 860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2016-44 | longest | en | 0.981207 |
http://mathisradical.com/how-to-simplify-radical-expressions/radical-worksheets/factor-tree-worksheets.html | 1,498,634,056,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322873.10/warc/CC-MAIN-20170628065139-20170628085139-00023.warc.gz | 260,924,222 | 14,825 | Algebra Tutorials!
Wednesday 28th of June
Home Exponential Decay Negative Exponents Multiplying and Dividing Fractions 4 Evaluating Expressions Involving Fractions The Cartesian Coordinate System Adding and Subtracting Fractions with Like Denominators Solving Absolute Value Inequalities Multiplying Special Polynomials FOIL Method Inequalities Solving Systems of Equations by Graphing Graphing Compound Inequalities Solving Quadratic Equations by Completing the Square Addition Property of Equality Square Roots Adding and Subtracting Fractions The Distance Formula Graphing Logarithmic Functions Fractions Dividing Mixed Numbers Evaluating Polynomials Power of a Product Property of Exponents Terminology of Algebraic Expressions Adding and Subtracting Rational Expressions with Identical Denominators Solving Exponential Equations Factoring The Difference of 2 Squares Changing Fractions to Decimals Solving Linear Equations Using Patterns to Multiply Two Binomials Completing the Square Roots of Complex Numbers Methods for Solving Quadratic Equations Conics in Standard Form Solving Quadratic Equations by Using the Quadratic Formula Simplifying Fractions 2 Exponential Notation Exponential Growth The Cartesian Plane Graphing Linear Functions The Slope of a Line Finding Cube Roots of Large Numbers Rotating Axes Common Mistakes With Percents Solving an Equation That Contains a Square Root Rational Equations Properties of Common Logs Composition of Functions Using Percent Equations Solving Inequalities Properties of Exponents Graphing Quadratic Functions Factoring a Polynomial by Finding the GCF The Rectangular Coordinate System Adding and Subtracting Fractions Multiplying and Dividing Rational Expressions Improper Fractions and Mixed Numbers Properties of Exponents Complex Solutions of Quadratic Equations Solving Nonlinear Equations by Factoring Solving Quadratic Equations by Factoring Least Common Multiples http: Solving Exponential Equations Solving Linear Equations Multiplication Property of Equality Multiplying Mixed Numbers Multiplying Fractions Reducing a Rational Expression to Lowest Terms Literal Numbers Factoring Trinomials Logarithmic Functions Adding Fractions with Unlike Denominators Simplifying Square Roots Adding Fractions Equations Quadratic in Form Dividing Rational Expressions Slopes of Parallel Lines Simplifying Cube Roots That Contain Variables Functions and Graphs Complex Numbers Multiplying and Dividing Fractions 1 Composition of Functions Intercepts of a Line Powers http: Multiplying Two Numbers with the same Tens Digit and whose Ones Digits add up to 10 Factoring Trinomials Exponents and Polynomials Decimals and their Equivalent Fractions Negative Integer Exponents Adding and Subtracting Mixed Numbers Solving Quadratic Equations Theorem of Pythagoras Equations 1 Subtracting Fractions Solving Quadratic Equations by Graphing Evaluating Polynomials Slope Angles and Degree Measure
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Author Message
Clga
Registered: 21.10.2003
From: Igloo
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oc_rana
Registered: 08.03.2007
From: egypt,alexandria
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Jrahan
Registered: 19.03.2002
From: UK
Posted: Friday 29th of Dec 20:26 I fully agree with what was just said . Algebrator has always come to my rescue, be it an assignment or be it my preparation for the midterm exams, Algebrator has always helped me do well in math . It really helped me on topics like difference of cubes, point-slope and adding numerators. I would highly recommend this software.
Svizes
Registered: 10.03.2003
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Posted: Saturday 30th of Dec 16:47 I would advise trying out Algebrator. It not only helps you with your math problems, but also gives all the necessary steps in detail so that you can enhance the understanding of the subject.
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http://gmatclub.com/forum/one-hour-after-yolanda-started-walking-from-x-to-y-a-28692.html?fl=similar | 1,485,021,189,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281162.88/warc/CC-MAIN-20170116095121-00193-ip-10-171-10-70.ec2.internal.warc.gz | 111,319,809 | 38,784 | One hour after Yolanda started walking from X to Y a : Quant Question Archive [LOCKED]
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# One hour after Yolanda started walking from X to Y a
post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Joined: 23 Aug 2004
Posts: 90
Followers: 1
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One hour after Yolanda started walking from X to Y a [#permalink]
### Show Tags
23 Apr 2006, 10:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
One hour after Yolanda started walking from X to Y a distance of 45 miles, Bob started walking along the same road from Y to X. If Yolanda's walking rate was 3 miles per hour and Bob's was 4 miles per hour, how many miles had Bob walked when they met?
a) 24
B) 23
C) 22
D) 21
E) 19.5
I need some quick way to solve such questions!
Thanks!
Manager
Joined: 20 Nov 2004
Posts: 108
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### Show Tags
23 Apr 2006, 10:55
After one hour, Yolanda has walked 3 mls. So there are
42 mls left. Since they walk in opposite directions, the
velocities of both add up until they meet. So they will
meet after 42 mls / 7 mph = 6 h. And after 6 h, Bob
has walked 6 h * 4 mph = 24 mls.
23 Apr 2006, 10:55
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# One hour after Yolanda started walking from X to Y a
post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 622 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-04 | latest | en | 0.913533 |
http://stackoverflow.com/questions/13796782/networkx-random-geometric-graph-implementation-using-k-d-trees?answertab=votes | 1,448,711,272,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398452385.31/warc/CC-MAIN-20151124205412-00325-ip-10-71-132-137.ec2.internal.warc.gz | 218,422,084 | 17,878 | # NetworkX Random Geometric Graph Implementation using K-D Trees
So it is clear with NetworkX that they use an algorithm in n^2 time to generate a random geometric graph. They say there is a faster algorithm possible with the use of K-D Trees. My question is how would one go about attempting to implement the K-D Tree version of this algorithm? I am not familiar with this data structure, nor would I call myself a python expert. Just trying to figure this out. All help is appreciated, thanks!
`````` def random_geometric_graph(n, radius, dim=2, pos=None):
G=nx.Graph()
G.name="Random Geometric Graph"
if pos is None:
# random positions
for n in G:
G.node[n]['pos']=[random.random() for i in range(0,dim)]
else:
nx.set_node_attributes(G,'pos',pos)
# connect nodes within "radius" of each other
# n^2 algorithm, could use a k-d tree implementation
nodes = G.nodes(data=True)
while nodes:
u,du = nodes.pop()
pu = du['pos']
for v,dv in nodes:
pv = dv['pos']
d = sum(((a-b)**2 for a,b in zip(pu,pv)))
return G
``````
-
can you fix up the indentation in your code, it is difficult to read and I don't want to guess at the indent levels. – tcaswell Dec 10 '12 at 15:39
scipy has a KD-tree implementation docs.scipy.org/doc/scipy/reference/generated/… – tcaswell Dec 10 '12 at 15:41
Here is a way that uses the scipy KD-tree implementation mentioned by @tcaswell above.
``````import numpy as np
from scipy import spatial
import networkx as nx
import matplotlib.pyplot as plt
nnodes = 100
r = 0.15
positions = np.random.rand(nnodes,2)
kdtree = spatial.KDTree(positions)
pairs = kdtree.query_pairs(r)
G = nx.Graph()
pos = dict(zip(range(nnodes),positions))
nx.draw(G,pos)
plt.show()
``````
- | 461 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2015-48 | latest | en | 0.798045 |
https://casinoonlinewithbonus.com/in-poker-what-is-a-flush/ | 1,718,422,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00658.warc.gz | 140,616,702 | 40,732 | Select Page
# In poker what is a flush
## Understanding the Basics: In Poker, What is a Flush?
In the game of poker, a flush is a hand that consists of five cards of the same suit, regardless of their numerical value. It is considered one of the stronger hands in poker, although it is not as powerful as a straight flush or a royal flush. The flush falls in the middle of the hand rankings, below a full house but above a straight.
To form a flush, a player must have five cards of the same suit, such as all hearts or all spades. The specific order of the cards does not matter, as long as they are all of the same suit.
For example, a hand with the cards 2, 5, 7, 9, and King of hearts would be considered a flush.
The significance of a flush in poker lies in its ability to beat certain other hands. It is ranked higher than a straight, which is a hand that consists of five consecutive cards of any suit. This means that if two players both have a flush, the player with the higher-ranking cards within the flush will win the pot. In case of a tie, the pot is split between the players.
The likelihood of obtaining a flush depends on various factors, including the number of players at the table and the number of cards that have been dealt.
The more players there are, the higher the chances of someone having a flush. However, the probability can also be influenced by the number of cards of the same suit that are already on the table.
When playing a flush, it is important to consider the strength of the flush in relation to the community cards. For example, if there are four cards of the same suit on the table and you hold one card of that suit, you have a flush draw. This means that if the next card dealt is of the same suit, you will have a flush.
In such cases, it may be worth considering a more aggressive betting strategy to maximize your potential winnings.
In conclusion, a flush in poker is a hand consisting of five cards of the same suit. It is ranked higher than a straight but lower than a full house. Understanding the basics of a flush and its significance in hand rankings is essential for any poker player. Additionally, being aware of the probability of obtaining a flush and employing appropriate strategies can greatly improve your chances of success in the game.
## Hand Rankings: Exploring the Significance of a Flush in Poker
Hand rankings are an essential aspect of poker, and understanding the significance of a flush within these rankings is crucial for any player. A flush is a strong hand that falls below a full house but ranks higher than a straight. It consists of five cards of the same suit, regardless of their numerical value. This means that the cards can be in any order as long as they share the same suit.
The importance of a flush in hand rankings lies in its ability to beat certain other hands.
It is ranked higher than a straight, which is a hand that consists of five consecutive cards of any suit. This means that if two players both have a flush, the player with the higher-ranking cards within the flush will win the pot. However, if the highest-ranking cards within the flush are the same for both players, then the pot is split.
The strength of a flush is determined by the highest card within the hand. For example, a flush with an Ace as the highest card is more valuable than a flush with a King as the highest card. In the event of a tie, the next highest card within the flush is compared, and so on until a winner is determined.
The lowest possible flush is an Ace, 2, 3, 4, and 5 of the same suit, while the highest possible flush is a 10, Jack, Queen, King, and Ace of the same suit, also known as a royal flush.
When playing a flush, it is important to consider the community cards on the table. If there are already three cards of the same suit on the table, and you hold two cards of that suit in your hand, you have a flush draw. This means that you need one more card of the same suit to complete your flush. It is crucial to calculate the odds of getting that card and make decisions based on the potential payout versus the risk involved.
Strategically, playing a flush requires careful consideration.
While it is a strong hand, it can still be beaten by higher-ranking hands such as a full house or a straight flush. It is important to assess the strength of your flush in relation to the community cards and the actions of your opponents. If the board shows a possibility of a higher-ranking hand, it may be wise to fold or play cautiously to avoid losing a significant amount of chips.
In conclusion, a flush is a hand in poker that consists of five cards of the same suit. It ranks higher than a straight but lower than a full house. Understanding the significance of a flush within hand rankings is crucial for strategic decision-making. Considering the strength of the flush in relation to the community cards and the actions of opponents is essential for successful gameplay.
## Strategies and Tips for Playing a Flush in Poker
Playing a flush in poker requires strategic thinking and careful decision-making. While it is a strong hand, there are several strategies and tips to consider to maximize its potential.
One important strategy when playing a flush is to assess the strength of your hand in relation to the community cards. If the board shows a possibility of a higher-ranking hand, such as a straight or a full house, it may be wise to proceed with caution. In such cases, it is crucial to consider the likelihood of your opponents having a stronger hand and adjust your betting accordingly.
Another strategy is to be aware of the number of players in the game.
The more players there are, the higher the chances of someone having a flush. This means that if you have a flush, it may be more difficult to win the pot. In such situations, it is important to consider the size of your bets and the potential risks involved.
Bluffing can also be an effective strategy when playing a flush. By representing a stronger hand, you can potentially force your opponents to fold, allowing you to win the pot without having to show your cards.
However, bluffing should be used sparingly and selectively, as it carries the risk of being called and losing a significant amount of chips.
Position is another crucial aspect to consider when playing a flush. Being in a late position gives you an advantage, as you have more information about your opponents’ actions. This allows you to make more informed decisions and potentially extract more value from your flush.
When playing a flush, it is important to be aware of the odds of completing your hand. If you have a flush draw, meaning you have four cards of the same suit and are waiting for the fifth card, it is essential to calculate the probability of getting that card.
In addition to these strategies, it is important to manage your bankroll effectively when playing a flush. While it is a strong hand, it does not guarantee a win. Therefore, it is crucial to set limits on your bets and not overcommit your chips.
In conclusion, playing a flush in poker requires strategic thinking and careful decision-making. Assessing the strength of your hand, considering the number of players, utilizing bluffing techniques, and being aware of your position are all important strategies to employ. Additionally, calculating the odds of completing your hand and managing your bankroll effectively are essential for success when playing a flush.
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http://www.wyzant.com/answers/rational_expressions | 1,369,243,343,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702127714/warc/CC-MAIN-20130516110207-00045-ip-10-60-113-184.ec2.internal.warc.gz | 818,173,619 | 7,713 | Search 69,948 tutors
# 1/x + 1/9 / 1xsquare - 1/3
1/x + 1/9 / 1xsquare - 1/3
# Combine the rational expressions
Combine the rational expressions.. Reduce your answer to lowest terms 2/x2 + 5x + 6 - 4/x2 + 4x + 3 + 3/x2 + 3x +2
# I need help on these questions
x2+10x+25/4x2+17x-15
y=6/4x+24
# Can this be further simplyfied?
Adding and Subtracting Rational Expressions 5 over 6ab - 7 over 8a after getting common denominators i get 20 - 21b over 24ab can this be further simpified?
# How do you multiply rational expressions?
m/m2-1 * m2+2m-3/my
# How is the process for simplifying and multiplying fractions similar to simplifying or multiplying rational expressions?
How is the process for simplifying 2x(x+1)/4(x+1) similar? Different?
# How is adding, subtracting, multiplying and dividing fractions similar to doing
How is adding, subtracting, multiplying and dividing fractions similar to doing the same with rational expressions. can you give me an example of each?
# How is adding, subtracting, multiplying and dividing fractions similar to doing the same with rational expressions.
How is adding, subtracting, multiplying and dividing fractions similar to doing the same with rational expressions.
# Rational Expressions and Equations
P(x) = 90(1 + 1.5x)/ 1 + 0.5x
# how do you make the denominator equal to 0
how do you get the values of x?
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19 subjects | 470 | 1,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2013-20 | latest | en | 0.843023 |
https://boardgames.stackexchange.com/posts/35566/revisions | 1,566,371,908,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315811.47/warc/CC-MAIN-20190821065413-20190821091413-00428.warc.gz | 397,290,620 | 18,052 | Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
There are other cards which break the movement rules and those cards fully describe the changes to the rules.
Go back 3 spaces
This card clearly describes an movement that is outside of the normal rules and does not use the word advance so it can't be confused with normal movement.
Official Tournament Rules where passing go a second time by this card is not mentioned even though it does mention passing go a second time from this space
GO Each time a player’s token lands on or passes over GO, whether by throw of the dice or by drawing a card, the Banker pays the player a \$200 salary. The \$200 is paid only once each time around the board. However, if a player passing GO on the throw of a dice lands 2 spaces beyond it on Community Chest, or 7 spaces beyond it on Chance, and draws the “Advance to GO” card, he/she collects \$200 for passing GO the first time and another \$200 for reaching it the second time by instructions on the card.
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
There are other cards which break the movement rules and those cards fully describe the changes to the rules.
Go back 3 spaces
This card clearly describes an movement that is outside of the normal rules and does not use the word advance so it can't be confused with normal movement.
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
There are other cards which break the movement rules and those cards fully describe the changes to the rules.
Go back 3 spaces
This card clearly describes an movement that is outside of the normal rules and does not use the word advance so it can't be confused with normal movement.
Official Tournament Rules where passing go a second time by this card is not mentioned even though it does mention passing go a second time from this space
GO Each time a player’s token lands on or passes over GO, whether by throw of the dice or by drawing a card, the Banker pays the player a \$200 salary. The \$200 is paid only once each time around the board. However, if a player passing GO on the throw of a dice lands 2 spaces beyond it on Community Chest, or 7 spaces beyond it on Chance, and draws the “Advance to GO” card, he/she collects \$200 for passing GO the first time and another \$200 for reaching it the second time by instructions on the card.
4 added the go back 3 spaces card.
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
There are other cards which break the movement rules and those cards fully describe the changes to the rules.
Go back 3 spaces
This card clearly describes an movement that is outside of the normal rules and does not use the word advance so it can't be confused with normal movement.
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
There are other cards which break the movement rules and those cards fully describe the changes to the rules.
Go back 3 spaces
This card clearly describes an movement that is outside of the normal rules and does not use the word advance so it can't be confused with normal movement.
3 formating
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania
2. B. & O.
3. Short Line
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
Look closely at the card and it tells you exactly what to do.
Pay owner twice the rental to which he/she is entitled. If railroad is unowned, you may buy it from the bank.
Advance means you are moving forward on the board so this will always move you to the railroad that you run into. In your example it means that you would stop at Pennsylvania.
From what it sounds like you are reading way to much into a simple card. As in the game of monopoly there is no way to go backwards so you never look behind you when you are trying to calculate distance.
To break it down from the chance spot that is in the light blues the ranking of railroads in terms of distance is
1. Pennsylvania (8 spaces away with normal movement)
2. B. & O. (18 spaces away with normal movement)
3. Short Line (28 spaces away with normal movement)
4. Reading (38 spaces away with normal movement)
The only time you can pass go with that card is on the chance spot right after the Short Line spot on the board.
Always need to remember that when playing a board game (unless specifically stated) you calculate distance based on the direction you move.
As for the comment about the go to jail card it does not instruct you to advance to jail. That card is specifying changes to the movement rules on the card because it wants you to treat it differently then other cards that direct you to move to a location
Go to jail
Go directly to jail
Do not pass go
Do not collect \$200
2 Updated to clarify
1 | 3,318 | 15,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-35 | latest | en | 0.95938 |
http://mathhelpforum.com/advanced-math-topics/201407-fun-graph-pf-increasing-population-size-mice-print.html | 1,527,386,437,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00482.warc.gz | 190,028,086 | 3,413 | # Fun graph pf the increasing population size of mice
• Jul 27th 2012, 05:13 AM
YeeP
Fun graph pf the increasing population size of mice
I could really use some help on this one, as I cannot remember what this kind of equation is called or how it can be done.
Here is the data I have inhand, which I am eventually trying to put into a chart in Excel.
best case worst case reproduce litters 5 10 sex ratio 2 3 females litter size 5 6 gestation period 19 21 days female life 2 3 years
female age to reproduce 4 weeks
So, step one was breaking the time down to days.
female life is 730 - 1095 days
female age to reproduce 28 days
By the "best / worst" at the top, I am trying to say the best case scenario (assuming we do not want a lot of mice), versus the worst case scenario.
The calculation is where I am having trouble. Once the female is 28 days old, it will give birth to 5 to 6 pups, 2 or 3 of which are female. The gestation period is 19 - 21 days, so nothing can be produced during this point of time. When the offspring are born, the system begins again, 2 or 3 of those females become 28 days old, then begin to reproduce in the same manner. Of course each female needs to stop reproducing at 730 - 1095 days... (Worried)
I would not expect this to be viewable in one line, being that I gave a range of numbers, it would probably have to be two, but the calc should be the same.
If someone could please help me with figuring out this calculation, which I then need to move into Excel and graph, it would be most appreciated. Also, can you tell me what this calc is called? I know I have done this before, but it has been a long time.
Thank you so much,
Ryan
• Jul 27th 2012, 05:15 AM
YeeP
Re: Fun graph pf the increasing population size of mice
Other fun facts that can be applied to this graph once it is created.... (kinda gross but that is the point)
• An average mouse weighs 15g
• Average mouse is 3.3” long
• Average mouse is approximately 1” tall
• In six months, a mouse can eat about 2 pounds of food and produce some 9,000 droppings
• Average dropping is ¼” long | 549 | 2,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-22 | latest | en | 0.958528 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-02-105732-X&chapter=8&stp=yes&headerFile=7 | 1,386,392,562,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053558/warc/CC-MAIN-20131204131733-00040-ip-10-33-133-15.ec2.internal.warc.gz | 366,397,623 | 5,911 | 1. Write the number of tens and ones in 16.
A. 1 hundred 6 ones B. 6 tens 1 one
C. 1 hundred 6 tens D. 1 ten 6 ones
Hint
2. Compare. Use >, <, or =.
1,030 ___ 988
A. both > and < B. =
C. > D. <
Hint
3. Which addition sentence and multiplication sentence represents 2 groups of 5?
A. 2 + 2 + 2 + 2 + 2 = 10;
2 × 5 = 10
B. 5 + 5 = 10;
2 × 2 = 10
C. 5 + 5 = 10;
2 × 5 = 10
D. 10 + 0 = 10;
5 × 2 = 10
Hint
4. Find the sum.
A. \$7 B. \$80
C. \$710 D. \$70
Hint
5. Find the sum.
278 + 187 = ___
A. 365 B. 465
C. 455 D. 355
Hint
6. Round 48 to the nearest ten.
A. 50 B. 40
C. 45 D. 58
Hint
7. What are the steps you would take to round 428 to the nearest ten?
A. Notice that the first two digits are 4 and 2, and then round to 420. B. Notice that the first digit is 4, and then round to 400.
C. Look at a number line and see whether 428 is closer to 400 or 500. D. Look at a number line and see whether 428 is closer to 420 or 430.
Hint
8. Find the difference.
A. 284 B. 216
C. 384 D. 316
Hint
9. Write an addition sentence for the model.
A. 5 + 5 + 5 = 20 B. 5 + 5 + 5 + 5 + 5 = 25
C. 5 + 5 + 5 = 15 D. 5 + 5 + 5 + 5 = 20
Hint
10. There were 7 cars parked on each side of the street. After some of the cars were driven away only 3 were left. How many cars were driven away?
A. 4 cars B. 21 cars
C. 10 cars D. 11 cars
Hint
11. Identify a pattern. Then find the missing numbers.
30, 40, 50, 60, ___ , ___
Hint
12. Use models to divide. Write a related multiplication fact.
A. 8; 2 × 8 = 18 B. 6; 6 × 3 = 18
C. 9; 2 × 9 = 18 D. 9; 9 + 9 = 18
Hint
13. Use models if needed to solve 42 ÷ 7.
A. 7 B. 4
C. 6 D. 8
Hint
14. Lucilla paid \$8 for a notebook, \$3 for a pen, and \$36 for a bookpack. Which number sentence shows how much she spent altogether?
A. \$8 + \$3 + d = \$20; \$4 B. \$20 – \$8 – \$3 = d; \$4
C. (\$d × 3) + \$8 = \$20; \$4 D. \$20 – \$8 – \$d = \$3; \$4
Hint
15. Which number sentence is shown by the model?
A. 16 + 11 = ___ B. ___ − 11 = 16
C. 16 − 11 = ___ D. ___ − 16 = 11
Hint
16. Since they began playing baseball, Shane always hits 4 fewer homeruns each season than Lyle. Write the function rule.
A. Δ B. Δ − 4
C. Δ + 4 D. Δ × 4
Hint
17. Rosa is saving her allowance. Predict how much money she will have saved at week 4.
Rosa's Savings Week Total Saved 1 \$3 2 \$6 3 \$9
A. \$14 B. \$15
C. \$11 D. \$12
Hint
18. Find the missing number and identify the property.
(6 + 9) + 4 = (9 + __) + 4
A. 6; Commutative Property of Addition B. 9; Commutative Property of Addition
C. 6; Associative Property of Addition D. 6; Identity Property of Addition
Hint
19. Hector’s dog weighs 35 pounds. Lisa’s dog weighs 28 pounds. Find the difference in the weights of the dogs.
A. 14 pounds B. 26 pounds
C. 6 pounds D. 16 pounds
Hint
20. Find the difference.
A. 3,008 B. 3,012
C. 3,018 D. 3,027
Hint | 1,120 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2013-48 | latest | en | 0.815248 |
https://socratic.org/questions/how-do-you-write-y-4x-2-8x-6-into-vertex-form | 1,653,571,455,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00258.warc.gz | 596,596,767 | 5,702 | # How do you write y = 4x^2-8x-6 into vertex form?
$y = 4 {x}^{2} - 8 x - 6 = 4 {\left(x - 1\right)}^{2} - 10$.
In general $a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right) .$ | 117 | 234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-21 | latest | en | 0.474359 |
http://forum.arduino.cc/index.php?topic=147722.msg1119755 | 1,477,041,801,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718034.77/warc/CC-MAIN-20161020183838-00205-ip-10-171-6-4.ec2.internal.warc.gz | 96,190,967 | 14,619 | Go Down
### Topic: Designing optocoupler circuit for arduino input (Read 13890 times)previous topic - next topic
#### xpleria
##### Feb 08, 2013, 06:01 pm
I'm designing an optocoupler circuit whose output will be give to one of the arduino input pins. I need to know for what IC current of the optocoupler I need to design the circuit. I'm concerned with the power loss and want to keep it as min as possible.
I'm using 4n35 Optocoupler.
Here's the signal flow,
230V -> Bridge rectifier + Filter -> R + R + optocoupler Input -> optocoupler output (CTR = 1) ->Arduino
Intially I thought of using 10mA as IC = IF which gives me 230V/10m=23kohm (12k + 12k standard)
But this gives me an I2 loss of 1.2 + 1.2 Watts
In case I use 1mA as IC = IF, I need a resistor of 230k (120k + 120k standard)
with a power loss of 0.1 + 0.1 Watts
1. How low current can I use? (Mainly to decide R1)
2. I don't think this current affects the arduino and it is concerned with the optocoupler. If I use the circuit below, then what VCC and resistor value R2 should I use?
#### MarkT
#1
##### Feb 08, 2013, 06:13 pm
I think fairly low currents can be used - this is something to test with a low voltage circuit first, but
a larger collector resistor for the phototransistor will be needed for lower currents - eventually
noise will become a problem which lower currents and higher resistances, and you will need to
debounce the input to the Arduino (or use a schmitt-trigger gate to clean up the output from
the opto-isolator) - mains waveforms are too slowly varying to drive logic inputs cleanly.
By the way on the mains side you need to use proper high-voltage design (no live conductors
exposed to the touch, plenty of space between traces, and the resistor on the input to the
photo isolator will need to be rated for high voltage (in fact use two in series to reduce the
risk of disaster should one resistor fail).
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#### Docedison
#2
##### Feb 08, 2013, 06:21 pm
That is not a great idea... it looks good on paper BUT Line transients can kill the opto, I've tried it and every easy protection method with very little success. Over time Every opto-isolator I used died. The trick to success is to use a neon light and an LDR shielded from ambient light. For 230 V mains power I would use a 220K series resistor and an NE2H neon light to control an LDR.
Bob
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#### xpleria
#3
##### Feb 08, 2013, 07:01 pm
Need to do some thinking.
Any specific method to couple Ne2H and the LDR? Any special care to be taken?
#### purehunter
#4
##### Feb 09, 2013, 09:21 am
I suggest, to use an additional capacitor in series with the resistor, that limits the current as well, but it does not produce as much heat. It does not just convert current to heat loss. The impedance [ohms] you can calculate with 1/(2*Pi*50*C). Some 100 nF should do. To protect the opto you just insert a Zener in parallel to the opto-LED. There are special surge supressor diodes for this usage. Your dealer should be able to help to find the right one. For a quick'n'dirty solution you could also use 3 or 4 normal diodes in series, and this (in forward direction) in parallel to the opto input.
Be careful!
#### Docedison
#5
##### Feb 09, 2013, 09:34 am
I used black heatshrink tubing. It worked well. The only thing to take care for is that it not leak any light and an ohm-meter will tell you when the LDR is light tight.
Bob
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#### dc42
#6
##### Feb 09, 2013, 10:21 am
I'd go with the circuit you originally intended, but I'd put the bridge rectifier after the resistors, design for 1mA, and use several resistors in series. Choose resistors with a high voltage rating - metal film are usually higher than carbon film. For example 4 resistors, each rated 500V or more. That way, the circuit should be able to cope with large transients.
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#### xpleria
#7
##### Feb 10, 2013, 05:09 pm
Thank you all for your replies.
I believe capacitor in series with the resistors are used since they resist change in voltage and this may limit transient voltage to some limit.
Quote
To protect the opto you just insert a Zener in parallel to the opto-LED. There are special surge supressor diodes for this usage. Your dealer should be able to help to find the right one. For a quick'n'dirty solution you could also use 3 or 4 normal diodes in series, and this (in forward direction) in parallel to the opto input.
I need to either use the zener diode or the surge suppressor diode. Not both right?
One more question. I'll need to make 16 such identical units for the demonstration of my project. What is the optimal solution for my problem, not too expensive, yet protective? (Combining elements of the above suggestions)
#### dc42
#8
##### Feb 10, 2013, 05:41 pmLast Edit: Feb 12, 2013, 03:58 pm by dc42 Reason: 1
I believe capacitor in series with the resistors are used since they resist change in voltage and this may limit transient voltage to some limit.
From the perspective of transient suppression, a capacitor is worse than a resistor, because it will pass the a short transient without attenuating it. I suggest you use just resistors.
Quote
To protect the opto you just insert a Zener in parallel to the opto-LED. There are special surge supressor diodes for this usage. Your dealer should be able to help to find the right one. For a quick'n'dirty solution you could also use 3 or 4 normal diodes in series, and this (in forward direction) in parallel to the opto input.
Personally, I don't believe you need any transient suppression if you use high value resistors (i.e. design for 1mA) with a high enough voltage rating (I suggest at least 1000V total voltage rating, preferably more). I take note of Docedison's post, but without knowing exactly what setup he was using, I can can't comment on why he experienced difficulties. But if you do want to add transient suppression then a zener or TVS diode across the opto isolator is not sufficient unless you put a resistor in series with the opt isolator as well. See attached schematic.
One more question. I'll need to make 16 such identical units for the demonstration of my project. What is the optimal solution for my problem, not too expensive, yet protective? (Combining elements of the above suggestions)
If you want transient suppression, then you can't get much simpler than this schematic. Without the zener or TVS diode and 4.7K resistor, it's even simpler and should still survive any transients that don't cause the resistors to break down.
[EDIT: I goofed - the bridge rectifier in the schematic below needs to be rotated 90 degrees clockwise.]
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#### xpleria
#9
##### Feb 11, 2013, 03:49 pm
Thanks for the neat reply dc42.
Will try and implement that. Thanks again for the schematic (The bridge rectifier diode directions need to be changed).
#### mmoscz
#10
##### Feb 12, 2013, 03:45 pmLast Edit: Feb 12, 2013, 04:15 pm by mmoscz Reason: 1
Try use this..
Now, I using one capacitor to four optocouplers each one with one resistor of 1/8W, and work fine to me...
This circuirt work with 110V too, but to my need I draw the circuirt only with 220V...
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#### purehunter
#11
##### Feb 14, 2013, 05:22 pm
Hi this circuit is no good idea as it is, mmoscz!
This circuit MUST be completed with diodes antiparallel to the opto-LED, or they will be destroyed by the negative voltage. LED's are sensitive against negative voltage.
Additionally (surge-) suppressor diodes would help to protect against bursts on the line which are caused by most electrical machines an by dimmers as well as just switching anything on or off.
#### purehunter
#12
##### Feb 14, 2013, 05:38 pm
There is another weak point in the circuit of mmoscz:
ALWAYS connect the transistors at the inputs with their emitter to GND and the collector to the input! This is more safe against disturbances.
The resistor is not necessary, just use the command
digitalWrite(Pin#,INPUT);
digitalWrite(Pin#,HIGH);
with this you enable the internal pull up resistor.
If it is absolutely required to go your way due to any other reason, you need to use much lesser resistor value.
#### xpleria
#13
##### Feb 15, 2013, 03:11 amLast Edit: Feb 15, 2013, 03:17 am by xpleria Reason: 1
If I use the internal pull up resistor, then I don't need to connect the Vcc as well right? Just connect the emitter to ground n collector directly to input pin?
In case I need to use an entire port as input with pull up resistor, then will this code work,
Code: [Select]
`DDRL = 0x00;PORTL = 0xFF;`
#### dc42
#14
##### Feb 15, 2013, 09:22 am
Correct on both counts.
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• Emilia is a 3 year old girl who loves cookies. She eats them all day long and can’t go a day without them. One day she was bad and her mom told her she can't have anymore cookies. She is looking up at the cookie jar that sits on the table and is determined to get it.
• Emilia waits until her mom goes to the other room. She is standing 29 feet away looking at the table with the cookie jar at an angle of 55°. She wonders how high the jar is, and to do so she uses trig. She finds out that Tan(55) * 29 = 41.4. The cookie jar is 41 feet high.
• 55°
• 29 ft
• X
• Unfortunately, she realizes that the cookie jar is too high for her to reach. She stands on a chair to get closer to the height of the table. Now the cookie jar is 15 feet high and her angle to the table is 53°. She uses trig to figure out how far away she is. She finds out that 15/Tan(53) = 11.3. Now she is 11 feet away.
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• Sadly, Emilia is still far away. She decides to move the chair closer and stand on her tippy-toes. She realizes that she can't reach the jar so she grabs a stick to try to bring it closer. The cookie jar is 10 feet high and the stick is 15 feet long. Her angle to the table is 50°. She finds out that Cos(50) * 15 = 9.6. The jar is 10 feet away.
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• 10 ft
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In following question, a number series is given with one term missing.
Choose correct alternative that will same pattern and fill in the blank spaces.
7, 6, 10 ,27, 104, __
###### Option:
A. 520
B. 420
C. 515
D. 525
Justification:
Here,
7 * 1 - 1 = 6
6 * 2 - 2 = 10
10 * 3 - 3 = 27
27 * 4 - 4 = 104
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## Question
http://www.codewars.com/kata/534d0a229345375d520006a0/train/javascript
Write a function that determines if given number is a power of two. A power of two means a number of the form 2^n where n is an integer, i.e. the result of exponentiation with number two as the base and integer n as the exponent. I.e. 1024 is a power of two: it 2^10.
```isPowerOfTwo(4096) // -> true
isPowerOfTwo(333) // -> false
```
Pay attention: hidden tests are using extremmely big numbers
## My Solution
```function isPowerOfTwo(n){
if(n==2){return true;}
else if(n%2==1){return false;}
else {return isPowerOfTwo(n/2);}
}
```
처음엔 재귀함수를 돌 때 앞에 `return`을 쓰지 않으니
마지막에 불리우는 `return`문에서 함수를 빠져나가는게 아니라 재귀함수를 빠져나가버렸다.
그래서 계속 `undefined`가 나왔다.
## @BinaryPanda ‘s Solution
```function isPowerOfTwo(n) {
return n && !(n & (n - 1));
}
```
이진법 연산이다. 이런 사람들 보면 내가 미워진다.
```For example:
1 is 000001
2 is 000010
4 is 000100
8 is 001000
16 is 010000
32 is 100000
```
So, we need a way to verify that only a single digit is set. A bitwise and (the single &) will compare the digits in the positions they are in. So 10 & 01 is 00, but 11 & 10 is 10. Only if both spots contain a 1, does the result contain a 1.
If a number is a power of two, subtracting one will result in something like this:
```8 = 001000
7 = 000111
```
7 & 8 will be 0, because they share no 1’s in the same spot. Then by using a !, we invert the answer.
What if the number is not a power of two? Like what about 7?
```7 = 000111
6 = 000110
7 & 6 = 000110 (6)
```
## @PandaCoder’s Solution
```function isPowerOfTwo(n){
return Math.log(n)/Math.log(2) % 1==0;
}
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How many perfect squares (Posted on 2006-04-25)
Suppose a and b are positive integers. We all know that aČ+2ab+bČ is a perfect square. Give an example where also aČ+ab+bČ is a perfect square. How many such examples exist?
See The Solution Submitted by Salil Rating: 3.0000 (2 votes)
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re(3): Partial Solution | Comment 5 of 19 |
(In reply to re(2): Partial Solution by tomarken)
The formula does seem to work. I haven't worked out how it works however; I'd like to see the derivation or proof.
The formula does not, however, provide all the base pairs. While it provides (7,8), it does not provide (7,33). It doesn't provide those base pairs where neither member is prime, such as (16,39) or (65,88).
But it does show there are infinitely many base pairs.
Posted by Charlie on 2006-04-25 11:03:53
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InvHypergeo(s,D,M) - no Crystal ball distribution
Inverse hypergeometric equations
The Inverse Hypergeometric distribution InvHypergeo(s,D,M) models the total number of trials one would have before achieving the sth success in a hypergeometric sampling where there are D individuals of interest (their selection is a 'success') in a population of size M. Four examples of the Inverse Hypergeometric distribution are shown below:
#### Uses
It should be used in any situation where there is hypergeometric sampling and one is asking the question: "How many samples will I need to get s successes?", or alternatively (if you subtract the number of success s) "How many failures will I see before I observe s successes?".
##### Example
The number of cards one needs to turn over to see three hearts will be Inverse Hypergeometric distributed. If the total number of cards is 54, of which 13 are hearts, the number of cards one needs to turn over is InvHypergeo(3,13,54).
#### Generation
The Inverse Hypergeometric distribution is not directly available with Crystal Ball, but can be easily constructed via the Discrete distribution as shown in the spreadsheet Image Removed Image AddedHyperGeoTrials. This model allows one to maintain the advantages of Latin Hypercube sampling. | 290 | 1,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.900486 |
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Kilocalorie
# Kilocalorie - Kilocalorie energy balance and weight control...
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Kilocalorie, energy balance and weight control 1. Kilocalorie a. Define: amount of heat required to raise the amount of 1L to 1 degree Celcius (1kc - ->1L- 1 degree Celcius) b. Determined by bomb calorimeter c. Determining individual k-calorie needs? c.i. Basil metabolic rate (BMR) c.i.1. Metabolism c.i.1.a. Process of food to gain energy c.i.1.b. Food becomes the fuel for our function c.i.2. “basil” c.i.2.a. Minimum energy requirement for life c.i.2.b. Taken when awake and flat on yo back c.i.2.c. Less energy needed when laying than standing c.i.2.d. Determined by 1kg of body weight c.i.3. How to determine c.i.3.a. Convert lbs into kg c.i.3.b. Take kg and multiply by 24 = BMR c.i.4. Variables c.i.4.a. Gender c.i.4.a.i. Male c.i.4.a.ii. Female c.i.4.a.ii.1. Has lower metabolism c.i.4.a.ii.2. Makes 10% decrease of value c.i.4.a.ii.3. Has more fat c.i.4.b. Age : When reached age 30, metabolism decreases 0.5% each year c.i.4.c. State of health c.i.4.c.i.
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Yes. Instead of applying $\quad \log_{e}e^{k}=k$, we apply $\quad \log_{10}e^{k}=k\cdot\log_{10}e$. (sample answer)
Taking the common logaritm on both sides, we would obtain $\log e^{0.06t}=\log 1500\qquad$ ... apply the power rule, $0.06t\log e==\log 1500\qquad$ ... divide with $(0.06\log e)$ $t=\displaystyle \frac{\log 1500}{0.06\log e}$ So, yes, we could have applied $\log$ instead of $\ln$. | 162 | 476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-16 | latest | en | 0.661867 |
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posted by .
Solid carbon dioxide (dry ice) goes from solid to gas at -78C. The molar heat of sublimation is 25.3 kj/mol. In one experiment 34.5g of dry ice are placed in an insulated isolated container with 500 g of water at 25C. What temp of the water is expected after all bubbles are gone? Specific heat of water is 4.184 J/g*C ??
• Chem -
I suppose we assume that the gas, after forming, does not cool the water on its way out of the system.
moles CO2 = 34.5g/44 = ??
?? moles CO2 x 25.3 kJ/mol = xx kJ to sublime the dry ice. Convert to joules.
[mass water x specific heat water x (Tfinal-25)] + xxJOULES = 0
Solve for Tfinal.
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10. ### PHYSIC
125 g of dry ice (solid CO2) is dropped into a beaker containing 500 g of 66°C water. The dry ice converts directly to gas, leaving the solution. When the dry ice is gone, the final temperature of the water is 29°C. What is the heat …
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A measuring instrument shows the length to be 508 feet. Back to Top To calculate the relative error use the following way:Observe the true value (x) and approximate measured value (xo). You pace from one tree to another and estimate that they're 18 feet apart. Please select a newsletter. weblink
Relative ErrorProblems Related Concepts Relative Error Formula Absolute Error Formula Formula for Sampling Error formula for relative density Relations Margin Error Type I Error and Type II Error Physical Chemistry and continue reading below our video How Does Color Affect How You Feel? Leave the relative error in fraction form, complete the division to render it in decimal form, or multiply the resulting decimal form by 100 to render your answer as a percentage. Please try again.
Relative Percent Error Equation
For this reason, it is more useful to express error as a relative error. Should the accepted or true measurement NOT be known, the relative error is found using the measured value, which is considered to be a measure of precision. up vote 10 down vote favorite 3 How do I calculate relative error when the true value is zero?
App preview Similar Apps:Loading suggestions...Used in these spaces:Loading... When weighed on a defective scale, he weighed 38 pounds. (a) What is the percent of error in measurement of the defective scale to the nearest tenth? (b) If Millie, the Any measurements within this range are "tolerated" or perceived as correct. Absolute Error Vs Relative Error which is the absolute error?
Simply substitute the equation for Absolute Error in for the actual number. How To Calculate Absolute And Relative Error Chemistry Flag as... The solution is to weigh the absolute error by the inverse of a yardstick signal, that has a similar fall-off properties to the signals of interest, and is positive everywhere. http://www.wikihow.com/Calculate-Relative-Error About this wikiHow 101reviews Click a star to vote Click a star to vote Thanks for voting!
Many scientific tools, like precision droppers and measurement equipment, often has absolute error labeled on the sides as "+/- ____ " 3 Always add the appropriate units. Absolute Deviation Calculator they could both be the smallest possible measure, or both the largest. Can an opponent folding make you go from probable winner to probable loser? Becomean Author!
How To Calculate Absolute And Relative Error Chemistry
In it, you'll get: The week's top questions and answers Important community announcements Questions that need answers see an example newsletter By subscribing, you agree to the privacy policy and terms The relative error of the measurement is 2 mph / 60 mph = 0.033 or 3.3%More About Experimental Error Show Full Article Related This Is How To Calculate Percent Error What Relative Percent Error Equation Note that absolute error is reported in the same units as the measurement.Alternatively, you may have a known or calculated value and you want to use absolute error to express how Absolute Error And Relative Error Formula Accuracy is a measure of how close the result of the measurement comes to the "true", "actual", or "accepted" value. (How close is your answer to the accepted value?) Tolerance is
Did you mean ? have a peek at these guys Paper Boat Creative, Getty Images By Anne Marie Helmenstine, Ph.D. Observed Value True Value RelatedPercentage Calculator | Scientific Calculator | Statistics Calculator In the real world, the data measured or used is normally different from the true value. Should I include him as author? Absolute Error And Relative Error Examples
Plural of "State of the Union" Is there any historical significance to the Bridge of Khazad-dum? No scientific study is ever perfectly error free -- even Nobel Prize winning papers and discoveries have a margin or error attached. Absolute Error: Absolute error is simply the amount of physical error in a measurement. check over here Since the measurement was made to the nearest tenth, the greatest possible error will be half of one tenth, or 0.05. 2.
Measuring instruments are not exact! Relative Standard Deviation Calculator b.) The relative error in the length of the field is c.) The percentage error in the length of the field is 3. The precision of a measuring instrument is determined by the smallest unit to which it can measure.
HomePhysicsRelative Error Formula Top Relative Error Formula Many a times it happens that there will approximately some error in the instruments due to negligence in measuring precisely.
EDIT Edit this Article Home » Categories » Education and Communications » Subjects » Mathematics ArticleEditDiscuss Edit ArticlewikiHow to Calculate Relative Error Two Methods:Calculating Absolute ErrorCalculating Relative ErrorCommunity Q&A Absolute error Train carriages in the Czech Republic Can I travel inside the US with a digital copy of my passport and visa? We don't know the actual measurement, so the best we can do is use the measured value: Relative Error = Absolute Error Measured Value The Percentage Error is the Relative How To Find Percent Relative Error After mixing and matching, her test tube contains 32 grams of substrate.
Email us at [email protected] Powered by Mediawiki. c.) the percentage error in the measured length of the field Answer: a.) The absolute error in the length of the field is 8 feet. http://integerwireless.com/relative-error/absolute-vs-relative-error.php Still, understanding where error comes from is essential to help try and prevent it:[5] Human error is the most common.
Chemistry Chemistry 101 - Introduction to Chemistry Chemistry Tests and Quizzes Chemistry Demonstrations, Chemistry Experiments, Chemistry Labs & Chemistry Projects Periodic Table and the Elements Chemistry Disciplines - Chemical Engineering and The system returned: (22) Invalid argument The remote host or network may be down. It is always the same problem with that. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the
Cookies make wikiHow better. Something which is not terminal or fatal but lifelong Multiple-Key Sorting Do I need to cite an old theorem, if I've strengthened it, wrote my own theorem statement, with a different Ways of Expressing Error in Measurement: 1. This is your absolute error![2] Example: You want to know how accurately you estimate distances by pacing them off.
I know the true parameter value (\$x_{true}\$), and I have simulation data from which I infer an estimate of the parameter (\$x_{test}\$). Relative error is expressed as fraction or is multiplied by 100 and expressed as a percent.Relative Error = Absolute Error / Known ValueFor example, a driver's speedometer says his car is going In many situations, the true values are unknown. The accepted value for her experiment was 34 grams. | 1,403 | 7,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-39 | latest | en | 0.824454 |
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1 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 221 le-tex Linear and Quadratic Discriminant Analysis, Logistic Regression, and Partial Least Squares Regression In this chapter, we review, for the most part, linear methods for classification. The only exception is quadratic discriminant analysis, a straightforward generalization of a linear technique. These methods are best known for their simplicity. A linear decision boundary is easy to understand and visualize, even in many dimensions. An example of such a boundary is shown in Figure 11.1 for Fisher iris data. Because of their high interpretability, linear methods are often the first choice for data analysis. They can be the only choice if the analyst seeks to discover linear relationships between variables and classes. If the analysis goal is maximization of the predictive power and the data do not have a linear structure, nonparametric nonlinear methods should be favored over simple interpretable techniques. Linear discriminant analysis (LDA), also known as Fisher discriminant, has been a very popular technique in particle and astrophysics. Quadratic discriminant analysis (QDA) is its closest cousin Discriminant Analysis Suppose we observe a sample drawn from a multivariate normal distribution N(μ, Σ ) with mean vector μ and covariance matrix Σ.ThedataareD-dimensional, and vectors, unless otherwise noted, are column-oriented. The multivariate density is then 1 P(x) D p (2π)D jσ j exp 1 2 (x μ)t Σ 1 (x μ), (11.1) where jσ j is the determinant of Σ. Suppose we observe a sample of data drawn from two classes, each described by a multivariate normal density 1 P(xjk) D p (2π)D jσ k j exp 1 2 (x μ k )T Σ 1 k (x μ k ) (11.2) Statistical Analysis Techniques in Particle Physics, First Edition. Ilya Narsky and Frank C. Porter WILEY-VCH Verlag GmbH & Co. KGaA. Published 2014 by WILEY-VCH Verlag GmbH & Co. KGaA.
2 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 222 le-tex Linear Classification Figure 11.1 Class boundary obtained by linear discriminant analysis for Fisher iris data. The square covers one observation of class versicolor and two observations of class virginica. for classes k D 1, 2. Recall that Bayes rule gives P(kjx) D π k P(xjk) P(x) (11.3) fortheposteriorprobabilityp(kjx) of observing an instance of class k at point x. The unconditional probability P(x) in the denominator does not depend on k.the prior class probability π k was introduced in Chapter 9; we discuss its role in the discriminant analysis below. Let us take a natural logarithm of the posterior odds: P(k D 1jx) log P(k D 2jx) D log π 1 1 π 2 2 log jσ 1j jσ 2 j C x T Σ 1 1 μ 1 Σ 1 2 μ x T Σ 1 1 Σ 1 2 x 1 2 μ T 1 Σ 1 1 μ 1 μ T 2 Σ 1 2 μ 2. (11.4) The hyperplane separating the two classes is obtained by equating this log-ratio tozero.thisisaquadraticfunctionofx, hencequadratic discriminant analysis. If the two classes have the same covariance matrix Σ 1 D Σ 2,thequadraticterm disappears and we obtain linear discriminant analysis. Fisher (1936) originally derived discriminant analysis in a different fashion. He searched for a direction q maximizing separation between two classes, q T 2 μ S(q) D 1 μ 2. (11.5) q T Σ q This separation is maximized at q D Σ 1 (μ 1 μ 2 ). The terms not depending on x in (11.4) do not change the orientation of the hyperplane separating the two distributions they only shift the boundary closer to one class and further away from the other. The formulation by Fisher is therefore equivalent to (11.4) for LDA.
3 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 223 le-tex 11.1 Discriminant Analysis 223 To use this formalism for K > 2 classes, choose one class, for example the last one, for normalization. Compute posterior odds log[p(kjx)/p(kjx)] for k D 1,...,K 1. The logarithm of the posterior odds is additive, that is, log[p(ijx)/p( j jx)] D log[p(ijx)/p(kjx)] log[p( j jx)/p(kjx)] for classes i and j.thecomputedk 1 log-ratios give complete information about the hyperplanes of separation. If we need to compute the posterior probabilities, we require that P K kd1 P(kjx) D 1 and obtain estimates of P(kjx) from the log-ratios. Thesametrickisusedinothermulticlassmodelssuchasmultinomiallogistic regression. For prediction on new data, the class label is assigned by choosing the class with the largest posterior probability. In this formulation, the class prior probabilities merely shift the boundaries between the classes without changing their orientations (for LDA) or their shapes (for QDA). They are not used to estimate the class means or covariance matrices; hence, they can be applied after training. Alternatively, we could ignore the prior probabilities and classify observations by imposing thresholds on the computed log-ratios. These thresholds would be optimized using some physics-driven criteria. Physicists often follow the second approach. As discussed in Chapter 9, classifying into the class with the largest posterior probability P(yjx) minimizes the classification error. If the posterior probabilities are accurately modeled, this classifier is optimal. If classes indeed have multivariate normal densities, QDA is the optimal classifier. If classes indeed have multivariate normal densities with equal covariance matrices, LDA is the optimal classifier. Most usually, we need to estimate the covariance matrices empirically. LDA is seemingly simple, but this simplicity may be deceiving. Subtleties in LDA implementation can change its result dramatically. Let us review them now Estimating the Covariance Matrix Under the LDA assumptions, classes have equal covariance matrices and different means. Take the training data with known class labels. Let M be an N K class membership matrix for N observations and K classes: m nk D 1ifobservationn is from class k and 0 otherwise. First, estimate the mean for each class in turn, nd1 Oμ k D m nkx n nd1 m nk. (11.6) Then compute the pooled-in covariance matrix. For example, use a maximum likelihood estimate, OΣ ML D 1 KX NX m nk (x n Oμ N k )(x n Oμ k ) T. (11.7) kd1 nd1 Vectors x n and Oμ k are D 1(column-oriented),and(x n Oμ k )(x n Oμ k ) T is therefore a symmetric D D matrix. This maximal likelihood estimator is biased. To remove
4 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 224 le-tex Linear Classification the bias, apply a small correction: OΣ D N N K O Σ ML. (11.8) Elementary statistics textbooks derive a similar correction for a univariate normal distribution and include a formula for an unbiased estimate, S 2 D nd1 (x n Nx) 2 /(N 1), of the variance, σ 2.Thestatistic(N 1)S 2 /σ 2 is distributed as χ 2 with N 1 degrees of freedom. We use N K instead of N 1becausewehaveK classes. Think of it as losing one degree of freedom per linear constraint. In this case, there are K linear constraints for K class means. In physics analysis, datasets are usually large, N K. For unweighted data, this correction can be safely neglected. For weighted data, the problem is a bit more involved. The weighted class means are given by Oμ k D nd1 m nkw n x n nd1 m nkw n. (11.9) The maximum likelihood estimate (11.7) generalizes to OΣ ML D KX kd1 nd1 NX m nk w n (x n Oμ k )(x n Oμ k ) T. (11.10) Above, we assume that the weights are normalized to sum to one: nd1 w n D 1. The unbiased estimate is then OΣ D OΣ ML 1 P K kd1 W (2) k W k, (11.11) where W k D nd1 m nkw n is the sum of weights in class k and W (2) k D nd1 m nkwn 2 is the sum of squared weights in class k. For class-free data K D 1, this simplifies to OΣ D OΣ ML /(1 nd1 w n 2 ). If all weights are set to 1/N, (11.11) simplifies to (11.8). In this case, the corrective term nd1 w n 2 attains minimum, and the denominator in (11.11) is close to 1. If the weights are highly nonuniform, the denominator in (11.11) can get close to zero. For LDA with two classes, this bias correction is, for the most part, irrelevant. Multiplying all elements of the covariance matrix by factor a is equivalent to multiplying x T Σ 1 (μ 1 μ 2 )by1/a. This multiplication does not change the orientation of the hyperplane separating the two classes, but it does change the posterior class probabilities at point x. Instead of using the predicted posterior probabilities directly, physicists often inspect the ROC curve and select a threshold on classification scores (in this case, posterior probabilities) by optimizing some function of true positive and false positive rates. If we fix the false positive rate, multiply the logratio at any point x by the same factor and measure the true positive rate, we will obtain the same value as we would without multiplication.
5 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 225 le-tex 11.1 Discriminant Analysis 225 Unfortunately, this safety mechanism fails for QDA, multiclass LDA, and even LDA with two classes if the covariance matrix is estimated as a weighted combination of the individual covariance matrices, as described in Section We refrain from recommending the unbiased estimate over the maximum likelihood estimate or the other way around. We merely point out this issue. If you work with highly nonuniform weights, you should investigate the stability of your analysis procedure with respect to weighting Verifying Discriminant Analysis Assumptions The key assumptions for discriminant analysis are multivariate normality (for QDA and LDA) and equality of the class covariance matrices (for LDA). You can verify these assumptions numerically. Two popular tests of normality proposed in Mardia (1970) are based on multivariate skewness and kurtosis. The sample kurtosis is readily expressed in matrix notation Ok D 1 N NX nd1 h (x n Oμ) T OΣ 1 (x n Oμ)i 2, (11.12) where Oμ and OΣ are the usual estimates of the mean and covariance matrix. Asymptotically, Ok has a normal distributionwith mean D(DC2) and variance 8D(DC2)/N for the sample size N and dimensionality D. A large observed value indicates a distribution with tails heavier than normal, and a small value points to a distribution with tails lighter than normal. A less rigorous but more instructive procedure is to inspect a quantile-quantile (QQ) plot of the squared Mahalanobis distance (Healy, 1968). If X is a random vector drawn from a multivariatenormaldistribution with mean μ and covariance Σ, its squared Mahalanobis distance (X μ) T Σ 1 (X μ) hasaχ 2 distribution with D degrees of freedom. We can plot quantiles of the observed Mahalanobis distance versus quantiles of the χ 2 D distribution. A departure from a straight line would indicate the lack of normality, and extreme points would be considered as candidates for outliers. In practice we know neither μ nor Σ and must substitute their estimates. As soon as we do, we, strictly speaking, can no longer use the χ 2 D distribution, although it remains a reasonable approximation for large N. No statistical test is associated with this approach, but visual inspection often proves fruitful. Equalityof the class covariancematrices can be verified by a Bartlett multivariate test described in popular textbooks such as Andersen (2003). Formally, we test hypothesis H 0 W Σ 1 D... D Σ K against H 1 W at least two Σ 0 s are different. The test statistic, 2logV D (N K)logj OΣj KX (n k 1) log j OΣ k j, (11.13) kd1
6 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 226 le-tex Linear Classification resembles a log-likelihood ratio for the pooled-in unbiased estimate OΣ and unbiased class estimates OΣ k. Here, n k is the number of observations in class k. This formula would need to be modified for weighted data. When the sizes of all classes arecomparableandlarge, 2logV can be approximated by a χ 2 distribution with (K 1)D(D C 1)/2 degrees of freedom (see, for example, Box, 1949). For small samples, the exact distribution can be found in Gupta and Tang (1984). H 0 should be rejected if the observed value of 2logV is large. Intuitively, 2logV measures the lack of uniformity of the covariance matrices across the classes. The pooled-in estimate is simply a sum over class estimates (N K) OΣ D P K kd1 (n k 1) OΣ k.if the sum of several positive numbers is fixed, their product (equivalently, the sum of their logs) is maximal when the numbers are equal. This test is based on the same idea. The Bartlett test is sensitive to outliers and should not be used in their presence. The tests described here are mostly of theoretical value. Practitioners often apply discriminant analysis when its assumptions do not hold. The ultimate test of any classification model is its performance. If discriminant analysis gives a satisfactory predictive power for nonnormal samples, don t let the rigor of theory stand in your way. Likewise, you can verify that QDA improves over LDA by comparing the accuracies of the two models using one of the techniques reviewed in Chapter Applying LDA When LDA Assumptions Are Invalid Under the LDA assumptions, all classes have multivariate normal distributions with different means and the same covariance matrix. The maximum likelihood estimate (11.10) is equal to the weighted average of the covariance matrix estimates per class: OΣ ML D KX W k OΣ k. (11.14) kd1 Above, W k D nd1 m nkw n is the sum of weights in class k. As usual, we take nd1 w n D 1. In practice, physicists apply LDA when none of the LDA conditions holds. The class densities are not normal and the covariance matrices are not equal. In these circumstances, you can still apply LDA and obtain some separation between the classes. But there is no theoretical justification for the pooled-in covariance matrix estimate (11.14). It is tempting to see how far we can get by experimenting with the covariance matrix estimate. Let us illustrate this problem on a hypothetical example. Suppose we have two classes with means μ 1 D 1/2 0 1/2 μ 2 D 0 (11.15)
7 Frank C Porter and Ilya Narsky: Statistical Analysis Techniques in Particle Physics Chap. c /9/9 page 227 le-tex 11.1 Discriminant Analysis 227 and covariance matrices 2 1 Σ 1 D 1 2 Σ 2 D (11.16) The two covariance matrices are inverse to each other, up to some constant. If we set the covariance matrix for LDA to (Σ 1 C Σ 2 )/2, the predicted line of optimal separation is orthogonal to the first coordinate axis and the optimal classification is given by x 1.IfwesetthecovariancematrixforLDAtoΣ 1, the optimal classification is 2x 1 x 2.IfwesetthecovariancematrixforLDAtoΣ 2, the optimal classification is 2x 1 C x 2. If we had to estimate the pooled-in covariance matrix on a training set, we would face the same problem. If classes 1 and 2 were represented equally in the training data, we would obtain (Σ 1 C Σ 2 )/2. If the training set were composed mostly of observations of class 1, we would obtain Σ 1. Similarly for Σ 2. Let us plot ROC curves for the three pooled-in matrix estimates. As explained in the previous chapter, a ROC curve is a plot of true positive rate (TPR) versus false positive rate (FPR), or accepted signal versus accepted background. Take class 1 to be signal and class 2 to be background. The three curves are shown in Figure Your choice of the optimal curve (and therefore the optimal covariance matrix) would be defined by the specifics of your analysis. If you were mostly concerned with background suppression, you would be interested in the lower left corner of the plot and choose Σ 2 as your estimate. If you goal were to retain as much signal as possible at a modest background rejection rate, you would focus on the upper right corner of the plot and choose Σ 1. If you wanted the best overall quality of separation measured by the area under the ROC curve, you would choose (Σ 1 C Σ 2 )/2. It is your analysis, so take your pick! If we took this logic to the extreme, we could search for the best a in the linear combination OΣ D a OΣ 1 C (1 a) OΣ 2 by minimizing some criterion, perhaps FPR at fixed TPR. If you engage in such optimization, you should ask yourself if LDA is the right tool. At this point, you might want to give up the beloved linearity and switch to a more flexible technique such as QDA. In this example, QDA beats LDA at any FPR, no matter what covariance matrix estimate you choose. Figure 11.2 ROC curves for LDA with three estimates of the covariance matrix and QDA.
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The figure shows an interactive graph of $y = f(t)$. (It may take a minute to load -- be patient.) You can change the value of $x$ in the graph by clicking and dragging the red dot along the horizontal $t$-axis. Assume that the lines in each piece of $f$ continue beyond the graphing window. Define a function that accumulates signed area between the $t$-axis and the graph of $y = f(t)$ by $\displaystyle F(x) = \int_0^x f(t) \, dt$ (a) For what values of $x$ does $F(x) = 0$? $x \approx$ (b) For what values of $x$ is $F(x) \geq 0$? All $x$ in the interval (c) For what values of $x$ does $F(x) = 6$? $x \approx$ (d) Why is $F(x)$ positive when $x$ is negative? A. Because accumulation functions like $F(x)$ are always positive B. Because integrals are always positive C. Because integrating right-to-left is negative D. Because $f(x)$ is negative and integrating right-to-left is negative E. None of the above Graph of $y = f(t)$
You can earn partial credit on this problem. | 318 | 1,135 | {"found_math": true, "script_math_tex": 21, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.810926 |
https://toph.co/p/floor-no-20 | 1,653,237,755,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00562.warc.gz | 655,821,320 | 11,953 | Practice on Toph
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Floor No. 20
By Sk_Sabit · Limits 1s, 512 MB
You are at a 12 story building called ECE Bhaban. You have just entered into a lift which has took you to a mysterious floor, Floor no. 20. In fact, you have got stuck into the lift and there is no way out of it except unlocking a Mysterious Lock. In order to unlock it, you have to solve a Puzzle.
The puzzle consists of a Binary Matrix of dimension $N \times M$. A binary Matrix has 0 or 1 in each of it’s cells and it’s called Beautiful when each of its rows has an odd number of ones and each of its columns has an even number of ones.
Now, to unlock the mysterious lock, you have to transform the given matrix into a Beautiful one. In one move, you can flip any cell value of the matrix (flipping a value means, if there is a $0$, after the move there will be a $1$, and vice versa). But the time is short, you are suffocating inside the lift. So, you have to do it with a $Minimum$ number of moves or report if it’s impossible to unlock the mysterious lock.
The rows are numbered from $1$ to $N$ (top to bottom) and the columns are numbered from $1$ to $M$ (left to right).
You have to print the minimum number of moves required to unlock the lock and the corresponding moves. If there are multiple solution, you can print any of them. If there is no solution, print “-1” (without quotes).
Input
The first line of input consists of two space separated integers $N$ and $M$ ($1 \leq N, M \leq 1000$), representing the number of rows and the number of columns of the matrix respectively.
Next $N$ line contains $M$ characters each (each of which is either 0 or 1) representing the rows of the matrix.
Output
The first line of output will contain a single integer $K$, the minimum number of moves required to unlock the lock or “-1” (without quotes) if it’s impossible to unlock.
If it’s possible to solve the puzzle, then next $K$ lines will print the moves. It will contain two space separated integers each, representing the co-ordinate of the cell which has been flipped on that move.
The output format will be of this form:
$K$
$X_{1}$ $Y_{1}$
$X_{2}$ $Y_{2}$
$X_{k}$ $Y_{k}$
here $X_{i}$ and $Y_{i}$ represents the row index of the cell (from top) and column index of the cell (from left) respectively.
($1 \leq X_{i} \leq N, 1 \leq Y_{i} \leq M$)
It is guaranteed that if there is a solution, then the number of moves of the solution will not exceed N x M.
Samples
InputOutput
4 3
101
011
100
110
3
1 1
2 1
4 1
After applying the 3 moves from output, the resultant matrix looks like this-
001
111
100
010
number of ones in rows are respectively 1, 3, 1, 1. All these values are odd.
number of ones in columns are respectively 2, 2, 2. All these values are even.
InputOutput
1 2
01
-1
There is no way to make this matrix beautiful.
Statistics
73% Solution Ratio
Zobayer_AbedinEarliest, 2M ago
nh_nayeemFastest, 0.0s
Zobayer_AbedinLightest, 131 kB
Zobayer_AbedinShortest, 1033B | 854 | 3,123 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 25, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-21 | longest | en | 0.925427 |
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# Call option/stock price/annual risk free rate
1. Suppose you believe that Sherwin Williams Co.'s stock price is going to increase from its current level of \$22.50 sometime during the next 5 months. For \$435.75 you can buy a 5-month call option giving you the right to buy 100 shares at a price of \$25 per share. If you buy this option for \$435.75 and Sherwin Williams Co.'s stock price actually rises to \$50, what would your pre-tax net profit be?
2. The current price of Samsung Electronics' stock is \$22, and at the end of one year its price will be either \$27 or \$17. The annual risk-free rate is 4.5%, based on daily compounding. A 1-year call option on the stock, with an exercise price of \$22, is available. Based on the binominal model, what is the option's value?
3. The current price of PPG Industries' stock is \$45, the annual risk-free rate is 5.5%, and a 1-year call option with a strike price of \$65 sells for \$7.00. What is the value of a put option, assuming the same strike price and expiration date as for the call option?
4. Suppose you believe that Renesas Electronics Corp's stock price is going to decline from its current level of \$82.50 sometime during the next 5 months. For \$463.63 you could buy a 5-month put option giving you the right to sell 100 shares at a price of \$85 per share. If you bought this option for \$463.00 and Renesas Electronics Corp's stock price actually dropped to \$48, what would your pre-tax net profit be?
5. An analyst wants to use the Black-Scholes model to value call options on the stock of Kia Motors when the price of the stock is \$45, the strike price of the option is \$30 and the option matures in 3 months (t = 0.25). The standard deviation of the stock's returns is 0.40 and the variance is 0.16 while the risk-free rate is 4.5%.
What is the value of the call option if, when given this information, the analyst then calculated the following necessary components of the Black-Scholes model and N(d1) and N(d2) represent areas under a standard normal distribution function:
• d1 = 1.613
• d2 = 1.413
• N(d1) = 0.94667
• N(d2) = 0.92123
#### Solution Summary
The call option/stock price/annual risk free rates are examined.
\$2.19 | 579 | 2,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-50 | longest | en | 0.925341 |
https://numbermatics.com/n/88209/ | 1,642,402,982,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00591.warc.gz | 455,323,595 | 6,292 | # 88209
## 88,209 is an odd composite number composed of two prime numbers multiplied together.
What does the number 88209 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 21 divisors.
88209 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of twenty-one divisors.
## Prime factorization of 88209:
### 36 × 112
(3 × 3 × 3 × 3 × 3 × 3 × 11 × 11)
See below for interesting mathematical facts about the number 88209 from the Numbermatics database.
### Names of 88209
• Cardinal: 88209 can be written as Eighty-eight thousand, two hundred nine.
### Scientific notation
• Scientific notation: 8.8209 × 104
### Factors of 88209
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 8
• Sum of prime factors: 14
### Divisors of 88209
• Number of divisors d(n): 21
• Complete list of divisors:
• Sum of all divisors σ(n): 145369
• Sum of proper divisors (its aliquot sum) s(n): 57160
• 88209 is a deficient number, because the sum of its proper divisors (57160) is less than itself. Its deficiency is 31049
### Bases of 88209
• Binary: 101011000100100012
• Base-36: 1W29
### Squares and roots of 88209
• 88209 squared (882092) is 7780827681
• 88209 cubed (882093) is 686339028913329
• 88209 is a perfect square number. Its square root is 297
• The cube root of 88209 is 44.5147869893
### Scales and comparisons
How big is 88209?
• 88,209 seconds is equal to 1 day, 30 minutes, 9 seconds.
• To count from 1 to 88,209 would take you about thirty minutes.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 88209 cubic inches would be around 3.7 feet tall.
### Recreational maths with 88209
• 88209 backwards is 90288
• 88209 is a Harshad number.
• The number of decimal digits it has is: 5
• The sum of 88209's digits is 27
• More coming soon!
MLA style:
"Number 88209 - Facts about the integer". Numbermatics.com. 2022. Web. 17 January 2022.
APA style:
Numbermatics. (2022). Number 88209 - Facts about the integer. Retrieved 17 January 2022, from https://numbermatics.com/n/88209/
Chicago style:
Numbermatics. 2022. "Number 88209 - Facts about the integer". https://numbermatics.com/n/88209/
The information we have on file for 88209 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 88209, math, Factors of 88209, curriculum, school, college, exams, university, Prime factorization of 88209, STEM, science, technology, engineering, physics, economics, calculator, eighty-eight thousand, two hundred nine.
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Some bits of this website may not work unless you switch it on. | 894 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | latest | en | 0.854714 |
https://sohcahtoa.org.uk/oldblog/maths/spreadsheets-to-talk-about/index.html | 1,713,724,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00307.warc.gz | 472,856,767 | 4,589 | Exploring Mathematics with Spreadsheets by Healy and Sutherland was published in 1991. It contains ideas for pair and group work in Maths using basic spreadsheet functions. There are photocopyable masters to give to students and some teaching notes. The two pupils on the front cover are using an early version of MS Excel on what looks like a Mac Classic. The logic was to use pair work as a way of encouraging talking about their thinking as they investigate the problem or situation that they have been allocated. A nostalgic feature of the book is the comparison of spreadsheet software in the teachers notes, does anyone else remember Grasshopper, Viewsheet and Multiplan?
An early activity is ‘find the formula’ where one student puts a simple formula like =B3+2 into a cell when the other student isn’t looking. Then the other student can change the number in B3 and try to work out the formula. More advanced activities include finding the dimensions of a rectangle that maximise the area for a given perimeter (including graphing the way the area depends on the length of a side) and explorations of sequences (arithmetic, geometric and Fibonacci style).
The punchline is that all the activities start with a blank spreadsheet. The students have to decide on the formulas, the layout and have to interpret the tables of numbers and to decide when a graph might be useful! It is this constructive use of the spreadsheet that keeps me using a few of these worksheets once or twice a year for half an hour or so. These activities work fine on OpenOffice Calc and on the Gnumeric spreadsheet, so you can use them on the Asus EeePC as well as the cheaper Elonex One. The Elonex looks as if it has AbiWord and Gnumeric as the ‘office’ components.
As you might have guessed, I’m having a bit of a clear out, and I came across an article from the June 1999 issue of Computer Shopper about using spreadsheets for technical applications. Again, the spreadsheets build up from basic formulas – in this case using the trig functions to simulate amplitude modulation and using a simple Euler method to solve a famous set of three coupled non-linear differential equations.
I built the Lorenz attractor spreadsheet in OpenOffice (using the 17 inch monitor) and reproduced the graphs shown in the article (having noticed the typo in the Lorenz equations). You can download my OpenOffice spreadsheet that integrates the Lorenz equations [500Kb, ODS format].
Moral: anything you develop for the humble spreadsheet program that uses basic functions and graphing is future proof, and students might find building the models informative. | 536 | 2,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.943746 |
https://coderanch.com/t/235511/certification/sin-cos-tan | 1,506,035,956,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687938.15/warc/CC-MAIN-20170921224617-20170922004617-00115.warc.gz | 651,867,658 | 7,052 | This week's book giveaway is in the JavaScript forum.We're giving away four copies of Cross-Platform Desktop Applications: Using Node, Electron, and NW.js and have Paul Jensen on-line!See this thread for details.
Win a copy of Cross-Platform Desktop Applications: Using Node, Electron, and NW.js this week in the JavaScript forum!
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# sin(), cos(), & tan()
Mike Cunningham
Ranch Hand
Posts: 130
In the Mike Meyer's Passport to Java it states:
The trig functions sin(), cos(), & tan(), all return a double value in radians, not degrees. You can use the method toDegrees() to convert the result.
When someone is looking for the value of something using one of these functions....are they looking for an answer in radians or degrees? Also, what's the benefit of being able to convert it to the other type of value?
- Mike
Jim Hall
Ranch Hand
Posts: 162
These functions do not return a radian value. They return a double value which represents the ratio of sides of a right triangle. The arguments to the methods must be in radians.
sin = o/h
cos = a/h
tan = o/a
where
o = length of opposite side
a = length of adjacent side
h = length of hypotenuse side
The methods asin(), acos(), atan() return a double value in radians that represents the corresponding angle. This number can then be converted into degrees using toDegrees().
Mike Cunningham
Ranch Hand
Posts: 130
So, if I was asked to give the cos() value for the following variable:
...which would be correct:
- Mike
Jane Griscti
Ranch Hand
Posts: 3141
Hi Mike,
Radians represent angular units of measurment based on the value of PI for measuring circles and arcs.
This physics presentation gives a bit of an explanation for their use.
Radians are commonly used in calculations with results being converted to degrees. Astronomers commonly convert between degrees and radians.
If you see an answer to a trig question that shows a result of 30.5 or 280.0 then you'll know that choice is incorrect; radians values won't exceed PI.
Hope that helps.
------------------
Jane Griscti
Sun Certified Programmer for the Java� 2 Platform
Co-author Mike Meyers' Java 2 Certification Passport
Jane Griscti
Ranch Hand
Posts: 3141
Doubt you'll see a question that would ask you to compute a cos() or sin() value except on a physics exam
------------------
Jane Griscti
Sun Certified Programmer for the Java� 2 Platform
Co-author Mike Meyers' Java 2 Certification Passport
Ranch Hand
Posts: 5040
Mike:
System.out.println(Math.sin(angle));
// = 1 this represents a ratio as Jim pointed out.
That doesn't mean that your second println stmt would fail.
Its just a magic with numbers. Your code doesn't deal with
units, its just us humans. So, your second stmt will take an input of 1 (radian) and convert it to degrees and output that
value which would be 57.??? (degrees).
regds.
- satya
ps:
Wasn't there a similar bug in some spaceship launch that
caused a mishap. Confusion between degrees and radians
or was it units...
With a little knowledge, a cast iron skillet is non-stick and lasts a lifetime. | 776 | 3,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-39 | latest | en | 0.840235 |
https://numberworld.info/11203112221 | 1,679,491,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00712.warc.gz | 503,463,269 | 3,878 | # Number 11203112221
### Properties of number 11203112221
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
29bc1ed1d
Base 32:
sin(11203112221)
-0.091780071793106
cos(11203112221)
0.9957793020653
tan(11203112221)
-0.09216908967956
ln(11203112221)
23.139457453522
lg(11203112221)
10.0493386863
sqrt(11203112221)
105844.75528339
Square(11203112221)
1.2550972343632E+20
### Number Look Up
Look Up
11203112221 which is pronounced (eleven billion two hundred three million one hundred twelve thousand two hundred twenty-one) is a very great number. The cross sum of 11203112221 is 16. If you factorisate 11203112221 you will get these result 7 * 1600444603. The number 11203112221 has 4 divisors ( 1, 7, 1600444603, 11203112221 ) whith a sum of 12803556832. The number 11203112221 is not a prime number. The number 11203112221 is not a fibonacci number. The number 11203112221 is not a Bell Number. The number 11203112221 is not a Catalan Number. The convertion of 11203112221 to base 2 (Binary) is 1010011011110000011110110100011101. The convertion of 11203112221 to base 3 (Ternary) is 1001220202122001211021. The convertion of 11203112221 to base 4 (Quaternary) is 22123300132310131. The convertion of 11203112221 to base 5 (Quintal) is 140420444042341. The convertion of 11203112221 to base 8 (Octal) is 123360366435. The convertion of 11203112221 to base 16 (Hexadecimal) is 29bc1ed1d. The convertion of 11203112221 to base 32 is ads3r8t. The sine of the number 11203112221 is -0.091780071793106. The cosine of the figure 11203112221 is 0.9957793020653. The tangent of the figure 11203112221 is -0.09216908967956. The root of 11203112221 is 105844.75528339.
If you square 11203112221 you will get the following result 1.2550972343632E+20. The natural logarithm of 11203112221 is 23.139457453522 and the decimal logarithm is 10.0493386863. You should now know that 11203112221 is impressive number! | 712 | 2,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.698428 |
https://www.coursehero.com/file/179457/electric20fields202/ | 1,519,503,803,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815934.81/warc/CC-MAIN-20180224191934-20180224211934-00296.warc.gz | 848,264,521 | 86,974 | {[ promptMessage ]}
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electric%20fields%202
# electric%20fields%202 - Electric field and its work on...
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1 Electric field and its work on charges We introduced the concept of electric field and this formula: F = q · E What can you associate force with? Motion or acceleration from F = m a . Work, from W = F · X (a review on vector dot product here). Example with a uniform electric field: E X Y A B C D The work to the field E does to move the charge from A to B is: W AB = F ·AB = q E · X = q EX, where X is the vector from A to B. From A to D: W AD = F ·AD = q E · Y = q EX = W AB , where Y is the vector from A to D. And this answer is true no matter the path is ABD, ACD or AD. So the electric force F = q E is conservative. (a review on conservative force). q X
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2 Electric Potential Electric potential of an electric field. Electric field exerts force on a charge inside it. This force moves the charge and does work to it. The work the electric force does not depend on the path, but only the start and end points of the charge the electric force is conservative. For a charge in a conservative force field, one can define a potential that
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Ask a homework question - tutors are online | 411 | 1,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-09 | latest | en | 0.894099 |
https://www.ncl.ac.uk/undergraduate/degrees/module/?code=MAS2706 | 1,610,752,078,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00341.warc.gz | 915,482,382 | 25,059 | # MAS2706 : Linear Algebra & Coding Theory
• Offered for Year: 2020/21
• Module Leader(s): Dr David Kimsey
• Lecturer: Professor Sarah Rees
• Owning School: Mathematics, Statistics and Physics
• Teaching Location: Newcastle City Campus
##### Semesters
Semester 1 Credit Value: 10 Semester 2 Credit Value: 10 ECTS Credits: 10
#### Aims
To provide students with an introduction to modern abstract linear algebra. Building on their existing knowledge of matrix methods, students will experience the benefits of an abstract and rigorous mathematical theory for the deeper understanding of a mathematical subject. To explain the necessity for error correcting codes, to establish their general properties, to show how to construct linear and cyclic codes and to gain practice in their use.
Module summary
Linear algebra is a fundamental subject that pervades many areas of modern mathematics. On the one hand it is often convenient to replace a complicated problem by a linear approximation which is easier to solve. On the other hand, linear algebra has beautiful applications in coding theory, projective geometry, and many other areas of mathematics and statistics.
Initially linear algebra aims to solve systems of linear equations. In the first year courses this led naturally to matrix algebra. In MAS2701 abstraction and generalisation are pushed one level further with the formal introduction of vector spaces and linear maps as a replacement for real n-dimensional space and matrices, respectively. This allows us to consider analogous problems in different settings simultaneously and eventually makes explanations easier and faster. We will need to introduce notions of dimension and basis in this general setting. A guiding question is how to transform matrices (or linear maps) to a simple form in which essential properties can be immediately read off.
Error-correcting codes are at the heart of the digital revolution. They are used to store music on CDs and video on DVDs; to send data across telecommunications networks; and to broadcast digital television. In practice, a digital signal may be degraded in transit by many factors - cosmic rays, fluctuations in power supplies, even (in the case of a CD) dust and scratches - so that some 0s are changed to 1s and vice versa; error-correcting codes are designed to rectify this. We work with words, binary strings of some standard length n. Certain words are designated as codewords, and the signal is converted to a sequence of codewords before transmission. At the receiving end, each word is examined as it arrives, and, if it turns out to be a non-codeword (indicating that the signal has been degraded), it is replaced by the nearest codeword. This explains why small imperfections on a CD do not affect the quality of the sound that you hear. We shall concentrate on a particularly nice class of codes called linear codes, a beautiful application of elementary linear algebra. Here errors can be corrected automatically by simple matrix operations. In particular, we shall investigate cyclic codes, linear codes based on polynomials.
#### Outline Of Syllabus
Vector spaces, span and bases, linear maps and their properties, eigenvectors and eigenvalues, inner product spaces, change of basis, diagonalisation.
General properties of codes. Perfect codes. Linear codes. Parity-check matrices and syndrome decoding. Hamming codes. Extensions of codes. Cyclic codes.
#### Teaching Methods
Please note that module leaders are reviewing the module teaching and assessment methods for Semester 2 modules, in light of the Covid-19 restrictions. There may also be a few further changes to Semester 1 modules. Final information will be available by the end of August 2020 in for Semester 1 modules and the end of October 2020 for Semester 2 modules.
##### Teaching Activities
Category Activity Number Length Student Hours Comment
Scheduled Learning And Teaching ActivitiesLecture91:009:00Synchronous On-Line Material
Structured Guided LearningLecture materials361:0036:00Non-Synchronous Activities
Guided Independent StudyAssessment preparation and completion301:0030:00N/A
Scheduled Learning And Teaching ActivitiesLecture91:009:00Present in Person
Structured Guided LearningStructured non-synchronous discussion181:0018:00N/A
Scheduled Learning And Teaching ActivitiesDrop-in/surgery41:004:00Office Hour or Discussion Board Activity
Guided Independent StudyIndependent study941:0094:00N/A
Total200:00
##### Teaching Rationale And Relationship
Non-synchronous online materials are used for the delivery of theory and explanation of methods, illustrated with examples, and for giving general feedback on assessed work. Present-in-person and synchronous online sessions are used to help develop the students’ abilities at applying the theory to solving problems and to identify and resolve specific queries raised by students, and to allow students to receive individual feedback on marked work. Students who cannot attend a present-in-person session will be provided with an alternative activity allowing them to access the learning outcomes of that session. In addition, office hours/discussion board activity will provide an opportunity for more direct contact between individual students and the lecturer: a typical student might spend a total of one or two hours over the course of the module, either individually or as part of a group.
Alternatives will be offered to students unable to be present-in-person due to the prevailing C-19 circumstances.
Student’s should consult their individual timetable for up-to-date delivery information.
#### Assessment Methods
Please note that module leaders are reviewing the module teaching and assessment methods for Semester 2 modules, in light of the Covid-19 restrictions. There may also be a few further changes to Semester 1 modules. Final information will be available by the end of August 2020 in for Semester 1 modules and the end of October 2020 for Semester 2 modules.
The format of resits will be determined by the Board of Examiners
##### Exams
Description Length Semester When Set Percentage Comment
Written Examination1202A80Alternative assessment - class test
##### Other Assessment
Description Semester When Set Percentage Comment
Written exercise1M8written exercises
Written exercise2M12written exercises
##### Assessment Rationale And Relationship
A substantial formal examination is appropriate for the assessment of the material in this module. The course assessments will will allow the students to develop their problem solving techniques, to practise the methods learnt in the module, to assess their progress and to receive feedback; these assessments have a secondary formative purpose as well as their primary summative purpose. | 1,320 | 6,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-04 | longest | en | 0.908193 |
https://www.jiskha.com/questions/742/hey-this-is-my-question-at-a-temperature-of-300-c-and-a-pressure-of-40-5-mpa-90-0-mol | 1,656,329,556,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00696.warc.gz | 904,210,541 | 5,808 | # chem.
Hey this is my question:
At a temperature of 300°C and a pressure of 40.5 MPa, 90.0 mol of H2(g) and
80.0 mol of N2(g) are injected into a reaction vessel. When equilibrium is
established, 37.0 mol of NH3(g) are present. The number of moles of H2(g) present
in this equilibrium mixture is
This is my equation so far. How do I go from here?
N2(g) + 3H2(g) >> 2 NH3(g)
initially: 80.0 mol 90.0 mol 0
at equilib:80.0 – x 90.0 – 3x 0 + 2x
0+2x=37.0 mol given
So solve for x and use for 90-3x = ??
ok, I got 20 as the H2 (g) present, would this be right?
Hardly.
If 2x = 37, then x = 37/2 =18.5
Then 90-3x=90-3(18.5)=90-55.5 = 34.5. Check my thinking. Check my math.
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. Set up your RICE or ICE table;
(R) 3H2(g) + N2(g) <-> 2NH3(g)
(I) 90 80 0
(C) 90-3x 80-x +2x
(E) 90-3x 80-x 20
Looking at NH3(g) you can calculate that x = 10, therefore 90-3(10)=60. 60M or (60mol/L) is the answer. I hope that formatting sticks and this makes sense.
1. 👍
2. 👎
3. ℹ️
4. 🚩
## Similar Questions
Still need help? You can ask a new question. | 448 | 1,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-27 | latest | en | 0.877702 |
http://mathoverflow.net/feeds/question/26389 | 1,371,615,786,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707187122/warc/CC-MAIN-20130516122627-00082-ip-10-60-113-184.ec2.internal.warc.gz | 163,658,583 | 3,485 | Eigenvectors of a certain big upper triangular matrix - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:23:07Z http://mathoverflow.net/feeds/question/26389 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix Eigenvectors of a certain big upper triangular matrix Michael Hardy 2010-05-29T22:16:28Z 2012-05-07T12:37:58Z <p>I'm looking at this matrix:</p> <p><code>$$\begin{pmatrix} 1 & 1/2 & 1/8 & 1/48 & 1/384 & \dots \\ 0 & 1/2 & 1/4 & 1/16 & 1/96 & \dots \\ 0 & 0 & 1/8 & 1/16 & 1/64 & \dots \\ 0 & 0 & 0 & 1/48 & 1/96 & \dots \\ 0 & 0 & 0 & 0 & 1/384 & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$</code></p> <p>The first row contains the reciprocals of the double factorials <code>$$2, \qquad 2 \cdot 4, \qquad 2 \cdot 4 \cdot 6, \qquad 2 \cdot 4 \cdot 6 \cdot 8, \qquad \dots$$</code> Each row is a <em>shift</em> of a <em>scalar multiple</em> of the first row, and the scalar multiple is in each case itself a reciprocal of a double factorial, so that the main diagonal is the same as the first row. A consequence is that each column is proportional to the corresponding row of Pascal's triangle. E.g. the last column shown is proportional to <code>$$1, 4, 6, 4, 1.$$</code> This matrix is the matrix of coefficients in the "inversion formulas" section of <a href="http://ncatlab.org/nlab/show/trigonometric+identities+and+the+irrationality+of+pi" rel="nofollow">this rant</a> that I wrote.</p> <p>I found the first three eigenvectors: <code>$$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 5 \\ -14 \\ 21 \\ 0 \\ 0 \\ \vdots \end{pmatrix}$$</code> Meni Rosenfeld pushed this through some software and found that up to the 40th eigenvalue, the signs of the components of the eigenvectors alternate.</p> <p>Can anything of interest be said about the eigenvectors?</p> <p>Can anything of interest be said about this matrix?</p> http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix/26431#26431 Answer by Wadim Zudilin for Eigenvectors of a certain big upper triangular matrix Wadim Zudilin 2010-05-30T07:59:30Z 2010-06-01T12:47:48Z <p>As far as I understand your construction, your matrix is $$\operatorname{diag}\biggl(1,\frac12,\frac18,\dots,\frac1{2^nn!},\dots\biggr) \cdot\exp\begin{pmatrix} 0 & \frac12 & 0 & 0 & 0 & \dots \cr 0 & 0 & \frac12 & 0 & 0 & \dots \cr 0 & 0 & 0 & \frac12 & 0 & \dots \cr 0 & 0 & 0 & 0 & \frac12 & \dots \cr \dots & \dots & \dots & \dots & \dots & \dots \end{pmatrix},$$ a diagonal matrix times the exponential of a nilpotent matrix. In your question you discuss some properties of truncations of your infinite matrix, finite $n\times n$ matrices. This corresponds to the truncations of the above diagonal and nilpotent matrices. I've never seen such matrices "in work" but this one could be a nice example of understanding the alteration property of entries of its eigenvectors. In view of the other response, this could be a good point of generalising the previous results in this area.</p> http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix/26436#26436 Answer by Hans Lundmark for Eigenvectors of a certain big upper triangular matrix Hans Lundmark 2010-05-30T10:16:12Z 2010-05-30T10:16:12Z <p>Your matrix is totally nonnegative (i.e., all minors are nonnegative). This is because it can be factorized as the matrix of binomial coefficients (which is totally nonnegative by the <a href="http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/" rel="nofollow">Karlin–McGregor–Lindström–Gessel–Viennot lemma</a>) times a diagonal matrix with positive entries $1/(2k)!!$ on the diagonal.</p> <p>For an oscillatory matrix (i.e., a totally nonnegative matrix such that some power of it is totally positive), there is a theorem by Gantmacher & Krein which says that the eigenvalues are real and simple, and the eigenvector corresponding to the $k$th largest eigenvalue has $k-1$ sign changes. (Theorem 5.3 in <a href="http://www.cambridge.org/us//catalogue/catalogue.asp?isbn=0521194083" rel="nofollow">Pinkus, <em>Totally positive matrices</em></a>.)</p> <p>Unfortunately that doesn't apply here, since a power of an upper triangular matrix is upper triangular, so that some minors (below the diagonal) are always zero; hence your matrix is not oscillatory. But perhaps it is possible to use similar ideas to prove the sign changes in your case?</p> http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix/96205#96205 Answer by Dan Fodor for Eigenvectors of a certain big upper triangular matrix Dan Fodor 2012-05-07T11:32:06Z 2012-05-07T12:37:58Z <p>Might be a wild intuition , I'd say the eigenvalues are the entries of the first row , and that the eigenvector coresponding to the $nth$ eigenvalue ,$k$ is made by adjoining a column of zeroes to the eigenvector coresponding to the eigenvector coresponding to the same eigenvalue for the first $n*n$ minor of the matrix . </p> <p>Example :</p> <p>for eigenvaue $1$ we take the matrix $\begin{pmatrix} 1 \end{pmatrix}$ ,the eigenvetor corresponding to $1$ is $\begin{pmatrix} 1 \end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr 0 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix}$ as the first eigevector . </p> <p>for eigenvaue $1/2$ we take the matrix $\begin{pmatrix} 1& 1/2 \cr 0 & 1/2\end{pmatrix}$ ,the eigenvetor corresponding to $1/2$ is $\begin{pmatrix} 1 \cr -1\end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr -1 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix}$ as the second eigevector . </p> <p>The eigenvector coresponding to $1/8$ for $\begin{pmatrix} 1& 1/2 & 1/8 \cr 0 & 1/2 & 1/4 \cr 0 & 0 & 1/8 \end{pmatrix}$ is $\begin{pmatrix} 5 \cr -14 \cr 21 \cr \end{pmatrix}$, you get the ideea .Also , the eigenvectors span the entire space , ie if a possibly infinite (but convergent) sum of eigenvectors is $\vec 0$ then the coefficients of those vectors are $0$ . </p> <p>Here is an explicit formula for the eigenvectors :first select $M_n$ , the $n*n$ truncation of the matrix and calculate $M_n - I*v_n$ , the nt'h eigenvalue . Example : for n=3 , we obtain \begin{pmatrix} 7/8 & 1/2 & 1/8 \cr 0 & 3/8 & 1/4 \cr 0 & 0 & 0 \end{pmatrix} . Now let $S$ be the $(n-1)*(n-1)$ truncation of that , ie \begin{pmatrix} 7/8 & 1/2 \cr 0 & 3/8 & \end{pmatrix} Calculate $S^{-1}$ = \begin{pmatrix} 8/7 & -32/21 \cr 0 & 8/3 & \end{pmatrix} , now multiply $S^{-1}$ with the truncation of the last column of $M_n$ , \begin{pmatrix} 1/8 \cr 1/4 \end{pmatrix} You obtain \begin{pmatrix} -5/21 \cr 2/3 \cr \end{pmatrix} . Concatenating $-1$ to that , you obtain \begin{pmatrix} -5/21 \cr 2/3 \cr -1 \end{pmatrix} , the third eigenvector ,or the nt'h eigenvector in the general case .</p> | 2,486 | 7,251 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2013-20 | latest | en | 0.597891 |
https://numberworld.info/1001202021 | 1,618,295,179,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00199.warc.gz | 544,148,507 | 3,941 | # Number 1001202021
### Properties of number 1001202021
Cross Sum:
Factorization:
3 * 3 * 631 * 176299
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
tqq8b5
sin(1001202021)
-0.89312356471748
cos(1001202021)
-0.44981140286384
tan(1001202021)
1.9855511866333
ln(1001202021)
20.724467136098
lg(1001202021)
9.0005217175925
sqrt(1001202021)
31641.776514602
Square(1001202021)
### Number Look Up
Look Up
1001202021 which is pronounced (one billion one million two hundred two thousand twenty-one) is a very unique figure. The cross sum of 1001202021 is 9. If you factorisate 1001202021 you will get these result 3 * 3 * 631 * 176299. 1001202021 has 12 divisors ( 1, 3, 9, 631, 1893, 5679, 176299, 528897, 1586691, 111244669, 333734007, 1001202021 ) whith a sum of 1448480800. 1001202021 is not a prime number. The number 1001202021 is not a fibonacci number. 1001202021 is not a Bell Number. 1001202021 is not a Catalan Number. The convertion of 1001202021 to base 2 (Binary) is 111011101011010010000101100101. The convertion of 1001202021 to base 3 (Ternary) is 2120202221022222100. The convertion of 1001202021 to base 4 (Quaternary) is 323223102011211. The convertion of 1001202021 to base 5 (Quintal) is 4022301431041. The convertion of 1001202021 to base 8 (Octal) is 7353220545. The convertion of 1001202021 to base 16 (Hexadecimal) is 3bad2165. The convertion of 1001202021 to base 32 is tqq8b5. The sine of the figure 1001202021 is -0.89312356471748. The cosine of 1001202021 is -0.44981140286384. The tangent of the figure 1001202021 is 1.9855511866333. The root of 1001202021 is 31641.776514602.
If you square 1001202021 you will get the following result 1002405486854484441. The natural logarithm of 1001202021 is 20.724467136098 and the decimal logarithm is 9.0005217175925. I hope that you now know that 1001202021 is special number! | 727 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.712052 |
https://flatearthsoc.com/flat-earth-memes-expensive-flat-earth-news-blog-2017.html | 1,563,213,368,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00202.warc.gz | 394,565,986 | 7,493 | As for flight paths and what Appears to be the silly way for a ball earth but makes sense for a flat earth, it reminds me of the child quiz. There is a spider in the corner of the room on the floor and he wants to get to the opp corner on the ceiling. Which is the quickest path? We instantly say, across the floor and up the wall join. BUT, if we flatten the room we then draw a straight line, we find the quickest path is diagonally up one wall and then diagonally across the ceiling, which Looks longer but is best.
#### The only explanation which has been given of this phenomenon is the refraction caused by the earth’s atmosphere. This, at first sight, is a plausible and fairly satisfactory solution; but on carefully examining the subject, it is found to be utterly inadequate; and those who have recourse to it cannot be aware that the refraction of an object and that of a shadow are in opposite directions. An object by refraction is bent upwards; but the shadow of any object is bent downwards, as will be seen by the following very simple experiment. Take a plain white shallow basin, and place it ten or twelve inches from a light in such a position that the shadow of the edge of the basin touches the centre of the bottom. Hold a rod vertically over and on the edge of the shadow, to denote its true position. Now let water be gradually poured into the basin, and the shadow will be seen to recede or shorten inwards and downwards; but if a rod or a spoon is allowed to rest, with its upper end towards the light, and the lower end in the bottom of the vessel, it will be seen, as the water is poured in, to bend upwards–thus proving that if refraction operated at all, it would do so by elevating the moon above its true position, and throwing the earth’s shadow downwards, or directly away from the moon’s surface. Hence it is clear that a lunar eclipse by a shadow of the earth is an utter impossibility.
164) Analysis of many interior videos from the “International Space Station,” have shown the use of camera-tricks such as green-screens, harnesses and even wildly permed hair to achieve a zero-gravity type effect. Footage of astronauts seemingly floating in the zero-gravity of their “space station” is indistinguishable from “vomit comet” Zero-G airplane footage. By flying parabolic maneuvers this Zero-G floating effect can be achieved over and over again then edited together. For longer uncut shots, NASA has been caught using simple wires and green screen technology.
```43.) The circumstances which attend bodies which are caused merely to fall from a great height prove nothing as to the motion or stability of the Earth, since the object, if it be on a thing that is in motion, will participate in that motion; but, if an object be thrown, upwards from a body at rest, and, again, from a body in motion, the circumstances attending its descent will be very different. In the former case, it will fall, if thrown vertically upwards, at the place from whence it was projected; in the latter case, it will fall behind the moving body from which it is thrown will leave it in the rear. Now, fix a gun, muzzle upwards, accurately, in the ground; fire off a projectile; and it will fall by the gun. If the Earth traveled eleven hundred miles a minute, the projectile would fall behind the gun, in the opposite direction to that of the supposed motion. Since, then, this is NOT the case, in fact, the Earth's fancied motion is negatived and we have a proof that the Earth is not a, globe.
```
97) NASA and modern astronomy say the Earth is a giant ball tilted back, wobbling and spinning 1,000 mph around its central axis, traveling 67,000 mph circles around the Sun, spiraling 500,000 mph around the Milky Way, while the entire galaxy rockets a ridiculous 670,000,000 mph through the Universe, with all of these motions originating from an alleged “Big Bang” cosmogenic explosion 14 billion years ago. That’s a grand total of 670,568,000 mph in several different directions we’re all supposedly speeding along at simultaneously, yet no one has ever seen, felt, heard, measured or proven a single one of these motions to exist whatsoever.
```If we move away from an elevated object on or over a plain or a prairie, the height of the object will apparently diminish as we do so. Now, that which is sufficient to produce this effect on a small scale is sufficient on a large one; and traveling away from an elevated object, no matter how far will cause the appearance in question - the lowering of the object. Our modern theoretical astronomers, however, in the case of the apparent lowering of the North Star as we travel southward, assert that it is evidence that the Earth is globular! But as it is clear that an appearance which is fully, accounted for on the basis of known facts cannot be permitted to figure as evidence in favor of that which is only a supposition, it follows that we rightfully order it to stand down, and make way for a proof that the Earth is not a globe.
```
```Another container ship made its way outward, as shown in Figure 11, a photograph taken through the supports of the pier at Virginia Beach. You can clearly read the name of the shipping company, Maersk Line, on the turquoise hull. What appears to be stains under the letters are the beginnings of an inferior mirage of the letters. Instead of a level of gray containers immediately above the hull, the layer of containers right above the hull on this ship appear a deep red. As with the other ship, in each succeeding photograph this ship is farther away, as evidenced by the decreasing apparent sizes of the containers and the ship.
```
The Ball Earth, heliocentric model is the foundation for Satan’s Babylon. He has brainwashed the world to believe in a Sun centered universe (Pagan Sun worship) that is the foundation for Evolution, the Big Bang, Atheism, the NWO, the alien deception, and many false religions. He is the father of lies (John 8:44)! The Flat Earth Truth reveals Satan’s deceptions and reveals that only God, the great Designer, could have created and designed such a perfect masterpiece as the Flat Earth.
166) The “geostationary communications satellite” was first created by Freemason science-fiction writer Arthur C. Clarke and supposedly became science-fact just a decade later. Before this, radio, television, and navigation systems like LORAN and DECCA were already well-established and worked fine using only ground-based technologies. Nowadays huge fibre-optics cables connect the internet across oceans, gigantic cell towers triangulate GPS signals, and ionospheric propagation allows radio waves to be bounced all without the aid of the science-fiction best-seller known as “satellites.”
The Flat Earth Society is a group actively promoting the Flat Earth Movement worldwide. Descending from Samuel Shenton's International Flat Earth Research Society, and the Universal Zetetic Society before it, we continue the age-old tradition of questioning the Round Earth doctrine and challenging authorities. Acknowledging the link between various unconventional beliefs, the Society also occasionally engages in other unconventional debates, striving to provide a voice for all free thinkers and Zeteticists.
When astronomers assert that it is "necessary" to make "allowance for curvature" in canal construction, it is, of course, in order that, in their idea, a level cutting may be had, for the water. How flagrantly, then, do they contradict themselves when the curved surface of the Earth is a "true level!" What more can they want for a canal than a true level? Since they contradict themselves in such an elementary point as this, it is an evidence that the whole thing is a delusion, and we have a proof that the Earth is not a globe.
88.) If we could – after our minds had once been opened to the light of Truth – conceive of a globular body on the surface of which human beings could exist, the power – no matter by what name it be called – that would hold them on would, then, necessarily, have to be so constraining and cogent that they could not live; the waters of the oceans would have to be as a solid mass, for motion would be impossible. But we not only exist, but live and move; and the water of the ocean skips and dances like a thing of life and beauty! This is a proof that the Earth is not a globe.
39) Practical distance measurements taken from “The Australian Handbook, Almanack, Shippers’ and Importers’ Directory” state that the straight line distance between Sydney and Nelson is 1550 statute miles. Their given difference in longitude is 22 degrees 2’14”. Therefore if 22 degrees 2’14” out of 360 is 1550 miles, the entirety would measure 25,182 miles. This is not only larger than the ball-Earth is said to be at the equator, but a whole 4262 miles greater than it would be at Sydney’s southern latitude on a globe of said proportions.
#### 3) The natural physics of water is to find and maintain its level. If Earth were a giant sphere tilted, wobbling and hurdling through infinite space then truly flat, consistently level surfaces would not exist here. But since Earth is in fact an extended flat plane, this fundamental physical property of fluids finding and remaining level is consistent with experience and common sense.
If we could - after our minds had once been opened to the light of Truth - conceive of a globular body on the surface of which human beings could exist, the power - no matter by what name it be called - that would hold them on would, then, necessarily, have to be so constraining and cogent that they could not live; the waters of the oceans would have to be as a solid mass, for motion would be impossible. But we not only exist, but live and move; and the water of the ocean skips and dances like a thing of life and beauty! This is a proof that the Earth is not a globe.
33.) If the Earth were a globe, people – except those on the top – would, certainly, have to be "fastened" to its surface by some means or other, whether by the "attraction" of astronomers or by some other undiscovered and undiscoverable process! But, as we know that we simply walk on its surface without any other aid than that which is necessary for locomotion on a plane, it follows that we have, herein, a conclusive proof that Earth is not a globe.
150) If Earth were a spinning ball it would be impossible to photograph star-trail time-lapses turning perfect circles around Polaris anywhere but the North Pole. At all other vantage points the stars would be seen to travel more or less horizontally across the observer’s horizon due to the alleged 1000mph motion beneath their feet. In reality, however, Polaris’s surrounding stars can always be photographed turning perfect circles around the central star all the way down to the Tropic of Capricorn.
26.) If the Earth were a globe, it would, if we take Valentia to be the place of departure, curvate downwards, in the 1665 miles across the Atlantic to Newfoundland, according to the astronomers' own tables, more than three hundred miles; but, as the surface of the Atlantic does not do so – the fact of its levelness having been clearly demonstrated by Telegraph Cable surveyors, – it follows that we have a grand proof that Earth is not a globe.
If we stand on the sands of the sea-shore and watch a ship approach us, we shall find that she will apparently "rise" - to the extent, of her own height, nothing more. If we stand upon an eminence, the same law operates still; and it is but the law of perspective, which causes objects, as they approach us, to appear to increase in size until we see them, close to us, the size they are in fact. That there is no other "rise" than the one spoken of is plain from the fact that, no matter how high we ascend above the level of the sea, the horizon rises on and still on as we rise, so that it is always on a level with the eye, though it be two-hundred miles away, as seen by Mr. J. Glaisher, of England, from Mr. Coxwell's balloon. So that a ship five miles away may be imagined to be "coming up" the imaginary downward curve of the Earth's surface, but if we merely ascend a hill such as Federal Hill, Baltimore, we may see twenty-!five miles away, on a level with the eye - that is, twenty miles level distance beyond the ship that we vainly imagined to be " rounding the curve," and "coming up!" This is a plain proof that the Earth is not a globe.
25.) The surveyor's plans in relation to the laying of the first Atlantic Telegraph cable, show that in 1665 miles – from Valentia, Ireland, to St . John's, Newfoundland – the surface of the Atlantic Ocean is a LEVEL surface – not the astronomers' "level," either! The authoritative drawings, published at the time, are a standing evidence of the fact, and form a practical proof that Earth is not a globe. | 2,787 | 12,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-30 | latest | en | 0.946593 |
https://stats.stackexchange.com/questions/273251/is-my-method-for-determining-any-sort-of-correlation-between-an-ordinal-variable/273293 | 1,585,722,125,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505366.8/warc/CC-MAIN-20200401034127-20200401064127-00255.warc.gz | 719,183,403 | 32,832 | Is my method for determining any sort of correlation between an ordinal variable and a continuous variable correct?
This data concerns an experiment my startup ran. We measured users' Interpupillary distance (the distance between their pupils), and then asked them to rate how satisfied they felt utilizing our Virtual Reality headsets.
Satisfaction was measured on a scale of 1 - 4 (1 = Very Dissatisfied, 4 = Very Satisfied). I had 50 samples.
So here I have one ordinal variable (Satisfaction), and one Continuous variable (Interpupillary distance). I decided to do a Spearman's rank correlation test: using that, I got a value of Spearman's Rank of -0.6935. This suggests a relatively strong negative correlation. However, when you graph my data, it makes no sense at all: here are graphed the ranks of each variable (x axis = IPD, y axis = satisfaction):
So, what exactly is going on? Am I using the wrong test? I am certain I did my calculations to arrive to my value of spearman's rank correctly.
Would greatly appreciate someone's help, thanks!
• What about the graph "makes no sense"? – Kodiologist Apr 12 '17 at 4:38
• How does the graph show a strong monotonic relationship? – Shak Apr 12 '17 at 4:53
• I wouldn't say it's strong exactly, but you can see that there are more 2s and fewer 4s on the y-axis as you go from left to right. I think as a general rule people tend to overestimate the effect of a correlation coefficient. If you generate some simulated data, it might help you realign your intuition. – Kodiologist Apr 12 '17 at 5:07
This isn't so much a problem with calculation as it is about sample size and knowing the limitations of correlation.
All correlation is attempting to do, is measure how close points are to an imaginary best fit line.
When you use Spearman's rank correlation, the line is drawn after ranking the data instead of the original scale.
If one of your variables have 0 variance, correlation will be undefined.
Consider the same variable, but introduce a single outlier. Suddenly it will show up as a measurable correlation, which is absurd based on your knowledge that it is simply a single fluctuation.
In your case, the satisfaction scores end up doing this. Even if by pure chance a few respondents rated you 4 or 2, it shows up as an apparent correlation.
The effect of this can only be minimized if your ordinal variable has maybe 10 levels instead of just 4. With a greater number of unique values and a larger sample this will then start approximating a continuous variable in which case correlation might actually tell you something useful.
You are probably better off running ANOVA in your case, but with only 50 samples I doubt that will lead to any reliable result either.
For ordinal correlations you should make both variables ordinal. In your data you could do this by rounding the IPD variable down to the next decade (0-9 --> 0, 10-19 --> 10, ...).
Looking at your data plot I see that - regardless of the IPD - all users are satisfied (= 3). I would not expect that the variables show a (a) significant and (b) meaningful correlation. | 700 | 3,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-16 | latest | en | 0.942873 |
https://oeis.org/A250497/internal | 1,716,121,600,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00215.warc.gz | 403,219,099 | 3,319 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
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A250497 Number of (n+2)X(1+2) 0..1 arrays with nondecreasing medians of every three consecutive values in every row and column 1
%I #4 Nov 24 2014 09:17:53
%S 512,2744,13824,68921,314432,1367631,5832000,24137569,97336000,
%T 385828352,1505060136,5783534875,21952000000,82426462208,306454670488,
%U 1129412837944,4129885013184,14994306423179,54087346107432
%N Number of (n+2)X(1+2) 0..1 arrays with nondecreasing medians of every three consecutive values in every row and column
%C Column 1 of A250504
%H R. H. Hardin, <a href="/A250497/b250497.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 8*a(n-1) -20*a(n-2) +44*a(n-3) -190*a(n-4) +292*a(n-5) -242*a(n-6) +1552*a(n-7) -933*a(n-8) -728*a(n-9) -7846*a(n-10) -2828*a(n-11) +2191*a(n-12) +22952*a(n-13) +13106*a(n-14) -556*a(n-15) -37208*a(n-16) -18168*a(n-17) +6972*a(n-18) +33492*a(n-19) -4164*a(n-20) -5916*a(n-21) -10326*a(n-22) +9744*a(n-23) -1676*a(n-24) +5224*a(n-25) -6310*a(n-26) +2776*a(n-27) -2345*a(n-28) +2396*a(n-29) -1876*a(n-30) +1244*a(n-31) -1086*a(n-32) +704*a(n-33) -410*a(n-34) +212*a(n-35) -109*a(n-36) +40*a(n-37) -14*a(n-38) +4*a(n-39) -a(n-40)
%e Some solutions for n=4
%e ..1..0..1....1..0..0....1..1..0....0..0..1....1..0..0....1..0..1....0..0..0
%e ..0..1..0....0..0..0....0..1..0....0..0..1....0..0..1....0..1..1....1..1..0
%e ..0..1..0....1..1..0....0..0..1....0..1..1....0..0..0....0..0..1....1..0..1
%e ..0..1..0....1..1..1....1..1..0....1..0..1....0..0..0....0..0..0....1..1..1
%e ..1..0..0....0..1..0....0..1..0....0..0..1....0..1..1....0..1..1....0..1..0
%e ..0..1..0....1..0..1....0..1..0....1..1..1....1..1..1....0..0..1....1..0..1
%K nonn
%O 1,1
%A _R. H. Hardin_, Nov 24 2014
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Last modified May 19 08:13 EDT 2024. Contains 372666 sequences. (Running on oeis4.) | 1,005 | 2,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.464552 |
https://calculatorguru.net/converter/microinch-to-rod-us/ | 1,719,249,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00692.warc.gz | 125,065,168 | 33,161 | Microinch to Rod (US) converter | μin to rd conversion
# Microinch to Rod (US) converter| μin to rd conversion
Are you struggling with converting Microinch to Rod (US)? Don’t worry! Our online “Microinch to Rod (US) Converter” is here to simplify the conversion process for you.
Here’s how it works: simply input the value in Microinch. The converter instantly gives you the value in Rod (US). No more manual calculations or headaches – it’s all about smooth and effortless conversions!
Think of this Microinch (μin ) to Rod (US) (rd) converter as your best friend who helps you to do the conversion between these length units. Say goodbye to calculating manually over how many Rod (US) are in a certain number of Microinch – this converter does it all for you automatically!
## What are Microinch and Rod (US)?
In simple words, Microinch and Rod (US) are units of length used to measure the size or distance of something. It helps us understand the length of objects, spaces, or dimensions. The short form of Microinch is “μin” and the short form for Rod (US) is “rd”
In everyday life, we use length units to express the size of anything in various contexts, such as measuring furniture, determining the length of a room, or specifying the dimensions of an object. Microinch and Rod (US) are also two common units of length.
## How to convert from Microinch to Rod (US)?
If you want to convert between these two units, you can do it manually too. To convert from Microinch to Rod (US) just use the given formula:
rd = Value in μin * 5.050494949E-9
here are some examples of conversion,
• 2 μin = 2 * 5.050494949E-9 rd = 0.000000010100989898 rd
• 5 μin = 5 * 5.050494949E-9 rd = 0.000000025252474745 rd
• 10 μin = 10 * 5.050494949E-9 rd = 0.00000005050494949 rd
### Microinch to Rod (US) converter: conclusion
Here we have learn what are the length units Microinch (μin ) and Rod (US) (rd)? How to convert from Microinch to Rod (US) manually and also we have created an online tool for conversion between these units.
Microinch to Rod (US) converter” or simply μin to rd converter is a valuable tool for simplifying length unit conversions. By using this tool you don’t have to do manual calculations for conversion which saves you time. | 571 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.887406 |
https://dir.md/wiki/Logical_disjunction?host=en.wikipedia.org | 1,652,841,680,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00786.warc.gz | 270,795,755 | 4,629 | # Logical disjunction
Classical disjunction is a truth functional operation which returns the truth value "true" unless both of its arguments are "false". Its semantic entry is standardly given as follows:[4]
In systems where logical disjunction is not a primitive, it may be defined as[5]
Operators corresponding to logical disjunction exist in most programming languages.
The `or` operator can be used to set bits in a bit field to 1, by `or`-ing the field with a constant field with the relevant bits set to 1. For example, `x = x | 0b00000001` will force the final bit to 1, while leaving other bits unchanged.[citation needed]
Many languages distinguish between bitwise and logical disjunction by providing two distinct operators; in languages following C, bitwise disjunction is performed with the single pipe operator (`|`), and logical disjunction with the double pipe (`||`) operator.
Logical disjunction is usually short-circuited; that is, if the first (left) operand evaluates to `true`, then the second (right) operand is not evaluated. The logical disjunction operator thus usually constitutes a sequence point.
In a parallel (concurrent) language, it is possible to short-circuit both sides: they are evaluated in parallel, and if one terminates with value true, the other is interrupted. This operator is thus called the parallel or.
Although the type of a logical disjunction expression is boolean in most languages (and thus can only have the value `true` or `false`), in some languages (such as Python and JavaScript), the logical disjunction operator returns one of its operands: the first operand if it evaluates to a true value, and the second operand otherwise.[citation needed]
The Curry–Howard correspondence relates a constructivist form of disjunction to tagged union types.[citation needed]
This inference has sometimes been understood as an entailment, for instance by Alfred Tarski, who suggested that natural language disjunction is ambiguous between a classical and a nonclassical interpretation. More recent work in pragmatics has shown that this inference can be derived as a conversational implicature on the basis of a semantic denotation which behaves classically. However, disjunctive constructions including Hungarian vagy... vagy and French soit... soit have been argued to be inherently exclusive, rendering ungrammaticality in contexts where an inclusive reading would otherwise be forced.[1]
Similar deviations from classical logic have been noted in cases such as free choice disjunction and simplification of disjunctive antecedents, where certain modal operators trigger a conjunction-like interpretation of disjunction. As with exclusivity, these inferences have been analyzed both as implicatures and as entailments arising from a nonclassical interpretation of disjunction.[1]
In many languages, disjunctive expressions play a role in question formation. For instance, while the following English example can be interpreted as a polar question asking whether it's true that Mary is either a philosopher or a linguist, it can also be interpreted as an alternative question asking which of the two professions is hers. The role of disjunction in these cases has been analyzed using nonclassical logics such as alternative semantics and inquisitive semantics, which have also been adopted to explain the free choice and simplification inferences.[1]
In English, as in many other languages, disjunction is expressed by a coordinating conjunction. Other languages express disjunctive meanings in a variety of ways, though it is unknown whether disjunction itself is a linguistic universal. In many languages such as Dyirbal and Maricopa, disjunction is marked using a verb suffix. For instance, in the Maricopa example below, disjunction is marked by the suffix šaa.[1] | 771 | 3,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.89035 |
https://www.intechopen.com/books/recent-studies-in-perturbation-theory/green-function | 1,560,989,065,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999066.12/warc/CC-MAIN-20190619224436-20190620010436-00187.warc.gz | 773,494,726 | 143,038 | Open access peer-reviewed chapter
# Green Function
By Jing Huang
Submitted: October 18th 2016Reviewed: February 21st 2017Published: June 14th 2017
DOI: 10.5772/68028
## Abstract
Both the scalar Green function and the dyadic Green function of an electromagnetic field and the transform from the scalar to dyadic Green function are introduced. The Green function of a transmission line and the propagators are also presented in this chapter.
### Keywords
• Green function
• boundary condition
• scatter
• propagator
• convergence
## 1. Introduction
In 1828, Green introduced a function, which he called a potential, for calculating the distribution of a charge on a surface bounding a region in Rn in the presence of external electromagnetic forces. The Green function has been an interesting topic in modern physics and engineering, especially for the electromagnetic theory in various source distributions (charge, current, and magnetic current), various construct conductors, and dielectric. Even though most problems can be solved without the use of Green functions, the symbolic simplicity with which they could be used to express relationships makes the formulations of many problems simpler and more compact. Moreover, it is easier to conceptualize many problems; especially the dyadic Green function is generalized to layered media of planar, cylindrical, and spherical configurations.
## 2. Definition of Green function
### 2.1. Mathematics definition
For the linear operator, there are: L^x=f(t), t > 0;
x(t)|t=0=y0;x(n)(t)|t=0=ynE1
Rewriting Eq. (1) as:
L^x=f(t)δ(tt)dtE2
Defining the Green function as:
L^G(t,t)=δ(tt)E3
So, the solution of Eq. (1) is:
x(t)=f(t)G(t,t)dtE4
We give several types of Green functions [1]
L^=−(d2dt2+2γddt+ω02) G(t,t′)=12π∫−∞+∞exp[−i(t−t′)k]k2+2iγk−ω02dk L^=−[f0(t)d2dt2+f1(t)ddt+f2(t)) G(t,t′)=-Ψ1(t)Ψ2(t′)−Ψ2(t)Ψ1(t′)f0(t′)[Ψ1(t′)Ψ˙2(t′)−Ψ˙1(t′)Ψ2(t′)] L^=−ddt[(1−t2)ddt] G(t,t′)=12+∑n=1∞1n(n+1)⋅2n+12Pn(t)Pn(t′)
## 3. The scalar Green function
### 3.1. The scalar Green function of an electromagnetic field
The Green function of a wave equation is the solution of the wave equation for a point source [2]. And when the solution to the wave equation due to a point source is known, the solution due to a general source can be obtained by the principle of linear superposition (see Figure 1).
This is merely a result of the linearity of the wave equation, and that a general source is just a linear superposition of point sources. For example, to obtain the solution to the scalar wave equation in V in Figure 1
(2+k2)ϕ(r)=s(r)E5
we first seek the Green function in the same V, which is the solution to the following equation:
(2+k2)g(r,r')=δ(r-r')E6
Given g (r, r′), φ(r) can be found easily from the principle of linear superposition, since g (r, r′) is the solution to Eq. (5) with a point source on the right-hand side. To see this more clearly, note that an arbitrary source s(r) is just
s(r)=dr's(r')δ(rr')E7
which is actually a linear superposition of point sources in mathematical terms. Consequently, the solution to Eq. (5) is just
ϕ(r)=Vdr'g(r,r')s(r')E8
which is an integral linear superposition of the solution of Eq. (6). Moreover, it can be seen that g(r, r′) ≡ g(r′, r,) from reciprocity irrespective of the shape of V.
To find the solution of Eq. (6) for an unbounded, homogeneous medium, one solves it in spherical coordinates with the origin at r'. By so doing, Eq. (6) becomes
(2+k2)g(r)=δ(x)δ(y)δ(z)E9
But due to the spherical symmetry of a point source, g(r) must also be spherically symmetric. Then, for r ≠ 0, adopt the proper coordinate origin (the vector r is replaced by the scalar r), the homogeneous, spherically symmetric solution to Eq. (9) is given by
g(r)=c1eikrr+c2eikrrE10
Since sources are absent at infinity, physical grounds then imply that only an outgoing solution can exist; hence,
g(r)=ceikrrE11
The constant c is found by matching the singularities at the origin on both sides of Eq. (9). To do this, we substitute Eq. (11) into Eq. (9) and integrate Eq. (9) over a small volume about the origin to yield
ΔVdVceikrr+ΔVdVk2ceikrr=1E12
Note that the second integral vanishes when ∆V → 0 because dV = 4πr2dr. Moreover, the first integral in Eq. (12) can be converted into a surface integral using Gauss theorem to obtain
limr04πr2ddrceikrr=1E13
or c = 1/(4π).
The solution to Eq. (6) must depend only on rr′. Therefore, in general,
g(r,r')=g(rr')=eik(rr')4π(rr')E14
implying that g(r, r') is translationally invariant for unbounded, homogeneous media. Consequently, the solution to Eq. (5), from Eq. (9), is then
ϕ(r)=Vdr'eik(rr')4π(rr')s(r')E15
Once ϕ(r) and n^ϕ(r)are known on S, then ϕ(r′) away from S could be found
ϕ(r)=SdSn^[g(r,r)ϕ(r)ϕ(r)g(r,r)]E16
### 3.2. The scalar Green functions of one-dimensional transmission lines
We consider a transmission line excited by a distributed current source, K(x), as sketched in Figure 2. The line may be finite or infinite, and it may be terminated at either end with impedance or by another line [3]. For a harmonically oscillating current source K(x), the voltage and the current on the line satisfy the following pair of equations:
dV(x)dx=iωLI(x)E17
dI(x)dx=iωCV(x)+K(x)E18
L and C denote, respectively, the distributed inductance and capacitance of the line.
By eliminating I(x) between Eq. (17) and Eq. (18), there is
d2V(x)dx2+k2V(x)=iωLK(x)E19
where k=ωLCdenotes the propagation constant of the line. Eq. (19) has been designated as an inhomogeneous one-dimensional scalar wave equation.
The Green function pertaining to a one-dimensional scalar wave equation of the form of Eq. (19), denoted by g(x, x′), is a solution of the Eq. (9). The solution for g(x, x′) is not completely determined unless there are two boundary conditions which the function must satisfy at the extremities of the spatial domain in which the function is defined. The boundary conditions which must be satisfied by g(x, x′) are the same as those dictated by the original function which we intend to determine, namely, V(x) in the present case. For this reason, the Green functions are classified according to the boundary conditions, which they must obey. Some of the typical ones (for the transmission line) are illustrated in Figure 3.
In general, the subscript 0 designates infinite domain so that we have outgoing waves at x±, often called the radiation condition. Subscript 1 means that one of the boundary conditions satisfies the so-called Dirichlet condition, while the other satisfies the radiation condition. When one of the boundary conditions satisfies the so-called Neumann condition, we use subscript 2. Subscript 3 is reserved for the mixed type. Actually, we should have used a double subscript for two distinct boundary conditions. For example, case (b) of Figure 3 should be denoted by g01, indicating that one radiation condition and one Dirichlet condition are involved. With such an understanding, the simplified notation should be acceptable.
In case (d), a superscript becomes necessary because we have two sets of line voltage and current (V1, I1) and (V2, I2) in this problem, and the Green function also has different forms in the two regions. The first superscript denotes the region where this function is defined, and the second superscript denotes the region where the source is located.
Let the domain of x corresponds to (x1, x2). The function g(x, x′) in Eq. (9) can represent any of the three types, g0, g1, and g2, illustrated in Figures 3a–c, respectively. The treatment of case (d) is slightly different, and it will be formulated later.
(a) By multiplying Eq. (19) by g(x, x′) and Eq. (9) by V(x) and taking the difference of the two resultant equations, we obtain
x1x2[V(x)d2g0(x,x)dx2g0(x,x)d2V(x,x)dx2]dx=x1x2V(x)δ(xx)dxiωLx1x2K(x)g0(x,x)dxE20
The first term at the right-hand side of the above equation is simply V(xl), and the term at the left-hand side can be simplified by integration by parts, which gives
V(x)=iωLx1x2g0(x,x)K(x)dxE21
If we use the unprimed variable x to denote the position of a field point, as usually is the case, Eq. (21) can be changed to [4]
V(x)=iωLx1x2g(x,x)K(x)dx=iωLx1x2g0(x,x)K(x)dxE22
The last identity is due to the symmetrical property of the Green function. The shifting of the primed and unprimed variables is often practiced in our work. For this reason, it is important to point out that g(x′, x), by definition, satisfies the Eq. (9).
The general solutions for Eq. (9) in the two regions (see Figure 3a) are
g0(x,x)={i/(2k)eik(xx),xxi/(2k)eik(xx),xxE23
The choice of the above functions is done with the proper satisfaction of boundary conditions at infinity. At x = x', the function must be continuous, and its derivative is discontinuous.
They are: [g0(x,x)]x0x+0=0, and [dg0(x,x)dx]x0x+0=-1
The physical interpretation of these two conditions is that the voltage at x' is continuous, but the difference of the line currents at x' must be equal to the source current.
(b) The choice of this type of function is done with the proper satisfaction of boundary conditions. At x = x', the function must be continuous, its derivative is discontinuous, and a Dirichlet condition is satisfied at x = 0.
g1(x,x)={ι/(2κ)[eικ(xx)eικ(x+x)],xxι/(2κ)[eικ(xx)eικ(x+x)],0xxE24
In view of Eq. (24), it can be interpreted as consisting of an incident and a scattered wave; that is
g1(x,x)=g0(x,x)+g1s(x,x)E25
where g1s(x,x)=i2keik(x+x).
Such a notion is not only physically useful, but mathematically it offers a shortcut to finding a composite Green function. It is called as the shortcut method or the method of scattering superposition.
(c) Similarly, the method of scattering superposition suggests that we can start with
g2(x,x)=g0(x,x)+AeikxE26
To satisfy the Neumann condition at x = 0, we require
[dg0(x,x)dx+ikAeikx]x=0=0E27
Hence
A=i2keikxE28
g2(x,x)=i/(2k){eik(xx)+eik(x+x),xxeik(xx)+eik(x+x),0xxE29
(d) In this case, we have two differential equations to start with
d2V1(x)dx2+k12V1(x)=iωL1K1(x),x0E30
d2V2(x)dx2+k22V2(x)=0,x0E31
It is assumed that the current source is located in region 1 (see Figure 3d). We introduce two Green functions of the third kind, denoted by g(11) (x, x') and g(21) (x, x'). g(21), the first number of the superscript corresponds to the region where the function is defined. The second number corresponds to the region where the source is located; then
d2g(11)(x,x)dx2+k12g(11)(x,x)=δ(xx),x0E32
d2g(21)(x,x)dx2+k22g(21)(x,x)=0,x0E33
At the junction corresponding to x = 0, g(11) and g(21) satisfy the boundary condition that
g(11)(x,x)x=0=g(21)(x,x)x=0E34
1L1dg(11)(x,x)dxx=0=1L2dg(21)(x,x)dxx=0E35
The last condition corresponds to the physical requirement that the current at the junction must be continuous. Again, by means of the method of scattering superposition, there are
g(11)(x,x)=g0(x,x)+gs(11)(x,x)=i2k1{eik1(xx)+Reik1(xx),xxeik1(xx)+Reik1(x+x),0xxE36
g(21)(x,x)=i2k1Tei(k2xk1x),x0E37
The characteristic impedance of the lines, respectively, is
z1=(L1C1)1/2,z2=(L2C2)1/2E38
By the boundary condition, there are
R=z2z1z2+z1,T=2z2z2+z1E39
Example: Green function solution of nonlinear Schrodinger equation in the time domain [5].
The nonlinear Schrodinger equation including nonresonant and resonant nonlinear items is:
Az+i2β22At216β33At3=a2A+i3k08nAeffχNR(3)|A|2A+ik0g(ω0)[1if(ω0)]2nAeffAtχR(3)(tτ)|A(τ)|2dτE40
Where A is the field, β2 and β3 are the second and third order dispersion, respectively. A(z) is the fiber absorption profile. k0=ω0/c, ω0 is the center frequency. Aeff is the effective core area. n is the refractive index.
f(ω1+ω2+ω3)=2(ω1+ω2+ω3)(1|Γ|)2(ω1+ω2+ω3)22|Γ|+|Γ|2E41
g(ω1+ω2+ω3)=[2(ω1+ω2+ω3)22|Γ|+|Γ|2]E42
where g(ω1 + ω2 + ω3) is the Raman gain and f(ω1 + ω2 + ω3) is the Raman nongain coefficient. Г is the attenuation coefficient.
The original nonlinear part is divided into the nonresonant and resonant susceptibility items χNR(3)and χR(3). The solution has the form:
A(z,t)=ϕ(t)eiEzE43
Then, there is:
12β22φt2+i6β33φt33k08nAeffχNR(3)|φ|2φk0g(ωs)[1if(ωs)]2nAeffφ+χN(3)(tτ)|φ(τ)|dτ=EφE44
Let:
H^0(t)=12β22t2+i6β33t3E45
V^(t)=3k08nAeffχNR(3)|φ|k0g(ωs)[1if(ωs)]2nAeff+χR(3)(tτ)|φ(τ)|2dτE46
and taking the operator V^(t)as a perturbation item, the eigenequation n=2kinn!βnnφTn=Eφis
12β22φT2+i6β33φT3=EφE47
Assuming E = 1, we get the corresponding characteristic equation:
12β2r2+β36r3=EE48
Its characteristic roots are r1,r2,r3. The solution can be represented as:
φ=c1φ1+c2φ2+c3φ3E49
where ϕm=exp(irmt),m=1,2,3, and c1,c2,c3 are determined by the initial pulse. The Green function of Eq. (47) is:
(EH^0(t))G0(t,t)=δ(tt)E50
Constructing the Green function as:
G0(t,t)={a1φ1+a2φ2+a3φ3,t>tb1φ1+b2φ2+b3φ3,t<tE51
At the point t = t′, there are:
a1φ1(t)+a2φ2(t)+a3φ3(t)=b1φ1(t)+b2φ2(t)+b3φ3(t)E52
a1φ1(t)+a2φ2(t)+a3φ3(t)=b1φ1(t)+b2φ2(t)+b3φ3(t)E53
a1φ1(t)+a2φ2(t)+a3φ3(t)b1φ1(t)b2φ2(t)b3φ3(t)=6i/β3E54
It is reasonable to let b1 = b2 = b3 = 0, then:
a1=φ2φ˙3φ˙2φ3W(t),a2=φ3φ˙1φ˙3φ1W(t),a3=φ1φ˙2φ˙1φ2W(t)E55
W(t)=|φ1φ2φ3φ1(1)φ2(1)φ3(1)φ1(2)φ2(2)φ3(2)|E56
Finally, the solution of Eq. (44) can be written with the eigenfunction and Green function:
φ(t)=φ(t)+G0(t,t)V(t)φ(t)dt=ϕ(t)+G0(t,t,E)V(t)ϕ(t)dt+dtG0(t,t,E)V(t)G0(t,t,E)V(t)φ(t)dt=ϕ(t)+G0(t,t,E)V(t)ϕ(t)dt+dtG0(t,t,E)V(t)G0(t,t,E)V(t)ϕ(t)dt++dtG0(t,t)V(t)G0(t,t)V(t)dttimes lG0(tl,tl+1)V(tl+1)φ(tl+1)dtl+1E57
The accuracy can be estimated by the last term of Eq. (57).
## 4. The dyadic Green function
### 4.1. The dyadic Green function for the electromagnetic field in a homogeneous isotropic medium
The Green function for the scalar wave equation could be used to find the dyadic Green function for the vector wave equation in a homogeneous, isotropic medium [3]. First, notice that the vector wave equation in a homogeneous, isotropic medium is
××E(r)k2E(r)=iωμJ(r)E58
Then, by using the fact that ××E(r)=-2E+Eand that E=ρ/ε=J/iωε, which follows from the continuity equation, we can rewrite Eq. (58) as
2E(r)k2E(r)=iωμ[I^+k2]J(r)E59
where I^is an identity operator. In Cartesian coordinates, there are actually three scalar wave equations embedded in the above vector equation, each of which can be solved easily in the manner of Eq. (4). Consequently,
E(r)=iωμVdr'g(r'r)[I^+''k2]J(r)E60
where g(r′r)is the unbounded medium scalar Green function. Moreover, by using the vector identities gf=fg+gfand gF=gF+(g)F, it can be shown that
Vdr'g(r'r)'f(r')=Vdr''g(r'r)f(r')E61
and
Vdr'['g(r'r)]'J(r')=Vdr'J(r')''g(r'r)E62
Hence, Eq. (60) can be rewritten as
E(r)=iωμVdr'J(r')[I^+''k2]g(r'r)E63
It can also be derived using scalar and vector potentials.
Alternatively, Eq. (63) can be written as
E(r)=iωμVdr'J(r')G^e(r',r)E64
where
G^e(r)=[I¯+''k2]g(r'r)E65
is a dyad known as the dyadic Green function for the electric field in an unbounded, homogeneous medium. (A dyad is a 3 × 3 matrix that transforms a vector to a vector. It is also a second rank tensor). Even though Eq. (64) is established for an unbounded, homogeneous medium, such a general relationship also exists in a bounded, homogeneous medium. It could easily be shown from reciprocity that
J1(r),G^e(r,r'),J2(r')=J2(r),G^e(r,r'),J1(r')=J1(r),G^et(r,r'),J2(r')E66
where
Ji(r),G^e(r,r'),Jj(r')=VVdr'drJi(r')G^e(r',r)Jj(r)E66a
is the relation between Ji and the electric field produced by Jj. Notice that the above equation implies [6]
G^et(r',r)=G^e(r,r')E66b
Then, by taking transpose of Eq. (66b), Eq. (64) becomes
E(r)=iωμVdr'G^e(r,r')J(r')E67
Alternatively, the dyadic Green function for an unbounded, homogeneous medium can also be written as
G^e(r,r')=1k2[××I^g(rr')I^δ(rr')]E68
By substituting Eq. (67) back into Eq. (58) and writing
J(r)=dr'I^δ(rr')J(r')E69
we can show quite easily that
××G^e(r,r')-k2G^e(r,r')=I^δ(rr')E70
Equation (64) or (67), due to the ∇∇ operator inside the integration operating on g(r′r), has a singularity of 1/|r′r|3 when r′r. Consequently, it has to be redefined in this case for it does not converge uniformly, specifically, when r is also in the source region occupied by J(r). Hence, at this point, the evaluation of Eq. (67) in a source region is undefined.
And as the vector analog of Eq. (16)
E(r)=SdS[n×E(r)×G^e(r,r)+iωμn×H(r)G^e(r,r)]E71
### 4.2. The boundary condition
The dyadic Green function is introduced mainly to formulate various canonical electromagnetic problems in a systematic manner to avoid treatments of many special cases which can be treated as one general problem [3, 7, 8]. Some typical problems are illustrated in Figure 4 where (a) shows a current source in the presence of a conducting sphere located in air, (b) shows a conducting cylinder with an aperture which is excited by some source inside the cylinder, (c) shows a rectangular waveguide with a current source placed inside the guide, and (d) shows two semi-infinite isotropic media in contact, such as air and “flat” earth with a current source placed in one of the regions.
Unless specified otherwise, we assume that for problems involving only one medium such as (a), (b), and (c) the medium is air, then the wave number k is equal to ω(μ0ε0)1/2=2π/λ. The electromagnetic fields in these cases are solutions of the wave Eq. (62) and
××H(r)k2H(r)=×J(r)E72
The fields must satisfy the boundary conditions required by these problems.
In general, using the notations G^eand G^mto denote, respectively, the electric and the magnetic dyadic Green functions; they are solutions of the dyadic differential equations
××G^e(r,r')-k2G^e(r,r')=I^δ(rr')E73
××G^m(r,r')-k2G^m(r,r')=×[I^δ(rr')]E74
is the same as Eq. (70), and there is
G^m=×G^eE75
(a) and (b): Electric dyadic Green function (the first kind, using the subscript 1 denotes G^e1,G^m1, and the subscript “0” represents the free-space condition that the environment does not have any scattering object) is required to satisfy the dyadic Dirichlet condition on Sd, namely,
n×G^e1=0,n×G^m1=0E76
So, for (a)
E(r)=drJ(r)G^e(r,r)E77
and for (b)
(c) the electric dyadic Green function is required to satisfy the dyadic boundary condition on Sd, namely,
n××G^e2=0n××G^m2=0E79
H(r)=drJ(r)×G^e(r,r)E80
(d) For problems involving two isotropic media such as the configuration shown in Figure 4d, there are two sets of fields [9]. The wave numbers in these two regions are denoted by k1=ω(μ1ε1)1/2and k2=ω(μ2ε2)1/2. There are four functions for the dyadic Green function of the electric type and another four functions for the magnetic type, denoted, respectively, by G^e11G^e12G^e21and G^e22, and G^m11G^m12G^m21and G^m22. The superscript notation in G^e11means that both the field point and the source point are located in region 1. For G^e21, it means that the field point is located in region 1 and the source point is located in region 2. A current source is located in region 1 only, and the two sets of wave equations are
××E1(r)k2E1(r)=iωμ1J1(r)E81
××H1(r)k2H1(r)=×J1(r)E82
and
××E2(r)k2E2(r)=0E83
××H2(r)k2H2(r)=0E84
There are
××G^e11(r,r')-k12G^e11(r,r')=I^δ(rr')E85
××G^e21(r,r')-k22G^e21(r,r')=0E86
At the interface, the electromagnetic field and the corresponding dyadic Green function satisfy the following boundary conditions
n×[G^e11G^e21]=0E87
n×[×G^e11/μ1×G^e21/μ2]=0E88
The electric fields are
E1(r)=iωμ1drJ(r)G^e11(r,r)E89
E2(r)=iωμ2drJ(r)G^e21(r,r)E90
## 5. Vector wave functions, L, M, and N
The vector wave functions are the building blocks of the eigenfunction expansions of various kinds of dyadic Green functions. These functions were first introduced by Hansen [1012] in formulating certain electromagnetic problems.Three kinds of vector wave functions, denoted by L, M, and N, are solutions of the homogeneous vector Helmholtz equation. To derive the eigenfunction expansion of the magnetic dyadic Green functions that are solenoidal and satisfy with the vector wave equation, the L functions are not needed. If we try to find eigenfunction expansion of the electric dyadic Green functions then the L functions are also needed.
A vector wave function, by definition, is an eigenfunction or a characteristic function, which is a solution of the homogeneous vector wave equation ××Fκ2F=0.
There are two independent sets of vector wave functions, which can be constructed using the characteristic function pertaining to a scalar wave equation as the generating function. One kind of vector wave function, called the Cartesian or rectilinear vector wave function, is formed if we let
F=×(Ψ1c)E91
where ψ1 denotes a characteristic function, which satisfies the scalar wave equation
2Ψ+κ2Ψ=0E92
And c denotes a constant vector, such as x, y, or z. For convenience, we shall designate c as the piloting vector and Ψ as the generating function. Another kind, designated as the spherical vector wavefunction, will be introduced later, whereby the piloting vector is identified as the spherical radial vector R.
Actually, substituting Eq. (91) into Eq. (92), it is
×[c(2Ψ1+κ2Ψ1)]=0E93
The set of functions so obtained
M1=×(Ψ1c)E94
N2=1κ××(Ψ2c)E95
L3=(Ψ3)E96
Ψ2,Ψ3 denote the characteristic functions which also satisfy (92) but may be different from the function used to define M1.
In the following, the expressions for the dyadic Green functions of a rectangular waveguide will be derived asserting to the vector wave functions. The method and the general procedure would apply equally well to other bodies (cylindrical waveguide, circular cylinder in free space, and inhomogeneous media and moving medium).
Figure 5 shows the orientation of the guide with respect to the rectangular coordinate system, and we will choose the unit vector z to represent the piloting vector c.
The scalar wave function
Ψ=(Acoskxx+Bsinkxx)(Ccoskyy+Dsinkyy)eihzE97
where kx2+ky2+h2=κ2.
the constants kx and ky should have the following characteristic values
kx=mπa,m=0,1,E98
ky=nπb,n=0,1,E99
The complete expression and the notation for the set of functions M, which satisfy the vector Dirichlet condition are
Memn(h)=×[Ψemnz]=(kyCxSyx+kxCySxy)eihzE100
where Sx=sinkxx,Cx=coskxx, Sy=sinkyy,Cy=coskyy. The subscript “e” attached to Memn is an abbreviation for the word “even,” and “o” for “odd.”
In a similar manner
Nomn=1κ(ihkxCxSyx+ihkyCySxy+(kx2+ky2)SxSyz)eihzE101
It is obvious that Memn represents the electric field of the TEmn mode, while Nomn represents that of the TMmn mode.
In summary, the vector wave functions, which can be used to represent the electromagnetic field inside a rectangular waveguide, are of the form
Me(o)mn=×[Ψe(o)mnz]E102
Ne(o)mn=1κ××[Ψe(o)mnz]E103
Then
G^m2(R,R)=+dhm,n(2δ0)κπab(kx2+ky2)[a(h)Nemn(h)Memn(h)+b(h)Momn(h)Nomn(h)]E104
where a(h)=b(h)=1κ2k2, h=±(k2kx2ky2)1/2and δ0={1,m=0orn=00,m0,n0.
M', N', m', n', h' denote another set of values, which may be distinct or the same as M, N, m, n, h.
## 6. Retarded and advanced Green functions
Green function is also utilized to solve the Schrödinger equation in quantum mechanics. Being completely equivalent to the Landauer scattering approach, the GF technique has the advantage that it calculates relevant transport quantities (e.g., transmission function) using effective numerical techniques. Besides, the Green function formalism is well adopted for atomic and molecular discrete-level systems and can be easily extended to include inelastic and many-body effects [13, 14].
(A) The definitions of propagators
The time-dependent Schrödinger equation is:
iħ|Ψ(t)t=H^|Ψ(t)E105
The solution of this equation at time t can be written in terms of the solution at time t′:
|Ψ(t)=U^(t,t)|Ψ(t)E106
where U^(t,t)is called the time-evolution operator.
For the case of a time-independent Hermitian Hamiltonian H^, so that the eigenstates |Ψn(t)=eiEnt/ħ|Ψnwith energies En are found from the stationary Schrödinger equation
H^|Ψn=En|ΨnE107
The eigenfunctions |Ψnare orthogonal and normalized, for discrete energy levels 1:
Ψm|Ψn=δmnE108
and form a complete set of states (I^is the unity operator)
nΨn|Ψn=1E109
The time-evolution operator for a time-independent Hamiltonian can be written as
U^(tt)=ei(tt)H^/ħE110
This formal solution is difficult to use directly in most cases, but one can obtain the useful eigenstate representation from it. From the identity U^=U^I^and (107), (109), (110) it follows that
U^(tt)=nei/ħEn(tt)|ΨnΨn|E111
which demonstrates the superposition principle. The wave function at time t is
|Ψ(t)=U^(t,t)|Ψ(t)=nei/ħEn(tt)Ψn|Ψ(t)|ΨnE112
where Ψn|Ψ(t)are the coefficients of the expansion of the initial function |Ψ(t)on the basis of eigenstates.
It is equivalent and more convenient to introduce two Green operators, also called propagators, retarded G^R(t,t)and advanced G^A(t,t):
G^R(t,t)=iħθ(tt)U^(t,t)=iħθ(tt)ei(tt)H^/ħE113
G^A(t,t)=iħθ(tt)U^(t,t)=iħθ(tt)ei(tt)H^/ħE114
so that at t > t′ one has
|Ψ(t)=iħG^R(tt)|Ψ(t)E115
while at t < t′ it follows
|Ψ(t)=iħG^A(tt)|Ψ(t)E116
The operators G^R(t,t)at t > t′and G^A(t,t)at t < t′ are the solutions of the equation
[iħtH^]G^R(A)(t,t)=I^δ(tt)E117
with the boundary conditions G^R(t,t)=0at t < t′, G^A(t,t)=0at t > t′. Indeed, at t > tEq. (118) satisfies the Schrödinger equation Eq. (105) due to Eq. (117). And integrating Eq. (117) from t=tηto t=t+ηwhere η is an infinitesimally small positive number η=0+, one gets
G^R(t+η,t)=1iħI^E118
giving correct boundary condition at t = t′. Thus, if the retarded Green operator G^R(t,t)is known, the time-dependent wave function at any initial condition is found (and makes many other useful things, as we will see below).
For a time-independent Hamiltonian, the Green function is a function of the time difference τ=t-t, and one can consider the Fourier transform
G^R(A)(E)=+G^R(A)(τ)eiEτ/ħdτE119
This transform, however, can not be performed in all cases, because G^R(A)(E)includes oscillating terms eiEτ/ħ. To avoid this problem we define the retarded Fourier transform
G^R(E)=limη0++G^R(τ)ei(E+iη)τ/ħdτE120
G^A(E)=limη0++G^A(τ)ei(Eiη)τ/ħdτE121
where the limit η → 0 is assumed in the end of calculation. With this addition, the integrals are convergent. This definition is equivalent to the definition of a retarded (advanced) function as a function of complex energy variable at the upper (lower) part of the complex plain.
Applying this transform to Eq. (117), the retarded Green operator is
G^R(E)=[(E+iη)I^H^]1E122
The advanced operator G^A(E)is related to the retarded one through
G^A(E)=G^R+(E)E123
Using the completeness propertyn|ΨnΨn|=1, there is
G^R(E)=n|ΨnΨn|(E+iη)I^H^E124
and
G^R(E)=n|ΨnΨn|E-En+iηE125
Apply the ordinary inverse Fourier transform to G^R(E), the retarded function becomes
G^R(τ)=-+G^R(E)eiEτ/ħdE2πħ=iħθ(τ)neiEnτ/ħ|ΨnΨn|E126
Indeed, a simple pole in the complex E plain is at E=Eniη, the residue in this point determines the integral at τ > 0 when the integration contour is closed through the lower half-plane, while at τ < 0 the integration should be closed through the upper half-plane and the integral is zero.
The formalism of retarded Green functions is quite general and can be applied to quantum systems in an arbitrary representation. For example, in the coordinate system Eq. (124) is
G^R(r,r,E)=nr|ΨnΨn|rE-En+iη=nΨn(r)Ψn(r)E-En+iηE127
(B) Path integral representation of the propagator
In the path integral representation, each path is assigned an amplitude eidtL, L is the Lagrangian function. The propagator is the sum of all the amplitudes associated with the paths connecting xa and xb (Figure 6). Such a summation is an infinite-dimensional integral.
The propagator satisfies
iG(xb,tb,xa,ta)=dxiG(xb,tb,x,t)iG(x,t,xa,ta)E128
Let us divide the time interval [ta, tb] into N equal segments, each of length Δt=(tbta)/N.
iG(xb,tb,xa,ta)=dx1dxNj=1NiG(xj,tj,xj1,tj1)=ANjdxjexp[iΔtL(tj,xj+xj12,xjxj12)]=D(x)eidtL(t,x,x˙)E129
where ln[iG(xj,tj,xj1,tj1)]=iΔtL(tj,xj+xj12,xjxj12).
Example: LC circuit-based metamaterials
In this section, we will use the relationship of current and voltage in the LC circuit to build the propagator of the LC circuit field coupled to an atom.
Figure 7 shows the LC-circuit.The following are valid:
I=dqdtE130
V=qC=LdIdtE131
Thus:
Cd2xdt2=xLE132
where x=LI, I is the current, V is the voltage, q is the charge quantity, L and C are the inductance and capacitance, respectively. Eq. (132) is equal to a harmonic, and the Lagrangian operator is:
L0(x,x˙)=12g(ε˙2ΩLC2ε2)E133
The Lagrangian operator describing the bipole is:
L0(x,x˙)=m2x˙2mΩ022x2E134
where x is the coordinate of the bipole, ε is the LC field, m is the mass of an electron, and e is the unit of charge. g=1c, and ΩLC=1LC. Defining their action items as:
SLC=dt[12g(ε˙2ΩLC2ε2)]E135
And
S0=dt[m2(x˙2Ω02x2)]E136
Taking the coupling effect (exε) into account, the Green function of the coupled system is:
G(x,ε)=DxDεeiSLC+iS0+idt[exε]E137
Where x represents the series coordinates x1,x2,…,and so on and ɛ represents ɛ1,ɛ2,…., and so on.
## 7. The recent applications of the Green function method
### 7.1. Convergence
In the Green function, the high oscillation of Bessel/Hankel functions in the integrands results in quite time-consuming integrations along the Sommerfeld integration paths (SIP) which ensures that the integrands can satisfy the radiation condition in the direction normal to the interface of a medium. To facilitate the evaluation, the method of moments (MoM) [15], the steepest descent path (SDP) method, and the discrete complex image method (DCIM) [16, 17] are very important methods.
The technique for locating the modes is quite necessary for accurately calculating the spatial Green functions of a layered medium. The path tracking algorithm can obtain all the modes for the configuration shown in Figure 8, even when region 2 is very thick [18]. Like the method in Ref. [19], it does not involve a contour integration and could be extended to more complicated configurations.
The discrete complex image method (DCIM) has been shown to deteriorate sharply for distances between source and observation points larger than a few wavelengths [20]. So, the total least squares algorithm (TLSA) is applied to the determination of the proper and improper poles of spectral domain multilayered Green’s functions that are closer to the branch point and to the determination of the residues at these poles [21].
The complex-plane for the determination of proper and improper poles is shown in Figure 9. Since half the ellipse is in the proper sheet of the -plane and half the ellipse is in the improper sheet, the poles will not only correctly capture the information of the proper poles but will also capture the information of those improper poles that are closer to the branch point kρ = k0.
For the 2-D dielectric photonic crystals as shown in Figure 10, the integral equation is written in terms of the unknown equivalent current sources flowing on the surfaces of the periodic 2-D cylinders. The method of moments is then employed to solve for the unknown current distributions. The required Green function of the problem is represented in terms of a finite summation of complex images. It is shown that when the field-point is far from the periodic sources, it is just sufficient to consider the contribution of the propagating poles in the structure [22]. This will result in a summation of plane waves that has an even smaller size compared with the conventional complex images Green function. This provides an analyzed method for the dielectric periodic structures.
Others, since the Gaussian function is an eigenfunction of the Hankel transform operator, for the microstrip structures, the spectral Green’s function can be expanded into a Gaussian series [23]. By introducing the mixed-form thin-stratified medium fast-multiple algorithm (MF-TSM-FMA), which includes the multipole expansion and the plane wave expansion in one multilevel tree, the different scales of interaction can be separated by the multilevel nature of the the fast multipole algorithm [24].
The vector wave functions, L, M, and N, are the solutions of the homogeneous vector Helmholtz equation. They can also be used for the analyses of the radiation in multilayer and this method avoids the finite integration in some cases.
### 7.2. Multilayer structure
The volume integral equation (VIE) can analyze electromagnetic radiation and scattering problems in inhomogeneous objects. By introducing an “impulse response” Green function, and invoking Green theorem, the Helmholtz equation can be cast into an equivalent volume integral equation including the source current or charges distribution. But the number of unknowns is typically large and the equation should be reformulated if there are in contrast both permittivity and permeability. At present, it is utilized to analyse the general scatterers in layered medium [25, 26].
When the inhomogeneity is one dimension, the Green function can be determined analytically in the spectral (Fourier) domain, and the spatial domain counterpart can be obtained by simply inverse Fourier transforming it.
Surface integral equation (SIE) method is another powerful method to handle electromagnetic problems. Similarly, by introducing the Green function, the Helmholtz equation can be cast into an equivalent surface integral equation, where the unknowns are pushed to the boundary of the scatterers [27].
Despite the convergence problem, the locations of the source and observation point may cause the change of Green function form, for example, for a source location either inside or outside the medium, the algebraic form of the Green functions changes as the receiver moves vertically in the direction of stratification from one layer to another [28].
First, we introduce the full-wave computational model [29]. A multilayer structure involving infinitely 1-D periodic chains of parallel circular cylinders in any given layer can be constructed as shown in Figure 11. Each layer consists of a homogeneous slab within which the circular cylinders are embedded. This is the typical aeronautic situation with fiber-reinforced four-layer pile (with fibers orientated at 0°, 45°, −45°, and 90°), but any other arrangement is manageable likewise.
In the multilayered photonic crystals, the Rayleig’s method and mode-matching are combined to produce scattering matrices. An S-matrix-based recursive matrix is developed for modeling electromagnetic scattering. Field expansions and the relationship between expansion coefficients are given.
There is a mix treatment for the inhomogeneous and homogeneous multilayered structure [30]. As shown in Figure 12, a substrate is divided into two regions. The top region is laterally inhomogeneous and for the finite-difference method (FDM) or the finite element method (FEM), the volume integral equation, is used. The bottom region is layerwise homogeneous, and the boundary-element methods (BEM) are used. The two regions are connected such as a BEM panel is associated with an FEM node on the interface.
A Green function was derived for a layerwise uniform substrate and was then used in a layerwise nonuniform substrate with additional boundary conditions applied to the interface. Given that the lateral inhomogeneity is local, volume meshing is used only for the local inhomogeneous regions, BEM meshing is applied to the surfaces of these local regions.
For a field (observation) point in the jth layer and a source point in the kth layer, the Green function has the form:
Gjku,l=Gjk,0u,l+m=0m+n0n=0cmnϕjku,labεkγmn×cosmπxfacosnπyfbcosmπxsacosnπysbE138
where the superscripts u and l indicate the upper and lower solutions, respectively, depending on whether the field point (or observation point) is above or below the source point. a and b are the substrate dimensions in the x- and y- directions, respectively, and more details can be found in Refs. [31, 32].
The electromagnetic field in a multilayer structure can be efficiently simplified by the assumption that the multilayer is grounded by a perfect electric conducto (PEC) plane [33, 34]. When the source and the field points are assumed to be inside the dielectric slab, in a layered medium as shown in Figure 13, by applying the boundary conditions, the 1-D Green functions is
Gx(x,x0;λx1,λx2)=(GxPMC+GxPEC)/2E139
where PMC represents the perfect magnetic conductor. The simplified Green function form can be deduced to the cae of (b).
The three-dimensional (3-D) Green function for a continuous, linearly stratified planar media, backed by a PEC ground plane, can also be expressed in terms of a single contour integral involving one-dimensional (1-D) green function. The constructure is shown in Figure 14.
The general formulation for a single electric current element has been worked out in detail in Ref. [35] which is based on the appropriate information from Ref. [36].
## How to cite and reference
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Jing Huang (June 14th 2017). Green Function, Recent Studies in Perturbation Theory, Dimo I. Uzunov, IntechOpen, DOI: 10.5772/68028. Available from:
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# Funnel
##### Stage: 4 Challenge Level:
You may have heard the term 'Calculus', it's a big idea in mathematics and so powerful that I find it hard to imagine mathematics without it. It's main technique, differentiation, makes problems like this fairly easy - with a little care over the algebra. So if you are thinking about continuing with mathematics beyond Stage 4, learning about differentiation would be one of the big benefits that await you.
But differentiation is not really Stage 4 mathematics, though very close, so we don't use it in solutions because that would be unfair to students who haven't seen the idea before, it would be like suddenly changing into a language you haven't had a chance to learn yet.
If that's got you interested why not take a look at Vicky Neale's article : Introduction to Differentiation
And thank-you to David and Berny from Gordonstoun and to Naren from Loughborough Grammar School who sent in great solutions using that technique .
In fact the use of differentiation did more than get an answer. It found the result that the cone needed to have its height 1.41 times bigger than its radius, like the Stage 4 result below, but found it in this more interesting form.
And the value of that is that it suggests a new direction to pursue with this problem : it seems so neat.
Why the square root of 2 ? Is there a connection with the diagonal of a square ? Or is it something else ?
That's maybe a little bit on from us at Stage 4 so here's a way to solve a problem like this using Stage 4 mathematics. It's really 'trial and improvement' but using a spreadsheet to make the calculation effortless. .
We are going to use :
• the radius, which we'll adjust to get nearer and nearer to the answer,
• the height, which will be determined by our choice of radius, so that we get the chosen target volume
• the slant length, which we'll find using r and h and Pythagoras,
• and the surface area, for which there's a great little formula.
The volume of a cone is one third the volume of a cylinder with the same base and height.
If we took the target volume to be 1 litre, 1000 ml, then the formula that connects h and r is
The slant length (s), using Pythagoras, is
And the surface area of a cone is
plus the base if you need it - here we don't.
Incidentally, if you don't know where that surface area formula comes from it may be good to take a moment to look at that. Flatten the curved surface out to get a sector (how do you know it's a sector ?). The radius will be the cone's slant length, so you can calculate the area of the whole circle. To know the proportion that the sector is of that circle compare the sector arc, which is the cone's base circumference, with the circumference of this new circle, radius s.
Back to the funnel and using as little plastic as possible. Take a look at this spreadsheet : Funnel
Can you see what each column does ? Click on a cell and check the formula.
• The first column has increasing radius values which you can control.
• The next column calculates the height, because we knew the volume was 1000 ml
• The radius and height are then used to calculate the slant length.
• And the final column uses the radius and the slant length to calculate the surface area, which we want to be as small as possible
There's even a graph so you can have some sense for how surface area varies as the radius value ranges across your chosen interval. | 1,024 | 4,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2016-40 | longest | en | 0.950238 |
http://blogs.sas.com/content/iml/2011/10/12/maximum-likelihood-estimation-in-sasiml.html | 1,496,084,066,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612537.91/warc/CC-MAIN-20170529184559-20170529204559-00045.warc.gz | 63,448,131 | 15,291 | A popular use of SAS/IML software is to optimize functions of several variables. One statistical application of optimization is estimating parameters that optimize the maximum likelihood function. This post gives a simple example for maximum likelihood estimation (MLE): fitting a parametric density estimate to data.
### Which density curve fits the data?
If you plot a histogram for the SepalWidth variable in the famous Fisher's iris data, the data look normally distributed. The normal distribution has two parameters: μ is the mean and σ is the standard deviation. For each possible choice of (μ, σ), you can ask the question, "If the true population is N(μ, σ), how likely is it that I would get the SepalWidth sample?" Maximum likelihood estimation is a technique that enables you to estimate the "most likely" parameters. This is commonly referred to as fitting a parametric density estimate to data.
Visually, you can think of overlaying a bunch of normal curves on the histogram and choosing the parameters for the best-fitting curve. For example, the following graph shows four normal curves overlaid on the histogram of the SepalWidth variable:
```proc sgplot data=Sashelp.Iris; histogram SepalWidth; density SepalWidth / type=normal(mu=35 sigma=5.5); density SepalWidth / type=normal(mu=32.6 sigma=4.2); density SepalWidth / type=normal(mu=30.1 sigma=3.8); density SepalWidth / type=normal(mu=30.5 sigma=4.3); run;```
It is clear that the first curve, N(35, 5.5), does not fit the data as well as the other three. The second curve, N(32.6, 4.2), fits a little better, but mu=32.6 seems too large. The remaining curves fit the data better, but it is hard to determine which is the best fit. If I had to guess, I'd choose the brown curve, N(30.5, 4.3), as the best fit among the four.
The method of maximum likelihood provides an algorithm for choosing the best set of parameters: choose the parameters that maximize the likelihood function for the data. Parameters that maximize the log-likelihood also maximize the likelihood function (because the log function is monotone increasing), and it turns out that the log-likelihood is easier to work with. (For the normal distribution, you can explicitly solve the likelihood equations for the normal distribution. This provides an analytical solution against which to check the numerical solution.)
### The likelihood and log-likelihood functions
The UNIVARIATE procedure uses maximum likelihood estimation to fit parametric distributions to data. The UNIVARIATE procedure supports fitting about a dozen common distributions, but you can use SAS/IML software to fit any parametric density to data. There are three steps:
1. Write a SAS/IML module that computes the log-likelihood function. Optionally, since many numerical optimization techniques use derivative information, you can provide a module for the gradient of the log-likelihood function. If you do not supply a gradient module, SAS/IML software automatically uses finite differences to approximate the derivatives.
2. Set up any constraints for the parameters. For example, the scale parameters for many distributions are restricted to be positive values.
3. Call one of the SAS/IML nonlinear optimization routines. For this example, I call the NLPNRA subroutine, which uses a Newton-Raphson algorithm to optimize the log-likelihood.
#### Step 1: Write a module that computes the log-likelihood
A general discussion of log-likelihood functions, including formulas for some common distributions, is available in the documentation for the GENMOD procedure. The following module computes the log-likelihood for the normal distribution:
```proc iml; /* write the log-likelihood function for Normal dist */ start LogLik(param) global (x); mu = param[1]; sigma2 = param[2]##2; n = nrow(x); c = x - mu; f = -n/2*log(sigma2) - 1/2/sigma2*sum(c##2); return ( f ); finish;```
Notice that the arguments to this module are the two parameters mu and sigma. The data vector (which is constant during the optimization) is specified as the global variable x. When you use an optimization method, SAS/IML software requires that the arguments to the function be the quantities to optimize. Pass in other parameters by using the GLOBAL parameter list.
It's always a good idea to test your module. Remember those four curves that were overlaid on the histogram of the SepalWidth data? Let's evaluate the log-likelihood function at those same parameters. We expect that log-likelihood should be low for parameter values that do not fit the data well, and high for parameter values that do fit the data.
```use Sashelp.Iris; read all var {SepalWidth} into x; close Sashelp.Iris; /* optional: test the module */ params = {35 5.5, 32.6 4.2, 30.1 3.8, 30.5 4.3}; LogLik = j(nrow(params),1); do i = 1 to nrow(params); p = params[i,]; LogLik[i] = LogLik(p); end; print Params[c={"Mu" "Sigma"} label=""] LogLik;```
The log-likelihood values confirm our expectations. A normal density curve with parameters (35, 5.5) does not fit the data as well as the other parameters, and the curve with parameters (30.5, 4.3) fits the curve the best because its log-likelihood is largest.
#### Step 2: Set up constraints
The SAS/IML User's Guide describes how to specify constraints for nonlinear optimization. For this problem, specify a matrix with two rows and k columns, where k is the number of parameters in the problem. For this example, k=2.
• The first row of the matrix specifies the lower bounds on the parameters. Use a missing value (.) if the parameter is not bounded from below.
• The second row of the matrix specifies the upper bounds on the parameters. Use a missing value (.) if the parameter is not bounded from above.
The only constraint on the parameters for the normal distribution is that sigma is positive. Therefore, the following statement specifies the constraint matrix:
```/* mu-sigma constraint matrix */ con = { . 0, /* lower bounds: -infty < mu; 0 < sigma */ . .}; /* upper bounds: mu < infty; sigma < infty */```
#### Step 3: Call an optimization routine
You can now call an optimization routine to find the MLE estimate for the data. You need to provide an initial guess to the optimization routine, and you need to tell it whether you are finding a maximum or a minimum. There are other options that you can specify, such as how much printed output you want.
```p = {35 5.5};/* initial guess for solution */ opt = {1, /* find maximum of function */ 4}; /* print a LOT of output */ call nlpnra(rc, result, "LogLik", p, opt, con);```
The following table and graph summarize the optimization process. Starting from the initial condition, the Newton-Raphson algorithm takes six steps before it reached a maximum of the log-likelihood function. (The exact numbers might vary in different SAS releases.) At each step of the process, the log-likelihood function increases. The process stops when the log-likelihood stops increasing, or when the gradient of the log-likelihood is zero, which indicates a maximum of the function.
You can summarize the process graphically by plotting the path of the optimization on a contour plot of the log-likelihood function. I used SAS/IML Studio to visualize the path. You can see that the optimization travels "uphill" until it reaches a maximum.
### Check the optimization results
As I mentioned earlier, you can explicitly solve for the parameters that maximize the likelihood equations for the normal distribution. The optimal value of the mu parameter is the sample mean of the data. The optimal value of the sigma parameter is the unadjusted standard deviation of the sample. The following statements compute these quantities for the SepalWidth data:
```OptMu = x[:]; OptSigma = sqrt( ssq(x-OptMu)/nrow(x) );```
The values found by the NLPNRA subroutine agree with these exact values through seven decimal places.
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Distinguished Researcher in Computational Statistics
Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of PROC IML and SAS/IML Studio. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
1. Attached are two very compact ways to do this using PROC OPTMODEL in SAS/OR. Note that both LogLikelihood and LogLikelihood2 differ from your LogLik by an additive constant but of course yield the same optimal parameters. Download the OPTMODEL program.
• Hi Rob, The code you provided was very helpful for me. I used it with my data and it worked perfect. I am now trying to do the same thing but for Poisson model instead of Normal. I wrote the following code but I am not sure whether it is correct. what's the best way to validate my code? I also need to estimate the parameters of the Lognormal model; do you have any code that would help me do so?
proc optmodel;
set OBS;
num x {OBS};
read data sashelp.iris into OBS=[_N_] x= SepalWidth;
num n = card(OBS);
var lambda>=0 init 0;
max LogLikelihood = sum {i in OBS}(x[i] * log(lambda)) - n * lambda;
/*max LogLikelihood1 = sum {i in OBS}(x[i] * log(lambda)) - log(x[i]) * lambda;
max LogLikelihood2 = sum {i in OBS} log(PDF('poisson',x[i],lambda));
LogLikelihood1 and LogLikelihood2 didn't work*/
solve ;
print lambda;
print LogLikelihood;
num optlambda = (sum {i in OBS} x[i]) / n;
print optlambda;
quit;
2. Hi Dr. Rick Wicklin, after running the MLE following above code, if I want to make a data table with the parameters (MU, SIGMA) estimated, how can I do? Thank you very much. Regards, Yong
• Ajewole Kenny on
Hi Rick Wicklin, pls I have an issue which I will like you help me out on my research.
Actually, I'm working on Polynomial regression, with a particular data, two methods have been used which are Numerical and OLS method with little difference in their results.. Now, I'm trying to contribute using MLE for the same Data.
How can I go abt that on SAS ?
I need a SAS code for this pls help me....
I must present this work some days left...
• Many standard models are solvable using PROC GENMOD, which computes MLE. For more sophisticated models, PROC NLMIXED is the SAS procedure that solves general MLE. You can post details of your question, along with sample data, at the SAS Support Communities.
3. Dr. Rick Wicklin.. When I have a likelihood function where I have to estimate 3 parameters mu,sigma and Pc (proabability) , can I use the same code ?
For example :likelihood = Pc ∫ f (x) dx where f is a normal distribution with unknown mu and sigma
Also, Can the CDF function be used in the function f that you have defined.
• Yes, you should be able to use similar code. Yes, you can call the CDF function from within the likelihood function.
4. Hi Rick
I am trying to create a vector in which each value is the result of an optimization relevant to the row. To be clearer:
First, I have a column vector called n, composed of the numbers 50 to 40000 in increments of 1
Second, I have a vector P defining the objective function, which is composed of two inputs: known n and unknown S where S needs to be optimized to maximize each row value of P. P is defined by:
P = 50*(S##.45)#(n##1.1)-40000*n-S*n-S##1.35#n##.5
Therefore, I need one optimized value in the S vector per row value of n.
I have tried creating a do-loop in IML but as you know modules cannot be made to start on loops.
I have tried to create a macro but had issues.
Any suggestions on the best way forward?
• PS I did manage to get it right outside of IML. However, this macro do-loop with Proc NLP takes a seriously long time once I loop between 50 and 40000, and I had to disable the log which kept on filling up. Still wondering if there's an easier way in IML :-)
options nonotes nosource nosource2 errors=0;
Data Estimates;
Input S 12.5;
Datalines;
9999.99999
;
Run;
%Macro simul;
%do i = 50 %to 90;
Proc NLP noprint outest=Parms(where=(_TYPE_='PARMS'));
Max P0;
Decvar S;
P0 = 50*(S**.45)*(&i**1.1)-40000-(S*&i)-(S**1.35)*(&i**.5);
Run;
Data Estimates;
Set Estimates Parms(keep=S);
Run;
%end;
%mend;
%simul;
• This is a rational expression in the variable S. Use calculus. Take the derivative w/r/t S and set it equal to zero. Solve for the root numerically by using the FROOT function.
• What do you mean "modules cannot be made to start on loops." I am not familiar with any restrictions on calling modules within loops or with looping inside a module.
5. Hyung S. Hwang on
Hi~~~
This blog's posts are really helpful to my job, thank you^^
I'm working on an statistical MLE job using IML's call NLPQN module. The situation is that initial values are very critical to the stability of the solution and I need to make a simulation with different initial values of more than 100 times. I have looked through the helpdoc or manuals for the way how to store or extract objective function values printed on log window by setting option vector, and however failed to find out _._. The simulation took a long time for the completetion, so I could not watch the objective function value printed in log window in each time of simulation path.
So, I need to get the objective function value at the MLE solution point and store it in a "data set" and use it later, for example, the selection among the simulation results.
If possible, then hessian matrix can be stored in a daa set?
6. ASudipta Banerjee on
Hi Rob, can you please share some real world problems where MLE is used?Is it only used to measure the value of mean and sigma of the dataset of the closest fit bell curve? What happens if the data is the dependent variable is driven by multiple factors?
• Who's Rob? MLE is explained in the Wikipedia article that I linked to in the 2nd paragraph. It is a general method of choosing parameters in a model, including regression models.
7. Hi again. Can you do this for uniform distribution too between 0 and 1? I don't understand how to incorporate the theta aspect.
Thanks
8. Hi dr. Rick Wicklin, i've read this article and I saw that you have used the iteration history (of the call nlprna) for the last graph. My question is trivial: how can i save the iteration history of the call nlprna in dataset and what is the procedure who you have used for the last graph? tnx.
• Good question. I use ODS OUTPUT to save the iteration history and then used the ContourPlot in IMLPlus to visualize the path. I have added a link in the article to download the program. If you prefer regular SAS, make a similar image by using a GTL template.
9. Hi Mr Rick Wicklin,
I am studying my statistics thesis about max Extreme Value distribution.
I try to estimate parameters of I. type (Gumbell) distribution via Maximum Likelihood Method on SAS. Is it possible to apply this on SAS ? If yes, could you please share with me related codes ?
Thanks! | 3,528 | 14,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-22 | latest | en | 0.827044 |
https://ai.stackexchange.com/questions/34577/calculating-the-value-of-a-state-in-an-optimal-policy-analytically-and-iterative | 1,723,079,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00167.warc.gz | 67,887,004 | 39,586 | # calculating the value of a state in an optimal policy analytically and iteratively
I am watching the lecture by Abbeel on MDPs and Reinforcement Learning. The setup of the problem is the classic gridworld with optimal policy (and corresponding values of states) pictured below.
The parameters of the problem here are: there is a 0.8 chance that you go to your chosen action and 0.1 chance you go to a direction perpendicular to it. For example, if you choose "right", then you have a 0.8 chance of going right and a 0.1 chance of going up, and another 0.1 chance of going down. If the action makes the agent 'bump' into the wall or 'leave' the world, then the agent stays in the same place.
My question is this: Since the problem above shows the optimal policy (having solved/demonstrated the solution via the Value Iteration algorithm), is it possible that I solve for the values of the states manually (analytically), knowing that the optimal policy is give above? I expected that if I setup the (system of) equations, I should be able to recover the values that are printed in the figure above.
I represented the states of the gridworld as follows. $$X$$ is just a place holder since it is a blocked state.
$$\begin{array}{cccc} H & D & A & +1 \\ G &X & B & -1 \\ F & E & C & J \end{array}$$
Knowing that $$V^{\pi}(s) = \displaystyle\sum_{a} \pi(a|S) \displaystyle\sum_{s'} P(s'|s,a) [r_{t+1} + \gamma V^{\pi}(s')]$$, $$\gamma = 0.9$$, there are no rewards awarded for every step except with entering the terminal states $$+1$$ and $$-1$$.
The actions are already deterministic at this point, so I set $$\pi(a|s) = 1$$. For simplicity of notation, I just wrote $$A$$ as the value of state $$A$$ in the following equations:
At state $$A$$: $$A = 0.8[1+0] + 0.1[0.9A] + 0.1[0.9B]$$. Meaning: 0.8 chance of going right, receiving a reward of 1, and the value of the terminal state is 0. 0.1 change of going up and therefore ending up in the same state $$A$$. Finally, 0.1 chance of going down, ending up in state $$B$$.
Another example: $$H = 0.8[0.9D] + 0.1[0.9H] + 0.1[0.9G]$$ Meaning: 0.8 chance to move right to D, 0.1 chance to bump the upper wall and stay in H, another 0.1 chance to move down the lower state G.
There are a total of 9 equations of this form. When I solved for the unknowns $$A,B,\cdots,J$$, why am I not getting the values placed here in this figure?
Edit: I found an error that made the values way above 1. The error is now fixed. The values are now below 1, which is good. But why am I not getting the values indicated in this figure?
In the gridworld setting, you can replicate the lecture's results, by defining the reward function $$R_t(s,a) = R(s)$$, meaning that the reward function simply aggregates only on the current state and ignores the selected action and the next state, as well. With this modification, the state value of a terminal state always (after the 1st iteration) equals its reward (+1 or -1). | 823 | 2,955 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.897137 |
http://poj.org/showmessage?message_id=152041 | 1,660,578,117,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00277.warc.gz | 40,600,895 | 2,844 | Online JudgeProblem SetAuthorsOnline ContestsUser
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主要是规律 附代码
Posted by mickeyandkaka at 2010-11-13 23:58:55 on Problem 2590
```计算差的绝对值
1 1
2 2
3~4 3
5~6 4
7~9 5
。。。
a1=1
an-a1=(i+1)/2（2<=i<=n)的和；
#include <stdio.h>
#include <stdlib.h>
int fun(int n);
int main()
{
int i,n,a,b;
scanf("%d",&n);
while(n--)
{
scanf("%d %d",&a,&b);
printf("%d\n",fun(abs(b-a)));
}
return 0;
}
int fun(int n)
{
int i,a=1;
if(n==1) return 1;
else if(n==0) return 0;
else
{
for(i=2;;i++)
{
a+=(i+1)/2;
if(a>=n) break;
}
}
return i;
}```
Followed by: | 288 | 715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-33 | latest | en | 0.23764 |
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