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# Allegro A4988 and Arduino (3)
In the last part of my tutorial about the A4988 driver, I’m going to explain how to build a controller to adjust speed and rotation direction of a stepper motor.
#### Schematics
I used the same setup introduced in a previous post:
• an Arduino Uno;
• a Pololu A4988 driver mounted on a breadboard;
• a NEMA17Â stepper motor.
#### GUI
Using the controller, you can adjust the speed (from 0 to 70 RPM, revolutions per minute) and the rotation direction.
On the LCD are displayed the actual speed, direction and a progress bar:
The 5 available buttons are used to control the motor:
• UP and DOWN adjust the speed;
• LEFT and RIGHT adjust the direction;
• SELECTÂ activates the emergency stop (speed = 0).
#### Continuous rotation
To achieve a continuous rotation, your sketch must send the step commands to the Pololu driver at constant frequency.
Let’s see an example: for a speed of 60RPM and for a motor with 200 steps / revolution, Arduino must send to the driver (60*200)/60 = 200 commands / sec or a command every 5ms.
To ensure an accurate timing, I’m going to use the interrupts generated by the Timer1 as explained in this article.
From the Playground you can download a convenient library to configure that Timer:
```#include [...] Timer1.initialize(100); Timer1.attachInterrupt(timerIsr);```
With the instructions above, you configure the Timer1 to trigger an interrupt every 100 millisecondi (0.1ms) and to invoke, at each interrupt, the timerIsr() function.
The “temporal distance” between a step command and the next one depends on the current speed (in the example above for a speed of 60RPM the distance is 5ms). Given that the possible speeds are only 15 (from 0 to 70RPM with a 5RPM increment), I calculated that distance (in ticks of 0.1ms) for the different speeds and stored the values in an array:
```const int speed_ticks[] = { -1, 600, 300, 200, 150, 120, 100, 86, 75, 67, 60, 55, 50, 46, 43};```
Within the timerIsr() function a counter is incremented and – if it has reached the ticks for the current speed – a step command is sent to the driver:
```void timerIsr() { if(actual_speed == 0) return; tick_count++; if(tick_count == ticks) { // make a step digitalWrite(PIN_STEP, HIGH); digitalWrite(PIN_STEP, LOW); tick_count = 0; } }```
In the next page, I’ll explain the rest of the sketch and show an example of its working…
#### Related Posts
Tuesday August 20th, 2013 - 07:42 AM
Hi Luca!
Thank you very much. But there is one big prolema, I have it hangs does not work, the engine is spinning and the buttons are not controlled.
Do “reset” button is working again controlled, then again hanging is not controlled by buttons.
Using the Arduino Uno, tried Arduino Mega 2560 also does not work.
Monday August 26th, 2013 - 05:15 AM
37 int previous_time;
change to
37 unsigned long previous_time;
Sunday November 17th, 2013 - 09:33 AM
Hi !
Can i use your scheme with EasyDriver 4.4 ?
Example here http://bildr.org/2011/06/easydriver/
I think that 3 inputs from arduino is the same ..?!
Sunday November 17th, 2013 - 10:38 AM
Hi, it should work because the EasyDriver has the same STEP/DIR interface as the controller I used.
Let me know!
Sunday December 15th, 2013 - 03:24 PM
Caro Luca,
Vorrei usare il tuo programma per collegare i seguenti componenti
un Arduino Uno;
un driver Pololu DRV8825 montato su breadboard;
un motore stepper NEMA 23
Potresti darmi qualche suggerimento per i collegamenti?
Grazie
Paolo
Monday December 16th, 2013 - 02:14 PM
Ciao Paolo,
sostanzialmente i collegamenti sono gli stessi del mio esempio perché il driver DRV8825 è pin compatibile con quello usato da me (chip A4988).
Thursday January 9th, 2014 - 07:14 PM
Salve Luca, in anzi tutto complimenti per i tutorial sono veramente interessanti e molto chiari.
vorrei usare il tuo programma facendogli qualche modifica ma non ci riesco, potresti darmi una mano?
Le modifiche sono le seguenti:
1 dovrei pilotare in microsecondi anzi che in milisecondi
2 sul display lcd anzi che visualizzare la velocità in RPM dovrei visualizzare il ritardo tra un passo e l’altro
Spero che tu possa darmi una mano perché non ci riesco proprio.
Grazie.
Tuesday January 14th, 2014 - 06:51 PM
Ciao Antonio
partiamo dalla prima richiesta perché mi “preoccupa”… difficilmente riuscirai ad avere passi della durata di microsecondi a causa dell’inerzia e della corrente; volevi ottenere una elevata velocità ?
Wednesday January 15th, 2014 - 09:24 AM
Ciao Luca Grazie per avermi risposto, no non è per aumentare la velocitÃ
Ma per avere più scelta.
Mi spiego meglio
Sto usando
Nema17 a 12v 1,2A 200 passi
Arduino 2009
Display lcd con tasti
DRV8825
Il tutto assemblato su una montatura equatoriale di un telescopico.
Il motore deve far si che il telescopio compia un giro completo di 360 gradi in 24 ore sul asse AR, secondo la demoltiplica che ho usato per aumentare i giri e diminuire lo sforzo, il motore deve compiere un giro i circa 65 secondi e fin che vado a passi interi o a mezzi passi tutto ok ma se piloto a micropassi diciamo a 1/16 o 1/32 di passo con i millisecondi non riesco ad ottenere la rotazione nel tempo giusto.
Praticamente dovrei impostare diverse velocità di rotazione aumentando di un secondo per ogni velocità e con i microsecondi dovrei riuscirci.
Spero di essermi spiegato bene e grazie per la pazienza.
Ciao
Monday January 27th, 2014 - 10:05 PM
Ciao Antonio
un giro in 65 secondi vuol dire che devi fare 200 passi in 65 secondi. Se anche vai a 32 microsteps, si parla sempre di 6400 “impulsi” ogni 65 secondi quindi circa 100 impulsi al secondo = 10ms a impulso… come mai ti serve invece lavorare con interrupt più frequenti?
Monday March 3rd, 2014 - 08:04 PM
Ciao Luca,
Per prima cosa vorrei farTi i complimenti per questo progetto che formidabile!
Forse ho sbagliato qualche cosa perchè ho comprato tutti i componenti hardware da Te usati, ma il programma non mi funziona bene.
Infatti all’inizio il motorino va che è una meraviglia, ma se dopo un pó che , per esempio va a 45 rpm , decido di cambiare la velocità o invertire il senso in cui il motorino gira , accade che la tastiera della LCD keypad shield si blocca ed il motore non si puó più controllare.
Potresti dirmi cosa sbaglio?
Grazie in anticipo
Marco
Friday April 25th, 2014 - 05:16 PM
This code… is beautiful! You have no idea how useful this was (as I have many steppers and A4988’s lying around from various Repraps!)
Thank you very much for posting a very concise and understandable set of tutorials!
😀
Thursday July 10th, 2014 - 08:41 AM
Buongiorno Luca
Ho provato il tuo tutorial, ma il motore invece che girare effettua un passo avanti e uno indietro rimanendo sempre nella stessa posizione. Vorrei capire se e’ il motore che e’ rotto o la pcb della Pololu.
grazie mille
saluti
Monday July 14th, 2014 - 03:24 PM
ciao Luca
è possibile che sia collegato sbagliato al driver? uno step avanti e uno indietro sembra indicare che gli avvolgimenti del motore non sono collegati bene…
Wednesday January 21st, 2015 - 01:13 PM
Hi Luca!
Greetings from Berlin 🙂
I was very glad to have seen your project, very great! I use it for my automatic Camera Slider
I have downloaded and install the sketsch, it works but only in one direction. What’s wrong with it?
lg. Alex
Wednesday January 21st, 2015 - 08:03 PM
Hi Alex,
the “arrow” on the LCD changes its direction? Check the connection between PIN3 of Arduino and the DIR pin of the stepper driver…
Thursday January 22nd, 2015 - 12:30 AM
ciao luca!
volevo chiederti riusciresti a fare una variante del firmware dove e possibile usare un lcd 16×2 con convertitore ic2 ? ho chiedo troppo saluti antony attendo risposta !
Saturday January 24th, 2015 - 03:54 PM
wow 🙂
you are right, the pin has bad contact
now it moves in all directions
one question…
at 70 RPM is the motor rotation to slow for my slider, how can i more speedup the turns of the motor?
at higher speed i can´t change the direction the motor stops and hums
Saturday January 31st, 2015 - 10:11 AM
Hi Alex! what did you change in the sketch to add higher speeds?
Tuesday February 3rd, 2015 - 12:10 PM
i have change folowing lines:
// debounce time (milliseconds)
#define DEBOUNCE_TIME 70
// lookup table speed – ticks (interrupts)
const int speed_ticks[] = {-1, 4000, 300, 200, 100, 80, 70, 60, 65, 60, 50, 48, 35, 25, 20};
(4000 is for the timelapse operation)
int unsigned long previous_time;
// init the timer1, interrupt every 0.1ms
Timer1.initialize(23);
Timer1.attachInterrupt(timerIsr);
when i change the timer1 value to higher the motor freaks out at higher speeds/turns.. can it be, that the motordriver is not fast enough?
Wednesday February 4th, 2015 - 09:53 PM
Hi Alex: ticks is the number of “interrupts” between one step and the next one, so try reducing the last value and at the same time reduce also the Timer1.initialize() value
Saturday February 21st, 2015 - 06:13 PM
Hi, Luka!
Thank you very much for your project, everything works fine. I have only one question, how to make the rotation of the stepper motor (FL57STH76-1006 1.8*) faster and reduce vibration?
I changed the code and it helped a little:
// init the timer1, interrupt every 0.05 ms
Timer1.initialize(18);
Timer1.attachInterrupt(timerIsr);
Tuesday February 24th, 2015 - 09:06 PM
Hi Ruslan,
you can also decrease the values in the speed_ticks[] vector… for example if you configure Timer1 to trigger every 0.02ms (`Timer1.initialize(20)`) a speed_ticks[] value of 100 makes the motor rotate 1 step / 2ms that is (for a 200 steps/rev motor) a revolution every 0.4 seconds = 150RPM
Thursday February 26th, 2015 - 07:33 PM
Luca, many thanks!!!
I played around with “values in the speed_ticks[] vector” and reached the speed limit for my motor. Now I am happy 🙂
Originally, my project was based on this: http://www.minie.airiclenz.com/hardware-2/hardware/ . Now I have 2 codes for timelapse and for video footage.
Monday April 13th, 2015 - 09:28 PM
Ciao come posso adattare il tuo fantastico progetto alla mia motor shield r3 e keypad lcd? ho reso tutto impilabile e perfettamente compatibile facendo tagliando A0 etc etc.
Grazie del futuro supporto.
Monday April 13th, 2015 - 09:35 PM
Ciao Paolo, purtroppo la r3 non consente un controllo “facile” dei motori stepper rispetto al driver che ho usato io quindi penso (non ho la shield) che richieda diverse modifiche perché funzioni.
Monday April 13th, 2015 - 10:02 PM
Ok, ti ringrazio davvero molto per la tempestiva risposta! incredibile!
Grazie Molte.
Tuesday April 21st, 2015 - 12:22 PM
Monday September 7th, 2015 - 09:43 PM
Hello
I apologize for my English.
I have a problem with the functioning of Arduino.
I have Arduino UNO R3 + LCD Shield. After starting set the speed of 5 RPM for 1 minute Arduino error. Arduino freezes. What is wrong ?
Thanks Pidrman
Tuesday September 8th, 2015 - 07:22 AM
Hi! after exactly one minute or is the time random? Could be some noise coming from the motor…
Wednesday January 27th, 2016 - 09:13 PM
Luca,
I need to control a bipolar, 2 phase stepper.
I would like to use your setup which I find it brilliant…
But… I need really low RPM…..0.25 – 0.5 – 1 rpm…
Is this achievable with this project ?
Sunday January 31st, 2016 - 01:31 PM
Hi Florin, if you increase the delay between two interrupts (`Timer1.initialize(xxx)`) you can achieve lower RPMs
Friday June 3rd, 2016 - 10:37 AM
Ciao Luca! Bel lavoro.. io ho i tuoi stessi pezzi, LCD pololu etc.. mi spieghi i collegamenti? In particolare da dove prendi i segnali di uscita per il motore? Sono due?
Wednesday June 29th, 2016 - 05:48 PM
hey, might want to declare prev_time as unsigned long in arduino 1.
Friday July 7th, 2017 - 10:19 PM
Hi luke, very nice project,i am happy to found you. but i need your help my LCD keypad shield is freezes after several minutes and the engine can no longer control.
Could you tell me what is wrong?
Thanks
Saturday July 8th, 2017 - 09:30 AM
Hi, it’s very difficult to tell you what’s happening… it seems that Arduino freezes: does it happen at the same time? I mean every n minutes or is it random?
Saturday July 8th, 2017 - 02:52 PM
Its happen randomly, all key buttons didn’t work. except, restart button to make lcd keypad works again but when i pressed restart button,the lcd is flickering and stepper motor make strange motion
Thanks Luke
Thursday July 13th, 2017 - 08:46 PM
Ma che bravo! Mi sta tornando la voglia di rimettermi a giocare con Arduino
Wednesday August 23rd, 2017 - 06:09 PM
Ciao Luca,
complimenti per i tuoi tutorial!
Mi servirebbe arrivare intorno ai 1000rpm. E’ possibile con questo setup?
Grazie.
Thursday August 24th, 2017 - 08:50 AM
Ciao! 1000 rpm significa (motore da 200 passi / giro) un impulso del timer ogni 5uS… in teoria (se il motore lo supporta) si può fare modificando `Timer1.initialize(5);`
Thursday August 24th, 2017 - 10:14 AM
Ahh ho capito, grazie mille Luca. Ora devo capire se tiene il carico, senza scaldare troppo, per qualche ora di rotazione continua.
Monday September 4th, 2017 - 01:08 PM
Ciao Luca, complimenti per la guida. ti vorrei chiedere come dovrei comportarmi per creare sketch per muovere a piccoli passi un motore nema 17 fino a compiere un giro completo (e poi fermarsi)..ho già impostato i microstep a 1/16..mi servirebbe per test di fotogrammetria, quindi che si muovesse a piccoli step e una pausa lunga tra di essi. grazie in anticipo
Thursday September 7th, 2017 - 01:36 PM
grazie Federico, per fare quello che serve a te basta che scrivi uno sketch che fa compiere al motore il numero esatto di step che compongono un giro… se il motore è da 220 step/giro e hai impostato un microstepping 1/16 significa fare uno sketch che esegue 3520 (220*16) “avanzamenti”.
Monday September 25th, 2017 - 03:47 PM
Ciao, intanto Complimenti, ottimo tutorial e grazie per la condivisione.
Volevo chiederti avrei esigenza di spazio e vorrei collegare questo lcd keybord ad arduino nano, devo per farlo comunque collegare tutti i pin?
Tipo come quando sovrapponi la scheda ad arduino uno ?
Oppure posso collegare solo alcuni pin ?
Se si, quali devo collegare?
Tuesday September 26th, 2017 - 08:52 AM
Ciao Angelo, ti rimando al sito ufficiale del produttore dello shield. In sintesi, “Pins 4, 5, 6, 7, 8, 9 and 10 are used to interface with the LCD. Just one Analog Pin 0 is used to read the five pushbuttons.” quindi almeno questi devi sicuramente collegarli.
Saturday March 24th, 2018 - 11:25 AM
sono alfeo dovrei fare una centralina per acquario che piloti pompa luce e ossigenatore la parte rele allo stato solido ce lo gia mi occorre la parte di programmazione un orologio con programmazione ora e giorno e accensione e spegnimento volevo un consiglio per il programma utilizzando arduino grazie
Sunday March 25th, 2018 - 09:43 AM
ciao Alfeo, la tua esigenza è molto comune… se cerchi con google “arduino timer” o simili trovi centinaia di progetti pronti da utilizzare
Thursday January 24th, 2019 - 09:47 AM
Ciao Luca, anche io come altri utenti ho il problema che si freeza tutto e si può solo resettare arduino.
La cosa è random e accade anche senza collegare driver e motore, ho provato a fare la stessa cosa anche usando lcd su i2c ma idem, funziona tutto, ma dopo pochi secondi si blocca tutto
Friday February 1st, 2019 - 02:37 PM
ciao! molto strano, ho appena riprovato lo sketch e va tranquillamente per decine di minuti…
Friday January 25th, 2019 - 08:44 PM
redefinition of ‘void setup()’
Friday February 1st, 2019 - 02:37 PM
I’ve just tested my “cntrSpeedDir.ino” sketch and it compiles perfectly… did you download it from my github repo?
Friday February 22nd, 2019 - 07:27 PM
Im trying to get your sketch to work with a TB6600 driver controller but having 0 luck. It just keeps locking up regardless of what speed I run at. Not having any knowledge of Arduino programming or C+ this has been frustrating to say the least. Im hoping you may be able to point me in the right direction for obtaining information on the changes needed to make your sketch work with my hardware. Im sure you are really busy and if you dont have the time I completely understand. Im very grateful for the information and the sketch you have provided the public.
Hardware
Single Axis TB6600 0.2-5A CNC Two-phase hybrid Driver Controller Stepper Motor
Dual Shaft Nema 23 CNC Stepper Motor Bipolar 3Nm(425oz.in) 4.2A 57x114mm 4 Wires
30:1 Geared Speed Reducer Nema23 Planetary Gearbox
1602 Serial Blue Backlight LCD Display Keypad 4 Arduino Uno R3 Mega 2560 Shield
ARDUINO UNO R3
Thank You
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Web Results
## Speed - Wikipedia
en.wikipedia.org/wiki/Speed
In everyday use and in kinematics, the speed of an object is the magnitude of its velocity ); it is thus a scalar quantity. The average speed of an object in an interval of time is the distance trav...
## Is velocity of an object moving in circular path changes continuously ...
www.quora.com/Is-velocity-of-an-object-moving-in-circular-path-changes-continuously-Explain
Because velocity is a vector quantity and in circular motion, the path is tangential which means ... Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio.
## Physical Science Midterm Flashcards
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Oct 15, 2012 ... downward acceleration of a freely falling object near the Earth's surface. .... rate at which an object covers distance relative to a frame of ...
## Difference between Speed and Velocity | Speed vs Velocity
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Key Difference: Speed is the rate at which an object covers a distance. Velocity not only determines the speed of the object in motion but also the direction of the ...
## Speed and Velocity - The Physics Classroom
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Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving ...
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Distance: refers to “how much ground an object has covered” during its motion. Speed: ... Speed can be thought of as the rate at which an object covers distance.
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Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving ...
## speed - ScienceFix - Science Fix
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Jul 25, 2010 ... Speed is the rate at which an object covers a distance. How do you know if that rate changes or not? The video shows how to determine if the ...
## Free Falling Object Motion - NASA
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May 5, 2015 ... Free falling object accelerates at a constant 9.8 m/sec^2. ... Time = 0, Accel = 9.8, Velocity = 0.0, Distance = 0.0 ... The remarkable observation that all free falling objects fall at the same rate was first proposed by Galileo, nearly ...
## Motion and Speed - Web Formulas
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When an object covers equal distance in equal time interval is called uniform ... For non-uniform motion, the rate of motion is described by their average speed.
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# [seqfan] Re: nice base-dependent sequence
Robert G. Wilson v rgwv at rgwv.com
Fri May 28 23:36:54 CEST 2010
```Et al,
I also like it.
Here is what I have written in Mathematica. But I am not happy with
it at all; particular the function 'gQ'.
lst = {0};
f[n_] := Block[{k = Max[lst[[-1]], 1]}, While[! gQ[k, n], k++]; k];
gQ[k_, n_] := Block[{i = 1, len = Floor[Log[10, k] + 1]},
s = Union[ FromDigits@# & /@ Flatten[ Table[ Partition[ IntegerDigits at k,
{j}, 1], {j, len}], 1]];
If[s[[1]] == 0, s = Rest at s];
While[i < n + 1 && Union[ IntegerQ@# & /@ (s/i)][[-1]] == True, i++ ];
i == n + 1];
Do[a = f at n; AppendTo[lst, f at n]; Print[{n, a}], {n, 100}]
I wrote it about a half an hour ago and so far I have only been able to
compute to f(32). This is far to so to create a nice b-text file.
I would really appreciate any one who has meaningful shortcuts.
Sincerely, Bob.
--------------------------------------------------
From: "Jonathan Post" <jvospost3 at gmail.com>
Sent: Friday, May 28, 2010 12:43 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: nice base-dependent sequence
> I like it too. But so long as someone's programming it, why not do so
> base j for bases other than 10, and show us the upper left corner of
> the array
> A[j,n] = n-th smallest integer k such that k or one of its substrings
> (base j) is divisible by every integer in {1, 2, ..., n}?
>
> Jonathan Vos Post
>
> On Fri, May 28, 2010 at 9:10 AM, N. J. A. Sloane <njas at research.att.com>
> wrote:
>> Dear Seq Fans, This just caught my eye. Anyone care to
>> produce a b-file?
>>
>> %I A177834
>> %S A177834
>> 1,2,6,12,45,54,56,56,245,504,1440,1440,5044,5044,10456,10569,11704,
>> %T A177834
>> 11704,11704,13608,13608,13608,26460,26460,198007,258064,264600,264600,
>> %U A177834
>> 475440,475440,1754608,1754608,2258064,2258064,2646004,2646004,2992520
>> %N A177834 Smallest integer k such that k or one of its substrings
>> (regarded as an integer) is divisible by every integer from {1,2,...,n}
>> %e A177834 a(8)=56 because 56 is divisible by 1,2,4,7,8; 5 is divisible
>> by 5; 6 is divisible by 3 and 6. Therefore the set {1,2,3,4,5,6,7,8} is
>> covered by the divisors. 56 is the smallest number with this property.
>> %K A177834 nonn,base,nice,new
>> %O A177834 1,2
>> %A A177834 Martins Opmanis (askola(AT)latnet.lv), May 14 2010
>> %E A177834 Edited by N. J. A. Sloane, May 28 2010.
>>
>> Neil
>>
>>
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
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# stats
Multivariate analysis of variance (MANOVA) table
Since R2023b
## Syntax
``s = stats(maov)``
``s = stats(maov,TestStatistic=testStat)``
``[s,H,E] = stats(___)``
## Description
example
````s = stats(maov)` returns a MANOVA table for the `manova` object `maov`. The MANOVA table contains statistics for the model terms, error, and total. ```
````s = stats(maov,TestStatistic=testStat)` specifies the test statistics to display in the MANOVA table.```
example
````[s,H,E] = stats(___)` also returns a table `H` of the hypothesis matrices for the MANOVA model terms, and the error matrix `E` used to perform the MANOVA, using any of the input argument combinations in the previous syntaxes.```
## Examples
collapse all
Load the `fisheriris` data set.
`load fisheriris`
The column vector `species` contains iris flowers of three different species: setosa, versicolor, and virginica. The matrix `meas` contains four types of measurements for the flower: the length and width of sepals and petals in centimeters.
Perform a one-way MANOVA with `species` as the factor and the measurements in `meas` as the response variables.
`maov = manova(species,meas);`
`maov` is a `manova` object that contains the results of the one-way MANOVA. Display the corresponding MANOVA table.
`s = stats(maov)`
```s=3×8 table Source DF TestStatistic Value F DFNumerator DFDenominator pValue _______ ___ _____________ ______ ______ ___________ _____________ __________ Factor1 2 pillai 1.1919 53.466 8 290 9.7422e-53 Error 147 Total 149 ```
The small p-value for `species` indicates that the flower species has a statistically significant effect on at least one of the flower measurements.
Load the `carsmall` data set.
`load carsmall`
The variable `Model_Year` contains data for the year a car was manufactured, and the variable `Cylinders` contains data for the number of engine cylinders in the car. The `Acceleration`, `Displacement`, and `Weight` variables contain data for car acceleration, displacement, and weight.
Use the `table` function to create a table from the data in `Model_Year`, `Cylinders`, `Acceleration`, `Displacement`, and `Weight`.
`tbl = table(Model_Year,Cylinders,Acceleration,Displacement,Weight,VariableNames=["Year" "Cylinders" "Acceleration" "Displacement" "Weight"]);`
Perform a two-way MANOVA using the table variables `Year` and `Cylinders` as factors, and the `Acceleration`, `Displacement`, and `Weight` variables as response variables.
`maov = manova(tbl,"Acceleration,Displacement,Weight ~ Cylinders + Year")`
```maov = 2-way manova Acceleration,Displacement,Weight ~ 1 + Year + Cylinders Source DF TestStatistic Value F DFNumerator DFDenominator pValue _________ __ _____________ _______ ______ ___________ _____________ _________ Year 2 pillai 0.11134 1.8471 6 188 0.092099 Cylinders 2 pillai 0.96154 29.012 6 188 1.891e-24 Error 95 Total 99 Properties, Methods ```
`maov` is a two-way `manova` object that contains the results of the two-way MANOVA. The small p-value for `Cylinders` indicates that enough evidence exists to conclude that `Cylinders` has a statistically significant effect on the mean response vector.
Return the hypothesis and error matrices for the MANOVA model terms.
`[~,H,E] = stats(maov)`
```H=3×2 table Year Cylinders ___________________________________ __________________________________ 33.703 -327.34 3443.7 278.01 -13017 -90619 -327.34 4835.3 -30382 -13017 7.1228e+05 4.9601e+06 3443.7 -30382 3.5753e+05 -90619 4.9601e+06 3.4541e+07 ```
```E = 3×3 107 × 0.0001 -0.0002 0.0021 -0.0002 0.0109 0.0451 0.0021 0.0451 1.3656 ```
The variables in the table `H` correspond to the MANOVA model terms of the same name. Each variable contains the hypothesis matrix for its corresponding MANOVA model term. The error matrix `E` contains the irreducible error for the MANOVA model. You can use `H` and `E` to perform hypothesis tests that are not supported by MATLAB® or Statistics and Machine Learning Toolbox™.
## Input Arguments
collapse all
MANOVA results, specified as a `manova` object. The properties of `maov` contain the response data and factor values used by `stats` to calculate the statistics in the MANOVA table.
MANOVA test statistics, specified as `maov.TestStatistic`, `"all"`, or one or more of the following values.
ValueTest NameEquation
`"pillai"` (default)Pillai's trace
`$V=trace\left({Q}_{h}{\left({Q}_{h}+{Q}_{e}\right)}^{-1}\right)=\sum {\theta }_{i},$`
where θi values are the solutions of the characteristic equation Qhθ(Qh + Qe) = 0. Qh and Qe are, respectively, the hypotheses and the residual sum of squares product matrices.
`"hotelling"`Hotelling-Lawley trace
`$U=trace\left({Q}_{h}{Q}_{e}^{-1}\right)=\sum {\lambda }_{i},$`
where λi are the solutions of the characteristic equation |QhλQe| = 0.
`"wilks"`Wilk's lambda
`$\Lambda =\frac{|{Q}_{e}|}{|{Q}_{h}+{Q}_{e}|}=\prod \frac{1}{1+{\lambda }_{i}}.$`
`"roy"`Roy's maximum root statistic
`$\Theta =\mathrm{max}\left(eig\left({Q}_{h}{Q}_{e}^{-1}\right)\right).$`
If you specify `testStat` as `"all"`, `stats` calculates all the test statistics in the table above.
Example: `TestStatistic="hotelling"`
Data Types: `char` | `string` | `cell`
## Output Arguments
collapse all
MANOVA table, returned as a table. In addition to rows for the error and total, `s` contains t rows per model term, where t is the number of test statistics in `maov.TestStatistic`. The table `s` also has the following columns:
• `Source` — MANOVA model term
• `DF` — Degrees of freedom for the term in `Source`
• `TestStatistic` — Name of the test statistic used to calculate the F-statistic in the column `F` and the p-value in the column `pValue`
• `Value` — Value of the test statistic named in `TestStatistic`
• `F` — Value of the F-statistic corresponding to the test statistic named in `TestStatistic`
• `DFNumerator` — Degrees of freedom for the numerator of the F-statistic
• `DFDenominator` — Degrees of freedom for the denominator of the F-statistic
• `pValue`p-value for the F-statistic
Data Types: `table`
Hypothesis matrices used to compute the F-statistics for the MANOVA model terms, returned as a table of matrices. Each column of `H` corresponds to a MANOVA model term in `maov.Formula`. For more information about `H`, see Qh in Multivariate Analysis of Variance for Repeated Measures.
Data Types: `table`
MANOVA model error matrix used to compute the F-statistics for the MANOVA model terms, returned as a numeric matrix. For more information about `E`, see Qe in Multivariate Analysis of Variance for Repeated Measures.
Data Types: `single` | `double`
## Version History
Introduced in R2023b
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Structural Equation Modeling (SEM)
What is a latent variable? What is an observed (manifest) variable? How does SEM handle measurement errors? Why does SEM have an advantage over regression and path analysis when it comes to multiple indicators? What are the two submodels in a structural equation model? What are their functions? How are observed correlations linked to the parameters of a structural equation model (via the diagram, that is)? Why can't we conclude cause and effect from structural equation models where there is no manipulation of variables? What is the danger of model fiddling?
Path analysis is a special case of SEM. Path analysis contains only observed variables, and has a more restrictive set of assumptions than SEM. Most of the models that you will see in the literature are SEM rather than path analyses. The main difference between the two types of models is that path analysis assumes that all variables are measured without error. SEM uses latent variables to account for measurement error.
Latent Variables
A latent variable is a hypothetical construct that is invoked to explain observed covariation in behavior. Examples in psychology include intelligence (a.k.a. cognitive ability), Type A personality, and depression. In the psychometrics literature, latent variables are also called factors, and have a rich history of statistical developments in the literature on factor analysis. The basic idea is that a latent variable or factor is an underlying cause of multiple observed behaviors. For example, suppose we have a test of vocabulary. The variance in response to each item in the test reflects individual differences in verbal ability across test takers, plus some error. In factor analysis, we would represent such a test as
Where V stands for the latent variable vocabulary, X1 through X4 stand for the items in the test, and e1 through e4 stand for measurement errors (unreliability) in each item. [Technical digression - the errors in factor analysis are actually measurement error plus stray causes, but the implications are the same for our purposes.] Points to notice:
1. We are assuming a single latent variable (factor) that corresponds to vocabulary.
2. The factor is assumed to cause the observed correlation among the 4 items.
3. The correlation among the items would be 1.0 except for the errors. [Recall the beginning of the course when we talked about what happens to correlations when we add random error to the observations.]
It turns out that the expected correlation between two variables that share a single factor is equal to the product of the two paths from that factor (analogous to a spurious effect in path analysis).
For example, suppose the path from V to X1 is .8 and the path from V to X2 is .9. Then the expected correlation between X1 and X2 is .72 (.72 = .8*.9). Recall that the correlation between X1 and X2 would be 1.0 if there were no measurement error. Therefore, measurement error accounts for the difference between the correlation of .72 and 1.0. That is, the correlation is reduced toward zero by the inclusion of less than perfectly reliable measures. Observed variables typically have some measurement error associated with them, and so their correlations with other variables are attenuated (too close to zero) due to the presence of this measurement error. Latent variables, on the other hand, are not directly measured and do not have measurement error associated with them. This is kind of a big deal because if we can estimate the association of latent variables, we can estimate the relations among variables without measurement error.
[Technical digression. There are ways to account for unreliability of measures for both correlation and regression coefficients, and to do so without estimating paths to assumed latent variables. That is, we can estimate the effects of unreliability directly with our observed variables. Both positions (with and without latent variables) have some merit.]
The other advantage of latent variables is that multiple indicators of the same construct are naturally handled with a structural equation model. In a regression model, multiple indicators cause collinearity problems and small increments in variance accounted for.
LISREL Notation
Most people follow the notation used the computer program LISREL (LInear Structural RELations) to talk about structural equation models. We can use the following diagram to illustrate the main points:
The latent variables or factors are indicated by circles. The observed variables are indicated by squares. The observed exogenous variables are labeled X. The latent exogenous variables are labeled ksi (x ). The observed endogenous variables are labeled Y; the latent endogenous variables are labeled eta (h ). The paths from the latent to the observed variables are labeled lamda (l ). The paths from the exogenous to the endogenous variables are labeled gamma (G ). The paths from the endogenous variables to other endogenous variables are labeled beta (b ). The correlations among the exogenous variables are labeled phi (f ). Finally, there are three kinds of errors. One kind of error is a stray cause of the latent endogenous variables, called psi (z ). There are also errors of the observed variables. For the observed exogenous variables, these errors are called delta (d ) and for the observed endogenous variables, these errors are called epsilon (e ). Each of these is labeled in the figure. (Not all labels for paths from the latent to the observed variables are included in the figure.)
Just as in path analysis, the diagram for the SEM shows the assumed casual relations. If the parameters of the model are identified, a covariance matrix or a correlation matrix can be used to estimate the parameters of the model, one parameter corresponding to each arrow in the diagram.
A Numerical Example
I am going to make up numbers for structural (path) coefficients for the model. Then I will present them in matrices that correspond to their LISREL notation. Finally I will combine them according the set of equations implied by the model to generate a correlation matrix. The path diagram looks like this:
There are two parts to a structural equation model, the structural model and the measurement model.
For the structural model, the equations look like this in matrix form:
This is an equation for predicting the values of endogenous variables (DVs). It says that the DVs are a function of the endogenous effects on themselves (the beta-eta part) plus the effects of the exogenous variables on the endogenous variables (gamma times ksi) plus the stray causes.
In our example, we have:
h 1 0 0 0 0 h 1 g 11 g 12 x 1 z 1 h 2 0 0 0 0 h 2 0 g 22 x 2 z 2 h 3 = b 31 b 32 0 0 h 3 0 0 z 3 h 4 0 0 b 43 0 h 4 0 0 z 4 h = b h + G x + z
If we substitute the numbers from the diagram, it looks like this:
h 1 0 0 0 0 h 1 .69 .03 x 1 z 1 h 2 0 0 0 0 h 2 0 .63 x 2 z 2 h 3 = .83 -.27 0 0 h 3 0 0 z 3 h 4 0 0 .59 0 h 4 0 0 z 4 h = b h + G x + z
Notice that beta (b ) and gamma (G ) are sets of parameters (path coefficients). The other entries --eta (h ), ksi (x ) and psi (z )-- are latent variables. This is the structural part of the model. It indicates how the latent variables are related.
The other part of the model is the measurement model. The measurement model indicates how the latent variables related to the observed variables. In our example, there are two parts, one for the exogenous variables and one for the endogenous variables:
X1 l 11 l 12 x 1 d 1 X2 = l 21 l 22 x 2 d 2 X3 l 31 l 32 + d 3 X4 l 41 l 42 d 4 X5 l 51 l 52 d 5 X = L X x + d
Our parameters were:
X1 .9 0 x 1 d 1 X2 = .9 0 x 2 d 2 X3 .9 0 + d 3 X4 0 .9 d 4 X5 0 .9 d 5 X = L X x + d
Notice that we have the path coefficients (L X) leading from the latent exogenous variables to the observed variables. These are the parameters of the model. The other items are variables.
For the endogenous variables, we have:
Y1 l 11 l 12 l 13 l 14 h 1 e 1 Y2 = l 21 l 22 l 23 l 24 h 2 e 1 Y3 l 31 l 32 l 33 l 34 h 3 + e 1 Y4 l 41 l 42 l 43 l 44 h 4 e 1 Y5 l 51 l 52 l 53 l 54 e 1 Y6 l 61 l 62 l 63 l 64 e 1 Y7 l 71 l 72 l 73 l 74 e 1 Y8 l 81 l 82 l 83 l 84 e 1 Y = L Y h + e
Our parameters were:
Y1 .9 0 0 0 h 1 e 1 Y2 = .9 0 0 0 h 2 e 1 Y3 0 .9 0 0 h 3 + e 1 Y4 0 .9 0 0 h 4 e 1 Y5 0 0 .9 0 e 1 Y6 0 0 .9 0 e 1 Y7 0 0 0 .9 e 1 Y8 0 0 0 .9 e 1 Y = L Y h + e
Here again we are looking at the paths that relate the latent endogenous variables to the observed endogenous variables.
To link the parameters of the model to the observed correlation matrix, note first that the correlation matrix can be split into 4 pieces:
X with X (5 by 5 in our example) X with Y (5 by 8 in our example) Y with X (8 by 5 in our example; this is the transpose of the upper right portion.) Y with Y (8 by 8 in our example).
To find the observed correlations in each of the four part of the correlation matrix, we need a different expression (well, the upper right and lower left are really the same). In one section we only have X variables, in one we only have Y variables, and in the other two, we need an expression for both X and Y. It turns out that the expressions for the correlations are:
X with X X with Y Y with X Y with Y
These equations may look a bit ugly (the sort of equation that only a mother could love), but remember that all we have here are a few matrices to add, subtract, multiply or invert. All of these except the error terms (theta & psi) are paths. Let's look at a couple of examples.
First, let's look at the correlations of the 5 observed X variables.
X with X
.9 0 .19 .9 0 1 .5 .9 .9 .9 0 0 .19 .9 0 .5 1 0 0 0 .9 .9 .19 0 .9 .19 0 .9 .19 L X f L 'X + q d
Note: the column for theta (q d ) is actually a diagonal matrix, that looks like this:
0.19 0 0 0 0 0 0.19 0 0 0 0 0 0.19 0 0 0 0 0 0.19 0 0 0 0 0 0.19
With a matrix of this order, we can add it to the product of the prior three matrices. It wouldn't fit on the same page in the table in its diagonal form, so I showed it as a column. The result of multiplying and adding the above matrices is the correlation matrix of the observed X variables:
X X1 X2 X3 X4 X5 X1 1 X2 .81 1 X3 .81 .81 1 X4 .405 .405 .405 1 X5 .405 .405 .405 .81 1
The correlations among the observed variables that belong to the same latent variable are all .81, because the paths from the latent variables to the observed variables are all .9, and we multiply the paths to get the expected correlation (.9*.9 = .81). The expected correlations among the observed variables with different latent variables are each equal to the path from the observed variable to the latent variable times the correlation of latent variables times the path from the latent variable to the other observed variable, that is .9*.5*.9 = .81*.5 = .405. Look at the path diagram to see how this works in the model.
One of the combinations of X and Y is
Y with X
The first part of the equation:
.9 0 0 0 -1 .9 0 0 0 0 .9 0 0 1 0 0 0 0 0 0 0 0 .9 0 0 0 1 0 0 - 0 0 0 0 0 0 .9 0 0 0 1 0 .83 -.27 0 0 0 0 .9 0 0 0 0 1 0 0 .59 0 0 0 0 .9 0 0 0 .9 L Y (I - b ) -1 8 x 4 4 x 4 Res: 8 x 4
The second part of the equation
.69 .03 1 .5 .9 .9 .9 0 0 0 .63 .5 1 0 0 0 .9 .9 0 0 0 0 G f L 'X 4 x 2 2 x 2 2 x 5 Res: 4 x 5
Total result: 8 x 5 (Y by X)
If we carry out the matrix operations, we get:
X1 X2 X3 X4 X5 Y1 .57 .57 .57 .30 .30 Y2 .57 .57 .57 .30 .30 Y3 .26 .26 .26 .51 .51 Y4 .26 .26 .26 .51 .51 Y5 .41 .41 .41 .11 .11 Y6 .41 .41 .41 .11 .11 Y7 .24 .24 .24 .07 .07 Y8 .24 .24 .24 .07 .07
Remember that there are three observed X variables (X1 to X3) for the first latent X, and the other two X variables refer to the second latent X. The Y variables come is pairs, one pair for each latent Y. To find the correlation between each X and each Y, we trace from one to the other, multiplying coefficients on a tracing and adding across tracings. (See the path diagram.) For example, suppose we want to know the correlation between X1 and Y1. There are two ways to get there - one direct path and one through the correlated cause.
For the direct path, we have .9*.69*.9 = .5589. For the indirect path, we have .9*.50*.03*.9 = .01215. When we add them together, we have .57105. Within rounding error, this is .57, which is what is shown in the table. The point of this exercise is to show you that SEM, just like path analysis, has a path diagram that implies a set of equations. The equations imply a set of correlations that can be tested against data that you collect. On the other hand, you can collect data (correlations) that will allow you to estimate the parameters contained in the equations.
Causal Modeling Revisited
Can you draw causal conclusions from an ordinary regression equation if the data were not collected from a design where the independent variable was manipulated?
(No, no, nooo, no.)
Path analysis is just a series of regressions applied sequentially to data. Can you draw causal conclusions from path analysis where no variables were manipulated? [It's just regression...]
SEM is basically just path analysis with latent variables. Can you draw causal inferences without manipulation in SEM? Incidentally, strictly speaking, manipulation doesn't allow one to draw causal inferences. It just helps to rule out LOTS of alternative explanations. In SEM we can fit the data with parameters, but we rarely to never rule out alternative explanations.
SEM: the good, the bad and the ugly
If you work in an area in which nonexperimental designs are common (industrial / organizational psychology and clinical psychology to name two), you must study SEM because it is widely used and is becoming required by reviewers for data analysis. Never mind that the reviewers are pretty much clueless about how to do this properly or even what it means. You still have to do it. All I have been able to do is introduce you to SEM concepts at an elementary level. You should take a course in it.
The big advantage of SEM. SEM allows for tests of theoretical propositions in nonexperimental data. You can test whether the factor structure of job satisfaction or the relations among personality characteristics is the same in the U.S. versus Japan with this. You can test whether a reciprocal path accounts for a specific relation in your data better than does a one-way causal flow. You can test quantitative predictions (e.g., the theory says that the path is .80) against data. This is powerful stuff.
What most people don't recognize. The power of the technique is rarely realized. It is very strong when you have parameters in mind and you want to test them against the data. People generally do not have any parameters in mind; they estimate them instead. To identify the problem (that is, to solve the identification problem so that all parameters have unique values), researchers assume that some parameters are zero in the population. The test of the model becomes a test of whether those parameters are zero in the population. This is generally not true (they are not zero in the population) and generally not very interesting (we are really interested in the parameters we estimated but did not test). That's not why this stuff was developed, but that's how everyone uses it. So far as I can tell, the situation is unlikely to change.
Model Fiddling. People develop a model, collect data, estimate the parameters, look at the fit statistics to see how well their data fit the model, and discover that the data fit the model rather poorly. This is roughly analogous to spending a year doing an experiment and having nonsignificant results. When this happens, it is definitely not okay. What do you do? Well, people begin fiddling with the model to make it fit. This usually takes a minimum of 1 month, and the resulting good-fitting model is generally not something you would have specified in advance. For this reason, I try to avoid doing SEM when I can. In my opinion, we generally don't have enough information to use the technique properly, that is, to conduct specific tests of meaningful hypotheses. However, my views on SEM are less enthusiastic than those of most researchers, who feel that its advantages outweigh its problems.
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Bus230AMinitest8AnsKey
# Bus230AMinitest8AnsKey - Net Book Value at end of Year 1...
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Bus 230A Minitest 8 Answer Key 1 b 2 b Combined Price of Land and Building \$260,000 \$260,000 Building at 70% 0.7 Buiding Value \$182,000 (182,000) Land Value \$78,000 3 a Equipment cost \$154,000 Less: Salvage Value (6,000) Depreciable Base \$148,000 Depreciation = 148,000/8yrs = \$18,500 Per Year 4 a If straight depreciation is 1/5 per year (20%) then 200% of straight-line is 2/5 per year (40%). First year depreciation is \$150,000 x 40% = \$60,000 Cost \$150,000 Less: First year's depreciation (60,000)
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Unformatted text preview: Net Book Value at end of Year 1 \$90,000 5 c 6 d 7 d 8 c Truck cost \$36,000 Less: Accumulated Depreciation (20,000) Net Book Value \$16,000 Sales price \$14,000 Less: Net Book Value (16,000) Loss on Sale \$(2,000) 9 a 10 c Extra Credit Problems 11 & 12: 11 b Net Income \$2,176,000 Add: Interest Expense 192,000 Total \$2,368,000 Return on Assets = (Net income + interest expense)/average total assets = 2,368,000/((19,042,000+21,209,000)/2) = 11.77% 12 c...
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## This note was uploaded on 04/16/2008 for the course BUS 230A taught by Professor S during the Spring '08 term at Sonoma.
Ask a homework question - tutors are online
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# What is a driven oscillation?
## What is a driven oscillation?
Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system. In such a case, the oscillator is compelled to move at the frequency νD = ωD/2π of the driving force.
### What is the formula for oscillation?
Key Equations
Relationship between frequency and period f=1T
Angular frequency of a physical pendulum ω=√mgLI
Period of a physical pendulum T=2π√ImgL
Period of a torsional pendulum T=2π√Iκ
Newton’s second law for harmonic motion md2xdt2+bdxdt+kx=0
What is meant by driven oscillation give an example for it?
If a sinusoidal driving force is applied at the resonant frequency of the oscillator, then its motion will build up in amplitude to the point where it is limited by the damping forces on the system.
What is damped and driven oscillation?
If a frictional force ( damping ) proportional to the velocity is also present, the harmonic oscillator is described as a damped oscillator. Driven harmonic oscillators are damped oscillators further affected by an externally applied force F(t).
## What is a driven harmonic oscillator?
If an external time-dependent force is present, the harmonic oscillator is described as a driven oscillator. Mechanical examples include pendulums (with small angles of displacement), masses connected to springs, and acoustical systems.
### What is the formula of frequency of oscillation?
The frequency f = 1/T = ω/2π of the motion gives the number of complete oscillations per unit time. It is measured in units of Hertz, (1 Hz = 1/s).
What is the formula of SHM?
simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. That is, F = −kx, where F is the force, x is the displacement, and k is a constant.
What are damped oscillations class 12 physics?
A damped oscillation means an oscillation that fades away with time. Examples include a swinging pendulum, a weight on a spring, and also a resistor – inductor – capacitor (RLC) circuit.
## What are forced oscillations Class 12 HSC?
When a body oscillates by being influenced by an external periodic force, it is called forced oscillation. Here, the amplitude of oscillation, experiences damping but remains constant due to the external energy supplied to the system.
### What is differential equation of motion for damped oscillation?
Fd = – pvWhere,v is the magnitude of the velocity of the object and p, the viscous damping coefficient, represents the damping force per unit velocity. The negative sign indicates that the force opposes the motion, tending to reduce velocity. In other words, the viscous damping force is a retarding force.
How do you calculate frequency of oscillation?
What is the equation of motion for a dridriven oscillator?
Driven Oscillator Consider a one-dimensional simple harmonic oscillator with a variable external force acting, so the equation of motion is L = 1 2 m x ˙ 2 − 1 2 k x 2 + x F t. (Landau “derives” this as the leading order non-constant term in a time-dependent external potential.)
## What is the equation of motion of a simple harmonic oscillator?
Consider a one-dimensional simple harmonic oscillator with a variable external force acting, so the equation of motion is L = 1 2 m x ˙ 2 − 1 2 k x 2 + x F t.
### What is a driven oscillator give an example?
Driven Oscillator Example. Since a resonant excitation will continue to contribute energy to the system at its natural frequency, the amplitude of oscillation will continue to grow until that energy is dissipated by the damping forces, or until something destructive happens to the system.
What is a damped driven oscillator?
Driven Oscillator. If a damped oscillator is driven by an external force, the solution to the motion equation has two parts, a transient part and a steady-state part, which must be used together to fit the physical boundary conditions of the problem. The initial behavior of a damped, driven oscillator can be quite complex.
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# VBA / Formatting Textbox To Number With 3 Decimal Places
Nov 21, 2012
I need to have textBox2. formatted to a number with 3 decimal places.
The code below is executed from the command button.
Code:
Private Sub CommandButton1_Click()
If IsNumeric(Me.TextBox1.Value) Then
Me.TextBox2.Value = Me.TextBox1.Value / 25.4
End If
End Sub
I found the code below but I cannot seem to figure it out
Code:
Private Sub TextBox2_Change()
TextBox2.Text = Format(Number, "0.000")
End Sub
## Define Number Of Decimal Places To Specific Textboxes On Multi Textbox Useform
May 16, 2012
I have a userform with 35 text boxes on which display data from a worksheet based on a selection made from a combobox.
My problem is that TextBox11 and TextBoxes 15-35 all need to show a value to 2 decimal places i.e 360.00.
I have found the following line of code:
Code:
For Each ctrl In Controls
If TypeName(ctrl) = "TextBox" Then ctrl.Value = Format(ctrl.Value, ".00")
Next ctrl
Which does the job but unfortunatley it applies it to every single textbox on the form (not just No's 11,15-35) which is a problem as some of the boxes contain dates so instead of 09/05/2012 i actually get 41038.00.
My question is how to I modifiy the above code (or is there an alternative code?) to only apply to the required textboxes?
## Take A Number With Several Decimal Places And Round It Up To Two Decimal Places
Feb 27, 2014
I need a formula to take a number with several decimal places and round it up to two decimal places to either .33, .66 or, .00 if its above .66.
For example, 4.23423423423423 will be 4.33
4.43453453533434 will be 4.66
4.8353453453 will be 5.00
Lets say the number is in cell A1. What formula would do this?
## Paste Two Decimal Number In Excel Without Extra Decimal Places Appearing
Aug 13, 2009
I have a vba macro that takes data from one workbook and pastes it into another workbook. In doing this I have declared a few variables of type single (I only need two decimal precision). However, when I copy the values from the cells on the source workbook and paste them into the target workbook, the numbers end up having 12 decimal places. Ultimately, this extra precision causes my totals to be off by .01 or more after a while. I have tried rounding the number as I pull it off the source workbook into the variable, but that didn't matter. How do I solve this problem? Code for pulling data from source workbook:...
## Restrict TextBox Entries To X Decimal Places
Oct 12, 2007
I have following for change event in text boxes to only allow numerics e.g.
Private Sub txtGBP10_Change()
If (Not IsNumeric(txtGBP10.Value) And (txtGBP10.Value <> "")) Then
txtGBP10.Value = Left(txtGBP10.Value, Len(txtGBP10.Value) - 1)
End If
End Sub
Private Sub txtShare10_Change()
If (Not IsNumeric(txtShare10.Value) And (txtShare10.Value <> "")) Then
txtShare10.Value = Left(txtShare10.Value, Len(txtShare10.Value) - 1)
End If
End Sub
Can I add some code so that the user can only add numeric entries to 2 DP (txtGBP) or 3DP(txtShare)
## Formatting Of A Cell To 2 Decimal Places
Feb 24, 2010
if I change the formatting of a cell to 2 decimal places, it appears as two decimal places (as it should) for example \$88.88888 will show as \$88.88
However, when I use this data in another application that displays this data it will display as 88.88888 still. I need to actually take the value and truncate it to 88.88 eg 8.8888888 will become 8.88
I have been using trunc by hand and wanted to try and find out if there is a way that I could write some sort of macro to do this for me each time.
## Excel 2010 :: Convert Text To Number And Format Number Without 2 Decimal Places?
Oct 23, 2011
I have a problem that when I try to convert text to number and format the number without 2 decimal places as seen on the link I have given below, Instead of 1607.947, I get 1607947. I have Excel 2010 loaded. The details are in below picture.
[URL]
## Count Number Of Decimal Places
Jan 21, 2005
I want to show (in a cell) how much decimals there are after a comma.
## How To Format A Number Into 2 Decimal Places
Jun 5, 2014
Is there a way to format a number into 2 decimal places and when you select the cell you wont see the true value?
For example:
316.2696
I still see the 316.2696 when I click the cell. Instead of 316.27 only.
## Format Number Of Decimal Places To Show
Mar 24, 2009
Is there a way to format a cell based upon a condition? If the cell value is <1, I want to show two decimal places. If the cell is >1, I want to show zero decimal places. I tried to use the conditional formatting, but there is no option for this.
## VBA Function To Find Number Of Decimal Places?
Dec 27, 2011
I need a function to find the number of decimal places of a certain number (in this specific case doubles)
The first solution would be something like this:
Code:
'returns the number of decimal places within a double
Public Function getDecPlaces(inputNum As Double) As Long
Dim ndx As Long
ndx = InStr(1, inputNum, ".")
If ndx > 0 Then
getDecPlaces = Len\$(CStr(inputNum)) - ndx
End If
End Function
But i feel there is likely a much better way of doing this..
## Limit Cell Entry To Two Decimal Places Or Whole Number?
May 29, 2014
content from form is captured via formulas to a CIMLoad format on another tab
Users keep entering four to six decimal places in weight value
I need to limit the cell to either whole numbers, or a maximum of 2 decimal places
The cell is formatted to two decimal places, but when you activate the cell it shows the full value they entered, and so does my CIMLoad
How to I limit the value in this cell to whole numbers or numbers with a maximum of two decimal places?
## Number Showing Extra Decimal Places In Formula Bar
Aug 22, 2009
If you look at the attached file in Column E line 47 you will see that it displays
373.97. However if you look in the formula bar it shows 373.9694. I am trying to make it so that the formula bar shows the same number as in the cell. Any idea on how to do this? I have tried everything I know how to do, which is not much.
## Excel 2007 :: Changing Number To Add Decimal Places
Jan 30, 2013
I am using Excel 2007 and importing a list of coordinates. The coordinates are in a text file and separated by commas, so I'm opening the file as a delimitted file. All of the coordinates import correctly except one. The value is supposed to 52530.6372, but Excel is automatically changing it to 52530.6371999999
If I manually type a "2" to replace the "1999999", it changes it back. If I type "3" it stays as a 3.
I added a round function to force it to round to 4 decimal places and that seemed to work until I then tried using the number with text commands. When I used an ampersand to insert the number into a text string, the additional decimal places are back, even with a round function!
## Variable Type Changing Number Of Decimal Places
May 14, 2014
Here's what I'm doing: I'm using a macro to assign a cell value to a variable then set another cell value to the variable instead of copy/paste (because even pasting values only was affecting other formulas in the file for some reason)
The problem: in using the macro, the number being 'copied' is acquiring a few extra decimal places IE 38334.61 is the original number and 38334.609375 is what I end up with. The numbers come from a CSV with only 2 decimal places and I checked by adding decimal places in the format so it's not a formatting/visible digits issue. Since the values are hour meter readings, the extra decimal places end up with very small values outside 0-24 hrs which messes with sorting and usage %. The values are so small all of them together in a month add up to a fraction of a cent but it's one of those things that bugs my OCD by not being right.
My VBA book explains the min/max capabilities, content type, memory bits but not fiddly details like this.
So my question is this: right now I'm declaring the clipboard-substitue variable as an single, is there a different one that would work better without adding anything? Criteria are: numeric, 6 digits before the decimal, 2 after, all positive values. (Ie 123456.12)
EDIT: fixed my senior moment.
## Removing Decimal Point While Maintaining # Of Decimal Places
Jun 10, 2009
I need to convert a column of numbers currently formatted with 2 decimal places e.g. 112.12 to 4 decimal places (without the decimal point). I need the end result to be 1121200. I've tried a few different suggestions given on the forum previously but can't seem to retain the 4 decimal places that I require.
## Decimal Places And INT?
Aug 5, 2013
I was going to use the following to test that a number has no more than two decimal places,
Code:
If Int(ActiveCell.Offset(0, 8) * 100) ActiveCell.Offset(0, 8) * 100 Then
However it is rather mysteringly failing when activecell = 16.99 or 17.99; on testing it appears that vba is evaluating Int(ActiveCell.Offset(0, 8) * 100) to equal 1698 or 1798
## Set Decimal Places By Cell Value
Jun 22, 2009
I am converting values from SAE to Metric and wish to retain the decimal places of the SAE value before converting.
My code below converts nicely but doesnt maintain decimal places.
## Set Decimal Places In Cell?
Jan 16, 2010
What do I need to add to this code to set the decimal places to zero?
## Concatenate And Decimal Places
Aug 2, 2006
I am facing a problem with Excel's Concatenate function. I am trying to make a text string with numbers from a cell. =CONCATENATE( "price paid= ", D23). D23 is a numreric cell, formatted for one decimal place. However, the text produced by above function is showing me two decimal places of the number in D23.
## Fix Decimal Places On Entry
Sep 4, 2006
I am creating a Daily Cash Count worksheet for a business to do their daily closeout paperwork. What I am trying to do is automatically apply decimal formatting to a cell. Examples, if they count 60 cents in nickles and enter 60 in the cell, it automatically converts it to .60. And if they count 7.50 cents in quarters and enter 750 in cell it automatically would convert it to 7.50.
## Round To More Than 30 Decimal Places
Oct 16, 2006
I'm trying to work out a formula in excel which requires me to use Pi to over 30 decimal places. Excel will only let me have 30 decimal places of Pi whether I copy and paste it as a number or use its Pi function and even then seems to round up. Is there any way I can get more than 30 decimal places for this calculation? If not in excel then can anyone suggest another programme that may be capable of this? You can post here or contact me at [email=" Deleted by Jack in the UK[/EMAIL]
## Consistent Decimal Places
Nov 30, 2006
Is it possible to apply some kind of formatting to a range, that will force all numbers entered into that range to have the same number of decimal places as the cell with the maximum number of decimal places?
## Fixed Decimal Places
Jun 6, 2007
Everytime I enter a number, excel automatically converts it to a decimal number.
I type 1 in a cell (or formula box)
Excel returns the value as 0.1
I think it may have to do with the FIX box highlighted to the bottom of the screen. (To the right side of the NUM lock). (I can't paste a screen dump to show the FIX box position). how to remove the FIX box, and/or change excel so that I can enter in numbers normally.
## Formatting Number With Two Decimal Values
Jul 10, 2014
Some of my numeric values are with single decimals .....
Eg:
1542.2
9856.5
659855.9
2589.7
When I format in Excel with Number with 2 decimals my results remain the same
But the result which is need is:
1542.20
9856.50
659855.90
2589.70
## Roundup Function - Two Decimal Places
Mar 13, 2014
I'm on Excel 2010 and I need to round up a complete price list to a 0.90 decimal.
Here is the formula that I have in my cell : =((J5-(J5*N\$2))*M5) I need to round the result to XXX.90
## Show As Many Or As Little Decimal Places In Any Cell
Dec 11, 2008
I am doing some calculations via Excel, and I have found that I get different results using a calculator. I know that by formatting the number, I can show as many or as little decimal places as I like in any cell, but can anyone tell me how many decimal places Excel actually uses when it's carrying out calculations? I have a feeling that the Excel results I'm getting might be more accurate than my calculator ones.
## Using The Accounting Format With No Decimal Places
Jan 15, 2009
I am using Excel 2003. I am attempting to use the Accounting format with numbers that should not have any decimal places (although what is entered might have a decimal place). The numbers line up fine on the right, however, the dollar signs on the left are not lining up. It looks something like:
## Converting Units And Decimal Places.
Nov 23, 2009
I have a simple spreadsheet that allows the user to enter a dimension in metric or inches. I want to display the other units in the adjacent cell. In cell A1, the units are "Metric" or "Inch" in a pull down list. In cell A2, the value is entered.
In cell A3 i want to show the value in the other units. So if A1 is Metric, then take A2 and divide by 25.4. And if A1 is Inch, then take A2 and multiply by 25.4.
Also, if A1 is Inch, then display 2 decimal places in A3, and if A1 is Metric, then display 3 decimal places in A3. Is this possible?
## Rounding Decimal Places In Increments?
Feb 11, 2013
How do I write a formula to round the decimal places in a number in set increments. For example;
a) 14.28 to the nearest 0.25 would result 14.25
b) 1.99 to the nearest 0.1 would result 2.0
c)2.97 to the nearest 0.25 would result 3.00
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Excel discount
Hi,
I have a column that I would like to apply a percentage discount too in excel, I'd like the user to be able to change the percentage discount by simply entering a new number in a cell, say for example today he wants to add a 10% discount to all prices in column D, when he types in the number the new prices in column D now have the discounted price.
also would it be possible for him to change the price in one cell of column D while always keeping the discounted price? Say he says today he feels like the toaster will 300\$, he types in 300 in one of the cells in column D, then the 300 gets mutiplied by the discounted number in the other cell?
Thank you,
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Commented:
It would be simple if you can use a separate column for the discounted price. Then you can just use a formula like:
=D2*(1-\$G\$1%)
where he enters say 10 in G1.
If you really want the original values in column D to change (I would not recommend it), you would need code.
0
Finance AnalystCommented:
Assuming Base Price in D2 and discount as percentage in F1, in E2:
=D2*(1+\$F\$1)
If user wants option available to apply discount to some lines and not others, have column with Y or N option, eg Column G:
=D2+IF(G2="Y",D2*\$F\$1,0)
Thanks
Rob H
0
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# Question about Recombination Redshift values
1. Dec 13, 2015
### Dr. Strange
Wikipedia gives the redshift of recombination at z = 1500 (roughly 4,000 K). The Plank report list a 'redshift for which the optical length equals unity' and gives a value of 1090. What is the difference between the two?
2. Dec 13, 2015
### Chronos
z=1500 is approximated based on achieving a 50% ionization fraction under the Saha equation. This calculation, however, assumes the universe was in a state of thermal equilibrium, which it was not. Under the Peebles equation, which does not assume equilibrium, the 50% ionization fraction is achieved around z=1210. The value of z=1090 merely represents the ratio between the reionization temperature of hydrogen and the current measured temperature of the CMB.
3. Dec 14, 2015
### Dr. Strange
Sorry, I didn't follow all of that. In layman's terms, which value represents the red-shift where the electrons travelled freely through the universe. Also, if it isn't asking too much, could you also give a layman's description of optical depth, τ. I see it's a parameter to ΛCDM and if you Google it, you get tons of information about various attempts to measure and constrain it, but I can't find a conceptual guide to tell me what this parameter is physically.
4. Dec 14, 2015
### Buzz Bloom
5. Dec 14, 2015
### Dr. Strange
6. Dec 14, 2015
### Buzz Bloom
Hi Dr. Strange:
You question requires expertise way over my head. The two people who have been particularly helpful in answering my questions about recombination are
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52,158 Pages
## Job Trait Overview
• Game Description: Increases the rate at which TP is gained.
• Job Traits are always active.
• Further Notes:
• Each upgrade of this trait further increases character's rate of TP gain.
• The amount of additional TP gained is based on this trait's Trait Value; Store TP+10 is equal to a 10% increase in TP gain per swing, meaning the actual increase to TP/hit will vary based on the weapon's base TP/hit.
• TP/hit from Store TP is calculated as (Base TP/hit) * (1 + (Total Store TP / 100))
• Base TP/hit varies depeding on the amount of delay the weapon you are using has. Refer to this list below:
Delay Range: Equation 0 - 180: 5.0 + [(Delay - 180) * 1.5] / 180 180 - 450: 5.0 + [(Delay - 180) * 6.5] / 270 450 - 480: 11.5 + [(Delay - 450) * 1.5] / 30 480 - 530: 13.0 + [(Delay - 480) * 1.5] / 50 530 - 999: 14.5 + [(Delay - 530) * 3.5] / 470
For example, Hagun has a 450 base delay, so its base TP per hit is: 5.0 + [(450 - 180) * 6.5] / 270 = 11.5 TP
• When calculating TP after accounting for Store TP, FFXI drops all decimal places after the first. For example, 11.49% adjusted TP per hit will count as exactly 11.4% TP per hit.
• For example, suppose a player without this trait lands an attack and gets back 10%TP.
• If a player with Store TP I lands the same attack, he/she will get back 11%TP (10% more).
• If a player with Store TP II lands the same attack, he/she will get back 11.5%TP (15% more).
• Another example, suppose a player without this trait lands an attack and gets back 15%TP.
• If a player with Store TP I lands the same attack, he/she will get back 16.5%TP(10% more).
• If a player with Store TP II lands the same attack, he/she will get back 17.2%TP (15% more).
• Additional hits from multi-hit weapon skills or double/triple attacks are also affected by Store TP with the same method.
• For example, any additional hit with Store TP II will be (1 TP/hit) * (1 + (15 / 100)), or 1.1 TP/hit (when appropriately rounded down).
#### Store TP I
Trait Value: 10
Trait Value +5~20 (Varies by Maneuver)
#### Store TP II
• Obtained: Samurai Level 30
• Trait Value: 15
#### Store TP III
• Obtained: Samurai Level 50
• Trait Value: 20
#### Store TP IV
• Obtained: Samurai Level 70
• Trait Value: 25
#### Store TP V
• Obtained: Samurai Level 90
• Trait Value: 30
## Merits that Enhance this Trait
• This trait can be enhanced by the Samurai Group 1 Merit Store TP Effect. Each merit gives +2 Store TP, for a maximum of +10.
• Combined with Samurai's Store TP job trait, these merits can increase base Store TP up to 40 (without gear).
## Blue Magic Spells that Activate this Trait
Set any two of the following Blue Magic spells to get this trait:
Level AvailableSpellSet Point Cost
48Sickle Slash4
69Tail Slap4
85Fantod1
95Sudden Lunge4
## Phantom Rolls that Activate this Trait
Level AvailableSpellStore TP Value
37Samurai Roll-5 to +50
## Ninjutsu that Enhance this trait
Level AvailableSpellStore TP Value
93Kakka: Ichi+10
## Food that Enhances this Trait
Shrimp Sushi (+2) Shrimp Sushi +1 (+2) Arrabbiata (+5) Arrabbiata +1 (+5) Peperoncino (+5) Peperoncino +1 (+6) Carbonara (+6) Carbonara +1 (+6) Nero di Seppia (+6) Nero di Seppia +1 (+6) Pescatora (+6) Pescatora +1 (+6) Vongole Rosso (+6) Vongole Rosso +1 (+6) Spaghetti Marinara (+7) Ambrosia (+7)
## Equipment that Enhances this Trait
Name Type Level Bonus
Katana Obi Waist 30 Latent Effect: +1
Rajas Ring Rings 30 +5
Fidelity Mantle Back 30 Pet:+3
Hikazu Hara-Ate Body 35 +1
Hikazu Sune-Ate Feet 35 +1
Haraldr's Mufflers Neck 40 +1, Ice: +5
Ecphoria Ring Rings 49 +1
Cancer Subligar Legs 50 +1
Shinimusha Haidate Legs 50 +6
Shinimusha Hara-Ate Body 50 +9
Bushido Cape Back 60 +1
Chivalrous Chain Neck 60 +1
Fourth Mantle Back 60 +2
Omokage Great Katana 60 +2
Rose Strap Grip 60 +4, Campaign: +20
Cobra Harness Body 68 +6
Cobra Leggings Feet 68 +4
Cobra Mittens Hands 68 +3
Cobra Subligar Legs 68 +3
Acantha Shavers Hand-to-Hand 69 +1
Danzo Tekko Hands 69 Campaign: +5
Haten Earring Earrings 69 Latent Effect: +2
Malagigi's Trousers Legs 69 Campaign: +5
Rainmaker Axe 69 +1
Attila's Earring Earrings 70 +1
Engetsuto Polearm 70 +1
Hachiman Domaru Body 70 +6
Hachiman Domaru +1 Body 70 +7
Hachiman Hakama Legs 70 +3
Hachiman Hakama +1 Legs 70 +4
Hachiman Kote Hands 70 +8
Hachiman Kote +1 Hands 70 +9
Hachiman Sune-Ate Feet 70 +5
Hachiman Sune-Ate +1 Feet 70 +6
Kosetsusamonji Great Katana 70 Assault: +2
Nobushi Kyahan Feet 70 +4
White Tathlum Ammunition 70 +2
Airy Buckler Shield 71 +2
Aurum Cuirass Body 72 +7
Cinquedea Dagger 72 Latent Effect +5
Enkidu's Subligar Legs 72 +5
Mekki Shakki Staff 73 +5
Almah Torque Neck 74 +2
Labrys Axe 74 -5
Myochin Haidate +1 Legs 74 +4
Saotome Domaru Body 74 +3
Brutal Earring Earrings 75 +1
Futsuno Mitama Great Katana 75 +8
Hachiryu Haramaki Set Full Set 75 +20
Pachipachio Great Katana 75 +2
Psi Ring Rings 75 Salvage: +5
Rindomaru Great Katana 75 (varies) 5~16
Saotome Domaru +1 Body 75 +5
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DISCOVER
# Activities for Investigating Magnets
Updated April 17, 2017
Experiments with magnets appeal to children of many ages. Invisible magnetic forces amaze children by "magically" moving metal objects before their eyes. Demonstrate attraction, repulsion, and polarity by using common classroom supplies and introduce the corresponding science terms with each activity. Allow children plenty of time to conduct their own investigations with magnets.
## Magnet Strength
Devise a simple test to measure the strength of a magnet. The KidsGen website suggests using a rubber band to attach two metal washers to the end of a ruler. Attach a magnet to the opposite end of the ruler. Tape a cardboard tube to the table, then lay the ruler across it like a see-saw, positioning the end with the magnet up in the air. Hold a second magnet directly over the first, slowly moving it down until the ruler tips over. Without moving the second magnet, use another ruler to measure its height above the table. Repeat the experiment with different magnets to compare their strengths. Magnets that must be held very close to the first magnet in order to tip the ruler are weaker than those that can tip the ruler from father away.
## Chaotic Magnets
According to the "Exploratorium" website, magnets can create movement in chaotic ways yet simultaneously follow an orderly pattern. Explore this apparent contradiction by creating a magnetic pendulum. Place four circular magnets on top of each other in a stack to align their poles. Mark each magnet to indicate its north pole. Attach a ring stand to a clamp, then tie one of the magnets to the end of the ring stand so that it hangs freely. Place the other magnets on the base of the ring stand in a triangle shape, making sure the north poles are on top. Raise the clamp so that the hanging magnet cannot touch the other magnets or the base. Push the hanging magnet and observe the interactions of the various magnetic fields. Move one of the magnets to a slightly different position on the ring stand base to create a new pattern of movement.
## Magnetic Cars
The "Kids Science Experiments" website describes a demonstration involving toy cars. Place bar magnets on top of two toy cars and secure them in place with a rubber band. Position the cars so that the magnets face each other, leaving a small space in the middle. If the like poles of the magnets point toward each other, the magnetic repulsion will push the cars apart. If the opposite poles of the magnets face each other, the magnetic attraction will pull the cars together.
## Magnetic Maze
Design a magnetic maze, as suggested by the "Kids Science Experiments" website. Draw a maze on a thin piece of cardboard. Place a paper clip, metal washer, or safety pin on top of the maze. Hold a magnet under the cardboard and use it to move the metal object through the maze. Alternatively, pour a small amount of iron filings onto the cardboard and move the magnet to guide them through the maze.
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# Lexicographically smallest string obtained after concatenating array
Given n strings, concatenate them in an order that produces the lexicographically smallest possible string.
Examples:
```Input : a[] = ["c", "cb", "cba"]
Output : cbacbc
Possible strings are ccbcba, ccbacb,
cbccba, cbcbac, cbacbc and cbaccb.
Among all these strings, cbacbc is
the lexicographically smallest.
Input : a[] = ["aa", "ab", "aaa"]
Output : aaaaaab```
One might think that sorting the given strings in the lexicographical order and then concatenating them produces the correct output. This approach produces the correct output for inputs like [“a”, “ab”, “abc”]. However, applying this method on [“c”, “cb”, “cba”] produces the wrong input and hence this approach is incorrect.
The correct approach is to use a regular sorting algorithm. When two strings a and b are compared to decide if they have to be swapped or not, do not check if a is lexicographically smaller than b or not. Instead check if appending b at the end of a produces a lexicographically smaller string or appending a at the end of b does.
This approach works because we want the concatenated string to be lexicographically small, not the individual strings to be in the lexicographical order.
Steps to solve this problem:
1. sort the string a from start index to end index.
2. declare an answer string as blank.
3. iterate through i=0 till n:
Implementation:
## C++
## Java
`// Java code to find the lexicographically``// smallest string` `class` `GFG {`` ` `// function to sort the``// array of string``static` `void` `sort(String a[], ``int` `n)``{`` ` ` ``//sort the array`` ``for``(``int` `i = ``0``;i < n;i++)`` ``{`` ``for``(``int` `j = i + ``1``;j < n;j++)`` ``{`` ` ` ``// comparing which of the`` ``// two concatenation causes`` ``// lexicographically smaller`` ``// string`` ``if``((a[i] + a[j]).compareTo(a[j] + a[i]) > ``0``)`` ``{`` ``String s = a[i];`` ``a[i] = a[j];`` ``a[j] = s;`` ``}`` ``}`` ``}``}`` ` `static` `String lexsmallest(String a[], ``int` `n)``{`` ` ` ``// Sort strings`` ``sort(a,n);` ` ``// Concatenating sorted strings`` ``String answer = ``""``;`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``answer += a[i];` ` ``return` `answer;``}` `// Driver code``public` `static` `void` `main(String args[])``{`` ``String a[] = {``"c"``, ``"cb"``, ``"cba"``};`` ``int` `n = ``3``;`` ``System.out.println(``"lexicographically smallest string = "`` ``+ lexsmallest(a, n));` `}``}` `// This code is contributed by Arnab Kundu`
## Python 3
`# Python 3 code to find the lexicographically``# smallest string``def` `lexSmallest(a, n):`` ``# Sort strings using above compare()`` ``for` `i ``in` `range``(``0``,n):`` ``for` `j ``in` `range``(i``+``1``,n):`` ``if``(a[i]``+``a[j]>a[j]``+``a[i]):`` ``s``=``a[i]`` ``a[i]``=``a[j]`` ``a[j]``=``s` ` ``# Concatenating sorted strings`` ``answer ``=` `""`` ``for` `i ``in` `range``( n):`` ``answer ``+``=` `a[i]`` ``return` `answer` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` ` ``a ``=` `[ ``"c"``, ``"cb"``, ``"cba"` `]`` ``n ``=` `len``(a)`` ``print``(lexSmallest(a, n))` `# This code is contributed by vibhu karnwal` ` `
## C#
`// C# code to find ``// the lexicographically``// smallest string``using` `System;` `class` `GFG {`` ` `// function to sort the``// array of string``static` `void` `sort(String []a, ``int` `n)``{`` ` ` ``//sort the array`` ``for``(``int` `i = 0;i < n;i++)`` ``{`` ``for``(``int` `j = i + 1;j < n;j++)`` ``{`` ` ` ``// comparing which of the`` ``// two concatenation causes`` ``// lexicographically smaller`` ``// string`` ``if``((a[i] + a[j]).CompareTo(a[j] +`` ``a[i]) > 0)`` ``{`` ``String s = a[i];`` ``a[i] = a[j];`` ``a[j] = s;`` ``}`` ``}`` ``}``}`` ` `static` `String lexsmallest(String []a, ``int` `n)``{`` ` ` ``// Sort strings`` ``sort(a,n);` ` ``// Concatenating sorted `` ``// strings`` ``String answer = ``""``;`` ``for` `(``int` `i = 0; i < n; i++)`` ``answer += a[i];` ` ``return` `answer;``}` `// Driver code``public` `static` `void` `Main()``{`` ``String []a = {``"c"``, ``"cb"``, ``"cba"``};`` ``int` `n = 3;`` ``Console.Write(``"lexicographically smallest string = "`` ``+ lexsmallest(a, n));` `}``}` `// This code is contributed by nitin mittal`
## Javascript
``
Output
`cbacbc`
Complexity Analysis:
• Time complexity : The above code runs in O(M * N * logN) where N is number of strings and M is maximum length of a string.
• Auxiliary Space: O(n)
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CHAPTER VIII.
ON TERRESTRIAL MAGNETISM.
465.] Our knowledge of Terrestrial Magnetism in derived from the study of the distribution of magnetic force on the earth's surface at any one time, and of the changes in that distribution at different times.
The magnetic force at any one place and time is known when its three coordinates are known. These coordinates may be given in the form of the declination or azimuth of the force, the dip or inclination to the horizon, and the total intensity.
The most convenient method, however, for investigating the general distribution of magnetic force on the earth's surface is to consider the magnitudes of the three components of the force,
${\displaystyle X=H\cos \delta }$, directed due north, } (1) ${\displaystyle Y=H\sin \delta }$, directed due west, ${\displaystyle Z=H\tan \theta }$, directed vertically downwards,
where ${\displaystyle H}$ denotes the horizontal force, ${\displaystyle \delta }$ the declination, and ${\displaystyle \theta }$ the dip.
If ${\displaystyle V}$ is the magnetic potential of the earth's surface, and if we consider the earth a sphere of radius ${\displaystyle a}$, then
${\displaystyle X={\frac {1}{a}}{\frac {dV}{dl}}}$, ${\displaystyle Y={\frac {1}{a\cos l}}{\frac {dV}{d\lambda }}}$, ${\displaystyle Z={\frac {dV}{dr}}}$
(2)
where ${\displaystyle l}$ is the latitude, and ${\displaystyle \lambda }$ the longitude, and ${\displaystyle r}$ the distance from the centre of the earth.
A knowledge of ${\displaystyle V}$ over the surface of the earth may be obtained from the observations of horizontal force alone as follows.
Let ${\displaystyle V_{0}}$ be the value of ${\displaystyle V}$ at the true north pole, then, taking the line-integral along any meridian we find,
${\displaystyle V=a\int _{\frac {\pi }{2}}^{l}Xdl+V_{0}}$,
(3)
for the value of the potential on that meridian at latitude ${\displaystyle l}$.
Thus the potential may be found for any point on the earth's surface provided we know the value of ${\displaystyle X}$, the northerly component at every point, and ${\displaystyle V_{0}}$, the value of ${\displaystyle V}$ at the pole.
Since the forces depend not on the absolute value of ${\displaystyle V}$ but on its derivatives, it is not necessary to fix any particular value for ${\displaystyle V_{0}}$.
The value of ${\displaystyle V}$ at any point may be ascertained if we know the value of ${\displaystyle X}$ along any given meridian, and also that of ${\displaystyle Y}$ over the whole surface.
Let
${\displaystyle V_{l}=a\int _{\frac {\pi }{2}}^{l}Xdl+V_{0}}$
(4)
where the integration is performed along the given meridian from the pole to the parallel ${\displaystyle l}$, then
${\displaystyle V=V_{l}+a\int _{\lambda _{0}}^{\lambda }Y\cos ld\lambda }$,
(5)
where the integration is performed along the parallel ${\displaystyle l}$ from the given meridian to the required point.
These methods imply that a complete magnetic survey of the earth's surface has been made, so that the values of ${\displaystyle X}$ or of ${\displaystyle Y}$ or of both are known for every point of the earth's surface at a given epoch. What we actually know are the magnetic components at a certain number of stations. In the civilized parts of the earth these stations are comparatively numerous; in other places there are large tracts of the earth's surface about which we have no data.
Magnetic Survey.
466.] Let us suppose that in a country of moderate size, whose greatest dimensions are a few hundred miles, observations of the declination and the horizontal force have been taken at a considerable number of stations distributed fairly over the country.
Within this district we may suppose the value of ${\displaystyle V}$ to be represented with sufficient accuracy by the formula
${\displaystyle V=V_{0}+a(A_{1}l+A_{2}\lambda +{\frac {1}{2}}B_{1}l^{2}+B_{2}l\lambda +{\frac {1}{2}}B_{3}\lambda ^{2}+\&c.)}$,
(6)
whence
${\displaystyle X=A_{1}+B_{1}l+B_{2}\lambda }$,
(7)
${\displaystyle Y\cos l=A_{2}+B_{2}l+B_{3}\lambda }$.
(8)
Let there be ${\displaystyle n}$ stations whose latitudes are ${\displaystyle l_{1}}$, ${\displaystyle l_{2}}$, ...&c. and longitudes ${\displaystyle \lambda _{1}}$, ${\displaystyle \lambda _{2}}$, &c., and let ${\displaystyle X}$ and ${\displaystyle Y}$ be found for each station.
Let
${\displaystyle l_{0}={\frac {1}{n}}\sum (l)}$, and ${\displaystyle \lambda _{0}={\frac {1}{n}}\sum (\lambda )}$,
(9)
${\displaystyle l_{0}}$ and ${\displaystyle \lambda _{0}}$ may be called the latitude and longitude of the central station. Let
${\displaystyle X_{0}={\frac {1}{n}}\sum (X)}$, and ${\displaystyle Y_{o}\cos l_{o}={\frac {1}{n}}\sum (Y\cos l)}$,
(10)
then ${\displaystyle X_{0}}$ and ${\displaystyle Y_{0}}$ are the values of ${\displaystyle X}$ and ${\displaystyle Y}$ at the imaginary central station, then
${\displaystyle X=X_{0}+B_{1}(l-l_{0})+B_{2}(\lambda -\lambda _{0})}$,
(11)
${\displaystyle Y\cos l=Y_{0}\cos l_{0}+B_{2}(l-l_{0})+B_{3}(\lambda -\lambda _{0})}$.
(12)
We have ${\displaystyle n}$ equations of the form of (11) and ${\displaystyle n}$ of the form (12). If we denote the probable error in the determination of ${\displaystyle X}$ by ${\displaystyle \xi }$, and that of ${\displaystyle Y\cos l}$ by ${\displaystyle \eta }$, then we may calculate ${\displaystyle \xi }$ and ${\displaystyle \eta }$ on the supposition that they arise from errors of observation of ${\displaystyle H}$ and ${\displaystyle \delta }$.
Let the probable error of ${\displaystyle H}$ be ${\displaystyle h}$, and that of ${\displaystyle \delta }$, ${\displaystyle d}$, then since
${\displaystyle dX=\cos \delta \centerdot dH-H\sin \delta \centerdot d\delta }$,
${\displaystyle \xi ^{2}=h^{2}\cos ^{2}\delta +d^{2}H^{2}\sin ^{2}\delta }$.
Similarly
${\displaystyle \eta ^{2}=h^{2}\sin ^{2}\delta +d^{2}H^{2}\cos ^{2}\delta }$.
If the variations of ${\displaystyle X}$ and ${\displaystyle Y}$ from their values as given by equations of the form (11) and (12) considerably exceed the probable errors of observation, we may conclude that they are due to local attractions, and then we have no reason to give the ratio of ${\displaystyle \xi }$ to ${\displaystyle \eta }$ any other value than unity.
According to the method of least squares we multiply the equations of the form (11) by ${\displaystyle \eta }$, and those of the form (12) by ${\displaystyle \xi }$ to make their probable error the same. We then multiply each equation by the coefficient of one of the unknown quantities ${\displaystyle B_{1}}$, ${\displaystyle B_{2}}$, or ${\displaystyle B_{3}}$ and add the results, thus obtaining three equations from which to find ${\displaystyle B_{1}}$, ${\displaystyle B_{2}}$, and ${\displaystyle B_{3}}$.
${\displaystyle P_{1}=B_{1}b_{1}+B_{2}b_{2}}$,
${\displaystyle (\eta ^{2}P_{2}+\xi ^{2}Q_{1})=B_{1}\eta ^{2}b_{2}+B_{2}(\xi ^{2}b_{1}+\eta ^{2}b_{3})+B_{3}\xi ^{2}b_{2}}$,
${\displaystyle Q_{2}=B_{2}b_{2}+B_{3}b_{3}}$;
in which we write for conciseness,
${\displaystyle b_{1}=\sum (l^{2})-nl_{0}^{2}}$, ${\displaystyle b_{2}=\sum (l\lambda )-nl_{0}\lambda _{0}}$, ${\displaystyle b_{3}=\sum (\lambda ^{2})-n\lambda _{0}^{2}}$,
${\displaystyle P_{1}=\sum (lX)-nl_{0}X_{0}}$, ${\displaystyle Q_{1}=\sum (lY\cos l)-nl_{0}Y_{0}\cos l_{0}}$,
${\displaystyle P_{2}=\sum (\lambda X)-n\lambda _{0}X_{0}}$, ${\displaystyle Q_{2}=\sum (\lambda Y\cos l)-n\lambda _{0}Y_{0}\cos l_{0}}$.
By calculating ${\displaystyle B_{1}}$, ${\displaystyle B_{2}}$, and ${\displaystyle B_{3}}$, and substituting in equations (11) and (12), we can obtain the values of ${\displaystyle X}$ and ${\displaystyle Y}$ at any point within the limits of the survey free from the local disturbances which are found to exist where the rock near the station is magnetic, as most igneous rocks are.
Surveys of this kind can be made only in countries where magnetic instruments can be carried about and set up in a great many stations. For other parts of the world we must be content to find the distribution of the magnetic elements by interpolation between their values at a few stations at great distances from each other.
467.] Let us now suppose that by processes of this kind, or by the equivalent graphical process of constructing charts of the lines of equal values of the magnetic elements, the values of ${\displaystyle X}$ and ${\displaystyle Y}$, and thence of the potential ${\displaystyle V}$, are known over the whole surface of the globe. The next step is to expand ${\displaystyle V}$ in the form of a series of spherical surface harmonics.
If the earth were magnetized uniformly and in the same direction throughout its interior, ${\displaystyle V}$ would be an harmonic of the first degree, the magnetic meridians would be great circles passing through two magnetic poles diametrically opposite, the magnetic equator would be a great circle, the horizontal force would be equal at all points of the magnetic equator, and if ${\displaystyle H_{0}}$ is this constant value, the value at any other point would be ${\displaystyle H=H_{0}\cos l^{\prime }}$, where ${\displaystyle l^{\prime }}$ is the magnetic latitude. The vertical force at any point would be ${\displaystyle Z=2H_{0}\sin l^{\prime }}$, and if ${\displaystyle \theta }$ is the dip, ${\displaystyle \tan \theta =2\tan l^{\prime }}$.
In the case of the earth, the magnetic equator is defined to be the line of no dip. It is not a great circle of the sphere.
The magnetic poles are defined to be the points where there is no horizontal force or where the dip is 90°. There are two such points, one in the northern and one in the southern regions, but they are not diametrically opposite, and the line joining them is not parallel to the magnetic axis of the earth.
468.] The magnetic poles are the points where the value of ${\displaystyle V}$ on the surface of the earth is a maximum or minimum, or is stationary.
At any point where the potential is a minimum the north end of the dip-needle points vertically downwards, and if a compass-needle be placed anywhere near such a point, the north end will point towards that point.
At points where the potential is a maximum the south end of the dip-needle points downwards, and the south end of the compass-needle points towards the point.
If there are ${\displaystyle p}$ minima of ${\displaystyle V}$ on the earth s surface there must be ${\displaystyle p-1}$ other points, where the north end of the dip-needle points downwards, but where the compass-needle, when carried in a circle round the point, instead of revolving so that its north end points constantly to the centre, revolves in the opposite direction, so as to turn sometimes its north end and sometimes its south end towards the point.
If we call the points where the potential is a minimum true north poles, then these other points may be called false north poles, because the compass-needle is not true to them. If there are ${\displaystyle p}$ true north poles, there must be ${\displaystyle p-1}$ false north poles, and in like manner, if there are ${\displaystyle q}$ true south poles, there must be ${\displaystyle q-1}$ false south poles. The number of poles of the same name must be odd, so that the opinion at one time prevalent, that there are two north poles and two south poles, is erroneous. According to Gauss there is in fact only one true north pole and one true south pole on the earth s surface, and therefore there are no false poles. The line joining these poles is not a diameter of the earth, and it is not parallel to the earth s magnetic axis.
469.] Most of the early investigators into the nature of the earth s magnetism endeavoured to express it as the result of the action of one or more bar magnets, the position of the poles of which were to be determined. Gauss was the first to express the distribution of the earth s magnetism in a perfectly general way by expanding its potential in a series of solid harmonics, the coefficients of which he determined for the first four degrees. These coefficients are 24 in number, 3 for the first degree, 5 for the second, 7 for the third, and 9 for the fourth. All these terms are found necessary in order to give a tolerably accurate representation of the actual state of the earth' magnetism.
To find what Part of the Observed Magnetic Force is due to External and what to Internal Causes.
470.] Let us now suppose that we have obtained an expansion of the magnetic potential of the earth in spherical harmonics, consistent with the actual direction and magnitude of the horizontal force at every point on the earth' surface, then Gauss has shewn how to determine, from the observed vertical force, whether the magnetic forces are due to causes, such as magnetization or electric currents, within the earth s surface, or whether any part is directly due to causes exterior to the earth's surface.
Let ${\displaystyle V}$ be the actual potential expanded in a double series of spherical harmonics,
${\displaystyle V=A_{1}{\frac {r}{a}}+\mathrm {\&c.} +A_{i}\left({\frac {r}{a}}\right)^{i},{}+B_{1}\left({\frac {r}{a}}\right)^{-2}+\mathrm {\&c.} +B_{i}\left({\frac {r}{a}}\right)^{-(i+1)}}$.
The first series represents the part of the potential due to causes exterior to the earth, and the second series represents the part due to causes within the earth.
The observations of horizontal force give us the sum of these series when ${\displaystyle r=a}$, the radius of the earth. The term of the order ${\displaystyle i}$ is
${\displaystyle V_{i}=A_{i}+B_{i}}$.
The observations of vertical force give us
${\displaystyle Z={\frac {dV}{dr}}}$,
and the term of the order ${\displaystyle i}$ in ${\displaystyle aZ}$ is
${\displaystyle aZ_{i}=iA_{i}-(i+1)B_{i}}$
.
Hence the part due to external causes is
${\displaystyle A_{i}={\frac {(i+1)V_{i}+aZ_{i}}{2i+1}}}$,
and the part due to causes within the earth is
${\displaystyle B_{i}={\frac {iV_{i}-aZ_{i}}{2i+1}}}$.
The expansion of ${\displaystyle V}$ has hitherto been calculated only for the mean value of ${\displaystyle V}$ at or near certain epochs. No appreciable part of this mean value appears to be due to causes external to the earth.
471.] We do not yet know enough of the form of the expansion of the solar and lunar parts of the variations of ${\displaystyle V}$ to determine by this method whether any part of these variations arises from magnetic force acting from without. It is certain, however, as the calculations of MM. Stoney and Chambers have shewn, that the principal part of these variations cannot arise from any direct magnetic action of the sun or moon, supposing these bodies to be magnetic[1].
472.] The principal changes in the magnetic force to which attention has been directed are as follows.
I. The more Regular Variations.
(1) The Solar variations, depending on the hour of the day and the time of the year.
(2) The Lunar variations, depending on the moon's hour angle and on her other elements of position.
(3) These variations do not repeat themselves in different years, but seem to be subject to a variation of longer period of about eleven years.
(4) Besides this, there is a secular alteration in the state of the earth' magnetism, which has been going on ever since magnetic observations have been made, and is producing changes of the magnetic elements of far greater magnitude than any of the variations of small period.
II. The Disturbances.
473.] Besides the more regular changes, the magnetic elements are subject to sudden disturbances of greater or less amount. It is found that these disturbances are more powerful and frequent at one time than at another, and that at times of great disturbance the laws of the regular variations are masked, though they are very distinct at times of small disturbance. Hence great attention has been paid to these disturbances, and it has been found that disturbances of a particular kind are more likely to occur at certain times of the day, and at certain seasons and intervals of time, though each individual disturbance appears quite irregular. Besides these more ordinary disturbances, there are occasionally times of excessive disturbance, in which the magnetism is strongly disturbed for a day or two. These are called Magnetic Storms. Individual disturbances have been sometimes observed at the same instant in stations widely distant.
Mr. Airy has found that a large proportion of the disturbances at Greenwich correspond with the electric currents collected by electrodes placed in the earth in the neighbourhood, and are such as would be directly produced in the magnet if the earth-current, retaining its actual direction, were conducted through a wire placed underneath the magnet.
It has been found that there is an epoch of maximum disturbance every eleven years, and that this appears to coincide with the epoch of maximum number of spots in the sun.
474.] The field of investigation into which we are introduced by the study of terrestrial magnetism is as profound as it is extensive.
We know that the sun and moon act on the earth's magnetism. It has been proved that this action cannot be explained by supposing these bodies magnets. The action is therefore indirect. In the case of the sun part of it may be thermal action, but in the case of the moon we cannot attribute it to this cause. Is it possible that the attraction of these bodies, by causing strains in the interior of the earth, produces (Art. 447) changes in the magnetism already existing in the earth, and so by a kind of tidal action causes the semidiurnal variations?
But the amount of all these changes is very small compared with the great secular changes of the earth's magnetism.
What cause, whether exterior to the earth or in its inner depths, produces such enormous changes in the earth's magnetism, that its magnetic poles move slowly from one part of the globe to another? When we consider that the intensity of the magnetization of the great globe of the earth is quite comparable with that which we produce with much difficulty in our steel magnets, these immense changes in so large a body force us to conclude that we are not yet acquainted with one of the most powerful agents in nature, the scene of whose activity lies in those inner depths of the earth, to the knowledge of which we have so few means of access.
1. Professor Hornstein of Prague has discovered a periodic change in the magnetic elements, the period of which is 26.33 days, almost exactly equal to that of the synodic revolution of the sun, as deduced from the observation of sun-spots near his equator. This method of discovering the time of rotation of the unseen solid body of the sun by its effects on the magnetic needle is the first instalment of the repayment by Magnetism of its debt to Astronomy. Akad., Wien, June 15, 1871. See Proc. R. S., Nov. 16, 1871.
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# GMAT test centre
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06 Dec 2011, 21:47
Hi, I had read in one of the posts that one should not start or end the test until the invigilator comes and logs on. Can't remember what was the post about. It helps in getting extra couple of minutes, the debrief said.
Could you please throw some light on how to start the test at the test centre in a way which gives you maximum time during the test and breaks.
Thanks
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### Show Tags
07 Dec 2011, 11:31
chigiwigi wrote:
Hi, I had read in one of the posts that one should not start or end the test until the invigilator comes and logs on. Can't remember what was the post about. It helps in getting extra couple of minutes, the debrief said.
Could you please throw some light on how to start the test at the test centre in a way which gives you maximum time during the test and breaks.
Thanks
See this mba.com page for details:
http://www.mba.com/the-gmat/test-day.aspx
The test will not START until a proctor logs you in for the test (you will have screens to go through before the test begins).
The test will END either when you finish or when time runs out. You will have little input on beginning the test, but the proctor will log in to get you back in from breaks (if you use too much time, the test timer will begin on its own and you still need the proctor to log in and let you resume).
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07 Dec 2011, 22:48
Thanks Mohater.
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08 Dec 2011, 09:59
Everything is timed, and they say you shouldn't write anything on the laminated paper. However I was able to write some quick notes before such as my timing strategy and some common Pythagorean theorems.
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1. ## Riemann Integrable Function
Suppose f and g are Riemann Integrable Functions on[a,b] and f(x)=g(x) if x not equal to c.
show that : $\displaystyle \int ^a_b f= \int ^a_b g$
2. I googled away to glory and maybe this will help:
Darboux theory
EDIT: Can the experienced answerers shed light on this problem?
3. My advice would be to split the interval up into the following parts:
$\displaystyle [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.
The integrals on the last 2 parts are the same. For the region in the middle since $\displaystyle f,g$ are assumed Riemann integrable they are bounded. So say $\displaystyle |f|\le F, |g|\le G$. Thus you have that
$\displaystyle \int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|$
4. where the $c$ coming from?
5. Originally Posted by hkito
where the $c$ coming from?
Well c was given in the problem!
Do you understand the question?
6. Well I don't see any $\displaystyle c$ in the problem. It's just saying that for $\displaystyle c$ not equal to x. For me it's not clear enough.
7. It says that as long as x does not equal c you know that f(x)=g(x). However, you are not guaranteed that f(c)=g(c).
8. Okay,
so the problem is show that $\displaystyle \int f=\int g$ given that $\displaystyle f(x)=g(x)$ except at $\displaystyle x=c$
9. Originally Posted by putnam120
My advice would be to split the interval up into the following parts:
$\displaystyle [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.
The integrals on the last 2 parts are the same. For the region in the middle since $\displaystyle f,g$ are assumed Riemann integrable they are bounded. So say $\displaystyle |f|\le F, |g|\le G$. Thus you have that
$\displaystyle \int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|$
Since $\displaystyle f(x)=g(x)$ on $\displaystyle [a,c-\epsilon]\cup[c+\epsilon]$
you just need to show that $\displaystyle \int_{c-\epsilon}^{c+\epsilon}f=\int_{c-\epsilon}^{c+\epsilon}g$
For that you since you know that f and g are integrable, so $\displaystyle f-g$ is bounded , say by some$\displaystyle M$.
$\displaystyle \int_{c-\epsilon}^{c+\epsilon}|f(x)-g(x)|dx\leq M\int_{c-\epsilon}^{c+\epsilon}dx=2M\epsilon$
And you make $\displaystyle \epsilon$ go to zero
10. There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable". That's very nice for this problem, because if $\displaystyle h=f-g$, then $\displaystyle h=\left\{\begin{array}{lr}k&:x=c\\0&:x\neq c\end{array}\right\}$
Calculate the lower Riemann sum of $\displaystyle h$. Let $\displaystyle a=x_1<x_2<...<x_n=b$ be a partition of $\displaystyle [a,b]$ and let $\displaystyle [x_{i-1},x_i]$ be the segment containing $\displaystyle c$. $\displaystyle \min h(x)$ on $\displaystyle [x_{i-1},x_i] = 0$, because it contains a point that is not $\displaystyle c$, so therefore the lower Riemann sum and the whole integral is zero.
11. Originally Posted by redsoxfan325
There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable".
If I remember correctly this can be extended to saying that if a function is discontinuous on a set of measure 0 then it is Riemann-integrable.
12. Originally Posted by putnam120
If I remember correctly this can be extended to saying that if a function is discontinuous on a set of measure 0 then it is Riemann-integrable.
(If this is true - it seems to be, but I don't have a proof lol) then it has to be with Lebesgue measure, isn't it ?
13. Yes sorry for the ambiguity.
14. We can prove something stronger. If $\displaystyle f$ is integrable and $\displaystyle g$ is a function that differs at countably many points distinct from $\displaystyle f$ such that they converge then $\displaystyle g$ is integrable with same value as $\displaystyle f$. The nice thing is that the proof is similar as in the finite case. Because when you have convergence it means that within a a small enough neighborhood of the limit all the points lie there. So you form a partition with thin enough rectangles around these values.
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PDA
View Full Version : Need some math help
Uncle O
12-27-2012, 07:38 PM
I need to find 2 points on this sketch and can't seem to get past a major brain fart....I have been staring at it and .....
Anybody have a cad program and care to help me out ?
Need the 2 points shown as +, at the top where the 30 deg line would intersect to the zero line if it ran all the way up and to the right where it hits the flat on the side.. 0, 0, is the center of the sketch.
It is a .778 dia. .66 across all flats
http://i170.photobucket.com/albums/u278/Ohboy_album/img026_zps9c00b6d1.jpg
Jpfalt
12-27-2012, 08:00 PM
Can't happen except in one case. That is if the vertical flat is of zero length and the angle is 90 degrees.
From the X at the center to the lower right + sign is the hypotenuse of a right triangle.
The formula c squared = a squared plus bsquared. C is the hypotenuse, a is the distance from X to the right flat. b is the vertical height of the vertical flat.
If the width really is .66 and the angle really is 60 degrees, then the length of the vertial flat will be .33 and the vertical heigth will be .165 from X to the lower right +.
However, the distance from X to the 60 degree angle flat will be .381 instead of .33. The height to the upper + would be .762.
Since c squared equals a squared, then b must equal zero.
The Artful Bodger
12-27-2012, 08:01 PM
Consider a triangle from the center to the mid point of the flat and out to the 30 degree corner. That would be a right angle triangle (90,60,30 degrees) with a height of .33.
Cheers
Tony Ennis
12-27-2012, 08:05 PM
I get (.33, .136) and (0, .707)
That assumes the shape is an octagon, and 60 degrees was intended.
Toolguy
12-27-2012, 08:15 PM
What's shown in the sketch is an octagon, but 60 degree sides would make a hexagon. With a hexagon the x on the right side would be on the horizontal centerline.
Jpfalt
12-27-2012, 08:35 PM
I looked at this again and realized the angle from X to lower right + could be 15 degrees from horizontal.
The height of lower right + would be .088"
The height of the upper + would be .660
Hopefuldave
12-27-2012, 08:47 PM
Hi Uncle O,
*not sure* what you're aiming at! I'm guessing it's an 8-sided part, and the 30* angle is *not* tangent to the next flat around?
If the + on the right is on the corner of the flats a bit of jommetry should find that, like so:
the angle between the axis and the corner is half of 360/8, so 22.5*
so... the X coordinate is half the A/F distance 0.330"
Tan of 22.5* is 0.4142 = opposite (the Y coordinate) / adjacent (half the A/F) = Y/0.33
multiply both sides by 0,33, Y coordinate = 0.4142 x 0.330 = 0.137 (to 3 sig' fig's)
So the right-hand + is at 0.330, 0.137
To get the top + position, draw a right-angled triangle with a 30* angle at the top, the opposite side from the edge of the flats we just worked out horizontally to meet the vertical line - this will be 0.330" long (half the A/F dimension). The hypotenuse goes from our new + point to the edge of the flat and the adjacent from the intersection of the vertical line through the axis to the 30* angle.
tan 30* is 0.5773 = opposite over adjacent, so...
0.5773 = 0.330 (the half-A/F dimension) / adjacent side
multiply both sides by (adjacent) gives 0.5773 x adjacent = 0.330 (opposite)
hmmm, not there yet!
divide both sides by tan 30* (0.5773) gives adjacent = 0.330 (opposite) / 0.5773
We we now know the adjacent side is 0.330 / 0.5773 = 0.5716"
to get the position of the upper +, we have to add the Y-coordinate of the first (right-hand) +, so:
x = 0 (as it's on the vertical from the axis), Y = 0.5716 + 0.137 = 0.7086"
If I can attach it, diagram illustrates....
http://s979.photobucket.com/albums/ae279/hopefuldave/
Hope this helps, rather than confuses!
Dave H. (the other one)
P.S. - just noticed for the first time that tan 22.5 is root(2) - 1... I'll remember that :)
Machine
12-27-2012, 09:05 PM
http://i289.photobucket.com/albums/ll224/my-pics-are-here/drwng_zps2a51b00c.jpg
Uncle O
12-27-2012, 11:57 PM
http://i289.photobucket.com/albums/ll224/my-pics-are-here/drwng_zps2a51b00c.jpg
Thanks for everybodies attempts to help, I have made these before, it's been a quite awhile tho.....
I need to find d as shown above, and from there I will be good. In the a.m. I will put a protractor to the corner of that flat ( on the print ) and take it from there. I will post results later.
Thanks again.
Tony Ennis
12-28-2012, 12:57 AM
I get d = .136. See my post above.
Jpfalt
12-28-2012, 03:14 AM
I stick by my calc that d=.088.
To check it out I laid it out in Solidworks and got the same answer.
I hereby declare victory and will now run like hell.
Jaakko Fagerlund
12-28-2012, 03:43 AM
Edit: Scratch that
Paul Alciatore
12-28-2012, 03:46 AM
I have to agree with Machine. Your drawing is too rough to really know what you are asking. It could be a regular octagon, meaning 8 equal sides and 8 equal angles or it could be just a plain octagon, meaning it has 8 sides but they are not necessairly equal and the angles may not be equal. Then again, the top and bottom seem to be highly curved so it could be a figure with six flat sides and two curved ones. Or perhaps something else.
And is your 60 degree line congruent with one of the sides or is it at a different angle and only touches it at the corner marked with a cross.
In any case, you do not give enough details/dimensions to figure the location of the points. I would suggest a better sketch and more dimensions.
Jaakko Fagerlund
12-28-2012, 03:55 AM
Edited totally
The piece is round with six flats on it. Dimensions as shown, the points are here:
http://i4.aijaa.com/t/00167/11495638.t.jpg (http://aijaa.com/NWhgnj)
Highest point is 0.66, the rightmost point is 0.088 above Y axis
Barrington
12-28-2012, 08:30 AM
The CAD solution is good, but here's a trig solution:-
http://i564.photobucket.com/albums/ss82/MrBarrington/geometry-1.png
AB = 0.33
AC = AB*cos(30) = 0.33*0.8660254 = 0.2857884
BC = AB*sin(30) = 0.33*0.5 = 0.165
DB = AC
ED = DB/tan(30) = 0.495
DA = BC
EA = ED+DA = 0.495 + 0.165 = 0.66
FG = 0.33
EF = FG/tan(30) = 0.5715768
GH = FA
GH = EA-EF = 0.66 - 0.5715768 = 0.0884232
Cheers
.
Uncle O
12-28-2012, 10:37 PM
Well, I found a sample piece in my box that I had made last time around. The small flat on the side was pretty close to .180 long, so I figured the .088 dimension was going to be good. I know that sketch looked really bad, I had not intended on publishing it , just using it to work with after I had gotten home. Also when I put a protractor to that point, it reads 20 degrees, which I think I recall going thru last time I worked on these. My trig knowledge is poor, I must admit. Usually I can trod my way thru with the help of my little trig table booklet.....Anyway, the .088 dimension worked nicely, as you know. My job was to just do the flats.
Thanks again .
http://i170.photobucket.com/albums/u278/Ohboy_album/001_zps867923e3.jpg
http://i170.photobucket.com/albums/u278/Ohboy_album/005_zps4c287a98.jpg
http://i170.photobucket.com/albums/u278/Ohboy_album/006_zps7627e6ca.jpg
Toolguy
12-28-2012, 10:54 PM
That looks like a diamond pin for locating a part on a fixture. It probably gets heat treated, then ground on centers to a .625 shank and .750 top if it's going to be an inch size.
Uncle O
12-28-2012, 11:11 PM
Yes, it is a locator pin, no heat treat, as it is ampco. Grind as shown.
A test piece .
http://i170.photobucket.com/albums/u278/Ohboy_album/003_zps86e5ded0.jpg
http://i170.photobucket.com/albums/u278/Ohboy_album/002_zpsf9e38e80.jpg
Jaakko Fagerlund
12-29-2012, 05:27 AM
Got me curious: How is that used to locate a part?
Rosco-P
12-29-2012, 09:40 AM
Got me curious: How is that used to locate a part?
Part with a round hole used as a locating feature drops over the diamond pin more easily than a round pin. Common jig and fixture design technique.
Toolguy
12-29-2012, 10:57 AM
You have to have 2 holes in the part. You use a round pin to locate the datum hole and the diamond pin turned 90 degrees to the centerline of the 2 holes. The round pin locates the part lengthwise and both pins locate the part side to side. It is much easier to get parts on and off this arrangement than with 2 round pins and the parts are still located accurately.
Jaakko Fagerlund
12-29-2012, 01:16 PM
Part with a round hole used as a locating feature drops over the diamond pin more easily than a round pin. Common jig and fixture design technique.
Hmm, clever :) Haven't run in to them as all the jigs I've made were for die cast aluminum parts which don't have precise round holes in them.
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# Multiple COUNTIF Statements With Different Criteria
Dec 21, 2007
I'm a marketer trying to find all calls within a week that are longer than 300 seconds. I'd prefer not to use a pivot table because of client issues.
=COUNTIF(A:A,1)-COUNTIF(F:F,"
## CountIF() On Multiple Criteria ..
Jan 9, 2010
i want to use Countif() when 3 ceritera are true
ceritera
--------
1) Branch Name
2) Status
can this be done by any mean
## Using Countif With Multiple Corresponding Criteria
Feb 12, 2010
what I'm trying to do is to make a logbook for a machining center. Each part has an op10 and an op20, essentially front and back. And each part number falls into the category of OS or FS. I've used AND logic to make tables in hidden columns to be used by a countif statement to determine my totals. I.e. to determine if a scroll is completed, op20 has a a value of 1 AND column C is "OS".
I use
=IF(AND(A9=1;C9="OS");1;" ")
Then I countif criteria is 1 in the column i created with that statement.
That works just fine. Now what I want to do is to be able to create daily totals of OS and FS by simply modifying a variable date in a formula. So I'd like to essentially say: Countif Column C =OS and Corresponding column D = 1, and corresponding Shift date = 10.02.12(date to be variable). I'm at a wall here. Is there any way to do this somewhat simply?
## Countif Multiple Criteria.
Oct 9, 2008
I'm trying to count multiple criteria from a second page in a work book, all the formulas i've looked up and tried do not seem to work... here's the formulas i've tried. DKOBULAR is the name of the 2nd page. D is the column used for the different resolves.
=COUNTIF(DKOBULAR!D:D="resolveA")+COUNTIF(DKOBULAR!D:D="resolveB")+COUNTIF(DKOBULAR!D:D="resolveC")+ COUNTIF(DKOBULAR!D:D="resolveD")
=COUNT(IF(DKOBULAR!D:D="resolveA",IF(DKOBULAR!D:D="resolveB",IF(DKOBULAR!D:D="resolveC",IF(DKOBULAR! D:D="resolveD")))))
## Multiple CountIf Criteria
May 19, 2009
In Excel 2003, I need a countif to check for 2 criteria: (1) the left function looking for the value "Territory" in column A and (2) value > 0 in column G. I only want to count the rows where both the criteria are met. I have tried different combinations of countif including "and" in the formula, but I cannot get it to work. What is the proper syntax?
## Countif Multiple Criteria
Aug 16, 2007
I need to create a formula which counts the number of times a username appears in column X based on a given value in Column Y. This data will not be static - will need to be refreshed regularly. Countif does not support multiple criteria - what is the best way to create this formula?
## Countif Formulas With Multiple Criteria?
May 26, 2014
I need a formula to count cells E7:E500 that say "Submitted" but only if cells R7:R500 say "PPB".
My original formula below is counting all cells except for the ones that say "PPB"
=COUNT(IF(E7:E500="Submitted",IF(R7:R50="PPB",'E7:E500)))
## Multiple Criteria Countif Or Sumproduct
Sep 16, 2009
I haven't been this deep into excel before. The deeper I look, the more potential I recognize, the more amazed I get. That being said, I have come to a tough count issue. Let me attempt to explain as precisely as possible.
My current worksheet is large but I am only particularly concerned with two columns of information (Regions) and (Days). The logic I am attempting is something along the lines of Count If Region = East, or West, and Days is greater than 0, less than 60.
I am open to any and all suggestions on how to tackle this situation. I have been able to achieve similar counts by using pivot tables but the dynamic nature of these two columns presents some difficulties that my “new user” mind has been unable to work through.
## Using Countif(s) To Count Multiple Criteria
Nov 25, 2013
I have a bunch of cells that i want to count in a table based on 2 criteria. The first would be checking whether or not it matches a certain text which i can do. Now i have a column for "Completed Date" and "Deadline Date". Each deadline date is different. I know i need some sort of array function in there to compare all the completed dates vs deadline dates but i just can't think of it. This will be the second criteria.
## Countif Array With Multiple Criteria?
Sep 13, 2012
=COUNTIF(A\$2:T\$1001,V2 & W2 & X2 & Y2)
I want to copy and paste this formula down column Z.
I want to count or add 1, only when a row of 20 cells (A to T) contains all 4 criteria in V W X Y.
## COUNTIF With Multiple Range And Criteria
Mar 6, 2008
I need to add all cells within multiple ranges where the characters "A", "S" and "D" occur.
=SUMPRODUCT(COUNTIF(INDIRECT({"C22:C25","C28:C32"}),{"A","S","D"}))
## COUNTIF Function With Multiple Criteria
Jun 7, 2006
I need to count rows that meet 2 criteria.
I have seen this help page
http://www.ozgrid.com/Excel/count-if.htm
but that counts rows with "criteria 1" OR "criteria 2"...
I need to count rows that fulfill "criteria 1" AND "criteria 2"
ie - count the rows that have todays date AND a cell that says "COMPLETE"
ideally it would be as easy as "=countif(A:F,"today()","COMPLETE") but that doesn't work... any way around this???
## Multiple Criteria For Countif Functions
Nov 11, 2006
I need to create a formula that counts the number of times that an age range appears within a column. In column G, there is a list of ages based on a demographic collection. The ages range from 13 to 50+. I want to designate another cell to count the number of times the characters between 13 and 18 occur within that column. I have =COUNTIF(G8:G20,"13") How do I add "14", "15", "16", "17", and "18"?
## Excel 2003 :: COUNTIF With Multiple Criteria
Jan 26, 2013
How many tickets are created and fixed by a user for a certain period. See below. I've tried countif but it doesn't seem to be working for many criteria.
Start Date
11/1/2012
End Date
11/31/2012
Assigned To
Created
Fixed
[Code] .....
## Countif Function Work On Multiple Criteria
Apr 15, 2009
Can countif function work on multiple criteria to look on?
this are the criteria.
SA SL 1.0SA SL 0.5SA VL 1.0SA VL 0.5SA SLWOP 1.0SA SLWOP 0.5SA VLWOP 1.0SA VLWOP 0.5SUSPRD
## Excel 2003 :: How To Use Countif For Multiple Range And Criteria
Jan 1, 2014
How to use countif in Excel 2003, for multiple range and criteria ? Can 'Nested ifs' be used? If so, a sample of such ifs
## Countif Multiple Conditions: Look At Two Columns And Set A Criteria To Count
Jan 15, 2010
I want to be able to look at two columns and set a criteria to count. I want to look at column A and if its blank then look at column B and if it has a value of more than 0 then count.
A B
1 1.00
2 Yes 4.78
3
4 5.00
5 Yes 4.89
6 11.99
So this example would count 3
## 2003: COUNTIF/SUMPRODUCT, Multiple Criteria W/Wildcard
Nov 24, 2008
I'm trying to write this but it returns a 0 when I know there are 3 records that match this criteria: =SUMPRODUCT(('Invoice-Detail'!J2:J50="NewJob_Post.NET")*('Invoice-Detail'!H2:H50="KY_*")). I think the problem is in the wildcard character. I don't know if I should be using COUNTIF or SUMPRODUCT or something else?
## Multiple IF Statements (skip The Next Two Statements Or The Result Will Be Changed Again)
Nov 22, 2008
I have three IF statements as below. the problem is if the first statement is true I want it to skip the next two statements or the result will be changed again.
## Consolidating Lists And Nested SumIf(AND / Countif(AND Statements
Jun 10, 2014
I am trying to produce a report of supplier transactions sorted by area code. I have a spreadsheet of data consisting of Supplier Code, Supplier Name, Transaction Amount and Area Code.
I want to be able to firstly seperate the transactions by area, then also consolidate the data so it shows one row per supplier with total amount spent on that supplier and a transaction count on that supplier. I know how to sumif the transaction total and countif the transaction count.
However I have problems consolidating the suppliers in to one row per unique supplier and I also have problems nesting an AND statement in to the sumif/countif statements. I need additional criteria in the forumula to only count transactions in a specified area code.
I've attached an example spreadsheet to make it alot clearer. See results sheet in this workbook.
## Formula Link Break For Countif And Averageif Statements?
Jun 5, 2013
I have simple Countif and Averageif formulas that I would like to have linked to variable data in cell E6 on the Assumptions (2) tab however, the formula is not taking the cell as the criteria and only taking a hard number when I am using >=.
The example would be: I3 needs to be able to use the data changed in E6 for the >= number instead of the hardkey 29...
## Multiple If Statements With Multiple True And One False
Aug 6, 2014
I am trying to get a function in a sheet where it has two possible truth outcomes and one false outcome. Both statements false statement goes back to
'Production Metrics 2'!E11/12
Here are the two If statements as they appear in the formula bar now. I need to combine them so they both work and don't cancel each out out.
=IF('Production Metrics 2'!E11/12>2500,2500,'Production Metrics 2'!E11/12) If c10+b10>=4500,0,'Production Metrics 2'!E11/12
## COUNTIF With Two Criteria
Nov 29, 2009
I use Excel 2003 and I have the following problem:
I have 3 columns,
A containing a list of employees (MICHAEL, BOB, MIKE, etc.)
B containing their work starting hour (8.00, 8.30, 9.00, etc.)
C containing the possible employee absence reason (ILLNESS, HOLIDAY, INJURY, etc.)
I would like to write a formula that counts the number of employees who have a work starting hour within 7.00 and are not absent. A possible table is this one:
NAME START ABSENCE
MICHAEL 6.30
BOB 8.30
MIKE 9.00 HOLIDAY
BRIAN 7.00
TOM 6.30 ILLNESS
The formula I'm looking for should calculate "2" (because MICHAEL and BRIAN are the only 2 employees starting work hour within 7.00 and not absent). As I have thought it could be useful, in another worksheet I have inserted: in A column the list of all the starting work hours: 0.00 (A2), 0.30 (A3), 1.00 (A4), 1.30 (A5), ... 7.00 (A16), 7.30 (A17), ... 23.30 (A49).
in B column the list of all the absence reasons: ILLNESS (B2), HOLIDAY (B3), INJURY (B4). I have defined 2 names, the first called EARLY_MORNING (that I have associated to the range from 0.00 to 7.00 of work starting hours column, that is A2:A16), the second called ABSENCE_REASON (that I have associated to the range (B2:B4) of absence reasons' column). What kind of formula can I write to obtain what I want (using the 2 names EARLY_MORNING and ABSENCE_REASONS defined in the other worksheet)?
## Countif Using 2 Criteria
Mar 7, 2008
I have a spreadsheet with a list of months numbers and average turnarounds. Each row represents a different factor, so there are multiple rows for each month.
eg
Month Turnaround Time
10 5.2
10 6.7
11 1.1
9 8.3
11 5.4
10 6.1
etc
What I am after is something that will count the number of instances where the turnaround time is above a certain limit (eg 6.0) for each month.
## COUNTIF Criteria In VBA
Jun 9, 2006
What vba function can I use that checks criteria much like SUMIF or COUNTIF uses? In other words, say you want to write COUNTIF that only includes visible cells...
Public Function CountVIf(rng As range, criteria As String)
Dim cell As range, cmd As String
For Each cell In rng
If cell.RowHeight <> 0 And cell.ColumnWidth <> 0 Then
cmd = "COUNTIF(" & cell.Address & ",""" & criteria & """)"
CountVIf = CountVIf + Evaluate(cmd)
End If
Next cell
End Function
How can I do this without having to rely on Evaluate("COUNTIF...."?
## Two COUNTIF Criteria
Jul 18, 2006
What I need to do is count the number of “Cs” in a column based on a date in another column but in the same row. I have tried something similar to this: COUNTIF(\$C\$1:\$C\$20,"=today()")+COUNTIF(\$E\$1:\$E\$20,"=complete") but is does not work. If the date in the column is less than or equal to a date specified in a cell in another worksheet, I want it to count the C in the row (if there is one).
## COUNTIF With Mutiple Criteria
Apr 6, 2009
In Row 3, starting at cell C3, I have a list of Stores in the format Store A (Town 1), Store A (Town 2), Store B (Town 1) etc
In Rows C4 to CA7 I have 1 or -1
I want to count the number of instances of 1 for Store A
I have tried the following formula, which is returning 0
{=SUM((C3:CA3="Store A *")*(C4:CA7=1))}
## Using Countif() Formula But With 2 Criteria
Jul 24, 2008
is there any way to put 2 criteria into a countif() statment?
Say A1 to A100 are filled with information
I'd like to count if things are equal to 4 and 5
I know this isn't proper syntax:
=countif(A1:A100, "4" && "5")
I'm trying to get something like that to work.
## COUNTIF Based On Two Criteria
Sep 9, 2007
I have data in two columns. Column B has calculated string values of "BMAJOR" and "BMINOR" and column C has date values.
I am trying to get a count of how many occurences of "BMAJOR" or "BMINOR" occur for every week starting today. I have column AJ which returns today() and column AK which returns today()+7. I can individually count number of occurences of dates that fall within these dates with the following formula. This for row 2.
=COUNTIF('Decision Tree'!\$C\$4:\$C\$215,"
## Countif With Two Ranges And Two Criteria
Jan 10, 2008
I am using a work sheet where I want to count if Column A has one criteria and column B has another
example
column B states is used to track contracts it can be vendor
column L tracks status it can be open, received, or closed
I want to count the cell if the vendor name matches and the status is open
also if that is possible is the same possible with 3 ranges and criterias?
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https://www.lumoslearning.com/llp/lesson-plan-resources.php?grade=grade-7&subject=math&lessonname=Significance+of+points+on+graphs+of+proportions
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# Grade 7 Math - Significance of points on graphs of proportions
Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate.
## TOPICS RELATED TO SIGNIFICANCE OF POINTS ON GRAPHS OF PROPORTIONS
### Why will the graph of every proportional relationship include the Point 0 0?
It's miles a straight line, however it does no longer intersect the point (zero, zero). the connection is notproportional, so there isn't a consistent of proportionality. it is no longer proportionalbecause the road does no longer pass through the starting place. consequently the ratio of x to y willbe one-of-a-kind for 2 special factors on the road.
### What points must be on a graph for it to be proportional?
In a graph, the points (1, r) and (0,0) need to seem on the line of the proportionalrelationship.
### What is proportional on a graph?
Which means as x increases, y will increase and as x decreases, y decreases-and that the ratio among them usually remains the same. the graph of the proportionalrelationship equation is a straight line through the starting place.
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https://kr.mathworks.com/matlabcentral/cody/players/3785922-bug-me/solved
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Cody
# Bug Me
Rank
Score
1 – 13 of 13
#### Problem 19. Swap the first and last columns
Created by: Cody Team
#### Problem 4. Make a checkerboard matrix
Created by: Cody Team
#### Problem 6. Select every other element of a vector
Created by: Cody Team
#### Problem 8. Add two numbers
Created by: Cody Team
#### Problem 3. Find the sum of all the numbers of the input vector
Created by: Cody Team
#### Problem 149. Is my wife right?
Created by: the cyclist
Tags fun, stupid, silly
#### Problem 9. Who Has the Most Change?
Created by: Cody Team
#### Problem 26. Determine if input is odd
Created by: Cody Team
#### Problem 5. Triangle Numbers
Created by: Cody Team
Tags math, triangle
#### Problem 17. Find all elements less than 0 or greater than 10 and replace them with NaN
Created by: Cody Team
#### Problem 7. Column Removal
Created by: Cody Team
#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
Created by: Cody Team
Tags basic, colon, x=1:10
#### Problem 1. Times 2 - START HERE
Created by: Cody Team
Tags intro, math, basics
1 – 13 of 13
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https://codereview.stackexchange.com/questions/93552/printing-all-factorizations-of-a-number
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# Printing all factorizations of a number
For example, if we have 120, the answer should contain (2,2,2,3,5),(2,60),(4,30),...
Here is my not-so-good attempt
for (int[] x : getPossibleProducts(120)) {
System.out.println(Arrays.toString(x));
}
public static List<String> getPossibleProducts2(int n) {
List<String> possprod = new ArrayList<String>();
if (isPrime(n)) {
return possprod;
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
for (String y : getPossibleProducts2(n / i)) {
}
}
}
return possprod;
}
public static List<int[]> getPossibleProducts(int n) {
List<String> l = new ArrayList<String>();
Set<int[]> set = new HashSet<int[]>();
List<String> possprod = getPossibleProducts2(n);
for (String x : possprod) {
String[] y = x.split("x");
Arrays.sort(y);
if (!l.contains(Arrays.stream(y).reduce("", (a, b) -> a + b))) {
l.add(Arrays.stream(y).reduce("", (a, b) -> a + b));
int[] z = stringArraytoIntArray(y);
}
}
return new ArrayList<int[]>(set);
}
public static int[] stringArraytoIntArray(String[] a) {
int[] arr = new int[a.length];
for (int i = 0; i < a.length; i++) {
arr[i] = Integer.parseInt(a[i]);
}
return arr;
}
public static boolean isPrime(int i2) {
if (i2 == 2)
return true;
if (i2 % 2 == 0)
return false;
for (int i = 3; i * i <= i2; i += 2)
if (i2 % i == 0)
return false;
return true;
}
Here is the output:
[15, 8]
[10, 3, 4]
[30, 4]
[2, 3, 4, 5]
[12, 2, 5]
[10, 2, 2, 3]
[10, 12]
[120]
[15, 2, 2, 2]
[15, 2, 4]
[3, 40]
[4, 5, 6]
[2, 2, 5, 6]
[20, 6]
[2, 2, 30]
[2, 20, 3]
[2, 60]
[2, 2, 2, 3, 5]
[24, 5]
[10, 2, 6]
[3, 5, 8]
It is not immediately obvious that getPossibleProducts2() finds all factorizations and getPossibleProducts() removes the duplicates, so the names could be chosen better. (And I would use "factorizations" instead of "possibleProducts").
Your code is ineffective for two reasons:
• The first method finds all factorizations with all possible permutations (e.g. $120 = 2 \cdot 60 = 60 \cdot 2$) so that you have to find the duplicates in a second step.
As an example, for $n = 1024$, the first method computes 364176 factorizations, which are then reduced to 627 unique factorizations in the second step.
• You use strings to represents the factorizations (e.g. "2x60", "60x2") which then have to be split into integer arrays again.
To find duplicate factorizations, you split the string (e.g. "60x2") to an integer array ["60", "2"], sort the factors (["2", "60"]) then concatenate the factors again ("260"). This is used as a hash key to eliminate the duplicates. This is quite computing intensive. In particular the hash key
Arrays.stream(y).reduce("", (a, b) -> a + b))
is compute twice, and StringBuilder might be a more effective way to concatenate the array elements.
More important, it does not work: For $n = 130$, the factorizations $2 \cdot 65$ and $5 \cdot 26$ are both mapped to the same hash key "265", so that the output of your program is
[2, 65]
[13, 2, 5]
[10, 13]
[130]
without the factorization [5, 26].
Generally a better method would be to determine the factorizations without duplicates in the first step. This can be done recursively as follows:
• Find the smallest factor f of the given number n.
• Compute all factorizations of the n / f where the smallest factor is at least f.
This automatically leads to factorizations with the factors in increasing order and without duplicates.
As a side effect, the isPrime() function is not needed.
Here is a possible implementation. (Please take this as a demonstration of the algorithm only. Java is not my first language, so there may be more idiomatic ways to do implement this.)
public static List<int[]> factorizations(int n) {
return factorizations(n, 2, new ArrayList<Integer>());
}
public static List<int[]> factorizations(int n, int minFactor, ArrayList<Integer> previousFactors) {
List<int[]> result = new ArrayList<int[]>();
if (n == 1) {
// This is where the recursion terminates. Convert the list of
// all previously found factors to an int[] array
// (found here: http://stackoverflow.com/a/23945015/1187415).
} else {
for (int factor = minFactor; factor <= n; factor++) {
if (n % factor == 0) {
// factor is a factor of n. Append it to (a clone of)
// the previously found factors ...
ArrayList<Integer> factors = new ArrayList<Integer>(previousFactors);
// ... and (recursively) find all factorizations of n / factor
// with factors greater or equal to factor:
result.addAll( factorizations(n / factor, factor, factors) );
}
}
}
return result;
}
Example:
for (int[] f : factorizations(130)) {
System.out.println(Arrays.toString(f));
}
produces the output
[2, 5, 13]
[2, 65]
[5, 26]
[10, 13]
[130]
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https://www.universetoday.com/category/habitability/page/2/
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White Dwarfs Could Support Life. So Where are All Their Planets?
Astronomers have found plenty of white dwarf stars surrounded by debris disks. Those disks are the remains of planets destroyed by the star as it evolved. But they’ve found one intact Jupiter-mass planet orbiting a white dwarf.
Are there more white dwarf planets? Can terrestrial, Earth-like planets exist around white dwarfs?
Continue reading “White Dwarfs Could Support Life. So Where are All Their Planets?”
What’s the Best Way to Find Planets in the Habitable Zone?
Despite the fact that we’ve discovered thousands of them, exoplanets are hard to find. And some types are harder to find than others. Naturally, some of the hardest ones to find are the ones we most want to find. What can we do?
Keep working on it, and that’s what a trio of Chinese scientists are doing.
Continue reading “What’s the Best Way to Find Planets in the Habitable Zone?”
If Earth is Average, We Should Find Extraterrestrial Life Within 60 Light-Years
In 1960, while preparing for the first meeting on the Search for Extraterrestrial Intelligence (SETI), legendary astronomer and SETI pioneer Dr. Frank Drake unveiled his probabilistic equation for estimating the number of possible civilizations in our galaxy – aka. The Drake Equation. A key parameter in this equation was ne, the number of planets in our galaxy capable of supporting life – aka. “habitable.” At the time, astronomers were not yet certain other stars had systems of planets. But thanks to missions like Kepler, 5523 exoplanets have been confirmed, and another 9,867 await confirmation!
Based on this data, astronomers have produced various estimates for the number of habitable planets in our galaxy – at least 100 billion, according to one estimate! In a recent study, Professor Piero Madau introduced a mathematical framework for calculating the population of habitable planets within 100 parsecs (326 light-years) of our Sun. Assuming Earth and the Solar System are representative of the norm, Madau calculated that this volume of space could contain as much as 11,000 Earth-sized terrestrial (aka. rocky) exoplanets that orbit within their stars’ habitable zones (HZs).
Continue reading “If Earth is Average, We Should Find Extraterrestrial Life Within 60 Light-Years”
When did the First Continents Appear in the Universe?
On Earth, continents are likely necessary to support life. Continents ‘float’ on top of the Earth’s viscous mantle, and heat from the planet’s core keeps the mantle from solidifying and locking the continents into place.
The core is hot because of the presence of radioactive elements that came from neutron star collisions. It should be possible to calculate when the first continents formed in the Universe.
So that’s what one researcher did.
Continue reading “When did the First Continents Appear in the Universe?”
Is it Time for a New Definition of “Habitable?”
Things tend to move from the simple to the complex when you’re trying to understand something new. This is the situation exoplanet scientists find themselves in when it comes to the term ‘habitable.’ When they were discovering the first tranche of exoplanets, the term was useful. It basically meant that the planet could have liquid water on its surface.
But now that we know of over 5,000 confirmed exoplanets, the current definition of habitable is showing its age.
Continue reading “Is it Time for a New Definition of “Habitable?””
Forget the Habitable Zone – We Need to Find the Computational Zone
Astronomers are currently searching for signs of life in the “habitable zones” of nearby stars, which is defined as the band around a star where liquid water can potentially exist. But a recent paper argues that we need to take a more nuanced and careful approach, based not on the potential for life, but the potential for computation.
Continue reading “Forget the Habitable Zone – We Need to Find the Computational Zone”
Even the Calmest Red Dwarfs are Wilder than the Sun
There’s something menacing about red dwarfs. Human eyes are accustomed to our benevolent yellow Sun and the warm light it shines on our glorious, life-covered planet. But red dwarfs can seem moody, ill-tempered, and even foreboding.
For long periods of time, they can be calm, but then they can flare violently, flashing a warning to any life that might be gaining a foothold on a nearby planet.
Continue reading “Even the Calmest Red Dwarfs are Wilder than the Sun”
Do Red Dwarfs Provide Enough Sunlight for Plants to Grow?
To date, 5,250 extrasolar planets have been confirmed in 3,921 systems, with another 9,208 candidates awaiting confirmation. Of these, 195 planets have been identified as “terrestrial” (or “Earth-like“), meaning that they are similar in size, mass, and composition to Earth. Interestingly, many of these planets have been found orbiting within the circumsolar habitable zones (aka. “Goldilocks zone”) of M-type red dwarf stars. Examples include the closest exoplanet to the Solar System (Proxima b) and the seven-planet system of TRAPPIST-1.
These discoveries have further fueled the debate of whether or not these planets could be “potentially-habitable,” with arguments emphasizing everything from tidal locking, flare activity, the presence of water, too much water (i.e., “water worlds“), and more. In a new study from the University of Padua, a team of astrobiologists simulated how photosynthetic organisms (cyanobacteria) would fare on a planet orbiting a red dwarf. Their results experimentally demonstrated that oxygen photosynthesis could occur under red suns, which is good news for those looking for life beyond Earth!
Continue reading “Do Red Dwarfs Provide Enough Sunlight for Plants to Grow?”
Venus is Like an Exoplanet that’s Right Next Door
We’re lucky to have a neighbour like Venus, even though it’s totally inhospitable, wildly different from the other rocky planets, and difficult to study. Its thick atmosphere obscures its surface, and only powerful radar can penetrate it. Its extreme atmospheric pressure and high temperatures are barriers to landers or rovers.
It’s like having a mysterious exoplanet next door.
Continue reading “Venus is Like an Exoplanet that’s Right Next Door”
Earth-Sized Planet Found At One of the Lightest Red Dwarfs
Astronomers have found another Earth-sized planet. It’s about 31 light-years away and orbits in the habitable zone of a red dwarf star. It’s probably tidally locked, which can be a problem around red dwarf stars. But the team that found it is optimistic about its potential habitability.
Continue reading “Earth-Sized Planet Found At One of the Lightest Red Dwarfs”
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# Integrating a logarithm
## Main Question or Discussion Point
Hello
So I have a problem, which is to use integration by parts to integrate...
$$\int^{1}_{0}(1-x) ln (1-x) dx$$
The way I have been working is it to seperate it out into just...
$$\int^{1}_{0}ln (1-x) dx - \int^{1}_{0}x ln (1-x) dx$$
and then integrating by parts on each of these seperatele, but for instance if I integrate by parts the first bit, I get...
$$[xln(1-x)]^{1}_{0} + \int^{1}_{0}x \frac{1}{1-x}$$
And im thinking the first part to this doesn't make sense, because $$ln (0)$$ is a mathematical nono. So im confused with regard to this problem - has anybody any decent suggestions on how to do this?
Thanks :)
## Answers and Replies
The integral isn't proper (you're working right up to a pole!) so you must take limit to 1, not just put in the values.
Integral is proper because the function $$(1-x)\ln(1-x)$$ is continuous and bounded on $$[0,1)$$ (it has finite limit when $$x\rightarrow 1$$).
$$\int_0^1(1-x)\ln(1-x)\,dx=\left.-\frac{(1-x)^2\ln(1-x)}{2}\right|_0^1-\int_0^1\frac{1-x}{2}\,dx=\left.\frac{(1-x)^2}{4}\right|_0^1=-\frac{1}{4}.$$
Your problem is in decomposing the finite value in the form of substraction of two infinite values.
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# Thread: circular to oscillatory motion
1. ## circular to oscillatory motion
Hi all
I'm from Bioengineering and unfortunately not too expert on the mechanical side. I'm making a device for which I need a mechanism to transform circular motion into oscillatory motion. I found the mechanism in the picture below:
leva_oscillante_1.gif
In my device, the oscillating 'stick' would cover an angle of 170 degrees.
My questions are:
Is the oscillatory motion actually sinusoidal? (ie is the velocity/position actually sinusoidal, or are they just approximated to sinusoids due to the small angle shown in the picture?)
If it is/can be sinusoidal, is there a straight forward way to go about to find the specs (size and position of rotating wheel...) of the system?
Suggestions on different mechanisms I could use are very welcome. (I thought of using a cam but the oscillating 'stick' would in my case probably be jumpy due to the high velocity and low weight needed).
Thanks everyone!!
2. Kinny,
The motion will not be pure sinusoidal, the small angles of the connecting rod will skew the output motion away from a pure sine wave. Why don't you graphically lay out the mechanism and plot the output motion vs. crank angle. From that you should be able to determine if the motion will still work for you. I can tell you that if you connect the con rod to the lever and the crank with scotch yokes, the result will be sinusoidal. Or, for a more rigorous analysis, write the equation of motion for the output, only a bit more complicated than a graphical analysis, but still simple geometry.
Timelord
3. For a quick understanding of a "scotch yoke" see: http://en.wikipedia.org/wiki/Scotch_yoke
4. Actually, a slight adjustment in Timelord's and jalbert's answers. Since the motion of the rod is connected to a fixed point on the disc's surface the motion is not truly sinusoidal. Depends on two things: (1) the actual geometry of the connections, and (2) how accurate you really want to be. If the connection was truly as shown in the scotch yoke link above the horizontal position of the rod would be determined by the horizontal position of the connection on the disc. Actually it is dependent on both the horizontal and the vertical position of the connection on the disc. The greater the ratio between the length of the connector arm the the radius of motion of the connection point the closer the motion is to truly sinusoidal. Just to be anal about it...
5. I rarely counter jboggs inputs, but in this case he is apparently using your arrangement or a piston and crank arrangement as a reference for a scotch yoke design. A true scotch yoke as accurately shown in my above Wikipedia reference does in fact have a true sinusoidal stroke.
6. JAlberts,
I agree, jboggs is wrong, a scotch yoke gives true sinusoidal motion. I also said that true sinusoidal motion of the output will result only if both ends of the con rod are connected with scotch yokes.
jboggs FYI: By carefully reading your post I get the impression you may have misinterpreted the illustration. In the illustration of a scotch yoke in JAlberts link, its not shown, but the horizontal slider/con rod is constrained in one direction i.e. in this case, it can only slide horizontally, so it only reads the horizontal motion of the crank pin, which is by definition sinusoidal. If I wrong about this misinterpretation, I apologize.
Timelord
7. Apparently we have had a "failure to communicate". My comments about not being truly sinusoidal were aimed at the OP's illustration, not the scotch yoke diagram. I should have been more clear. The OP's diagram is not truly sinusoidal, the scotch yoke diagram is.
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## Engineering
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problem 1: Consider an MFSK scheme with fc = 250 kHz, fc = 25 kHz and M = 8 (L = 3 bits)
a) Make a frequency assignment for each of the eight possible 3-bit data combinations.
b) We wish to apply FHSS to this MFSK scheme with k = 2; that is, the system will hop among four various carrier frequencies.
Expand the outcomes of part a) to show that 4×8 = 32 frequency assignments.
problem 2: describe the fundamental principle of the fast and slow FHSS.
problem 3: Describe the concept of FHSS by using MFSK.
problem 4: How can a DSSS signal can be produced by using BPSK?
problem 5: Describe the concept of CDMA with an illustration.
problem 6: Describe CDMA for direct sequence spread spectrum.
problem 7: How does a single bit error distinct from a burst error?
problem 8: Describe the concept of the redundancy in error detection.
problem 9: How can a parity bit detect a damaged data unit?
problem 10: What is the fundamental difference between even parity and odd parity?
Electrical & Electronics, Engineering
• Category:- Electrical & Electronics
• Reference No.:- M97274
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# Model Eksponensial dan Logaritma
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Information about Model Eksponensial dan Logaritma
Education
Published on February 25, 2014
Author: snmpsimamora
Source: slideshare.net
## Description
MA205 Matematika Teknik I (3 sks) Dosen: Ir. Sihar, MT. Departemen Sistem Komputer – Fak. Teknik Bandung 2003 Eksponensial dan Logaritma
Referensi: Ayres, F., Mendelson, E. Calculus - 5th edition. Schaum's Series. McGraw-Hill. 1999. Edwards, L. Calculus-9th edition. Cengage Learning. 2010. Walker, J. Fundamentals of Physics - 9th edition. John Wiley & Sons. 2003.
Bilangan Euler (e) = 2.718281828459045 Didapatkan dari: e= 1 n (1 ) n
Bilangan Euler (e) = 2.718281828459045 Juga bisa didapatkan dari: e= n 1 1 1 1 1 1 ... 1 2 x1 3x 2 x1 4 x3 x 2 x1 0 n!
<script language=JavaScript> c=Math.log(10000); b=Math.log(10); a=c/b; document.write(a); </script> Postulat: ab c b log c log c log b y e a a x 25 20 15 Series1 10 5 0 -4 -2 0 2 4
Postulat: e 2.71828 ln y x ln e y ln x x 1 ln e 1 ln y x ln e ln y ln x x e ln y ln x ln y y ex 2 e 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2 -2.5 -3 2 4 6 8 Series1
1 y x x 0; x 3 2 1 0 -5 0 -1 -2 -3 -4 5 10 Series1
y x x2
Gerak Lurus Beraturan, Gerak Lurus Berubah Beraturan, Gerak Melingkar Berubah Beraturan
Gerak Lurus Beraturan, Gerak Lurus Berubah Beraturan, Gerak Melingkar Berubah Beraturan
Postulat Integral: n k .x dx k (n 1) .x n c.......... n 1 ... k .dx k .x c.......... k 1 dx x kons tan ta ln x c e x dx e x c Postulat merupakan representasi suatu teori yang dijabarkan dalam model matematika
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# is x^4 + y^4 > z^4? 1. x^2 + y^2 > z^2 2. x+y > z
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is x^4 + y^4 > z^4? 1. x^2 + y^2 > z^2 2. x+y > z [#permalink]
### Show Tags
06 Nov 2005, 21:52
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is x^4 + y^4 > z^4?
1. x^2 + y^2 > z^2
2. x+y > z
Kudos [?]: 12 [0], given: 0
VP
Joined: 22 Aug 2005
Posts: 1111
Kudos [?]: 124 [0], given: 0
Location: CA
### Show Tags
06 Nov 2005, 23:18
E.
1) false if
x=sqrt(2)
y=sqrt(2)
z=sqrt(3)
true for all positive integers
insufficient
2)
x=y=2
z=3
false
for x=2,y=3,z=3 true
insufficient
together:
x=y=sqrt(2)
z=sqrt(3)
false
x=y=z=1
true
insuffucient
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Senior Manager
Joined: 19 Feb 2005
Posts: 486
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Location: Milan Italy
### Show Tags
06 Nov 2005, 23:19
unless someone else has a quick way to do this, I'll go ahead with E.
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Director
Joined: 27 Jun 2005
Posts: 501
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Location: MS
### Show Tags
06 Nov 2005, 23:48
E
is x^4 + y^4 > z^4?
1. x^2 + y^2 > z^2
x=5,y=4,z=6 as 5^2+4^2>6^2 but 5^4+4^4 <6^4
x=3,y=1,z=2 as 3^2+1^2>2^2 and 3^4+1^4 > 2^4
so Insuff
2. x+y > z
x=3,y=2 so 3+2>4 but 3^4+2^4<4^4
x=9,y=10 so 9+10 > 4 and 9^4+10^4 > 4^4
so Insuff
now taking both together
x=9,y=10,z=5 satisfy both 1 and 2 and also gives yes
but
x=5,y=4,z=6 satisfy both 1 and 2 but gives No
so Insuff
....certainly took more than 2 minutes ...initally i was trying to solve the question by using equation and inequalites ...then gave up and tired with value substituion ........
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Manager
Joined: 19 Sep 2005
Posts: 110
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### Show Tags
07 Nov 2005, 10:29
2^2 + 2^2 is not greater than 3^2
8 < 9
so the example you used for statement 1 is not usable.
Anyone got a quicker way to solve this then picking numbers? I'm afraid I'd spend all day on something like this.
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Current Student
Joined: 28 Dec 2004
Posts: 3345
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Location: New York City
Schools: Wharton'11 HBS'12
### Show Tags
07 Nov 2005, 11:29
This is E...
and (1)
try numbers for X, Y and Z such as sqrt(x), Sqrt(y) and Sqrt(z)
if x=sqrt(2)=y
z=sqrt(3)
x^2 + y^2 > z^2
2+2=4 > 3
but
2^2+2^2<3^2 OK
Insuff
(d) use the same number x+y>z but we still have 2 possibilites...
always try fractions, positive, negative and sqrt numbers to verify these DS questions....no other shorter and sweeeter way...
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07 Nov 2005, 11:29
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Problem: An equilibrium mixture of the following reaction was found to have [SO3] = 0.411 M and [O2] = 0.205 M at 600°C What is the concentration of SO2?2 SO2(g) + O2(g) ⇌ 2 SO3(g)Keq = 4.14 at 600 °C
🤓 Based on our data, we think this question is relevant for Professor Gillespie's class at FIU.
Problem Details
An equilibrium mixture of the following reaction was found to have [SO3] = 0.411 M and [O2] = 0.205 M at 600°C What is the concentration of SO2?
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
Keq = 4.14 at 600 °C
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SELF 32 S.B. Karavashkin and O.N. Karavashkina
3. The law of vectors summation with Lorentz transformation In this item we will check the validity of the law of vectors summation in IRFs which is known to be the basic law both in studying of the trajectories of bodies motion and for finding the first and following derivatives. In general case, when differentiating, we always seek the difference in vectors - it means that the validity of vectors summation law in fact defines the legacy of differentiating and integrating in space in which we perform this operation. It is also known and long ago proven that in classical formalism, in limits of IRFs, this law is absolutely true. “Equations of mechanics have an identical form in any inertial reference frames” [10, p. 73]. With it, “the constancy (invariance) of laws of mechanics in Galilean transformations is the content of the relativity principle of the Newtonian mechanics” [ibidem]. It is easy to show it again. Let we have two Galilean reference frames K and K' mutually moving along the common axis x with the speed v, and let with respect to the frame K there moves some material point with velocity (ux, uy) . Then we can check, whether the law of vectors summation is true, determining the resulting vector in K' by two ways: transforming the vector into the given coordinate system and finding its module and the inclination angle to the coordinate axes, and transforming the vector projections and also finding the module and the inclination angle of the resulting vector in the dotted coordinate axes. If both calculations come to the same result, the law in this mathematical formalism is true. If the results are different, the law in this formalism is violated. Such check is common, as it does not require the equal modules either retained inclination angles of vectors in transformation, but only clears, whether the formalism is closed. This has to provide the result of calculation independently of the way in which we calculated. This principle of closed formalism undoubtedly has to remain true in IRFs, where the space is not warped. In classical formalism it is checked very simple. In direct transformation of the vector of speed
(39)
whence
(40)
On the other hand, transforming the x-projection of speed, we have
(41)
and accordingly for the y-projection
(42)
From this, (41) and (42) fully coincide with (40), which corroborates the law of vectors summation true in Galilean RFs. In relativistic mechanics, with all outward simplicity of the check, it is not so obvious. To show it, let us first determine the module of velocity of the point in the RF K' with direct transformation of vector. In this case the point moves in the plane, so in accordance with the law of relativistic speeds summation
(43)
Thus,
(44)
Now let us try to yield (44) through transformation of projections of into K' and determining the module of resulting vector. For the x-projection of speed
(45)
and for y-projection, accordingly,
(46)
As in transformation the y-projection gains the projection in x' ,
(47)
Comparing (44) and (47), we see different results. This means, in relativistic formalism the law of vectors summation in transformation from one IRF to another is not true. With it, we may not state that only the way of direct transformation of vector is true but the way of transformation of projections of vector is untrue, as we might say, basing on the standard relativistic statements like the following. “But we cannot use one kinematics for one amount of physical phenomena and another - for another amount, requiring the invariance in relation to Galilean transformation for mechanics and invariance in relation to Lorentz transformation for electrodynamics. However, we know that the first transformation is the limiting case of that second, just the case in which the constant c is infinitely large. Accordingly, following Einstein, we will premise that classical mechanics is not strongly true but rather needs to be some modified. The laws of new mechanics have to appear invariant with respect to Lorentz transformation” [11, p. 261]. To estimate the degree of Born’s statement, we have to draw our attention that the very his statement has a stipulation to check, whether Galilean and Lorentz transformations are compatible. If we agree that the invariance of Lorentz transformations generally has not to be identical to Galilean invariance, we at the same time, tending the value of speed of light to infinity, have to yield (44) and (47) coinciding. But we yield in the first case
(48)
and in the second
(49)
Thus, in the limit passing to Galilean RF, we also do not achieve the equal results of transformation; this evidences of untrue law of vector summation in formalism containing Lorentz transformation. Noting this feature, we already may not establish any relations in vectorial projections, may not differentiate, may not consider the law of angular momentum conservation for a system of bodies, as the very angular momentum is a vectorial value and in the course of proof we have to sum the momenta of bodies which the system contains. Also, in relativistic formalism the study of dynamic behaviour of bodies loses any sense, as the result of modelling will depend not on the regularities of phenomena but on the chosen way of calculation.
Contents: / 29 / 30 / 31 / 32 / 33 / 34 / 35 / 36 / 37 / 38 / 39 /
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Math May 2005 Exam Paper C1
# Math May 2005 Exam Paper C1
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N23491A
This publication may only be reproduced in accordance with London Qualifications copyright policy.©2005 London Qualifications Limited.Paper Reference(s)
6663/01
Edexcel GCE
Core Mathematics C1Advanced Subsidiary
Monday 23 May2005
MorningTime: 1 hour 30 minutes
Materials required for examination Items included with question papers
Mathematical Formulae (Green)
Nil
Calculators may NOT be used in this examination.
Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, theunit title (Core Mathematics C1), the paper reference (6663), your surname, initials andsignature.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.There are 10 questions in this question paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answerswithout working may gainno credit.
© Science Exam Papers
N23491A
2
1.
(
a
)Write down the value of
31
8.
(1)
(
b
)Find the value o
32
8
.
(2)2.
Given that
y
= 6
x
–
2
4
x
,
x
≠ 0,
(
a
)find
x y
dd,
(2)
(
b
)find
y
d
x
.
(3)3.
x
2
–8
x
29
(
x
+
a
)
2
+
b
,where
a
and
b
are constants.(
a
)Find the value o
a
and the value of
b
.
(3)
(
b
)Hence, or otherwise, show that the roots o
x
2
–8
x
–29 = 0are
c
5, where
c
and
are integers to be found.
(3)
© Science Exam Papers
N23491A
3
Turn over4.Figure 1
Figure 1 shows a sketch of the curve with equation
y
= f(
x
). The curve passes through theorigin
O
and through the point (6, 0). The maximum point on the curve is (3, 5).On separate diagrams,sketch the curve with equation(
a
)
y
= 3f(
x
),
(2)
(
b
)
y
= f(
x
+ 2).
(3)
On each diagram, show clearly the coordinates of the maximum point and of each point at whichthe curve crosses the
x
-axis.
5
.
Solve the simultaneous equations
x
–2
y
= 1,
x
2
+
y
2
= 29.
(6)
y x
(6, 0)
O
(3, 5)
© Science Exam Papers
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# You're Going To Have To Think! Why Interval Arithmetic Won't Cure Your Floating Point Blues
So far we have discussed floating point arithmetic[1], fixed point arithmetic[2], arbitrary precision arithmetic[3], rational arithmetic[4] and computer algebra[5] and have found that, with the notable exception of the unfortunately impossible to achieve infinite precision that the last of these seems to promise, they all suffer from the problems that floating point arithmetic is so often criticised for.
No matter what number representation we use, we shall in general have to think carefully about how we use them.
We must design our algorithms so that we keep approximation errors in check, only allowing them to grow as quickly as is absolutely unavoidable.
We must ensure that these algorithms can, as far as is possible, identify their own modes of failure and issue errors if and when they arise.
Finally, we must use them cautiously and with full awareness of their numerical properties.
Now, all of this sounds like hard work, so it would be nice if we could get the computer to do the analysis for us.
Whilst it would be extremely difficult to automate the design of stable numerical algorithms, thankfully there is a numeric type that can keep track of the errors as they accumulate.
### Interval Arithmetic
Recall that, assuming we are rounding to nearest, the basic floating point arithmetic operations are guaranteed to introduce a proportional error of no greater than $$1 \pm \frac{1}{2} \epsilon$$. It’s not wholly unreasonable to assume that more complex floating point functions built into our chips and libraries will also do no worse than $$1 \pm \frac{1}{2} \epsilon$$.
This suggests that it might be possible to represent a number with upper and lower bounds of accuracy and propagate the growth of these bounds through arithmetic operations.
To capture the upper and lower bound of the result of a calculation accurately we need to perform it twice; once rounding towards minus infinity and once rounding towards plus infinity. Unfortunately, JavaScript provides no facility for manipulating the rounding mode.
We shall, instead, propagate proportional errors at a rate of $$1 \pm 1\frac{1}{2} \epsilon$$ since we shall be able to do this without switching the IEEE rounding mode. Specifically, we can multiply a floating point result by $$1+\epsilon$$ to get an upper bound and multiply it by $$1-\epsilon$$ to get a lower bound. The rate of error propagation results from widening the $$1 \pm \frac{1}{2} \epsilon$$ proportional error in the result by this further $$1 \pm \epsilon$$ to yield the bounds.
Naturally, this will not work correctly for denormalised numbers since they have a zero rather than a one before the decimal point, so we shall have to treat them as a special case. Specifically, rather than multiplying by $$1 \pm \epsilon$$, we shall add and subtract the smallest non-zero floating point number, conveniently provided by ak.MIN_VALUE.
Before we apply rounding error to the upper and lower bounds of the result a calculation we shall want them to represent the largest and the smallest possible result respectively. We therefore define the basic arithmetic operations as follows.
\begin{align*} (a,b)+(c,d) &= (a+c,b+d)\\ (a,b)-(c,d) &= (a-d,b-c)\\ (a,b)\times(c,d) &= (\min(a \times c, a \times d, b \times c, b \times d), \max(a \times c, a \times d, b \times c, b \times d))\\ (a,b)\div(c,d) &= (\min(a \div c, a \div d, b \div c, b \div d), \max(a \div c, a \div d, b \div c, b \div d)) \end{align*}
As an example, consider the square root of 2. Working with 3 accurate decimal digits of precision, this would be represented by the interval
$(1.413,1.415)$
Squaring this result would yield
$(1.413 \times 1.413 - 0.001, 1.415 \times 1.415 + 0.001)$
or
$(1.996, 2.003)$
which clearly straddles the exact result of 2 and furthermore gives a good indication of the numerical error.
Unfortunately, as is often the case, the devil is in the details.
The first thing we need to decide is whether the bounds are open or closed; whether the value represented is strictly between the bounds or can be equal to them. If we choose the former we cannot represent those numbers which are exact in floating point, like zero for example, so we shall choose the latter.
However, if either of the bounds is infinite we should prefer to consider them as open so as to simplify the rules of our interval arithmetic. To ensure consistency we shall map a lower bound of $$+\infty$$ to the maximum floating point value and an upper bound of $$-\infty$$ to the negated maximum floating point value.
A consequence of treating infinite bounds as open is that multiplying them by zero can yield zero rather than NaN. In particular, we have as special case of
$(-\infty, \infty) \times (0, 0) = (0, 0)$
and hence
$\frac{(0,0)}{(0,0)} = (-\infty, \infty)$
This might seem a bit odd, but it makes perfect sense once we realise that (-∞, ∞) is pronounced any number.
Making certain assumptions we can consistently, albeit not particularly mathematically soundly, define division by intervals containing zero or either infinity as follows. Given
\begin{align*} 0 & < a \leqslant b\\ 0 & < c\\ 0 & < d\\ 0 & < e\\ 0 & < f \end{align*}
we define
\begin{align*} (-\infty,\infty) &= \frac{(a,b)}{(0,0)} = \frac{(-b,-a)}{(0,0)} = \frac{(-c,d)}{(0,0)} = \frac{(0,0)}{(0,0)}\\ &= \frac{(a,b)}{(-e,f)} = \frac{(-b,-a)}{(-e,f)} = \frac{(-c,d)}{(-e,f)} = \frac{(0,0)}{(-e,f)}\\ &= \frac{(-c,d)}{(0,f)} = \frac{(-c,d)}{(-e,0)}\\ (0,\infty) &= \frac{(a,b)}{(0,f)} = \frac{(-b,-a)}{(-e,0)} = \frac{(0,d)}{(0,f)} = \frac{(-c,0)}{(-e,0)}\\ (-\infty,0) &= \frac{(a,b)}{(-e,0)} = \frac{(-b,-a)}{(0,f)} = \frac{(0,d)}{(-e,0)} = \frac{(-c,0)}{(0,f)} \end{align*}
and
\begin{align*} (0,\infty) &= \frac{(d,\infty)}{(f,\infty)} = \frac{(-\infty,-c)}{(-\infty,-e)}\\ (-\infty,0) &= \frac{(d,\infty)}{(-\infty,-e)} = \frac{(-\infty,-c)}{(f,\infty)} \end{align*}
### An Interval Class
Listing 1 gives the definition of an interval number class.
Listing 1: ak.interval
ak.INTERVAL_T = 'ak.interval';
function Interval(){}
Interval.prototype = {TYPE: ak.INTERVAL_T, valueOf: function(){return ak.NaN;}};
ak.interval = function() {
var i = new Interval();
var arg0 = arguments[0];
var state = {lb: 0, ub: 0};
constructors[ak.nativeType(arg0)](state, arg0, arguments);
if(isNaN(state.lb) || isNaN(state.ub)) {
state.lb = ak.NaN;
state.ub = ak.NaN;
}
if(state.lb=== ak.INFINITY) state.lb = ak.MAX_VALUE;
if(state.ub===-ak.INFINITY) state.ub = -ak.MAX_VALUE;
i.lb = function() {return state.lb;};
i.ub = function() {return state.ub;};
i.mid = function() {return 0.5*(state.lb+state.ub);};
i.toNumber = i.mid;
i.toString = function() {return '[' + state.lb + ',' + state.ub + ']';};
i.toExponential = function(d) {
return '['+state.lb.toExponential(d)+','+state.ub.toExponential(d)+']';
};
i.toFixed = function(d) {
return '['+state.lb.toFixed(d)+','+state.ub.toFixed(d)+']';
};
i.toPrecision = function(d) {
return '['+state.lb.toPrecision(d)+','+state.ub.toPrecision(d)+']';
};
return Object.freeze(i);
};
var constructors = {};
As usual, the class only provides property accessor and type conversion methods, which all have very simple implementations. The lb and ub properties are the defining elements, being the lower and upper bounds of the interval respectively.
The state is initialised by dispatching calls to a constructors object based on the types of the arguments after which it is validated by propagating NaNs to both bounds and handling infinites as described above.
### ak.interval Constructors
The constructors are similar in structure to those of our other classes, as shown in listing 2.
Listing 2: ak.interval Constructors
constructors[ak.NUMBER_T] = function(state, x, args) {
var arg1 = args[1];
constructors[ak.NUMBER_T][ak.nativeType(arg1)](state, x, arg1);
};
constructors[ak.NUMBER_T][ak.NUMBER_T] = function(state, lb, ub) {
var i = sort(Number(lb), Number(ub));
state.lb = i.lb;
state.ub = i.ub;
};
constructors[ak.NUMBER_T][ak.UNDEFINED_T] = function(state, x) {
var y = Number(x);
var i = widen(y, y)
state.lb = i.lb;
state.ub = i.ub;
};
constructors[ak.OBJECT_T] = function(state, obj) {
var lb = ak.nativeType(obj.lb)===ak.FUNCTION_T ? Number(obj.lb()) : Number(obj.lb);
var ub = ak.nativeType(obj.ub)===ak.FUNCTION_T ? Number(obj.ub()) : Number(obj.ub);
var i = sort(lb, ub);
state.lb = i.lb;
state.ub = i.ub;
};
Note that if the object is constructed with a pair of numbers, either explicitly or via properties of an initialising object, they are assumed to be the exact bounds of the interval and are put into the correct order with a call to sort. If constructed with a single number, it is assumed to be an inexact value and we add our conservative error estimates with a call to widen
The definitions of sort and widen are given in listing 3.
Listing 3: ak.interval sort And widen
var EPS_SUB = 1-ak.EPSILON;
var EPS_SUP = 1+ak.EPSILON;
function sort(x, y) {
return x<=y ? {lb: x, ub: y} : {lb: y, ub: x};
}
function widenLB(lb) {
if(lb>ak.MIN_NORMAL) lb *= EPS_SUB;
else if(lb<-ak.MIN_NORMAL) lb *= EPS_SUP;
else lb -= ak.MIN_VALUE;
return lb
}
function widenUB(ub) {
if(ub>ak.MIN_NORMAL) ub *= EPS_SUP;
else if(ub<-ak.MIN_NORMAL) ub *= EPS_SUB;
else ub += ak.MIN_VALUE;
return ub;
}
function widen(lb, ub) {
var i = sort(lb, ub);
i.lb = widenLB(i.lb);
i.ub = widenUB(i.ub);
return i;
}
The sort function is simple enough, but widen could bear some further explanation. We have already noted that denormalised numbers must accumulate errors additively rather than multiplicitively, which we do in the final clauses of the widening functions.
The tricky bit is to make sure that the lower bound always widens towards minus infinity and the upper bound towards plus infinity. We do this by carefully choosing whether to multiplicatively widen normal valued bounds towards or away from zero depending on their signs.
As usual, overloaded functions do most of the work and, as usual, neg is amongst the simplest, as shown in listing 4.
Listing 4: ak.interval neg
function neg(x) {
return ak.interval(-x.ub(), -x.lb());
}
Since sign manipulation is exact, this function introduces no additional error to the interval.
Unusually, abs isn't one of the simplest, as shown in listing 5.
Listing 5: ak.interval abs
function abs(x) {
if(x.lb()>=0) return x;
if(x.ub()<=0) return ak.interval(-x.ub(), -x.lb());
return ak.interval(0, Math.max(-x.lb(), x.ub()));
}
The subtlety here is that if we have a negative lower bound and a positive upper bound, then the smallest absolute value in the interval is zero and not the smaller absolute bound.
In comparison, addition and subtraction are relatively simple, as illustrated by listing 6.
Listing 6: ak.interval add And sub
function add(lhs, rhs) {
return ak.interval(widen(lhs.lb()+rhs.lb(), lhs.ub()+rhs.ub()));
}
function sub(lhs, rhs) {
return ak.interval(widen(lhs.lb()-rhs.ub(), lhs.ub()-rhs.lb()));
}
Things start getting a little tricky when it comes to multiplication, as shown in listing 7.
Listing 7: ak.interval mul
function mul(lhs, rhs) {
var ll, lu, ul, uu, lb, ub;
if(isNaN(lhs.lb()) || isNaN(rhs.lb())) return ak.interval(ak.NaN, ak.NaN);
ll = lhs.lb() * rhs.lb();
lu = lhs.lb() * rhs.ub();
ul = lhs.ub() * rhs.lb();
uu = lhs.ub() * rhs.ub();
if(isNaN(ll)) ll = 0;
if(isNaN(lu)) lu = 0;
if(isNaN(ul)) ul = 0;
if(isNaN(uu)) uu = 0;
lb = Math.min(Math.min(ll, lu), Math.min(ul, uu));
ub = Math.max(Math.max(ll, lu), Math.max(ul, uu));
return ak.interval(widen(lb, ub));
}
Now this is essentially the scheme described above where the bounds are set to the minimum and maximum of all possible products. However, since we are treating infinities as open bounds, we must take care to ensure that when multiplied by zero they yield zero rather than NaN. Since the only multiplicative operations that result in NaN are those involving a NaN or two or those of an infinity and zero, by handling the former with an early return we can safely replace those that arise from the latter with zeros.
Division is similarly based on that scheme, albeit with a few more corner cases, as illustrated in listing 8.
Listing 8: ak.interval div
function div(lhs, rhs) {
var ll, lu, ul, uu, lb, ub;
if(isNaN(lhs.lb()) || isNaN(rhs.lb())) return ak.interval(ak.NaN, ak.NaN);
if(rhs.lb()>0 || rhs.ub()<0) {
ll = lhs.lb() / rhs.lb();
lu = lhs.lb() / rhs.ub();
ul = lhs.ub() / rhs.lb();
uu = lhs.ub() / rhs.ub();
if(isNaN(ll)) ll = ak.INFINITY;
if(isNaN(lu)) lu = -ak.INFINITY;
if(isNaN(ul)) ul = -ak.INFINITY;
if(isNaN(uu)) uu = ak.INFINITY;
lb = Math.min(Math.min(ll, lu), Math.min(ul, uu));
ub = Math.max(Math.max(ll, lu), Math.max(ul, uu));
return ak.interval(widen(lb, ub));
}
if((rhs.lb() < 0 && rhs.ub() > 0)
|| (rhs.lb()===0 && rhs.ub()===0)
|| (lhs.lb() < 0 && lhs.ub() > 0)) {
lb = -ak.INFINITY;
ub = ak.INFINITY;
}
else if((rhs.lb()===0 && lhs.lb()>=0)
|| (rhs.ub()===0 && lhs.ub()<=0)) {
lb = 0;
ub = ak.INFINITY;
}
else if((rhs.lb()===0 && lhs.ub()<=0)
|| (rhs.ub()===0 && lhs.lb()>=0)) {
lb = -ak.INFINITY;
ub = 0;
}
return ak.interval(lb, ub);
}
Once again, we handle NaNs with an early return to simplify later cases.
If the right hand side of the division does not include zero, if the upper bound is negative or the lower bound is positive, we have a similar implementation to that of multiplication. The only possible source of NaNs is the division of one infinity by another and, given that infinities represent open bounds and that arbitrarily large numbers divided by arbitrarily large numbers result in arbitrarily large numbers, they should be replaced with infinities. Since ak.interval ensures that lower bounds cannot be +∞ and upper bounds cannot be -∞ we can easily identify which NaNs should correspond to which infinities.
The last three cases handle division by zero in what I hope is a reasonably consistent fashion. If the right hand side straddles, or is identically equal to, zero then the result of the division can be any number, as represented by (-∞, +∞). Similarly, if the left hand side straddles zero and the right hand side includes zero then the result could be any number. Finally, if the left and right hand sides represent any number of a given sign, including zero, we return the relavent semi-infinite interval.
Listing 9 shows the full complement of ak.interval overloads.
ak.overload(ak.abs, ak.INTERVAL_T, abs);
The complete implementation of ak.interval can be found in Interval.js.
### Using ak.interval
Note that the only comparisons provided are equality and inequality since intervals can overlap, or even be subsets and supersets of one another, making it rather difficult to define exactly what it means to say one is less than or greater than another.
Program 1 illustrates the use ak.interval of objects for calculations, specifically for solving quadratic equations using the formula we learnt as children; that
$a x^2 + b x + c = 0$
has solutions
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Program 1: Using ak.interval
The functions solve and quad use the ak.calc RPN calculator, which in turn uses the overloaded functions, to compute the solutions to the quadratic equation and the value of the left hand side for a given argument.
I think that it's pretty clear that the results of these calculations both include the exact results and capture the uncertainty introduced by working with finite precision.
Note that we use the two argument ak.interval constructors for integers that we know can be exactly represented.
### Aliasing
Unfortunately interval arithmetic is not entirely foolproof. One problem is that it can yield overly pessimistic results if an interval appears more than once in an expression. For example, consider the result of multiplying the interval $$(-1, 1)$$ by itself.
Ignoring rounding error, doing so yields
\begin{align*} (-1, 1) \times (-1, 1) &= \left(\min(-1 \times -1, -1 \times 1, 1 \times -1, 1 \times 1), \max(-1 \times -1, -1 \times 1, 1 \times -1, 1 \times 1)\right)\\ &= (-1, 1) \end{align*}
rather than the correct result of $$(0, 1)$$.
This can be particularly troublesome if such expressions appear as the denominator in a division. For example, given
\begin{align*} x &= (-1, 1)\\ y &= (\tfrac{1}{2}, 1)\\ z &= (0, 1) \end{align*}
and again ignoring rounding errors, we have
\begin{align*} \frac{z}{x \times x + y} &= \frac{(0, 1)}{(-1, 1) + (\tfrac{1}{2}, 1)}\\ &= \frac{(0, 1)}{(-\tfrac{1}{2}, 2)}\\ &= (-\infty, \infty) \end{align*}
rather than
\begin{align*} \frac{z}{x \times x + y} &= \frac{(0, 1)}{(0, 1) + (\tfrac{1}{2}, 1)}\\ &= \frac{(0, 1)}{(\tfrac{1}{2}, 2)}\\ &= (0, 2) \end{align*}
If we wish to ensure that such expressions yield as accurate results as possible we shall have to rearrange them so that we avoid aliasing. For example, rather than
$x \times x + 3x - 1$
we should prefer
$(x + 1.5)^2 - 3.25$
Provided we keep in mind the fact that interval arithmetic can be pessimistic we can still use it naively to give us a warning of possible precision loss during a calculation.
Unfortunately there is a much bigger problem that we must consider.
### Precisely Wrong
Consider the finite difference approximation to the derivative
$\frac{\mathrm{d}f}{\mathrm{d}x} \approx \frac{f(x + \delta) - f(x)}{\delta}$
Recall that such calculations are sensitive to cancellation error when $$\delta$$ is very small. By using our interval type we might identify if such errors have occurred.
Listing 10 illustrates how we might do so for the derivative of the exponential function at 1 with integer i leading binary zeros in $$\delta$$.
Listing 10: An Approximate Derivative
var d = Math.pow(2, -i);
var x = ak.interval(1, 1);
var xd = ak.interval(1+d, 1+d);
var df = ak.calc(xd,'e^',x,'e^','-');
var dfdx = ak.interval(df.lb()/d, df.ub()/d);
Note that because we have chosen d to be a power of two then, so long as it's no smaller than the floating point epsilon, 1+d is exactly representable which is why we can safely use an exact interval to represent it. In the final line we are exploiting the fact that a division by a negative power of 2 is also exact.
Program 2 plots the number of accurate binary digits in the finite difference approximation in black and the number of precise binary digits in the finite difference formula in red against the position of the most significant binary digit in $$\delta$$.
Program 2: Finite Difference Precision
Clearly there’s a linear relationship between the number of leading zeros of $$\delta$$ and the lack of precision in the finite difference calculation. Unfortunately, it is initially in the opposite sense to that between the number of leading zeros and the accuracy of the approximation.
We must therefore be extremely careful to distinguish between these two types of error. If we consider precision alone we are liable to very precisely calculate the wrong number.
So, whilst intervals are an extremely useful tool for ensuring that errors in precision do not grow to significantly impact the accuracy of a calculation, they cannot be used blindly. As has consistently been the case, we shall have to think carefully about our calculations if we wish to have confidence in their results.
### References
[1] You're Going To Have To Think! Why [Insert Technique Here] Won't Cure Your Floating Point Blues, www.thusspakeak.com, 2013.
[2] You're Going To Have To Think! Why Fixed Point Arithmetic Won’t Cure Your Floating Point Blues, www.thusspakeak.com, 2013.
[3] You're Going To Have To Think! Why Arbitrary Precision Arithmetic Won’t Cure Your Floating Point Blues, www.thusspakeak.com, 2013.
[4] You're Going To Have To Think! Why Rational Arithmetic Won’t Cure Your Floating Point Blues, www.thusspakeak.com, 2013.
[5] You're Going To Have To Think! Why Computer Algebra Won’t Cure Your Floating Point Blues, www.thusspakeak.com, 2013.
Based upon an article I wrote for ACCU's Overload magazine.
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Cody
# Problem 5. Triangle Numbers
Solution 535935
Submitted on 27 Nov 2014 by Adam
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
%% n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
%% n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
%% n = 30; t = 465; assert(isequal(triangle(n),t))
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An Entity of Type: book, from Named Graph: http://dbpedia.org, within Data Space: dbpedia.org
In mathematical logic, model theory is the study of the relationship between formal theories (a collection of sentences in a formal language expressing statements about a mathematical structure), and their models (those structures in which the statements of the theory hold). The aspects investigated include the number and size of models of a theory, the relationship of different models to each other, and their interaction with the formal language itself. In particular, model theorists also investigate the sets that can be defined in a model of a theory, and the relationship of such definable sets to each other.As a separate discipline, model theory goes back to Alfred Tarski, who first used the term "Theory of Models" in publication in 1954.Since the 1970s, the subject has been shaped deci
Property Value
dbo:abstract
• In mathematical logic, model theory is the study of the relationship between formal theories (a collection of sentences in a formal language expressing statements about a mathematical structure), and their models (those structures in which the statements of the theory hold). The aspects investigated include the number and size of models of a theory, the relationship of different models to each other, and their interaction with the formal language itself. In particular, model theorists also investigate the sets that can be defined in a model of a theory, and the relationship of such definable sets to each other.As a separate discipline, model theory goes back to Alfred Tarski, who first used the term "Theory of Models" in publication in 1954.Since the 1970s, the subject has been shaped decisively by Saharon Shelah's stability theory. Compared to other areas of mathematical logic such as proof theory, model theory is often less concerned with formal rigour and closer in spirit to classical mathematics.This has prompted the comment that "if proof theory is about the sacred, then model theory is about the profane".The applications of model theory to algebraic and diophantine geometry reflect this proximity to classical mathematics, as they often involve an integration of algebraic and model-theoretic results and techniques. The most prominent scholarly organization in the field of model theory is the Association for Symbolic Logic. (en)
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• In mathematical logic, model theory is the study of the relationship between formal theories (a collection of sentences in a formal language expressing statements about a mathematical structure), and their models (those structures in which the statements of the theory hold). The aspects investigated include the number and size of models of a theory, the relationship of different models to each other, and their interaction with the formal language itself. In particular, model theorists also investigate the sets that can be defined in a model of a theory, and the relationship of such definable sets to each other.As a separate discipline, model theory goes back to Alfred Tarski, who first used the term "Theory of Models" in publication in 1954.Since the 1970s, the subject has been shaped deci (en)
rdfs:label
• Model theory (en)
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# Tessellating Capitals
## Tesselating Capitals
I know that when doing symmetry we often have a look at CAPITAL LETTERS.
We could try tessellating certain ones, deciding on good ones to try and those that would be no good.
A simple one to start with is the letter C - see what has happened when I've tried to tessellate them:
Not hard!
Slightly different we have:
Do you see how they are different?
Now try another one or two for yourself!
Capital F also seems a good one, but maybe it's a little harder.
I must admit I chose the sizes to be a bit easier!
I decided to lay two of them down like this:
Then playing about with putting them together I got:
This I was quite happy with, so I thought that having shown you, you could try some for yourself.
Please send us your tessellations. You could scan a drawing into the computer and send it that way, or you might like to take a digital photo of the tessellations.
### Why do this problem?
This is a really good activity to engage all children. Those who are less motivated and find mathematics difficult will be able to have a go (and may well surprise you) but there is also enjoyment and intrigue here for the enthusiastic and mature mathematician!
The problem is good for developing pupils' ability to visualise, and depending on their experience, you may also want to use this as an opportunity to encourage them to use correct mathematical vocabulary relating to transformations.
### Possible approach
This is a sheet with four letters, though you could use other ones too.
Other simple C's I found were;
and
Just to carry on here is an H tessellation.
It's also good to look at what seem like very simple [maybe too simple, we think!] letters and find many ways of tessellating them. To show you the sort of thing I mean look at these variations on tessellating the simple letter L. I used four different orientations and coloured them to help.
### Key questions
The following questions might be useful in coaxing children to think more deeply, and perhaps differently, about what they are doing:
Could this tessellation be continued for ever? Why or why not?
What is the single repeating unit in the tessellation?
How would you describe the tessellation to someone else?
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Related Articles
# 3D Visualisation of Merge Sort using Matplotlib
• Last Updated : 17 Aug, 2020
Visualizing algorithms makes it easier to understand them by analyzing and comparing the number of operations that took place to compare and swap the elements. 3D visualization of algorithms is less common, for this we will use matplotlib to plot bar graphs and animate them to represent the elements of the array.
Approach :
1. We will generate an array with random elements.
2. The algorithm will be called on that array and yield statement will be used instead of return statement for visualization purposes.
3. We will yield the current states of the array after comparing and swapping. Hence the algorithm will return a generator object.
4. Matplotlib animation will be used to visualize the comparing and swapping of the array.
5. We will then plot the graph, which will return an object of Poly3dCollection using which further animation will be done.
`# import all the modules``import` `matplotlib.pyplot as plt``from` `matplotlib.animation ``import` `FuncAnimation``from` `mpl_toolkits.mplot3d ``import` `axes3d``import` `matplotlib as mp``import` `numpy as np``import` `random`` ` `# merge sort function to divide the array``def` `mergesort(A, start, end):`` ``if` `end <``=` `start:`` ``return`` ` ` ``mid ``=` `start ``+` `((end ``-` `start ``+` `1``) ``/``/` `2``) ``-` `1`` ` ` ``# yield from statement is used to `` ``# yield the array from the merge function`` ``yield` `from` `mergesort(A, start, mid)`` ``yield` `from` `mergesort(A, mid ``+` `1``, end)`` ``yield` `from` `merge(A, start, mid, end)`` ` `# function to merge the array``def` `merge(A, start, mid, end):`` ``merged ``=` `[]`` ``leftIdx ``=` `start`` ``rightIdx ``=` `mid ``+` `1`` ` ` ``while` `leftIdx <``=` `mid ``and` `rightIdx <``=` `end:`` ``if` `A[leftIdx] < A[rightIdx]:`` ``merged.append(A[leftIdx])`` ``leftIdx ``+``=` `1`` ``else``:`` ``merged.append(A[rightIdx])`` ``rightIdx ``+``=` `1`` ` ` ``while` `leftIdx <``=` `mid:`` ``merged.append(A[leftIdx])`` ``leftIdx ``+``=` `1`` ` ` ``while` `rightIdx <``=` `end:`` ``merged.append(A[rightIdx])`` ``rightIdx ``+``=` `1`` ` ` ``for` `i ``in` `range``(``len``(merged)):`` ``A[start ``+` `i] ``=` `merged[i]`` ``yield` `A`` ` `# function to plot bars``def` `showGraph():`` ` ` ``# for random unique values`` ``n ``=` `int``(``input``(``"enter array size\n"``))`` ``a ``=` `[i ``for` `i ``in` `range``(``1``, n ``+` `1``)]`` ``random.shuffle(a)`` ``datasetName ``=` `'Random'`` ` ` ``# generator object returned by the function`` ``generator ``=` `mergesort(a, ``0``, ``len``(a)``-``1``)`` ``algoName ``=` `'Merge Sort'`` ` ` ``# style of the chart`` ``plt.style.use(``'fivethirtyeight'``)`` ` ` ``# set colors of the bars`` ``data_normalizer ``=` `mp.colors.Normalize()`` ``color_map ``=` `mp.colors.LinearSegmentedColormap(`` ``"my_map"``,`` ``{`` ``"red"``: [(``0``, ``1.0``, ``1.0``),`` ``(``1.0``, .``5``, .``5``)],`` ``"green"``: [(``0``, ``0.5``, ``0.5``),`` ``(``1.0``, ``0``, ``0``)],`` ``"blue"``: [(``0``, ``0.50``, ``0.5``),`` ``(``1.0``, ``0``, ``0``)]`` ``}`` ``)`` ` ` ``fig ``=` `plt.figure()`` ``ax ``=` `fig.add_subplot(projection ``=` `'3d'``)`` ` ` ``# z values and posistions of the bars `` ``z ``=` `np.zeros(n)`` ``dx ``=` `np.ones(n)`` ``dy ``=` `np.ones(n)`` ``dz ``=` `[i ``for` `i ``in` `range``(``len``(a))]`` ` ` ``# Poly3dCollection returned into variable rects`` ``rects ``=` `ax.bar3d(``range``(``len``(a)), a, z, dx, dy, dz, `` ``color ``=` `color_map(data_normalizer(``range``(n))))`` ` ` ``# setting and x and y limits equal to the length of the array`` ``ax.set_xlim(``0``, ``len``(a))`` ``ax.set_ylim(``0``, ``int``(``1.1` `*` `len``(a)))`` ``ax.set_title(``"ALGORITHM : "` `+` `algoName ``+` `"\n"` `+` `"DATA SET : "` `+` ` ``datasetName, fontdict ``=` `{``'fontsize'` `: ``13``, `` ``'fontweight'` `: ``'medium'``, `` ``'color'` `: ``'#E4365D'``})`` ` ` ``# text to plot on the chart`` ``text ``=` `ax.text2D(``0.1``, ``0.95``, "", horizontalalignment ``=``'center'``, `` ``verticalalignment ``=``'center'``,`` ``transform ``=` `ax.transAxes, color ``=``"#E4365D"``)`` ``iteration ``=` `[``0``]`` ` ` ``# animation function to be repeatedly called`` ``def` `animate(A, rects, iteration):`` ` ` ``# to clear the bars from the Poly3DCollection object`` ``ax.collections.clear()`` ``ax.bar3d(``range``(``len``(a)), A, z, dx, dy, dz, `` ``color ``=` `color_map(data_normalizer(``range``(n))))`` ``iteration[``0``] ``+``=` `1`` ``text.set_text(``"iterations : {}"``.``format``(iteration[``0``]))`` ` ` ``# animate function is called here and the generator object is passed`` ``anim ``=` `FuncAnimation(fig, func ``=` `animate,`` ``fargs ``=``(rects, iteration), frames ``=` `generator, interval ``=` `50``,`` ``repeat ``=` `False``)`` ``plt.show()`` ` `showGraph()`
Output :
For array size 20
Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.
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My Personal Notes arrow_drop_up
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# How can eight quants learn their average salary without revealing individual salaries? [closed]
Here is a problem from section 2.3 (Thinking Out of the Box) section of the popular book by Xinfeg Zhou titled 'Quant salary':
### Quant Salary
Eight quants from different banks are getting together for drinks. They are all interested in knowing the average salary of the group. Nevertheless, being cautious and humble individuals, everyone prefers not to disclose his or her salary to the group. Can you come up with a strategy for the quants to calculate the average salary without knowing other people's salaries?
The book provides the following solution:
Solution: This is a light-hearted problem with more than one answer. One approach is for the first quant to choose a random number, add it to his/her salary and give it to the second quant. The second quant will add his/her own salary to the result and give it to the third quant; and so on; the eighth quant will add his/her own salary to the result and give it back to the first quant. Then the first quant will deduct the 'random' number from the total and divide the 'real' total by 8 to yield the average salary.
The author mentions that there are alternative solutions to this problem. What are some alternative approaches?
### Further Thoughts
The problem of secure multiparty communication is one of the interesting problems in the field of cryptography and security. While my knowledge of the same is limited, I wonder if there has been some work done on the design of an algorithm which can perform computation on an encrypted version of its inputs and decrypting the outputs can reveal the result of the computation without revealing any information about the inputs. That is, can we design an algorithm to which every quant can submit an encrypted version of their salaries, get the output and infer the average salary from the output either by decrypting it with a key known to any number of the quants or otherwise, which would allow them to infer the average salary without revealing any information about the individual salaries?
• See also: Homomorphic Encryption, an open research topic today. Commented Jun 26 at 21:09
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# RADIO FREQUENCY LINES OR LOW LOSS LINES
RADIO FREQUENCY LINES OR LOW LOSS LINES
A LOW LOSS TRANSMISSION LINE IS ONE FOR WHICH
R<<шl
AND G<<шC
Thus Z=R+jшL- jшL
And Y=G+jшC – jшC
(a) Characteristic impedance,
Z0=R+jшL/G+jшC
= jшL/jшC
Z0= L/C
Z0 is pure resistance for low-loss line.
(b) Propagation constant
Ỳ= (R+jшL)(G+jшC)
= (jшL)(jшC)
Ỳ=jш LC
As Ỳ=α+jβ
By comparing real and imaginary parts of above two equations, we get
Attenuation constant, α=0
And phase shift constant, β=ш LC
Let us also discuss another concept called BALUN
BALUN consists of two words bal(short form of balanced) and up (short form of unbalanced).It is a form of a quarter wave transformer matching balancedline to an unbalanced transmission circuit.this problem arises when an ungrounded antenna is being fed by a coaxial cable.The cable from the dipole antenna offers reasonably high output impedance which does not match with the input impedance of the TV receiver.A transformer is thus required that matches the output impedance a parallel wire cable with input impedance of the TV set.
This transformer is known as balun and has got few turns of winding on a ferrite core. Balun is also called as Bazooka.
This entry was posted in Transmission Lines and tagged , , . Bookmark the permalink.
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Special Issue
## Modeling and Control of Complex Networked Systems
View this Special Issue
Research Article | Open Access
Volume 2014 |Article ID 730652 | https://doi.org/10.1155/2014/730652
Chao Liu, Rong Li, "Matthew Effect of the Random Drift on the Evolution of Cooperation", Mathematical Problems in Engineering, vol. 2014, Article ID 730652, 8 pages, 2014. https://doi.org/10.1155/2014/730652
# Matthew Effect of the Random Drift on the Evolution of Cooperation
Accepted07 Jan 2014
Published16 Feb 2014
#### Abstract
The effect of the random drift on the evolutionary prisoner’s dilemma game is studied on regular lattices. A new evolutionary rule is proposed, which stochastically combines the deterministic rule with the random drift rule. It is found that the random drift has an effect on the evolutionary dynamics depending on the values of the temptation-to-defect and the probability of the random drift. When the random drift occurs with low probabilities, which interests us more, a phenomenon of the Matthew effect on the evolution of cooperation is found. Explanations of this phenomenon are deduced through the analysis on the dynamics and pattern formations of the PDG system.
#### 1. Introduction
Since the introduction of evolutionary game theory by Smith and Price [1, 2], one of the important issues is to understand the emergence and evolution of cooperation [36]. Among the many games, the most studied example is the prisoner’s dilemma game (PDG), as it provides a simple example of the difficulties of cooperation [710]. The standard PDG is described by the following set of rules. When two players play a PDG, each of them can choose to cooperate (C) or defect (D). Each player will gain a payoff depending jointly on his choice and the opponent’s choice. A cooperator receives when playing with a cooperator and when playing with a defector while a defector earns when playing with a defector, and against a cooperator, with . Given this payoff ordering, in a well-mixed (unstructured) population where each agent interacts with all other agents (or a representative sample of the population composition), defectors are fitter and thus the fraction of cooperators asymptotically vanishes.
However, cooperation is widespread in many biological, social, and economic systems [11, 12]. One of the proposed mechanisms to explain this phenomenon is network reciprocity [13, 14]. For example, early pioneering work on the PDG in two-dimensional square lattices made the observation that, unlike in unstructured populations, cooperators and defectors can coexist in the lattice indefinitely [15, 16]. Simply said, the clustering of cooperators in the lattice could provide high enough fitness to the cooperator nodes exposed to invasion, to the extent of preserving cooperators from evolutionary extinction, even when defection is favoured by the one-shot (two-players) game analysis [17]. A recent and powerful approach to studying such questions is provided by evolutionary graph theory [18].
All the game models incorporate some kind of evolutionary dynamics, which also play a crucial role in the results. As for [15, 16], the well-known promotion of cooperation enforced by spatial lattices is linked to a particular best-takes-over update rule, in which each individual node plays with its immediate neighbours each time step accumulating a payoff; then it updates its strategy by imitating the one of the highest payoff in its neighbourhood, including itself.
The best-takes-over update rule is a nonstochastic imitation strategy. However, for real dynamical systems, external disturbances or system errors are generally inevitable. Therefore, it is worth considering the dependence of the promotion of cooperation on the evolutionary rules and its robustness against perturbations. Indeed, previous work has pursued this enquiry [1921], replacing the deterministic update strategy with some stochastic imitative rules, finding greatly reduced cooperation levels.
In these game models, players are viewed as rational, who update their strategy by copying, within certain constraints, the strategy of those others that are doing better, or in game theoretical terms that are obtaining higher payoffs from the game. Then, what difference will be there in the evolution of cooperation if players occasionally behave irrationally, updating strategy with no concern of their payoffs. In this paper, we will explore this problem by combining the best-takes-over rule with the random drift reproduction rule in the prisoner’s dilemma game.
#### 2. The Model
We study the PDG with pure strategies: the players can either defect (D) or cooperate (C). The same as in [15, 16], players are disposed on two-dimensional square lattices, and each player interacts only with its nearest neighbors and collects profits depending on the payoff parameters. Following common studies, the PDG is rescaled such that it depends on a single parameter; that is, the parameters are chosen to be , , and , representing the advantage of defectors over cooperators (or the temptation-to-defect). After each round, the players are allowed to inspect their neighbors’ payoffs and, according to the comparison, determine their strategies to adopt in the next round. Two alternative update rules determining the transformation of each player’s strategy are described below.(I)Best-takes-over: in each generation, an individual node imitates the strategy of one of its neighbors (including the node itself) that received the highest payoff in the last round. The best-takes-over rule is a deterministic rule according to which the individual with the highest gain in a given neighborhood reproduces with certainty.(II)Random drift: whenever an individual node is updated, one neighbor is drawn at random among all its neighbors (including node itself), and node imitates the strategy of node with certainty regardless of the payoff of node or in the last round. Thus, for the random drift update rule, reproduction ability or fitness is not correlated with payoff but is correlated with the proportion of one strategy in the neighborhood.
To investigate how the random drift affects evolutionary games, we propose a new evolutionary rule which stochastically combines the deterministic rule with the random drift rule. A parameter of probability is used in the choosing of the two alternative rules. That is, in each generation, an individual node updates its strategy by the best-takes-over rule with probability and by the random drift rule with probability .
Our simulations are carried out on the regular 8-neighbored square lattices. Initially, the cooperative and defective strategies are randomly distributed among the players with equal probability . A synchronous updating scheme is adopted. Equilibrium frequencies of cooperators are obtained by averaging over 1000 generations after a transient time of 50000 generations. Each data point results from an average over 100 realizations of the initial conditions.
#### 3. Results and Discussion
Figure 1 shows the varying frequency of cooperation versus temptation-to-defect on and regular lattice, respectively. represents the density of cooperators. The random drift update rule is employed with different probabilities , where means the pure deterministic imitative dynamics. As shown in Figure 1, the random drift has obvious influence on the evolution of cooperation. Moreover, the influence depends on the value of the parameter .
Despite the huge gap of the two lattices in size, the simulation results are qualitatively similar. In the following, we mainly analyze the results on the lattice as an example. Based on the different influence of the random drift, the range of parameter can be roughly divided into three regions. The first region is the small value region . In this parameter region, the phenomenon of stochastic resonance is observed. For small temptation-to-defect , the frequency of cooperation is high under the pure best-takes-over mechanism. However, it is still raised when the random drift is applied with small probabilities, even to an all-C state. The second parameter region is the moderate value region . In this region, whether the random drift favours cooperation depends on the value of the probability . When the random drift update rule is applied with probability , the density of cooperators is higher than the one with the pure deterministic rule. However, when the probability rises to , the frequency of cooperation decreases lower than the deterministic one. is the large parameter region. In this region, the evolutionary survival of cooperators is hindered by the random drift update rule. In particular, the parameter region is worth of more attention. It is the chaotic parameter region under the deterministic mechanism, where the cooperators and defectors can coexist indefinitely; see [16] for details. As the random drift is applied, the chaotic evolution of the dynamic system is broken down, leading to an all-D state.
To have a more clear knowledge of the influence of the random drift on the evolution of cooperation, the variation of as a function of the probability is given in Figure 2. Because of the qualitatively similarity, we still focus on the results of the lattice. The three values of the temptation-to-defect are from the three parameter regions mentioned above. As shown in Figure 2, for all the three values of , the random drift is not favourable to the maintenance of cooperation when the probability is large enough. In particular, an all-D state prevails to form a flat valley in each of the three curves. When reaches , the equilibrium frequency of cooperation fluctuates around the original frequency (50%). It is because, in this situation, the PDG evolves entirely under the random drift mechanism, for which payoffs of nodes do not count and the chance of survival for cooperators and defectors are equal. In the case , the frequency of cooperation decreases monotonously with the increasing probability of the random drift. In the case or , where is relatively small, an earlier raised and later decreased frequency of cooperation is found. Similar results are found on the lattice.
Thus, in fact, the random drift has an influence on the evolutionary PDG depending on the values of the temptation-to-defect and the probability . When is high, which means that the random drift occurs frequently, it tends to depress cooperation for all values of . When the random drift occurs only with low probabilities, which interests us more, a conclusion can be drawn as follows: when the temptation for players to defect is relatively small, with cooperators prevailing in the original PDG, the applied random drift tends to further enhance cooperation; under the opposite conditions which is cooperation unfavourable, the random drift tends to further depress cooperation. In other words, for the cases with small , the effect of the random drift on the evolution of cooperation is the so-called Matthew effect, which makes the rich richer.
How come this kind of Matthew effect arises when the random drift is applied to the deterministic imitative mechanism with small probability? The dynamics and the pattern formation of the PDG system may shed light on the explanation.
Setting and starting from the same initial distribution of cooperators, Figure 3 displays the dynamics of the PDG system with and without the influence of the random drift on the lattice. Under the pure deterministic imitation mechanism, the system settles at a period 2 oscillator after a short-term fluctuation as shown in Figure 3(a). In Figure 3(b), when the random drift update rule is applied with a probability , the cooperation is obviously suppressed and the system soon converges to an all-D state.
The corresponding pattern formations of the above two evolutionary processes are displayed in Figures 4 and 5, respectively. Figure 4(a) shows the initial pattern formation of the PDG system, in which cooperators and defectors randomly distributed in the lattice with equal probability. At the second time generation, the density of cooperators drops dramatically due to the high temptation to defectors. After that, the frequency of cooperators grows up gradually for the reason that the square shaped C-clusters can survive and even grow in this region of parameter , as displayed in Figures 4(b)4(f). At last, the system regularly switches between two different pattern formations as shown in Figures 4(g)4(i), corresponding to the period 2 oscillator in Figure 3(a). As for Figure 5, although starting from the same initial configuration, cooperators vanish soon. From Figure 5(b) to Figure 5(c), one node (node (7, 17)) of a square shaped C-cluster turns to a defector due to the effect of the random drift, which leads to the breakdown of the entire square shaped C-cluster and eventually to the extermination of all cooperators.
The dynamics of the PDG system with temptation-to-defect is displayed in Figure 6. Under the deterministic imitative mechanism, the dynamical system quickly settles at a period 2 oscillator with a high average equilibrium frequency of cooperators. When the random drift update rule is employed with a probability , the phenomenon of intermittent period oscillations appears in the dynamics, where the period oscillators become unstable due to the influence of the randomness. Finally, the system reaches an all-C state.
Figures 7 and 8 illustrate the dynamics of Figure 6 with pattern formations. As shown in Figures 7(c)7(f), for the small value of , defectors can only survive and remain stable in straight or oblique lines. At the same time, as shown at the bottom-left parts of Figures 7(c)7(f), a single defector () will grow to form a square shaped -cluster () and then return to a single defector () in the next time generation, which explains the period 2 oscillation of the dynamical system in Figure 7. However, when the random drift is applied with probability , both the stable lines and the oscillators of defectors can be violated by the randomness, resulting in the enhancement of cooperators.
At the beginning, the two systems have similar pattern formations as shown in Figures 7(a)7(c) and Figures 8(a)8(c). Figures 8(d)8(f) show a process of the breakdown of a line shaped -cluster. In Figure 8(d), node (13, 10) and node (14, 10) form a line shaped -cluster, which is stable under the deterministic mechanism. Due to the random drift update strategy, node (13, 11), a neighbour of this line shaped -cluster, turns from a cooperator to a defector. Then an arrow shaped -cluster is formed in Figure 8(e), which is vulnerable under the deterministic imitative mechanism and eaten up by cooperators as shown in Figure 8(f). Figures 8(g)8(l) display the vanishing of a oscillator of defectors. In Figures 8(g)8(i), a oscillator of defectors is shown at the upper center. In Figure 8(j), affected by the randomness, the square shaped -cluster turns to a line shaped -cluster with two defectors instead of a single defector, which terminates the oscillation of defectors. After that, the line shaped -cluster remains stable for several time generations until it turns to an arrow shaped -cluster and vanishes in the next time generation as shown in Figures 8(k)8(l).
Now we are ready to explain the Matthew effect of the random drift on the evolutionary PDG. Despite the population of its opponent, some particular shaped C-clusters and -clusters can survive and maintain under the deterministic imitative mechanism. However, these equilibrium states strictly depend on the shapes of these clusters and thus are vulnerable to disturbance. The randomness brought by the employed random drift update rule can change the shapes of these clusters, leading to the breakdown of these clusters and the phenomenon of the Matthew effect.
#### 4. Conclusions
Evolutionary dynamics are affected by population structure and update rules. Spatial or network structure facilitates the clustering of strategies, which represents a mechanism for the evolution of cooperation. However, whether the evolutionary dynamics is robust to disturbance deserves further studies. In this paper, the problem is explored by combining the deterministic imitative rule with the random drift reproduction rule in the prisoner’s dilemma game on regular lattices. It is found that the employed random drift has an effect on the evolutionary dynamics depending on the values of the temptation-to-defect and the probability . We are more interested in the cases where the random drift occurs only with low probabilities. Then, the random drift has an apparent Matthew effect on the evolution of cooperation: when the temptation for players to defect is relatively small, with cooperators prevailing in the original PDG, the applied random drift tends to further enhance the frequency of cooperators; under the opposite conditions which is cooperation unfavourable, the random drift tends to further depress cooperation. Analysis on the dynamics and pattern formation of the PDG system implies that the advantage brought by the clustering of strategies is diluted by the random drift, leading to the phenomenon of the Matthew effect.
#### Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
#### Acknowledgments
The authors would like to thank Dr. Yu-Zhong Chen and Professor Ying-Cheng Lai for helpful discussions. This work was supported by the National Science Foundation of China under Grant no. 11101256 and Research and Innovation Project of Shanghai Municipal Education Commission (no. 14YZ149).
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Copyright © 2014 Chao Liu and Rong Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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# Thomas Sykes
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# Properties
Label 315.6.a.c Level $315$ Weight $6$ Character orbit 315.a Self dual yes Analytic conductor $50.521$ Analytic rank $1$ Dimension $2$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$315 = 3^{2} \cdot 5 \cdot 7$$ Weight: $$k$$ $$=$$ $$6$$ Character orbit: $$[\chi]$$ $$=$$ 315.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$50.5209032411$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{65})$$ Defining polynomial: $$x^{2} - x - 16$$ x^2 - x - 16 Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 35) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \frac{1}{2}(1 + \sqrt{65})$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - \beta q^{2} + (\beta - 16) q^{4} + 25 q^{5} - 49 q^{7} + (47 \beta - 16) q^{8} +O(q^{10})$$ q - b * q^2 + (b - 16) * q^4 + 25 * q^5 - 49 * q^7 + (47*b - 16) * q^8 $$q - \beta q^{2} + (\beta - 16) q^{4} + 25 q^{5} - 49 q^{7} + (47 \beta - 16) q^{8} - 25 \beta q^{10} + ( - 97 \beta + 349) q^{11} + (53 \beta - 315) q^{13} + 49 \beta q^{14} + ( - 63 \beta - 240) q^{16} + ( - 251 \beta + 105) q^{17} + ( - 86 \beta + 358) q^{19} + (25 \beta - 400) q^{20} + ( - 252 \beta + 1552) q^{22} + (902 \beta - 230) q^{23} + 625 q^{25} + (262 \beta - 848) q^{26} + ( - 49 \beta + 784) q^{28} + (945 \beta - 3415) q^{29} + (924 \beta - 660) q^{31} + ( - 1201 \beta + 1520) q^{32} + (146 \beta + 4016) q^{34} - 1225 q^{35} + ( - 1260 \beta - 3822) q^{37} + ( - 272 \beta + 1376) q^{38} + (1175 \beta - 400) q^{40} + (3818 \beta - 2796) q^{41} + (922 \beta - 14022) q^{43} + (1804 \beta - 7136) q^{44} + ( - 672 \beta - 14432) q^{46} + (1575 \beta + 9857) q^{47} + 2401 q^{49} - 625 \beta q^{50} + ( - 1110 \beta + 5888) q^{52} + (454 \beta + 27564) q^{53} + ( - 2425 \beta + 8725) q^{55} + ( - 2303 \beta + 784) q^{56} + (2470 \beta - 15120) q^{58} + ( - 5184 \beta - 27208) q^{59} + (5706 \beta - 28776) q^{61} + ( - 264 \beta - 14784) q^{62} + (1697 \beta + 26896) q^{64} + (1325 \beta - 7875) q^{65} + ( - 4568 \beta - 20388) q^{67} + (3870 \beta - 5696) q^{68} + 1225 \beta q^{70} + ( - 5304 \beta - 37720) q^{71} + ( - 4192 \beta - 4670) q^{73} + (5082 \beta + 20160) q^{74} + (1648 \beta - 7104) q^{76} + (4753 \beta - 17101) q^{77} + (17635 \beta - 34715) q^{79} + ( - 1575 \beta - 6000) q^{80} + ( - 1022 \beta - 61088) q^{82} + (3924 \beta - 56876) q^{83} + ( - 6275 \beta + 2625) q^{85} + (13100 \beta - 14752) q^{86} + (13396 \beta - 78528) q^{88} + (5722 \beta + 15964) q^{89} + ( - 2597 \beta + 15435) q^{91} + ( - 13760 \beta + 18112) q^{92} + ( - 11432 \beta - 25200) q^{94} + ( - 2150 \beta + 8950) q^{95} + (13943 \beta - 55141) q^{97} - 2401 \beta q^{98} +O(q^{100})$$ q - b * q^2 + (b - 16) * q^4 + 25 * q^5 - 49 * q^7 + (47*b - 16) * q^8 - 25*b * q^10 + (-97*b + 349) * q^11 + (53*b - 315) * q^13 + 49*b * q^14 + (-63*b - 240) * q^16 + (-251*b + 105) * q^17 + (-86*b + 358) * q^19 + (25*b - 400) * q^20 + (-252*b + 1552) * q^22 + (902*b - 230) * q^23 + 625 * q^25 + (262*b - 848) * q^26 + (-49*b + 784) * q^28 + (945*b - 3415) * q^29 + (924*b - 660) * q^31 + (-1201*b + 1520) * q^32 + (146*b + 4016) * q^34 - 1225 * q^35 + (-1260*b - 3822) * q^37 + (-272*b + 1376) * q^38 + (1175*b - 400) * q^40 + (3818*b - 2796) * q^41 + (922*b - 14022) * q^43 + (1804*b - 7136) * q^44 + (-672*b - 14432) * q^46 + (1575*b + 9857) * q^47 + 2401 * q^49 - 625*b * q^50 + (-1110*b + 5888) * q^52 + (454*b + 27564) * q^53 + (-2425*b + 8725) * q^55 + (-2303*b + 784) * q^56 + (2470*b - 15120) * q^58 + (-5184*b - 27208) * q^59 + (5706*b - 28776) * q^61 + (-264*b - 14784) * q^62 + (1697*b + 26896) * q^64 + (1325*b - 7875) * q^65 + (-4568*b - 20388) * q^67 + (3870*b - 5696) * q^68 + 1225*b * q^70 + (-5304*b - 37720) * q^71 + (-4192*b - 4670) * q^73 + (5082*b + 20160) * q^74 + (1648*b - 7104) * q^76 + (4753*b - 17101) * q^77 + (17635*b - 34715) * q^79 + (-1575*b - 6000) * q^80 + (-1022*b - 61088) * q^82 + (3924*b - 56876) * q^83 + (-6275*b + 2625) * q^85 + (13100*b - 14752) * q^86 + (13396*b - 78528) * q^88 + (5722*b + 15964) * q^89 + (-2597*b + 15435) * q^91 + (-13760*b + 18112) * q^92 + (-11432*b - 25200) * q^94 + (-2150*b + 8950) * q^95 + (13943*b - 55141) * q^97 - 2401*b * q^98 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - q^{2} - 31 q^{4} + 50 q^{5} - 98 q^{7} + 15 q^{8}+O(q^{10})$$ 2 * q - q^2 - 31 * q^4 + 50 * q^5 - 98 * q^7 + 15 * q^8 $$2 q - q^{2} - 31 q^{4} + 50 q^{5} - 98 q^{7} + 15 q^{8} - 25 q^{10} + 601 q^{11} - 577 q^{13} + 49 q^{14} - 543 q^{16} - 41 q^{17} + 630 q^{19} - 775 q^{20} + 2852 q^{22} + 442 q^{23} + 1250 q^{25} - 1434 q^{26} + 1519 q^{28} - 5885 q^{29} - 396 q^{31} + 1839 q^{32} + 8178 q^{34} - 2450 q^{35} - 8904 q^{37} + 2480 q^{38} + 375 q^{40} - 1774 q^{41} - 27122 q^{43} - 12468 q^{44} - 29536 q^{46} + 21289 q^{47} + 4802 q^{49} - 625 q^{50} + 10666 q^{52} + 55582 q^{53} + 15025 q^{55} - 735 q^{56} - 27770 q^{58} - 59600 q^{59} - 51846 q^{61} - 29832 q^{62} + 55489 q^{64} - 14425 q^{65} - 45344 q^{67} - 7522 q^{68} + 1225 q^{70} - 80744 q^{71} - 13532 q^{73} + 45402 q^{74} - 12560 q^{76} - 29449 q^{77} - 51795 q^{79} - 13575 q^{80} - 123198 q^{82} - 109828 q^{83} - 1025 q^{85} - 16404 q^{86} - 143660 q^{88} + 37650 q^{89} + 28273 q^{91} + 22464 q^{92} - 61832 q^{94} + 15750 q^{95} - 96339 q^{97} - 2401 q^{98}+O(q^{100})$$ 2 * q - q^2 - 31 * q^4 + 50 * q^5 - 98 * q^7 + 15 * q^8 - 25 * q^10 + 601 * q^11 - 577 * q^13 + 49 * q^14 - 543 * q^16 - 41 * q^17 + 630 * q^19 - 775 * q^20 + 2852 * q^22 + 442 * q^23 + 1250 * q^25 - 1434 * q^26 + 1519 * q^28 - 5885 * q^29 - 396 * q^31 + 1839 * q^32 + 8178 * q^34 - 2450 * q^35 - 8904 * q^37 + 2480 * q^38 + 375 * q^40 - 1774 * q^41 - 27122 * q^43 - 12468 * q^44 - 29536 * q^46 + 21289 * q^47 + 4802 * q^49 - 625 * q^50 + 10666 * q^52 + 55582 * q^53 + 15025 * q^55 - 735 * q^56 - 27770 * q^58 - 59600 * q^59 - 51846 * q^61 - 29832 * q^62 + 55489 * q^64 - 14425 * q^65 - 45344 * q^67 - 7522 * q^68 + 1225 * q^70 - 80744 * q^71 - 13532 * q^73 + 45402 * q^74 - 12560 * q^76 - 29449 * q^77 - 51795 * q^79 - 13575 * q^80 - 123198 * q^82 - 109828 * q^83 - 1025 * q^85 - 16404 * q^86 - 143660 * q^88 + 37650 * q^89 + 28273 * q^91 + 22464 * q^92 - 61832 * q^94 + 15750 * q^95 - 96339 * q^97 - 2401 * q^98
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
4.53113 −3.53113
−4.53113 0 −11.4689 25.0000 0 −49.0000 196.963 0 −113.278
1.2 3.53113 0 −19.5311 25.0000 0 −49.0000 −181.963 0 88.2782
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$3$$ $$-1$$
$$5$$ $$-1$$
$$7$$ $$1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 315.6.a.c 2
3.b odd 2 1 35.6.a.b 2
12.b even 2 1 560.6.a.l 2
15.d odd 2 1 175.6.a.d 2
15.e even 4 2 175.6.b.d 4
21.c even 2 1 245.6.a.c 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
35.6.a.b 2 3.b odd 2 1
175.6.a.d 2 15.d odd 2 1
175.6.b.d 4 15.e even 4 2
245.6.a.c 2 21.c even 2 1
315.6.a.c 2 1.a even 1 1 trivial
560.6.a.l 2 12.b even 2 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{2} + T_{2} - 16$$ acting on $$S_{6}^{\mathrm{new}}(\Gamma_0(315))$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2} + T - 16$$
$3$ $$T^{2}$$
$5$ $$(T - 25)^{2}$$
$7$ $$(T + 49)^{2}$$
$11$ $$T^{2} - 601T - 62596$$
$13$ $$T^{2} + 577T + 37586$$
$17$ $$T^{2} + 41T - 1023346$$
$19$ $$T^{2} - 630T - 20960$$
$23$ $$T^{2} - 442 T - 13172224$$
$29$ $$T^{2} + 5885 T - 5853350$$
$31$ $$T^{2} + 396 T - 13834656$$
$37$ $$T^{2} + 8904 T - 5978196$$
$41$ $$T^{2} + 1774 T - 236091496$$
$43$ $$T^{2} + 27122 T + 170086856$$
$47$ $$T^{2} - 21289 T + 72995224$$
$53$ $$T^{2} - 55582 T + 768990296$$
$59$ $$T^{2} + 59600 T + 451339840$$
$61$ $$T^{2} + 51846 T + 142927344$$
$67$ $$T^{2} + 45344 T + 174936944$$
$71$ $$T^{2} + 80744 T + 1172746624$$
$73$ $$T^{2} + 13532 T - 239780284$$
$79$ $$T^{2} + 51795 T - 4382959400$$
$83$ $$T^{2} + 109828 T + 2765333536$$
$89$ $$T^{2} - 37650 T - 177665240$$
$97$ $$T^{2} + 96339 T - 838817066$$
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https://www.blog.trainindata.com/variance-stabilizing-transformations-in-machine-learning/
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Select Page
Variance stabilizing transformations in machine learning
by | May 31, 2022 | Feature Engineering, Machine Learning
You’ve probably heard that before training machine learning models, data scientists transform random variables to change their distribution into something closer to the normal distribution.
But, why do we do this? Which variables should we transform? Which transformations should we use? And, do we need to transform variables to train any machine learning algorithm?
Why do we transform random variables?
Many statistical methods used in data analysis and supervised machine learning models make assumptions about the data. For example, to draw conclusions from a linear regression model, many assumptions must be true. Some of the assumptions are the following:
• The values of the dependent variable (that is, the target) are independent.
• There is a linear relationship between the target and the independent (predictor) variables.
• The residuals, that is, the difference between the predictions and the values of the target, are normally distributed and centered at zero.
• The residuals have constant variance.
Many people, myself included, confuse the last assumption with the idea that all the predictor variables have to be normally distributed. But this is not the case. What needs to be normally distributed and centered at zero are the residuals, which means that any difference between the predictions and the target is just random.
What happens if the assumptions are not met?
When the assumptions are not met, the conclusions drawn from the data analysis or from the linear regression models, might not be reliable.
Fortunately, we can correct the failure in the assumptions by transforming the variables prior to the analysis. This would improve the performance and reliability of the models.
We can transform the target variable itself when its distribution is skewed. Transforming the predictor variables, very often helps meet the model assumptions when the raw data does not. The transformations that data scientists apply to predictor variables and the target are called variable stabilizing transformations, and I will explain what that means in the coming sections.
When to apply variance stabilizing transformations?
You probably guessed that variable transformations are usually applied when we analyze data through linear statistical tests like ANOVA and when training linear regression models.
In other words, there is no need to transform variables when training non-linear models like decision tree based algorithms, nearest neighbors, or neuronal networks. What are variance stabilizing transformations?
Variable transformations
Variable transformation consists of replacing the original variable values with a function of that variable. Transforming variables with mathematical functions helps reduce variable skewness, therefore improving the value spread, and sometimes unmasks linear and additive relationships between predictors and target.
Commonly used mathematical transformations include the logarithm, reciprocal, power, and square root transformations, as well as the Box-Cox and Yeo-Johnson transformations. These transformations are commonly referred to as “variance stabilizing transformations”. Variance stabilizing transformations intend to bring the distribution of the variable to a more symmetric, or in other words, Gaussian shape.
In other words, a variance-stabilizing transformation is a function f that turns all possible values of y into other values y’=f(y) in such a way that the variance of y’ remains constant.
For some distributions like Poisson or the binomial distribution, exact variance stabilization is not possible, so we say that the transformations are “approximate variance stabilizing” transformations.
Many variance stabilizing transformations were discussed and analyzed in the context of Poisson distributions, where the variance of the variable is equal to the mean. Hence, the larger the mean, the larger the variance.
The transformation of the variables aims to obtain values, such that their variance is independent of their mean, that is, variables with a constant variance. Thus, variance stabilizing transformations.
In the following paragraphs, we will discuss the following variance stabilizing transformations:
• Logarithm
• Reciprocal
• Square root
• Arcsin
• Power
• Box-Cox
• Yeo-Johnson
Logarithm transformation
The logarithm function is a powerful transformation for dealing with positive data with a right-skewed distribution (observations accumulate at lower values of the variable).
If y is the variable, then the logarithmic transformation is log(y).
Candidate variables for the log transformation are those like income, or salary, which are continuous variables and tend to show a heavy accumulation of observations towards smaller values. In other words, most people earn little, just a few earn a lot.
In particular, if we take the variable Median Income from the California housing data set from Scikit-learn, we see that it is continuous and right-skewed:
Yet, after the logarithm transformation, we observe more widely spread and evenly distributed values:
Reciprocal transformation
The reciprocal function is defined as 1/y, where y is the random variable. It is a transformation with a drastic effect on the variable distribution.
The reciprocal transformation is useful when we have ratios, that is, values resulting from the division of two variables. Classical examples are variables like population density, that is, people per area, or house occupancy, that is, the number of occupants per house.
When we calculate the inverse of these variables, we pass from a representation of people per area to area per person, or occupants per house to houses per occupant. The transformed data still make sense (to humans), and it tends to show a better spread of values.
If you don’t believe me, take a look at the histogram of house occupancy from the California housing data set from Scikit-learn, a highly skewed variable:
And have a look at the distribution of the same variable after the reciprocal data transformation:
You can see how the reciprocal transformation dramatically improved the spread of values and even transformed a discrete variable into a continuous one.
One caveat of the reciprocal or inverse transformation is that it is not defined for the value 0. So if our variables contain zeroes… well, we should try something else.
Square root transformation
We mentioned previously that variance stabilizing transformations are discussed quite often in the context of Poisson distributions.
The square root transformation (√x) is a variance stabilizing transformation that transforms variables with a Poisson distribution (counts) into variables with an approximately standard Gaussian distribution. The Anscombe transformation (√(x+3/8)) and the Freeman-Tukey transformation (√x + √(x+1)) are variations of the square root transformation that also achieve variance stabilization.
The square root transformation is a form of power transformation where the exponent is 1/2 and is only defined for positive values. We will discuss general power transformations in the coming paragraphs.
There are many variables with Poisson distributions. For example the number of credit cards or bank accounts per person, the number of children per family, or the number of pets. Those variables are naturally counts. So, in these cases, a square root transformation could be suitable to stabilize the variance.
Now, if we transform these variables using a Poisson distribution, we won’t see the same clear changes that we would see with continuous variables. However, we will see that the observations are more evenly distributed along the diagonal in Q-Q plots.
Example Poisson distribution before the transformation:
The same distribution after the square root transformation:
Note how the observations are more evenly distributed along the red line in the precedent image.
Arcsin transformation
Before diving into generalized power transformations, let’s have a quick look at the arcsin transformation. The arcsin transformation, also called the arcsin square root transformation, or angular transformation, takes the form of arcsin(sqrt(x)) where x is a real number between 0 and 1.
The arcsin square root transformation helps in dealing with probabilities, percentages, and proportions. It aims to stabilize the variance of the variable and return more evenly distributed (Gaussian-looking) values.
As you can imagine, there are plenty of examples of variables that could be suitable candidates for the arcsin transformation, like those from the breast cancer dataset from Scikit-learn.
You can see how a bunch of these variables show skewed distributions in their raw state:
And after the arcsin transformation the values are more evenly distributed:
This was probably a transformation that was out of your radar, and truth be told, it is rarely used. But here it is, a function that has been widely studied in the past.
Power transformations
While variance-stabilizing transformations for some parametric families of distributions, such as the Poisson and binomial distributions, are well-known, some methods of data analysis rely on trial and error, such as searching through power transformations to find a suitable fixed transformation.
Power functions are mathematical formulations like this: X = X^lambda where lambda can take any value.
The square and cube root transformations are special cases of power transformations where lambda is 1/2 or 1/3, respectively. The reciprocal transformation is also a power transformation where lambda is -1. So hey, all along we’ve been discussing power transformations!
The challenge in choosing a power transformation resides in finding a suitable value for the parameter lambda that returns variables whose values are more evenly distributed.
We discussed the special cases of the square root and reciprocal transformation in the previous paragraphs because they are suitable for specific variables, or better say, specific distributions. In reality, we don’t really manually try exponents to see which one works best, because the Box-Cox transformation, which we will discuss in the next paragraph, automatically finds the parameter lambda for us.
As general guidance, if data is right-skewed (i.e. more observations around lower values), use lambda <1. If data is left-skewed (i.e. more observations around higher values), use lambda >1.
Box-Cox transformation
The Box-Cox transformation is a generalization of the power family of transformations, and it is defined by:
where X is the variable and λ is the transformation parameter.
The Box-Cox transformation can be used for transformations that we discussed before, including no transformation (λ = 1), the logarithm (λ = 0), the reciprocal (λ = -1), the square root (when λ = 0.5), and the cube root.
In the Box-Cox transformation, several values of the parameter λ are evaluated using maximum likelihood, and the λ that returns the best transformation is selected. The Box-Cox transformation is usually the preferred choice for machine learning practitioners, because it is not necessary to think about which transformation to apply to which variable.
The only caveat with the Box-Cox transformation is that it was designed only for positive variables. So, if your variables contain negative values, you can either shift the distribution by adding a constant, or use the Yeo-Johnson transformation.
Yeo-Johnson transformation
The Yeo-Johnson transformation is an extension of the Box-Cox transformation that is no longer constrained to positive values. In other words, the Yeo-Johnson transformation can be used on variables with zero and negative values as well as positive values.
The Yeo-Johnson transformation is defined as follows:
In short, if the variable X is strictly positive, then, the Yeo-Johnson transformation is the same as the Box-Cox power transformation of X + 1. If X is strictly negative, then the Yeo-Johnson transformation is the Box-Cox transformation of (-X + 1) but with power 2 — λ.
If the variable has positive and negative values, then the transformation is a mixture of these 2 functions, so different powers are used for the positive and negative values of the variable. If you ask me, it’s a bit of a mess, but as long as it works…
Why, which, how and when to transform variables?
We started the blog post with the following questions:
• Why do we transform random variables?
• Which variables should we transform?
• Which transformations should we use for variance stabilization?
• And, do we need to transform variables to train any machine learning algorithm?
By now, I think we have answers to all of these questions.
Why do we transform variables?
In order to make data meet the assumptions of certain statistical models, typically analysis of variance (ANOVA) and linear regression models, and thus be able to draw accurate or reliable conclusions from the data analysis.
Which variables should we transform?
In general, those random variables which show distributions that are not normal.
Which transformations should we use?
There are a bunch of transformations that we can use. If we are keen to understand our transformed variables, we might prefer to do some data analysis and select which transformation to apply to which variable based on what we discussed throughout the blog post. For example, we would apply square root to counts, arcsin to fractions, and reciprocal to ratios. We would apply log to variables with observations accumulating on lower values, and for everything else, other power transformations.
In practice, to speed things up, we just go for Box-Cox or Yeo-Johnson, which consider all of the above transformations, and choose the transformation automatically. But beware, automation does not always resolve the issue!
Sometimes, applying transformations blindly creates an issue. So it is always good practice to plot the variables after the transformation, and be sure that we have obtained the expected result. I know, I sound like grandma.
Take a look for example at the following figure taken from the scikit-learn documentation:
Applying transformations to variables that are already normally distributed does not really change the distribution (see the lilac plots), so there is no need to do that.
On the other hand, the Box-Cox and Yeo-Johnson transformations may not return Gaussian shaped distributions after the transformation, as in the extreme examples of the bimodal (green) and uniform (black) distributions.
Where am I going with this? Although it may be tempting to automatically transform variables and train models, it may be worth taking the time to analyze the transformations and understand the data in our datasets and what we are feeding to our models.
Finally, do we need to transform variables to train any machine learning algorithm?
No. These transformations were studied and designed for their use with linear models. So, if you want to train non-linear models like decision tree-based algorithms or nearest neighbors, you might as well skip this step.
Python implementation of variance stabilizing transformations
I wrote a lot about variance stabilizing transformations, but I haven’t really shown you how to implement these transformations in Python, have I?
Applying these transformations with Python is really easy. We can do so with Numpy as follows:
``````Import numpy as np
data[“variable_log”] = np.log(data[“variable_original”])
``````
For the reciprocal, we would use `np.reciprocal()`, for the square root `np.sqrt()`, and for the Power `np.exp(data[“variable_original”], lambda)`, where lambda is the desired exponent of the transformation.
For BoxCox and Yeo-Johnson, we would use scipy.stats:
``````import scipy.stats as stats
X_tf[“new_var”], param = stats.boxcox(X[“original_var”])
X_tf[“new_var”], param = stats.yeojohnson(X[“original_var”])
``````
where param is the suitable lambda found by the transformation.
Yet, with Numpy and scipy.stats we need to modify one variable at a time. We can transform various variables simultaneously by utilizing Scikit-learn, or the library for which I am the maintainer: Feature-engine.
So for example, if we want to apply the Box-Cox transformation with Feature-engine, we would do the following:
``````from feature_engine.transformation import BoxCoxTransformer
boxcox = BoxCoxTransformer()boxcox.fit(X_train)
train_transformed = boxcox.transform(X_train)
test_tranformed = boxcox.transform(X_test)
``````
With Scikit-learn we would do:
``````from sklearn.preprocessing import PowerTransformer
transformer = PowerTransformer(method=”box-cox”, standardize=False)
boxcox.fit(X_train)
train_transformed = boxcox.transform(X_train)
test_tranformed = boxcox.transform(X_test)
``````
There are differences between the Scikit-learn and Feature-engine implementations which I discuss in this article.
I also highlight the differences between Numpy, scipy.stats, Scikit-learn and Feature-engine in our online course and book.
References
For more information on variable transformation and variance stabilizing transformation, check the following resources:
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2008 Compressible Flow Midterm Exam
November 1st 2008
15:00 — 17:00
ANSWER ALL 6 PROBLEMS; ALL PROBLEMS HAVE EQUAL VALUE; NO NOTES OR BOOKS; TAKE $\gamma=1.4$ IN ALL CASES; USE GAS TABLES THAT WERE DISTRIBUTED.
05.22.14
Question #1
Consider the following nozzle which is designed to yield an efficiency essentially of 1.0:
Air flowing in the nozzle is driven by a large reservoir in which the pressure is 30 bar and the temperature $75^\circ$C. A pitot tube measurement in the exit plane gives a pressure of 23.3 bar. Determine the exit Mach number, the exit pressure and sketch out the Mach number distribution along the length of the nozzle from the throat to the exit.
Question #2
A continuous-running supersonic wind tunnel has the usual configuration of a converging-diverging nozzle, followed by a short, straight test section, then by a converging-diverging diffuser culminating in mechanical recompression beyond the diffuser exit. The wind tunnel is intended to be run at close to “ideal” operating conditions with an intended diffuser exit Mach number of about $M_{\rm e}\approx 0.3$ before mechanical recompression. The wind tunnel segment itself is instrumented in the following way: at the nozzle throat, which is 5 $\rm cm^2$, there is a pressure tap (i.e, a tiny hole perpendicular to the nozzle throat wall and connected to a pressure sensor). The test section itself is equipped with a pitot tube. In a typical operational run at effectively “ideal” operating conditions, the following pressures were recorded:
pressure tap at nozzle throat: 483.4 kPa
pitot tube pressure in test section: 616.9 kPa
Based on these measurements determine:
(a) the wind tunnel dimensions in terms of nozzle throat, test section, diffuser throat and exit areas (b) the pressure and Mach number distributions from the nozzle entrance to diffuser exit (including the pressure and Mach number at the nozzle entrance and diffuser exit)
Question #3
To obtain hypersonic flow in the test section of a wind tunnel, a shock tunnel followed by an hypersonic nozzle is generally used, as shown below.
Initially, air is at a pressure and temperature of 1 atm and 305 K throughout the whole assembly. After the shock tunnel is started, a normal shock is formed and travels towards the nozzle section. The speed of the shock with respect to the laboratory frame is of 1747 m/s. Since $A_1\gg A_2$, the nozzle section effectively acts like a solid wall, from which the incident shock reflects. The idea of the shock tunnel is to use the high pressure behind the reflected shock as a reservoir to drive the hypersonic nozzle. After shock reflection, the gas is essentially stationary. Calculate the pressure and temperature after the reflected shock assuming normal reflection. What are the resulting stagnation pressure and temperature?
Question #4
A fixed geometry converging-diverging intake diffuser is designed for shock-free operation at $M_\infty=1.6$. Determine the minimum flight Mach number $M_\infty$ to first achieve “choking” at the throat in the take-off sequence. Further, calculate the % mass spill at $M_\infty=0.8,$ 1.0, 1.4, and 1.6 (i.e. before the shock is swallowed). Determine the “overspeed” Mach number necessary to “swallow” the shock.
1. ${M}_3=0.52$, $P_3=19.4~{\rm bar}$. 2. ${M}_{\rm N}=1$, ${M}_{\rm T}=2.1$, $A_{\rm T}= 9.18~{\rm cm^2}$, $A_{\rm D}=7.43~{\rm cm^2}$, ${M}_{{\rm D}x}=1.84$, ${M}_{{\rm D}y}= 0.608$, $A_{\rm e}=12.85~{\rm cm^2}$ 3. $T_y= 3626~{\rm K}$, $P_y=192~{\rm atm}$, $P_\circ= 192~{\rm atm}$, $T_\circ= 3626~{\rm K}$. 4. ${M}_1= 0.55$, ${\rm spill} = 17\%$, ${\rm spill} = 20\%$, ${\rm spill} = 14.6\%$, ${\rm spill} = 10.6\%$, ${M}_{1x}= 2.17$.
$\pi$
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# Distance between Brazzaville (BZV) and Kigali (KGL)
Flight distance from Brazzaville to Kigali (Maya-Maya Airport – Kigali International Airport) is 1040 miles / 1674 kilometers / 904 nautical miles. Estimated flight time is 2 hours 28 minutes.
Driving distance from Brazzaville (BZV) to Kigali (KGL) is 2092 miles / 3367 kilometers and travel time by car is about 48 hours 54 minutes.
## Map of flight path and driving directions from Brazzaville to Kigali.
Shortest flight path between Maya-Maya Airport (BZV) and Kigali International Airport (KGL).
## How far is Kigali from Brazzaville?
There are several ways to calculate distances between Brazzaville and Kigali. Here are two common methods:
Vincenty's formula (applied above)
• 1040.025 miles
• 1673.758 kilometers
• 903.757 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1039.011 miles
• 1672.127 kilometers
• 902.876 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Maya-Maya Airport
City: Brazzaville
Country: Congo (Brazzaville)
IATA Code: BZV
ICAO Code: FCBB
Coordinates: 4°15′6″S, 15°15′10″E
B Kigali International Airport
City: Kigali
Country: Rwanda
IATA Code: KGL
ICAO Code: HRYR
Coordinates: 1°58′7″S, 30°8′22″E
## Time difference and current local times
The time difference between Brazzaville and Kigali is 1 hour. Kigali is 1 hour ahead of Brazzaville.
WAT
CAT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 153 kg (338 pounds).
## Frequent Flyer Miles Calculator
Brazzaville (BZV) → Kigali (KGL).
Distance:
1040
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1040
Round trip?
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## Enter Simultaneous Equations:
<-- Equation 1
<-- Equation 2
##### Use Cramers method to solve:
0.1d + 0.25q = 2.25
d + q = 12
##### Check Format
Equation 1 is in the correct format.
##### Check Format
Equation 2 is in the correct format.
Set up standards equations
Standard equation 1 = ax + by = c and Standard equation 2 = dx + ey = f.
##### Find a, b, c in ax + by = c
0.1d + 0.25q = 2.25
a = 0.1, b = 0.25, c = 2.25
##### Find d, e, f in dx + ey = f
d + q = 12
d = 1, e = 1, f = 12
##### Step 1, calculate Delta (Δ):
Δ = a * e - b * d
Δ = (0.1 * 1) - (0.25 * 1)
Δ = 0.1 - 0.25
Δ = -0.15
##### Step 2, calculate the numerator for d
Numerator(d) = c * e - b * f
Numerator(d) = (2.25 * 1) - (0.25 * 12)
Numerator(d) = 2.25 - 3
Numerator(d) = -0.75
##### Step 3, calculate the numerator for q
Numerator(q) = a * f - c * d
Numerator(q) = (0.1 * 12) - (2.25 * 1)
Numerator(q) = 1.2 - 2.25
Numerator(q) = -1.05
##### Evaluate and solve:
d = Numerator(d) Δ
d = -0.75 -0.15
d = 5
q = Numerator(q) Δ
q = -1.05 -0.15
q = 7
q = 7
##### How does the Simultaneous Equations Calculator work?
Free Simultaneous Equations Calculator - Solves a system of simultaneous equations with 2 unknowns using the following 3 methods:
1) Substitution Method (Direct Substitution)
2) Elimination Method
3) Cramers Method or Cramers Rule Pick any 3 of the methods to solve the systems of equations 2 equations 2 unknowns
This calculator has 2 inputs.
### What 1 formula is used for the Simultaneous Equations Calculator?
Δ = a * e - b * d
For more math formulas, check out our Formula Dossier
### What 7 concepts are covered in the Simultaneous Equations Calculator?
cramers rule
an explicit formula for the solution of a system of linear equations with as many equations as unknowns
eliminate
to remove, to get rid of or put an end to
equation
a statement declaring two mathematical expressions are equal
simultaneous equations
two or more algebraic equations that share variables
substitute
to put in the place of another. To replace one value with another
unknown
a number or value we do not know
variable
Alphabetic character representing a number
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# What does “identity map $id$” mean?
What does "identity map $id$" mean in this context? Two metrics $d_1$ and $d_2$ on $X$ are said to be Lipschitz equivalent if the identity map $id\colon (X,d_1)\to (X,d_2)$ is bilipschitz.
-
The identity map always means $Id(x)=x$, in every context.
In this one, note that the underlying set $X$ is the same, so there is sense to speak about the identity map. However the metrics may be different, which would imply a different topology.
The identity map need not be continuous when going from one topology to another (e.g. $x_0$ is an isolated point in the range, but not in the domain), but in a sense it measures some compatibility between the topologies and in this case the metrics.
-
At least in finite dimensional vector spaces, the norms are equivalent, yielding equivalent topologies. Thus, $Id$ is continuous. – robjohn Nov 9 '11 at 8:29
The symbol 'id' is often used to refer to the identity map, which is the map which takes every element of a set $X$ to itself:
\begin{align} \textrm{id} : & X \to X \\ & x \mapsto x \end{align}
or, phrased differently, id is the map from $X$ to $X$ that satisfies $\textrm{id}(x)=x$ for every $x\in X$.
-
It is the map $X\to X$ given by the formula $x\mapsto x$.
It is the identity map of the underlying set $X$ of the two metric spaces you consider. There is not really an identity map from $(X,d_1)$ to $(X,d_2)$ because these are not the same metric spaces.
-
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Skip to main content
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.4: Absolute Value Inequalities
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In the last section, we solved absolute value equations. In this section, we turn our attention to inequalities involving absolute value.
## Solving |x| < a
The solutions of $|x|<a$ again depend upon the value and sign of the number a. To solve |x| < a graphically, we must determine where the graph of the left-hand side lies below the graph of the right-hand side of the inequality |x| < a. There are three cases to consider.
• Case I: a < 0
In this case, the graph of y = a lies strictly below the x-axis. As you can see in Figure $$\PageIndex{1}$$(a), the graph of y = |x| never lies below the graph of y = a. Hence, the inequality |x| < a has no solutions.
• Case II: a = 0
In this case, the graph of y = 0 coincides with the x-axis. As you can see in Figure $$\PageIndex{1}$$(b), the graph of y = |x| never lies strictly below the x-axis. Hence, the inequality |x| < 0 has no solutions.
• Case III: a > 0
In this case, the graph of y = a lies strictly above the x-axis. In Figure $$\PageIndex{1}$$(c), the graph of y = |x| and y = a intersect at x = −a and x = a. In Figure $$\PageIndex{1}$$(c), we also see that the graph of y = |x| lies strictly below the graph of y = a when x is in-between −a and a; that is, when −a < x < a.
In Figure $$\PageIndex{1}$$(c), we’ve dropped dashed vertical lines from the points of intersection of the two graphs to the x-axis. On the x-axis, we’ve shaded the solution of |x| < a, that is, −a < x < a.
This discussion leads to the following key property.
property 1
The solution of |x| < a depends upon the value and sign of a.
• Case I: a < 0
The inequality |x| < a has no solution.
• Case II: a = 0
The inequality |x| < 0 has no solution.
• Case III: a > 0
The inequality |x| < a has solution set {x : −a < x < a}.
Let’s look at some examples.
Example $$\PageIndex{1}$$
Solve the inequality |x| < −5 for x.
Solution
The graph of the left-hand side of |x| < −5 is the “V” of Figure $$\PageIndex{1}$$(a). The graph of the right-hand side of |x| < −5 is a horizontal line located 5 units below the x-axis. This is the situation shown in Figure $$\PageIndex{1}$$(a). The graph of y = |x| is therefore never below the graph of y = −5. Thus, the inequality |x| < −5 has no solution.
An alternate approach is to consider the fact that the absolute value of x is always nonnegative and can never be less than −5. Thus, the inequality |x| < −5 has no solution.
Example $$\PageIndex{2}$$
Solve the inequality |x| < 0 for x.
Solution
This is the case shown in Figure $$\PageIndex{1}$$(b). The graph of y = |x| is never strictly below the x-axis. Thus, the inequality |x| < 0 has no solution.
Example $$\PageIndex{3}$$
Solve the inequality |x| < 8 for x.
Solution
The graph of the left-hand side of |x| < 8 is the “V” of Figure $$\PageIndex{1}$$(c). The graph of the right-hand side of |x| < 8 is a horizontal line located 8 units above the x-axis. This is the situation depicted in Figure $$\PageIndex{1}$$(c). The graphs intersect at (−8, 8) and (8, 8) and the graph of y = |x| lies strictly below the graph of y = 8 for values of x in-between −8 and 8. Thus, the solution of |x| < 8 is −8 < x < 8.
It helps the intuition if you check the results of the last example. Note that numbers between −8 and 8, such as −7.75, −3 and 6.8 satisfy the inequality,
$|-7.75|<8 \qquad \text { and } \quad|-3|<8 \quad \text { and } \quad|6.8|<8$
while values that do not lie between −8 and 8 do not satisfy the inequality. For example, none of the numbers −9.3, 8.2, and 11.7 lie between −8 and 8, and each of the following is a false statement.
$|-9.3|<8 \quad \text { and } \qquad|8.2|<8 \qquad \text { and } \qquad|11.7|<8 \quad \text { (all are false) }$
If you reflect upon these results, they will help cement the notion that the solution of |x| < 8 is all values of x satisfying −8 < x < 8.
Example $$\PageIndex{4}$$
Solve the inequality |5 − 2x| < −3 for x.
Solution
If the inequality were |x| < −3, we would not hesitate. This is the situation depicted in Figure $$\PageIndex{1}$$(a) and the inequality |x| < −3 has no solutions. The reasoning applied to |x| < −3 works equally well for the inequality |5 − 2x| < −3. The left-hand side of this inequality must be nonnegative, so its graph must lie on or above the x-axis. The right-hand side of |5 − 2x| < −3 is a horizontal line located 3 units below the x-axis. Therefore, the graph of y = |5 − 2x| can never lie below the graph of y = −3 and the inequality |5 − 2x| < −3 has no solution.
We can verify this result with the graphing calculator. Load the left- and right-hand sides of |5 − 2x| < −3 into Y1 and Y2, respectively, as shown in Figure $$\PageIndex{2}$$(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure $$\PageIndex{2}$$(b).
As predicted, the graph of y = |5 − 2x| never lies below the graph of y = −3, so the inequality |5 − 2x| < −3 has no solution.
Example $$\PageIndex{5}$$
Solve the inequality |5 − 2x| < 0 for x.
Solution
We know that the left-hand side of the inequality |5 − 2x| < 0 has the “V” shape indicated in Figure $$\PageIndex{1}$$(b). The graph “touches” the x-axis when |5 − 2x| = 0, or when
\begin{aligned} 5-2 x &=0 \\-2 x &=-5 \\ x &=\frac{5}{2} \end{aligned}
However, the graph of y = |5 − 2x| never falls below the x-axis, so the inequality |5 − 2x| < 0 has no solution.
Intuitively, it should be clear that the inequality |5−2x| < 0 has no solution. Indeed, the left-hand side of this inequality is always nonnegative, and can never be strictly less than zero.
Example $$\PageIndex{6}$$
Solve the inequality |5 − 2x| < 3 for x.
Solution
In this example, the graph of the right-hand side of the inequality |5 − 2x| < 3 is a horizontal line located 3 units above the x-axis. The graph of the left-hand side of the inequality has the “V” shape shown in Figure $$\PageIndex{3}$$(b) and (c). You can use the intersect utility on the graphing calculator to find the points of intersection of the graphs of y = |5 − 2x| and y = 3, as we have done in Figures $$\PageIndex{3}$$(b) and (c). Note that the calculator indicates two points of intersection, one at x = 1 and a second at x = 4.
The graph of y = |5 − 2x| falls below the graph of y = 3 for all values of x between 1 and 4. Hence, the solution of the inequality |5 − 2x| < 3 is the set of all x satisfying 1 < x < 4; i.e. {x : 1 < x < 4}.
Expectations:
We need a way of summarizing this graphing calculator approach on our homework paper. First, draw a reasonable facsimile of your calculator’s viewing window on your homework paper. Use a ruler to draw all lines. Complete the following checklist.
• Label each axis, in this case with x and y.
• Scale each axis. To do this, press the WINDOW button on your calculator, then report the values of xmin, xmax, ymin, and ymax on the appropriate axis.
• Label each graph with its equation.
• Drop dashed vertical lines from the points of intersection to the x-axis. Shade and label the solution set of the inequality on the x-axis.
Following the guidelines in the above checklist, we obtain the image in Figure $$\PageIndex{4}$$.
Algebraic Approach. Let’s now explore an algebraic solution of the inequality |5 − 2x| < 3. Much as |x| < 3 implies that −3 < x < 3, the inequality
$|5-2 x|<3$
requires that
$-3<5-2 x<3$
We can subtract 5 from all three members of this last inequality, then simplify.
\begin{aligned}-3-5 &<5-2 x-5<3-5 \\ &-8<-2 x<-2 \end{aligned}
Divide all three members of this last inequality by −2, reversing the inequality symbols as you go.
$4>x>1$
We prefer that our inequalities read from “small-to-large,” so we write
$1<x<4$
This form matches the order of the shaded solution on the number line in Figure $$\PageIndex{4}$$, which we found using the graphing calculator.
The algebraic technique of this last example leads us to the following property.
Property 8
If a > 0, then the inequality |x| < a is equivalent to the inequality −a < x < a.
This property provides a simple method for solving inequalities of the form |x| < a. Let’s apply this algebraic technique in the next example.
Example $$\PageIndex{7}$$
Solve the inequality |4x + 5| < 7 for x.
Solution
The first step is to use Property 8 to write that $|4 x+5|<7$
is equivalent to the inequality
$-7<4 x+5<7$
From here, we can solve for x by first subtracting 5 from all three members, then dividing through by 4.
$\begin{array}{l}{-12<4 x<2} \\ {-3<x<\frac{1}{2}}\end{array}$
We can sketch the solution on a number line.
And we can describe the solution in both interval and set-builder notation as follows.
$\left(-3, \frac{1}{2}\right)=\left\{x :-3<x<\frac{1}{2}\right\}$
Assuming that a > 0, the inequality $$|x| \leq a$$ requires that we find where the absolute value of x is either “less than” a or “equal to” a. We know that |x| < a when −a < x < a and we know that |x| = a when x = −a or x = a. Thus, the solution of $$|x| \leq a$$ is the “union” of these two solutions.
This argument leads to the following property.
Property 10
If $$a > 0$$, then the inequality $$|x| \leq a$$ is equivalent to the inequality $$−a \leq x \leq a$$.
Example $$\PageIndex{8}$$
Solve the inequality $$5 − 3|x − 4| \geq −4$$ for x.
Solution
At first glance, the inequality $5-3|x-4| \geq-4$ has a form quite dissimilar from what we’ve done thus far. However, let’s subtract 5 from both sides of the inequality.
$-3|x-4| \geq-9$
Now, let’s divide both sides of this last inequality by −3, reversing the inequality sign.
$|x-4| \leq 3$
Aha! Familiar ground. Using Property 10, this last inequality is equivalent to
$-3 \leq x-4 \leq 3$
and when we add 4 to all three members, we have the solution.
$1 \leq x \leq 7$
We can sketch the solution on a number line
And we can describe the solution with interval and set-builder notation.
$[1,7]=\{x : 1 \leq x \leq 7\}$
## Solving |x| > a
The solutions of |x| > a again depend upon the value and sign of a. To solve |x| > a graphically, we must determine where the graph of y = |x| lies above the graph of y = a. Again, we consider three cases.
• Case I: a < 0
In this case, the graph of y = a lies strictly below the x-axis. Therefore, the graph of y = |x| in Figure $$\PageIndex{5}$$(a) always lies above the graph of y = a. Hence, all real numbers are solutions of the inequality |x| > a.
• Case II: a = 0
In this case, the graph of y = 0 coincides with the x-axis. As shown in Figure $$\PageIndex{5}$$(b), the graph of y = |x| will lie strictly above the graph of y = 0 for all values of x with one exception, namely, x cannot equal zero. Hence, every real number except x = 0 is a solution of |x| > 0. In Figure $$\PageIndex{5}$$(b), we’ve shaded the solution of |x| > 0, namely the set of all real numbers except x = 0.
• Case III: a > 0
In this case, the graph of y = a lies strictly above the x-axis. In Figure $$\PageIndex{5}$$(c), the graph of y = |x| intersects the graph of y = a at x = −a and x = a. In Figure $$\PageIndex{5}$$(c), we see that the graph of y = |x| lies strictly above the graph of y = a if x is less than −a or greater than a.
In Figure $$\PageIndex{5}$$(c), we’ve dropped dashed vertical lines from the points of intersection to the x-axis. On the x-axis, we’ve shaded the solution of |x| > a, namely the set of all real numbers x such that x < −a or x > a.
This discussion leads to the following property.
Property 12
The solution of |x| > a depends upon the value and sign of a.
• Case I: a < 0
All real numbers are solutions of the inequality |x| > a.
• Case II: a = 0
All real numbers, with the exception of x = 0, are solutions of |x| > 0.
• Case III: a > 0
The inequality |x| > a has solution set {x : x < −a or x > a}.
Example $$\PageIndex{9}$$
State the solution of each of the following inequalities.
$\text { a. }|x|>-5 \qquad \text { b. }|x|>0 \qquad \text { c. }|x|>4$
Solution
a. The solution of |x| > −5 is all real numbers.
b. The solution of |x| > 0 is all real numbers except zero.
c. The solution of |x| > 4 is the set of all real numbers less than −4 or greater than 4.
Example $$\PageIndex{10}$$
Solve the inequality |4 − x| > −5 for x.
Solution
The left-hand side of the inequality |4 − x| > −5 is nonnegative, so the graph of y = |4 − x| must lie above or on the x-axis. The graph of the right-hand side of |4 − x| > −5 is a horizontal line located 5 units below the x-axis. Therefore, the graph of y = |4 − x| always lies above the graph of y = −5. Thus, all real numbers are solutions of the inequality |4 − x| > −5.
We can verify our thinking with the graphing calculator. Load the left- and righthand sides of the inequality |4 − x| > −5 into Y1 and Y2, respectively, as shown in Figure $$\PageIndex{6}$$(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure $$\PageIndex{6}$$(b).
As predicted, the graph of y = |4 − x| lies above the graph of y = −5 for all real numbers.
Intuitively, the absolute value of any number is always nonnegative, so |4−x| > −5 for all real values of x.
Example $$\PageIndex{11}$$
Solve the inequality |4 − x| > 0 for x.
Solution
As we saw in Figure $$\PageIndex{6}$$(b), the graph of y = |4 − x| lies on or above the x-axis for all real numbers. It “touches” the x-axis at the “vertex” of the “V,” where $|4-x|=0$
This can occur only if
\begin{aligned} 4-x &=0 \\-x &=-4 \\ x &=4 \end{aligned}
Thus, the graph of y = |4 − x| is strictly above the x-axis for all real numbers except x = 4. That is, the solution of |4 − x| > 0 is {x : x 6= 4}.
Example $$\PageIndex{12}$$
Solve the inequality |4 − x| > 5 for x.
Solution
In this example, the graph of the right-hand side of |4 − x| > 5 is a horizontal line located 5 units above the x-axis. The graph of y = |4 − x| has the “V” shape shown in Figure $$\PageIndex{6}$$(c). You can use the intersect utility on the graphing calculator to approximate the points of intersection of the graphs of y = |4 − x| and y = 5, as we have done in Figure $$\PageIndex{7}$$(c) and (d). The calculator indicates two points of intersection, one at x = −1 and a second at x = 9.
The graph of y = |4 − x| lies above the graph of y = 5 for all values of x that lie either to the left of −1 or to the right of 9. Hence, the solution of |4 − x| > 5 is the set {x : x < −1 or x > 9}.
Following the guidelines established in Example $$\PageIndex{6}$$, we create the image shown in Figure $$\PageIndex{8}$$ on our homework paper. Note that we’ve labeled each axis, scaled each axis with xmin, xmax, ymin, and ymax, labeled each graph with its equation, and shaded and labeled the solution on x-axis.
Algebraic Approach. Let’s explore an algebraic solution of |4 − x| > 5. In much the same manner that |x| > 5 leads to the conditions x < −5 or x > 5, the inequality
$|4-x|>5$
requires that
$4-x<-5 \qquad \text { or } \qquad 4-x>5$
We can solve each of these independently by first subtracting 4 from each side of the inequality, then multiplying both sides of each inequality by −1, reversing each inequality as we do so.
$\begin{array}{rllrrl}{4-x} & {<} & {-5} & {\text { or }} & {4-x} & {>} & {5} \\ {-x} & {<} & {-9} && {-x} & {>} & {1} \\ {x} & {>} & {9} && {x} & {<} & {-1}\end{array}$
We prefer to write this solution in the order
$x<-1 \qquad \text { or } \qquad x>9$
as it then matches the order of the graphical solution shaded in Figure $$\PageIndex{8}$$. That is, the solution set is {x : x < −1 or x > 9}.
The algebraic technique of this last example leads to the following property.
Property 17
If a > 0, then the inequality |x| > a is equivalent to the compound inequality x < −a or x > a.
This property provides a simple algebraic technique for solving inequalities of the form |x| > a, when a > 0. Let’s concentrate on this technique in the examples that follow.
Example $$\PageIndex{13}$$
Solve the inequality |4x − 3| > 1 for x.
Solution
The first step is to use Property 17 to write that $|4 x-3|>1$ is equivalent to
$4 x-3<-1 \qquad \text { or } \qquad 4 x-3>1$
We can now solve each inequality independently. We begin by adding 3 to both sides of each inequality, then we divide both sides of the resulting inequalities by 4.
$\begin{array}{rrlrrl}{4 x-3} & {<} & {-1} & {\text { or }} & {4 x-3} & {>} & {1} \\ {4 x} & {<} & {2} && {4 x} & {>} & {4} \\ {x} & {<} & {\frac{1}{2}} & &{x} & {>} & {1}\end{array}$
We can sketch the solutions on a number line.
And we can describe the solution using interval and set-builder notation.
$(-\infty, 1 / 2) \cup(1, \infty)=\{x : x<1 / 2 \text { or } x>1\}$
Again, let a > 0. As we did with $$|x| \leq a$$, we can take the union of the solutions of |x| = a and |x| > a to find the solution of $$|x| \geq a$$. This leads to the following property.
Definition
If a > 0, then the inequality $$|x| \geq a$$ is equivalent to the inequality $$x \leq −a$$ or $$x \geq a$$.
Example $$\PageIndex{14}$$
Solve the inequality $$3|1 − x| − 4 \geq |1 − x|$$ for x.
Solution
Again, at first glance, the inequality $3|1-x|-4 \geq|1-x|$
looks unlike any inequality we’ve attempted to this point. However, if we subtract |1 − x| from both sides of the inequality, then add 4 to both sides of the inequality, we get
$3|1-x|-|1-x| \geq 4$
On the left, we have like terms. Note that 3|1−x|−|1−x| = 3|1−x|−1|1−x| = 2|1−x|. Thus,
$2|1-x| \geq 4$
Divide both sides of the last inequality by 2.
$|1-x| \geq 2$
We can now use Property 19 to write
$1-x \leq-2 \quad \text { or } \qquad 1-x \geq 2$
We can solve each of these inequalities independently. First, subtract 1 from both sides of each inequality, then multiply both sides of each resulting inequality by −1, reversing each inequality as you go.
$\begin{array}{rllrrl}{1-x } & {\leq} & {-2} & {\text { or }} & {1-x } & {\geq} & { 2} \\ {-x } & {\leq} & {-3} && {-x } & {\geq } & {1} \\ {x} & { \geq } & {3} & &{x} & { \leq} & {-1}\end{array}$
We prefer to write this in the order
$x \leq-1 \qquad \text { or } \qquad x \geq 3$
We can sketch the solutions on a number line.
And we can describe the solutions using interval and set-builder notation.
$(-\infty,-1] \cup[3, \infty)=\{x : x \leq-1 \text { or } x \geq 3\}$
## Revisiting Distance
If a and b are any numbers on the real line, then the distance between a and b is found by taking the absolute value of their difference. That is, the distance d between a and b is calculated with d = |a − b|. More importantly, we’ve learned to pronounce the symbolism |a − b| as “the distance between a and b.” This pronunciation is far more useful than saying “the absolute value of a minus b.”
Example $$\PageIndex{15}$$
Solve the inequality |x − 3| < 8 for x.
Solution
This inequality is pronounced “the distance between x and 3 is less than 8.” Draw a number line, locate 3 on the line, then note two points that are 8 units away from 3.
Now, we need to shade the points that are less than 8 units from 3.
Hence, the solution of the inequality |x − 3| < 8 is $(-5,11)=\{x :-5<x<11\}$
Example $$\PageIndex{16}$$
Solve the inequality |x + 5| > 2 for x.
Solution
First, write the inequality as a difference.
$|x-(-5)|>2$
This last inequality is pronounced “the distance between x and −5 is greater than 2.” Draw a number line, locate −5 on the number line, then note two points that are 2 units from −5.
Now, we need to shade the points that are greater than 2 units from −5.
Hence, the solution of the inequality |x + 5| > 2 is
$(-\infty,-7) \cup(-3, \infty)=\{x : x<-7 \quad \text { or } \quad x>-3\}$
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# oracle 笔试题
ORACLE笔试题
1.在Oracle中，以下不属于集合操作符的是（ ）。
A. UNION B. SUM C. MINUS D. INTERSECT
2.在Oracle中，执行下面的语句：
SELECT ceil(-97.342),
floor(-97.342),
round(-97.342),
trunc(-97.342)
FROM dual;
A.ceil() B. floor() C. round(0) D. trunc()
3.以下哪个命令可以被用来从表 STATE中 drop 列 UPDATE_DT （ ）。
A. ALTER TABLE STATE DROP COLUMN UPDATE_DT;
B. ALTER TABLE STATE REMOVE COLUMN UPDATE_DT;
C. DROP COLUMN UPDATE_DT FROM STATE;
D. 你不能从这个表中DROP该列.
4.哪个命令用来创建一个primary key constraint pk_books 在表 books, 列 ISBN上，请选择一个（ ）。
A. create primary key on books(ISBN);
B. create constraint pk_books primary key on books(ISBN);
C. alter table books add constraint pk_books primary key (ISBN);
D. alter table books add primary key (ISBN);
5.以下哪行有错（ ）。
1 X :=Y +200;
2 IF X <10 THEN
3 Y :=30;
4 ELSEIF X <40 THEN
5 Y :=20;
6 END IF;
7 END IF;
A. Line 2 B. Line 3 C. Line 4 D. Line 5
6.一个VIEW被以下语句创建，请问在该VIEW上可进行哪个操作（ ）
CREATE VIEW USA_STATES
AS SELECT *FROM STATE
WHERE CNT_CODE =1
A. SELECT B. SELECT , UPDATE C. SELECT , DELETE D. SELECT , INSERT
7.下面哪个语句是表示将T表中第3~5行数据列出来的SQL语句（）。
A. select * from t where rownum <=5
minus
select * from t where rownum <=2;
B. select * from t where rownum <=5
intersect
select * from t where rownum <=2;
C. select * from t where rownum <=5
union
select * from t where rownum <=2;
D. select * from t where rownum <=5
union all
select * from t where rownum <=2;
8.表EMPLOYEES的结构为:
(
EMP_ID NUMBER(4) NOT NULL
LAST_NAME VARCHAR2(30) NOT NULL
FIRST_NAME VARCHAR2(30)
DEPT_ID NUMBER(2)
JOB_CAT VARCHAR2(30)
SALARY NUMBER(8,2)
)
A. SELECT dept_id, MIN(salary), MAX(salary) FROM employees
WHERE MIN(salary) < 5000 AND MAX(salary) > 15000;
B. SELECT dept_id, MIN(salary), MAX(salary) FROM employees
WHERE MIN(salary) < 5000 AND MAX(salary) > 15000 GROUP BY dept_id;
C. SELECT dept_id, MIN(salary), MAX(salary) FROM employees
HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;
D. SELECT dept_id, MIN(salary), MAX(salary) FROM employees
GROUP BY dept_id HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;
E. SELECT dept_id, MIN(salary), MAX(salary) FROM employees
GROUP BY dept_id, salary HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;
9.在oracle中，下面哪个函数与｜｜运算符有相同的功能（ ）。
A、 ltrim B、contact C、substr D、instr
10.函数floor(-2345.67)返回结果是（ ）。
A. 2345.67 B. 2346 C.-2346 D. -2345
11. Oracle数据库中，下面（）可以作为有效的列名。（选择一项）
A. Column B. 123_NUM C. NUM_#123 D. #NUM123
12.表TEST数据如下，P_ID为上级ID：
CREATE TABLE TEST AS
SELECT 1 ID ,NULL P_ID ,'a' NAME FROM DUAL UNION
SELECT 2 ID ,1 P_ID ,'b' NAME FROM DUAL UNION
SELECT 3 ID ,1 P_ID ,'c' NAME FROM DUAL UNION
SELECT 4 ID ,2 P_ID ,'d' NAME FROM DUAL UNION
SELECT 5 ID ,2 P_ID ,'e' NAME FROM DUAL UNION
SELECT 6 ID ,4 P_ID ,'f' NAME FROM DUAL UNION
SELECT 7 ID ,4 P_ID ,'g' NAME FROM DUAL UNION
SELECT 8 ID ,1 P_ID ,'h' NAME FROM DUAL;
A. SELECT * FROM TEST CONNECT BY PRIOR P_ID=ID START WITH ID=2;
B. SELECT * FROM TEST CONNECT BY PRIOR ID=P_ID START WITH ID=2;
C. SELECT * FROM TEST CONNECT BY PRIOR P_ID=ID START WITH P_ID=2;
D. SELECT * FROM TEST CONNECT BY PRIOR ID=P_ID START WITH P_ID=2;
13. 表EMPLOYEES结构为
(
EMPLOYEE_ID NUMBER(6) not null primary key,
FIRST_NAME VARCHAR2(20) unique,
LAST_NAME VARCHAR2(25),
EMAIL VARCHAR2(25) not null,
PHONE_NUMBER VARCHAR2(20),
HIRE_DATE DATE,
JOB_ID VARCHAR2(10),
SALARY NUMBER(8,2),
COMMISSION_PCT NUMBER(2,2),
MANAGER_ID NUMBER(6),
DEPARTMENT_ID NUMBER(4)
)
A.select count(*) from EMPLOYEES
B.select count(EMPLOYEE_ID) from employees
C.select count(FIRST_NAME) from employees
D.select count(EMAIL) from employees
E.select count(JOB_ID) from employees
1. 有EMP表一张，字段为ID,NAME，ID有重复数据，请写出有重复数据的id、NAME语句，并写出没有重复数据的ID、name语句。
2. 去除字符串“ aabb ”两边的空格(函数)。
3. student表字段为（classno,name,birthday）,将所有3班出生日期小于1981年5月12日的记录删除。
4. student表字段为（classno，name,score）
Dept（部门表）： dno（部门号），dname（部门名）
Emp（员工表）：eno（员工号），ename（员工姓名）,esex（性别：男士/女士），sal（工资）,dno（部门号）,join_date（入职时间），elevel（级别：1,2,3…）
1. 列出部门为“销售部”，入职时间为2011年12月1日之前的所有男员工工资大于5000元的信息，并按入职时间降序、工资升序进行展示，展示内容为：部门名，员工姓名，工资，入职时间。
2. 列出员工人数最多的部门员工平均工资。
3. 写出SQL实现入职超过三年的员 工工资增加10%、级别加1。
4. 列出各部门员工工资在2000到3000之间不同性别人数，展示字段如下：部门、男士人数、女士人数。
posted @ 2015-09-09 16:43 她妈妈不喜欢我 阅读(...) 评论(...) 编辑 收藏
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jox51
Member
2019 Donor
VIP
The below indicator is a replica of the ADX system that Chris Moody coded for TradingView at the link below. Feel free to edit and use as you wish. He uses the Gann Swing Hi Low indicator, I used the PSAR indicator as an alternate. Some images attached below.
Code:
``````#
# This study attempts to replicate Chris Moody's ADX System for TOS. All credit goes to him. This study incorporates the GannHiLo study to determine the swing hi's and lows.
# Take a look at the video here where he explains this system.
# This study attempts to catch the beginning of a swing move in a trending underlying. What you want to do is waiting for crossovers of the DI+/DI- respectively while the ADX is beginning to Trend above 20 while the ParabolicSAR is in a supportive Swing High or Low. The buy/sell signals will alert you to this. Use your own discretion.
# When the ADX is over 40, the trend is exhausted and you should be looking at exiting your positions.
# Gann High Low
# Converted by theelderwand
input HPeriod= 13;
input LPeriod= 21;
def HLd = if close > SimpleMovingAvg(high, HPeriod)[1]
then 1
else (if close< SimpleMovingAvg(low, LPeriod)[1]
then -1
else 0);
def HLv = if HLd != 0 then HLd else 0;
plot HiLo = if HLv == -1
then SimpleMovingAvg(high, HPeriod)
else SimpleMovingAvg(low, LPeriod);
HiLo.AssignValueColor(if HLv==-1 then Color.Red else Color.Green);
#Below code is for ADX and DMI confirmation
input length = 14;
input averageType = AverageType.WILDERS;
def hiDiff = high - high[1];
def loDiff = low[1] - low;
def plusDM = if hiDiff > loDiff and hiDiff > 0 then hiDiff else 0;
def minusDM = if loDiff > hiDiff and loDiff > 0 then loDiff else 0;
def ATR = MovingAverage(averageType, TrueRange(high, close, low), length);
def "DI+" = 100 * MovingAverage(averageType, plusDM, length) / ATR;
def "DI-" = 100 * MovingAverage(averageType, minusDM, length) / ATR;
def DX = if ("DI+" + "DI-" > 0) then 100 * AbsValue("DI+" - "DI-") / ("DI+" + "DI-") else 0;
def ADX = MovingAverage(averageType, DX, length);
def sell = ADX > 20 && ADX > ADX[1] && "DI-" crosses above "DI+" && HLv==-1;
# Plot Signals
plot sellsignal = sell;
sellsignal.SetPaintingStrategy(PaintingStrategy.BOOLEAN_ARROW_DOWN);
sellsignal.SetDefaultColor(Color.CYAN);
sellsignal.SetLineWeight(1);``````
https://tos.mx/PRaxsci
DMIMod code below
https://tos.mx/CUaFybK
Incorporated the GannHiLow study to stay true to the original system.
Thanks @BenTen for pointing me to the GannHiLo study on this site.
Last edited:
diazlaz
Well-known member
2019 Donor
VIP
Should we port the Gann Swing Hi Low indicator, If it hasn't already. Let me know if you think it's worth it?
jox51
Member
2019 Donor
VIP
I would say so, that way it remains true to the original system
jox51
Member
2019 Donor
VIP
Just updated the code to incorporate the Gann High Low study. Enjoy.
diazlaz
Well-known member
2019 Donor
VIP
Hi @jox51 -awesome ty! - the buy and sell plots are not plotting not sure if the HLv criteria is valid I changed the buy from <> to != and it started the buy signal, haven't seen a sell condition triggered yet. Have you compared GANN to the PSAR? Something I added to backlog to test. thanks for the conversion.
jox51
Member
2019 Donor
VIP
Tested it, and it appears to be working. I did a scan but not many buy signals show up.
Last edited:
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# Distance between Charleston, SC (CHS) and Grand Junction, CO (GJT)
Flight distance from Charleston to Grand Junction (Charleston International Airport – Grand Junction Regional Airport) is 1645 miles / 2647 kilometers / 1429 nautical miles. Estimated flight time is 3 hours 36 minutes.
Driving distance from Charleston (CHS) to Grand Junction (GJT) is 1941 miles / 3124 kilometers and travel time by car is about 33 hours 1 minutes.
## Map of flight path and driving directions from Charleston to Grand Junction.
Shortest flight path between Charleston International Airport (CHS) and Grand Junction Regional Airport (GJT).
## How far is Grand Junction from Charleston?
There are several ways to calculate distances between Charleston and Grand Junction. Here are two common methods:
Vincenty's formula (applied above)
• 1644.894 miles
• 2647.200 kilometers
• 1429.374 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1641.649 miles
• 2641.978 kilometers
• 1426.554 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Charleston International Airport
City: Charleston, SC
Country: United States
IATA Code: CHS
ICAO Code: KCHS
Coordinates: 32°53′54″N, 80°2′25″W
B Grand Junction Regional Airport
City: Grand Junction, CO
Country: United States
IATA Code: GJT
ICAO Code: KGJT
Coordinates: 39°7′20″N, 108°31′37″W
## Time difference and current local times
The time difference between Charleston and Grand Junction is 2 hours. Grand Junction is 2 hours behind Charleston.
EDT
MDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 189 kg (416 pounds).
## Frequent Flyer Miles Calculator
Charleston (CHS) → Grand Junction (GJT).
Distance:
1645
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1645
Round trip?
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Thanks: 0
# Thread: Intermediate value theorem 2
1. If:
f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)
,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η
Proof.
1)Let S be a set of all xε[a,b] such that :f(x)<η.
Since a belongs to S and b is an upper bound then S has a Supremum
2)Let ξ= SupS
3) Will show that : a) $f(\xi)\geq\eta$ ,b) $f(\xi)\leq\eta$ and hence : $f(\xi)=\eta$
3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a
3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$
I have no idea how he gets $f(\xi)\geq \eta$ in part 3b and $f(\xi)\leq\eta$ in part 3a
2.
3. Originally Posted by solakis
If:
f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)
,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η
Proof.
1)Let S be a set of all xε[a,b] such that :f(x)<η.
Since a belongs to S and b is an upper bound then S has a Supremum
2)Let ξ= SupS
3) Will show that : a) $f(\xi)\geq\eta$ ,b) $f(\xi)\leq\eta$ and hence : $f(\xi)=\eta$
3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a
3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$
I have no idea how he gets $f(\xi)\geq \eta$ in part 3b and $f(\xi)\leq\eta$ in part 3a
3a) $\displaystyle f(\xi) \leq \eta$
Pick $\displaystyle \epsilon = \eta - f(\xi)$ as you should know
f is continuous at ξ that means for any $\displaystyle \epsilon >0$ there exist $\displaystyle \delta >0$
such that whenever $\displaystyle \mid x - \xi \mid < \delta$ we have $\displaystyle \mid f(x) - f(\xi) \mid < \epsilon$
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
$\displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi)$
$\displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi)$
$\displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta$ for
$\displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta$
but $\displaystyle \xi$ is an upper bound for S, $\displaystyle \xi < \xi +\frac{\delta}{2}$
and
$\displaystyle \xi + \frac{\delta}{2} \in S$ since
$\displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta$ but $\displaystyle \xi$ is an upper bound for S contradiction!! so $\displaystyle \eta - f(\xi)$ is not positive so still two choices less than or equal zero
Lets see if $\displaystyle f(\xi) > \eta$
f is continuous at $\displaystyle \xi$ , now pick $\displaystyle \epsilon = f(\xi) - \eta$
there exist delta
$\displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta$ for $\displaystyle \mid x - \xi \mid < \delta$
$\displaystyle -f(\xi) + \eta < f(x) - f(\xi)$
$\displaystyle \eta < f(x)$ for $\displaystyle \xi - \delta < x < \xi + \delta$
$\displaystyle \xi - \delta$ is in S the definition of the upper bound which means $\displaystyle f \left(\xi - \delta \right)$ should be less than $\displaystyle \eta$ not larger contradiction so our epsilon is not positive
This idea from Wiki see
Originally Posted by Amer
3a) $\displaystyle f(\xi) \leq \eta$
Pick $\displaystyle \epsilon = \eta - f(\xi)$ as you should know
f is continuous at ξ that means for any $\displaystyle \epsilon >0$ there exist $\displaystyle \delta >0$
such that whenever $\displaystyle \mid x - \xi \mid < \delta$ we have $\displaystyle \mid f(x) - f(\xi) \mid < \epsilon$
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
$\displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi)$
$\displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi)$
$\displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta$ for
$\displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta$
but $\displaystyle \xi$ is an upper bound for S, $\displaystyle \xi < \xi +\frac{\delta}{2}$
and
$\displaystyle \xi + \frac{\delta}{2} \in S$ since
$\displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta$ but $\displaystyle \xi$ is an upper bound for S contradiction!! so $\displaystyle \eta - f(\xi)$ is not positive so still two choices less than or equal zero
Lets see if $\displaystyle f(\xi) > \eta$
f is continuous at $\displaystyle \xi$ , now pick $\displaystyle \epsilon = f(\xi) - \eta$
there exist delta
$\displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta$ for $\displaystyle \mid x - \xi \mid < \delta$
$\displaystyle -f(\xi) + \eta < f(x) - f(\xi)$
$\displaystyle \eta < f(x)$ for $\displaystyle \xi - \delta < x < \xi + \delta$
$\displaystyle \xi - \delta$ is in S the definition of the upper bound which means $\displaystyle f \left(\xi - \delta \right)$ should be less than $\displaystyle \eta$ not larger contradiction so our epsilon is not positive
This idea from Wiki see
The proof you proposed is completely different from the proof i wrote in my OP.
Sorry but i am not asking for a proof of the IVT but an explanation of the OP's proof
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Hands-on with dplyr
Dmitry Grapov
Introduction
This is meant to be an introduction to dplyr which covers dplyr basics, gets into a little bit of programming with dplyr and ends with brief mention of some gotchas and a benchmarking comparison to base for the split-apply strategy. You might also find Rstudio's Data Wrangling Cheat Sheet featuring dplyr useful (this is also where I borrowed some of the images used in this tutorial).
The dplyr package from Hadley Wickham is plain awesome. It features consistent and succinct syntax, is computationally fast and getting better with every release. The `dplyr` package has replaced many common more verbose R idioms which I had to previously rely upon for most common data analysis tasks.
For example, many data analysis tasks involve the procedure of splitting the data set based on a grouping variable and then applying a function to each of the groups (split-apply). Lets say I want to calculate the median values for a few parameters for cars with different numbers of cylinders using the mtcars data set.
Set up the data for the example.
```#some data prep
data(mtcars)
data<-mtcars
data\$cyl<-factor(data\$cyl)```
Split-lapply-apply in `base`:
```#select some variable of interest
vars<-c("mpg","wt","qsec")
tmp.data<-data[,colnames(data)%in%vars]
#split the data on the number of cylinders
big.l<-split(tmp.data,data\$cyl)
#apply some function of interest to all columns
results<-lapply(big.l, function(x) apply(x,2,median))
#bind results and add splitting info
data.frame(cyl=names(results),do.call("rbind",results))```
``````## cyl mpg wt qsec
## 4 4 26.0 2.200 18.900
## 6 6 19.7 3.215 18.300
## 8 8 15.2 3.755 17.175
``````
Now the same process using `dplyr`:
```suppressPackageStartupMessages(library(dplyr))
#variables of interest
vars<-c("mpg","wt","qsec")
data %>% group_by(cyl) %>% select(one_of(vars)) %>% summarise_each(funs(median(.)))```
``````## Source: local data frame [3 x 4]
##
## cyl mpg wt qsec
## 1 4 26.0 2.200 18.900
## 2 6 19.7 3.215 18.300
## 3 8 15.2 3.755 17.175
``````
Switching from `base` to `dplyr` for data manipulation feels a little like this:
`base`
`dplyr`
Each of the individual `dplyr` verbs are discussed in more detail below, but the use of `%>%` or the pipe operator is worth mentioning now. The `%>%` operator is imported from `magrittr` and for the purpose of this tutorial we can simply think of it as `then`. From the cheatsheet referenced above:
Overview of common dplyr functions and verbs
I highly recommend that you take a look at the `dplyr` vignetts for more detailed description of all of this packages capabilities. One immediate addition in `dplyr` you might notice is `tbl_df` which is a local data frame and mostly behaves like the classical `data.frame` but is more convenient for working with large data.
`tbl_df(mtcars)`
``````## Source: local data frame [32 x 11]
##
## mpg cyl disp hp drat wt qsec vs am gear carb
## 1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
## 2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
## 3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
## 4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
## 5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
## 6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
## 7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
## 8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
## 9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
## 10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
## .. ... ... ... ... ... ... ... .. .. ... ...
``````
```# control the number of rows
print(tbl_df(mtcars),n=5)```
``````## Source: local data frame [32 x 11]
##
## mpg cyl disp hp drat wt qsec vs am gear carb
## 1 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
## 2 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
## 3 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
## 4 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
## 5 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
## .. ... ... ... ... ... ... ... .. .. ... ...
``````
You can can make sure all columns are output to the screen using `options(dplyr.width = Inf)`. `glimpse` is another useful function which is an analogue of `str` but tries to show you more of the data.
`str(mtcars)`
``````## 'data.frame': 32 obs. of 11 variables:
## \$ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
## \$ cyl : num 6 6 4 6 8 6 8 4 4 6 ...
## \$ disp: num 160 160 108 258 360 ...
## \$ hp : num 110 110 93 110 175 105 245 62 95 123 ...
## \$ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
## \$ wt : num 2.62 2.88 2.32 3.21 3.44 ...
## \$ qsec: num 16.5 17 18.6 19.4 17 ...
## \$ vs : num 0 0 1 1 0 1 0 1 1 1 ...
## \$ am : num 1 1 1 0 0 0 0 0 0 0 ...
## \$ gear: num 4 4 4 3 3 3 3 4 4 4 ...
## \$ carb: num 4 4 1 1 2 1 4 2 2 4 ...
``````
`glimpse(mtcars)`
``````## Observations: 32
## Variables:
## \$ mpg (dbl) 21.0, 21.0, 22.8, 21.4, 18.7, 18.1, 14.3, 24.4, 22.8, 19....
## \$ cyl (dbl) 6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8, 8, 8, 8, 8, 8, 4, 4, ...
## \$ disp (dbl) 160.0, 160.0, 108.0, 258.0, 360.0, 225.0, 360.0, 146.7, 1...
## \$ hp (dbl) 110, 110, 93, 110, 175, 105, 245, 62, 95, 123, 123, 180, ...
## \$ drat (dbl) 3.90, 3.90, 3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.9...
## \$ wt (dbl) 2.620, 2.875, 2.320, 3.215, 3.440, 3.460, 3.570, 3.190, 3...
## \$ qsec (dbl) 16.46, 17.02, 18.61, 19.44, 17.02, 20.22, 15.84, 20.00, 2...
## \$ vs (dbl) 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, ...
## \$ am (dbl) 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, ...
## \$ gear (dbl) 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, ...
## \$ carb (dbl) 4, 4, 1, 1, 2, 1, 4, 2, 2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, ...
``````
For the purpose of this tutorial we will be mostly working with data.frames, however it should be noted that dplyr syntax abstracts away the need to specify the kind of object being manipulated and most everything we will cover can also be applied to interact with a variety of database objects.
The most common `dplyr` functions also referred to as verbs are as follows (see more in the introduction vignette):
• filter() and slice()
• arrange()
• select() and rename()
• mutate() and transmute()
• summarise()
• group_by()
The following commands will be demonstrated using the hflights data set.
`suppressPackageStartupMessages(library(hflights))`
``````## Warning: package 'hflights' was built under R version 3.1.3
``````
`(flights <- tbl_df(hflights))`
``````## Source: local data frame [227,496 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
Which contains 227496 records for 21 variables for flights departing Houston airport for 2011.
Filter
Use `filter` to keep or select rows matching some criteria or condition(s).
`base`
`flights[flights\$Month == 1 & flights\$DayofMonth == 1, ]`
``````## Source: local data frame [552 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 1 6 728 840 AA 460
## 3 2011 1 1 6 1631 1736 AA 1121
## 4 2011 1 1 6 1756 2112 AA 1294
## 5 2011 1 1 6 1012 1347 AA 1700
## 6 2011 1 1 6 1211 1325 AA 1820
## 7 2011 1 1 6 557 906 AA 1994
## 8 2011 1 1 6 1824 2106 AS 731
## 9 2011 1 1 6 654 1124 B6 620
## 10 2011 1 1 6 1639 2110 B6 622
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
```#could have also used subset
subset(flights, Month == 1 & DayofMonth == 1)```
``````## Source: local data frame [552 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 1 6 728 840 AA 460
## 3 2011 1 1 6 1631 1736 AA 1121
## 4 2011 1 1 6 1756 2112 AA 1294
## 5 2011 1 1 6 1012 1347 AA 1700
## 6 2011 1 1 6 1211 1325 AA 1820
## 7 2011 1 1 6 557 906 AA 1994
## 8 2011 1 1 6 1824 2106 AS 731
## 9 2011 1 1 6 654 1124 B6 620
## 10 2011 1 1 6 1639 2110 B6 622
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
`#be wary of using subset programmatically: http://stackoverflow.com/questions/9860090/in-r-why-is-better-than-subset`
`dplyr`
```#comma is the same as an ampersand (&)
filter(flights, Month == 1, DayofMonth == 1)```
``````## Source: local data frame [552 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 1 6 728 840 AA 460
## 3 2011 1 1 6 1631 1736 AA 1121
## 4 2011 1 1 6 1756 2112 AA 1294
## 5 2011 1 1 6 1012 1347 AA 1700
## 6 2011 1 1 6 1211 1325 AA 1820
## 7 2011 1 1 6 557 906 AA 1994
## 8 2011 1 1 6 1824 2106 AS 731
## 9 2011 1 1 6 654 1124 B6 620
## 10 2011 1 1 6 1639 2110 B6 622
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
```# use pipe(|) for or
filter(flights, Month == 1 | DayofMonth == 1)```
``````## Source: local data frame [25,769 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
We can also include any of the following operators in filter.
Slice is a variant of filter used to extract rows based on position.
`base`
`flights[1:10,]`
``````## Source: local data frame [10 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
`dplyr`
`slice(flights, 1:10)`
``````## Source: local data frame [10 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
Arrange
Order data based on specified columns.
`base`
`flights[order(flights\$Month),]`
``````## Source: local data frame [227,496 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
`dplyr`
`arrange(flights,Month)`
``````## Source: local data frame [227,496 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
```#decreasing order
arrange(flights,desc(Month))```
``````## Source: local data frame [227,496 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 12 15 4 2113 2217 AA 426
## 2 2011 12 16 5 2004 2128 AA 426
## 3 2011 12 18 7 2007 2113 AA 426
## 4 2011 12 19 1 2108 2223 AA 426
## 5 2011 12 20 2 2008 2107 AA 426
## 6 2011 12 21 3 2025 2124 AA 426
## 7 2011 12 22 4 2021 2118 AA 426
## 8 2011 12 23 5 2015 2118 AA 426
## 9 2011 12 26 1 2013 2118 AA 426
## 10 2011 12 27 2 2007 2123 AA 426
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
```#break ties using more columns
arrange(flights,desc(Month),DayOfWeek)```
``````## Source: local data frame [227,496 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 12 19 1 2108 2223 AA 426
## 2 2011 12 26 1 2013 2118 AA 426
## 3 2011 12 5 1 558 926 AA 466
## 4 2011 12 12 1 609 921 AA 466
## 5 2011 12 19 1 603 913 AA 466
## 6 2011 12 26 1 558 912 AA 466
## 7 2011 12 5 1 1206 1311 AA 865
## 8 2011 12 12 1 1339 1436 AA 865
## 9 2011 12 19 1 1203 1314 AA 865
## 10 2011 12 26 1 1200 1318 AA 865
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int)
``````
Select
Select columns from the data.
`base`
`flights[,colnames(flights)%in%c("Month","DayOfWeek")]`
``````## Source: local data frame [227,496 x 2]
##
## Month DayOfWeek
## 1 1 6
## 2 1 7
## 3 1 1
## 4 1 2
## 5 1 3
## 6 1 4
## 7 1 5
## 8 1 6
## 9 1 7
## 10 1 1
## .. ... ...
``````
`dplyr`
`select(flights,Month,DayOfWeek)`
``````## Source: local data frame [227,496 x 2]
##
## Month DayOfWeek
## 1 1 6
## 2 1 7
## 3 1 1
## 4 1 2
## 5 1 3
## 6 1 4
## 7 1 5
## 8 1 6
## 9 1 7
## 10 1 1
## .. ... ...
``````
```#select using a dynamic variable
variables<-c("Month","DayOfWeek")
select(flights,one_of(variables))```
``````## Source: local data frame [227,496 x 2]
##
## Month DayOfWeek
## 1 1 6
## 2 1 7
## 3 1 1
## 4 1 2
## 5 1 3
## 6 1 4
## 7 1 5
## 8 1 6
## 9 1 7
## 10 1 1
## .. ... ...
``````
```#remove variables
select(flights,one_of(variables),-Month)```
``````## Source: local data frame [227,496 x 1]
##
## DayOfWeek
## 1 6
## 2 7
## 3 1
## 4 2
## 5 3
## 6 4
## 7 5
## 8 6
## 9 7
## 10 1
## .. ...
``````
Select also provides many regular expression wrappers.
Use `rename` to change column names.
`rename(flights,diverted=Diverted)`
``````## Source: local data frame [227,496 x 21]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), diverted (int)
``````
Mutate and Transmute
Use theses verbs to create a new column variable which in the case of `mutate` will be added to the or created as a stand-alone variables `transmute`.
Lets calculate the wait time based on the difference between `ArrTime` and `DepTime`.
`base`
```#transmute like
``````## [1] 100 100 150 110 102 144
``````
```#mutate like
``````## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## TailNum ActualElapsedTime AirTime ArrDelay DepDelay Origin Dest Distance
## 1 N576AA 60 40 -10 0 IAH DFW 224
## 2 N557AA 60 45 -9 1 IAH DFW 224
## 3 N541AA 70 48 -8 -8 IAH DFW 224
## 4 N403AA 70 39 3 3 IAH DFW 224
## 5 N492AA 62 44 -3 5 IAH DFW 224
## 6 N262AA 64 45 -7 -1 IAH DFW 224
## TaxiIn TaxiOut Cancelled CancellationCode Diverted wait
## 1 7 13 0 0 100
## 2 6 9 0 0 100
## 3 5 17 0 0 150
## 4 9 22 0 0 110
## 5 9 9 0 0 102
## 6 6 13 0 0 144
``````
`dplyr`
```#stand alone
transmute(flights,diff = ArrTime - DepTime)```
``````## Source: local data frame [227,496 x 1]
##
## diff
## 1 100
## 2 100
## 3 150
## 4 110
## 5 102
## 6 144
## 7 150
## 8 99
## 9 111
## 10 110
## .. ...
``````
```#added to data
mutate(flights,diff = ArrTime - DepTime)```
``````## Source: local data frame [227,496 x 22]
##
## Year Month DayofMonth DayOfWeek DepTime ArrTime UniqueCarrier FlightNum
## 1 2011 1 1 6 1400 1500 AA 428
## 2 2011 1 2 7 1401 1501 AA 428
## 3 2011 1 3 1 1352 1502 AA 428
## 4 2011 1 4 2 1403 1513 AA 428
## 5 2011 1 5 3 1405 1507 AA 428
## 6 2011 1 6 4 1359 1503 AA 428
## 7 2011 1 7 5 1359 1509 AA 428
## 8 2011 1 8 6 1355 1454 AA 428
## 9 2011 1 9 7 1443 1554 AA 428
## 10 2011 1 10 1 1443 1553 AA 428
## .. ... ... ... ... ... ... ... ...
## Variables not shown: TailNum (chr), ActualElapsedTime (int), AirTime
## (int), ArrDelay (int), DepDelay (int), Origin (chr), Dest (chr),
## Distance (int), TaxiIn (int), TaxiOut (int), Cancelled (int),
## CancellationCode (chr), Diverted (int), diff (int)
``````
Many dplyr functions will let you use newly create variables in the same function which is creating the variable in the first place.
`transmute(flights,diff = ArrTime - DepTime, ratio = ArrTime/diff, ratio2 = diff/ratio)`
``````## Source: local data frame [227,496 x 3]
##
## diff ratio ratio2
## 1 100 15.00000 6.666667
## 2 100 15.01000 6.662225
## 3 150 10.01333 14.980027
## 4 110 13.75455 7.997356
## 5 102 14.77451 6.903782
## 6 144 10.43750 13.796407
## 7 150 10.06000 14.910537
## 8 99 14.68687 6.740715
## 9 111 14.00000 7.928571
## 10 110 14.11818 7.791372
## .. ... ... ...
``````
The `mutate_each` function can be used to apply a function to every column in the dataframe. Lets bin each column into quartiles using the `ntile` function.
`glimpse(mutate_each(flights,funs(ntile(.,n=4))))`
``````## Observations: 227496
## Variables:
## \$ Year (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ Month (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ DayofMonth (int) 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2,...
## \$ DayOfWeek (int) 3, 4, 1, 1, 2, 2, 3, 3, 4, 1, 1, 2, 2, 3, 3,...
## \$ DepTime (int) 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 2, 2,...
## \$ ArrTime (int) 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,...
## \$ UniqueCarrier (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ FlightNum (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ TailNum (int) 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3,...
## \$ ActualElapsedTime (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ AirTime (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ ArrDelay (int) 1, 1, 1, 3, 2, 2, 2, 1, 4, 4, 4, 3, 1, 2, 1,...
## \$ DepDelay (int) 2, 3, 1, 3, 3, 2, 2, 1, 4, 4, 4, 4, 2, 1, 2,...
## \$ Origin (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ Dest (int) 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,...
## \$ Distance (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ TaxiIn (int) 3, 3, 2, 4, 4, 3, 4, 3, 4, 3, 4, 1, 3, 2, 3,...
## \$ TaxiOut (int) 2, 1, 3, 4, 1, 2, 3, 2, 4, 4, 4, 2, 2, 3, 1,...
## \$ Cancelled (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ CancellationCode (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
## \$ Diverted (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
``````
Here is a list of some other `dplyr` convenience functions.
Summarise
Carry out a function on the data frame returning a single value.
Lets calculate the range for wait times and use this as an opportunity to involve `%>%`.
`transmute(flights,diff = ArrTime - DepTime) %>% summarise(.,min=min(diff,na.rm=TRUE),max=max(diff,na.rm=TRUE))`
``````## Source: local data frame [1 x 2]
##
## min max
## 1 -2322 1350
``````
Either we just discovered time travel or we should have been referencing the day and time in our calculation of the difference between arrival and departure. Lets count how many calculations we may have screwed up.
`(bad<-transmute(flights,diff = ArrTime - DepTime) %>% filter(diff<0) %>% count(.))`
``````## Source: local data frame [1 x 1]
##
## n
## 1 2718
``````
An error rate of 1.2 %may or may not be acceptable.
We can use `summarise_each` to apply a function to every column in the data set. Lets calculate the median and median absolute deviation for every numeric variable.
```vars<-colnames(flights)[sapply(flights,is.numeric)]
``````## Source: local data frame [1 x 32]
##
## Year_median Month_median DayofMonth_median DayOfWeek_median
## 1 2011 7 16 4
## Variables not shown: DepTime_median (int), ArrTime_median (dbl),
## FlightNum_median (dbl), ActualElapsedTime_median (dbl), AirTime_median
## (dbl), ArrDelay_median (dbl), DepDelay_median (int), Distance_median
## (dbl), TaxiIn_median (dbl), TaxiOut_median (int), Cancelled_median
``````
Here are some additional functions which can be used with summarise.
Group_by
Break the data sets into groups of rows.
`group_by` adds the final piece of the puzzle we need to execute the split-apply strategy to our hearts content. This function becomes very powerful when combined with the previously discussed `dplyr` verbs. For example lets calculate which day of the week has the most cancellations.
```flights %>% group_by(DayOfWeek) %>%
select(Cancelled) %>% summarise_each(funs(canceled=sum(.,na.rm=TRUE),
total=n(),
percent_cancelled=round(canceled/total*100,1)))```
``````## Source: local data frame [7 x 4]
##
## DayOfWeek canceled total percent_cancelled
## 1 1 344 34360 1.0
## 2 2 369 31649 1.2
## 3 3 396 31926 1.2
## 4 4 616 34902 1.8
## 5 5 663 34972 1.9
## 6 6 272 27629 1.0
## 7 7 313 32058 1.0
``````
We can also use `group_by` to generate groups using more than one variable. For example lets calculate the median `AirTime` times by `Distance` and `TailNum`.
```(slowest<-flights %>% group_by(Distance,TailNum) %>%
select(AirTime) %>% summarise_each(funs(mean(.,na.rm=TRUE))))```
``````## Source: local data frame [44,698 x 3]
## Groups: Distance
##
## Distance TailNum AirTime
## 1 79 N14940 30.00000
## 2 79 N14943 29.00000
## 3 79 N17928 22.00000
## 4 127 NaN
## 5 127 N11106 29.00000
## 6 127 N11107 24.00000
## 7 127 N11155 26.00000
## 8 127 N11187 25.00000
## 9 127 N11189 26.00000
## 10 127 N11535 27.33333
## .. ... ... ...
``````
Lets identify the 3 slowest and fastest planes based on `AirTime` for some arbitrary `Distance`.
```#choose arbitrary distance
tmp<-slowest %>% na.omit(.) %>%
filter(.,Distance==781) %>%
arrange(AirTime)
#not clear why, but the results can't be bound directly
``````## Source: local data frame [3 x 3]
## Groups: Distance
##
## Distance TailNum AirTime
## 1 781 N275WN 85
## 2 781 N453WN 87
## 3 781 N725SW 88
``````
`tmp %>% tail(.,3) %>% arrange(desc(.))`
``````## Source: local data frame [3 x 3]
## Groups: Distance
##
## Distance TailNum AirTime
## 1 781 N474WN 155
## 2 781 N16646 127
## 3 781 N14653 139
``````
A common data analysis task might be to carry out some group-wise normalization or adjustments of the data. For example we may want to calculate the day of the week with the slowest flights, but also adjust for differences between individual planes. To do this we will start by calculating the average speed for each plane.
```#calculate speed
flights<-flights %>% mutate(hrs=AirTime/60, speed=Distance/hrs)
(averages<-flights %>%
group_by(TailNum) %>%
select(.,speed) %>%
summarise_each(funs(mean(.,na.rm=TRUE))) %>%
rename(.,mean_speed=speed))```
``````## Source: local data frame [3,320 x 2]
##
## TailNum mean_speed
## 1 NaN
## 2 N0EGMQ 462.9232
## 3 N10156 441.9362
## 4 N10575 413.5374
## 5 N11106 436.8909
## 6 N11107 437.8388
## 7 N11109 434.9298
## 8 N11113 438.9109
## 9 N11119 441.4825
## 10 N11121 436.0297
## .. ... ...
``````
Next lets express the overall speed for each plane as ratio to the mean plane speed. To do this we will use one of the powerful join capabilities in `dplyr`.
We will join with the original data set based on `TailNum` and calculate the plane-adjusted measure of speed.
```right_join(flights,averages,by="TailNum") %>%
mutate(norm_speed = speed / mean_speed) %>%
group_by(DayOfWeek) %>%
select(contains("speed")) %>%
summarise_each(funs(mean(.,na.rm=TRUE)))```
``````## Source: local data frame [7 x 4]
##
## DayOfWeek speed mean_speed norm_speed
## 1 1 419.2984 420.5830 0.9965149
## 2 2 418.2220 420.4509 0.9942130
## 3 3 419.2926 420.4236 0.9968347
## 4 4 420.3597 420.3783 0.9995784
## 5 5 419.9916 420.4688 0.9985305
## 6 6 427.3059 422.7704 1.0107539
## 7 7 423.7673 421.4198 1.0053550
``````
So if things worked out like we expected it looks like Saturday flights are fastest and Tuesday the slowest.
Using dplyr programmatically
Most of the examples up to this point featured using `dplyr` in interactive mode. However there are variants of nearly every verb which are best suited for use inside other functions. To see what theses are take a look at `verb_` versions of each function (e.g. `summarise_`).
Benchmarking split-apply in base and dplyr
Finally I will wrap with a relatively non-sophisticated benchmarking head-to-head comparison of `base` and `dplyr` speed for the almighty split-apply strategy.
Lets set up the data.
```rows<-10000
cols<-100
groups<-100
samples<-rows/groups
tmp.data<-data.frame(matrix(rnorm(rows),rows,cols))
tmp.data\$group<-rep(1:groups,each=samples)```
This data set has 10^{4} rows, 100 columns and 100 groups with 100 samples each.
`base`
```ptm <- Sys.time()
#split the data on the number of cylinders
big.l<-split(tmp.data,tmp.data\$group)
#apply some function of interest to all columns
results<-lapply(big.l, function(x) apply(x,2,median))
#bind results and add splitting info
results<-data.frame(group=names(results),do.call("rbind",results))
#elapsed time
(bd<-Sys.time()-ptm )```
``````## Time difference of 0.5802019 secs
``````
`dplyr`
```ptm <- Sys.time()
results<-tmp.data %>% group_by(group) %>% summarise_each(funs(median(.)))
#elapsed time
``````## Time difference of 0.3492 secs
``````
Wow I just saved 0.2310019 seconds of my life!
Create a benchmark visualization comparing base to dplyr for differing number of groups, rows and columns. Uncomment the code in the appendix below and modify as needed to re-run the benchmark.
```load(file="benchmark results")
#create a plot
library(reshape2)
library(ggplot2)
tmp.data<-melt(res,id.vars=c("rows","columns","groups","samples")) %>% mutate(seconds=value*60)
ggplot(tmp.data, aes(y=seconds,x=groups,group=variable,color=variable)) + geom_line() +geom_point()+ facet_grid(rows ~ columns) +scale_y_log10()```
The plot above shows the calculation time for 10 replications in seconds (y-axis) for calculating the median of varying number of groups (x-axis), rows (y-facet) and columns (x-facet).
Gotchas
1. In `dplyr` rownames are a second class citizen and are not stored.
2. Even though nearly every `dplyr` tutorial features pipes (`%>%`) it may be easier to learn both `dplyr` and `%>%` separately.
3. Most `dplyr` functions only work on objects coercible to ~ `data.frames`. Lots of my debugging sessions start with trying to understand the data structure of objects I am passing to `dplyr`.
Appendix
```#
# #set up functions to time
# base_fun<-function(data){
# #split the data on the number of cylinders
# big.l<-split(data,data\$group)
#
# #apply some function of interest to all columns
# results<-lapply(big.l, function(x) apply(x,2,median))
#
# #bind results and add splitting info
# data.frame(group=names(results),do.call("rbind",results))
# }
#
# dplyr_fun<-function(data){
#
# data %>% group_by(group) %>% summarise_each(funs(median(.)))
# }
#
# #benchmark function
# benchmark_fun<-function(rows,cols,groups){
#
# #set up data
# samples<-floor(rows/groups)
# tmp.data<-data.frame(matrix(rnorm(rows),rows,cols))
# tmp.data\$group<-rep(1:groups,length.out=rows)
#
# #base
# base.time<-system.time(replicate(10,base_fun(tmp.data)))
#
# #dplyr
# dplyr.time<-system.time(replicate(10,dplyr_fun(tmp.data)))
#
# data.frame(rows=rows,columns=cols,groups=groups,samples=samples,base=signif(base.time["elapsed"]/10,3),dplyr=signif(dplyr.time["elapsed"]/10,3))
#
# }
#
# #run benchmarks
# len<-5
# groups<-seq(5, 100,length.out=len) %>% signif(.,0)
# rows<-seq(100, 10000,length.out=len) %>% signif(.,0)
# cols<-seq(10, 100,length.out=len) %>% signif(.,0)
#
# #benchmarks
# results<-list()
# counter<-1
# for(i in 1:length(groups)){
# .group<-groups[i]
# for(j in 1:length(rows)){
# .row<-rows[j]
# for(k in 1:length(cols)){
# .col<-cols[k]
# results[[counter]]<- benchmark_fun(.row,.col,.group)
# counter<-counter+1
# }
# }
# }
#
#
# res<-do.call("rbind",results)
# save(res,file="benchmark results")
# #create a plot
# library(reshape2)
# library(ggplot2)
# tmp.data<-melt(res,id.vars=c("rows","columns","groups","samples")) %>% mutate(seconds=value*60)
#
# ggplot(tmp.data, aes(y=seconds,x=groups,group=variable,color=variable)) + geom_line() +geom_point()+ facet_grid(columns ~ rows) +scale_y_log10()```
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the coefficient of friction between a block of weight 15N and a horizontal board on which it is resting is 3.0find the horizontal force which will just remove the block and the force acting at an angle of 30 with the horizontal which will just move the block.
• physics -
In the first case, the horizontal force needed to move the block is the (static) friction coefficient multiplied by the weight. That would be 45.0 N.
In the second case, the required friction force F is given by the formula
(Weight - Fsin30)*mu = F cos30
F[sin30*mu + cos30]= 15*mu
F = 15*mu/[1.5 + 0.866] = 19.0 N
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## All About These 15 Worksheets
These Decimal Addition worksheets are designed to help students master the concept of adding decimal numbers. The worksheets include a variety of creative and engaging exercises to help students build their skills and confidence in decimal addition.
Each worksheet in the series includes a mix of problems with varying degrees of difficulty, designed to challenge and engage students at different skill levels. The problems require students to add decimal numbers with different numbers of decimal places and to carry over decimals when necessary.
This series is most suitable for students in grades 5 to 7. It can be used in the classroom as a supplement to the math curriculum or at home as a way for parents to help their children practice and reinforce their math skills.
Overall, these Decimal Addition worksheets provide a comprehensive and engaging way for students to develop their skills and confidence in adding decimal numbers. With these worksheets, students can build a solid foundation in math and develop the skills they need to succeed in higher-level math courses.
## How to Add Decimal Values
Adding decimal values is similar to adding whole numbers, with the added step of aligning decimal points. Here’s a step-by-step guide to adding decimal values that teachers can share with their students:
1. Align Decimal Points – Write the numbers vertically so that the decimal points line up with each other. This makes it easier to add digits in the same place value.
2. Add Zeros, if Necessary – If one decimal has more digits after the decimal point than the other, add zeros to the shorter decimal so that they have the same number of digits. This will make adding the decimals simpler. Remember that adding zeros to the right of a decimal number does not change its value.
3. Start Adding From the Rightmost Digit – Begin with the rightmost digits (the smallest place value) and add them together. If the sum is 10 or greater, carry the 1 to the next place value on the left.
4. Continue Adding Digits – Continue adding the digits in each place value from right to left, carrying any value of 10 or greater to the next place value on the left.
5. Place the Decimal Point – Once you have added all the digits, place the decimal point in the sum directly below the aligned decimal points of the numbers being added.
6. Simplify, if Necessary – If the sum has trailing zeros after the decimal point (e.g., 4.50), you can remove them to simplify the result (e.g., 4.5).
Here’s an example of adding 3.25 and 0.879:
Step 1: Align decimal points
Step 2: Add zeros, if necessary
Start adding from the rightmost digit: 9 (0+9)
12 (5+7, carry the 1)
Step 4: 11 (1 + 2 + 8, carry the 1)
Step 5: Place the decimal point
Simplify, if necessary: No trailing zeros in this case, so the sum remains 4.129.
The sum of 3.25 and 0.879 is 4.129.
## How to Master Decimal Addition
Learning to add decimals is an essential skill for students, and it can be taught effectively using a step-by-step approach. Here’s a suggested process for teaching students how to add decimals:
• Ensure understanding of decimals – Before students begin learning to add decimals, make sure they have a solid understanding of decimals and their place values (tenths, hundredths, etc.). They should also be comfortable with the concept of fractions and how they relate to decimals.
• Review basic addition – Students should have a strong foundation in basic addition with whole numbers before attempting to add decimals.
• Align the decimals – Teach students that when adding decimals, it’s crucial to align the decimal points in a vertical arrangement. This ensures that each place value lines up correctly. Instruct students to write one number below the other, aligning the decimal points, and then draw a horizontal line below the numbers.
• Add zeros if necessary – To make it easier to add decimals with different numbers of decimal places, students can add zeros to the end of the shorter decimal number. This ensures that each decimal place has a corresponding number to add.
• Add from right to left – Students should start adding the numbers from the right-most column (smallest place value) and move to the left, just as they would with whole numbers. They should carry over any values when necessary.
• Place the decimal point – In the final answer, the decimal point should be placed directly below the decimal points in the original numbers, maintaining its position relative to the place values.
• Practice with examples – Provide students with numerous examples of decimal addition problems. Start with simple problems and gradually increase the difficulty level as they become more comfortable with the process.
• Use manipulatives or visual aids – To help students grasp the concept of adding decimals, use manipulatives like base-ten blocks or grid paper to visually represent the addition process.
• Incorporate real-life scenarios – Encourage students to apply their decimal addition skills to real-life situations, such as calculating prices, measuring distances, or working with time. This helps them understand the practical applications of decimal addition and reinforces their learning.
• Review and practice regularly – Ensure that students regularly review and practice adding decimals to build their confidence and fluency.
By using this step-by-step approach and incorporating a variety of teaching methods, students can effectively learn to add decimals and develop a strong foundation in this essential mathematical skill.
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# 11 DIXIE STATE UNIVERSITY DEPARTMENT OF EDUCATION ELEMENTARY 3rd SEMESTER LESSON PLAN TEMPLATE (3/3/14) Teacher Candidate: April
l Talbot
Contextual Factors: 26 Students 15 Boys 11 Girls 3 IEP 6 High Level Learners 2 ELL 2 Speech 6 Low Level Learners Classroom Environment: Desks are arranged in groups of five or six. They are placed randomly throughout the classroom. Whiteboard is on the east end of the room. The rug instruction area is on the west wall. Technology available.
## CONTEXTUAL FACTORS (classroom factors)
WALK-AWAY (As a result of this lesson, what do I want the students to know, understand, and be able to do?)
Standard 3: Physical Science. Students will gain an understanding of Physical Science through the study of the properties of Materials. Objective 2: Compare and contrast the differences in how different materials respond to change. a. Model physical changes of various materials. b. Investigate and provide evidence that matter is not destroyed or created through changes. Content Walk-Away:
Lesson 1: I will be able to identify matter and the three states, solid, liquid, and gas.
Language Walk-Away:
I will be able to describe and write the three states of matter, whether they are solid, liquid, or gas.
Vocabulary:
Matter, Mass, State, Solid, Liquid, Gas. ASSESSMENT EVIDENCE (What evidence do I need to show the students
have learned the Walk-Away?)
## Modifications/Accomo dations (ELL, IEP, GATE, etc.)
Give extra help for low level learners. Use pictures to describe vocabulary for William and Aaliyah. Ask the High Level Learners some higher level questions to
Formative Evidence (checking for understanding throughout the lesson): Listen and observe students describe the three states of matter, whether they are a solid, liquid, or gas. Content Walk-Away Evidence (Summative): Students will be able to identify the three states of matter, whether they are a solid, liquid, or gas. Language Walk-Away Evidence (Summative):
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I will be able to describe and write the three states of matter, whether they are solid, liquid, or gas.
stretch their thinking Keep an eye on where Treysen and Hailey sit so that they will focus on the lesson more. Make sure that Damian and William do not sit next to each other.
## ACTIVE LEARNING PLAN
Activate/Building Background Knowledge Have you ever wondered what is all around us? What our desks, pencils, and paper is made of? Remember, last week when we were learning about gravity? We talked about matter? Ask students by the raise of hands who remembers what matter is? Explain to the students that today they will be learning all about matter. Define the vocabulary words matter, mass, state, solid, liquid, gas. SIOP1-Content objectives, 2-Language objectives, 3-Content appropriate 7-Linked to background, 8-Linked to past learning, 9-Key vocabulary Focus Lesson (I do it)
5 min.
Give students the pretest on matter. View presentation on the PowerPoint. Explain and show the students the three states of matter, solid, liquid, and gas. SIOP, 4-Supplementary materials, 5Adaptation of content, 6-Meaningful activities, 10-Appropriate speech, 11-Clear explanation, 12-Variety of techniques
Modification/accommodations: (ELL, IEP, GATE, etc.) Help lower level students describe matter and the three states, by giving prompts when needed. Ask the high level and students about each state, whether its solid, liquid, or gas.
Formative assessment: Learning Goal Students can identify the three states of matter. Success Criteria Students will able to explain what matter is, and the three states of matter. Assessment Strategy Observe students when they describe matter and the three states.
Modification/accommodations: Use real objects of each state of matter. Guided Instruction (We do it) Have students move to the rug. Use popsicle sticks to call on students to come and draw from a mystery bag. As students draw an item from the mystery bag have them identify whether the item is a solid, liquid, or gas. Encourage group discussion by having the other students ask questions about the item. SIOP 16-Opportunity for interaction, 18-Wait time, 19-Opportunity for L1 students, 20-Hands-on materials, 29-Feedback Formative Assessment: Learning Goal Students can identify the three states of matter. Success Criteria Students will able to explain the three states of matter. Assessment Strategy Observe students when answering questions about the objects.
7 min.
Collaborative/Cooperative (You do it together) Students will sort pictures of the three states of matter. At their desks students will work together and sort the pictures according to the three states solids, liquids, or gas. SIOP 17-Grouping supports objectives, 21Activities to apply content/language knowledge, 23-Content objective supported, 24-Language objective
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5 min.
## supported, 25-Students engaged, 26-Pacing, 29-Feedback
Formative Assessment: Learning Goal Students can identify and place the pictures under the correct state, solid, liquid, or gas.
Success Criteria Students explore the three states of matter with their table group.
Assessment Strategy Listen and observe the students compare the three states of matter.
## Modification/accommodations: Give extra help to the lower and learners if needed.
10 min.
Independent (You do it alone) Students will write about three states of matter. This will be done in workshops. SIOP 22-Language skills: reading, writing, listening, speaking, 30-Assessment Summative Assessment: Student worksheet Modification/accommodations: Help the lower level students by making sure they understand the directions. Explain further if necessary.
5 min.
Closure/Review of walk-aways, vocabulary, and essential questions (Note: Closure includes student interactions, reflection, and/or demonstrations.) Have the students place their completed journals on the desk in a neat pile and then meet back at the rug. Ask the students what is matter, and can they identify the three states of matter. Inform the students that tomorrow we will continue to learn about matter and how matter can change from one state to another. We will be making predictions and observations. SIOP 27-Review vocabulary, 28-Review concepts
SIOP Indicators (Add SIOP number and description within the lesson plan) Preparation: 1-Content objectives, 2-Language objectives, 3-Content appropriate, 4-Supplementary materials, 5-Adaptation of content, 6-Meaningful activities Building Background: 7-Linked to background, 8-Linked to past learning, 9-Key vocabulary Comprehensive Input: 10-Appropriate speech, 11-Clear explanation, 12-Variety of techniques Interaction: 16-Opportunity for interaction, 17-Grouping supports objectives, 18-Wait time, 19-Opportunity for L1 students Practice/Application: 20-Hands-on materials, 21-Activities to apply content/language knowledge, 22-Language skills: reading, writing, listening, speaking Lesson Delivery: 23-Content objective supported, 24-Language objective supported, 25-Students engaged, 26-Pacing Review/Assessment: 27-Review vocabulary, 28-Review concepts, 29-Feedback, 30-Assessment
TEACHING NOTES
What materials do I need to have ready? PowerPoint, mystery box filled with objects that are different states of matter. What is the approximate time needed for this lesson? 35 minutes.
## REFLECTION AFTER LESSON
How can I use the assessment data to reflect on & evaluate the outcomes of teaching and learning? How can I transfer what I learned from teaching this lesson to future teaching? What was effective and not effective? What goals can I set to improve my practice and student learning?
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I can use the assessment data to reflect on my teaching of how well the students understood the three states of matter. If students are having trouble identifying the three states of matter I can give prompts when needed. Especially for my lower learners that need more time to verbalize the information about the three states of matter. I need to work on the pacing of my lesson.
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## What is the mass in kg of Pluto?
Bulk parameters
Pluto Ratio (Pluto/Earth)
Mass (1024 kg) 0.01303 0.0022
Volume (1010 km3) 0.702 0.0065
## How much is a pound on Pluto?
The surface gravity on Pluto is about 1/12th the surface gravity on Earth. For example, if you weigh 100 pounds on Earth, you would weigh 8 pounds on Pluto.
How do you calculate your weight on Pluto?
One easy way to calculate your weight on other planets is to multiply your weight on Earth by the ratio between gravity on another planet and Earth. Your mass is the same no matter what planet you’re on, but your weight changes because gravity is different.
### How do you calculate your mass in kilograms?
In physics the standard unit of weight is Newton, and the standard unit of mass is the kilogram. On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.
### What is the density and mass of Pluto?
1.88 g/cm³Pluto / Density
What is on Pluto?
Pluto’s surface is composed of a mixture of frozen nitrogen, methane, and carbon monoxide ices. The dwarf planet also has polar caps and regions of frozen methane and nitrogen. Pluto has three known moons, Hydra, Nix, and Charon.
#### How do you calculate weight and mass on other planets?
We calculate weight by multiplying mass by the gravity on the surface of the planet. So, if you know your weight on Earth and the surface gravity on Earth, you can calculate your mass. You can then calculate your weight on any other planet by using the surface gravity of that planet in the same equation.
#### How do we calculate mass?
Mass is always constant for a body. One way to calculate mass: Mass = volume × density. Weight is the measure of the gravitational force acting on a mass.
What is the density of Pluto in kg m3?
Density: 2000 kg/m3 (how does this compare with other objects in the solar system?) Orbital Period: 248 years!
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The pattern of the back-seamed sleeve
In the series of made to measure patterns, I would like to introduce a type of sleeve that is often used for coats or trench coats and it is not very complicated to achieve. This is the back-seamed sleeve which is articulated and fitted thanks to an elbow line seam.
Before you get to work, make sure you have the following measurements: armhole height and armhole length. If you do not know how to get them, go back to the post “the pattern for a made to measure sleeve” to find explanations.
So let’s start to draw the frame:
1) Draw a line AD parallel to the straight line. AD = length of the sleeve
2) Perpendicular to AD, draw a line EF = (7/10 of the armhole length) / 2
3) Draw the rectangle EFF’E ‘as in figure 1, with AE = DE’ and AF = DF’
4) Place AB = armhole height – 1 / 8th of this measurement, draw GK parallel to EF
5) Place AC = elbow length and draw HN parallel to EF
Let’s build the sleeve cap.
1) On EE’, place I such that EI = 1/3 of EG
2) On FF ‘, place J such that FJ = ½ of FK
3) On the one side and the other of B, place BB’= 1.5 cm and BB”= 1 cm
4) Place II’= 2 cm, parallel to EF
5) Draw IAJB’B”I’ in a curve
Let’s continue with the back seam.
1) Place HH’= 1 cm and E’L = 5 cm
2) Extend HL to L’ so that the L’D forms a straight angle in L’.
3) Extend L’D in M to get L’M = ½ desired width of the sleeve. M can be in or out of the frame
4) Place G’and G” on one side and the other of G such that G’G = GG”= 1 cm.
5) Draw IG”H and I’G’H’ in a curve and H’L in a straight line.
6) Check that IG”HL’= I’G’H’L’
Finish with the full pattern of the sleeve.
1) Fold on the JM line and mirror I’H’L’ and I’B’B”J
2) Retrace the bottom of the sleeve L’M by softening the angle with a French curve.
Your pattern finished, do not forget to place the mounting notches on the cap of the sleeve and at the elbow and to add the values of sewing.
Feel free to share your experience with this pattern from Esmod’s book “Become a pattern drafter, volume 2”. I found it perfectly suited to my last coat that I will try to show the pictures soon.
In the meantime, have a great holiday season and make a lot of sewing projects for 2019.
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# Mini IR Theremin
The Theory
The theory for how this musical instrument actually works will be split into two sections. This first seection will explain how the IR Proximity sensor takes and outputs data and how the microcontroller translates that data into usable distances units (centimeters or inches).
As discussed before the IR sensor has three connections, power, ground and output. The output will give us a specific analog voltage that correlates to the distance the sensor is away from an object. The voltage curve for Output Voltage to Detected Distance (for the 2Y0A21) can be see in the picture below.
Since we know the expected voltage output for certain distances we could brute-force the system and hard code the expected voltages but that wouldn't be the scientific approach. Both the 2Y0A21 and the GP2D120 have the same output curve.
Finding A Linear Model
To make it easier, some math people out there decided to use their brains and linearize the curve seen above and give us a nice equation that, for the valid portions of sensor data, has the same x,y data graph. Many different equations exist out in the wild, so I'll just pick one of them. The picture below shows you the graph of the formula function, it's the same as the output curve already seen above. Please note, if you choose to use the GP2D120 sensor for this project, the formula for that sensor is different.
The formula seen above uses a 10-bit A-to-D value, that number should always be between 600 and 50, representing the 3v to 0.4v range that the IR sensor can output. Also, notice that in the graph above voltage is used and not the raw A-to-D value like in the formula. The PIC will see the output voltage from the sensor as a 10-bit A-to-D value, never as a single 3v or 2v number.
What this tells us is that we have a formula that we can apply all input voltages to and get reliable results, instead of hardcoding different voltages into the system, which would work too, but would require more effort.
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## Wednesday, March 23, 2016
### Fixes for dotTree function for small trees
I have been working on fixing some issues with the phytools function dotTree that pertain to the scaling of dot sizes when the number of taxa in the trees is small.
The reason this issue arose is because the way the function currently works is by trying to maximize the use of vertical space in the plot for the circles. This works great when the number of tips is many. For instance:
library(phytools)
tree<-rtree(n=30)
X<-fastBM(tree,nsim=4)
dotTree(tree,X,length=8)
Or, for a discrete character:
Q<-matrix(c(-1,1,1,-1),2,2)
rownames(Q)<-colnames(Q)<-letters[1:2]
y<-sim.history(tree,Q)\$states
## Done simulation(s).
y
## t26 t28 t14 t23 t9 t20 t13 t7 t5 t22 t21 t16 t6 t11 t25 t8 t19 t15
## "a" "a" "b" "b" "a" "a" "a" "a" "b" "b" "a" "b" "a" "b" "a" "a" "b" "a"
## t12 t17 t4 t2 t30 t10 t3 t27 t18 t29 t1 t24
## "b" "b" "b" "b" "b" "a" "a" "b" "a" "b" "b" "b"
dotTree(tree,y)
Unfortunately, it starts to break down when the number of tips is few. For example:
tree<-rtree(n=8)
X<-fastBM(tree,nsim=6)
dotTree(tree,X,length=8)
y<-sim.history(tree,Q)\$states
## Done simulation(s).
dotTree(tree,y)
is obviously not working!
The main fix I implemented is to now first check the chosen circle radii sizes against the size of a character of the plotted tip labels, and then rescale these value sensibly.
The result seems to work pretty well - although I would be careful to watch out for bugs as it is new!
library(plotrix)
phylogram<-phytools:::phylogram
source("https://raw.githubusercontent.com/liamrevell/phytools/master/R/dotTree.R")
dotTree(tree,X,length=8,fsize=1.2)
dotTree(tree,y,method="phylogram",fsize=1.2)
It also works fine for larger trees as well:
tree<-rtree(n=40)
X<-fastBM(tree,nsim=6)
dotTree(tree,X,length=8,fsize=0.8)
This fix can already be installed from GitHub as follows:
library(devtools)
install_github("liamrevell/phytools")
That's it.
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## Sunday, December 30, 2012
### PR2023 INSTRUMENTATION AND CONTROL SYLLABUS | ANNA UNIVERSITY BE PRODUCTION ENGINEERING 6TH SEMESTER SYLLABUS REGULATION 2008 2011 2012-2013
Latest: TNEA 2014 Engineering Application Status, Counselling Date, Rank List
PR2023 INSTRUMENTATION AND CONTROL SYLLABUS | ANNA UNIVERSITY BE PRODUCTION ENGINEERING 6TH SEMESTER SYLLABUS REGULATION 2008 2011 2012-2013 BELOW IS THE ANNA UNIVERSITY SIXTH SEMESTER B.E PRODUCTION ENGINEERING DEPARTMENT SYLLABUS, TEXTBOOKS, REFERENCE BOOKS,EXAM PORTIONS,QUESTION BANK,PREVIOUS YEAR QUESTION PAPERS,MODEL QUESTION PAPERS, CLASS NOTES, IMPORTANT 2 MARKS, 8 MARKS, 16 MARKS TOPICS. IT IS APPLICABLE FOR ALL STUDENTS ADMITTED IN THE YEAR 2011 2012-2013 (ANNA UNIVERSITY CHENNAI,TRICHY,MADURAI, TIRUNELVELI,COIMBATORE), 2008 REGULATION OF ANNA UNIVERSITY CHENNAI AND STUDENTS ADMITTED IN ANNA UNIVERSITY CHENNAI DURING 2009
PR2023 INSTRUMENTATION AND CONTROL L T P C
3 0 0 3
UNIT I INTRODUCTION 9
Static and dynamic characteristics of measurement systems, standards and calibration,
error and uncertainty analysis, statistical analysis of data, and curve fitting.
UNIT II MECHANICAL MEASUREMENTS AND INDUSTRIAL
INSTRUMENTATION 10
Measurement of displacement, velocity (linear and rotational), acceleration, shock,
vibration, force torque power, strain, stress, pressure temperature.
UNIT III DATA DISPLAY AND RECORDING DEVICES 8
Data display-CRO,LED, LCD, magnetic tape recorders, x-y recorders, UV recorders,
Oscilloscope recorders, digital printers and data loggers.
UNIT IV CONTROL 9
Introduction to control systems, mathematical model of physical systems in transfer
function and state space forms, response of dynamic systems, concept of pole and zero
of a system, realization of transfer functions.
UNIT V STABILITY ANALYSIS 9
Stability criteria bode plots, routh and Nyquist criteria.
TOTAL: 45 PERIODS TEXT BOOKS:
1. B.C.Nakra, K.K.choudry, “Instrumentation, Measurement and analysis”, Tata McGraw
Hill 2002
2. J.J.Nagrath and Gopal, “control system engineering”, New age international (p) ltd.,
2000
REFERENCES:
1. C.S.Rangan, G.R.Sarma, VSV Mani, Instrumentation devices and systems”, Tata
McGraw Hill, 2000
2. A.K. Sowhney, “electricaland electronic measurement and instrumentation, “Dhanpat
rai & Cu, 2003.
3. Benjamin C.Kuo, “Automatic control system”, prentice hall of India pvt ltd.,2002
4. Ernest O.Doeblin, “measurement systems applications and design”, McGraw Hill
International editions, 1990
5. S.Renganathan, “transducer engineering”, Allied publishers, 1990.
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LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
Searching...
No Matches
## ◆ cpteqr()
subroutine cpteqr ( character compz, integer n, real, dimension( * ) d, real, dimension( * ) e, complex, dimension( ldz, * ) z, integer ldz, real, dimension( * ) work, integer info )
CPTEQR
Purpose:
``` CPTEQR computes all eigenvalues and, optionally, eigenvectors of a
symmetric positive definite tridiagonal matrix by first factoring the
matrix using SPTTRF and then calling CBDSQR to compute the singular
values of the bidiagonal factor.
This routine computes the eigenvalues of the positive definite
tridiagonal matrix to high relative accuracy. This means that if the
eigenvalues range over many orders of magnitude in size, then the
small eigenvalues and corresponding eigenvectors will be computed
more accurately than, for example, with the standard QR method.
The eigenvectors of a full or band positive definite Hermitian matrix
can also be found if CHETRD, CHPTRD, or CHBTRD has been used to
reduce this matrix to tridiagonal form. (The reduction to
tridiagonal form, however, may preclude the possibility of obtaining
high relative accuracy in the small eigenvalues of the original
matrix, if these eigenvalues range over many orders of magnitude.)```
Parameters
[in] COMPZ ``` COMPZ is CHARACTER*1 = 'N': Compute eigenvalues only. = 'V': Compute eigenvectors of original Hermitian matrix also. Array Z contains the unitary matrix used to reduce the original matrix to tridiagonal form. = 'I': Compute eigenvectors of tridiagonal matrix also.``` [in] N ``` N is INTEGER The order of the matrix. N >= 0.``` [in,out] D ``` D is REAL array, dimension (N) On entry, the n diagonal elements of the tridiagonal matrix. On normal exit, D contains the eigenvalues, in descending order.``` [in,out] E ``` E is REAL array, dimension (N-1) On entry, the (n-1) subdiagonal elements of the tridiagonal matrix. On exit, E has been destroyed.``` [in,out] Z ``` Z is COMPLEX array, dimension (LDZ, N) On entry, if COMPZ = 'V', the unitary matrix used in the reduction to tridiagonal form. On exit, if COMPZ = 'V', the orthonormal eigenvectors of the original Hermitian matrix; if COMPZ = 'I', the orthonormal eigenvectors of the tridiagonal matrix. If INFO > 0 on exit, Z contains the eigenvectors associated with only the stored eigenvalues. If COMPZ = 'N', then Z is not referenced.``` [in] LDZ ``` LDZ is INTEGER The leading dimension of the array Z. LDZ >= 1, and if COMPZ = 'V' or 'I', LDZ >= max(1,N).``` [out] WORK ` WORK is REAL array, dimension (4*N)` [out] INFO ``` INFO is INTEGER = 0: successful exit. < 0: if INFO = -i, the i-th argument had an illegal value. > 0: if INFO = i, and i is: <= N the Cholesky factorization of the matrix could not be performed because the leading principal minor of order i was not positive. > N the SVD algorithm failed to converge; if INFO = N+i, i off-diagonal elements of the bidiagonal factor did not converge to zero.```
Definition at line 144 of file cpteqr.f.
145*
146* -- LAPACK computational routine --
147* -- LAPACK is a software package provided by Univ. of Tennessee, --
148* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
149*
150* .. Scalar Arguments ..
151 CHARACTER COMPZ
152 INTEGER INFO, LDZ, N
153* ..
154* .. Array Arguments ..
155 REAL D( * ), E( * ), WORK( * )
156 COMPLEX Z( LDZ, * )
157* ..
158*
159* ====================================================================
160*
161* .. Parameters ..
162 COMPLEX CZERO, CONE
163 parameter( czero = ( 0.0e+0, 0.0e+0 ),
164 \$ cone = ( 1.0e+0, 0.0e+0 ) )
165* ..
166* .. External Functions ..
167 LOGICAL LSAME
168 EXTERNAL lsame
169* ..
170* .. External Subroutines ..
171 EXTERNAL cbdsqr, claset, spttrf, xerbla
172* ..
173* .. Local Arrays ..
174 COMPLEX C( 1, 1 ), VT( 1, 1 )
175* ..
176* .. Local Scalars ..
177 INTEGER I, ICOMPZ, NRU
178* ..
179* .. Intrinsic Functions ..
180 INTRINSIC max, sqrt
181* ..
182* .. Executable Statements ..
183*
184* Test the input parameters.
185*
186 info = 0
187*
188 IF( lsame( compz, 'N' ) ) THEN
189 icompz = 0
190 ELSE IF( lsame( compz, 'V' ) ) THEN
191 icompz = 1
192 ELSE IF( lsame( compz, 'I' ) ) THEN
193 icompz = 2
194 ELSE
195 icompz = -1
196 END IF
197 IF( icompz.LT.0 ) THEN
198 info = -1
199 ELSE IF( n.LT.0 ) THEN
200 info = -2
201 ELSE IF( ( ldz.LT.1 ) .OR. ( icompz.GT.0 .AND. ldz.LT.max( 1,
202 \$ n ) ) ) THEN
203 info = -6
204 END IF
205 IF( info.NE.0 ) THEN
206 CALL xerbla( 'CPTEQR', -info )
207 RETURN
208 END IF
209*
210* Quick return if possible
211*
212 IF( n.EQ.0 )
213 \$ RETURN
214*
215 IF( n.EQ.1 ) THEN
216 IF( icompz.GT.0 )
217 \$ z( 1, 1 ) = cone
218 RETURN
219 END IF
220 IF( icompz.EQ.2 )
221 \$ CALL claset( 'Full', n, n, czero, cone, z, ldz )
222*
223* Call SPTTRF to factor the matrix.
224*
225 CALL spttrf( n, d, e, info )
226 IF( info.NE.0 )
227 \$ RETURN
228 DO 10 i = 1, n
229 d( i ) = sqrt( d( i ) )
230 10 CONTINUE
231 DO 20 i = 1, n - 1
232 e( i ) = e( i )*d( i )
233 20 CONTINUE
234*
235* Call CBDSQR to compute the singular values/vectors of the
236* bidiagonal factor.
237*
238 IF( icompz.GT.0 ) THEN
239 nru = n
240 ELSE
241 nru = 0
242 END IF
243 CALL cbdsqr( 'Lower', n, 0, nru, 0, d, e, vt, 1, z, ldz, c, 1,
244 \$ work, info )
245*
246* Square the singular values.
247*
248 IF( info.EQ.0 ) THEN
249 DO 30 i = 1, n
250 d( i ) = d( i )*d( i )
251 30 CONTINUE
252 ELSE
253 info = n + info
254 END IF
255*
256 RETURN
257*
258* End of CPTEQR
259*
subroutine xerbla(srname, info)
Definition cblat2.f:3285
subroutine cbdsqr(uplo, n, ncvt, nru, ncc, d, e, vt, ldvt, u, ldu, c, ldc, rwork, info)
CBDSQR
Definition cbdsqr.f:233
subroutine claset(uplo, m, n, alpha, beta, a, lda)
CLASET initializes the off-diagonal elements and the diagonal elements of a matrix to given values.
Definition claset.f:106
logical function lsame(ca, cb)
LSAME
Definition lsame.f:48
subroutine spttrf(n, d, e, info)
SPTTRF
Definition spttrf.f:91
Here is the call graph for this function:
Here is the caller graph for this function:
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# Thread: for karl and johannes
1. Join Date
Apr 2010
Posts
423
## for karl and johannes
DOCUMENT 1 BY jOHANNES
As you wrote about GR; it is simple to say where GR is all about:
1. Physical laws must have the same form in any system of reference.
2. Tensors transform multilinear; thus when physical laws are expressed as tensors; they take the same form in any system of reference. So we express physical laws in tensors.
3. The path of an object is determined by the metric tensor.
4. The equivalence principle states that gravity is no more then acceleration (in simple words); so gravity must be embedded into the metric tensor, since the metric defines eventually acceleration.
5. We need a tensor equation that relates the metric tensor (that includes gravity) with the mass discribution (the cause of gravity). This is the Einstein Field Equation.
There are variations of GR; but the difference is an alternative field-equation.
6. An empty space gives the tensor (−1,+1,+1,+1) as solution; which is actually the SR metric. So SR is GR in an empty space; no gravity.
This is where GR is all about
********
DOCUMENT 2 BY JOHANNES
Hello Karl,
--- You have seen yourself on this site the confusion that is generated just by not understanding what your point #6 is. The confusion is further extended by their lack of understanding that the lorentz transforms are limited to describing how different observers describe an "event" depending on what their platforms relative velocities are. ---
you understand what relativity is all about.
As for time-dilation and length-contraction, they are observer based. They don't tell anything about what value the OTHER observer does measure. They only tell the difference between TWO measurements done by ONE observer involving for example a clock at rest and a moving clock or a rod at rest and a moving rod.
Time-dilation and length-contraction cannot be used to convert events from one observer to another; for that we need transformations.
Using time-dilation and length-contraction to convert events from one observer to another observer (thus not understanding time-dilation nor length-contraction) would leads to contradictions.
That is; if there is an observer O that has time T and there is another moving observer O' that has time T'; once you write T'=ζT; then you get the contradiction as T=ζT' as well; so this can only be true for ζ²=1; meaning within special relativity that v=0 as the only physical solution. Using however the observer that measures; we can write T'O=ζTO and TO'=ζT'O', where XY is the time X measured by observer Y. The observer supscript is normally dropped as it should be understood that time-dilation and length-contraction are based on measurements done by ONE observer.
The decay of muon is a simple observation that explains correct understanding of time-dilation. The lifetime of a muon is 2.22μs as observed in a laboratory. A cosmic ray that enters the atmosphere produces muons that are approaching the speed of light. The lifetime of those fast moving muons is about 11μs as observed by the observer in the laboratory. There is ONE observer that observes muons in the laboratory and fast moving muon in the atmosphere (due to cosmic rays). The ONE observer can use time-dilation to find the about 5 times larger life-time of the moving muon. The life-time in the muon frame remains however 2.22μs.
Any theory that uses time-dilation and length-contraction to convert events from one observer to another is by definition wrong, as it leads to contradictions. The only correct method to convert events from one observer to another is using transformations.
Best Regards - Johannes
***************************
i WANT TO TALK TO YOU ABOUT THESE TWO DOCUMENTS. i AGREE WITH EVERYTHING YOU SAY IF I STICK WITH THE TEXTBOOKS, BUT I AM NOT SURE THAT THE TEXTBOOKS ARE THE WAY TO GO WHEN WE ARE TALKING ABOUT REALITY VS OBSERVER CREATED REALITY. MY VIEW IS BIASED BY THE QUANTUM PICTURE AND I BELIEVE IT MUST BE WORKED IN.
Last edited by Dennis; 10-23-2010 at 05:34 AM.
2. Join Date
Apr 2010
Posts
423
## lorentz transforms
Let us take the muon example you quote. You agree with the textbooks on that!
-The decay of muon is a simple observation that explains correct understanding of time-dilation. The lifetime of a muon is 2.22μs as observed in a laboratory. A cosmic ray that enters the atmosphere produces muons that are approaching the speed of light. The lifetime of those fast moving muons is about 11μs as observed by the observer in the laboratory. There is ONE observer that observes muons in the laboratory and fast moving muon in the atmosphere (due to cosmic rays). The ONE observer can use time-dilation to find the about 5 times larger life-time of the moving muon. The life-time in the muon frame remains however 2.22μs.
************************************************** ********
I am going to place my laboratory in the exact center of the universe where there are no gravitational fields. I will not accept the existence of any other influences at that point.
1) I can then say that the clock rate of my muon produced in the laboratory is exactly rate1 = M1 x C^2 / h where energy1 =M1 x C^2.
*********
Now your muon falling toward the laboratory , since it is in motion relative to me at a constant rate, has obviously been produced in a non zero gravitational field and/or has been subjected to forces causing acceleration. Since a muon is a muon everywhere, i.e. its' rest mass is also M1, and that the motion indicates that its total energy2 is M1 x C^2 + ke2.
2) I can then say that it' clock rate, based on QM, the equivalence principle , and confirmed by experiments (C.Alley, et.al) is exactly
rate2 = {M1 x C^2 - 1/2(ke2)}/h . (note check me carefully against my article at this point.)
************************************************** *********************************
Thus your clock runs slower than mine. So If we assume space is fixed, and you are moving at V relative to me, the same unit measuring rod measured using my clock would appear shorter to us both. When we measure the unit rod using your clock we both say the rod appears longer.
*********************
2) A significant point here is that since we are using all of our resources (and we can because we have experimental confirmation) we know the rate at which the other's clock's runs is the ratio :
rate1/rate2 is = [M1 x C^2 / h ] / [{M1 x C^2 - 1/2(ke2)}/h ]
and we can see that :
a) The ratio of the rates are strictly a function of dV, a vector in fixed space, and
b) It is not true that we would see each other's clock running slower than ours because a vector is involved, and
c) That the dS in measurement of the unit measuring rod in a fixed space by observers traveling at a relative velocity dV and using their own clocks is observer driven.
*******************
3) And we can see that the Lorentz transform for length is not compatable to GR and QM which are compatable with each other. This is of course because the
Lorentz transforms allow no past history of acceleration to influence their results.
***********************
This is what I have been pointing out, to everyone, the difference between observer driven reality and "Reality".
Do you agree and what comments do you have on this. Where did I screw up? I always do , you know! I think this is the way to discuss relativity. Next, I want to look at Einstein's velocity transformation in light of the above.
Last edited by Dennis; 10-23-2010 at 07:40 AM.
3. Join Date
Mar 2010
Posts
2,025
## To Karl
Hello Karl,
Originally Posted by Dennis
Let us take the muon example you quote. You agree with the textbooks on that!
-
"agree with the textbooks" is a general remark. I do not know if ALL textbooks treat this phenomena the same way. And for the purpose of education, sometimes a simpler method is used in textbooks, that is formaly not correct.
To what textbook do you refer specificaly?
Regards - Johannes
4. Join Date
Apr 2010
Posts
423
## johannes
Oh, I had a book on electrodynamics out, it was by David Griffiths, "introduction ti electrodynamics" . The discussion on the lorentz transforms seem clearer than other texts, at least to me. It discusses the meson example also. Do you agree with all I said? You will have to go to QM or GR to get the 1/2 factor. --- keep me honest!---I figured it best to analyze the problem using everything and then consider the lorentz transforms which have no aprior knowledge. best---karl
5. Join Date
Apr 2010
Posts
423
## apriori knowledge
I just wanted to get this down on paper for you before I lost the thought since astrophysics is your ball game. If the relative velocities are a summed history of not only the clock rates of the forces applied but also the clock rate of the source then by making measurements from several observers(They know their clock rates at their gravitational positions) YOU COULD LEAST SQUARES TRIANGULATE IN ON THE CLOCK RATE, AND THUS THE MASS, of the known source, say a particular star and would not have to approximate it by the brightness! jUST A THOUGHT FOR THE DAY! BEST, SARGE, AN OLD AERIAL SURVEYOR!
Last edited by Dennis; 10-23-2010 at 08:51 AM.
6. Join Date
Apr 2010
Posts
423
## Johannes-system of reference
Just so I understand you clearly would you please give me a little lesson on your term "system of reference"? I assume You mean one inertial system referenced to another. Does the difference between the GR transformation and Newtonian
space enter in to this?
I admit that I have never thought of expressing Newtonian Space in terms of a 4 x 4 GR fundamental tensor with a zero instead of a -1.
Is there any way using tensors to show that mass in Newtonian space is the transform of space contraction. In other words to explain the reason why it is! Mentally I see the relationship in terms of geodesics but I can't explain it in any other terms than my serendipity produced in my article on QM. If there is, then you would have achieved a major modification to GR by showing that statistical QM is embedded in it and would have a formal unification.
These are things I am curious on but not skilled enough at. ---karl
P.S. I realize that with your effort on the other string you have little time to reply to me, and that is all right! I am basicly just talking to myself and throwing out an occasional idea to you. I know that the answer to my above paragraph lines in the extension of th conservation of momentum and energy laws to four space. I just don't know how to embed them in GR. I'm working on it and if I figure it out will give you another shot.
Last edited by Dennis; 10-24-2010 at 11:38 AM.
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# Golf Single Round Handicap Average Score Calculator
A key element of golf's appeal is its meticulous balance of challenge and fairness. This article will guide you through the concept of the single round handicap average score, its relevance in golf, how it's calculated, and its practical applications. Moreover, we will also touch upon notable individuals who have excelled in this area of golf.
🖹 Normal View 🗖 Full Page View Your Handicap Course Rating Slope Rating
Actual Differential = Factored Differential = Actual Average Round =
Please provide a rating, it takes seconds and helps us to keep this resource free for all to use
## Understanding the Single Round Handicap Average Score
The single round handicap average score reflects a golfer's potential ability in a single round, rather than over a prolonged period. It provides a snapshot of a golfer's potential performance on any given day and allows for dynamic adjustments based on recent scores.
## Calculation of the Single Round Handicap Average Score
The calculation involves a relatively straightforward formula that requires three pieces of information: the golfer's score for the round, the course rating, and the slope rating of the course.
Single Round Handicap = (Score - Course Rating) × 113 / Slope Rating
The difference between the golfer's score and the course rating is multiplied by a constant (113, representing a standard difficulty rating), then divided by the course's slope rating. The result estimates the golfer's performance level for that single round.
## Real-world Example
Let's say a golfer scores 90 on a course with a course rating of 72 and a slope rating of 130. The single round handicap would be calculated as:
Single Round Handicap = (90 - 72) × 113 / 130 = 15.78
This suggests that the golfer performed to a level of roughly 16 over par for that round.
## Notable Achievements
Focusing on single-round performance allows us to appreciate extraordinary feats in golf. For instance, in 1977, Al Geiberger was the first player in PGA Tour history to shoot a 59 in a single round, an achievement considered golf's "magic number." This impressive score demonstrates exceptional single-round performance and underlines the value of understanding and calculating single round handicaps.
As golf is a game of consistency and gradual improvement, understanding concepts like the single round handicap average score can aid players in their development and enjoyment of the sport.
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# Search by Topic
#### Resources tagged with Visualising similar to Semaphore:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to Semaphore
Compound transformations. Rotations. Symmetry. Reflections. Translations. Dynamic geometry. Investigations. Visualising.
### A Problem of Time
##### Stage: 4 Challenge Level:
Consider a watch face which has identical hands and identical marks for the hours. It is opposite to a mirror. When is the time as read direct and in the mirror exactly the same between 6 and 7?
### Screwed-up
##### Stage: 3 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Troublesome Dice
##### Stage: 3 Challenge Level:
When dice land edge-up, we usually roll again. But what if we didn't...?
### Reflecting Squarely
##### Stage: 3 Challenge Level:
In how many ways can you fit all three pieces together to make shapes with line symmetry?
### Turning Triangles
##### Stage: 3 Challenge Level:
A triangle ABC resting on a horizontal line is "rolled" along the line. Describe the paths of each of the vertices and the relationships between them and the original triangle.
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Coloured Edges
##### Stage: 3 Challenge Level:
The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?
### Pattern Power
##### Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
### Counting Triangles
##### Stage: 3 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Coke Machine
##### Stage: 4 Challenge Level:
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Can You Explain Why?
##### Stage: 3 Challenge Level:
Can you explain why it is impossible to construct this triangle?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
### Tilting Triangles
##### Stage: 4 Challenge Level:
A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates?
### Triangular Tantaliser
##### Stage: 3 Challenge Level:
Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles.
### Something in Common
##### Stage: 4 Challenge Level:
A square of area 3 square units cannot be drawn on a 2D grid so that each of its vertices have integer coordinates, but can it be drawn on a 3D grid? Investigate squares that can be drawn.
### Prime Magic
##### Stage: 2, 3 and 4 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
### Rolling Triangle
##### Stage: 3 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Cubic Net
##### Stage: 4 and 5 Challenge Level:
This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!
##### Stage: 4 Challenge Level:
Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Square It
##### Stage: 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Sliced
##### Stage: 4 Challenge Level:
An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron?
### Changing Places
##### Stage: 4 Challenge Level:
Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . .
### Flight of the Flibbins
##### Stage: 3 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Chess
##### Stage: 3 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Chords
##### Stage: 4 Challenge Level:
Two intersecting circles have a common chord AB. The point C moves on the circumference of the circle C1. The straight lines CA and CB meet the circle C2 at E and F respectively. As the point C. . . .
### Bent Out of Shape
##### Stage: 4 and 5 Challenge Level:
An introduction to bond angle geometry.
### Just Opposite
##### Stage: 4 Challenge Level:
A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square?
### John's Train Is on Time
##### Stage: 3 Challenge Level:
A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station?
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Three Frogs
##### Stage: 4 Challenge Level:
Three frogs hopped onto the table. A red frog on the left a green in the middle and a blue frog on the right. Then frogs started jumping randomly over any adjacent frog. Is it possible for them to. . . .
### Inside Out
##### Stage: 4 Challenge Level:
There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . .
### Circuit Training
##### Stage: 4 Challenge Level:
Mike and Monisha meet at the race track, which is 400m round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise. Where will they meet on their way around and will they ever. . . .
### Tetrahedra Tester
##### Stage: 3 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
### Coordinate Patterns
##### Stage: 3 Challenge Level:
Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead?
### Bands and Bridges: Bringing Topology Back
##### Stage: 2 and 3
Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Wari
##### Stage: 4 Challenge Level:
This is a simple version of an ancient game played all over the world. It is also called Mancala. What tactics will increase your chances of winning?
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
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longest
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en
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https://research.chalmers.se/publication/7350
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# Simulation of Irregular Waves and Wave Induced Loads on Wind Power Plants in Shallow Water Licentiatavhandling, 2003
The essay gives a short introduction to waves and discusses the problem with non-linear waves in shallow water and how they effect an offshore wind energy converter. The focus is on the realisation of non-linear waves in the time domain from short-term statistics in the form of a variance density spectrum of the wave elevation. For this purpose the wave transformation from deep water to the near to shore site of a wind energy farm at Bockstigen has been calculated with the use of SWAN (Simulating WAves Near Shore). The result is a wave spectrum, which can be used as input to the realisation. The realisation of waves is done by perturbation theory to the first and second-order. The properties calculated are the wave elevation, water particle velocity and acceleration. The wave heights from the second order perturbation equations are higher than those from the first order perturbation equations. This is also the case for the water particle kinematics. The increase of variance is significant between the first order and the second order realisation. The calculated wave elevation exhibits non-linear features as the peaks become sharper and the troughs flatter. The resulting forces are calculated using Morisons equation. For second order force and base moment there is an increase in the maximum values. The force and base moment are largest approximately at the zero up and down crossing of the wave elevation. This indicates an inertia dominated wave load. So far the flexibility and the response of the structure have not been taken into account. They are, however, of vital importance. For verification of the wave model the results will later on be compared with measurements at Bockstigen off the coast of Gotland in the Baltic Sea.
Non-linear wave models
ocean engineering
wind energy converter
perturbation theory
shallow water
## Författare
### Jenny Trumars
Chalmers, Institutionen för Vatten Miljö Transport, Vågor och hydraulik
1650-4143 (ISSN)
### Extreme non-linear wave forces on a monopile in shallow water
OMAE Offshore Mechanics and Arctic Engineering,; Vol. 22(2003)
Paper i proceeding
### The effect of wave non-linerarity on the forces on a wind turbine foundation in shallow water
Proceedings of the European Wind Energy Conference and Exhibition, 16 - 19 June, 2003, Madrid, Spain,; (2003)
Övrigt konferensbidrag
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en
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http://cboard.cprogramming.com/c-programming/147389-bubble-sort-algorithm-displaying-content.html
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text/html
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# Thread: bubble sort algorithm - displaying content
1. ## bubble sort algorithm - displaying content
Hello,
I am working on a bubble sort program and it is kicking my butt. I have been learning C for about 6 weeks now 1-2 times a week when I have the time away from work and family.
One of the items I am working on is a bubble sort program. I have listened to lecture after lecture, read and followed the tutorial (which did not work for me) on cprogramming.com and looked at multiple other examples.
I am not good at understanding techinese yet. I need some help displaying this. I have two programs. I am working in linux.
My first program generates psuedo random numbers. I pipe the generated numbers over to the find program which contains the bubblesort algorithm to look for a specfic number and display those numbers in order.
The numbers I am trying to sort are (using a seed for the psuedo random numbers):
17767
9158
39017
18547
56401
23807
37962
22764
7977
31949
22714
55211
16882
Now, the command for this is:
./generate 13 1
Now I use the pipe character to send those numbers to the find program as such ./generate 13 1 | ./find 9158
The find program has the bubble sort algorithm. I think the algorithm is correct but I cannot get this to diplay correctly to save my life.
Code:
```void
sort(int values[], int n)
{
// O(n^2) sort
for(int i=n-2;i>0;i--) {
for(int j=0;j<i;j++) {
if(values[j]>values[j+1]) {
int tempholder=values[j+1];
values[j+1]=values[j];
values[j]=tempholder;
printf("%d\n",values[j]);
}
}
}
return;
}```
Now it should display
7977
9158
16882
17767
18547
22714
22764
23807
31949
37962
39017
55211
56401
9158
18547
23807
37962
22764
7977
31949
22714
55211
23807
37962
22764
7977
31949
22714
22764
7977
31949
22714
22764
7977
22714
7977
22714
7977
22714
7977
7977
My two biggest issues:
1. I don't quite understand what I am doing with the bubblesort even after everything that I have read. I understand it probably 70% of the way.
2. I know that I have the printf() in the wrong place but I have tried it in numerous places to no avail so I figure maybe I have programmed something wrong.
Any help would be greatly appreciated.
Thank you,
Wayne
2. Print the array of numbers after they're sorted, not while you're sorting them. Take the printf right out of the sort routine. Print the array in the function that you call sort from, after you call it.
3. I have tried that already.
Code:
```void
sort(int values[], int n)
{
// TODO: implement an O(n^2) sort
for(int i=n-2;i>0;i--) {
for(int j=0;j<i;j++) {
if(values[j]>values[j+1]) {
int tempholder=values[j+1];
values[j+1]=values[j];
values[j]=tempholder;
}
printf("%d\n",values[j]);
}
}
return;
}```
Output
9158
17767
18547
39017
23807
37962
22764
7977
31949
22714
55211
9158
17767
18547
23807
37962
22764
7977
31949
22714
39017
9158
17767
18547
23807
22764
7977
31949
22714
37962
9158
17767
18547
22764
7977
23807
22714
31949
9158
17767
18547
7977
22764
22714
23807
9158
17767
7977
18547
22714
22764
9158
7977
17767
18547
22714
7977
9158
17767
18547
7977
9158
17767
7977
9158
7977
AND
Code:
```void
sort(int values[], int n)
{
// TODO: implement an O(n^2) sort
for(int i=n-2;i>0;i--) {
for(int j=0;j<i;j++) {
if(values[j]>values[j+1]) {
int tempholder=values[j+1];
values[j+1]=values[j];
values[j]=tempholder;
}
}
printf("%d\n",values[j]);
}
return;
}```
ERROR
make find
gcc -ggdb -std=c99 -Wall -Werror -o find find.c helpers.c -lcs50 -lm
helpers.c: In function 'sort':
helpers.c:46:26: error: 'j' undeclared (first use in this function)
helpers.c:46:26: note: each undeclared identifier is reported only once for each function it appears in
make: *** [find] Error 1
4. Originally Posted by scrfix
No you didn't. Printing them after the array has been sorted means completely outside all the loops involved in sorting. Eg;
Code:
```sort (values, n); // your actual sort call in main or wherever
for (i = 0; i < n; i++) printf("%d\n", values[i]); // completely separate event```
5. You also don't need a return statement at the end of a void function.
Returning is what a function just does when it has nowhere left to go.
6. Thank you so much for input and helping with that. I finally got it to print out in descending order. I am going to work on getting it in ascending order. I could not do exactly what you suggested because the function being called was actually in another file that was not included in to the first. I kept getting an error stating that it didn't know about the variables which was true so I added like so below and it is now working.
Code:
```void
sort(int values[], int n)
{
int i,j,k;
// TODO: implement an O(n^2) sort
for(i=n-2;i>0;i--) {
for(j=0;j<i;j++) {
if(values[j]>values[j+1]) {
int tempholder=values[j+1];
values[j+1]=values[j];
values[j]=tempholder;
}
}
}
for(k=n-2;k>0;k--) {
printf("%d\n",values[k]);
}
}```
7. I ran in to some issues sorting them realizing that I was missing some of the numbers that were being passed through and it wasn't sorting all of the numbers so I made a correction to the function. Now I don't know whether or not that is correct or not. All of the tutorials I read stated it was n-2 so that you didn't give a chance to outside the bounds of the array or get an unexpected bug in the function however I found that when doing n-2 I couldn't get all of the numbers sorted so I had to change it to n-1 and then correct the ascending and descending orders.
Code:
```void
sort(int values[], int n)
{
int i,j,k;
// TODO: implement an O(n^2) sort
for(i=n-1;i>0;i--) {
for(j=0;j<i;j++) {
if(values[j]>values[j+1]) {
int tempholder=values[j+1];
values[j+1]=values[j];
values[j]=tempholder;
}
}
}
for(k=n-1;k>=0;k--) {
if(k==n-1)
printf("\nDescending Order:\n");
printf("%d\n",values[k]);
}
for(k=0;k<n;k++) {
if(k==0)
printf("\n\nAscending Order:\n");
printf("%d\n",values[k]);
}
}```
8. n-1 is the highest allowable index. Since j is always LESS than i, and i is maximum n-1, your j+1 index will stay in bounds. So it looks okay.
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| 2.765625
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crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00549.warc.gz
| 563,724,926
| 7,163
|
# Search by Topic
#### Resources tagged with Making and proving conjectures similar to Plus Minus:
Filter by: Content type:
Age range:
Challenge level:
### There are 27 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures
### DOTS Division
##### Age 14 to 16 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Always a Multiple?
##### Age 11 to 14 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Regular Hexagon Loops
##### Age 11 to 14 Challenge Level:
Make some loops out of regular hexagons. What rules can you discover?
### What's Possible?
##### Age 14 to 16 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Maxagon
##### Age 11 to 14 Challenge Level:
What's the greatest number of sides a polygon on a dotty grid could have?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Consecutive Negative Numbers
##### Age 11 to 14 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Loopy
##### Age 14 to 16 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Triangles Within Triangles
##### Age 14 to 16 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours house numbers. Can you explain the patterns I noticed?
### To Prove or Not to Prove
##### Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Problem Solving, Using and Applying and Functional Mathematics
##### Age 5 to 18 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Few and Far Between?
##### Age 14 to 18 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### How Old Am I?
##### Age 14 to 16 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Curvy Areas
##### Age 14 to 16 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### A Little Light Thinking
##### Age 14 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Close to Triangular
##### Age 14 to 16 Challenge Level:
Drawing a triangle is not always as easy as you might think!
### Polycircles
##### Age 14 to 16 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
##### Age 14 to 16 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Triangles Within Pentagons
##### Age 14 to 16 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Triangles Within Squares
##### Age 14 to 16 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Helen's Conjecture
##### Age 11 to 14 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
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# How to Make a Bar Graph – Definition, Advantages & Examples
This post is also available in: हिन्दी (Hindi)
A graphical representation is a visual display of data and statistical results. It is more effective than presenting the data in the form of tables. There are many ways to represent the data graphically such as bar graph, double bar graph, pictograph, line graph, linear graph, histogram, pie chart, etc.
A bar chart or bar graph is a chart or graph that presents the data with rectangular bars with heights or lengths proportional to the values that they represent. For example, if you want to present information on sales of cars during a period, you draw rectangles of the same width and height corresponding to the numbers representing the sales figures.
Let’s understand what is a bar graph and how is it read and how to make a bar graph using examples.
## What is a Bar Graph?
A bar graph or a bar chart is a pictorial representation of grouped data. A bar graph shows the data with rectangular bars of the same width and the heights proportional to the values that they represent. The bars in the graph can be shown vertically or horizontally.
A bar graph is an excellent tool to represent data that are independent of one another and that do not need to be in any specific order while being represented. The bars give a visual display for comparing quantities in different categories.
The bar graphs have two lines, horizontal and vertical axis, also called the $x$ and $y$-axis along with the title, labels, and scale range.
Let’s consider an example of how does a bar graph looks like.
The bar graph shown above represents the sales of cars in a town during the period from 2000 to 2004. You can see the bar graph is drawn between two axes. The horizontal axis(or $x$-axis) shows the years and the vertical axis(or $y$-axis) shows the number of cars sold.
The rectangular bars drawn of the same width represent the sales of cars during the year corresponding to the bars. The length (or height) of each of the bars corresponds to the value(or the number of cars sold) during a year.
Let’s consider another example of the number of games played by tennis players in a particular year.
This bar graph also represents the same information related to sales of cars in a town during the period from 2000 to 2004. This bar graph is also drawn between two axes. The horizontal axis(or $x$-axis) shows the number of cars sold and the vertical axis(or $y$-axis) shows the years.
The only difference between these two bar graphs is that the first one is a vertical bar graph and the second one is a horizontal bar graph.
## Types of Bar Graphs
Bar Graphs are broadly classified into two types depending on the orientation of the bars(rectangles):
• Vertical Bar Graph
• Horizontal Bar Graph
The bars in bar graphs can be plotted horizontally or vertically, but the most commonly used bar graph is the vertical bar graph. Either of these two categories can be of the following two types.
• Grouped Bar Graph
• Stacked Bar Graph
### Vertical Bar Graphs
Vertical bar graphs are bar graphs where data is represented vertically in a graph or chart with the help of rectangular bars that show the measure of data. The rectangular bars are vertically drawn on the $x$-axis, and the $y$-axis shows the value of the height of the rectangular bars which represents the quantity of the variables written on the $x$-axis.
### Horizontal Bar Graphs
Horizontal bar graphs are bar graphs where data is represented horizontally in a graph or chart with the help of rectangular bars that show the measure of data. In this type, the variables or the categories of the data have to be written and then the rectangular bars are horizontally drawn on the y-axis and the x-axis shows the length of the bars equal to the values of different variables present in the data.
### Stacked Bar Graph
The stacked bar graph(or composite bar graph) divides the whole bar into different parts. In this, each part of a bar is represented using different colors to easily identify the different categories. It requires specific labeling to indicate the different parts of the bar. Thus, in a stacked bar graph every rectangular bar represents the whole, and each segment in the rectangular bar shows the different parts of the whole.
### Grouped Bar Graph
The grouped bar graph(or clustered bar graph) is used to show the discrete value for two or more categorical data. In this, rectangular bars are grouped by position for levels of one categorical variable, with the same colors showing the secondary category level within each group. It can be shown both vertically and horizontally.
## Properties of Bar Graph
These are some properties that make a bar graph unique and different from other types of graphs.
• All rectangular bars should have equal width and should have equal space between them.
• The rectangular bars can be drawn horizontally or vertically.
• The height of the rectangular bar is equivalent to the data they represent.
• The rectangular bars must be on a common base.
These are the advantages of bar graphs over the other types of graphs.
• Display relative numbers or proportions of multiple categories
• Summarize a large amount of data in a visual, easily interpretable form
• It makes trends easier to highlight than tables do
• Bar graphs help in studying patterns over a long period of time
• It is used to compare data sets. data sets are independent of each other
• It is the most widely used method of data representation. therefore, it is used in various fields
• Bar graphs estimates can be made quickly and accurately
• It permits visual guidance on the accuracy and reasonableness of calculations
• Bar graphs are very efficient in comparing two or three data sets
## How to Read a Bar Graph
The different steps to read a bar graph are given below:
Step 1: Check whether the given bar graph is a horizontal bar graph or a vertical bar graph.
Step 2: In the case of a vertical bar graph, the categories are present on the horizontal axis (or x-axis) and the data values are represented by the vertical bars(or rectangles).
In the case of a horizontal bar graph, the categories are present on the vertical axis (or y-axis) and the data values are represented by the horizontal bars(or rectangles).
Step 3: For each category,
In the case of a vertical bar graph note down the point on the $y$-axis corresponding to the height of a vertical bar
In the case of a horizontal bar graph note down the point on the $x$-axis corresponding to the length of a horizontal bar
### Examples
Let’s consider an example to understand how a bar graph is read.
Ex 1: Observe the bar graph and answer the following questions
• In which year maximum number of trees were planted by the eco-club?
• In which year the number of trees planted by the eco-club was minimum?
• In which two years, the same number of trees were planted. How many total trees were planted in those two years?
• How many total trees were planted by the eco-club between 2005 and 2010?
The longest bar(or rectangle) corresponds to the year 2008. Therefore, the maximum number of trees planted in the year 2008, and was 300.
The smallest bar(or rectangle) corresponds to the year 2009. Therefore, the minimum number of trees planted in the year 2009, and was 100.
The height of the bars corresponding to the years 2005 and 2010 is the same and is equal to 150. Therefore, between 2005 and 2010, each year 150 trees were planted each. The total number of trees planted in these two years was 150 + 150 = 300.
The number of trees planted each year is
During 2005, the number of trees planted = 150
During 2006, the number of trees planted = 250
During 2007, the number of trees planted = 200
During 2008, the number of trees planted = 300
During 2009, the number of trees planted = 100
During 2010, the number of trees planted = 150
Total number of trees planted by the eco-club between 2005 and 2010 were 150 + 250 + 200 + 300 + 100 + 150 = 1150
## How to Make a Bar Graph
The different steps to make a bar graph are given below:
Step 1: First, decide the title of the bar graph.
Step 2: Draw the horizontal axis and vertical axis. (For example, Favourite Fruit)
Step 3: Now, label the horizontal axis.
Step 4: Write the names on the horizontal axis, such as Apple, Mango, Pineapple, Orange, and Grapes.
Step 5: Now, label the vertical axis. (For example, the Number of Students)
Step 6: Finalise the scale range for the given data.
Step 7: Finally, draw the bar graph that should represent each category of the fruit with their respective numbers.
### Examples
Let’s consider an example to understand how a bar graph is drawn for the given data.
Ex 1: A group of students was asked about their favourite activities. The following table shows the data regarding the hobbies of the students. Draw a bar graph to represent the given data.
The first step is deciding the title of the bar graph. Let’s name our bar graph “Favourite Activities of Students”.
Next, draw the axes.
The horizontal axis represents the activity and the vertical axis represents the number of students.
To choose an appropriate scale, note down the minimum and maximum values in the data set.
The minimum value (Dance – Number of students = 5)
The maximum value (Drawing – Number of students = 30)
Since the maximum value is 30 and the minimum 5, so let’s mark the maximum point on the vertical axis as 35 starting from 0 with equal intervals of 5. This will fit all the values properly in the bar graph.
Finally, draw the bar graph that should represent each category of the activity with their respective numbers.
## Practice Problems
In the library of a school, there are books on the following subjects in given numbers: English — 150, History — 300, Math — 500, Science — 325. Draw a bar graph for the given data.
Draw a bar graph for the given data
The number of students studying in each of the five classes of a school is given below. Draw a bar graph to represent the numerical data.
From a pond, the following numbers of fish were caught on different days. Draw the bar graph of it.
## FAQs
### What is a bar graph explain with an example.
A bar graph can be defined as a graphical representation of data, quantities, or numbers using bars or strips. They are used to compare and contrast different types of data, frequencies, or other measures of distinct categories of data. For example, the graph on the right shows the number of trees planted by eco-club during the years 2005 – 2010.
### What is the difference between a bar graph and a histogram?
The major difference between a bar chart and a histogram is the bars of the bar chart are not just next to each other. In a histogram, the bars are adjacent to each other. In statistics, bar charts and histograms are important for expressing a huge or big number of data.
### How do you represent a bar graph?
The rectangular bars in a bar graph can be drawn horizontally or vertically. In a bar graph, horizontal (or vertical) rectangular bars should have equal width and space between them. The height of the rectangular bars in a bar chart is equivalent to the given data it represents.
## Conclusion
A bar graph shows the data with rectangular bars of the same width and the heights proportional to the values that they represent. The bars in the graph can be shown vertically or horizontally. These are very helpful in displaying relative numbers or proportions of multiple categories.
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Search a number
172034067 = 3173373217
BaseRepresentation
bin10100100000100…
…00100000010011
3102222201020101010
422100100200103
5323020042232
625023142003
74156160022
oct1220204023
9388636333
10172034067
1189121817
1249744903
132984510a
1418bc28b9
15101830cc
hexa410813
172034067 has 8 divisors (see below), whose sum is σ = 242871696. Its totient is φ = 107942912.
The previous prime is 172034041. The next prime is 172034069. The reversal of 172034067 is 760430271.
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 172034067 - 223 = 163645459 is a prime.
It is not an unprimeable number, because it can be changed into a prime (172034069) 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, 1686558 + ... + 1686659.
It is an arithmetic number, because the mean of its divisors is an integer number (30358962).
Almost surely, 2172034067 is an apocalyptic number.
172034067 is a gapful number since it is divisible by the number (17) formed by its first and last digit.
172034067 is a deficient number, since it is larger than the sum of its proper divisors (70837629).
172034067 is a wasteful number, since it uses less digits than its factorization.
172034067 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 3373237.
The product of its (nonzero) digits is 7056, while the sum is 30.
The square root of 172034067 is about 13116.1757764983. The cubic root of 172034067 is about 556.1664906491.
The spelling of 172034067 in words is "one hundred seventy-two million, thirty-four thousand, sixty-seven".
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Resposta rápida: Does a kite have a right angle?
Contents
The intersection of the diagonals of a kite form 90 degree (right) angles. This means that they are perpendicular. The longer diagonal of a kite bisects the shorter one.
Does a kite always have a right angle?
Thus the right kite is a convex quadrilateral and has two opposite right angles. If there are exactly two right angles, each must be between sides of different lengths. All right kites are bicentric quadrilaterals (quadrilaterals with both a circumcircle and an incircle), since all kites have an incircle.
Does a kite have a 90 degree angle?
The kites that are also cyclic quadrilaterals (i.e. the kites that can be inscribed in a circle) are exactly the ones formed from two congruent right triangles. That is, for these kites the two equal angles on opposite sides of the symmetry axis are each 90 degrees.
How many right angles does a kite have?
A kite has two pairs of equal sides. It has one pair of equal angles. The diagonals bisect at right angles.
Does a kite have 4 right angles?
No, because a rhombus does not have to have 4 right angles. Kites have two pairs of adjacent sides that are equal.
ЭТО ИНТЕРЕСНО: How long does a kitesurfing kite last?
What are the 5 properties of a kite?
Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Other important polygon properties to be familiar with include trapezoid properties, parallelogram properties, rhombus properties, and rectangle and square properties.
Is every kite a rhombus?
For example, kites, parallelograms, rectangles, rhombuses, squares, and trapezoids are all quadrilaterals. Kite: A quadrilateral with two pairs of adjacent sides that are equal in length; a kite is a rhombus if all side lengths are equal.
Do angles in a kite add up to 360?
Find An Angle In A Kite : Example Question #4
Explanation: … A kite is a polygon with four total sides (quadrilateral). The sum of the interior angles of any quadrilateral must equal: degrees degrees degrees. Additionally, kites must have two sets of equivalent adjacent sides & one set of congruent opposite angles.
How do you prove a kite?
How to Prove that a Quadrilateral Is a Kite
1. If two disjoint pairs of consecutive sides of a quadrilateral are congruent, then it’s a kite (reverse of the kite definition).
2. If one of the diagonals of a quadrilateral is the perpendicular bisector of the other, then it’s a kite (converse of a property).
Why is a rectangle not a kite?
A kite and a rectangle cannot be the same at any time. The reasons are: Two pairs of adjacent sides are equal in a kite, but not so in a rectangle. Two diagonals intersect at right angles in a kite, but not so in a rectangle.
Are opposite angles in a kite equal?
A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs). … The opposite angles at the endpoints of the cross diagonal are congruent (angle J and angle L).
ЭТО ИНТЕРЕСНО: Pergunta frequente: When were Chinese kites invented?
Does a trapezium have right angles?
The trapezoid has two right angles.
5 right
Rectangle
Do all parallelograms have 4 right angles?
Rectangle: A parallelogram with 4 right angles. Rhombus: A parallelogram with 4 sides with equal length.
Can a trapezium have 2 right angles?
Explanation: A trapezoid can have either 2 right angles, or no right angles at all.
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# Aceburic Acid [C6H10O4] Molecular Weight Calculation
The molecular weight of Aceburic Acid [C6H10O4] is 146.1418.
Aceburic acid [C6H10O4] is an organic compound of three elements: Carbon, Hydrogen, and Oxygen. The molecular weight of Aceburic acid [C6H10O4] is 146.1418 which can be calculated by adding up the total weight (atomic weight multiplied by their number) of Carbon, Hydrogen, and Oxygen.
# CALCULATION PROCEDURE: Aceburic Acid [C6H10O4] Molecular Weight Calculation
Step 1: Find out the chemical formula and determine constituent atoms and their number in an Aceburic Acid molecule.
You will know different atoms and their number in an Aceburic Acid molecule from the chemical formula. The chemical formula of Aceburic Acid is C6H10O4. From the chemical formula, you can find that one molecule of Aceburic Acid has six Carbon (C) atoms, ten Hydrogen (H) atoms, and four Oxygen (O) atoms.
Step 2: Find out the atomic weights of each atom (from the periodic table).
Atomic weight of Carbon (C): 12.0107 (Ref: Jlab-ele006)
Atomic weight of
Hydrogen (H) : 1.008 (Ref: Lanl-1)
Atomic weight of Oxygen (O) : 15.9994 (Ref: Jlab-ele008)
Step 3: Calculate the total weight of each atom in an Aceburic Acid molecule by multiplying its atomic weight by its number.
Number of Carbon atoms in Aceburic Acid: 6
Atomic weight of Carbon: 12.0107
Total weight of Carbon atoms in Aceburic Acid: 12.0107 x 6 = 72.0642
Number of Hydrogen atoms in Aceburic Acid: 10
Atomic weight of Hydrogen: 1.008
Total weight of Hydrogen atoms in Aceburic Acid: 1.008 x 10 = 10.08
Number of Oxygen atoms in Aceburic Acid: 4
Atomic weight of Oxygen: 15.9994
Total weight of Oxygen atoms in Aceburic Acid: 15.9994 x 4 = 63.9976
Step 4: Calculate the molecular weight of Aceburic Acid by adding up the total weight of all atoms.
Molecular weight of Aceburic Acid: 72.0642 (Carbon) + 10.08 (Hydrogen) + 63.9976 (Oxygen) = 146.1418
So the molecular weight of Aceburic Acid is 146.1418.
Aceburic Acid [C6H10O4] Molecular Weight Calculation
## REFERENCES:
• Lanl-1: https://periodic.lanl.gov/1.shtml
• Pubchem-Hydrogen : https://pubchem.ncbi.nlm.nih.gov/element/Hydrogen
• Jlab-ele006; https://education.jlab.org/itselemental/ele006.html
• Pubchem-6: https://pubchem.ncbi.nlm.nih.gov/element/6
• Jlab-ele008: https://education.jlab.org/itselemental/ele008.html
• Pubchem-8: https://pubchem.ncbi.nlm.nih.gov/element/8
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1. What is the largest number? 2. What is the smallest number? 3. Put the numbers in order from largest to smallest? 4. What is the sum of the two smallest numbers? 5. What is the difference of the largest and smallest number? 6.What is the sum of the largest number? 7. List the odd numbers. 8. List the even numbers. 9. Find a number with an even number in the hundreds place. 10. What the largest number in expanded form.
Student Instructions
### Number Sense
Use the numbers on the number sense chart to complete the weeks activity.
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manifold_tracing_custom_function.cpp
# Nerve_GIC
#include <iostream>
#include <gudhi/Coxeter_triangulation.h>
#include <gudhi/Functions/Function_Sm_in_Rd.h>
#include <gudhi/Implicit_manifold_intersection_oracle.h>
#include <gudhi/Manifold_tracing.h>
#include <gudhi/Coxeter_triangulation/Cell_complex/Cell_complex.h>
#include <gudhi/Functions/Linear_transformation.h>
#include <gudhi/IO/build_mesh_from_cell_complex.h>
#include <gudhi/IO/output_meshes_to_medit.h>
using namespace Gudhi::coxeter_triangulation;
/* A definition of a function that defines a 2d surface embedded in R^4, but that normally
* lives on a complex projective plane.
* In terms of harmonic coordinates [x:y:z] of points on the complex projective plane,
* the equation of the manifold is x^3*y + y^3*z + z^3*x = 0.
* The embedding consists of restricting the manifold to the affine subspace z = 1.
*/
struct Function_surface_on_CP2_in_R4 {
Eigen::VectorXd operator()(const Eigen::VectorXd& p) const {
// The real and imaginary parts of the variables x and y
double xr = p(0), xi = p(1), yr = p(2), yi = p(3);
Eigen::VectorXd result(cod_d());
// Squares and cubes of real and imaginary parts used in the computations
double xr2 = xr * xr, xi2 = xi * xi, yr2 = yr * yr, yi2 = yi * yi, xr3 = xr2 * xr, xi3 = xi2 * xi, yr3 = yr2 * yr,
yi3 = yi2 * yi;
// The first coordinate of the output is Re(x^3*y + y^3 + x)
result(0) = xr3 * yr - 3 * xr * xi2 * yr - 3 * xr2 * xi * yi + xi3 * yi + yr3 - 3 * yr * yi2 + xr;
// The second coordinate of the output is Im(x^3*y + y^3 + x)
result(1) = 3 * xr2 * xi * yr + xr3 * yi - 3 * xr * xi2 * yi - xi3 * yr + 3 * yr2 * yi - yi3 + xi;
return result;
}
std::size_t amb_d() const { return 4; };
std::size_t cod_d() const { return 2; };
Eigen::VectorXd seed() const {
Eigen::VectorXd result = Eigen::VectorXd::Zero(4);
return result;
}
Function_surface_on_CP2_in_R4() {}
};
int main(int argc, char** argv) {
// The function for the (non-compact) manifold
Function_surface_on_CP2_in_R4 fun;
// Seed of the function
Eigen::VectorXd seed = fun.seed();
// Creating the function that defines the boundary of a compact region on the manifold
// Defining the intersection oracle
auto oracle = make_oracle(fun, fun_sph);
// Define a Coxeter triangulation scaled by a factor lambda.
// The triangulation is translated by a random vector to avoid violating the genericity hypothesis.
double lambda = 0.2;
Coxeter_triangulation<> cox_tr(oracle.amb_d());
cox_tr.change_offset(Eigen::VectorXd::Random(oracle.amb_d()));
cox_tr.change_matrix(lambda * cox_tr.matrix());
// Manifold tracing algorithm
using Out_simplex_map = typename MT::Out_simplex_map;
std::vector<Eigen::VectorXd> seed_points(1, seed);
Out_simplex_map interior_simplex_map, boundary_simplex_map;
manifold_tracing_algorithm(seed_points, cox_tr, oracle, interior_simplex_map, boundary_simplex_map);
// Constructing the cell complex
std::size_t intr_d = oracle.amb_d() - oracle.cod_d();
Cell_complex<Out_simplex_map> cell_complex(intr_d);
cell_complex.construct_complex(interior_simplex_map, boundary_simplex_map);
// Output the cell complex to a file readable by medit
output_meshes_to_medit(3, "manifold_on_CP2_with_boundary",
build_mesh_from_cell_complex(cell_complex, Configuration(true, true, true, 1, 5, 3),
Configuration(true, true, true, 2, 13, 14)));
return 0;
}
A class that constructs the cell complex from the output provided by the class Gudhi::coxeter_triangu...
Definition: Cell_complex.h:38
A class that stores Coxeter triangulation of type . This triangulation has the greatest simplex quali...
Definition: Coxeter_triangulation.h:45
A class that assembles methods for manifold tracing algorithm.
Definition: Manifold_tracing.h:39
A class for the function that defines an m-dimensional implicit sphere embedded in the d-dimensional ...
Definition: Function_Sm_in_Rd.h:27
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# I need 3 cans of paint for each room and need to paint 20 rooms.
Discussion in 'Calculator Requests' started by math_celebrity, Mar 28, 2020.
Tags:
I need 3 cans of paint for each room and need to paint 20 rooms.
Calculate the Total Paint Needed
Total Paint Needed = Total Rooms * Cans of paint per room
Total Paint Needed = 20 * 3
Total Paint Needed = 60
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To install click the Add extension button. That's it.
The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.
4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
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Leo
Newton
Brights
Milds
# Singleton (mathematics)
In mathematics, a singleton, also known as a unit set[1] or one-point set, is a set with exactly one element. For example, the set ${\displaystyle \{0\}}$ is a singleton whose single element is ${\displaystyle 0}$.
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• Types of Sets - Universal, Null, and Singleton
• Singleton Sets | Includes Examples
• Singleton Set #shorts
• Lecture 4, Video 1: The Singleton Bound
• Singleton set
## Properties
Within the framework of Zermelo–Fraenkel set theory, the axiom of regularity guarantees that no set is an element of itself. This implies that a singleton is necessarily distinct from the element it contains,[1] thus 1 and {1} are not the same thing, and the empty set is distinct from the set containing only the empty set. A set such as ${\displaystyle \{\{1,2,3\}\}}$ is a singleton as it contains a single element (which itself is a set, however, not a singleton).
A set is a singleton if and only if its cardinality is 1. In von Neumann's set-theoretic construction of the natural numbers, the number 1 is defined as the singleton ${\displaystyle \{0\}.}$
In axiomatic set theory, the existence of singletons is a consequence of the axiom of pairing: for any set A, the axiom applied to A and A asserts the existence of ${\displaystyle \{A,A\},}$ which is the same as the singleton ${\displaystyle \{A\}}$ (since it contains A, and no other set, as an element).
If A is any set and S is any singleton, then there exists precisely one function from A to S, the function sending every element of A to the single element of S. Thus every singleton is a terminal object in the category of sets.
A singleton has the property that every function from it to any arbitrary set is injective. The only non-singleton set with this property is the empty set.
Every singleton set is an ultra prefilter. If ${\displaystyle X}$ is a set and ${\displaystyle x\in X}$ then the upward of ${\displaystyle \{x\}}$ in ${\displaystyle X,}$ which is the set ${\displaystyle \{S\subseteq X:x\in S\},}$ is a principal ultrafilter on ${\displaystyle X.}$[2] Moreover, every principal ultrafilter on ${\displaystyle X}$ is necessarily of this form.[2] The ultrafilter lemma implies that non-principal ultrafilters exist on every infinite set (these are called free ultrafilters). Every net valued in a singleton subset ${\displaystyle X}$ of is an ultranet in ${\displaystyle X.}$
The Bell number integer sequence counts the number of partitions of a set (), if singletons are excluded then the numbers are smaller ().
## In category theory
Structures built on singletons often serve as terminal objects or zero objects of various categories:
• The statement above shows that the singleton sets are precisely the terminal objects in the category Set of sets. No other sets are terminal.
• Any singleton admits a unique topological space structure (both subsets are open). These singleton topological spaces are terminal objects in the category of topological spaces and continuous functions. No other spaces are terminal in that category.
• Any singleton admits a unique group structure (the unique element serving as identity element). These singleton groups are zero objects in the category of groups and group homomorphisms. No other groups are terminal in that category.
## Definition by indicator functions
Let S be a class defined by an indicator function
${\displaystyle b:X\to \{0,1\}.}$
Then S is called a singleton if and only if there is some ${\displaystyle y\in X}$ such that for all ${\displaystyle x\in X,}$
${\displaystyle b(x)=(x=y).}$
## Definition in Principia Mathematica
The following definition was introduced by Whitehead and Russell[3]
${\displaystyle \iota }$${\displaystyle x={\hat {y}}(y=x)}$ Df.
The symbol ${\displaystyle \iota }$${\displaystyle x}$ denotes the singleton ${\displaystyle \{x\}}$ and ${\displaystyle {\hat {y}}(y=x)}$ denotes the class of objects identical with ${\displaystyle x}$ aka ${\displaystyle \{y:y=x\}}$. This occurs as a definition in the introduction, which, in places, simplifies the argument in the main text, where it occurs as proposition 51.01 (p.357 ibid.). The proposition is subsequently used to define the cardinal number 1 as
${\displaystyle 1={\hat {\alpha }}((\exists x)\alpha =\iota }$${\displaystyle x)}$ Df.
That is, 1 is the class of singletons. This is definition 52.01 (p.363 ibid.)
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0
How long will it take?
With an average acceleration of -1.2 m/s^2, how long will it take a cyclist to bring a bicycle with an initial velocity of 6.5 m/s to a complete stop?
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# Maxwell-boltzman distrubution
• E92M3
#### E92M3
If I have two particles that follows the maxwell velocity distribution:
$$\phi(v_i)dv_i=4 \pi v_i^2 \left ( \frac{m_iv_i}{2\pi kT} \right ) ^{3/2}e^{\frac{-m_iv_i^2}{2kT}}dv_i$$
Why is their combined distribution:
$$\phi(v)dv=4 \pi v^2 \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2}e^{\frac{-\mu v^2}{2kT}}dv$$
where mu is the reduced mass and v=v2-v1
I have these questions because I don't quite follow these derivations.
http://dissertations.ub.rug.nl/FILES/faculties/science/2007/a.matic/c2.pdf [Broken]
http://www.astro.psu.edu/users/rbc/a534/lec11.pdf
Namely, I not sure why the following holds:
$$\int_0^\infty \int_0^\infty \phi(v_1) \phi(v_2) v_1 v_2 \sigma dv_1 dv_2 = 4\pi \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2} \int_0^\infty v^3 \sigma e^{\frac{-\mu v^2}{2kT}}dv$$
Last edited by a moderator:
You have v=v1-v2. Let u=v1+v2. The new integration is straightforward as long as the integral limits are -∞ to ∞. You then need |v1v2| instead of v1v2 in the integrand. The u integration should leave you with what you want.
## What is the Maxwell-Boltzmann distribution?
The Maxwell-Boltzmann distribution is a probability distribution that describes the speed of particles in a gas at a given temperature. It is named after James Clerk Maxwell and Ludwig Boltzmann, who independently developed the concept in the late 19th century.
## What is the significance of the Maxwell-Boltzmann distribution?
The Maxwell-Boltzmann distribution is significant because it provides a mathematical relationship between the temperature of a gas and the average speed of its particles. This relationship has wide-ranging applications in fields such as thermodynamics, statistical mechanics, and atmospheric science.
## What factors affect the shape of the Maxwell-Boltzmann distribution curve?
The shape of the Maxwell-Boltzmann distribution curve is affected by two main factors: temperature and molecular mass. As temperature increases, the curve shifts to the right, indicating a higher average speed of particles. As molecular mass increases, the curve becomes narrower and taller, indicating a smaller range of speeds and a higher peak at the most probable speed.
## How is the Maxwell-Boltzmann distribution related to the ideal gas law?
The Maxwell-Boltzmann distribution is derived from the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and the number of molecules present. The distribution describes the probability of finding a particle at a given speed in an ideal gas, and is used to calculate various thermodynamic properties of gases.
## What are some real-world applications of the Maxwell-Boltzmann distribution?
The Maxwell-Boltzmann distribution has numerous applications in various fields of science and engineering. Some examples include predicting the speed distribution of molecules in a gas, analyzing the behavior of particles in a plasma, and understanding the distribution of kinetic energy in a chemical reaction. It is also used in the design of gas turbines, combustion engines, and other heat engines.
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Ex 7.2
Chapter 7 Class 9 Triangles
Serial order wise
This video is only available for Teachoo black users
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 7.2, 7 ABC is a right angled triangle in which A = 90 and AB = AC. Find B and C. Given AB = AC C = B In ABC, A + B + C = 180 90 + B + C = 180 90 + B + B = 180 2 B = 90 B = 45 B = C = 45
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mycollegehive
The number of radians in a 540-degree angle can be written a...
SAT-math-test SAT math
0
The number of radians in a 540-degree angle can be written as aπ, where a is a constant, What is the value of a ?
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University of Benin Nigeria
19 November 2020
School not set Nigeria
15 March 2021
0
To convert degree to radian, we use the formula:
θ . π/180
where θ is the value in degree
θ = 540
540 . π/180
= 3π
Therefore, a = 3
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Nigeria
15 March 2021
0
θ=π/180
θ=540
Substituting we have
540=π/180
Share
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Hypergeometric function
1. Nov 2, 2009
gholamghar
1. The problem statement, all variables and given/known data
write the following serie in the form of hypergeometric function:
f(x)=$$\sum$$$$\frac{(-1*(x^2))^n}{(2^n)(2n-1)(2n+1)(2n+3)}$$
n changes from 0 to $$\infty$$
2. Relevant equations
hypergeometric function:
3. The attempt at a solution
guys i have thought about this for 2 hours,and i had not any initial idea how to solve it,i appreciate if anyone could give me an initial idea that how should i solve this problem
Attached Files:
• NumberedEquation6.gif
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Last edited: Nov 2, 2009
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# Prove the Following Mathematic Form
#### Albert
##### Well-known member
$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$
prove :$A=k\times (k+1),\,\, where\,\, k\in N$
(A can be expressed as the multiplication of two consecutive positive integers)
Last edited:
#### M R
##### Active member
$$\displaystyle A=\frac{10^m-1}{9} \times 10^m +2 \times \frac{10^m-1}{9}$$
$$\displaystyle 9A=(10^m-1) \times 10^m +2 \times (10^m-1)$$
$$\displaystyle 9A=10^{2m} -10^m +2 \times 10^m-2$$
$$\displaystyle 9A=10^{2m} +10^m - 2$$
$$\displaystyle 9A=(10^m-1)(10^m +2)$$
Each factor on the right is a multiple of 3 and they differ by 3 so on dividing by 9 we get $$\displaystyle A=k(k+1)$$ where $$\displaystyle k = \frac{10^m-1}{3}$$ .
I nice start to my day, thank you
Hi MR
A nice solution
#### Albert
##### Well-known member
$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$
prove :$A=k\times (k+1),\,\, where\,\, k\in N$
(A can be expressed as the multiplication of two consecutive positive integers)
$A=\overbrace{ 11-------1 }^{m}\times 10^m+\underbrace{ 1-------1 }_{m}\times 2$
$=\overbrace{ 11-------1 }^{m}\times (10^m+2)=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 99-------9 }^{m}+3)$
$=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 33-------3 }^{m}\times 3+3)$
$=\overbrace{ 33-------3 }^{m}\times (\overbrace{ 33-------3 }^{m}+1)=k\times (k+1)$
($k=\overbrace{ 33-------3 }^{m}$)
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# Fermat's Last Theorem
The 1670 edition of Diophantus' Arithmetica includes Fermat's commentary, particularly his "Last Theorem" (Observatio Domini Petri de Fermat).
In number theory Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c satisfy the equation an + bn = cn for any integer value of n greater than 2. The cases n = 1 and n = 2 have been known to have infinitely many solutions since antiquity.[1]
This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The proof was described as a 'stunning advance' in the citation for his Abel Prize award in 2016.[2] The proof of Fermat's Last Theorem also proved much of the modularity theorem and opened up entire new approaches to numerous other problems and mathematically powerful modularity lifting techniques.
The unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof, it was in the Guinness Book of World Records as the "most difficult mathematical problem", one of the reasons being that it has the largest number of unsuccessful proofs.[3]
## Overview
### Pythagorean origins
The Pythagorean equation, x2 + y2 = z2, has an infinite number of positive integer solutions for x, y, and z; these solutions are known as Pythagorean triples. Around 1637, Fermat wrote in the margin of a book that the more general equation an + bn = cn had no solutions in positive integers, if n is an integer greater than 2. Although he claimed to have a general proof of his conjecture, Fermat left no details of his proof, and no proof by him has ever been found. His claim was discovered some 30 years later, after his death. This claim, which came to be known as Fermat's Last Theorem, stood unsolved in mathematics for the following three and a half centuries.
The claim eventually became one of the most notable unsolved problems of mathematics. Attempts to prove it prompted substantial development in number theory, and over time Fermat's Last Theorem gained prominence as an unsolved problem in mathematics.
### Subsequent developments and solution
With the special case n = 4 proved by Fermat himself, it suffices to prove the theorem for exponents n that are prime numbers[note 1]. Over the next two centuries (1637–1839), the conjecture was proved for only the primes 3, 5, and 7, although Sophie Germain innovated and proved an approach that was relevant to an entire class of primes. In the mid-19th century, Ernst Kummer extended this and proved the theorem for all regular primes, leaving irregular primes to be analyzed individually. Building on Kummer's work and using sophisticated computer studies, other mathematicians were able to extend the proof to cover all prime exponents up to four million, but a proof for all exponents was inaccessible (meaning that mathematicians generally considered a proof impossible, exceedingly difficult, or unachievable with current knowledge).
Entirely separately, around 1955, Japanese mathematicians Goro Shimura and Yutaka Taniyama suspected a link might exist between elliptic curves and modular forms, two completely different areas of mathematics. Known at the time as the Taniyama–Shimura–Weil conjecture, and (eventually) as the modularity theorem, it stood on its own, with no apparent connection to Fermat's Last Theorem. It was widely seen as significant and important in its own right, but was (like Fermat's theorem) widely considered completely inaccessible to proof.
In 1984, Gerhard Frey noticed an apparent link between these two previously unrelated and unsolved problems. An outline suggesting this could be proved was given by Frey. The full proof that the two problems were closely linked was accomplished in 1986 by Ken Ribet, building on a partial proof by Jean-Pierre Serre, who proved all but one part known as the "epsilon conjecture" (see: Ribet's Theorem and Frey curve).[2] In plain English, these papers by Frey, Serre and Ribet showed that if the Modularity Theorem could be proven for at least the semi-stable class of elliptic curves, a proof of Fermat's Last Theorem would also follow automatically. The connection is described below: any solution that could contradict Fermat's Last Theorem could also be used to contradict the Modularity Theorem. So if the modularity theorem were found to be true, then by definition no solution contradicting Fermat's Last Theorem could exist, which would therefore have to be true as well.
Although both problems were daunting problems widely considered to be "completely inaccessible" to proof at the time,[2] this was the first suggestion of a route by which Fermat's Last Theorem could be extended and proved for all numbers, not just some numbers. Also important for researchers choosing a research topic was the fact that unlike Fermat's Last Theorem the Modularity Theorem was a major active research area for which a proof was widely desired and not just a historical oddity, so time spent working on it could be justified professionally.[4] However, general opinion was that this simply showed the impracticality of proving the Taniyama–Shimura conjecture.[5] Mathematician John Coates' quoted reaction was a common one:
"I myself was very sceptical that the beautiful link between Fermat’s Last Theorem and the Taniyama–Shimura conjecture would actually lead to anything, because I must confess I did not think that the Taniyama–Shimura conjecture was accessible to proof. Beautiful though this problem was, it seemed impossible to actually prove. I must confess I thought I probably wouldn’t see it proved in my lifetime." [5]
On hearing that Ribet had proven Frey's link to be correct, English mathematician Andrew Wiles, who had a childhood fascination with Fermat's Last Theorem and had a background of working with elliptic curves and related fields, decided to try to prove the Taniyama–Shimura conjecture as a way to prove Fermat's Last Theorem. In 1993, after six years working secretly on the problem, Wiles succeeded in proving enough of the conjecture to prove Fermat's Last Theorem. Wiles's paper was massive in size and scope. A flaw was discovered in one part of his original paper during peer review and required a further year and collaboration with a past student, Richard Taylor, to resolve. As a result, the final proof in 1995 was accompanied by a smaller joint paper showing that the fixed steps were valid. Wiles's achievement was reported widely in the popular press, and was popularized in books and television programs. The remaining parts of the Taniyama–Shimura–Weil conjecture, now proven and known as the Modularity theorem, were subsequently proved by other mathematicians, who built on Wiles's work between 1996 and 2001. For his proof, Wiles was honoured and received numerous awards, including the 2016 Abel Prize.[6][7][8]
### Equivalent statements of the theorem
There are several alternative ways to state Fermat's Last Theorem that are mathematically equivalent to the original statement of the problem.
In order to state them, we use mathematical notation: let N be the set of natural numbers 1, 2, 3, ..., let Z be the set of integers 0, ±1, ±2, ..., and let Q be the set of rational numbers a/b where a and b are in Z with b≠0. In what follows we will call a solution to xn + yn = zn where one or more of x, y, or z is zero a trivial solution. A solution where all three are non-zero will be called a non-trivial solution.
• Original statement. With n, x, y, zN (meaning n, x, y, z are all positive whole numbers) and n > 2 the equation xn + yn = zn has no solutions.
Most popular treatments of the subject state it this way. In contrast, almost all math textbooks state it over Z:
• Equivalent statement 1: xn + yn = zn, where integer n ≥ 3, has no non-trivial solutions x, y, zZ.
The equivalence is clear if n is even. If n is odd and all three of x, y, z are negative then we can replace x, y, z with x, −y, −z to obtain a solution in N. If two of them are negative, it must be x and z or y and z. If x, z are negative and y is positive, then we can rearrange to get (−z)n + yn = (−x)n resulting in a solution in N; the other case is dealt with analogously. Now if just one is negative, it must be x or y. If x is negative, and y and z are positive, then it can be rearranged to get (−x)n + zn = yn again resulting in a solution in N; if y is negative, the result follows symmetrically. Thus in all cases a nontrivial solution in Z would also mean a solution exists in N, the original formulation of the problem.
• Equivalent statement 2: xn + yn = zn, where integer n ≥ 3, has no non-trivial solutions x, y, zQ.
This is because the exponent of x, y and z are equal (to n), so if there is a solution in Q then it can be multiplied through by an appropriate common denominator to get a solution in Z, and hence in N.
• Equivalent statement 3: xn + yn = 1, where integer n ≥ 3, has no non-trivial solutions x, yQ.
A non-trivial solution a, b, cZ to xn + yn = zn yields the non-trivial solution a/c, b/cQ for vn + wn = 1. Conversely, a solution a/b, c/dQ to vn + wn = 1 yields the non-trivial solution ad, cb, bd for xn + yn = zn.
This last formulation is particularly fruitful, because it reduces the problem from a problem about surfaces in three dimensions to a problem about curves in two dimensions. Furthermore, it allows working over the field Q, rather than over the ring Z; fields exhibit more structure than rings, which allows for deeper analysis of their elements.
• Equivalent statement 4 – connection to elliptic curves: If a, b, c is a non-trivial solution to xp + yp = zp , p odd prime, then y2 = x(xap)(x + bp) (Frey curve) will be an elliptic curve.[9]
Examining this elliptic curve with Ribet's theorem shows that it does not have a modular form. However, the proof by Andrew Wiles proves that any equation of the form y2 = x(xan)(x + bn) does have a modular form. Any non-trivial solution to xp + yp = zp (with p an odd prime) would therefore create a contradiction, which in turn proves that no non-trivial solutions exist.[10]
In other words, any solution that could contradict Fermat's Last Theorem could also be used to contradict the Modularity Theorem. So if the modularity theorem were found to be true, then it would follow that no contradiction to Fermat's Last Theorem could exist either. As described above, the discovery of this equivalent statement was crucial to the eventual solution of Fermat's Last Theorem, as it provided a means by which it could be 'attacked' for all numbers at once.
## Mathematical history
### Pythagoras and Diophantus
#### Pythagorean triples
In ancient times it was known that a triangle whose sides were in the ratio 3:4:5 would have a right angle as one of its angles. This was used in construction and later in early geometry. It was also known to be just one example of a general rule that any triangle where the length of two sides, each squared and then added together (32 + 42 = 9 + 16 = 25), equals the square of the length of the third side (52 = 25), would also be a right angle triangle. This is now known as the Pythagorean theorem, and a triple of numbers that meets this condition is called a Pythagorean triple – both are named after the ancient Greek Pythagoras. Examples include (3, 4, 5) and (5, 12, 13). There are infinitely many such triples,[11] and methods for generating such triples have been studied in many cultures, beginning with the Babylonians[12] and later ancient Greek, Chinese, and Indian mathematicians.[1] Mathematically, the definition of a Pythagorean triple is a set of three integers (a, b, c) that satisfy the equation[13]
#### Diophantine equations
Fermat's equation, xn + yn = zn with positive integer solutions, is an example of a Diophantine equation,[14] named for the 3rd-century Alexandrian mathematician, Diophantus, who studied them and developed methods for the solution of some kinds of Diophantine equations. A typical Diophantine problem is to find two integers x and y such that their sum, and the sum of their squares, equal two given numbers A and B, respectively:
Diophantus's major work is the Arithmetica, of which only a portion has survived.[15] Fermat's conjecture of his Last Theorem was inspired while reading a new edition of the Arithmetica,[16] that was translated into Latin and published in 1621 by Claude Bachet.[17]
Diophantine equations have been studied for thousands of years. For example, the solutions to the quadratic Diophantine equation x2 + y2 = z2 are given by the Pythagorean triples, originally solved by the Babylonians (c. 1800 BC).[18] Solutions to linear Diophantine equations, such as 26x + 65y = 13, may be found using the Euclidean algorithm (c. 5th century BC).[19] Many Diophantine equations have a form similar to the equation of Fermat's Last Theorem from the point of view of algebra, in that they have no cross terms mixing two letters, without sharing its particular properties. For example, it is known that there are infinitely many positive integers x, y, and z such that xn + yn = zm where n and m are relatively prime natural numbers.[note 2]
### Fermat's conjecture
Problem II.8 in the 1621 edition of the Arithmetica of Diophantus. On the right is the margin that was too small to contain Fermat's alleged proof of his "last theorem".
Problem II.8 of the Arithmetica asks how a given square number is split into two other squares; in other words, for a given rational number k, find rational numbers u and v such that k2 = u2 + v2. Diophantus shows how to solve this sum-of-squares problem for k = 4 (the solutions being u = 16/5 and v = 12/5).[20]
Around 1637, Fermat wrote his Last Theorem in the margin of his copy of the Arithmetica next to Diophantus's sum-of-squares problem:[21]
Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet. It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.[22][23]
After Fermat’s death in 1665, his son Clément-Samuel Fermat produced a new edition of the book (1670) augmented with his father’s comments.[24] Although not actually a theorem at the time (meaning a mathematical statement for which proof exists), the margin note became known over time as Fermat’s Last Theorem,[25] as it was the last of Fermat’s asserted theorems to remain unproved.[26]
It is not known whether Fermat had actually found a valid proof for all exponents n, but it appears unlikely. Only one related proof by him has survived, namely for the case n = 4, as described in the section Proofs for specific exponents. While Fermat posed the cases of n = 4 and of n = 3 as challenges to his mathematical correspondents, such as Marin Mersenne, Blaise Pascal, and John Wallis,[27] he never posed the general case.[28] Moreover, in the last thirty years of his life, Fermat never again wrote of his "truly marvelous proof" of the general case, and never published it. Van der Poorten[29] suggests that while the absence of a proof is insignificant, the lack of challenges means Fermat realised he did not have a proof; he quotes Weil[30] as saying Fermat must have briefly deluded himself with an irretrievable idea.
The techniques Fermat might have used in such a "marvelous proof" are unknown.
Taylor and Wiles's proof relies on 20th-century techniques.[31] Fermat's proof would have had to be elementary by comparison, given the mathematical knowledge of his time.
While Harvey Friedman's grand conjecture implies that any provable theorem (including Fermat's last theorem) can be proved using only 'elementary function arithmetic', such a proof need be ‘elementary’ only in a technical sense and could involve millions of steps, and thus be far too long to have been Fermat’s proof.
### Proofs for specific exponents
Fermat's infinite descent for Fermat's Last Theorem case n=4 in the 1670 edition of the Arithmetica of Diophantus (pp. 338–339).
#### Exponent = 4
Only one relevant proof by Fermat has survived, in which he uses the technique of infinite descent to show that the area of a right triangle with integer sides can never equal the square of an integer.[32][33] His proof is equivalent to demonstrating that the equation
has no primitive solutions in integers (no pairwise coprime solutions). In turn, this proves Fermat's Last Theorem for the case n = 4, since the equation a4 + b4 = c4 can be written as a4 + b4 = (c2)2.
Alternative proofs of the case n = 4 were developed later[34] by Frénicle de Bessy (1676),[35] Leonhard Euler (1738),[36] Kausler (1802),[37] Peter Barlow (1811),[38] Adrien-Marie Legendre (1830),[39] Schopis (1825),[40] Terquem (1846),[41] Joseph Bertrand (1851),[42] Victor Lebesgue (1853, 1859, 1862),[43] Theophile Pepin (1883),[44] Tafelmacher (1893),[45] David Hilbert (1897),[46] Bendz (1901),[47] Gambioli (1901),[48] Leopold Kronecker (1901),[49] Bang (1905),[50] Sommer (1907),[51] Bottari (1908),[52] Karel Rychlík (1910),[53] Nutzhorn (1912),[54] Robert Carmichael (1913),[55] Hancock (1931),[56] and Vrǎnceanu (1966).[57]
For other proofs for n=4 by infinite descent, see Infinite descent: Non-solvability of r2 + s4 = t4, Grant and Perella (1999),[58] Barbara (2007),[59] and Dolan (2011).[60]
#### Other exponents
After Fermat proved the special case n = 4, the general proof for all n required only that the theorem be established for all odd prime exponents.[61] In other words, it was necessary to prove only that the equation an + bn = cn has no integer solutions (a, b, c) when n is an odd prime number. This follows because a solution (a, b, c) for a given n is equivalent to a solution for all the factors of n. For illustration, let n be factored into d and e, n = de. The general equation
an + bn = cn
implies that (ad, bd, cd) is a solution for the exponent e
Thus, to prove that Fermat's equation has no solutions for n > 2, it would suffice to prove that it has no solutions for at least one prime factor of every n. Each integer n > 2 is divisible by 4 or by an odd prime number (or both). Therefore, Fermat's Last Theorem could be proved for all n if it could be proved for n = 4 and for all odd primes p.
In the two centuries following its conjecture (1637–1839), Fermat's Last Theorem was proved for three odd prime exponents p = 3, 5 and 7. The case p = 3 was first stated by Abu-Mahmud Khojandi (10th century), but his attempted proof of the theorem was incorrect.[62] In 1770, Leonhard Euler gave a proof of p = 3,[63] but his proof by infinite descent[64] contained a major gap.[65] However, since Euler himself had proved the lemma necessary to complete the proof in other work, he is generally credited with the first proof.[66] Independent proofs were published[67] by Kausler (1802),[37] Legendre (1823, 1830),[39][68] Calzolari (1855),[69] Gabriel Lamé (1865),[70] Peter Guthrie Tait (1872),[71] Günther (1878),[72] Gambioli (1901),[48] Krey (1909),[73] Rychlík (1910),[53] Stockhaus (1910),[74] Carmichael (1915),[75] Johannes van der Corput (1915),[76] Axel Thue (1917),[77] and Duarte (1944).[78] The case p = 5 was proved[79] independently by Legendre and Peter Gustav Lejeune Dirichlet around 1825.[80] Alternative proofs were developed[81] by Carl Friedrich Gauss (1875, posthumous),[82] Lebesgue (1843),[83] Lamé (1847),[84] Gambioli (1901),[48][85] Werebrusow (1905),[86] Rychlík (1910),[87] van der Corput (1915),[76] and Guy Terjanian (1987).[88] The case p = 7 was proved[89] by Lamé in 1839.[90] His rather complicated proof was simplified in 1840 by Lebesgue,[91] and still simpler proofs[92] were published by Angelo Genocchi in 1864, 1874 and 1876.[93] Alternative proofs were developed by Théophile Pépin (1876)[94] and Edmond Maillet (1897).[95]
Fermat's Last Theorem was also proved for the exponents n = 6, 10, and 14. Proofs for n = 6 were published by Kausler,[37] Thue,[96] Tafelmacher,[97] Lind,[98] Kapferer,[99] Swift,[100] and Breusch.[101] Similarly, Dirichlet[102] and Terjanian[103] each proved the case n = 14, while Kapferer[99] and Breusch[101] each proved the case n = 10. Strictly speaking, these proofs are unnecessary, since these cases follow from the proofs for n = 3, 5, and 7, respectively. Nevertheless, the reasoning of these even-exponent proofs differs from their odd-exponent counterparts. Dirichlet's proof for n = 14 was published in 1832, before Lamé's 1839 proof for n = 7.[104]
All proofs for specific exponents used Fermat's technique of infinite descent, either in its original form, or in the form of descent on elliptic curves or abelian varieties. The details and auxiliary arguments, however, were often ad hoc and tied to the individual exponent under consideration.[105] Since they became ever more complicated as p increased, it seemed unlikely that the general case of Fermat's Last Theorem could be proved by building upon the proofs for individual exponents.[105] Although some general results on Fermat's Last Theorem were published in the early 19th century by Niels Henrik Abel and Peter Barlow,[106][107] the first significant work on the general theorem was done by Sophie Germain.[108]
#### Sophie Germain
In the early 19th century, Sophie Germain developed several novel approaches to prove Fermat's Last Theorem for all exponents.[109] First, she defined a set of auxiliary primes θ constructed from the prime exponent p by the equation θ = 2hp + 1, where h is any integer not divisible by three. She showed that, if no integers raised to the pth power were adjacent modulo θ (the non-consecutivity condition), then θ must divide the product xyz. Her goal was to use mathematical induction to prove that, for any given p, infinitely many auxiliary primes θ satisfied the non-consecutivity condition and thus divided xyz; since the product xyz can have at most a finite number of prime factors, such a proof would have established Fermat's Last Theorem. Although she developed many techniques for establishing the non-consecutivity condition, she did not succeed in her strategic goal. She also worked to set lower limits on the size of solutions to Fermat's equation for a given exponent p, a modified version of which was published by Adrien-Marie Legendre. As a byproduct of this latter work, she proved Sophie Germain's theorem, which verified the first case of Fermat's Last Theorem (namely, the case in which p does not divide xyz) for every odd prime exponent less than 270,[109][110] and for all primes p such that at least one of 2p+1, 4p+1, 8p+1, 10p+1, 14p+1 and 16p+1 is prime (specially, the primes p such that 2p+1 is prime are called Sophie Germain primes). Germain tried unsuccessfully to prove the first case of Fermat's Last Theorem for all even exponents, specifically for n = 2p, which was proved by Guy Terjanian in 1977.[111] In 1985, Leonard Adleman, Roger Heath-Brown and Étienne Fouvry proved that the first case of Fermat's Last Theorem holds for infinitely many odd primes p.[112]
#### Ernst Kummer and the theory of ideals
In 1847, Gabriel Lamé outlined a proof of Fermat's Last Theorem based on factoring the equation xp + yp = zp in complex numbers, specifically the cyclotomic field based on the roots of the number 1. His proof failed, however, because it assumed incorrectly that such complex numbers can be factored uniquely into primes, similar to integers. This gap was pointed out immediately by Joseph Liouville, who later read a paper that demonstrated this failure of unique factorisation, written by Ernst Kummer.
Kummer set himself the task of determining whether the cyclotomic field could be generalized to include new prime numbers such that unique factorisation was restored. He succeeded in that task by developing the ideal numbers.
(Note: It is often stated that Kummer was led to his "ideal complex numbers" by his interest in Fermat's Last Theorem; there is even a story often told that Kummer, like Lamé, believed he had proven Fermat's Last Theorem until Lejeune Dirichlet told him his argument relied on unique factorization; but the story was first told by Kurt Hensel in 1910 and the evidence indicates it likely derives from a confusion by one of Hensel's sources. Harold Edwards says the belief that Kummer was mainly interested in Fermat's Last Theorem "is surely mistaken".[113] See the history of ideal numbers.)
Using the general approach outlined by Lamé, Kummer proved both cases of Fermat's Last Theorem for all regular prime numbers. However, he could not prove the theorem for the exceptional primes (irregular primes) that conjecturally occur approximately 39% of the time; the only irregular primes below 270 are 37, 59, 67, 101, 103, 131, 149, 157, 233, 257 and 263.
#### Mordell conjecture
In the 1920s, Louis Mordell posed a conjecture that implied that Fermat's equation has at most a finite number of nontrivial primitive integer solutions, if the exponent n is greater than two.[114] This conjecture was proved in 1983 by Gerd Faltings,[115] and is now known as Faltings's theorem.
#### Computational studies
In the latter half of the 20th century, computational methods were used to extend Kummer's approach to the irregular primes. In 1954, Harry Vandiver used a SWAC computer to prove Fermat's Last Theorem for all primes up to 2521.[116] By 1978, Samuel Wagstaff had extended this to all primes less than 125,000.[117] By 1993, Fermat's Last Theorem had been proved for all primes less than four million.[118]
However despite these efforts and their results, no proof existed of Fermat's Last Theorem. Proofs of individual exponents by their nature could never prove the general case: even if all exponents were verified up to an extremely large number X, a higher exponent beyond X might still exist for which the claim was not true. (This had been the case with some other past conjectures, and it could not be ruled out in this conjecture.)[119]
### Connection with elliptic curves
The strategy that ultimately led to a successful proof of Fermat's Last Theorem arose from the "astounding"[120]:211 Taniyama–Shimura–Weil conjecture, proposed around 1955—which many mathematicians believed would be near to impossible to prove,[120]:223 and was linked in the 1980s by Gerhard Frey, Jean-Pierre Serre and Ken Ribet to Fermat's equation. By accomplishing a partial proof of this conjecture in 1994, Andrew Wiles ultimately succeeded in proving Fermat's Last Theorem, as well as leading the way to a full proof by others of what is now the modularity theorem.
#### Taniyama–Shimura–Weil conjecture
Around 1955, Japanese mathematicians Goro Shimura and Yutaka Taniyama observed a possible link between two apparently completely distinct branches of mathematics, elliptic curves and modular forms. The resulting modularity theorem (at the time known as the Taniyama–Shimura conjecture) states that every elliptic curve is modular, meaning that it can be associated with a unique modular form.
The link was initially dismissed as unlikely or highly speculative, but was taken more seriously when number theorist André Weil found evidence supporting it, though not proving it; as a result the conjecture was often known as the Taniyama–Shimura–Weil conjecture. It became a part of the Langlands programme, a list of important conjectures needing proof or disproof.[120]:211–215
Even after gaining serious attention, the conjecture was seen by contemporary mathematicians as extraordinarily difficult or perhaps inaccessible to proof.[120]:203–205, 223, 226 For example, Wiles's doctoral supervisor John Coates states that it seemed "impossible to actually prove",[120]:226 and Ken Ribet considered himself "one of the vast majority of people who believed [it] was completely inaccessible", adding that "Andrew Wiles was probably one of the few people on earth who had the audacity to dream that you can actually go and prove [it]."[120]:223
#### Ribet's theorem for Frey curves
In 1984, Gerhard Frey noted a link between Fermat's equation and the modularity theorem, then still a conjecture. If Fermat's equation had any solution (a, b, c) for exponent p > 2, then it could be shown that the semi-stable elliptic curve (now known as a Frey-Hellegouarch[note 3])
y2 = x (x ap)(x + bp)
would have such unusual properties that it was unlikely to be modular.[121] This would conflict with the modularity theorem, which asserted that all elliptic curves are modular. As such, Frey observed that a proof of the Taniyama–Shimura–Weil conjecture might also simultaneously prove Fermat's Last Theorem.[122] By contraposition, a disproof or refutation of Fermat's Last Theorem would disprove the Taniyama–Shimura–Weil conjecture.
In plain English, Frey had shown that, if this intuition about his equation was correct, then any set of 4 numbers (a, b, c, n) capable of disproving Fermat's Last Theorem, could also be used to disprove the Taniyama–Shimura–Weil conjecture. Therefore if the latter were true, the former could not be disproven, and would also have to be true.
Following this strategy, a proof of Fermat's Last Theorem required two steps. First, it was necessary to prove the modularity theorem – or at least to prove it for the types of elliptical curves that included Frey's equation (known as semistable elliptic curves). This was widely believed inaccessible to proof by contemporary mathematicians.[120]:203–205, 223, 226 Second, it was necessary to show that Frey's intuition was correct: that if an elliptic curve were constructed in this way, using a set of numbers that were a solution of Fermat's equation, the resulting elliptic curve could not be modular. Frey showed that this was plausible but did not go as far as giving a full proof. The missing piece (the so-called "epsilon conjecture", now known as Ribet's theorem) was identified by Jean-Pierre Serre who also gave an almost-complete proof and the link suggested by Frey was finally proved in 1986 by Ken Ribet.[123]
Following Frey, Serre and Ribet's work, this was where matters stood:
• Fermat's Last Theorem needed to be proven for all exponents n that were prime numbers.
• The modularity theorem – if proved for semi-stable elliptic curves – would mean that all semistable elliptic curves must be modular.
• Ribet's theorem showed that any solution to Fermat's equation for a prime number could be used to create a semistable elliptic curve that could not be modular;
• The only way that both of these statements could be true, was if no solutions existed to Fermat's equation (because then no such curve could be created), which was what Fermat's Last Theorem said. As Ribet's Theorem was already proved, this meant that a proof of the Modularity Theorem would automatically prove Fermat's Last theorem was true as well.
#### Wiles's general proof
British mathematician Andrew Wiles.
Ribet's proof of the epsilon conjecture in 1986 accomplished the first of the two goals proposed by Frey. Upon hearing of Ribet's success, Andrew Wiles, an English mathematician with a childhood fascination with Fermat's Last Theorem, and a prior study area of elliptical equations, decided to commit himself to accomplishing the second half: proving a special case of the modularity theorem (then known as the Taniyama–Shimura conjecture) for semistable elliptic curves.[124]
Wiles worked on that task for six years in near-total secrecy, covering up his efforts by releasing prior work in small segments as separate papers and confiding only in his wife.[120]:229–230 His initial study suggested proof by induction,[120]:230–232, 249–252 and he based his initial work and first significant breakthrough on Galois theory[120]:251–253, 259 before switching to an attempt to extend horizontal Iwasawa theory for the inductive argument around 1990–91 when it seemed that there was no existing approach adequate to the problem.[120]:258–259 However, by the summer of 1991, Iwasawa theory also seemed to not be reaching the central issues in the problem.[120]:259–260[125] In response, he approached colleagues to seek out any hints of cutting edge research and new techniques, and discovered an Euler system recently developed by Victor Kolyvagin and Matthias Flach that seemed "tailor made" for the inductive part of his proof.[120]:260–261 Wiles studied and extended this approach, which worked. Since his work relied extensively on this approach, which was new to mathematics and to Wiles, in January 1993 he asked his Princeton colleague, Nick Katz, to help him check his reasoning for subtle errors. Their conclusion at the time was that the techniques Wiles used seemed to work correctly.[120]:261–265[126]
By mid-May 1993, Wiles felt able to tell his wife he thought he had solved the proof of Fermat's Last Theorem,[120]:265 and by June he felt sufficiently confident to present his results in three lectures delivered on 21–23 June 1993 at the Isaac Newton Institute for Mathematical Sciences.[127] Specifically, Wiles presented his proof of the Taniyama–Shimura conjecture for semistable elliptic curves; together with Ribet's proof of the epsilon conjecture, this implied Fermat's Last Theorem. However, it became apparent during peer review that a critical point in the proof was incorrect. It contained an error in a bound on the order of a particular group. The error was caught by several mathematicians refereeing Wiles's manuscript including Katz (in his role as reviewer),[128] who alerted Wiles on 23 August 1993.[129]
The error would not have rendered his work worthless – each part of Wiles's work was highly significant and innovative by itself, as were the many developments and techniques he had created in the course of his work, and only one part was affected.[120]:289, 296–297 However without this part proved, there was no actual proof of Fermat's Last Theorem. Wiles spent almost a year trying to repair his proof, initially by himself and then in collaboration with his former student Richard Taylor, without success.[130][131][132] By the end of 1993, rumours had spread that under scrutiny, Wiles's proof had failed, but how seriously was not known. Mathematicians were beginning to pressure Wiles to disclose his work whether or not complete, so that the wider community could explore and use whatever he had managed to accomplish. But instead of being fixed, the problem, which had originally seemed minor, now seemed very significant, far more serious, and less easy to resolve.[133]
Wiles states that on the morning of 19 September 1994, he was on the verge of giving up and was almost resigned to accepting that he had failed, and to publishing his work so that others could build on it and find the error. He adds that he was having a final look to try and understand the fundamental reasons why his approach could not be made to work, when he had a sudden insight that the specific reason why the Kolyvagin–Flach approach would not work directly also meant that his original attempts using Iwasawa theory could be made to work, if he strengthened it using his experience gained from the Kolyvagin–Flach approach. Fixing one approach with tools from the other approach would resolve the issue for all the cases that were not already proven by his refereed paper.[130][134] He described later that Iwasawa theory and the Kolyvagin–Flach approach were each inadequate on their own, but together they could be made powerful enough to overcome this final hurdle.[130]
"I was sitting at my desk examining the Kolyvagin–Flach method. It wasn't that I believed I could make it work, but I thought that at least I could explain why it didn’t work. Suddenly I had this incredible revelation. I realised that, the Kolyvagin–Flach method wasn't working, but it was all I needed to make my original Iwasawa theory work from three years earlier. So out of the ashes of Kolyvagin–Flach seemed to rise the true answer to the problem. It was so indescribably beautiful; it was so simple and so elegant. I couldn't understand how I'd missed it and I just stared at it in disbelief for twenty minutes. Then during the day I walked around the department, and I'd keep coming back to my desk looking to see if it was still there. It was still there. I couldn't contain myself, I was so excited. It was the most important moment of my working life. Nothing I ever do again will mean as much."
— Andrew Wiles, as quoted by Simon Singh[135]
On 24 October 1994, Wiles submitted two manuscripts, "Modular elliptic curves and Fermat's Last Theorem"[136] and "Ring theoretic properties of certain Hecke algebras",[137] the second of which was co-authored with Taylor and proved that certain conditions were met that were needed to justify the corrected step in the main paper. The two papers were vetted and published as the entirety of the May 1995 issue of the Annals of Mathematics. These papers established the modularity theorem for semistable elliptic curves, the last step in proving Fermat's Last Theorem, 358 years after it was conjectured.
### Subsequent developments
The full Taniyama–Shimura–Weil conjecture was finally proved by Diamond (1996), Conrad, Diamond & Taylor (1999), and Breuil et al. (2001) who, building on Wiles's work, incrementally chipped away at the remaining cases until the full result was proved.[138][139][140] The now fully proved conjecture became known as the modularity theorem.
Several other theorems in number theory similar to Fermat's Last Theorem also follow from the same reasoning, using the modularity theorem. For example: no cube can be written as a sum of two coprime n-th powers, n 3. (The case n = 3 was already known by Euler.)
## Exponents other than positive integers
### Reciprocal integers (inverse Fermat equation)
The equation can be considered the "inverse" Fermat equation. All solutions of this equation were computed by Lenstra in 1992.[141] In the case in which the mth roots are required to be real and positive, all solutions are given by[142]
for positive integers r, s, t with s and t coprime.
### Rational exponents
For the Diophantine equation with n not equal to 1, Bennett, Glass, and Székely proved in 2004 for n > 2, that if n and m are coprime, then there are integer solutions if and only if 6 divides m, and , and are different complex 6th roots of the same real number.[143]
### Negative exponents
#### n =–1
All primitive integer solutions (i.e., those with no prime factor common to all of a, b, and c) to the optic equation can be written as[144]
for positive, coprime integers m, k.
#### n =–2
The case n = –2 also has an infinitude of solutions, and these have a geometric interpretation in terms of right triangles with integer sides and an integer altitude to the hypotenuse.[145][146] All primitive solutions to are given by
for coprime integers u, v with v > u. The geometric interpretation is that a and b are the integer legs of a right triangle and d is the integer altitude to the hypotenuse. Then the hypotenuse itself is the integer
so (a, b, c) is a Pythagorean triple.
#### Integer n <–2
There are no solutions in integers for for integers n < –2. If there were, the equation could be multiplied through by to obtain , which is impossible by Fermat's Last Theorem.
## Base values other than positive integers
Fermat's last theorem can easily be extended to positive rationals:
can have no solutions with n > 2, because any solution could be rearranged as:
,
to which Fermat's Last Theorem applies.
## Monetary prizes
In 1816, and again in 1850, the French Academy of Sciences offered a prize for a general proof of Fermat's Last Theorem.[147] In 1857, the Academy awarded 3000 francs and a gold medal to Kummer for his research on ideal numbers, although he had not submitted an entry for the prize.[148] Another prize was offered in 1883 by the Academy of Brussels.[149]
In 1908, the German industrialist and amateur mathematician Paul Wolfskehl bequeathed 100,000 gold marks—a large sum at the time—to the Göttingen Academy of Sciences to offer as a prize for a complete proof of Fermat's Last Theorem.[150] On 27 June 1908, the Academy published nine rules for awarding the prize. Among other things, these rules required that the proof be published in a peer-reviewed journal; the prize would not be awarded until two years after the publication; and that no prize would be given after 13 September 2007, roughly a century after the competition was begun.[151] Wiles collected the Wolfskehl prize money, then worth $50,000, on 27 June 1997.[152] In March 2016, Wiles was awarded the Norwegian government's Abel prize worth €600,000 for "his stunning proof of Fermat’s Last Theorem by way of the modularity conjecture for semistable elliptic curves, opening a new era in number theory."[153] Prior to Wiles's proof, thousands of incorrect proofs were submitted to the Wolfskehl committee, amounting to roughly 10 feet (3 meters) of correspondence.[154] In the first year alone (1907–1908), 621 attempted proofs were submitted, although by the 1970s, the rate of submission had decreased to roughly 3–4 attempted proofs per month. According to F. Schlichting, a Wolfskehl reviewer, most of the proofs were based on elementary methods taught in schools, and often submitted by "people with a technical education but a failed career".[155] In the words of mathematical historian Howard Eves, "Fermat's Last Theorem has the peculiar distinction of being the mathematical problem for which the greatest number of incorrect proofs have been published."[149] In The Simpsons episode The Wizard of Evergreen Terrace Homer writes the equation on a blackboard, which appears to be a counterexample to Fermat's Last Theorem. The equation is incorrect but appears to be correct if it is tested on a hand held calculator that only displays, say, 10 significant figures.[156] In the March 27, 1989, episode of the American science-fiction TV series Star Trek: The Next Generation titled "The Royale" and set in the 24th century, Captain Jean-Luc Picard tells his first officer, Commander Riker, about his efforts at solving Fermat's Last Theorem, "still unsolved" 800 years after its first formulation. He concludes, "In our arrogance, we feel we are so advanced. And yet we cannot unravel a simple knot tied by a part-time French mathematician working alone without a computer."[157] (Andrew Wiles's insight leading to his breakthrough proof happened four months after the series ended.[158]) ## See also ## Footnotes 1. If the exponent "n" were not prime or 4, then it would be possible to write n either as a product of two smaller integers (n = P*Q) in which P is a prime number greater than 2, and then an = aP*Q = (aQ)P for each of a, b, and c—i.e., an equivalent solution would also have to exist for the prime power P that is smaller than n, as well; or else as n would be a power of 2 greater than four and writing n=4*Q, the same argument would hold. 2. For example, 3. This elliptic curve was first suggested in the 1960s by Yves Hellegouarch, but he did not call attention to its non-modularity. For more details, see Hellegouarch, Yves (2001). Invitation to the Mathematics of Fermat-Wiles. Academic Press. ISBN 978-0-12-339251-0. ## References 1. Singh, pp. 18–20. 2. Abel prize 2016 – full citation 3. "Science and Technology". The Guinness Book of World Records. Guinness Publishing Ltd. 1995. 4. Singh p. 144 quotes Wiles's reaction to this news: "I was electrified. I knew that moment that the course of my life was changing because this meant that to prove Fermat’s Last Theorem all I had to do was to prove the Taniyama–Shimura conjecture. It meant that my childhood dream was now a respectable thing to work on." 5. Singh p. 144 6. "Fermat's last theorem earns Andrew Wiles the Abel Prize". Nature. 15 March 2016. Retrieved 15 March 2016. 7. British mathematician Sir Andrew Wiles gets Abel math prize – The Washington Post 8. 300-year-old math question solved, professor wins$700k – CNN.com
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123. Ribet, Ken (1990). "On modular representations of Gal(Q/Q) arising from modular forms" (PDF). Inventiones Mathematicae. 100 (2): 431–476. Bibcode:1990InMat.100..431R. doi:10.1007/BF01231195. MR 1047143.
124. Singh, p. 205; Aczel, pp. 117–118.
125. Singh, pp. 237–238; Aczel, pp. 121–122.
126. Singh, pp. 239–243; Aczel, pp. 122–125.
127. Singh, pp. 244–253; Aczel, pp. 1–4, 126–128.
128. Aczel, pp. 128–130.
129. Singh, p. 257.
130. Singh, pp. 269–277.
131. A Year Later, Snag Persists In Math Proof 28 June 1994
132. Singh, pp. 175–185.
133. Aczel, pp. 132–134.
134. Singh p. 186–187 (text condensed).
135. Wiles, Andrew (1995). "Modular elliptic curves and Fermat's Last Theorem" (PDF). Annals of Mathematics. 141 (3): 443–551. doi:10.2307/2118559. JSTOR 2118559. OCLC 37032255.
136. Taylor R, Wiles A (1995). "Ring theoretic properties of certain Hecke algebras". Annals of Mathematics. 141 (3): 553–572. doi:10.2307/2118560. JSTOR 2118560. OCLC 37032255. Archived from the original on 27 November 2001.
137. Diamond, Fred (1996). "On deformation rings and Hecke rings". Annals of Mathematics. Second Series. 144 (1): 137–166. doi:10.2307/2118586. ISSN 0003-486X. MR 1405946.
138. Conrad, Brian; Diamond, Fred; Taylor, Richard (1999). "Modularity of certain potentially Barsotti-Tate Galois representations". Journal of the American Mathematical Society. 12 (2): 521–567. doi:10.1090/S0894-0347-99-00287-8. ISSN 0894-0347. MR 1639612.
139. Breuil, Christophe; Conrad, Brian; Diamond, Fred; Taylor, Richard (2001). "On the modularity of elliptic curves over Q: wild 3-adic exercises". Journal of the American Mathematical Society. 14 (4): 843–939. doi:10.1090/S0894-0347-01-00370-8. ISSN 0894-0347. MR 1839918.
140. Lenstra Jr. H.W. (1992). "On the inverse Fermat equation". Discrete Mathematics. 106–107: 329–331. doi:10.1016/0012-365x(92)90561-s.
141. Newman M (1981). "A radical diophantine equation". Journal of Number Theory. 13: 495–498. doi:10.1016/0022-314x(81)90040-8.
142. Bennett, Curtis D.; Glass, A. M. W.; Székely, Gábor J. (2004). "Fermat's last theorem for rational exponents". American Mathematical Monthly. 111 (4): 322–329. doi:10.2307/4145241. MR 2057186.
143. Dickson, pp. 688–691
144. Voles, Roger (July 1999). "Integer solutions of a−2 + b−2 = d−2". Mathematical Gazette. 83: 269–271.
145. Richinick, Jennifer (July 2008). "The upside-down Pythagorean Theorem". Mathematical Gazette. 92: 313–317.
146. Aczel, p. 69; Singh, p. 105.
147. Aczel, p. 69.
148. Koshy T (2001). Elementary number theory with applications. New York: Academic Press. p. 544. ISBN 978-0-12-421171-1.
149. Singh, pp. 120–125, 131–133, 295–296; Aczel, p. 70.
150. Singh, pp. 120–125.
151. Singh, p. 284
152. "The Abel Prize citation 2016" (PDF). The Abel Prize. The Abel Prize Committee. March 2016. Retrieved 16 March 2016.
153. Singh, p. 295.
154. Singh, pp. 295–296.
155. Singh, Simon (2013). The Simpsons and their Mathematical Secrets. London. pp. 35–36. ISBN 978-1-4088-3530-2.
156. Kevin Knudson (20 August 2015). "The Math Of Star Trek: How Trying To Solve Fermat's Last Theorem Revolutionized Mathematics". Forbes.
157. "Are mathematicians finally satisfied with Andrew Wiles's proof of Fermat's Last Theorem? Why has this theorem been so difficult to prove?". Scientific American. 21 October 1999. Retrieved 16 March 2016.
## Bibliography
• Aczel, Amir (30 September 1996). Fermat's Last Theorem: Unlocking the Secret of an Ancient Mathematical Problem. Four Walls Eight Windows. ISBN 978-1-56858-077-7.
• Dickson LE (1919). History of the Theory of Numbers. Volume II. Diophantine Analysis. New York: Chelsea Publishing. pp. 545–550, 615–621, 688–691, 731–776.
• Edwards, HM (1997). Fermat's Last Theorem. A Genetic Introduction to Algebraic Number Theory. Graduate Texts in Mathematics. 50. New York: Springer-Verlag.
• Friberg, Joran (2007). Amazing Traces of a Babylonian Origin in Greek Mathematics. World Scientific Publishing Company. ISBN 978-981-270-452-8.
• Kleiner I (2000). "From Fermat to Wiles: Fermat's Last Theorem Becomes a Theorem" (PDF). Elemente der Mathematik. 55: 19–37. doi:10.1007/PL00000079. Archived from the original (PDF) on 13 July 2010.
• Mordell LJ (1921). Three Lectures on Fermat's Last Theorem. Cambridge: Cambridge University Press.
• Panchishkin, Alekseĭ Alekseevich (2007). Introduction to Modern Number Theory (Encyclopedia of Mathematical Sciences. Springer Berlin Heidelberg New York. ISBN 978-3-540-20364-3.
• Ribenboim P (2000). Fermat's Last Theorem for Amateurs. New York: Springer-Verlag. ISBN 978-0-387-98508-4.
• Singh S (October 1998). Fermat's Enigma. New York: Anchor Books. ISBN 978-0-385-49362-8.
• Stark H (1978). An Introduction to Number Theory. MIT Press. ISBN 0-262-69060-8.
• Bell, Eric T. (6 August 1998) [1961]. The Last Problem. New York: The Mathematical Association of America. ISBN 978-0-88385-451-8.
• Benson, Donald C. (5 April 2001). The Moment of Proof: Mathematical Epiphanies. Oxford University Press. ISBN 978-0-19-513919-8.
• Brudner, Harvey J. (1994). Fermat and the Missing Numbers. WLC, Inc. ISBN 978-0-9644785-0-3.
• Edwards, H. M. (March 1996) [1977]. Fermat's Last Theorem. New York: Springer-Verlag. ISBN 978-0-387-90230-2.
• Faltings G (July 1995). "The Proof of Fermat's Last Theorem by R. Taylor and A. Wiles" (PDF). Notices of the American Mathematical Society. 42 (7): 743–746. ISSN 0002-9920.
• Mozzochi, Charles (7 December 2000). The Fermat Diary. American Mathematical Society. ISBN 978-0-8218-2670-6.
• Ribenboim P (1979). 13 Lectures on Fermat's Last Theorem. New York: Springer Verlag. ISBN 978-0-387-90432-0.
• van der Poorten, Alf (6 March 1996). Notes on Fermat's Last Theorem. WileyBlackwell. ISBN 978-0-471-06261-5.
• Saikia, Manjil P (July 2011). "A Study of Kummer's Proof of Fermat's Last Theorem for Regular Primes" (PDF). IISER Mohali (India) Summer Project Report.
• Daney, Charles (2003). "The Mathematics of Fermat's Last Theorem". Archived from the original on 3 August 2004. Retrieved 5 August 2004.
• The bluffer's guide to Fermat's Last Theorem
• Elkies, Noam D. "Tables of Fermat "near-misses" – approximate solutions of xn + yn = zn".
• Freeman, Larry (2005). "Fermat's Last Theorem Blog". Blog that covers the history of Fermat's Last Theorem from Fermat to Wiles.
• Hazewinkel, Michiel, ed. (2001) [1994], "Fermat's last theorem", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
• Ribet, Ken (1995). "Galois representations and modular forms" (PDF). Retrieved 2016-03-17. Discusses various material that is related to the proof of Fermat's Last Theorem: elliptic curves, modular forms, Galois representations and their deformations, Frey's construction, and the conjectures of Serre and of Taniyama–Shimura.
• Shay, David (2003). "Fermat's Last Theorem". Retrieved 14 January 2017. The story, the history and the mystery.
• Weisstein, Eric W. "Fermat's Last Theorem". MathWorld.
• O'Connor JJ, Robertson EF (1996). "Fermat's last theorem". Archived from the original on 4 August 2004. Retrieved 5 August 2004.
• "The Proof". The title of one edition of the PBS television series NOVA, discusses Andrew Wiles's effort to prove Fermat's Last Theorem.
• "Documentary Movie on Fermat's Last Theorem (1996)". Simon Singh and John Lynch's film tells the story of Andrew Wiles.
• Beal Fermat and Pythagora's Triplets (sic)
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Home
Search results “Data analysis and statistical methods”
17:59
01:02:58
Copyright Broad Institute, 2013. All rights reserved. The presentation above was filmed during the 2012 Proteomics Workshop, part of the BroadE Workshop series. The Proteomics Workshop provides a working knowledge of what proteomics is and how it can accelerate biologists' and clinicians' research. The focus of the workshop is on the most important technologies and experimental approaches used in modern mass spectrometry (MS)-based proteomics.
09:33
Seven different statistical tests and a process by which you can decide which to use. The tests are: Test for a mean, test for a proportion, difference of proportions, difference of two means - independent samples, difference of two means - paired, chi-squared test for independence and regression. This video draws together videos about Helen, her brother, Luke and the choconutties. There is a sequel to give more practice choosing and illustrations of the different types of test with hypotheses.
Views: 715025 Dr Nic's Maths and Stats
39:54
This tutorial provides an overview of statistical analyses in the social sciences. It distinguishes between descriptive and inferential statistics, discusses factors for choosing an analysis procedure, and identifies the difference between parametric and nonparametric procedures.
Views: 222932 The Doctoral Journey
01:05:31
Ethan Meyers, Hampshire College - MIT BMM Summer Course 2018
24:49
Views: 19316 S Manikandan
09:44
In this lecture, I show which types of statistical models should be used when; the most important decision concerns the explanatory variables: When these are continuous, the analysis type will be regression; however, when these are factors, then we will conduct an analysis of variance. Overall, I show that both analyses are special examples of what is called a Linear Statistical Model. I briefly introduce linear statistical models. Later lectures will cover this in greater detail.
Views: 37959 Christoph Scherber
45:32
Views: 845 Ekeeda
02:22:43
Views: 411436 ExcelIsFun
05:39
The process of doing statistical analysis follows a clearly defined sequence of steps whether the analysis is being done in a formal setting like a medical lab or informally like you would find in a corporate environment. This lecture gives a brief overview of the process.
Views: 47486 White Crane Education
20:17
Views: 29950 Simplilearn
10:05
Views: 7236 morgankenneth12
18:32
Views: 71667 David Russell
12:27
In this lecture, I provide a very basic introduction to quantitative data analysis and statistics. We begin by defining what "data" is, what a dataset looks like, and software tools for analyzing data.
Views: 3764 David Russell
53:55
Data download: http://www.windengineering.byg.dtu.dk/download The video introduces basic methods in statistics and three Matlab scripts that can be used to analyse measured data for example from wind tunnel testing. The scripts allow basic signal processing (detrending and digital filtering), assessment of probability and spectral densities (Matlab signal processing toolbox required!), the collection of maximum and minimum extremes from sub-series for extreme value analysis, correlation between two time series and the calculation of the joint probability density function. The video is used for education at the Technical University of Denmark (DTU) in course 11374 "Seismic and Wind Engineering" and for preparation of wind tunnel testing in civil engineering. For further information see www.windengineering.byg.dtu.dk or contact the author under [email protected]
Views: 12374 Holger Koss
02:03
“Fundamentals of Engineering Statistical Analysis” is a free online course on Janux that is open to anyone. Learn more at http://janux.ou.edu. Created by the University of Oklahoma, Janux is an interactive learning community that gives learners direct connections to courses, education resources, faculty, and each other. Janux courses are freely available or may be taken for college credit by enrolled OU students. Dr. Kash Barker is an Assistant Professor in the School of Industrial Engineering. Video produced by NextThought (http://nextthought.com). Copyright © 2000-2014 The Board of Regents of the University of Oklahoma, All Rights Reserved.
Views: 1975 Janux
01:39:12
Views: 2260 Florence HEP
06:02
www.ozanozcan.us
Views: 260397 ozanteaching
17:12
Qualitative research is a strategy for systematic collection, organization, and interpretation of phenomena that are difficult to measure quantitatively. Dr. Leslie Curry leads us through six modules covering essential topics in qualitative research, including what it is qualitative research and how to use the most common methods, in-depth interviews and focus groups. These videos are intended to enhance participants' capacity to conceptualize, design, and conduct qualitative research in the health sciences. Welcome to Module 5. Bradley EH, Curry LA, Devers K. Qualitative data analysis for health services research: Developing taxonomy, themes, and theory. Health Services Research, 2007; 42(4):1758-1772. Learn more about Dr. Leslie Curry http://publichealth.yale.edu/people/leslie_curry.profile Learn more about the Yale Global Health Leadership Institute http://ghli.yale.edu
Views: 153962 YaleUniversity
06:27
Views: 13018 Great Learning
30:52
Subject:Geography Paper: Quantitative techniques in geography
Views: 2077 Vidya-mitra
24:05
Views: 7454 Simplilearn
14:06
Use simple data analysis techniques in SPSS to analyze survey questions.
Views: 812787 Claus Ebster
13:14
Part 1
Views: 172712 Teresa Johnson
01:11:27
Presenter: Christopher Fonnesbeck Description This tutorial will introduce the use of Python for statistical data analysis, using data stored as Pandas DataFrame objects. Much of the work involved in analyzing data resides in importing, cleaning and transforming data in preparation for analysis. Therefore, the first half of the course is comprised of a 2-part overview of basic and intermediate Pandas usage that will show how to effectively manipulate datasets in memory. This includes tasks like indexing, alignment, join/merge methods, date/time types, and handling of missing data. Next, we will cover plotting and visualization using Pandas and Matplotlib, focusing on creating effective visual representations of your data, while avoiding common pitfalls. Finally, participants will be introduced to methods for statistical data modeling using some of the advanced functions in Numpy, Scipy and Pandas. This will include fitting your data to probability distributions, estimating relationships among variables using linear and non-linear models, and a brief introduction to Bayesian methods. Each section of the tutorial will involve hands-on manipulation and analysis of sample datasets, to be provided to attendees in advance. The target audience for the tutorial includes all new Python users, though we recommend that users also attend the NumPy and IPython session in the introductory track. Tutorial GitHub repo: https://github.com/fonnesbeck/statistical-analysis-python-tutorial Outline Introduction to Pandas (45 min) Importing data Series and DataFrame objects Indexing, data selection and subsetting Hierarchical indexing Reading and writing files Date/time types String conversion Missing data Data summarization Data Wrangling with Pandas (45 min) Indexing, selection and subsetting Reshaping DataFrame objects Pivoting Alignment Data aggregation and GroupBy operations Merging and joining DataFrame objects Plotting and Visualization (45 min) Time series plots Grouped plots Scatterplots Histograms Visualization pro tips Statistical Data Modeling (45 min) Fitting data to probability distributions Linear models Spline models Time series analysis Bayesian models Required Packages Python 2.7 or higher (including Python 3) pandas 0.11.1 or higher, and its dependencies NumPy 1.6.1 or higher matplotlib 1.0.0 or higher pytz IPython 0.12 or higher pyzmq tornado
Views: 72052 Enthought
08:18
01:23:02
Webinar 8: Methods of data analysis: Advanced and emerging methods of statistical analysis Tues 20th September 2016 Short talks within the webinar include: ---------------------------------------------------------------------------------- “Using more and more variables in statistical analysis” by Paul Lambert (est 20 mins) "Data Analysis Skills" by Alasdair Rutherford “The idea of multilevel modelling” by Paul Lambert (est. 10 mins) “Estimating and communicating uncertainty” by Alasdair Rutherford (est 20 mins) . The webinar includes a mix of presentation sessions and opportunities for online discussions, questions, clarifications, and information provision. ---------------------------------------------------------------------------------- Find details of our other webinars at http://thinkdata.org.uk/events/CSDPWebinars/ More information on the research, capacity building, and collaboration activities of the Think Data network can be found at http://www.thinkdata.org.uk The Scottish Civil Society Data Partnership project is run by the Universities of Stirling and St Andrews and the Scottish Council for Voluntary Organisations (SCVO). http://www.stir.ac.uk http://www.st-andrews.ac.uk http://www.scvo.org.uk Funded by the Economic and Social Research Council (ESRC) http://www.esrc.ac.uk/ ---------------------------------------------------------------------------------- Music: http://www.bensound.com/royalty-free-music
Views: 51 Think Data
18:36
Paper: Multivariate Analysis Module name: Introduction toMultivariate Analysis Content Writer: Souvik Bandyopadhyay
Views: 55139 Vidya-mitra
00:16
Views: 14 Stănescu
05:18
Tutorial introducing the idea of linear regression analysis and the least square method. Typically used in a statistics class. Playlist on Linear Regression http://www.youtube.com/course?list=ECF596A4043DBEAE9C Like us on: http://www.facebook.com/PartyMoreStudyLess Created by David Longstreet, Professor of the Universe, MyBookSucks http://www.linkedin.com/in/davidlongstreet
Views: 687563 statisticsfun
04:16
Views: 13254 The Audiopedia
24:57
Includes application examples, scales of measurement (nominal, ordinal, interval & ratio), qualitative versus quantitative data, cross-sectional versus time-series data, experimental versus observational data, and descriptive statistics versus statistical inference.
Views: 28977 Bharatendra Rai
01:00:37
Through real-world examples, webinar participants learn strategies for choosing appropriate outcome measures, methods for analysis and randomization, and sample sizes as well as tips for collecting the right data to answer your scientific questions.
Views: 8774 RhoInc1984
55:06
This session will provide information regarding descriptive statistics that are often used when reviewing assessment data. We will cover the statistics available in the Baseline reporting site and we will use example situations to identify which statistics should be used to answer the questions being asked. We will also provide an overview regarding levels of measurement that can help determine what types of statistics you are able to run on your data. - See more at: http://www2.campuslabs.com/support/training/basic-statistics-quantitative-analysis-i-5/#sthash.FDO5HA6i.dpuf
Views: 34682 Campus Labs
15:22
This video is the first in a series of six which cover best practice for analyzing spectra with multivariate data analysis. In this edition we introduce multivariate data analysis, or chemometrics, and why these powerful tools are especially useful for analyzing spectral data. The video gives examples of typical applications, discusses the benefits of Multivariate analysis over Univariate analysis, and gives an explanation of some important multivariate methods. The video concludes with a demonstration of spectral data being analyzed.
Views: 28407 Camo Analytics
00:27
Views: 17 Marie Dunning
01:00
Views: 26 Louise Whitmire
47:10
A step-by-step approach for choosing an appropriate statistcal test for data analysis.
15:49
R programming for beginners - This video is an introduction to R programming in which I provide a tutorial on some statistical analysis (specifically using the t-test and linear regression). I also demonstrate how to use dplyr and ggplot to do data manipulation and data visualisation. Its R programming for beginners really and is filled with graphics, quantitative analysis and some explanations as to how statistics work. If you’re a statistician, into data science or perhaps someone learning bio-stats and thinking about learning to use R for quantitative analysis, then you’ll find this video useful. Importantly, R is free. If you learn R programming you’ll have it for life. This video was sponsored by the University of Edinburgh. Find out more about their programmes at http://edin.ac/2pTfis2 This channel focusses on global health and public health - so please consider subscribing if you’re someone wanting to make the world a better place – I’d love to you join this community. I have videos on epidemiology, study design, ethics and many more.
01:10
06:51
Let's go on a journey through univariate analysis and learn about descriptive statistics in research!
Views: 46193 ChrisFlipp
01:04:01
What can text analysis tell us about society? Corpora of news, books, and social media encode human beliefs and culture. But it is impossible for a researcher to read all of today's rapidly growing text archives. My research develops statistical text analysis methods that measure social phenomena from textual content, especially in news and social media data. For example: How do changes to public opinion appear in microblogs? What topics get censored in the Chinese Internet? What character archetypes recur in movie plots? How do geography and ethnicity affect the diffusion of new language? In order to answer these questions effectively, we must apply and develop scientific methods in statistics, computation, and linguistics. In this talk I will illustrate these methods in a project that analyzes events in international politics. Political scientists are interested in studying international relations through *event data*: time series records of who did what to whom, as described in news articles. To address this event extraction problem, we develop an unsupervised Bayesian model of semantic event classes, which learns the verbs and textual descriptions that correspond to types of diplomatic and military interactions between countries. The model uses dynamic logistic normal priors to drive the learning of semantic classes; but unlike a topic model, it leverages deeper linguistic analysis of syntactic argument structure. Using a corpus of several million news articles over 15 years, we quantitatively evaluate how well its event types match ones defined by experts in previous work, and how well its inferences about countries correspond to real-world conflict. The method also supports exploratory analysis; for example, of the recent history of Israeli-Palestinian relations.
Views: 1143 Microsoft Research
08:32
Views: 4255 Ranywayz Random
00:12
00:15
Views: 5 Michael 2
47:12
Multivariate statistical techniques are the application of statistics to simultaneous observations and can include the analysis of more than one outcome (dependent) variable. Good multivariate analysis starts with exploratory and graphical analyses to reveal potential relations in the data and to highlight potential outliers. First, this presentation will discuss how to extend univariate and bivariate methods for graphical analysis to multivariate data, as well as methods unique to multivariate data. Second, multivariate outlier detection will be presented. Third, there will be a brief discussion of multivariate statistical analysis methods, such as multiple regression, principal component analysis, and cluster analysis, including examples and suggestions as to when one might want to use these techniques.
09:52
This video is part of the University of Southampton, Southampton Education School, Digital Media Resources http://www.southampton.ac.uk/education http://www.southampton.ac.uk/~sesvideo/
54:53
Free MATLAB Trial: https://goo.gl/yXuXnS Request a Quote: https://goo.gl/wNKDSg Contact Us: https://goo.gl/RjJAkE Learn more about MATLAB: https://goo.gl/8QV7ZZ Learn more about Simulink: https://goo.gl/nqnbLe ------------------------------------------------------------------------- Researchers and scientists have to commonly process, visualize and analyze large amounts of data to extract patterns, identify trends and relationships between variables, prove hypothesis, etc. A variety of statistical techniques are used in this data mining and analysis process. Using a realistic data from a clinical study, we will provide an overview of the statistical analysis and visualization capabilities in the MATLAB product family. Highlights include: • Data management and organization • Data filtering and visualization • Descriptive statistics • Hypothesis testing and ANOVA • Regression analysis
Views: 14807 MATLAB
03:36
This statistical analysis overview explains descriptive and inferential statistics. Watch more at http://www.lynda.com/Excel-2007-tutorials/business-statistics/71213-2.html?utm_medium=viral&utm_source=youtube&utm_campaign=videoupload-71213-0101 This specific tutorial is just a single movie from chapter one of the Excel 2007: Business Statistics course presented by lynda.com author Curt Frye. The complete Excel 2007: Business Statistics course has a total duration of 4 hours and 19 minutes and covers formulas and functions for calculating averages and standard deviations, charts and graphs for summarizing data, and the Analysis ToolPak add-in for even greater insights into data Excel 2007: Business Statistics table of contents: Introduction 1. Introducing Statistics 2. Learning Useful Excel Techniques 3. Summarizing Data Using Tables and Graphics 4. Describing Data Using Numerical Methods 5. Using Probability Distributions 6. Sampling Values from a Population 7. Testing Hypotheses 8. Using Linear and Multiple Regression Conclusion
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January 01, 2012 by SirThorfinn Hi, I'm relatively new to the Nerdkits and I just finished the LED Array and decided to create my own clock with it like i saw in the demonstration video. I don't know if there is allready a code for this or not and i'm definitely sure the way I coded it is not the most efficient way but i'm still learning with this and in my High schools computer science course. I took the most basic (in my opinion) approach to codeing it by essentially creating a ton of nested If and else/if statements. Also once I had the code and ran the clock after a while i noticed it lags behind real time and is slow by about 2 min every 15 min. I noticed that the code might be causeing the controller to take extra time to read the code then wait 1000 milliseconds so I wen't back to the code and changed the delay from 1000 milliseconds to 871 after doing some simple calculations. I'm still tring to find the perfect delay because its barley to fast now. If anyone can show me how to find a good clock delay or at least a less lengthy coding would be greatly appreciated, Thanks. Here is my code. Oh and this is a method i created inside the template so if you want to try it out you need the method declared in the main method but i'm sure you guys are way ahead of me there, :) ``````// MY MEATHOD!!! void clock() { //H = Hours //M = Min //AM = Display AM or PM //blink = used in ":" blink int H=10, M=36, S=0, AM=0, blink=0; int8_t offset=0; ledarray_blank(); while(1){ ledarray_blank(); offset=0; //Hours display if (H==0){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (H==1){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (H==2){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (H==3){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (H==4){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (H==5){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (H==6){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (H==7){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (H==8){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (H==9){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (H==10){ font_display('1', offset); offset += font_width('0')+1; delay_ms(0); font_display('0', offset); offset += font_width('1')+1; delay_ms(0); } else if (H==11){ font_display('1', offset); offset += font_width('0')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (H==12){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); H=0; if (AM=0){ AM++; } else { AM=0; } } //End of hours //Blinking ":" if (blink==0){ font_display(':', offset); offset += font_width(':')+1; delay_ms(0); blink=1; } else{ font_display(' ', offset); offset += font_width(':')+1; delay_ms(0); blink=0; } //Min Display if (M==0){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (M==1){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (M==2){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (M==3){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==3){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==4){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (M==5){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (M==6){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (M==7){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (M==8){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (M==9){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (M==10){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (M==11){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (M==12){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (M==13){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==14){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (M==15){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (M==16){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (M==17){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (M==18){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (M==19){ font_display('1', offset); offset += font_width('1')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (M==20){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (M==21){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (M==22){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (M==23){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==24){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (M==25){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (M==26){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (M==27){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (M==28){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (M==29){ font_display('2', offset); offset += font_width('2')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (M==30){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (M==31){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (M==32){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (M==33){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==34){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (M==35){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (M==36){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (M==37){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (M==38){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (M==39){ font_display('3', offset); offset += font_width('3')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (M==40){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (M==41){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (M==42){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (M==43){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==44){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (M==45){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (M==46){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (M==47){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (M==48){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (M==49){ font_display('4', offset); offset += font_width('4')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (M==50){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); } else if (M==51){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('1', offset); offset += font_width('1')+1; delay_ms(0); } else if (M==52){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('2', offset); offset += font_width('2')+1; delay_ms(0); } else if (M==53){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('3', offset); offset += font_width('3')+1; delay_ms(0); } else if (M==54){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('4', offset); offset += font_width('4')+1; delay_ms(0); } else if (M==55){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('5', offset); offset += font_width('5')+1; delay_ms(0); } else if (M==56){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('6', offset); offset += font_width('6')+1; delay_ms(0); } else if (M==57){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('7', offset); offset += font_width('7')+1; delay_ms(0); } else if (M==58){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('8', offset); offset += font_width('8')+1; delay_ms(0); } else if (M==59){ font_display('5', offset); offset += font_width('5')+1; delay_ms(0); font_display('9', offset); offset += font_width('9')+1; delay_ms(0); } else if (M==60){ font_display('0', offset); offset += font_width('0')+1; delay_ms(0); font_display('0', offset); offset += font_width('0')+1; delay_ms(0); M=0; H++; } offset += font_width('0'); //end min //AM or PM Display if (AM==1){ font_display('A', offset); offset += font_width('A')+1; delay_ms(0); } else{ font_display('P', offset); offset += font_width('P')+1; delay_ms(0); } //Checking if to add a Min if (S==60){ S=0; M++; } //Delay set at 871 because of the small amout of time the //Micro Controller takes to proccess. //This is not a perfect delay, still perfecting. delay_ms(871); S++; } } `````` And sorry again for the crazy lengthy code, lol. I redid my other thread because of the format accident... Hi SirThorfinn- In the "//hours display" section of your code, I would replace it with: ``````if (H==12) {H = 0;} lo_digit = (H % 10) + 0x30; hi_digit = (H / 10) + 0x30; font_display(hi_digit, offset); offset += font_width(hi_digit)+1; delay_ms(0); font_display(lo_digit, offset); offset += font_width(lo_digit)+1; delay_ms(0); `````` And in the "//Min display" section, I would use: ``````if (M==60) {M = 0; H++;} lo_digit = (M % 10) + 0x30; hi_digit = (M / 10) + 0x30; font_display(hi_digit, offset); offset += font_width(hi_digit)+1; delay_ms(0); font_display(lo_digit, offset); offset += font_width(lo_digit)+1; delay_ms(0); `````` Of course you have to declare "lo_digit" and "hi_digit" as uint8_t types. The reason this works is because when you use '0' in your code, the compiler automatically replaces it with its byte code equivalent 0x30 (48 in decimal). Likewise '1' becomes 0x31 etc. The other parts just use integer division and modulus (remainder) operators to dissect the two digit number. Oh and if you still have timing issues, watch the Nerdkit video called "Crystal Real Time Clock". It uses interrupts to accurately use the MCU timer. Good luck. Thanks, I'll see how it goes
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+0
# help help
0
5
1
+1351
The six faces of a cube are painted black. The cube is then cut into \$3^3\$ smaller cubes, all the same size.
(a) How many of the smaller cubes have exactly one black face?
(b) How many of the smaller cubes do not have any black faces?
(c) One of the small cubes is chosen at random, and rolled. What is the probability that when it lands, the face on the top is black?
Apr 23, 2024
#1
+128774
+1
6 cubes have 1 black face
12 have 2 black faces
8 have 3 black faces
1 has no black faces (in the center)
Probability of black face =
(6/27) ( 1/6) + (12/27)(2/6) + ( 8/27) (3/6) + (1/27) (0/6) = 1/3
(The first set of parentheses contains the probability of that type of cube is chosen; the second set of parentheses contains the probability of having a black side facing up; mult/add these together for the total probability
Apr 23, 2024
edited by CPhill Apr 23, 2024
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# Thread: What Math Classes to Take Before Grad Stats or "Math for Economists"?
1. Good post? |
## What Math Classes to Take Before Grad Stats or "Math for Economists"?
So ultimately I want to take Grad Micro and Statistics while at undergrad and was wondering what math classes I would have to take beforehand in order to be best prepared for those two. Right now I'm planning to take honors sections of Calc III and Linear Algebra but I'm not sure what the rest will entail? Maybe some real analysis? Any abstract algebra? Thanks
2. Good post? |
## Re: What Math Classes to Take Before Grad Stats or "Math for Economists"?
Not taken grad stats but I personally find real analysis and probability (plus linear algebra) most useful for grad micro and metrics. Optimization also somewhat helps. Hope this helps
3. Good post? |
## Re: What Math Classes to Take Before Grad Stats or "Math for Economists"?
Okay thank you. I'll look into trying to fit real analysis into my schedule at some point beforehand. By probability do you actually mean a probability class or something like math stats?
4. Good post? |
## Re: What Math Classes to Take Before Grad Stats or "Math for Economists"?
Your math or stats departments probably has a two semester sequence of (a) probability theory and (b) mathematical statistics. These are great classes to take as graduate school preparation.
5. Good post? |
## Re: What Math Classes to Take Before Grad Stats or "Math for Economists"?
I agree with tbe. In your two-semester probability and statistics sequence, probability will be harder than statistics. It is a good class to take.
How many more semesters do you have before you apply for graduate school?
Abstract algebra is not necessary, but if it's something you love, than go ahead and take it. If you want to look at other useful classes beyond Calc III & Honors linear algebra (and probability & statistics), I'd suggest in order of importance:
- Real analysis (The honors LA class should be proof-based, good preparation for this class).
Differential equations is also a good class, but I'm assuming you will have covered this topic in your calculus classes. This is also a computational class that is fairly easy to learn on your own, as well.
I think there are probably better courses to take than graduate statistics, in my opinion. Micro I is a good choice though. Do know that in your math for economists class, you will do a lot of optimization, so a solid grounding in calculus is important. In Micro, you will do proofs - the Honors LA and Real Analysis will help you here. Probability/Statistics will help you in metrics.
6. Good post? |
## Re: What Math Classes to Take Before Grad Stats or "Math for Economists"?
I'm a rising sophomore so I have plenty of time. Just want to plan it out so I know which classes to take when. The first year math for economists course covers constrained optimization so Kuhn-Tucker, envelope theorem, etc. The professor for that class said that real analysis will be sufficient as a prerequisite but after looking over the syllabus and material, a lot looks foreign. Would the proof based LA and Calc III be a good foundation to have before taking that course? Real Analysis is in the fall and because of timing I wouldn't be able to take it until fall of 2021.
7. Good post? |
## Re: What Math Classes to Take Before Grad Stats or "Math for Economists"?
I don't know your course schedule but here is one potential schedule if you cannot take RA until Fall 2021. I see nothing wrong with taking real analysis concurrently with math for economics, which is far more computational than you think, if it uses standard textbooks. It is going to look a bit foreign but as long as you have solid calculus and linear algebra foundations you will be fine - it is often just an extension of that material. I am assuming that your school has a two-semester sequence of probability and stats that starts in the fall.
Question: When do you want to take Micro I? This assumes you want to take it Senior Fall, but you may wish to take it in Junior Fall (it'll definitely be on your transcript, this will help your letters). I can DM you more if you give me more context.
Soph Fall: Honors LA / Calc III / Probability
Soph Spring: Diff Eq / Statistics
Jr Fall: Real Analysis / Math for Economics
Jr Spring: (advanced probability or topology or second semester real analysis or advanced linear algebra)
Sen Fall: Micro I
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# The three integers X, Y, and Z. Is their product XYZ = zero
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Yes I agree... Theorems are the keys. In my eyes, 0^0=1 appears as 1+1=2 almost an evidence .
'Series', more a french word?, are pratical and concrete examples of the use of 0^0=1
Why 0^0=1?
Since,
o e^0=1
o x^x=e^(x*ln(x))
o Lim x*ln(x) = 0 when x -> 0
Thus, 0^0=e^0=1
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This problem was sent (2003) by a famous legend on this forum stolyar
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Fig wrote:
Yes I agree... Theorems are the keys. In my eyes, 0^0=1 appears as 1+1=2 almost an evidence .
'Series', more a french word?, are pratical and concrete examples of the use of 0^0=1
Why 0^0=1?
Since,
o e^0=1
o x^x=e^(x*ln(x))
o Lim x*ln(x) = 0 when x -> 0
Thus, 0^0=e^0=1
Could you please provide any authentic source that describes 0^0 = 1 ???
Before doing that see these:
This is from IBM not from Windows. The below link (of IBM) also cites IEEE 854 standard.
http://www2.hursley.ibm.com/decimal/daops.html#refpower
-----------------------------------------------
Professor Keith Geddes
Symbolic Computation Group
School of Computer Science
University of Waterloo
200 University Avenue West
Waterloo ON N2L 3G1
E-mail: kogeddes@uwaterloo.ca
URL: http://www.uwaterloo.ca/~kogeddes
-----------------------------------------------
--------------------------------------------------------------------------------
The algorithm for integration of elementary functions is known as the Risch algorithm. It will either compute the integral as an elementary function or else prove that the integral cannot be expressed as an elementary function.
For the integrand x^x, the Risch algorithm first converts it to exp-log form. Namely the integrand is:
exp(x*log(x)) .
We give names to the functions appearing here, namely
theta[1] = log(x), theta[2] = exp(x*theta[1]) .
We know that theta[1] is "transcendental over Q(x)" and we know that theta[2] is "transcendental over Q(x,theta[1])".
We now have the problem:
integral theta[2] dx .
Note that the "outermost" function in the integrand is theta[2], which is an exponential function.
Moreover, the integrand is a polynomial in theta[2].
For such a case, the integration theory tells us that the integral, if it exists as an elementary function, must be a similar polynomial in
theta[2],
namely
integral theta[2] dx = q(x)*theta[2] .
(Note that we need to know the theory behind this fact.)
We then differentiate both sides of the above equation, which yields
theta[2] = q'(x)*theta[2] + q(x)*(theta[2])' .
Now in this case, from the definition of theta[2] we have
(theta[2])' = (exp(x*log(x)))'
= exp(x*log(x)) * (log(x) + 1)
= theta[2] * (theta[1] + 1)
Therefore, we have
theta[2] = q'(x)*theta[2] + q(x)*(theta[1]+1)*(theta[2]) .
Since theta[2] is "transcendental over Q(x,theta[1])", we can equate coefficients of theta[2] on each side which gives
1 = q'(x) + (theta[1]+1)*q(x) .
Also, since theta[1] is "transcendental over Q(x)", equating
coefficients of this equation as polynomials in theta[1] gives
1 = q'(x) + q(x)
0 = q(x)
which has no solution.
(The second equation tells us that q(x) = 0 and therefore q'(x) = 0, which makes the first equation become 1 = 0, which is a contradiction.)
CONCLUSION: (integral x^x dx) does not exist as an elementary function.
--------------------------------------------------------------------------------
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5
Kudos [?]: 124 [0], given: 39
I do not know whether this long dicussion is of a real benefit or not ,
but i would leave this question unanswered on the test day if it will take me to go through all of this allogarithms
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8
Kudos [?]: 102 [0], given: 0
ps_dahiya wrote:
Fig wrote:
Yes I agree... Theorems are the keys. In my eyes, 0^0=1 appears as 1+1=2 almost an evidence .
'Series', more a french word?, are pratical and concrete examples of the use of 0^0=1
Why 0^0=1?
Since,
o e^0=1
o x^x=e^(x*ln(x))
o Lim x*ln(x) = 0 when x -> 0
Thus, 0^0=e^0=1
Could you please provide any authentic source that describes 0^0 = 1 ???
Before doing that see these:
This is from IBM not from Windows. The below link (of IBM) also cites IEEE 854 standard.
http://www2.hursley.ibm.com/decimal/daops.html#refpower
-----------------------------------------------
Professor Keith Geddes
Symbolic Computation Group
School of Computer Science
University of Waterloo
200 University Avenue West
Waterloo ON N2L 3G1
E-mail: kogeddes@uwaterloo.ca
URL: http://www.uwaterloo.ca/~kogeddes
-----------------------------------------------
--------------------------------------------------------------------------------
The algorithm for integration of elementary functions is known as the Risch algorithm. It will either compute the integral as an elementary function or else prove that the integral cannot be expressed as an elementary function.
For the integrand x^x, the Risch algorithm first converts it to exp-log form. Namely the integrand is:
exp(x*log(x)) .
We give names to the functions appearing here, namely
theta[1] = log(x), theta[2] = exp(x*theta[1]) .
We know that theta[1] is "transcendental over Q(x)" and we know that theta[2] is "transcendental over Q(x,theta[1])".
We now have the problem:
integral theta[2] dx .
Note that the "outermost" function in the integrand is theta[2], which is an exponential function.
Moreover, the integrand is a polynomial in theta[2].
For such a case, the integration theory tells us that the integral, if it exists as an elementary function, must be a similar polynomial in
theta[2],
namely
integral theta[2] dx = q(x)*theta[2] .
(Note that we need to know the theory behind this fact.)
We then differentiate both sides of the above equation, which yields
theta[2] = q'(x)*theta[2] + q(x)*(theta[2])' .
Now in this case, from the definition of theta[2] we have
(theta[2])' = (exp(x*log(x)))'
= exp(x*log(x)) * (log(x) + 1)
= theta[2] * (theta[1] + 1)
Therefore, we have
theta[2] = q'(x)*theta[2] + q(x)*(theta[1]+1)*(theta[2]) .
Since theta[2] is "transcendental over Q(x,theta[1])", we can equate coefficients of theta[2] on each side which gives
1 = q'(x) + (theta[1]+1)*q(x) .
Also, since theta[1] is "transcendental over Q(x)", equating
coefficients of this equation as polynomials in theta[1] gives
1 = q'(x) + q(x)
0 = q(x)
which has no solution.
(The second equation tells us that q(x) = 0 and therefore q'(x) = 0, which makes the first equation become 1 = 0, which is a contradiction.)
CONCLUSION: (integral x^x dx) does not exist as an elementary function.
--------------------------------------------------------------------------------
Thanks a lot for your help
I understand the reasoning of the professor above. We cannot calculate the integral of x^x. But, I must say that i don't know what is the link to 0^0 on it?
The function x^x still exists. If u plot it, u will see that, when x tends to 0, x^x tends to 1.
I'm also sorry, I searched on Internet (google, etc...). And I have only found the impact of 0^0 through the example of 'series'.
Perhaps, I should send an email to one of my former teachers
I like also the point of view of Yezz.
About the useness of this discussion, at least and a bit selfish , I simply say me... On GDAY 0^0 is not definied as 1/x ! Got it Fig
Last edited by Fig on 12 Sep 2006, 12:35, edited 1 time in total.
CEO
Joined: 20 Nov 2005
Posts: 2910
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 19
Kudos [?]: 132 [0], given: 0
Thank god. This ended constructively.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5
Kudos [?]: 124 [0], given: 39
I LOVE HAPPY ENDINGS DONT YOU GUYS??
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8
Kudos [?]: 102 [0], given: 0
Yes
Finally, all can fall on u Yezz What a question do u bring us here ? Subject of a such long discussion here Loosing time and energy
More seriously, It's a good one Thx
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5
Kudos [?]: 124 [0], given: 39
Ride on mate glad u enjoyed it
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# Equivalent of Rs `aov(Y ~ B + Error(A/B), data=d)`
Is there a julia equivalent of Rโs way to compute classical anovas with random effects?
For example, is there an easy way to obtain the following result without calling R?
``````julia> using DataFrames, RCall
julia> df = DataFrame(group = [fill("A", 24); fill("B", 24); fill("C", 24)],
order = repeat([0, 0, 1, 1], 18),
id = vcat([repeat(repeat(4i+1:4(i+1), inner = 2), 3) for i in 0:2]...),
trial = repeat(vcat([fill(i, 8) for i in 1:3]...), 3),
treatment = repeat(["X", "Y"], 36),
data = rand(72)
)
72ร6 DataFrame
Row โ group order id trial treatment data
โ String Int64 Int64 Int64 String Float64
โโโโโโผโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
1 โ A 0 1 1 X 0.207635
2 โ A 0 1 1 Y 0.376889
3 โ A 1 2 1 X 0.172655
4 โ A 1 2 1 Y 0.711968
5 โ A 0 3 1 X 0.451885
6 โ A 0 3 1 Y 0.344559
7 โ A 1 4 1 X 0.909206
8 โ A 1 4 1 Y 0.0705808
9 โ A 0 1 2 X 0.924089
10 โ A 0 1 2 Y 0.207873
11 โ A 1 2 2 X 0.312672
โฎ โ โฎ โฎ โฎ โฎ โฎ โฎ
63 โ C 1 12 2 X 0.0685433
64 โ C 1 12 2 Y 0.732906
65 โ C 0 9 3 X 0.492607
66 โ C 0 9 3 Y 0.851092
67 โ C 1 10 3 X 0.162958
68 โ C 1 10 3 Y 0.190779
69 โ C 0 11 3 X 0.588648
70 โ C 0 11 3 Y 0.911885
71 โ C 1 12 3 X 0.169196
72 โ C 1 12 3 Y 0.87193
51 rows omitted
julia> categorical!(df, [:group, :order, :id, :trial, :treatment])
72ร6 DataFrame
Row โ group order id trial treatment data
โ Catโฆ Catโฆ Catโฆ Catโฆ Catโฆ Float64
โโโโโโผโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
1 โ A 0 1 1 X 0.207635
2 โ A 0 1 1 Y 0.376889
3 โ A 1 2 1 X 0.172655
4 โ A 1 2 1 Y 0.711968
5 โ A 0 3 1 X 0.451885
6 โ A 0 3 1 Y 0.344559
7 โ A 1 4 1 X 0.909206
8 โ A 1 4 1 Y 0.0705808
9 โ A 0 1 2 X 0.924089
10 โ A 0 1 2 Y 0.207873
11 โ A 1 2 2 X 0.312672
โฎ โ โฎ โฎ โฎ โฎ โฎ โฎ
63 โ C 1 12 2 X 0.0685433
64 โ C 1 12 2 Y 0.732906
65 โ C 0 9 3 X 0.492607
66 โ C 0 9 3 Y 0.851092
67 โ C 1 10 3 X 0.162958
68 โ C 1 10 3 Y 0.190779
69 โ C 0 11 3 X 0.588648
70 โ C 0 11 3 Y 0.911885
71 โ C 1 12 3 X 0.169196
72 โ C 1 12 3 Y 0.87193
51 rows omitted
julia> @rput df
72ร6 DataFrame
Row โ group order id trial treatment data
โ Catโฆ Catโฆ Catโฆ Catโฆ Catโฆ Float64
โโโโโโผโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
1 โ A 0 1 1 X 0.207635
2 โ A 0 1 1 Y 0.376889
3 โ A 1 2 1 X 0.172655
4 โ A 1 2 1 Y 0.711968
5 โ A 0 3 1 X 0.451885
6 โ A 0 3 1 Y 0.344559
7 โ A 1 4 1 X 0.909206
8 โ A 1 4 1 Y 0.0705808
9 โ A 0 1 2 X 0.924089
10 โ A 0 1 2 Y 0.207873
11 โ A 1 2 2 X 0.312672
โฎ โ โฎ โฎ โฎ โฎ โฎ โฎ
63 โ C 1 12 2 X 0.0685433
64 โ C 1 12 2 Y 0.732906
65 โ C 0 9 3 X 0.492607
66 โ C 0 9 3 Y 0.851092
67 โ C 1 10 3 X 0.162958
68 โ C 1 10 3 Y 0.190779
69 โ C 0 11 3 X 0.588648
70 โ C 0 11 3 Y 0.911885
71 โ C 1 12 3 X 0.169196
72 โ C 1 12 3 Y 0.87193
51 rows omitted
julia> R"""
summary(aov(data ~ group*trial*treatment + Error(id/(trial*treatment)), df))
"""
RObject{VecSxp}
Error: id
Df Sum Sq Mean Sq F value Pr(>F)
group 2 0.0273 0.01363 0.312 0.74
Residuals 9 0.3932 0.04368
Error: id:trial
Df Sum Sq Mean Sq F value Pr(>F)
trial 2 0.0944 0.04718 0.373 0.694
group:trial 4 0.0733 0.01833 0.145 0.963
Residuals 18 2.2776 0.12653
Error: id:treatment
Df Sum Sq Mean Sq F value Pr(>F)
treatment 1 0.0744 0.07444 1.776 0.2154
group:treatment 2 0.4652 0.23259 5.549 0.0269 *
Residuals 9 0.3773 0.04192
---
Signif. codes: 0 โ***โ 0.001 โ**โ 0.01 โ*โ 0.05 โ.โ 0.1 โ โ 1
Error: id:trial:treatment
Df Sum Sq Mean Sq F value Pr(>F)
trial:treatment 2 0.0986 0.04932 0.555 0.584
group:trial:treatment 4 0.7584 0.18960 2.134 0.118
Residuals 18 1.5995 0.08886
``````
Itโs a bit out of date but you could have a look as to which of the packages mentioned in the summary towards the end of the thread are still actively maintained & fit your use case.
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Thread: Total Differentiation with Nested, Unspecified Composite Functions
1. Total Differentiation with Nested, Unspecified Composite Functions
I have no idea if the thread title accurately explains the question I'm having, but it is the best I could do. Googled and searched the forum and couldn't find what I was looking for.
Basically, I'm familiar with total differentiation. For instance I'm fairly certain that for a function
$U=u(C(R),E(R))$,
the total derivative is
${dU \over dR} = U_CC_R + U_EE_R$.
But what if the composite functions are functions of many variables. For instance,
$U=u(C(Y,R_D),E(R_D,R_F))$.
My first attempt for the total differential w.r.t. $R_D$:
${dU \over dR} = U_CC_Y + U_CC_{R_D}+U_EE_{R_D}+U_EE_{R_F}$
My second attempt:
${dU \over dR} = U_CC_YY_{R_D} + U_CC_{R_D}+U_EE_{R_D}+U_EE_{R_F}{R_F}_{R_D}$
Are either correct? Any help is appreciated!
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# M & d fractions
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### M & d fractions
1. 1. Multiplying and Dividng Fractions <ul><li>Reciprocal </li></ul><ul><li>Multiplying Fractions </li></ul><ul><li>Dividing Fractions </li></ul>QUIZ: Multiplying Fractions QUIZ: Dividing Fractions
2. 2. Multiplying Fractions <ul><li>Change any mixed numbers into improper fractions. </li></ul><ul><li>Change any whole numbers into improper fractions by putting 1 under the whole number. </li></ul><ul><li>Cross cancel wherever possible. </li></ul><ul><li>Multiply horizontally. </li></ul><ul><li>Reduce if necessary. </li></ul>EXAMPLES
3. 3. Reciprocal To find the reciprocal of a fractions, flip it! You need the reciprocal when dividing fractions! EXAMPLES: 2 3 3 2 2½ = 5 2 2 5
4. 4. Dividing Fractions <ul><li>Change any mixed numbers into improper fractions. </li></ul><ul><li>Change any whole numbers into improper fractions by putting 1 under the whole number. </li></ul><ul><li>KEEP – SWITCH - FLIP </li></ul><ul><li>Cross cancel wherever possible. </li></ul><ul><li>Multiply horizontally. </li></ul><ul><li>Reduce if necessary. </li></ul>EXAMPLES
5. 5. Examples of multiplying fractions 1 1 2 3 1 2 1 3 X <ul><li>X 2 = 2 </li></ul><ul><li>1 X 3 = 3 </li></ul>Another Example 3 8 4 9 X 3 8 4 9 X
6. 6. Examples of multiplying fractions 2 ½ X ¼ 5 1 2 4 X 5 8
7. 7. Examples of dividing fractions 5 10 9 12 KEEP SWITCH FLIP 1 3 4 2 1 4 3 2 x 4 6 2 3 Another Example 5 12 9 10 X
8. 8. Examples of dividing fractions 3 ½ 2 ¼ 7 4 2 9 x 28 18 KEEP SWITCH FLIP 7 9 2 4 1 5 9 1 10 18
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Cody
# Problem 722. Make a run-length companion vector
Solution 175926
Submitted on 11 Dec 2012
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
%% x = [5 3 3 1 0 9 9 4 4 4 4 5 1 2 2]; r_correct = [1 1 2 1 1 1 2 1 2 3 4 1 1 1 2]; assert(isequal(run_length(x),r_correct))
Error: Too many input arguments.
2 Fail
%% x = ones(1,20); r_correct = 1:20; assert(isequal(run_length(x),r_correct))
Error: Too many input arguments.
3 Fail
%% x = [1 1 1 2 2 3 4 4 5 5 5]; r_correct = [1 2 3 1 2 1 1 2 1 2 3]; assert(isequal(run_length(x),r_correct))
Error: Too many input arguments.
4 Fail
%% x = 1:40; r_correct = ones(size(x)); assert(isequal(run_length(x),r_correct))
Error: Too many input arguments.
5 Fail
%% x = [-34 -17*ones(1,100)]; r_correct = [1 1:100]; assert(isequal(run_length(x),r_correct))
Error: Too many input arguments.
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# math
posted by on .
The heights of the five starters of a college basketball team are 6'6, 6'7, 6'9, 6'11 and 7'. What it the mean height of these players , in inches?
• math - ,
First, convert all the heights into inches. Add them all up. Then divide the sum by 5, which is how many heights there are. You then have your mean.
### Related Questions
More Related Questions
Post a New Question
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https://www.airmilescalculator.com/distance/csk-to-oxb/
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Distance between Cap Skirring (CSK) and Bissau (OXB)
Flight distance from Cap Skirring to Bissau (Cap Skirring Airport – Osvaldo Vieira International Airport) is 82 miles / 132 kilometers / 71 nautical miles. Estimated flight time is 39 minutes.
Driving distance from Cap Skirring (CSK) to Bissau (OXB) is 134 miles / 215 kilometers and travel time by car is about 3 hours 16 minutes.
Map of flight path and driving directions from Cap Skirring to Bissau.
Shortest flight path between Cap Skirring Airport (CSK) and Osvaldo Vieira International Airport (OXB).
How far is Bissau from Cap Skirring?
There are several ways to calculate distances between Cap Skirring and Bissau. Here are two common methods:
Vincenty's formula (applied above)
• 81.935 miles
• 131.861 kilometers
• 71.199 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 81.930 miles
• 131.853 kilometers
• 71.195 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Cap Skirring Airport
City: Cap Skirring
Country: Senegal
IATA Code: CSK
ICAO Code: GOGS
Coordinates: 12°24′36″N, 16°44′45″W
B Osvaldo Vieira International Airport
City: Bissau
Country: Guinea-Bissau
IATA Code: OXB
ICAO Code: GGOV
Coordinates: 11°53′41″N, 15°39′13″W
Time difference and current local times
There is no time difference between Cap Skirring and Bissau.
GMT
GMT
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 37 kg (82 pounds).
Frequent Flyer Miles Calculator
Cap Skirring (CSK) → Bissau (OXB).
Distance:
82
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
82
Round trip?
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http://www.ck12.org/physics/Motion/lesson/Motion-Middle-School/r24/
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Motion ( Read ) | Physics | CK-12 Foundation
You are viewing an older version of this Concept. Go to the latest version.
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Practice Motion
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# Motion
The wings of this hummingbird are moving so fast that they’re just a blur of motion. You can probably think of many other examples of things in motion. If you can’t, just look around you. It’s likely that you’ll see something moving, and if nothing else, your eyes will be moving. So you know from experience what motion is. No doubt it seems like a fairly simple concept. However, when you read this article, you’ll find out that it’s not quite as simple as it seems.
### Defining Motion
In science, motion is defined as a change in position. An object’s position is its location. Besides the wings of the hummingbird in opening image, you can see other examples of motion in the Figure below . In each case, the position of something is changing.
Q : In each picture in the Figure above , what is moving and how is its position changing?
A : The train and all its passengers are speeding straight down a track to the next station. The man and his bike are racing along a curving highway. The geese are flying over their wetland environment. The meteor is shooting through the atmosphere toward Earth, burning up as it goes.
### Frame of Reference
There’s more to motion than objects simply changing position. You’ll see why when you consider the following example. Assume that the school bus pictured in the Figure below passes by you as you stand on the sidewalk. It’s obvious to you that the bus is moving, but what about to the children inside the bus? The bus isn’t moving relative to them, and if they look at the other children sitting on the bus, they won’t appear to be moving either. If the ride is really smooth, the children may only be able to tell that the bus is moving by looking out the window and seeing you and the trees whizzing by.
This example shows that how we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion. For the children on the bus, if they use other children riding the bus as their frame of reference, they do not appear to be moving. But if they use objects outside the bus as their frame of reference, they can tell they are moving. The video at the URL below illustrates other examples of how frame of reference is related to motion.
Q : What is your frame of reference if you are standing on the sidewalk and see the bus go by? How can you tell that the bus is moving?
A : Your frame of reference might be the trees and other stationary objects across the street. As the bus goes by, it momentarily blocks your view of these objects, and this helps you detect the bus’ motion.
### Summary
• Motion is defined as a change of position.
• How we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion.
### Vocabulary
• frame of reference : Something that is not moving with respect to an observer that can be used to detect motion.
• motion : Change in position.
### Practice
Do the frame of reference activity at the following URL. Watch the introduction and then do the nine trials. Repeat any trial you answer incorrectly until you get the correct answer.
### Review
1. How is motion defined in science?
2. Describe an original example that shows how frame of reference influences the perception of motion.
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• ### Recent Forum Posts
#### 9 1 1
Rev. 1 3 : 1 8, Ivan Panin's translation,
Here is the wisdom, who hath understanding, let him
count the number of the beast, for
Scott Yesterday, 05:36 PM
#### Son of David
Matthew has the story of blind Bartimaeus also, but split into two stories and without mentioning "son of Timaeus Bartimaeus"
sylvius 09-26-2020, 04:05 AM
#### Son of David
the passage in Mark ends with:
καὶ ἀποκριθεὶς αὐτῷ
sylvius 09-25-2020, 10:56 AM
#### Son of David
"son of David"
one time more in Mark
Mark 12:35-37 in the King James Perversion:
And Jesus answered
sylvius 09-25-2020, 07:16 AM
#### 9 1 1
1 9 6 3 + 1 5 9 0 + 3 1 = 3 5 8 4
1 9 6 3 + 1 5 9 0 digits after the decimal point, leaves
3 1 digits to reach the first
Scott 09-25-2020, 06:32 AM
#### Son of David
I found this:
https://cage.ugent.be/~hs/polyhedra/dodeicos.html
So it appears that the edge of the icosahedron
sylvius 09-25-2020, 03:40 AM
#### Son of David
Hi,
hey that's weird
Son of David, Hebrew "ben-david" has gematria 66, like of RAM's wheel , "galgal"
sylvius 09-25-2020, 03:04 AM
#### 9 1 1
13 + 18 = 31
In Pi, the 31st occurrence of 31
completes at digit # 3 5 8 5
1 5 9 0 + 1 9 9 5 = 3 5 8 5
Scott 09-23-2020, 06:13 PM
• ## Charisma
There is no available content written by Charisma
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# The Azimuth Project delete 75 (Rev #6, changes)
Showing changes from revision #5 to #6: Added | Removed | Changed
This page is a blog article in progress, written by David Tanzer.
Please remember that blog articles need HTML, not Markdown.
guest post by David Tanzer</i>
## Lake Petri Net
Hi this is Rick from the Portland Gear Works, checking back in with you. It’s now time to brew that pot of coffee I mentioned last time (link). I’ve thumbed through the catalog of their articles, and chosen one. I don’t really want to state the title here, because what’s the point of starting out with some tall words, when I can tell you right now that it has to do with counting the number of balls in those Pachinko machines I mentioned called Petri nets. Well, now since you’re curious, the paper is The Large-Number Limit for Reaction Networks (part 1), by John Baez. But we won’t get to it in this post, because there are some basics that we need to know about first.
Now I didn’t get get it quite right when I called those Petri networks “Pachinko machines.” Instead of balls, they call the things that move around the network “tokens.” Also, more to the point, it’s not one big machine, but a network built up by connecting together lots of little machines. And the tokens aren’t necessarily part of a game. Each token represents something in the world. Imagine they have different colors, with each color for a different kind of thing. Suppose we have snakes, frogs and butterflies, and these are represented by tokens that are white, green and yellow, respectively.
Now the machines represent “processes” in the world, which eat take up in some tokens at their input, inputs, and then send cough out some kind of tokens at their output. outputs. Suppose You could imagine, for example example, there was a volcano magic called saucer Thor that with floats arms around and the legs, lake, which had tries a very specific kind of appetite. For every meal it needs to eat attract exactly one snake and one frog. So When it walks around the bottom of the lake, catches these one food of items, each, then presto, a transformation occurs: they shake hands, and stuffs are them immediately converted into the opening at the top. Then after cooking and digesting, this food is transformed into seven butterflies, butterflies. which As fly a out mnemonic of for all this, let’s call the volcano. saucer To Preston. summarize:
Thor: Each 1 one Snake of + these 1 “meals” Frog is –> represented 7 by Butterfly the following formula:
Now Preston: what 1 would Snake happen + if 1 our Frog lake –> started 7 out Butterfly with 507 snakes, 379 frogs, and 27 butterflies. After each meal, the volcano will have depleted the snake and frog population some more, and the butterfly population will be soaring. Sadly, a population imbalance would develop. To make things more fair, let’s suppose that Thor had a brother and a sister, each of which had a different eating habit.
Now what would happen if our lake started out with 507 snakes, 379 frogs, and 27 butterflies. After each meal, Preston will deplete the snake and frog population some more, and the butterfly population will be soaring. Sadly, a population imbalance would develop. To make things more fair, let’s suppose that he had a brother and a sister saucer, each of which had a different eating habit:
Minerva: 5 Butterfly –> 1 Frog
Evan: 5 Butterfly –> 1 Snake
Now Thor Preston will be in cooperation with both Evan and Minerva, because he will supply them with butterflies and they will supply him with frogs and snakes. On the other hand, the Minerva brother and the Evan sister will be competing for butterflies.
Now we can ask how the three populations will change over time, with all of the volcanoes working at the same time. This is a matter of “population dynamics.”
Well the answer depends on the feeding rates of each of the volcanoes. saucers. Suppose that Evan takes ten years between meals, whereas Minerva and Thor Preston take only ten seconds. Then Thor Preston will quickly deplete all of the snakes, and then run out of food, and so he will grind to a halt. Whatever butterflies are there will quickly be eaten used up by the Minerva, sister, who will also grind to a halt. There will only be frogs. When Then, when Evan is ready to eat again, all the food will be gone, and so the system will be stuck in the state of Eternal Frogs.
To make it more realistic, we should consider that the speed with which the volcanoes saucers move go from one meal to the next will depend upon how much of their specific nutrients are present in the lake. After all, if there are hardly any frogs present, then Thor Preston will have to spend a lot of time wading floating around in looking search for of them, and so there will be long search times between the meals. So the reaction rate of a saucer is a function of the input supplies. And it is an increasing function the more inputs present, the faster it goes.
So Now I will explain the standard model for how the feeding rate of depends a volcano is a function of the input food supplies. And it is an increasing function the more food present, the faster it eats. Depending on what function you choose for the feeding amount rate, of you food. will Suppose get that different results for how the populations lake evolve. But there is one big standard and one, square, which and we the frogs and snakes are relatively few and far between. Suppose that Preston can explain attract any creatures that are within 50 feet. He moves by analyzing making the hops process of by 500 which feet, Thor searches for food in a random direction each time. After each hop, if there is both a frog and a snake within 50 feet, then the lake. magic reaction takes place.
Suppose that the lake is big and square, and the frogs and snakes are relatively few and far between. Suppose that he can reach any creatures that are within a square that is 50 feet across, centered at the spout. He breaks the whole lake into grid of many squares cells that are 50 feet across, and he moves by jumping from one cell to another. Upon entering a cell, if that cell contains both a snake and a frog, then the meal takes place and the butterflies are generated, otherwise no meal takes place.
Suppose he makes 20 hops per hour, that half the cells contains snakes, and that half the cells contain frogs. Then, assuming that the snakes and the frogs are independently distributed across the lake, how many meals per hour would take place? Well, of the 20 cells that are visited per hour, on the average, only one quarter of them will have both a frog and a snake, and so there will be an average feeding rate of 5 meals per hour.
So the feeding rate of a volcano is proportional to the product of the numbers of its input species that are present.
What if Thor were to change his diet from Snake + Frog to Snake + Snake? Then he could only have a meal if there were two snakes found in a cell. Well if the odds of finding one snake in a cell are 50%, then the odds of finding two snakes will be 25%. So the feeding rate would still be 5 meals per hour. This example shows us that if a volcano eats two instances of a species, then its feeding rate will be proportional to the square of the number of tokens that are present for that species. You can generalize this to the case where it eats more than two instances of a species.
In closing, I’d like to point out that the terminology we have used here, which makes things pretty clear, is unfortunately not standard. Let’s see how to connect it with the terms that are used by “professionals” – which by the way are much less emotional and vivid.
First, the volcanoes are called processes, reactions, or transitions. Second, each meal of a volcano is called a firing event of the transition. The feeding rate of a volcano is referred to as the firing rate of the transition. The transitions could actually represent processes that have nothing to do with our beloved lake, like the splitting apart of a molecule into atoms. There the tokens would represent different types of chemical entities, and the transitions correspond to different kinds of chemical reactions. So a Petri net is really closely related to the idea of a reaction network.
One closing comment on the name “Petri net.” The name actually comes from the inventor of it, but a useful mnemonic is to visualize Lake Petri Net growing, like a microcosm, inside of a Petri dish.
TODO:
• describe rate coefficient
• clarify 50% odds versus 50% of the cells have a frog
category: blog
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Twenty Divided Into Six
Katie had a pack of twenty cards numbered from $1$ to $20$.
She arranged the cards into six unequal piles.
The numbers on the cards in each pile added to the same total.
What was the total and how could this be done?
Why do this problem?
This problem is one that can be accessed easily - everyone can make a start. It offers opportunities for learners to practise addition and subtraction, along with some multiplication and division, and requires a systematic approach.
Possible approach
You could start by asking the group to work on the problem in pairs with digit cards numbered from $1$ to $20$ without saying very much else at this stage. (You could print and cut out cards from this sheet if you do not have enough to go round.) Learners might find it useful to make jottings on mini-whiteboards or paper as they explore the problem.
After some time, draw the class together to find out how they are getting on. Invite some pairs to share their approach so far with the whole group. Some children might be using trial and improvement, some may have worked out what the total of each pile needs to be and then used trial and improvement. You may need to talk about how they work out the total of each pile if this does not come up naturally. Can they think of a quick way of doing it without a calculator?
They could then continue to work in pairs on the problem. After the initial calculations the problem is a fairly simple one of adding and building the piles but there are many ways of doing it. It would be interesting and instructive to listen to the way that the various pairs are working on the problem, and you may like to gather solutions as a whole class on the board.
In a plenary at the end of the lesson, you could talk about how they have found the different solutions and you may want to ask whether they think they have got them all. This might be a good opportunity to share ways of working systematically so that they could convince you they would be able to find every solution.
Key questions
What number must all the piles add to?
What is the total of each pile?
How do you know you have all the solutions?
Possible extension
Learners could find as many completely different solutions to this problem as possible and some children will be able to suggest a way to find them all.
Possible support
If you want to focus on finding all possibilities, some learners might benefit from using a calculator so they are not held up by the mental arithmetic.
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Which of the following is equal to 4 kilograms? A. 4,000 g B. 4,000 mg C. 4,000 cg D. 40 dag
Updated 10/11/2018 1:07:00 PM
s
Original conversation
User: Which of the following is equal to 4 kilograms? A. 4,000 g B. 4,000 mg C. 4,000 cg D. 40 dag
Weegy: 4,000 g is equal to 4 kilograms.
User: 8 gallons equals how many pints?
Weegy: There are 8 pints in a gallon, therefore 8 gallons is equal to 8 x 8 = 64 pints.
User: 8 gallons equals how many pints?
Weegy: There are 8 pints in a gallon, therefore 8 gallons is equal to 8 x 8 = 64 pints.
User: How many cubic feet are in 3 cubic yards?
Weegy: There are 81 cubic feet in 3 cubic yards.
User: If a sports shop sold 1,500 dozen golf balls last week, how many individual golf balls did the shop sell?
Updated 10/11/2018 1:07:00 PM
Rating
8
If a sports shop sold 1,500 dozen golf balls last week, the shop sold 18,000 individual golf balls.
1,500 x 12 = 18,000.
What would you have to do to change 10 cubic feet into cubic inches
Weegy: You have to multiply by 1728 to change 10 cubic feet into cubic inches. 1 cubic ft. = 1728 cu.inches. 10 cubic ft. = 1728 *10 cu.inches. 10 cubic ft. = 17280 cu.inches. User: How many yards are in 7 miles? Weegy: 1 miles = 1760 yards , 7 miles = 7 *1760 =12320 yards. User: 20,000 pounds equals how many English tons? Weegy: 20,000 pounds is equals 10 tons. User: What's the conversion factor used to convert miles to yards? Weegy: 1 mile = 1760 yards is the conversion factor used to convert miles to yards. (More)
What would you have to do to change 10 cubic feet into cubic inches
Weegy: You have to multiply by 1728 to change 10 cubic feet into cubic inches. 1 cubic ft. = 1728 cu.inches. 10 cubic ft. = 1728 *10 cu.inches. 10 cubic ft. = 17280 cu.inches. You have to multiply by 1728 to change 10 cubic feet into cubic inches. (More)
What's the volume of a block of ice measuring 3 meters long, 1.5 meters wide, and 2 meters high?
Weegy: The volume of a block of ice measuring 3 meters long, 1.5 meters wide, and 2 meters high, Volume= l*b*h = 3*1.5 *2 =9 cu.mt. (More)
Find the missing number 1:2 = 3: ____
Weegy: The missing number 1:2 = 3:6 Let the missing number be x. 1/2 = 3/x ; x = 3*2 = 6 (More)
Updated 10/11/2018 9:02:52 PM
A car recently sold for \$32,345. If there's a 6% sales tax on automobile sales, how much tax will be added to the price of the car?
Weegy: A car recently sold for \$32,345. If there's a 6% sales tax on automobile sales, \$1,940.70 will be added to the price of the car. x = \$32,345 * 6/100 , x = \$32,345 * 0.06 ,x = \$1,940.70 User: What is 6:12 in simplest form? Weegy: 6:12 in simplest form is 1:2. User: Paul can type 60 words per minute and Jennifer can type 80 words per minute. How does Paul's typing speed compare to Jennifer's? Weegy: Steve can complete the 100m dash in 10 seconds while Paul can run it in 12 seconds. Steve's time compare to Paul's. Steve / paul = 10/12 = 5/6. Steve is 5/6 as fast as Paul. (More)
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Weegy: -7 + N = 20 User: y - 12 = -10 Weegy: x + 5 = 2x User: y - 12 = -10 Weegy: x + 5 = 2x User: y + 1.05 = ...
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## Formula for calculating stock turnover rate
25 Jul 2019 Here's the formula to calculate the AI. │AI = (Beginning Inventory + Ending Inventory) ÷ 2. Average Inventory. When you know the values of
The inventory turnover formula in 3 simple steps. Inventory turnover is a ratio that measures the number of times inventory is sold or consumed in a given time period. Also known as inventory turns, stock turn, and stock turnover, the inventory turnover formula is calculated by dividing the cost of goods sold (COGS) by average inventory. Then, we calculate Inventory Turnover Ratio using Formula. Inventory Turnover Ratio = Cost of Goods Sold/ Average Inventory; Inventory Turnover Ratio = \$1,000,000 / \$3500000; Inventory Turnover Ratio = 0.29 ; As you can see Luxurious Furniture Company turnover is .29. Calculate turnover for your firm. Most businesses calculate the turnover rate at least annually. You can calculate the rate for a shorter period of time, such as a fiscal quarter (3 months). Assume that your total number of workers is 1,000 on January 1st. By December 31 of the same year, the total is 1,200. Inventory Turnover Formula Inventory Turnover = Cost of Goods Sold / Average Inventory for the Period To get an annual number, start with the total cost of goods sold for the fiscal year, then divide that by the average inventory for the same time period. Formula The inventory turnover ratio is calculated by dividing the cost of goods sold for a period by the average inventory for that period. Average inventory is used instead of ending inventory because many companies’ merchandise fluctuates greatly throughout the year.
## In this example, we define new hire turnover rate as the number of new employees who leave within a year. Your new hire turnover formula would look like this: A healthy turnover rate. Now that know how to calculate employee turnover rate using a basic formula, you can calculate your company’s turnover and come up with a number.
Here we will do the same example of the Inventory Turnover Ratio formula in Excel. It is very easy and simple. You need to provide the two inputs i.e Average Inventories and Cost of goods sold. You can easily calculate the Inventory Turnover Ratio using Formula in the template provided. In the first Example, First, we calculate Average Inventories the formula for calculating employee turnover rate Employee turnover is usually expressed as a turnover rate. In other words, how to calculate turnover rate is basically just percentage math. The formula for the inventory turnover ratio measures how well a company is turning their inventory into sales. The costs associated with retaining excess inventory and not producing sales can be burdensome. If the inventory turnover ratio is too low, a company may look at their inventory to appropriate cost cutting. How To Calculate Employee Turnover Rate Turnover Definition: Turnover is the number of employees that have to be replaced in a given period of time. Turnover rate is that value expressed as a percentage. Turnover Formula: (# of separations / average # of employees) x 100 = turnover rate These two parts are called the Net Turnover Rate, and the True Turnover Rate. The Net Turnover Rate: is the rate that is calculated and includes all of the staff who have left employment of the company. The reasoning behind them leaving is not taken into consideration and all numbers are taken into a count. The turnover rate formula is (Employee separations for the period) / (Average number of employees during the period). Some businesses use the word “termination” instead of separation. Both terms refer to a worker leaving the company. How to Calculate Inventory Turnover - Finding the Inventory Turnover Ratio Choose a time period for your calculation. Find your cost of goods sold for the time period. Divide your COGS by your average inventory. Use the …
### Inventory Turnover Formula. To calculate inventory turnover, divide the ending inventory figure into the annualized cost of sales. If the ending inventory figure is not a representative number, then use an average figure instead, such as the average of the beginning and ending inventory balances.
Formula: Inventory Turnover Ratio. Inventory turnover is calculated using following formula: Inventory turnover = Cost of goods sold / Average inventory
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https://printablesudokufree.com/printable-samurai-sudoku-medium/double-harakiri-sudoku-x-printable-samurai-sudoku-medium/
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# Double Harakiri Sudoku X | Printable Samurai Sudoku Medium
Double Harakiri Sudoku X | Printable Samurai Sudoku Medium
Printable Samurai Sudoku Medium – printable samurai sudoku medium, An entertaining activity that workouts the human brain and offers you feelings of fulfillment – that is an ideal outline from the amounts puzzle trend from Japan referred to as sudoku. This brilliant head activity is exciting to try out, and everyone can get it done. Taking part in may even support wait dementia by instructing your mind to believe in a new way.
## Printable Samurai Sudoku Medium
You have almost certainly noticed a sudoku within your neighborhood local newspaper. It is a puzzle that exercise routines the brain by permitting one to feel rationally concerning how to position phone numbers in containers in just a grid. The purpose of sudoku will not be to replicate any phone numbers whilst you are satisfying inside the grid. The problem is finding out which quantity suits which box. The sudoku grid has 9 cases; inside of every box are smaller sized containers which have 9 pieces each and every. A few of those squares will currently have amounts inside them.
### What exactly is a Sudoku?
Sudoku is actually a plausible puzzle video game, initially made in puzzle textbooks then offered in a great number of papers globally. Lots of people are postpone by visiting a grid with amounts inside, however the puzzle is not going to demand any arithmetic in any way – just deduction and reasoning.
Sudoku is a straightforward to understand reasoning-structured quantity position puzzle. The phrase Sudoku is quick for Su-ji wa dokushin ni kagiru meaning “the phone numbers has to be solitary”.
The beginnings of your Sudoku puzzle will be in the Swiss. Leonhard Euler produced “carré latin” from the 18h century which is just like a Sudoku puzzle but with no extra constraint about the belongings in specific territories. The 1st genuine Sudoku was posted in 1979 and was introduced by Howard Garns, an United States designer. Reality broad recognition were only available in Japan in 1986 soon after it absolutely was printed and because of the brand Sudoku by Nikoli.
Rules and Terms : A Sudoku puzzle is made up of 81 tissues that are split into 9 posts, lines and territories. The job is already to position the amounts from 1 to 9 in to the unfilled cellular material in such a manner that in each and every row, line and 3×3 place every amount presents itself just once. A Sudoku has at the very least 17 provided figures but usually you will find 22 to 30. Printable Samurai Sudoku Medium
## Printable Samurai Sudoku Medium
Sudoku Printable Grids – Canas.bergdorfbib.co | Printable Samurai Sudoku Medium
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Printable Sudoku Samurai! Give These Puzzles A Try, And You&#039;ll Be | Printable Samurai Sudoku Medium
Printable Sudoku | Printable Samurai Sudoku Medium
Sudoku Puzzles | Document Sample | Puzzles | Sudoku Puzzles, Puzzle | Printable Samurai Sudoku Medium
Printable Sudoku | Printable Samurai Sudoku Medium
Super Samurai Sudoku 13 Grids | Printable Samurai Sudoku Medium
Double Harakiri Sudoku X | Printable Samurai Sudoku Medium Uploaded by Lamont N. Nelson on Monday, May 6th, 2019 in category Printable Sudoku Free.
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