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# Category Archives: qbasic
## 03 QBasic Tutorial Decisions and loops
Learning Objectives
THIS SECTION IS BEING UPDATED… NOV 2015
At the end of this tutorial you will know:
How to use Loops eg conditional (pre-check, post-check), fixed to repeat actions
How to use conditional statements and logical operators to make decisions
Selection: Where you choose to do something or not!
If there is something you might want to do depending on a situation or condition use IF/ENDIF.
If there are one out of two actions you might want to do depending on a situation or condition us IF/ELSE/ENDIF
```IF condition1 THEN
[statementblock-1]
END IF```
```IF condition1 THEN
[statementblock-1]
[ELSE]
[statementblock-2]]
END IF```
```IF condition1 THEN
[statementblock-1]
[ELSEIF condition2 THEN
[statementblock-2]]...
[ELSE]
[statementblock-n]]
END IF```
condition1 Any expression that can be evaluated as
condition2 true (nonzero) or false (zero).
statementblock – One or more statements on one or more lines of basic.
Conditions
Examples
```DIM strName as STRING
CLS
PRINT "Who is a clever clogs? ";
INPUT strName
IF strName <> “Michael” THEN
PRINT strName “ is a thickie”
ENDIF
PRINT “Michael is clever”
END```
Test the program with this data
Input: Michael
Expected output:Michael is clever
Input: Peter Bob Bill Michael MICHAEL, michael, M TOM DAVE
Expected output: is a thickie ( Peter, Bob, Bill, Michael, MICHAEL, etc) Michael is clever
Using ALL the test data above and change strName <> “Michael” to:
• strName <= “Michael” strName >= “Michael”
• strName > “Michael”
• strName < “Michael”
• strName = “Michael”
• strName = “michael”
• lcase\$(strName) = “michael”
• lcase\$(strName) = “Michael”
• ucase\$(strName) = “michael”
• ucase\$(strName) = “MICHAEL”
Make a note of what happened here, you will be asked in class what occurred.
Program Code – Type it in
DIM strName as STRING
CLS
PRINT “Who is a clever clogs? “;
INPUT strName
IF strName <> “Michael” THEN
PRINT strName “ is a thickie”
ELSE
PRINT “Michael is clever”
ENDIF
END
Test the program with this data
Input: Michael
Expected out put:
Michael is clever
Input: Peter Bob Bill Michael MICHAEL, michael, M TOM DAVE
Expected output:
is a thickie ( Peter, Bob, Bill, Michael, MICHAEL, etc)
Michael is clever
Using ALL the test data above and change strName <> “Michael” to:
strName <= “Michael” strName >= “Michael”
strName > “Michael”
strName < “Michael”
strName = “Michael”
strName = “michael”
lcase\$(strName) = “michael”
lcase\$(strName) = “Michael”
ucase\$(strName) = “michael”
ucase\$(strName) = “MICHAEL”
Make a note of what happened here, you will be asked in class what occurred.
CASE
Executes one of several statement blocks depending on the value of an
expression.
SELECT CASE testexpression
CASE expressionlist1
[statementblock-1]
[CASE expressionlist2
[statementblock-2]]…
[CASE ELSE
[statementblock-n]]
END SELECT
■ testexpression Any numeric or string expression.
■ expressionlist1 One or more expressions to match testexpression.
■ expressionlist2 The IS keyword must precede any relational operators in an expression.
■ statementblocks One or more basic statements on one or more lines
‘Author Michael Fabbro
‘Date 7 December 2003
DIM strCommand AS STRING
DIM intStartLine AS INTEGER
CLS
LET IntStartLine = 0
LOCATE 2, 25: Print “Concrete Calc Main Menu”
LOCATE intStartLine + 6, 15: Print “[B]lock”
LOCATE intStartLine + 8, 15: Print “[C]ylinder”
LOCATE intStartLine + 10, 15: Print “[P]yramid”
LOCATE intStartLine + 12, 15: Print “Enter B,C,P”
LOCATE intStartLine + 12, 30: INPUT strCommand
‘convert to uppercase
strCommand = UCase\$(strCommand)
Select Case strCommand
Case Is = “B”
PRINT “You want a block”
Case Is = “C”
PRINT “You want a Cylinder”
Case Is = “P”
PRINT “You want a Pyramid”
Case Else
PRINT “Don’t know that option”
PRINT “Try again”
End Select
END
Test the program with this data
B “You want a block”, C You want a Cylinder”, P “You want a Pyramid”
Z W “Don’t know that option”
Test Results
STRING FUNCTIONS
LCASE\$(stringexpression\$)
UCASE\$(stringexpression\$)
Convert strings to all lowercase or all uppercase letters.
■ stringexpression\$ Any string expression.
Example:
strPlace = “THE string”
PRINT strPlace
PRINT LCASE\$( strPlace); ” in lowercase”
PRINT UCASE\$( strPlace); ” IN UPPERCASE”
LEFT\$(stringexpression\$,n%)
RIGHT\$(stringexpression\$,n%)
Return a specified number of leftmost or rightmost characters in a string.
■ stringexpression\$ Any string expression.
■ n% The number of characters to return, beginning
with the leftmost or rightmost string character.
Example:
strText = “Microsoft QBasic”
PRINT LEFT\$( strText, 5) ‘Output is: Micro
PRINT RIGHT\$( strText, 5) ‘Output is: Basic
MID\$(stringexpression\$,start%[,length%])
MID\$(stringvariable\$,start%[,length%])=stringexpression\$
The MID\$ function returns part of a string (a substring).
The MID\$ statement replaces part of a string variable with another string.
■ stringexpression\$ The string from which the MID\$ function returns a substring, or the replacement string used by the MID\$ statement. It can be any string expression.
■ start% The position of the first character in the substring being returned or replaced.
■ length% The number of characters in the substring. If the length is omitted, MID\$ returns or replaces all characters to the right of the start position.
■ stringvariable\$ The string variable being modified by the MID\$ statement.
Example:
strPlace = “Where is Paris?”
PRINT MID\$( strPlace, 10, 5) ‘Output is: Paris
strPlace = “Paris, France”
PRINT strPlace ‘Output is: Paris, France
MID\$( strPlace, 8) = “Texas ”
PRINT strPlace ‘Output is: Paris, Texas
Exercise 1
You know how to input numbers, calculate and print out the answer. Now add the case statement to this knowledge and write a program that will input two numbers and then a menu that will add, multiply, subtract or divide those number and print out the answer. You will need the following basic:
LET + – / *, DIM STRING INTEGER SINGLE, CASE, INPUT, PRINT, PRINT USING , CLS
Checking for errors in input
Avoiding negative numbers in calculations
Type this program in
DIM sngPi, sngRadius, sngArea as SINGLE
CLS
sngPi = 3.1415
INPUT “What is the radius of the circle? (-1 to end) “, sngRadius
sngArea = sngPi * sngRadius ^ 2
PRINT “The area of the circle is “, sngArea
PRINT
ENDIF
END
Test the program with this data
Output: The area of the circle is
Output: The area of the circle is 50.272
Error condition
Output: The area of the circle is
Output: The area of the circle is
Error condition
Test the program with this data
Test Results
Which is better?
Part 3 Loops/Iteration ‘Repeating things’ (Session 7-9)
There’s one big problem with the program. If you needed to try hundreds of radii, you must run the program over again. This is not practical. If we had some kind of a loop until we wanted to quit that just kept on repeating over and over it would be much more useful. Of course, QBasic has the means of performing this feat. Loop structures. They start with the statement DO, and end with the statement LOOP. You can LOOP UNTIL or WHILE , or DO UNTIL or WHILE a condition is true. Another option (which we will use) is to break out of the loop manually as soon as a condition is true. Lets revise the previous code:
Program Code – Type it in
DIM sngPi, sngRadius, sngArea as SINGLE
CLS
sngPi = 3.1415
DO ‘ Begin the loop here
INPUT “What is the radius of the circle? (-1 to end) “, sngRadius
sngArea = sngPi * sngRadius ^ 2
PRINT “The area of the circle is “, sngArea
PRINT
ENDIF
END
Test the program with this data
Output: The area of the circle is
Output: Program finishes
Output: The area of the circle is
Output: The area of the circle is
Test the program with this data
Test Results
Now we can end the program by entering -1 as the radius. The program checks the radius after the user inputs it and checks if it is -1. If it is, it exits the loop. If it isn’t it just keeps going it’s merry way. TRY changing
LOOP WHILE sngRadius <> -1 to
The Do statement repeats a block of statements depending on a boolean (true or false) condition. The condition is tested at the beginning if it appears on the Do or at the end of the loop if it appears on the Loop. The While keyword continues the loop while the condition is true. The Until keyword continues the loop while the condition is false (stops when the condition is true).
ANSWER THE FOLLOWING 3 QUESTION and you will know what type of loop you will need.
· If you want to repeat something and you know how many times you will do it before you start – used the FOR NEXT LOOP
· If you want to do something at least once and maybe repeat it again – use DO LOOP
· If you are not sure whether you want to do something nor how many times to repeat it – use the DO Loop
FOR loops
Use where you know how many times you want something to repeat.
The FOR statement repeats a block of statements for a specified number of times.
FOR variable=initialValue TO finalValue
Statements are repeated while the variable is at or between initial and final values. The variable is incremented by 1 each time.
NEXT variable
FOR variable=initialValue TO finalValue STEP
Statements are repeated while the variable is at or between initial and final values. The variable is incremented by increment each time.
NEXT variable
Description
Repeats a group of statements a specified number of times.
Syntax
For counter = start To end [Step step]
[statements]
[Exit For]
[statements]
Next
The For…Next statement syntax has these parts:
Part
Description
counter
Numeric variable used as a loop counter. The variable can’t be an array element or an element of a user-defined type.
start
Initial value of counter.
end
Final value of counter.
step
Optional: Amount counter is changed each time through the loop. If not specified, step defaults to one.
statements
One or more statements between For and Next that are executed the specified number of times.
DIM intNumOfTimes,intStart,intEnd AS INTEGER
DIM intIncremnt AS INTEGER
LET intStart = 1
LEt intEnd= 10
LET intIncremnt = 2
FOR intNumOfTimes = intStart TO intEnd STEP intIncremnt
Print intNumOfTimes
NEXT ntNumOfTimes
The step argument can be either positive or negative. The value of the step argument determines loop processing as follows:
Value
Loop executes if
Positive or 0
counter <= end Negative counter >= end
Once the loop starts and all statements in the loop have executed, step is added to counter. At this point, either the statements in the loop execute again (based on the same test that caused the loop to execute initially), or the loop is exited and execution continues with the statement following the Next statement.
Tip Changing the value of counter while inside a loop can make it more difficult to read and debug your code.
EXIT FOR
Exit For can only be used within a For Each…Next or For…Next control structure to provide an alternate way to exit. Any number of Exit For statements may be placed anywhere in the loop. Exit For is often used with the evaluation of some condition (for example, If…Then), and transfers control to the statement immediately following Next.
You can nest For…Next loops by placing one For…Next loop within another. Give each loop a unique variable name as its counter. The following construction is correct:
For I = 1 To 10
For J = 1 To 10
For K = 1 To 10
. . .
Next
Next
Next
DO LOOPS
Use this loop when your are not sure how many times you want repeat something, but you do know you want to do the action(s) at least once
Use a Do loop to execute a block of statements an indefinite number of times. There are several variations of the Do…Loop statement, but each evaluates a numeric condition to determine whether to continue execution. As with If…Then, the condition must be a value or expression that evaluates to False (zero) or to True (nonzero).
DO LOOP (Pre-test)
DO WHILE condition
The condition is tested at the beginning of every loop.
Statements are repeated while the condition is true
LOOP
DO UNTIL condition
The condition is tested at the beginning of every loop.
statements repeated while the condition is false
LOOP
Dim IntCount AS INTEGER
intCount = 1
DO WHILE intCount < 10
PRINT intCount
intCount = intCount + 1
Loop
END
Try the operators <=, < , > = and <> too
Dim IntCount AS INTEGER
intCount = 1
DO UNTIL intCount >= 10
PRINT intCount
intCount = intCount + 1
LOOP
END
Try the operators <=, < , > = and <> too
DO LOOP (post-test)
This variation of the Do…Loop statement executes the statements first and then tests condition after each execution. This variation guarantees at least one execution of statements:
DO
statements repeated while the condition is true
The condition is tested at the end of every loop.
LOOP WHILE condition
DO
statements repeated while the condition is false
The condition is tested at the end of every loop.
LOOP UNTIL condition
The loop can execute any number of times, as long as condition is nonzero or True.
Dim IntCount AS INTEGER
intCount = 1
DO
PRINT intCount
intCount = intCount + 1
LOOP WHILE intCount < 10
END
Try the operators <=, < , > = and <> too
This is another variation analogous to the previous one, except that they loop as long as condition is False rather than True.
Dim IntCount AS INTEGER
intCount = 1
DO
PRINT intCount
intCount = intCount + 1
LOOP UNTIL intCount >= 10
END
‘ Try the operators <=, < , > = and <> too
Exercises 2
Modify each of the above DO LOOPS so that the Statement block within the loop does the following:
A) Problem one
Input a number
when the loop finishes it prints out the total of all numbers input.
B) Problem 2
Input a number
Multiplies the number by itself (squared)
Adds the square to a total
When the loop finishes it prints out the total of all numbers input squared.
C) Problem 3
Modify the CASE statement menu program so that it is in a loop. Add an option so that the program quits the loop and finishes when “Q” is entered.
Example of a CGI program
This programs output is HTML, suitable for a web browser
Dim strAirport As String
Dim strTime As String
Dim strDestination As String
Cls
Do While 1 <> 2
RESTORE Start
INPUT strDestination
Print “” Print “” Print “Current flights” Print “” Print “”
Do Until UCase\$(strAirport) = “END”
If UCase\$(strAirport) = UCase\$(strDestination) Then
Print ”
Airport: “; strAirport; Tab(25); ” Time:
“; Tab(45); strTime
End If
Loop
Print ”
” Print “”
Loop
End
Start: Data “Rome”, “6:00”
Data “Rome”, “7:00”
Data “Rome”, “9:00”
Data “Rome”, “11:00”
Data “Rome”, “14:00”
Data “Rome”, “19:00”
Data “Rome”, “6:00”
Data “Paris”, “7:30”
Data “Paris”, “9:30”
Data “Paris”, “11:30”
Data “Paris”, “14:20”
Data “Paris”, “19:20”
Data “END”, “”
Download it here and type in Paris, then London and finally Rome
Testing
Black box and white-box are test design methods. Black-box test design treats the system as a “black-box”, so it does not explicitly use knowledge of the internal structure. Black-box test design is usually described as focusing on testing functional requirements.
Synonyms for Black-box include: behavioural, functional, opaque-box, and Closed-box. White-box test design allows one to peek inside the “box”, and it focuses specifically on using internal knowledge of the software to guide the selection of test data.
Synonyms for white-box include: structural, glass-box and clear-box.
While black box and white-box are terms that are still in popular use, many people prefer the terms “behavioural” and “structural”. Behavioural Test design is slightly different from black-box test design because the use of internal knowledge isn’t strictly forbidden, but it’s still discouraged. In practice, it hasn’t proven useful to use a single test design method. One has to use a mixture of different methods so that they aren’t hindered by the limitations of a particular one. Some call this “grey-box” or “translucent-box” test design.
It is important to understand that these methods are used during the test design phase, and their influence is hard to see in the tests once they’re implemented. Note that any level of testing (unit testing, system testing, etc.) can use any test design methods. Unit testing is usually associated with structural test design, but this is because testers usually don’t have well-defined requirements at the unit level to validate.
Black box testing
A software testing technique whereby the internal workings of the item being tested are not known by the tester. For example, in a black box test on a software design the tester only knows the inputs and what the expected outcomes should be and not how the program arrives at those outputs. The tester does not ever examine the programming code and does not need any further knowledge of the program other than its specifications.
The advantages of this type of testing include:
The test is unbiased because the designer and the tester are independent of each other.
The tester does not need knowledge of any specific programming languages.
The test is done from the point of view of the user, not the designer.
Test cases can be designed as soon as the specifications are complete.
The disadvantages of this type of testing include:
The test can be redundant if the software designer has already run a test case.
The test cases are difficult to design.
Testing every possible input stream is unrealistic because it would take a inordinate amount of time; therefore, many program paths will go untested.
White box testing
Also known as glass box, structural, clear box and open box testing. A software testing technique whereby explicit knowledge of the internal workings of the item being tested are used to select the test data. Unlike black box testing, white box testing uses specific knowledge of programming code to examine outputs. The test is accurate only if the tester knows what the program is supposed to do. He or she can then see if the program diverges from its intended goal. White box testing does not account for errors caused by omission, and all visible code must also be readable.
For a complete software examination, both white box and black box tests are required.
Source Webopedia
Test data
Test data should include:
1) Normal data, but avoid the same number for two different input values.
2) Extreme data
a. Extra high values
b. Low values
c. Negative
d. zero
3) Abnormal
a. Numbers where text strings expected
b. Text where numbers expected
c. No entry of data where data expected.
Some examples of Loops and selections
A history database, try these programs out with the following data:
First just press return then try: hitler, nero, bodica, capt kirk, moses, noah, dracula, dumbo
Look at the each of the programs and find out what each part does.
History 1 Uses IF
History 2 Uses CASE. This has some errors for you to correct first
Histroy 3 Final Version
Exercises 3 (Session 11-12) (suitable variable names, indention and comments)
Write and test the following programs using:
“PRINT USING” to format any answer.
LOCATE to position inputs and outputs.
a) Test data and expected results for a range of test values
b) Program listing include suitable variable names, indention and comments
c) Screen shots of the test results
a) Input and adds 3 Integer numbers together
b) Input and adds 3 floating point numbers together then divides by 3
c) Input and adds 3 floating numbers together then divides by 3 and then multiplies the answer by 10.
d) Calculate the circumference of a given circle.
e) Input hours worked (floating point) and manufacturing cost per hours (floating point) to calculate the total manufacturing cost of a product
f) Input 5 numbers and display the average.
g) Input hours worked and manufacturing cost per hours to calculate the total manufacturing cost for 5 products.
h) Input length and width then calculate the area of any rectangle in metre2.
i) Input length and width then calculate the area of any two rectangle in metre2.
j) A pool consists of two rectangles in metre2. One of the rectangles is the water-covered area. The other is the full size of pool area (Including water and paved edge. Subtract the total area for the water from the full pool size to produce the paved area. Print water and paved areas out.
· Paved area = Total area – water area
Using DO LOOPS
k) A program that prints out the 3 times table using a DO LOOP
l) A program that prints out the 3 times table using a FOR NEXT LOOP
m) Enter the names mary, bill, bob, jane, gill randomly. The program should say if the name is male or female
n) A program that add a series of number and stops when a negative number is entered. The program then displays the sum of the numbers. (Needs IF and DO LOOP)
o) A program that prints out the Times Table (up to 12 numbers in a column) for any number
Now do assignment 1 Week 12
## 05 SIPOS Design
SIPOS is a simple design process that allows you design a simple program and produce the code automatically, while also producing the final paper-based documentation. This methodology works well for BTEC Software development courses. If you do not skip a step and throughly test each stage, you will produce a working solution much faster than just ‘messing about and dabbling.’ with making the code first that newbies normally want to do….
(Input/Output Data and calculations/processes)
Learning Objectives
At the end of this tutorial you will know about:
• Understand how to design software and use a design tool
• Be able to design and create a program
• Technical documentation: requirements specification; other as appropriate structure charts, data dictionary
1) Work out the problem’s calculation backwards
2) Put calculations into the Process Column. Make sure you arrange them this way:
3) Check you have used brackets (BODMAS) if required, because multiply and divide is always done before add and subtract . e.g. A = 4 + 6 / 2 A = 7. A= ( 4 + 6)/2 A= 5
4) Work through the right hand side of the calculations and decide whether the variables are input, stored or a calculated output from a previous calculation.
5) Work through the input, output and stored variables and decide if they are:
INTEGER (int)- The largest integer value is 215= ±32768
LONG (lng) – The largest integer value is 231= ±2147483648
SINGLE (sng) – Single-precision floating-point variables can represent a number up to seven digits in length. The decimal point can be anywhere within those digits.
DOUBLE (dbl) – Double-precision floating-point variables can represent a number up to 15 digits in length. The decimal point can be anywhere within those digits.
Problem work out the cost of painting a ceiling
A) Work backwards from the final answer
PaintCost = Price per litre X Litres needed
Price is given, but litres needed require calculation.
Litres Needed = Tins Needed X SizeOfTin
Size of tin is given, but tins needed require calculation.
TinsNeeded = CeilingArea / CoverageOfTin
Coverage of a tin is given, but area needed require calculation.
CeilingArea = Width Of Room X Length Of Room
Both width and length will be input for each room
Place the calculation in reverse order into the SIPO table
Stored Input Processes Outputs CeilingArea = WidthOfRoom * LengthOfRoom TinsNeeded = CeilingArea / CoverageOfTin LitresNeeded = TinsNeeded X SizeOfTin PaintCost = TinPricePerLitre * LitresNeeded
B) Work out for each calculation, what is output and what needs to be input or previously stored.
Work right to left
CeilingArea = WidthOfRoom * LengthOfRoom
WidthOfRoom is INPUT
LengthOfRoom is INPUT
CeilingArea is Output
TinsNeeded = CeilingArea / CoverageOfTin
CeilingArea is already available as output
CoverageOfTin is to be stored in the program
TinsNeeded is output
LitresNeeded = TinsNeeded X SizeOfTin
TinsNeeded is already available as output
SizeOfTin is to be stored in program
LitresNeeded is output
PaintCost = TinPricePerLitre * LitresNeeded
TinPricePerLitre is to be stored in the program
LitresNeeded is already available as output
PaintCost is output
Stored Input Processes Outputs CoverageOfTin TinPricePerLitre SizeOfTin WidthOfRoom LengthOfRoom CeilingArea = WidthOfRoom * LengthOfRoom TinsNeeded = CeilingArea / CoverageOfTin LitresNeeded = TinsNeeded X SizeOfTin PaintCost = TinPricePerLitre* LitresNeeded CeilingArea TinsNeeded LitresNeeded PaintCost
C) Design Stage Work out varable type
Stored Input Processes Outputs CoverageOfTin TinPricePerLitre SizeOfTin WidthOfRoom LengthOfRoom CeilingArea = WidthOfRoom * LengthOfRoom TinsNeeded = CeilingArea / CoverageOfTin LitresNeeded = TinsNeeded X SizeOfTin PaintCost = TinPricePerLitre* LitresNeeded CeilingArea TinsNeeded LitresNeeded PaintCost
D) Use the table to produce the Structure chart
E) Produce a data dictionary
The can also be used to help design the Input and output screens. Note there are no whole numers and none of them exceed 7 digits. Single precision are therefore acceptable as variables
Item Variable Name Type Format Picture/Length sngCoverageOfTin sngPricePerLitre sngSizeOfTin sngWidthOfRoom sngLengthOfRoom dblCeilingArea sngTinsNeeded sngLitresNeeded sngPaintCost sngCoverageOfTin sngPricePerLitre sngSizeOfTin sngWidthOfRoom sngLengthOfRoom dblCeilingArea sngTinsNeeded sngLitresNeeded sngPaintCost single single single single single single single single single 99.99 99.99 99.9 99.99 99.99 99999.99 99.99 9999.99 9999.99
F) Create the Screen Design
G: Creating the program
You can use the SIPO table to produce the program
The input, output and stored variables can be used pasted at the beginning and turned into DIM statements
DIM sngCoverageOfTin AS SINGLE
DIM dblTinPricePerLitre AS SINGLE
DIM sngSizeOfTin AS SINGLE
REM Input variables
DIM sngWidthOfRoom AS SINGLE
DIM sngLengthOfRoom AS SINGLE
REM Calculatated variables
DIM dblCeilingArea AS SINGLE
DIM sngTinsNeeded AS SINGLE
DIM sngLitresNeeded AS SINGLE
DIM sngPaintCost AS SINGLE
The add the stored values and turn them into LETS
REM Stored values
LET sngCoverageOfTin= 40
LET sngTinPricePerLitre = 5.2
LET sngSizeOfTin = 2.5
INPUT sngWidthOfRoom
INPUT sngLengthOfRoom
LET dblCeilingArea = sngWidthOfRoom * sngLengthOfRoom
LET sngTinsNeeded = dblCeilingArea / sngCoverageOfTin
LET sngLitresNeeded = sngTinsNeeded * sngSizeOfTin
LET sngPaintCost = sngPricePerLitre * sngLitresNeeded
then print the answers out using the outputsdblCeilingArea
PRINT sngTinsNeeded
PRINTsngLitresNeeded
PRINT sngPaintCost
END
Note a better way to store values that don’t change in program is to use CONST to declares symbolic constants. The following comes from the QBASIC help file.
CONST constantname = expression [
constantname: The name of the constant. This name can consist of up to 40 characters and must begin with a letter. Valid characters are A-Z, 0-9, and period (.). expression
An expression that is assigned to the constant. The expression can consist of literals (such as 1.0), other constants, any arithmetic or logical operators except exponentiation (^), or a single literal string. Place the CONST after your DIMs. It is good practice to put the DIMs/CONSTs in alphabetical order.
Example:
CONST PI = 3.141593
CONST intCoverageOfTin= 40
CONST dblTinPricePerLitre = 5.2
CONST sngSizeOfTin = 2.5
## 04 QBasic Tutorial Procedures
SECTION 5 – DESIGNING APPLICATIONS (Linear V Modular)
Learning Objectives
At the end of this tutorial you will know about:
• How to use functions and procedures
It is not practical in real world terms to set up an application in one long list of code. Many early programming languages were purely linear, meaning that they started from one point on a list of code, and ended at another point. However, linear programming is not practical in a team environment. If one person could write one aspect of code, and another write another part of the program, things would be much more organized. QBasic contains the capability to meet these needs, called modular programming. You can break a program into different “modules” which are separate from the main program and yet can be accessed by any part of it. I highly recommend the use of separate modules in programming applications, although it is not a simple task to learn.
Procedures & Functions
These separate modules are also known as procedures in the QBasic environment. There are two types of procedures: subs and functions. Subs merely execute a task and return to the main program, which functions execute a task and return a value to the main program. An example of a sub might be a procedure which displays a title screen on the screen, while a function may be a procedure that returns a degree in degrees given a number in radians. Function procedures are also used in Calculus, so you Calculus people should already be familiar with functions.
Arguments
Procedures can accept arguments in what is called an argument list. Each argument in the argument list has a defined type, and an object of that type must be passed to the procedure when it is called. For example, the CHR\$ QBasic function accepts a numeric argument. The function itself converts this numeric argument into a string representation of the ASCII value of the number passed, and returns this one character string.
Procedures
Procedures in QBasic are given their own screen. When you enter the QBasic IDE, you are in the main procedure which can access all the others. Other procedures are created by typing the type of procedure (SUB or FUNCTION), the procedure name, followed by the complete argument list. You can view your procedures through the VIEW menu. Here is an example of a sub procedure which performs some operations for a program that will be using graphics, random numbers, and a logical plane.
SUB initProgram()
RANDOMIZE TIMER
SCREEN 12
WINDOW (0,0)-(100,100)
COLOR 15
END SUB
The only thing you need to type is SUB initProgram (), and the screen will be switched to that procedure. The END SUB is placed there for you, so the only thing you need to type then is the code within the sub. Try typing this out on your own to see how this works. This procedure is called by simply typing initProgram in the main procedure. An alternative method is CALL initProcedure (). Right here the parentheses are optional, but if you were to pass arguments to the procedure, parentheses would be required with the CALL statement. Now lets try passing an argument to a procedure. We will pass two arguments to a procedure called centre which are a string containing the text to be centreed, and the horizontal location on the screen at which you wish to centre it.
SUB centre( strText, sngHLointC )
LOCATE sngHLointC, 41 – (LEN(strText) / 2)
PRINT strText
END SUB
The first line after the sub declaration positions the starting point of the text at the horizontal location we passed at the second argument and vertical coordinate. The vertical coordinate is calculated by subtracting one half the screen’s width in characters (41) and half the LENgth of the text we passed as the first argument. We would call centre from the main procedure like this:
centre “Programmed by QP7”, 12
Or like this
CALL centre (“Programmed by QP7”, 12)
The Concrete Calculator
This program calculates the volume of a block using a procedure
Program Code – Type it in
DECLARE SUB subBlock ()
DECLARE SUB subcylinder ()
‘Program: Concrete Calculator
‘Author Michael Fabbro
‘Date 7 December 2003
‘This program will work out costings for quotes
Dim dblTotalVolume As Double
Dim intStartLine As Integer
‘Main section
intStartLine = 2
Call subBlock
End
Sub subBlock()
‘Procedure to calculate volume of a block
Dim sngHeight, sngWidth, sngLength As Single
Dim dblVolume As Double
Color 10, 1
Cls
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Height”
LOCATE intStartLine + 7, 5: Print “Width”
LOCATE intStartLine + 9, 5: Print “Length”
LOCATE intStartLine + 15, 15: Print “Volume”
LOCATE intStartLine + 5, 12: INPUT ” “, sngHeight
LOCATE intStartLine + 7, 12: INPUT ” “, sngWidth
LOCATE intStartLine + 9, 12: INPUT ” “, sngLength
‘Calculate Volume
dblVolume = sngHeight * sngWidth * sngLength
LOCATE intStartLine + 15, 24: Print USING “#####.##”; dblVolume;
End Sub
Test the program with this data Input: Expected output:
Results
The full program in modular form can be found here. A menu using procedures is used and INKEY\$ which reads a character from the keyboard.
IN KEY\$
■ INKEY\$ returns a null string if there is no character to return
■ For standard keys, INKEY\$ returns a 1-byte string containing the character read.
■ For extended keys, INKEY\$ returns a 2-byte string made up of the character (ASCII 0) and the keyboard scan code.
Example:
PRINT “Press Esc to exit…”
DO
LOOP UNTIL INKEY\$ = CHR\$(27) ’27 is the ASCII code for Esc.
Keycodes
```Key Code ║ Key Code ║ Key Code
c 1 ║ A 30 ║ Caps Lock 58
! or 1 2 ║ S 31 ║ F1 59
@ or 2 3 ║ D 32 ║ F2 60
# or 3 4 ║ F 33 ║ F3 61
\$ or 4 5 ║ G 34 ║ F4 62
% or 5 6 ║ H 35 ║ F5 63
^ or 6 7 ║ J 36 ║ F6 64
& or 7 8 ║ K 37 ║ F7 65
* or 8 9 ║ L 38 ║ F8 66
( or 9 10 ║ : or ; 39 ║ F9 67
) or 0 11 ║ " or ' 40 ║ F10
12 ║ ~ or ` 41 ║ F11 133
+ or = 13 ║ Left Shift 42 ║ F12 134
Bksp 14 ║ | or \ 43 ║ NumLock 69
Tab 15 ║ Z 44 ║ Scroll Lock 70
Q 16 ║ X 45 ║ Home or 7 71```
This is the full program
DECLARE SUB subBlock ()
DECLARE SUB subCylinder ()
‘Program: Concrete Calculator
‘Author Michael Fabbro
‘Date 7 December 2003
‘This program will work out costings for quotes
‘Glossary of Variables
‘intStartLine – start position of the inputscrenn
‘dblTotalVolume – running total of volume
‘Global Scope Variables
Dim dblTotalVolume As Double
Dim intStartLine As Integer
‘Local Variables
Dim strCommand As String
‘Main section
‘initialise variables
dblTotalVolume = 0
‘Set up input screen
intStartLine = 2
Color 10, 1
DO
CLS
LOCATE 2, 25: Print “Concrete Calc Main Menu”
LOCATE intStartLine + 6, 15: Print “[B]lock”
LOCATE intStartLine + 8, 15: Print “[C]ylinder”
LOCATE intStartLine + 10, 15: Print “[R]set Total”
LOCATE intStartLine + 12, 15: Print “[Q]uit”
LOCATE intStartLine + 14, 15: Print “Enter B,C,R,Q”
strCommand = “”
Do While strCommand = “”
‘note strCommand= ucase\$(inkey\$) is a better solution
strCommand = INKEY\$
Loop
‘note strCommand= ucase\$(inkey\$) above is a better solution
‘convert to uppercase
strCommand = UCase\$(strCommand)
Select Case strCommand
Case Is = “B”
Call subBlock
Case Is = “C”
Call subCylinder
Case Is = “R”
dblTotalVolume = 0
Case Is = “Q”
END
END SELECT
LOOP WHILE strCommand <> “Q”
Sub subBlock()
‘Procedure to calculate volume of a block
‘ Alphabetical Glossary of Local Variables
‘ sngJunk – used for pausing screen
‘sngLength – Length of Cylinder
‘sngWidth – Width of Block
‘sngHeight – height of block
‘dblVolume – Calulated Volume of Cylinder
‘Declare Local Variables
Dim sngRadius, sngWidth, sngLength As Single
Dim dblVolume As Double
Dim strJunk As String
‘input screen
Color 10, 1
Cls
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Height”
LOCATE intStartLine + 7, 5: Print “Width”
LOCATE intStartLine + 9, 5: Print “Length”
LOCATE intStartLine + 15, 15: Print “Volume”
LOCATE intStartLine + 5, 12: INPUT ” “, sngHeight
LOCATE intStartLine + 7, 12: INPUT ” “, sngWidth
LOCATE intStartLine + 9, 12: INPUT ” “, sngLength
‘Calculate Volume and output
dblVolume = sngHeight * sngWidth * sngLength
dblTotalVolume = dblTotalVolume + dblVolume
LOCATE intStartLine + 15, 24: Print USING “#####.##”; dblVolume;
Print ” Cubic Metres”
LOCATE intStartLine + 17, 24: Print “Total so far: “;
Print USING “#####.##”; dblTotalVolume;
Print ” Cubic Metres”
LOCATE intStartLine + 19, 24: INPUT “Press enter to continue”; strJunk
End Sub
Sub subCylinder()
‘Subroutine to calulate volume of a concrete cylinder
‘Alphabetical Glossary of Local Variables
‘sngJunk – used for pausing screen
‘sngLength – Length of Cylinder
‘dblVolume – Calulated Volume of Cylinder
‘Declare Local Variables
Dim strJunk As String
Dim dblVolume As Integer
Const conPi = 3.142
‘Procedure to calculate volume of a cylinder
Color 10, 1
CLS
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Radius”
LOCATE intStartLine + 7, 5: Print “Length”
LOCATE intStartLine + 15, 15: Print “Volume”
LOCATE intStartLine + 5, 12: INPUT ” “, sngRadius
LOCATE intStartLine + 7, 12: INPUT ” “, sngLength
‘Calculate Volume
dblVolume = conPi * sngRadius ^ 2 * sngLength
dblTotalVolume = dblTotalVolume + dblVolume
LOCATE intStartLine + 15, 24: Print USING “#####.##”; dblVolume;
Print ” Cubic Metres”
LOCATE intStartLine + 17, 24: Print “Total so far: “;
Print USING “#####.##”; dblTotalVolume;
Print ” Cubic Metres”
LOCATE intStartLine + 19, 24: INPUT “Press enter to continue”; strJunk
End Sub
There is one final concept which has proven to be very successful in programming: a message loop. With QBasic, you can construct a loop which runs for the length of the program, receives input from the user, and executes a message based on what the user does. We will construct a basic application which receives input from the user in the form of an arrow key, and moves a box on the screen based on the direction the user pressed. The arrow keys are different from normal inputted keys received with INKEY\$.
On the enhanced 101 keyboards which have arrow keys, INKEY\$ returns two values: the ASCII text representation of the key pressed, and the keyboard scan code of the key pressed. Since the arrow keys do not have an ASCII text representation, we must use the keyboard scan codes for them. The keyboard scan codes can be viewed in the HELP | CONTENTS section of the QBasic menus. For this program, we will have two procedures in addition to the main procedure. The first will initialize the program settings and position the character in his starting position. The other will move the guy in the direction which we pass to the function. The main procedure will call the sub procedures and contains the main message loop which retrieves input from the user. First of all, here is the code for the main procedure:
CONST UP = 1
CONST DOWN = 2
CONST LEFT = 3
CONST RIGHT = 4
TYPE objectType
intX AS INTEGER
intY AS INTEGER
END TYPE
DIM object AS objectType
initScreen
object.x = 41
object.y = 24
DO
SELECT CASE INKEY\$
CASE CHR\$(0) + CHR\$(72)
move UP, object
CASE CHR\$(0) + CHR\$(80)
move DOWN, object
CASE CHR\$(0) + CHR\$(75)
move LEFT, object
CASE CHR\$(0) + CHR\$(77)
move RIGHT, object
CASE CHR\$(32)
EXIT DO
END SELECT
LOOP
LOCATE 1,1: PRINT “Thank you for playing”
END
This code is fairly self explanatory with the exception of the SELECT CASE… END SELECT structure which I have not yet explained. This type of conditional testing format tests a condition, and several cases for that condition are then tested. In this case, we are seeing IF INKEY\$ = CHR\$(0) + CHR\$(72), IF INKEY\$ = CHR\$(0) + CHR\$(80), and so on. This is just a more legible format than IF…THEN…ELSE. Note that in the QuickBasic compiler, a CASE ELSE statement is required in the structure for what reason
I have no idea. The above code is the driver for the rest of the program. First some CONSTants are declared which remain constant for the duration of the program and in any module. A user defined type is declared to store the coordinates of the character.
Then an endless loop is executed, calling the appropriate procedure for the arrow key pressed until the user presses the space bar (CHR\$(32)). Here is the code for the initScreen procedure:
SUB initScreen ()
SCREEN 12
COLOR 9
WIDTH 80,50
LOCATE 24,41
PRINT CHR\$(1)
END SUB
The WIDTH 80,50 statement sets the screen text resolution to 80 columns and 50 rows. We then print a smiley face in the middle of the screen in a nice bright blue colour. Next we need to write the move procedure, and then we will be done with the program.
SUB move (way AS INTEGER, object AS objectType)
LOCATE object.y, object.x
PRINT CHR\$(0) ‘ erase previous image
SELECT CASE way
CASE UP
IF object.y > 1 THEN
object.y = object.y – 1
END IF
CASE DOWN
IF object.y < 49 THEN
object.y = object.y + 1
END IF
CASE LEFT
IF object.x > 1 THEN
object.x = object.x – 1
END IF
CASE RIGHT
IF object.x < 79 THEN
object.x = object.x + 1
END IF
END SELECT
LOCATE object.y, object.x
PRINT CHR\$(1) ‘ draw current image
END SUB
And that’s the whole program… confusing as it may be! Ideas should be going through your head about what you could do with this information. Entire games can be created with this simple construct.
There are more things to consider, but they are beyond the scope of this tutorial. If you were to design an application in QBasic, you would only need the information from this section and imagination. Programming takes knowledge of the language and a creative mind… programs are made by programmers with both. If you can develop a creative mind, then you can develop any program conceivable.
Validation Checking for Errors
This program checks the input of hours worked for erroneous data, i.e. working for more than 60 hours.
‘ Program to validate inputs ‘ Author Michael J Fabbro ‘ Date 16 December 2003 Dim sngHoursWorked As Integer Const cntRed = 12 Const cntBlue = 9 Const cntYellow = 14 Color cntYellow, cntBlue Cls Do LOCATE 5, 5: Print “Hours Worked: “; INPUT ” “, sngHoursWorked If sngHoursWorked >= 60 Then ‘ invalid hours input Color , cntRed LOCATE 7, 5: Print “Warning ‘con’ in progress” Do Loop While INKEY\$ = “” Color cntYellow, cntBlue Cls End If Loop While sngHoursWorked >= 60
This is a final example of Validation
‘ Program to validate inputs ‘ Author Michael J Fabbro ‘ Date 16 December 2003 Dim strTitle As String ‘ Modification of valid titles? Const strValidTitles = “Mr:Mrs:Ms:Master” Const cntRed = 12 Const cntBlue = 9 Const cntYellow = 14 Color cntYellow, cntBlue CLS DO LOCATE 5, 5: Print “Title: “; INPUT ” “, strTitle Select Case strTitle Case Is = “Mr” LOCATE 7, 5: Print “Male title known” Case Is = “Mrs” LOCATE 7, 5: Print “Female title known” Case Is = “Ms” LOCATE 7, 5: Print “Female title known” Case Else ‘ invalid title Color , cntRed LOCATE 7, 5: Print “Title not known” Do Loop While INKEY\$ = “” Color cntYellow, cntBlue Cls End Select Loop Until strTitle = “Mr” Or strTitle = “Ms” Or strTitle = “Mrs”
## 02 Tutorial – Software Development Basic QBasic
INTRODUCTION Session 1
Learning Objectives
At the end of this tutorial you will know about:
• The basic nature and features of a procedural programming language
• Variables – naming conventions and data types: text; integer; floating point; byte; date; Boolean. The benefits of appropriate choice of data type eg additional validation and efficiency of storage
• How and when to use assignment statements; input statements; output statements
SECTION 1 – VARIABLES
A variable, simply defined, is a name which can contain a value. Programming involves giving values to these names and presenting them in some form to the user. A variable has a type which is defined by the kind of value it holds. If the variable holds a number, it may be of integer, floating decimal, long integer, or imaginary. If the variable holds symbols or text, it may be a character variable or a string variable. These are terms you will become accustomed to as you continue programming.
Here are some examples of values a variable might contain:
• STRING “hello, this is a string”
• INTEGER 5
• LONG 92883
• SINGLE 39.2932
• DOUBLE 98334288.18
• INTEGER A 16-bit signed integer variable.
• LONG 32-bit signed integer variable.
• SINGLE single-precision 32-bit floating-point variable.
• DOUBLE double-precision 64-bit floating-point variable.
• STRING * n% fixed-length string variable n% bytes long.
• STRING variable-length string variable.
• INTEGER (int)- The largest integer value is 215= ± 32768. The smallest is 1, if you exclude zero.
• LONG (lnt) – The largest integer value is 231= ±2147483648. The smallest is 1, if you exclude zero.
• SINGLE (sng) – Single-precision floating-point variables can represent a number up to seven digits in length. The decimal point can be anywhere within those digits. The largest number is 9,999,999 the smallest is .0000001 if you exclude zero.
• DOUBLE (dbl) – Double-precision floating-point variables can represent a number up to 15 digits in length. The decimal point can be anywhere within those digits. The largest number is 99,999,999,999,999,999 the smallest is .000000000000001 if you exclude zero.
Variable types support by C++
Name Description Size* Range*
char Character or small integer. 1byte signed: -128 to 127
unsigned: 0 to 255
short int (short) Short Integer. 2bytes signed: -32768 to 32767
unsigned: 0 to 65535
int Integer. 4bytes signed: -2147483648 to 2147483647
unsigned: 0 to 4294967295
long int (long) Long integer. 4bytes signed: -2147483648 to 2147483647
unsigned: 0 to 4294967295
bool Boolean value. It can take one of two values: true or false. 1byte true or false
float Floating point number. 4bytes +/- 3.4e +/- 38 (~7 digits)
double Double precision floating point number. 8bytes +/- 1.7e +/- 308 (~15 digits)
long double Long double precision floating point number. 8bytes +/- 1.7e +/- 308 (~15 digits)
wchar_t Wide character. 2 or 4 bytes 1 wide character
## Declaration of variables
In order to use a variable in BASIC we must first declare it specifying which data type we want it to be. The syntax to declare a new variable is to write the specifier of the desired data type (like int, bool, float…) followed by a valid variable identifier. For example:
DIM intNumber AS INTEGER
Remember the largest integer value you can store in here is is 215= ± 32768. The smallest is 1, if you exclude zero. It will use up two bytes of RAM in the memory used by the program. It uses two bytes to store number
DIM lngNumber AS LONG
The largest integer value is now 231= ±2147483648. The smallest is 1, if you exclude zero. LONG uses four bytes to store a number.
Note
You choose which on the basis of speed and ram usage.
Think
This is done by using the DIM statement. Say you wanted to make a variable called intNumber which would contain an integer (whole number, no digits after the decimal point).
Then you would use that variable as an integer. The word DIM actually originates from the word Dimension, but you won’t see why until we discuss the topic of arrays.
SECTION 2 – INTERACTING WITH THE COMPUTER
You know what a variable is and how to control them, it’s time you learned some programming. QBasic (like all other languages) is set up using pre-defined statements according to the syntax specified for that statement. It may be helpful to look in the help index to learn a statement, although I’ve heard many complaint’s that the help index is too hard. Indeed it is too hard for new programmers, but as you learn more and more statements and their syntaxes, you’ll become accustomed to the index and use it as a casual reference. Lets make a program that prints some text on the screen. Type qbasic at the DOS prompt and enter the following program.
Program Code – Type it in
CLSPRINT “This text will appear on the screen”
END
Test the program with this data Input: NoneExpected out put: This text will appear on the screen Test results (Write results in the ox below)
Typing
QBasic allows to you define the ‘type’ of variable as number or characters like C++. It also allows loose typing, i.e where context tells you the type of data stored. Below is show how variables can be ‘typed,’ for loose typing all you need to do is give a variable a value:
name\$ =” fred
age=3
pay = 3.45
This means spelling errors in variable names can cause problems. JavaScript is prone to this. QBasic provides good practice in dealing with this problem.
COLOUR
Below are the 16 colours QBASIC uses –
00 = black01 = dark blue 02 = dark green 03 = dark cyan 04 = dark red 05 = dark purple 06 = orange brown 07 = gray 08 = dark gray09 = light blue 10 = light green 11 = light cyan 12 = light red 13 = magenta 14 = yellow 15 = bright white
CLS COLOR 12 PRINT “I’m light red text!” COLOR 14 PRINT “I’m yellow text!” END
Sets the screen display colours.
COLOR [foreground%] [,[background%] [,border%]] Screen mode 0 (text only)
COLOR [background%] [,palette%] Screen mode 1
COLOR [foreground%] Screen modes 4, 12, 13
COLOR [foreground%] [,background&] Screen modes 7-10
foreground% A number that sets the foreground screen colour.
background% A number that sets the background screen colour.
border% A colour attribute that sets the screen border colour.
palette% A number (0 or 1) that specifies which of two sets of colour attributes to use:
The available colour attributes and values depend on your graphics adapter and the screen mode set by the most recent SCREEN statement
LET and INPUT
You have learned how to use the PRINT and CLS commands. In this chapter, you will learn about variables and the INPUT command.
What are variables? Variables are “boxes” in the computer’s memory to store a value – be it a number, name, decimal, currency amount, or what have you. There are two main types of variables – numbers and “strings,” which are text variables.
The “numbers” category is further broken down into two areas.
Integer Numbers
Integers, they can be in the range of -32767 to 32767.
Long integers, they have a range of -2 billion to 2 billion. “Why not make all numbers long integers?” The memory of the computer, especially in QBasic, is limited – you should save as much space as possible. Only use long integers where they are necessary.
Floating-point” numbers.
These are decimal variables that can have very long decimal spaces.
Single types
Double types
You still may be in the dark as to how variables are used.
Variables are assigned a value using the LET command. For example:
LET intNumber = 123
This would assign a value of 123 to the short integer variable ” intNumber.” You can use maths functions while assigning variables, too. “LET intNumber = 4 * 12″ would make ” intNumber ” equal 48. You can increment variables like this:
LET intNumber = intNumber + 1
Or, you can make it ” intNumber = intNumber + 2, intNumber = number – 1,” and so on. You can also add two variables together using the same syntax, or sequence of commands. You need a way to output these variables to the screen if you want them to have any meaning to the user of the program.
You can use the PRINT command for this task easily:
PRINT intNumber
This would output to the screen the value assigned to “intNumber.” If you want to include text before the number, you must use this format:
Dim IntNumber AS INTEGER CLS LET intNumber = 100 PRINT “The number is”; intNumber END
This would output, when run:
The number is 100
“How do I ask the user for something?”
Like strings, You do this by using the INPUT command. :
DIM IntNumber AS INTEGER CLS INPUT intNumber PRINT “The number is”; intNumber END
This would give the user a prompt for a short integer to assign to “number” and then print it out. You can also supply a prompt for INPUT to use – like this:
DIM IntNumber AS INTEGER CLS INPUT “Enter a Number”; intNumber PRINT “You typed “; intNumber; “.” END
Exercise 1
Write a program to input the user’s name (STRING) and age (INTEGER) and output them using the PRINT command.
The first statement — CLS — stands for “clear screen.” It erases whatever was on the screen before it was executed. PRINT simply displays its argument to the screen at the current text cursor location. The argument in this case is the text enclosed in quotes. PRINT displays text within quotes directly, or it can display the value of a variable, like this:
Program Code – Type it in
DIM intA, intB AS INTEGERCLS
LET intA = 50
LET intB = 100
PRINT “The value of a is “; intA; ” and the value of b is “; intB
END
Test the program with this data Input: NoneExpected out put: The value of a is 50 and the value of b is 100. Test results (Write results in the box below)
This will yield the output; The value of a is 50 and the value of b is 100. The semicolons indicate that the next time something is printed, it will be right after where the last PRINT statement left off. Remember that PRINT prints literally what is inside quotes, and the value of the variable which is not in quotes. intA and intB are integers containing values in this example, and their values are printed using the PRINT statement.
Say you want to interact with the user now. You’ll need to learn a statement called INPUT. INPUT displays a prompt (the first argument) and assigns what the user types in to a variable (the second argument)
Program Code – Type it in
DIM strName as StringDIM intAge as Integer
CLS
INPUT “What is your name? “, strName
INPUT “How old are you? “, intAge
PRINT “So, “; strName; “, you are “; intAge; ” years old. That’s interesting.”
END
Test the program with this data Test 1 Input: Fred, 12 Expected out put: So, Fred you are 12 years old. That’s interesting. Test 2 (Write in your own Test data and expected results) Test results (Write results in the box below)
Test 1Test 2
This firsts asks the user for their name and assigns it to the string variable strName. Then the age is requested, and the result is printed in a sentence. Try it out! So what happens if you input I DON’T KNOW for the age prompt? You’ll get a weird message that says REDO FROM START. Why? The program is trying to assign a string (text) to an integer (number) type, and this makes no sense so the user is asked to do it over again.
Another cornerstone of programming is the conditional test. Basically, the program tests if a condition is true, and if it is, it does something. It looks like English so it’s not as hard as it sounds.
DIM intSelection as INTEGER CLS PRINT “1. Say hello” ‘ option 1 PRINT “2. Say nice tie” ‘ option 2 PRINT “Enter your selection ” INPUT intSelection IF intSelection = 1 THEN PRINT “hello” ENDIF IF intSelection = 2 THEN PRINT “nice tie” ENDIF END
The user is given a set of options, and then they input a value which is assigned to the variable selection%. The value of selection% is then tested, and code is executed based on the value. If the user pressed 1, it prints hello, but if they pressed 2, it prints nice tie. Also notice the text after the ‘ in the code. These are remark statements. Anything printed after a ‘ on a line does not affect the outcome of the program. Back to the actual code — but what if the user doesn’t input 1 or 2? What if they input 328? This must be taken into account as part of programming. You usually can’t assume that the user has half a brain, so if they do something wrong, you can’t screw up the program. So the ELSE statement comes into play. The logic goes like this:
IF the condition is true, THEN do something, but if the condition is anything ELSE, then do something else. The ELSE statement is used with IF…THEN to test if a condition is anything else.
DIM intNumber as INTEGERCLS INPUT “Press 1 if you want some pizza.”, intNumber IF intNumber = 1 THEN PRINT “Here’s your pizza” ELSE PRINT “You don’t get pizza” ENDIF END
Revision Session 2
Solving simple mathematical problems is one of the easiest tasks that you could do in BASIC. If you are more keen at writing a game or making graphics, you will have to be patient.:) We’ll deal about that later.
Before proceeding, you’ll have to understand some Basic concepts:
What is a program?
A program is a set of instructions that makes the computer work.
In most programs you will have to write you will have to think as follows:
What data do I need to give to the computer?
What operations will the computer have to perform with the data given?
What results will the program display
There are three parts in every task that we accomplish
Input –> Process –> Output
Input: What is needed
Process: What you need to do, calculate with the input
Output: The result obtained when the process has been done
Before beginning to code your program, it is important to write algorithms or pseudo codes in plain English that will outline what you want to do.
Algorithms
An algorithm is a set of precise instructions for solving a problem. Algorithms can be represented in plain English or in form of flowcharts
Here’s a simple algorithm for calculating the sum of 2 numbers
Get the first numberGet the second number Calculate the sum Display the sum
Let’s illustrate this concept using some mathematical examples:
We’ll be using colours to differentiate between the Input, Process and Output parts of the algorithm and program code.
Example 1: Finding the sum of 2 numbers Write a program that will display the sum of numbers 8 and 12 Algorithm Program Code Get the first number Get the second number Calculate the sum Write down the result DIM intNumber1, intNumber2 as INTEGERDIM intSum as INTEGER intNumber1 = 8 intNumber2 = 12 intSum = intNumber1 + intNumber2 PRINT intSum Output on Screen 20
You will have noticed that this program is not very explicit. The following code makes the presentation more understandable:
Enhanced Example 1: Finding the sum of 2 numbers Program Code DIM intNumber1, intNumber2, intSum AS INTEGERCLS intNumber1 = 8 intNumber2 = 12 PRINT “This program finds the sum of two numbers” PRINT “The first number is:” PRINT intNumber1 PRINT “The second number is:” PRINT intNumber2 intSum = intNumber1 + intNumber2 PRINT “The sum is:” ; PRINT intSum END Output on Screen This program finds the sum of two numbers”The first number is: 8 The second number is: 12 The sum is: 20
The use of semi-colons in the PRINT statement tells BASIC to keep the cursor on the same line.
Enhancing Program Presentation
It’s important for you to enhance you program by:
• Using meaningful variables
• Saying what is the purpose of the program
As you will notice later, programs may become huge, thus difficult to understand. The REM statement allows you to put remarks in your program. Any line starting with the REM statement is ignored at run-time. You can replace the REM statement by a single-quote ( ) symbol.
The two lines below are similar:
REM Program written by H.B.
‘ Program written by H.B.
The single-quote form of REM can also be placed at the end of a line
PRINT “Welcome everybody” ‘Welcome message
Using meaningful variables
The use single-letter variables may be easy to type especially if your typewriting speed is slow. However they may cause your program to be difficult to understand in the long-run. The use of meaningful variables makes your program easier to understand and maintain. Here is an illustration:
ix = 2000y = 10 z = x * (100 + y)/100
Could you guess the code above? Now see how it is easy when meaningful variables are used:
LET intOldSalary = 2000LET intPercentageIncrease = 10 LET sngNewSalary = intOldSalary * (100 + intPercentageIncrease )/100
Saying what is the purpose of the program Many novice programmers learning how to program, often neglect to say what is the purpose of their program. Example of an undocumented program:
LET a = 20LET b = 30 LET c = a * b PRINT “The product is “; c
As we said above, use the REM statement to put remarks, use meaningful variables and, at run-time, say what your program is doing.
REM This program calculates the product of 2 numbers: 20 and 30. LET intnumber1 = 20LET intNumber2 = 30LET intProduct = intNumber1 * intNumber2PRINT “This program calculates the product of 2 numbers”PRINT “The first number is”; intNmber1PRINT “The second number is”; intNumber2PRINT “The product is “; intProduct
Getting user input at run-time
Two commands or functions that allow you to get user input at run-time are:
The INPUT statement
The INKEY\$ function
The INPUT statement
INPUT [variable]
Example:
Example 1: Asking the user for his nameFinding the sum of 2 numbers Program Code DIM strYourname AS STRINGPRINT “Enter your name:”; INPUT strYourname PRINT “Hello”; strYourname; “. Have a nice day!” Output on Screen Enter your name: LoulouHello Loulou. Have a nice day!
The lines:
INPUT strYourname
can also be written as:
Personally I prefer to use the first option because it is easier to understand and because it is the standard procedure in most programming languages.
Example 2: Asking the user for his score in English and Math and calculating the sum Program Code DIM intEnglish AS INTEGERDIM intMath AS INTEGER DIM intSum AS INTEGER PRINT “Enter your score in English:”; INPUT intEnglish PRINT “Enter your score in Math:”; INPUT intMath LET intSum = intEnglish + intMath PRINT “The sum is:”; intSum Output on Screen Enter your score in English: 65Enter your score in Math: 59 The sum is: 124
Program Code – Type it in
DIM sngPi, sngRadius, sngArea as SINGLECLS
LET sngPi = 3.1415
LET sngArea = sngPi * sngRadius ^ 2
PRINT “The area of the circle is “, sngArea
END
Test data
First, we’re defining the variable pi. It’s a single number, which means that it can be a fairly large number with some decimal places. The exclamation mark tells QBasic that sngPi is of the single type. Next, the user is prompted for the radius of their circle. Then the area is calculated. The * means “times,” and the ^ (carrot) means “to the power of.” sngRadius ^ 2 means “radius squared.” This could also be written as:
Worked Examples
IPO TABLE For Block problem
Input Process Output Height (H) Calculate volume by multiplying H X W X L Volume Width (W) Length (L) Input controls/Variables Calculations Output Controls/variables sngHeight dblVolume = sngHeight * sngWidth * sngLength dblVolume sngWidth sngLength
Structure Chart For Block problem
Program Code – Type it in
‘Program: Concrete Calculator’Author Michael Fabbro
‘Date 7 December 2003
‘This program will work out costings for quotes
DIM sngPi, sngRadius, sngArea as SINGLE
Dim sngHeight, sngWidth, sngLength As Single
Dim dblVolume As Double
Dim intStartLine As Integer
‘Main section
‘Set up input screen
intStartLine = 2
Color 10, 1
Cls
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Height”
LOCATE intStartLine + 7, 5: Print “Width”
LOCATE intStartLine + 9, 5: Print “Length”
LOCATE intStartLine + 15, 15: Print “Volume”
LOCATE intStartLine + 5, 12: INPUT ” “, sngHeight
LOCATE intStartLine + 7, 12: INPUT ” “, sngWidth
LOCATE intStartLine + 9, 12: INPUT ” “, sngLength
‘Calculate Volume
dblVolume = sngHeight * sngWidth * sngLength
LOCATE intStartLine + 15, 24: Print USING “#####.##”; dblVolume;
END
Test data Write in some test data here
Expected Results
Exercises
• Create the test data and state the output the program should produce for the problems below.
• At least five different tests for each.
• Write the QBASIC code to solve the problem – USE MEANINGFUL VARIABLE NAMES AND COMMENTS.
• Print out copies of the program and the test results, keep them in a safe place as they are to be handed in.
1. Add two numbers input from the keyboard together and output the answer.
1. Subtract two numbers input from the keyboard together and output the answer.
1. Divide two numbers input from the keyboard together and output the answer.
1. Add three numbers input from the keyboard together and then divide the sum by 3.Output the answer to this calculation as the average. (Do the problem as two separate calculations not one)
1. Rewrite exercise using just one calculation.
1. Input the cost of an item and add 30% of its cost as profit to give the retail price. Output the retail price.
Print Using revisited
PRINT [#filenumber%,] USING formatstring\$; expressionlist [{; | ,}]
LPRINT USING formatstring\$; expressionlist [{; | ,}]
The PRINT with no arguments prints a blanks line so we can separate our answers.
Say you want to print something in a certain pre-defined format. Say you want to print a series of digits with only 2 places after the decimal point and a dollar sign before the first digit. To do this requires the PRINT USING statement, which is very handy in applications for businesses. The PRINT USING statement accepts two types of arguments. The first is a string which has already been defined. This is a special type of string, in that it contains format specifiers, which specify the format of the variables passed as the other arguments. Confused? You won’t be. Here’s a quick list of the most common format specifiers:
### digits
& Prints an entire string
\ \ Prints a string fit within the backslashes. Any thing longer is truncated
\$\$ Puts a dollar sign to the left of a number
. Prints a decimal point
, Prints a comma every third digit to the left of the decimal point.
• And these can be combined in a format string to make a user defined way to print something.
• So \$\$#,###.## will print a number with a dollar sign to the left of it. If the number has more than two decimal places, it is truncated to two. If it is more than four digits long to the left of the decimal place, it is also truncated to fit.
• To use a PRINT USING statement, you must first define the format string containing the format specifiers. Then you use PRINT USING, then the name of the format string, and variable values to fill the places defined in the format string.
Here’s a code example:
DIM intA AS INTEGERDIM strA AS STRING
intA = 123.4567
PRINT USING “###.##”; intA
PRINT USING “+###.####”; intA
strA = “ABCDEFG”
PRINT USING “!”; strA
PRINT USING “\ \”; strA
Program Code – Type it in
DIM strItemName AS STRINGDIM strFormat AS STRING
DIM intNumItems AS INTEGER
DIM sngItemCost, sngTotalCost AS SINGLE
CLS
‘ get user input
INPUT “Enter item name: “, strItemName
INPUT “How many items?: “, intNumItems
INPUT “What does one cost?: “, sngItemCost
CLS
‘ display inputs
strFormat = “\ \ #,### \$\$#,###.## \$\$#,###,###.##”
PRINT “Item Name Quantity Cost Total Cost ”
PRINT “————– ——– ———- ————–”
sngTotalCost = intNumItems * sngItemCost
PRINT USING strFormat; strItemName ; intNumItems ; sngItemCost; sngTotalCost
END
Test the program with this data
First, we get the item name, number of items, and cost per item from the user. Then we clear the screen and define the format string to be used. It contains a static length string (text that will be truncated if it is too long), up to 4 digits for the quantity, 4 digits and two decimals for the item cost, and 7 digits and two decimals for the total cost. Then we print out some column headers so we know what each value will represent, and some nice lines to go under the column headers. Then the total cost is calculated by multiplying the number of items by the item cost. Finally, the four variable’s values are displayed under the column headers using the PRINT USING statement.
Volume of a Cylinder
Volume of a cylinder is the area of the a circle making up the clynder times the height of the cylinder.
Volume = Area X height
Or
Or
Volume = pi X radius2 X height
IPO TABLE For Cylinder problem
Input Process Output Radius (R) Calculate volume by multiplying pi * radius * radius * Length Volume Length (L) Input controls/Variables Calculations Output Controls/variables sngHeight dblVolume = 3.142 * sngRadius * sngRadius * sngLength dblVolume sngRadius
Structure Chart For Cylinder problem
A Data Dictionary
Variable Name Description/Use Type Picture sngHeight Height of Cylinder Single 9999.99 sngRadius Radius of Cylinder Single 9999.99 dblVolume Volumeof Cylinder Double 999999.99
Program Code – Type it in
‘Program: Concrete Calculator’Author Michael Fabbro
‘Date 7 December 2003
‘This program will work out costings for quotes
Dim dblVolume, As DOUBLE
Dim intStartLine AS INTEGER
Const conPi = 3.142
‘Procedure to calculate volume of a cylinder
Color 10, 1
CLS
‘draw input screen
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Radius”
LOCATE intStartLine + 7, 5: Print “Length”
LOCATE intStartLine + 15, 15: Print “Volume”
LOCATE intStartLine + 5, 12: INPUT ” “, sngRadius
LOCATE intStartLine + 7, 12: INPUT ” “, sngLength
‘Calculate Volume
LET dblVolume = conPi * sngRadius ^ 2 * sngLength
LOCATE intStartLine + 15, 24: Print USING “#####.##”; dblVolume;
END
Test the program with this data
Pi Radius Volume 3.142 2 12.568 3.142 343 2155.412 3.142 5435435 34156274 3.142 0.000005 3.14E-05 3.142 -453 -2846.652 3.142 45.435 285.5135 3.142 boo #ERROR!
Enter the following programs and test they are working correctly
Exercise 3
Payroll Problem
Write an IPO table to process payroll. To calculate Pay multiply hours work time rate of pay.
Input Process Output Input controls/Variables Calculations Output Controls/variables
Now draw a Sructure Chart for the problem
Program Code – Type it in
‘Program: Payroll program’Author Michael Fabbro
‘Date 7 December 2003
DIM sngRate, shgHours,sngWage AS SINGLE
DIM strName As STRING
DIM intPayNum, intStartLine As Integer
‘Main section
‘Set up input screen
intStartLine = 2
Color 10, 1
CLS
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Name”
LOCATE intStartLine + 7, 5: Print “Payroll Number”
LOCATE intStartLine + 9, 5: Print “Pay Rate”
LOCATE intStartLine + 15, 15: Print “Hours”
LOCATE intStartLine + 5, 12: INPUT ” “, strName
LOCATE intStartLine + 7, 12: INPUT ” “, intPAyNum
LOCATE intStartLine + 9, 12: INPUT ” “, sngRate
LOCATE intStartLine + 11, 12: INPUT ” “, sngHours
‘Calculate Volume
LET sngWage = sngHours * sngRate
LOCATE intStartLine + 15, 24: Print USING “#####.##”; sngPay;
END
Test the program with this data
Expected Results
Enter the following programs and test they are working correctly using the supplied test data. Note some of these programs may not be working correctly. If the program is not working properly state on the listing you produced what was wrong with it and what you did to correct it.
Exercise 4
Using this sheet annotate the listing below describing what each line does.
‘Program: Payroll program
‘Author Michael Fabbro
‘Date 7 December 2003
DIM sngRate, shgHours,sngWage AS SINGLE
DIM strName As STRING
DIM intPayNum, intStartLine As Integer
‘Main section
‘Set up input screen
intStartLine = 2
Color 10, 1
CLS
LOCATE 2, 25: Print “Concrete Calculator”
LOCATE intStartLine + 5, 5: Print “Name”
LOCATE intStartLine + 7, 5: Print “Payroll Number”
LOCATE intStartLine + 9, 5: Print “Pay Rate”
LOCATE intStartLine + 15, 15: Print “Hours”
LOCATE intStartLine + 5, 12: INPUT ” “, strName
LOCATE intStartLine + 7, 12: INPUT ” “, intPAyNum
LOCATE intStartLine + 9, 12: INPUT ” “, sngRate
‘Calculate Volume
sngWage = sngHours * sngRate
LOCATE intStartLine + 15, 24: Print USING “#####.##”; sngPay;
END
Glossary of Basic found in this Chapter.
PRINT
Writes data to the screen or to a file. LPRINT prints data on the printer LPT1.
Print [#filenumber%,]; [expressionlist]; [{; | ,}]
LPRINT [expressionlist] [{; | ,}]
filenumber% The number of an open file. If you don’t specify a file number, PRINT writes to the screen.
expressionlist A list of one or more numeric or string expressions to print.
{; | ,} Determines where the next output begins:
; means print immediately after the last value.
, means print at the start of the next print zone.
Print zones are 14 characters wide.
TAB
Moves the text cursor to a specified print position.
TAB(column%)
column% The column number of the new print position.
Example:
Print Tab(25); “Text”
SPC
Skips a specified number of spaces in a PRINT or LPRINT statement.
SPC(n%)
n% The number of spaces to skip; a value in the range
o through 32, 767#
Example: Print “Text1”; Spc(10); “Text2”
CLS
Clears the screen.
Cls [{0 | 1 | 2}]
CLS Clears either the text or graphics viewport. If a graphics viewport has been set using VIEW), clears only the graphics viewport. Otherwise, clears the text viewport or entire screen.
CLS 0 Clears the screen of all text and graphics.
CLS 1 Clears the graphics viewport or the entire screen if no graphics viewport has been set.
CLS 2 Clears the text viewport.
REM
Allows explanatory remarks to be inserted in a program.
Rem remark
‘ remark
remark Any text.
Remarks are ignored when the program runs unless they contain
metacommands. A remark can be inserted on a line after an executable statement if it is preceded by the single-quote (‘) form of REM or if REM is preceded by a colon (:).
Example:
Rem This is a comment.
‘ This is also a comment.
Print “Test1” ‘This is a comment after a PRINT statement.
Print “Test2”: Rem This is also a comment after a PRINT statement.
LET
Assigns the value of an expression to a variable.
[LET] variable=expression
■ variable Any variable. Variable names can consist of up to 40 characters and must begin with a letter. Valid characters are A-Z, 0-9, and period (.).
■ expression Any expression that provides a value to assign.
■ Use of the optional LET keyword is not recommended. The
variable=expression assignment statement performs the same action
with or without LET.
LOCATE
Moves the cursor to a specified position on the screen.
CSRLIN returns the current row position of the cursor.
POS returns the current column position of the cursor.
LOCATE [row%] [,[column%] [,[cursor%] [,start% [,stop%]]]]
CSRLIN
POS(expression)
■ row% and column% The number of the row and column to which the
cursor moves.
■ cursor% Specifies whether the cursor is visible:
0 = invisible, 1 = visible
■ start% and stop% Integer expressions in the range 0 through 31
that specify the first and last cursor scan lines.
You can change the cursor size by changing the
cursor scan lines.
■ expression Any expression.
Task – Data Types Task 1 Data types include string/ text; integer; floating point; byte; date; Boolean other eg long, double single. Using examples from Microsoft Qbasic and other languages, describe the benefits of the appropriate choice of data types available to the programmer, eg additional validation, efficiency of storage, strong typing. Task 2 Write the following program for a mail order company. Input the purchase cost of an item add 30% to give the retail then add a delivery charge of 10% of the retail cost. Ensure your programs are documented with comments. You should submit a) Structure chart b) Test data and the results expected c) A listing of the program’s code with comments d) A screen grab of you program’s results with the test data. e) A Data dictionary of variables.
## 01 Introduction to Software Design using QBASIC
There are six sections to this tutorial:
1. Introduction
2. Basics ‘Basic’
3. Decisions and Loops
4. Procedures
5. The SIPOS Way
Introduction to Software Design using QBASIC
QBasic is a programming language designed for DOS (disk operating system) and is a very good language to learn and then migrate to modern day languages like VBA, VBSCRIPT, ASP, Visual Basic, C++, javascript and Java. You will learn to deal with more code, commands, and complex routines. Once you have learnt these basics, it easier to learn other more modern languages. You will find (QBasic.exe and QBasic.hlp) on DOS disks and Windows 98-95 CDs.
Typing
QBasic allows to you define the ‘type’ of variable as number or characters like C++. It also allows loose typing, i.e where context tells you the type of data stored. Below is show how variables can be ‘typed,’ for loose typing all you need to do is give a variable a value:
name\$ =” fred
age=3
pay = 3.45
This means spelling errors in variable names can cause problems. JavaScript is prone to this. QBasic provides good practice in dealing with this problem.
Warning
The tutorial can be used in standalone mode, but is designed to complement classroom teaching. Do not switch off and think you can do this course all by yourself at the last minute.
You can navigate the tutorial using the menu shown below. You can see it in the banner above.
Learning Objectives
At the end of this tutorial you will know about:
• The basic nature and features of a procedural programming language
• Variables – naming conventions and data types: text; integer; floating point; byte; date; Boolean. The benefits of appropriate choice of data type eg additional validation and efficiency of storage
• How and when to use assignment statements; input statements; output statements
• How to use Loops eg conditional (pre-check, post-check), fixed to repeat actions
• How to use conditional statements and logical operators to make decisions
• How and when to use assignment statements; input statements; output statements
• How to use functions and procedures
• Understand how to design software and use design tools
• Understand the software development life cycle
• Be able to design and create a program
• Technical documentation: requirements specification; other as appropriate to
structure charts, data dictionary
• Be able to document, test, debug and review a programmed solution
• Testing and debugging: test strategy; test plan structure eg test, date, expected
result, actual resultUser documentation: eg details of hardware platform required, loading instructions, user guide; getting help | 19,562 | 78,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.792903 |
https://uk.mathworks.com/matlabcentral/answers/102145-why-do-i-get-the-error-subscript-indices-must-either-be-real-positive-integers-or-logicals?s_tid=faqs_link | 1,631,817,041,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053717.37/warc/CC-MAIN-20210916174455-20210916204455-00252.warc.gz | 638,343,828 | 55,577 | # Why do I get the error 'Subscript indices must either be real positive integers or logicals. ' ?
1,254 views (last 30 days)
Why do I get the following error message:
ERROR: Subscript indices must either be real positive integers or logicals.
MathWorks Support Team on 17 Feb 2021
Edited: MathWorks Support Team on 17 Feb 2021
This error occurs when you attempt to index into an array using indices that are not positive integers or logical values. Here are some tips for common situations that cause this error message:
1) Double check that your indices are positive integers. Indices in MATLAB cannot be 0, and by default, start with 1.
2) If you are using logical indexing to index into an array, be sure that your index array is of type 'logical', and not a 'double' array of 1s and 0s. You can convert a 'double' array to a logical array before attempting to use logical indexing. For example:
A = [1 2 3 4; 5 6 7 8];
ind_double = [0 1 0 1; 0 1 0 1];
ind_logical = logical(ind_double);
A(ind_logical)
For an index array 'ind', you can check its data type using the 'whos' function:
whos ind
3) If you use floating-point arithmetic to compute an index array, then the array values may not be exact integers. The 'round' function is handy when you know you have an index value that is nearly the integer index that you want. For example,
A = [1 2 3 4; 5 6 7 8];
ind_float = 2.00001;
ind_int = round(ind_float);
A(ind_float)
Below is a way to check if an index array 'ind'' contains exact integer values. This command returns a 'logical' array, where 1 indicates the index value is an exact integer, and 0 indicates it is not.
ind == round(ind)
4) If you assign a variable to the same name as a built-in function in MATLAB, then you will overwrite that function and encounter the error when you attempt to call it. For example,
max = rand(5);
A = rand(5);
max(A)
In this event, rename your variable and clear the old one to proceed:
B = max;
clear max max(A)
For more information on indexing in MATLAB, see the following documentation page:
Walter Roberson on 25 Sep 2018
"Has this been changed in a recent update of Matlab? (I updated from 2013 to 2017). Before that I never had to convert from double to logical."
R2013a:
>> A=rand(1,3), A([1 0 0])
A =
0.8147 0.9058 0.1270
Subscript indices must either be real positive integers or logicals.
Same with R2010bSP1, which is the oldest MATLAB I happen to have installed at the moment.
If we examine the R14 documentation, https://www.mathworks.com/help/releases/R14/techdoc/matlab.html in the document for " Special Characters [ ] ( ) {} = ' . ... , ; : % ! @ " we see,
"The components of V must be integers to be used as subscripts. An error occurs if any such subscript is less than 1 or greater than the size of X."
This is not a new restriction, and No, there is no setting in MATLAB that could change this.
Antoine Pichot on 19 Nov 2015
Iaredi Sabinas's comment should be a valid answer.
It may happen when you have a variable named after an existing matlab function, like min or diff. Matlab thinks you are using the variable instead of the builtin function.
Stephen on 19 Nov 2015
It is already part of the answer: "Another common cause is that a variable has overwritten a function name and is thus shadowing the function. For example:..." and it then proceeds to give an example of how this can happen.
Schrecklich Er on 18 Mar 2017
Edited: Walter Roberson on 4 Apr 2017
I got this message while I was trying process a image using 'imread', i had the following structure:
for i = 1 : f
for j = 1 : c
B(i,j)=([A1(i,j)*A2(i,j)]);
B1(i,j)=(A2(i,j)/A1(i,j));
end
end
i got the same error message so until 30 minutes of research I just switch the letter 'i' for a 'k' and the error message disapeared, I think that the error was there because the letter 'i' is used for imaginary numbers.
just a little hint! Hope it is useful.
Farah Nadiah on 22 Apr 2016
how about this. the word that have star...how can i declare..if i run this prgoram it get error..thnx a lot
for i=0:1: maxrow-1
for j=0:1: maxcol-1
% inv([i, j]) = 255 - image([i, j]);
for k = 0: 1
*sto([i, j, k]) = image([i, j]);*
end
end
end
Walter Roberson on 22 Apr 2016
MATLAB indexing starts at 1, not at 0. You need to add 1 to all of your indices.
Also remember that sto([i, j, k]) is indexing sto at 3 locations, sto(i), sto(j), sto(k). It is not an index into a 3 dimensional array: that would be sto(i, j, k)
Pratyush Lohumi on 18 Mar 2017
Edited: Walter Roberson on 18 Mar 2017
for n = 0:ns
s(n) = (ns-n)/ns; %slip
Tmech(n) = ph*V1eq^2*R2/((s(n)*omegas)*((R1eq + R2/s(n))^2+(X1+X2)^2)); %Electromechanical torque
end %End of slip loop
Error: Subscript indices must either be real positive integers or logicals. (Line 2)
Query: Couldn't seem to rectify the mistake?
If someone could provide a valid explanation or a corrected code for this loop, that would really help my project.
Imen BOUGRINE on 4 Apr 2017
for m = 1:num_pulse_int
%Update sensor and target positions
[sensorpos,sensorvel] = sensormotion(1/prf);
[tgtpos,tgtvel] = tgtmotion(1/prf);
padmini kutturu on 26 Apr 2017
Edited: Walter Roberson on 26 Apr 2017
for j=1:n
y(j)= (T(j)-Ts)/(Tb-Ts);
h(j)=(1/427)*(-0.0717)*L*(Tb-Ts)*(y(j+1)-y(j-1)/2*deltas);
r(j)=(e*sigma*L^2*P*Ts^3)/(kAc)*(((T(j)/Ts)^3)+((T(j)/Ts)^2)+(T(j)/Ts)+1);
c(j)=1-(h(j)*deltas/2);
a(j)=-2+(r(j)*deltas^2);
b(j)=1+(h(j)*deltas/2);
end
Walter Roberson on 25 Sep 2018
Ram: you had a k-5 subscript with k starting at 1. You need to start k at 6 or more.
Somenone can help me. I got this erro when using this code below:
W2=[];%will contain watermark signal extracted from the image
for t=1:wmsz
W2 = [W2(D_w(IND(t,1),IND(t,2))/D(IND(t,1),IND(t,2)))*10] %watermark extraction
end
Walter Roberson on 4 Oct 2017
Let us match brackets. The number will be the count of open brackets "after" the character aligned with:
W2 = [W2(D_w(IND(t,1),IND(t,2))/D(IND(t,1),IND(t,2)))*10]
1 2 3 4 3 4 32 3 4 3 4 321 0
We see from this that W2 is being indexed by
D_w(IND(t,1),IND(t,2)) / D(IND(t,1),IND(t,2))
1 2 1 2 10 1 2 1 2 10
which contains a division. So for the index into W2 to be an integer, D_w(IND(t,1),IND(t,2)) would have to be an non-zero integer exact integer multiple of D(IND(t,1),IND(t,2)) . That condition is not impossible, but it is something I would tend to doubt.
Yago Veloso on 6 Oct 2017
Edited: Walter Roberson on 6 Oct 2017
Hi everyone! I get the same error in my code, I'm trying to connect a function to my main code, where the function supply my code with all variables of my system equation solution.
Below is part of my main code where I found this error:
[Y1des]= ANN (Z1,e1,s1,D);
[ ac, ao, at_coluna, kp, r1, r2, L, vj, ro_ar, visc_ar, k_ar, RA, P, t_ar, Tar_e, T_ar_ext, Urel_e, Urel, Uabs_e, cps, ro_p_ap, hparede, cal_lat, cpar, cpl, cpv, ro_p, dp, Ubs, u, Rep1, Nu, hp1, St, Rep2, hw1, e, ae1, aet, X1, t_seg, X, ae, hp, hw, d_xp, tt,z]= variaveis (h, Y1des, hi, uar, t_min);
[G ,u, Pvse, Pve, UAEsat, UAE, He]= prop_ar (ro_ar,uar,ac, ao, at_coluna,t_ar, Tar_e, Urel_e, P);
[vl, mg, mss]= prop_leito (ro_p_ap,ro_p,ro_ar,ac,hi,vj);
%%%Initial condition
Tp(1)= 298.15;
Tar_s(1)= Tar_e - ((1-exp(-X(1)))*(Tar_e-Tp(1)));
tar_s(1) = Tar_s(1)-273.15;
Here is the error message
Subscript indices must either be real positive
integers or logicals.
Error in Dif_finitas_plus_ANN (line 44)
Tar_s(1)= Tar_e -
((1-exp(-X(1)))*(Tar_e-Tp(1)));
Thanks for the help !!
Yago Veloso on 6 Oct 2017
Walter Roberson, thanks for the help! That's what was causing the error.
MarkusP on 5 Feb 2018
Edited: MarkusP on 5 Feb 2018
hello guys, maybe someone of you is able to help me...I tried to fix my problem with the solution above but it wasn`t possible.
I get the same error message if I`m runnung the following skript. The skript is for solving a problem with methods of characteristic.
l=1;
d=0.01;
rho=1000;
f=0.1;
a=1;
p0=2e3;
v0=(p0*2*d/(l*rho*f))^0.5;
n=101;
h=l/(n-1);
v(1:n)=v0;
p(1)=p0;
for i=2:n
p(i)=p(i-1)-f*rho*v0^2*h/(2*d);
end
dt=h/a;
tmax=3;
itmax=tmax/dt;
fhr=f*h/(2*a*d);
for it=1:itmax
t=it*dt;
for i=2:n-1
pa=p(i-1);
pb=p(i+1);
va=v(i-1);
vb=v(i+1);
pc(i)=a/2*((pa+pb)/a+rho*(va-vb)+fhr*(vb*abs(vb)-va*abs(va)));
vc(i)=0.5*((pa-pb)/(a*rho)+va+vb-fhr*(vb*abs(vb)+va*abs(va)));
end
pc(1)=p0;
vb=v(2);
pb=p(2);
vc(1)=vb+(pc(1)-pb)/(a*rho)-fhr*vb*abs(vb);
vc(n)=v0*valve(t);
va=v(n-1);
pa=p(n-1);
pc(n)=pa-rho*a(vc(n)-va)+f*h*rho/(2*d)*va*abs(va); %%error is in this line
vres(it,1:n)=vc(1:n);
pres(it,1:n)=pc(1:n);
p=pc;
v=vc;
end
My valve(t) function looks like this:
function vrel = valve(t)
if t<.1
vrel=1;
else
vrel=exp(-10*(t-.1));
end
Here is the error message:
Subscript indices must either be real positive integers or logicals.
Error in MOCwaterhammer (line 45)
pc(n)=pa-rho*a(vc(n)-va)+f*h*rho/(2*d)*va*abs(va);
Thanks for you help!
Torsten on 5 Feb 2018
Edited: Torsten on 5 Feb 2018
You forgot a multiplication sign:
pc(n)=pa-rho*a->*<-here(vc(n)-va)+f*h*rho/(2*d)*va*abs(va);
Best wishes
Torsten.
MarkusP on 5 Feb 2018
Oh the solution can be so simple...
Thank you so much!
Francisca Belart on 30 Aug 2018
just using round()solved it for me.Thanks sooo much!
Hamza saeed khan on 21 Dec 2018
I think it may help. I myself remove (t) from c(t) and m(t) as follows:
clc
close all
clear all
%let values we asigned as followed
A1=4
A2=2
f1=5
f2=6
t=0.1:0.001:1
c=20*A1*cos(2*pi)*f1
m=12*A2*cos(2*pi)*f2
y=c*m
plot(y,t,'r')
xlabel('c(t)')
ylabel('m(t)')
grid on
Best wishes Hamza Pakistani lol
¿How I put a range in for loop that starts in a negative number?
Something like that
for i=-10000:100:10000
delta_g(i,1)=((4*R1^3*sp*G)/3)*(z1/(i^2+z1^2)^(3/2));
end
Salu2
Stephen on 8 Oct 2019
Loop over indices, rather than looping over data.
Jamal Nasir on 31 Mar 2020
this error when a negative number or real number used as indecies of matrix
Sofia Santos on 8 May 2020
Hello, I tried using round() and convert double to logical, bue when I type "whos maximum" or "whos minimum" nothing is show. I want to now the time in minimum and the time in maximum for calculate de rise time.
Can you look ate my code and see if you can help? Thank you very much!!
data_org = importdata(sinais,'\t',7); %text file with 7 columms
EDA=data_org.data;
EDA= EDA(:,6); %I just want the values in the 6th column
T=N/fs;
t = (1:N)/fs;
minimum= min(EDAuS_filter);
maximum= max(EDAuS_filter);
amp=maximo-minimo; %signal amplitude
tmin=t(minimo);
tmax=t(maximo);
rise_time= tmax-tmin;
Sofia Santos on 8 May 2020
I'm not executing this in the context of a function. The idea is to create a script that reads the six column of each text file and then filter that signal and get the amplitude and time rise. But I'm getting troubles to find the corresponding t.
Seahawkgo on 27 May 2020
Edited: Walter Roberson on 27 May 2020
"Subscript indices must either be real positive integers or logicals."
Error in Chanprojectpart32 (line 36)
E(Zr)=mean(Zr);
=====================
for j=1:N-1
s1 = 0;
s2 = 0;
for k=1:No+1
Phi=(pi-(-pi))*rand-pi;%Phi-k
Fk = Fd.*(cos((2*pi*k)/M)); %Fk
Bk = pi/(No+1)*k; %k(pi/No+1)
..........
.........
y6 = cos(2*pi*Fd*t + Phi);
s1 = s1 + y1*y2;
s2 = s2 + y5*y2;
end
Zr = NN*s1 + MM*(y3*y6);
Zi = NN*s2 + MM*(y4*y6);
end
E(Zr)=mean(Zr);
E(Zi)=mean(Zi);
##### 2 CommentsShowHide 1 older comment
Seahawkgo on 30 May 2020
Thanks
Viktoriia Buliuk on 3 Aug 2020
Hello! I try to perform the transforn of obtained data from .csv file. Could you, please, help me. I have the following errors:
Subscript indices must either be real positive integers or logicals.
Error in walvet>@(tau)hh(tau)*(2/(3^(1/2)*pi^(1/4))*exp(-((tau-ttt(j))/a(i))^2/2)*(1-((tau-ttt(j))/a(i))^2))
Error in integralCalc/iterateArrayValued (line 156)
fxj = FUN(t(1)).*w(1);
[q,errbnd] = iterateArrayValued(u,tinterval,pathlen);
Error in integralCalc (line 103)
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in walvet (line 26)
Q = integral(F,-Inf,Inf, 'ArrayValued', true);
And the code is follofing:
type wet.csv
Test = importdata('wet.csv');
t = Test(:, 1);
h = Test(:, 2);
tt = max(t);
tr = min(t);
ttt = tr:4.0003*10^-11:tt;
hh = interp1(t, h, ttt);
figure(2);
plot(t, h, 'ro');
hold on;
plot(ttt, hh, 'g');
a = 0.3*10^-10:10^-11:10^-9;
for i = 1:length(a)
for j = 1:length(ttt)
F = @(tau) hh(tau)*(2/(3^(1/2)*pi^(1/4))*exp(-((tau-ttt(j))/a(i))^2/2)*(1 - ((tau-ttt(j))/a(i))^2));
Q = integral(F,-Inf,Inf, 'ArrayValued', true);
K = (1/(a(i)^(1/2)))*Q;
S = real(K);
end
end
contour (ttt, hh, a, S, 500)
Thank you very much!!!
##### 2 CommentsShowHide 1 older comment
Viktoriia Buliuk on 11 Aug 2020
Thanks a lot!
I defined hh = @(t_per) interp1(t, h, t_per)
Destaw Masresha on 13 Dec 2020
Edited: Walter Roberson on 13 Dec 2020
A=sort(fc);
B=A((c*d-n+1):c*d); % selete the top n of the variance
Subscript indices must either be real positive integers or logicals
Walter Roberson on 13 Dec 2020
What are c and d ? What is n ?
I speculate that you constructed c and d such that you expect c*d to always be an integer, but that it is not always exactly an integer. For example if c were 0.1:0.1:1 and d were 10, then you might expect that c*d is always an integer, but that would be false:
format long g
c = 0.1 : 0.1 : 1;
mat2str(c)
ans = '[0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1]'
d = 10;
mat2str(c * d)
ans = '[1 2 3 4 5 6 7 8 9 10]'
mat2str(c * d - (1:10))
ans = '[0 0 4.44089209850063e-16 0 0 0 0 0 0 0]'
c(3) - 0.3
ans =
5.55111512312578e-17
Lessons:
• 0.1 is not exactly representable in double precision
• multiplying a double precision number that is "mathematically" a fraction, by something that "mathematically" give you back an integer... doesn't always give you back an integer. 0.3 * 10 does not give 3
• numbers calculated by fractional increments on colon operators, such as the third value of 0.1:0.1:1, are not always exactly the same as the corresponding literal value you would expect. 0.1+0.1+0.1 does not give you the same value as you would get from writing 0.3
common fernando on 5 Apr 2021
Edited: Walter Roberson on 5 Apr 2021
hello guys, maybe someone of you is able to help me...I tried to fix my problem with the solution above but it wasn`t possible.
Why do I get the following error message:
Subscript indices must either be real positive integers or logicals.
Error in threeDFT (line 33)
x=v((j-1:j+N0-2)*dt);
code:
fs=50*512;
dt =1/fs;
N0=fs/50
tmax=1;
j_max=tmax*fs;
for j=1:j_max+1
x=v((j-1:j+N0-2)*dt);
Walter Roberson on 7 Apr 2021
The code you posted before was three months ago; I would think that you have developed the code further since that time.
neelu pareek on 11 Jul 2021
please someone help as i am getting this error for the first time although calculating and using gamma function many times in my work.
t=[0:.5:1];
eta=0.8;gamma=.03;p=.9;pi=.4;mu=.75,sigma=.7
S= mu+(( ((1-p)*(pi)^(mu))-(.5*(mu)^2*(eta)^(mu))-(mu*(sigma)^(mu)) )* (t.^( mu)/ gamma(mu + 1)))
mu =
0.7500
sigma =
0.7000
Subscript indices must either be real positive integers or logicals.
Walter Roberson on 11 Jul 2021
gamma=.03
gamma stops being a function and starts being a scalar. gamma(mu+1) would then be an attempt to index the scalar.
Alexandra Roxana on 18 Jul 2021
Hello, I'm trying to solve the heat equation in 2D using the finite difference method. I've also changed k=0 to k=1 and u(k+1) to u(k) but still no result. Are the initializations correct?
clc
clear all
alpha = 2;
L=50;
dx = 1;
dt = (dx^2)/(4*alpha);
gamma = (alpha^dt)/(dx^2);
itert = 1000;
u=zeros(itert,L,L);
uinit = 0;
utop=100;
uleft=0;
ubottom=0;
uright=0;
%Boundary conditions
u(:,L-1,:)=utop;
u(:,:,1)=uleft;
u(:,1,1)=ubottom;
u(:,:,L-1)=uright;
for k=0:itert-1
for i=1:L-1
for j=1:L-1
u(k+1,i,j) = gamma*(u(k,i+1,j) + u(k,i-1,j) + ...
u(k,i,j+1) + u(k,i,j-1) - 4*u(k,i,j)) + u(k,i,j);
end
end
end
display(u)
Alexandra Roxana on 18 Jul 2021
Thank you very much! Yes, it was *dt.
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# Marginal Cost: Firm Producing Hamburgers
APMACR-LAENOZ
Suppose the following data for a firm producing hamburgers:
• The firm can produce 1 million burgers for a total revenue of 1 million dollars and total input costs of 700,000 dollars.
• The same firm can produce 2 million burgers for a total revenue of 1.5 million dollars with total input costs of 1 million dollars.
Which of the following statements are true given the data above?
I. Profit per burger is greater at 2 million burgers of production than at 1 million burgers of production.
II. To produce at 2 million burgers instead of 1 million burgers of production the firm incurs marginal costs of 300,000 dollars.
III. When producing at 2 million burgers instead of 1 million burgers the marginal output is 1 million burgers.
A
I only
B
II only
C
III only
D
II and III only
E
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• 5
##### Re: Reasoning behind the Universal Accelerator
« Reply #60 on: December 27, 2021, 06:34:28 PM »
At a constant acceleration of 9.81 m/s2 over a single year we'd approach a significant fraction of the speed of light.
Right. So you've jumped into the middle of an in-depth discussion on relativity just to tell us you haven't read about UA, and that you don't understand the differences between classical mechanics and special relativity.
Don't do that.
I didn't. I'm new to this forum but I assure you that I have read about UA and understand the differences between classical Newtonian mechanics and special relativity. If I didn't I'd hardly be invoking an zenith-oriented Lorentz contraction and blue-shift of the starfield as a problem. I'd be objecting to breaking the light barrier or something equally daft.
So now that we have that out of the way perhaps you can explain how constant acceleration over any significant amount of time doesn't land you at a velocity where relativistic effects become horrifyingly obvious by way of being horrifyingly deadly. While you're at it, perhaps you can explain what's accelerating Earth, because dark energy isn't going to do the trick unless you're invoking some very strange interactions.
As I see it, classic gravitation on a disk is simpler, more elegant, and a better fit to the observed data.
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #61 on: December 27, 2021, 06:38:01 PM »
I'd be objecting to breaking the light barrier or something equally daft.
Yes, while you stopped short of repeating the "breaking the light barrier" cliché, you did go for something equally daft - "approaching relativistic velocities" without even defining the FoR you're talking about. That's why you got told off.
So now that we have that out of the way perhaps you can explain how constant acceleration over any significant amount of time doesn't land you at a velocity where relativistic effects become horrifyingly obvious by way of being horrifyingly deadly.
That's not how any of this works. You'll have to present a hypothetical observer from whose perspective the Earth would "land you" at relativistic velocities for us to even begin considering it. I will not "explain" why something you haven't defined hasn't been defined.
« Last Edit: December 27, 2021, 06:43:55 PM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### AFlat
• 5
##### Re: Reasoning behind the Universal Accelerator
« Reply #62 on: December 27, 2021, 07:41:13 PM »
I'd be objecting to breaking the light barrier or something equally daft.
Yes, while you stopped short of repeating the "breaking the light barrier" cliché, you did go for something equally daft - "approaching relativistic velocities" without even defining the FoR you're talking about. That's why you got told off.
I thought that the FoR was obvious from the discussion of an accelerating Earth. One observer on Earth looking outward from horizon to zenith. The observer is accelerating at 9.81 m/s2 by virtue of being pushed along by the accelerating Earth as per UA. The observer is accelerating relative to the background starfield with both observer and starfield at v = 0 at t0. tnow being no less than 100 years after t0, the observer's vnow relative to the starfield should be "approaching relativistic velocities". Somewhere north of 0.999 c by my back of the envelope calcs.
So now that we have that out of the way perhaps you can explain how constant acceleration over any significant amount of time doesn't land you at a velocity where relativistic effects become horrifyingly obvious by way of being horrifyingly deadly.
That's not how any of this works. You'll have to present a hypothetical observer from whose perspective the Earth would "land you" at relativistic velocities for us to even begin considering it. I will not "explain" why something you haven't defined hasn't been defined.
Honestly Pete, this is so basic I shouldn't have to explain. You, I, and everybody else on the Earth are non-hypothetical observers, all accelerating as per UA. The background starfield begins as stationary relative to us, so after even a year we should be able to perceive some pretty extreme relativistic distortions of the starfield, Assuming there was anybody left alive to observe anything.
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #63 on: December 27, 2021, 10:36:21 PM »
I thought that the FoR was obvious from the discussion of an accelerating Earth.
Oh, thank the Lord! You wouldn't believe how often people get these completely wr-
One observer on Earth looking outward from horizon to zenith. The observer is accelerating at 9.81 m/s2 by virtue of being pushed along by the accelerating Earth as per UA. The observer is accelerating relative to the background starfield with both observer and starfield at v = 0 at t0. tnow being no less than 100 years after t0, the observer's vnow relative to the starfield should be "approaching relativistic velocities". Somewhere north of 0.999 c by my back of the envelope calcs.
Oh. Oh dear.
So, this is why reading about UA would have been helpful. Sure, hypothetically, an observer that's somehow not subject to UA would see the Earth zoom away at ludicrous speeds after a year of somehow not being affected by UA. However, the "U" in "UA" is a bit of an issue there.
You can feasibly have an observer that's in relative proximity to the Earth, in which case they'll observe the Earth accelerating towards them together with the atmolayer until they reach terminal velocity. This will also not last anywhere near a year 100 years(?!), because falling is usually a rather temporary affair.
You can also have an observer that's outside of the Earth's area of influence, and thus affected by UA. From their perspective, the Earth is not accelerating.
Your hypothetical does make sense from a physics standpoint (a breath of fresh air, honestly), but it simply isn't very relevant to what's being discussed. It's like saying that if I somehow managed to propel myself to relativistic speeds relative to the Round Earth, I'd observe time dilation. Sure, I would. So what?
What saddens me particularly about your contribution here is that we just finished talking about why your observer isn't relevant. You said you've done your reading, but this turns out to have been untrue. You should have done so much better.
Honestly Pete, this is so basic I shouldn't have to explain.
I agree, and I was honestly excited when you seemed to know your stuff, but, alas, you managed to cock it up.
The background starfield begins as stationary relative to us
This incorrect assumption is at the core of your misunderstanding. There is no magical "background" that's unaffected by UA; Universal Acceleration is... universal. So, for your FoR to make sense, you have to introduce a hypothetical observer. One that, from an Earthly perspective, has unprecedented energy that somehow allows it to defy the nature of our universe. You will have to prove the existence of such an object before we can discuss its relevancy, but if such an object exists and you can slap a time-measuring device on it, then, by all means, I agree, you'd measure significant time dilation there.
« Last Edit: December 27, 2021, 10:51:14 PM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### AFlat
• 5
##### Re: Reasoning behind the Universal Accelerator
« Reply #64 on: December 29, 2021, 06:01:06 AM »
So, this is why reading about UA would have been helpful. Sure, hypothetically, an observer that's somehow not subject to UA would see the Earth zoom away at ludicrous speeds after a year of somehow not being affected by UA. However, the "U" in "UA" is a bit of an issue there.
Ironically, you have no idea how right you are.
Quote
This incorrect assumption is at the core of your misunderstanding. There is no magical "background" that's unaffected by UA; Universal Acceleration is... universal. So, for your FoR to make sense, you have to introduce a hypothetical observer. One that, from an Earthly perspective, has unprecedented energy that somehow allows it to defy the nature of our universe. You will have to prove the existence of such an object before we can discuss its relevancy, but if such an object exists and you can slap a time-measuring device on it, then, by all means, I agree, you'd measure significant time dilation there.
My incorrect assumption was that UA was supposed to work. If you accelerate everything universally and equally then it's indistinguishable from no effect whatsoever. If you and I and the Earth are all being accelerated at the same rate then the Earth doesn't exert any force at all, g = 0 m/s2, and we drift off into space. It's like being in a falling elevator.
Quote
What saddens me particularly about your contribution here is that we just finished talking about why your observer isn't relevant. You said you've done your reading, but this turns out to have been untrue. You should have done so much better.
Sorry, I clearly overestimated you and failed to understand how fundamentally broken UA was. But hey, kudos for adding epicycles that do nothing and being condescending throughout such an epic failure.
#### stack
• 3427
##### Re: Reasoning behind the Universal Accelerator
« Reply #65 on: December 29, 2021, 06:39:45 AM »
My incorrect assumption was that UA was supposed to work. If you accelerate everything universally and equally then it's indistinguishable from no effect whatsoever. If you and I and the Earth are all being accelerated at the same rate then the Earth doesn't exert any force at all, g = 0 m/s2, and we drift off into space. It's like being in a falling elevator.
I think the deal with UA is that whatever U force that is pushing up everything is shielded by the Earth...Up to a certain altitude. So picture the force like the wind, pushing upward from below the Earth. The wind pushes upward past the edges of the flat Earth disk and then curls inward to continue to push everything upwards over the entirety of the flat Earth disk somewhere below all celestial objects. That way, the earth is pushed up, the celestial bodies are pushed up along with it, yet we on terra firma are not pushed up if our feet leave the ground. We are shielded from the wind, so to speak.
#### AFlat
• 5
##### Re: Reasoning behind the Universal Accelerator
« Reply #66 on: December 29, 2021, 11:12:04 AM »
My incorrect assumption was that UA was supposed to work. If you accelerate everything universally and equally then it's indistinguishable from no effect whatsoever. If you and I and the Earth are all being accelerated at the same rate then the Earth doesn't exert any force at all, g = 0 m/s2, and we drift off into space. It's like being in a falling elevator.
I think the deal with UA is that whatever U force that is pushing up everything is shielded by the Earth...Up to a certain altitude. So picture the force like the wind, pushing upward from below the Earth. The wind pushes upward past the edges of the flat Earth disk and then curls inward to continue to push everything upwards over the entirety of the flat Earth disk somewhere below all celestial objects. That way, the earth is pushed up, the celestial bodies are pushed up along with it, yet we on terra firma are not pushed up if our feet leave the ground. We are shielded from the wind, so to speak.
That'd work, but it's implausable that a force that pervades the universe, accelerating everything from subatomic particles to galaxies, is going to even notice a little rock in its way. It rules out dark energy immediately as that stuff doesn't even slow down for ordinary matter. You're left with a mysterious force that accelerates everything uniformly except us, and does so regardless of distance.
There aren't any good candidates for that. Simpler to go with gravitation on a disk. You wouldn't even notice edge effects until you were well into the Antarctic rim.
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #67 on: December 29, 2021, 04:00:21 PM »
My incorrect assumption was that UA was supposed to work. If you accelerate everything universally and equally then it's indistinguishable from no effect whatsoever.
You forget that motion is relative. The UA article, which you should have read, explains what this acceleration is relative to.
If you and I and the Earth are all being accelerated at the same rate then the Earth doesn't exert any force at all, g = 0 m/s2, and we drift off into space. It's like being in a falling elevator.
Yes. Luckily, that's not what UA postulates.
I think the problem here is that you decided to read my comment as a personal insult or challenge. As a result, you rushed to reassert that you are righteous and just, and neglected to think that perhaps some basic knowledge of a subject would be helpful before you formed an opinion on it.
What I meant is only what I wrote, and nothing more: You've jumped into the middle of a discussion on UA without understanding what UA is, and you assumed a universal frame of reference which doesn't exist*. Don't do that.
If you feel insulted by your errors, try not to make them again.
* - n.b. I was wrong about this part, and I'm taking this opportunity to correct myself. I thought you simply assumed classical mechanics, but instead you chose to define a FoR that can't exist in UA, and which you could never hope to identify even if UA were false. A mistake "equally daft" to the one I suspected, but a distinct one still.
Simpler to go with gravitation on a disk. You wouldn't even notice edge effects until you were well into the Antarctic rim.
Ah, the classic RE'er approach. We're not looking for things that are "simple". We're looking for things which are true.
Nonetheless, you are of course wrong about your gravitational model being workable. RE'ers commonly assume that this is what we propose (you're not the first person to make things up on the spot and assume that it's what we think), so pop-sci outlets did a good job at ripping it to shreds.
« Last Edit: December 29, 2021, 04:16:17 PM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### ichoosereality
• 216
##### Re: Reasoning behind the Universal Accelerator
« Reply #68 on: December 30, 2021, 03:41:09 PM »
Dr. Edward Dowdye says that the medium of the Solar Corona bends light, not gravity. And the observations further away from the edge of the sun fails to match prediction.
http://beyondmainstream.org/nasa-scientist-says-coronas-bend-light-not-gravity/
So what? Anyone can "say" anything. Did he publish anything on this in any peer reviewed journal? Not that I can find.
If you can show beyond reasonable doubt that the journals are unbiased I'll consider your argument.
See this quote:
"Science today is locked into paradigms. Every avenue is blocked by beliefs that are wrong, and if you try to get anything published by a journal today, you will run against a paradigm and the editors will turn it down." -- Fred Hoyle, British Mathematician and Astronomer
Fred Hoyle thought that journals were biased and unwilling to publish certain topics.
Hoyle had no problems getting his ground breaking work on stellar nucleosynthesis published (in 1956). But he simply did not make a good case for the steady state theory or that flu was carried on particles in space and came to earth via solar winds. Statements like Hoyle's are invariably from folks who did not get their favored items published.
On the other hand he certainly did NOT think the earth was flat. There in lies the problem with appearling to an individual as your authority. Why accept his opinion on journals but reject his view of the standard model of our solar system and the galaxy? What is your basis for picking one but rejecting the other?
Journals are refereed by humans and humans are imperfect and biased so of course bias CAN influence what gets published. As Iceman points out, it is certainly more difficult to get published the more you are going against the current consensus view. But if that did not happen routinely, the consensus view would not have changed so much over the last 100 years. The better your data and analysis the easier it is. The increasing expansion of the universe made it in in record time due to their undeniable data. But even things with that are purely theoretical (e.g. string theory) can get in if they keep after it (the initial string theory papers were rejected for years but now it is widely though perhaps not univerally, accepted). Or continental drift or DNA based heredity and on and on.
I also don't see that any journal has refuted and contradicted him.
Contradicted him on what, his opinion that journals are biased for not publishing his pet paper? It would be extremely unusual (at least) for the editors of a journal to accept a paper countering an (unpublished) opinion.
« Last Edit: December 30, 2021, 07:44:01 PM by ichoosereality »
If "bendy light" were real the spot shape and power output of large solid-state lasers would vary depending on their orientation relative to the surface of the earth, but this is not observed thus bendy light is not real.
#### Rog
• 69
##### Re: Reasoning behind the Universal Accelerator
« Reply #69 on: December 31, 2021, 06:56:43 AM »
Quote
It really, really, really isn't. What you're looking for is Lorentz transformations and the velocity addition formula. In this case, since the bodies aren't moving at relativistic speeds relative to each other, the Galilean approach will yield a very good estimate with much less effort.
That is the velocity addition formula that I used. Just rearranged (as explained in the link) to calculate objects moving in different directions.
The only real difference in the velocity addition formulas for relativistic and non-relativistic velocities is that you have to use the reduction factor for relativistic speeds to keep from exceeding c. And I already showed in the link I provided how the Lorentz Transformation is embedded into the reduction factor. You can use the relativistic formula for non-relativistic speeds and get the same result, because the reduction factor will just be 1 (or very close to it).
John Norton has a really good explanation of it.
Quote
Indeed. As soon as you find a scenario in which two bodies are moving at relativistic speeds relative to one another, you'll be able to meaningfully consider time dilation. It's just that, so far, you haven't. Furthermore, I propose you will not be able to come up with an earthly scenario in which the speeds would even remotely approach relativistic speeds, but I'm happy for you to
I did. The earth is moving at a relativistic speed relative to our jumper.
If A= Earth B=Jumper C= Stationary observer the velocity addition formula will give you the velocity of the earth relative to the jumper. The velocity of the jumper to the earth would be the same magnitude, just opposite direction, so both are relativistic speeds if the earth has been accelerating at g for however long. So use either formula you want, but it would be nice to see some actual calculations.
If FE didn’t acknowledge that the earth was moving at relativistic speeds relative to an observer close to the surface of the earth there would be no need to explain how it is that an accelerating earth can never exceed c. While it is true that using the relativistic velocity addition formula will reduce the perceived velocity so that it doesn’t exceed c, it doesn’t reduce it to the point that a jumper wouldn’t be vaporized when he meets the ground. (While we are on the subject, the formula used in the wiki is the wrong formula. It should be be velocity addition formula, not the Lorentz Factor. And even that is calculated wrong). Also, time dilation occurs and has been measured at less than relativistic speeds. GPS has to take it into account. The Hafele–Keating experiment didn’t involve relativistic speeds.
Quote
Because the falling observer's velocity relative to the Earth will generally not exceed their terminal velocity. For a human, that would be something to the tune of 200km/h, or roughly 0.0000002c
The ‘falling observer” has no velocity. He begins inert (according to your own admission) for an “infinitesimal moment” but he he can’t acquire any velocity after that moment unless a force is applied. A body at rest stays at rest unless a force is applied. Also, remember, the point I made about being in a vacuum? That’s because I knew you would eventually bring this argument up. There is no drag or terminal velocity in a vacuum.
#### Clyde Frog
• 1043
• [kʰlaɪ̯d fɹɒg]
##### Re: Reasoning behind the Universal Accelerator
« Reply #70 on: December 31, 2021, 02:43:53 PM »
The jumper is moving at a relativistic velocity with respect to the Earth?? How do you figure?
#### Iceman
• 1823
• where there's smoke there's wires
##### Re: Reasoning behind the Universal Accelerator
« Reply #71 on: December 31, 2021, 02:59:13 PM »
Because, in The Book of Rog, when you’re standing on a chair you’re stationary, but as you step off it, the earth’s acceleration brings it to your feet at relativistic speeds, instantly vaporizing any soul unfortunate enough to do so.
Eventually he’s going to realize he keeps confounding relative velocity and relativistic speed, and/or that they can’t just be used interchangeably.
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #72 on: December 31, 2021, 03:35:45 PM »
John Norton has a really good explanation of it.
You don't need to look for more explanations. We all get what you're saying - RE'ers and FE'ers alike. It's just that what you said is ludicrously wrong, and you won't be able to progress until you've fixed your errors.
I already gave you a list. You just need to work through it.
I did. The earth is moving at a relativistic speed relative to our jumper.
Just stating it isn't enough. You're starting with a false assumption, which is why your reasoning breaks down.
C= Stationary observer
Stationary relative to what?
Eventually he’s going to realize he keeps confounding relative velocity and relativistic speed, and/or that they can’t just be used interchangeably.
Unlikely. He's been doing this for a very long time. He's actually done the whole "falling isn't a thing that happens, because time is the same as acceleration which is the same as velocity" schtick a few times before.
« Last Edit: December 31, 2021, 03:42:55 PM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### Rog
• 69
##### Re: Reasoning behind the Universal Accelerator
« Reply #73 on: January 02, 2022, 11:59:23 PM »
Quote
Stationary relative to what?
Just as I thought/ You really don’t understand how relative velocity works.
Quote
Put into words, the velocity of A with respect to C is equal to the velocity of A with respect to B plus the velocity of B with respect to C.
http://hyperphysics.phy-astr.gsu.edu/hbase/relmot.html
That’s the velocity addition formula, the very same one you said was the correct formula to use. You want to try and make it work somehow just using the relative velocity of the ground and the jumper, without an external reference frame but that’s not how the formula works. Not in relativistic or non-relativistic scenarios.
If A and C are the ground and the jumper, B would be an observer stationary relative to the earth. (standing on the ground) . I explained this in an earlier post. The velocity of the ground relative to jumper=the velocity of the ground relative to the observer + the velocity of the observer relative to the jumper. Using, j, g and o as subscripts:
Vgj=Vgo+Voj=0+.99c
Vgj=.99c
If I asked you what the relative velocity of two cars, one going 50mph and one going 30mph, would you ask me relative to what? I don’t think so because it is understood the 50mph and 30 mph are relative to the ground. The ground would be the “stationary observer”. But since the ground is one of the actors in our scenario, we have to introduce another external reference frame, that is independent of both moving objects.
To calculate the relative velocity of the cars, you use the velocity each car as measured (I) within their own reference frame(/i). You’d do the same with the jumper and the ground. There is no difference in the formulas for relativistic and non-relativistic scenarios so you would still use the velocity each observer measures within their own reference frame. except that for relativistic situations, you have to use the Lorentz Factor to prevent exceeding c. The jumper measures his velocity as zero in his own reference frame and the ground’s velocity according to FE is .99 (or something close to it) in its reference frame.
The “reduction factor” is just the Lorentz Factor. It will reduce the relative velocity so that it doesn’t exceed c, but it doesn’t reduce it to anything close to what would be considered “normal velocities”, Anytime dilation aside, anything meeting the ground would be vaporized.
This calculator uses the velocity addition formula with the Lorentz Factor the very same formula you said was the correct one Go ahead and use it with whatever values you want. It should be very easy to prove me wrong by doing and showing the calculations yourself, but so far you refuse to do so. I wonder why.
Note the symbols used “Let Vb be the be the velocity as seen by an external reference frame. That would be our “stationary observer”.
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c1
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #74 on: January 03, 2022, 12:19:51 AM »
Sorry, none of the bodies you specified are moving at anywhere close to c relative to any of them. It's great that you are now finally using the correct formula, but you can't just pull 0.99c out of your posterior and use it. You need to show your workings.
Alternatively, consider a thought experiment: jump off a chair. In your estimation, did you suddenly start turbozooming through the air at 0.99c relative to a friend that's observing you?
If I asked you what the relative velocity of two cars, one going 50mph and one going 30mph, would you ask me relative to what? I don’t think so because it is understood the 50mph and 30 mph are relative to the ground.
I already addressed this argument before you made it, and explained why you can't make this assumption during this discussion. If you simply paid attention, you'd save yourself a lot of futile typing.
To calculate the relative velocity of the cars, you use the velocity each car as measured (I) within their own reference frame(/i).
This continues to be nonsense. A car is not moving relative to itself. This has been explained to you before - just repeating the error isn't going to progress you.
Go ahead and use it with whatever values you want.
I already did. It's just that reasonable values are in the ballpark of hundreds of metres per second tops, and not 0.99c. Therein lies your tragic error.
« Last Edit: January 03, 2022, 12:33:17 AM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### Clyde Frog
• 1043
• [kʰlaɪ̯d fɹɒg]
##### Re: Reasoning behind the Universal Accelerator
« Reply #75 on: January 03, 2022, 12:38:36 AM »
An observer standing on the surface of the Earth is moving at 0.99c with respect to the surface of the Earth while still just simply standing on the surface of the Earth? That's an amazing thing to say. I have to be missing something here.
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #76 on: January 03, 2022, 12:43:40 AM »
No, no, Voj is likely the velocity of the observer relative to the jumper. The jumper just became ludicrously fast for, y'know, reasons.
I suspect he's confused by conservation of momentum. It seems that he thinks that the moment the jumper stops touching the Earth, his momentum reverts to what it was the last time he wasn't touching the Earth. In his mind, any acceleration that happened in between jumps just gets undone, so the jumper instantly vapourises in the atmolayer. Try reading through his posts again with that assumption in mind. A lot of it suddenly falls into place.
« Last Edit: January 03, 2022, 12:49:25 AM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### Rog
• 69
##### Re: Reasoning behind the Universal Accelerator
« Reply #77 on: January 03, 2022, 01:12:01 AM »
Quote
Sorry, none of the bodies you specified are moving at anywhere close to c relative to any of them. It's great that you are now finally using the correct formula, but you can't just pull 0.99c out of your posterior and use it. You need to show your workings.
The .99 comes from your own admission that the velocity of the earth would be “approximately, away from c”
Quote
It is not close to c. It is approximately c away from c.
I interpreted that to mean almost c, but not quite. If it means something else, please clarify.
Again, the whole discussion can be put to rest and you can very easily prove me wrong by providing the velocity that the ground would approach a jumper according to an external observer and providing justification for that number.
Quote
An observer standing on the surface of the Earth is moving at 0.99c with respect to the surface of the Earth while still just simply standing on the surface of the Earth? That's an amazing thing to say. I have to be missing something here
You are missing alot. Read it again I’ll translate. The velocity of the ground relative to the jumper=the velocity of the ground relative to the observer, which is zero plus the velocity of the observer relative to the jumper which is some undetermined number Pete is keeping secret, therefore the velocity of the ground relative to the jumper is Pete’ secret number.
Vgj=Vgo+Voj=0+ PSN
Vgj= PSN
I don’t understand why it is so difficult to answer the question “The external observer would see the ground approaching the jumper at X velocity (or close approximation) and this is why”.
I would answer that question by saying the external observer would see the ground approaching the jumper at whatever velocity the ground is moving as measured within it's own reference frame, provided it will always remain below c.
#### Pete Svarrior
• e
• Planar Moderator
• 15517
• (◕˽ ◕ ✿)
##### Re: Reasoning behind the Universal Accelerator
« Reply #78 on: January 03, 2022, 01:17:45 AM »
The .99 comes from your own admission that the velocity of the earth would be “approximately, away from c”
Quote
It is not close to c. It is approximately c away from c.
I interpreted that to mean almost c, but not quite. If it means something else, please clarify.
Read what you wrote in quotation marks, and the quote you included immediately afterwards. They are not the same thing. I said "c away from c". c-c=0
I also said it is not close to c. How did you get "almost c, but not quite" from that?
In the same post, just before my remark on the fact that the number is not close to c, I explicitly stated it's 0m/s. You know, a figure that's not 0.99c. There really was little room for misinterpretation there.
Again, the whole discussion can be put to rest and you can very easily prove me wrong by providing the velocity that the ground would approach a jumper according to an external observer and providing justification for that number.
I already did. At the time of the jumper leaving the chair, it would be 0. This is because they would (briefly) not be moving relative to the Earth, and consequently they would not be moving relative to any observer stood on the Earth. From that point onward, the jumper will accelerate over time until he reaches terminal velocity (or meets the ground).
whatever velocity the ground is moving as measured within it's own reference frame
This is still nonsense, no matter how many times you mindlessly repeat that phrase. Once again: the ground is not moving relative to itself. Measuring the velocity of anything relative to itself is a meaningless endeavour - you will always reach the answer of 0.
« Last Edit: January 03, 2022, 01:30:11 AM by Pete Svarrior »
P.S. All of us illiterate folks understood this the first time.
#### Clyde Frog
• 1043
• [kʰlaɪ̯d fɹɒg]
##### Re: Reasoning behind the Universal Accelerator
« Reply #79 on: January 03, 2022, 01:50:35 AM »
Rog this is an amazingly wonderful fail so please take a second to breathe and read. Velocity quite literally only has a definition as one body WITH RESPECT TO another body. If there is a person standing at a suspended elevation above a disc, or a giant ball, or on the top of a balcony on a space ship, or anywhere at all really where there is any sort of perceived downwards force acting on them in a gravity-like fashion, the velocity of both the Earth and the person in their own FoR is 0m/s. Neither body is moving at any percentage of c. In fact, they measure their instantaneous velocity to be exactly 0% of c, because light still moves away from them at exactly c. It doesn't matter how much time has passed. It doesn't matter how old the universe is. The velocity they measure between themselves (the observer on Earth and the one suspended above, about to jump) as 0/ms, while light continues to move away at c.
It seems weird. It's a wild thing to wrap your brain around. I mean, no matter how fast a car is driving, another car can always drive a little bit faster and notice that they are catching up with the driver in front of them. But no matter how fast any of those cars drive, every single driver always measures light moving exactly c away from them. None of them make any headway in trying to get closer to driving anywhere near as close as the light can move away from them, even in a hypothetical car that can accelerate at 10m/s/s forever. That weird car, even with its crazy infinite acceleration ability, would have a driver that would STILL always measure light moving away from them at exactly c.
« Last Edit: January 03, 2022, 01:53:23 AM by Clyde Frog » | 8,022 | 34,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.953494 |
http://experiment-ufa.ru/what-is-49-percent-of-99 | 1,529,536,139,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863939.76/warc/CC-MAIN-20180620221657-20180621001657-00623.warc.gz | 102,986,253 | 6,720 | # What is 49 percent of 99 - step by step solution
## Simple and best practice solution for 49% of 99. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the calculator fields your own values, and You will get the solution.
To get the solution, we are looking for, we need to point out what we know.
1. We assume, that the number 99 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 99 is 100%, so we can write it down as 99=100%.
4. We know, that x is 49% of the output value, so we can write it down as x=49%.
5. Now we have two simple equations:
1) 99=100%
2) x=49%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
99/x=100%/49%
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for what is 49% of 99
99/x=100/49
(99/x)*x=(100/49)*x - we multiply both sides of the equation by x
99=2.04081632653*x - we divide both sides of the equation by (2.04081632653) to get x
99/2.04081632653=x
48.51=x
x=48.51
now we have:
49% of 99=48.51
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https://www.vedantu.com/physics/angular-acceleration | 1,558,305,949,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255182.37/warc/CC-MAIN-20190519221616-20190520003616-00195.warc.gz | 979,657,366 | 42,731 | # Angular Acceleration
## Angular Acceleration - Units, Equation and Examples
Angular acceleration is the rate at which angular speed changes or change in angular speed in unit time. Since it is in three dimensions, it is a pseudovector. In SI units, it is estimated in radians per second squared (rad/s2) and is generally depicted by the Greek letter alpha (α).
$\alpha = \frac{{d\omega }}{{dt}},$
where ω (omega) is the vector depicting angular velocity.
The angular acceleration of a given particle α can be connected to the torque(τ) being applied through the following equation:
$I\alpha = \tau ,$
where α is the angular acceleration of the particle and "I" is its moment of inertia.
Angular acceleration is a quantitative articulation of the change in angular speed that a turning object experiences for each unit time and it is also sometimes called as rotational acceleration. It is a vector amount, comprising of a range segment and both of the two characterized headings or faculties.
The extent, or length, of the angular acceleration vector, is, in particular, relative to the rate of progress of angular speed, and is estimated in radians per second squared (radians per second) 2 or (radians * second) - 2. On the other hand, the angular acceleration magnitude can be communicated in degrees per second squared (degrees per second) 2 or (degrees * second) - 2
The course of the angular acceleration vector is opposite to the plane in which the turn happens. If the increase in angular speed seems clockwise as per an eyewitness, at that point the angular acceleration vector points away from the onlooker. If the increase in angular acceleration seems counterclockwise, at that point the angular acceleration vector points towards the spectator.
The angular acceleration vector does not necessarily point along the same direction as the angular speed vector. Consider a vehicle moving forward along an expressway at increasing speed. The angular acceleration vectors for each of the four tires point toward the left along the lines containing the wheel axles. In the event that the vehicle quits acceleration and keeps up a steady speed, the angular acceleration vectors vanish. In the event that the vehicle backs off going ahead, the vectors turn around their headings and point toward the right, i.e., along with the lines containing the wheel axles. On the off chance that the vehicle is put into a reverse mode and increases speed while going in reverse, the angular acceleration vectors point toward the right, i.e., along with the lines containing the axles. On the off chance that the retrogressive speed is steady, the angular acceleration vectors disappear. If the regressive speed diminishes, the angular acceleration vectors point toward the left along the lines containing the wheel axles.
How is angular acceleration determined?
The angular acceleration of a turning object is the rate at which the angular speed changes with respect to the time taken. It is the adjustment in the angular speed, isolated by the adjustment in time. The normal angular acceleration is the adjustment in the angular speed, partitioned by the adjustment in time. The angular acceleration is a pseudovector that focuses toward a path along the turn pivot. The extent of the angular acceleration is given by the equation beneath. The unit of angular acceleration is radians/s2
$\alpha = \frac{{\Delta \omega }}{{\Delta t}} = \frac{{{\omega _2} - {\omega _1}}}{{{t_2} - {t_1}}}$
Δω = change in angular speed (radians/s)
Δt = change in time (s)
ω1 = introductory angular acceleration (radians/s)
t1 = introductory time (s)
t2= last time (s)
What causes angular acceleration
In physics, when an all-encompassing article is pivoted, for example, a bar, plate, or 3D square, which has its mass appropriated through space, it is necessary to consider where the power is connected. This is where the concept of torque may be observed. Torque is a proportion of the volume of a power to cause a revolution. In physics terms, the torque applied on an item relies upon the power itself (its magnitude and direction) and where the power is applied. It goes from the entirely linear path of power as something that demonstrations in a straight line, (for example, when you drive a cooler up an incline) to its angular partner, torque.
Torque carries powers into the rotational world. Most items aren't simply tough and inflexible masses, so on the probability that they are pushed, they move as well as turn. For instance, in the event that you apply a power externally to a carousel, you don't move the carousel far from its present area — you cause it to begin turning.
The torque is a vector. The extent or magnitude of the torque discloses the capacity of the torque to create a revolution; all the more explicitly, the size of the torque is relative to the angular acceleration it produces. The bearing of the torque is along the hub of this angular acceleration.
Some important terms that are associated with angular acceleration:
• 1. Torque
• Torque, momentum, or moment of force is the variable which can be compared to linear force. The idea began with the investigations of Archimedes on the use of switches. Just as a linear force is a push or energy, a torque can be considered a twist to an article. The symbol for torque is normally ” τ”, the lowercase Greek letter tau. While being alluded to a moment of force, it is generally signified by M.
Since it is in three dimensions, the torque is a pseudovector. In the case of point particles, it is given by the cross result of the position vector (separate vector) and the power vector. The magnitude of torque of an inflexible body relies upon three variables: the force that is applied, the switch arm vector interfacing the root to the point of force application, and the angle between the power and switch arm vectors. In symbols:
Where,
τ = torque produced
F is the force vector perpendicular to the radius vector,
× refers to the cross product, which is characterized as sizes of the particular vectors times is the point between the power vector and the switch arm vector.
The SI unit for torque is N⋅m.
A force connected at a correct edge to a switch duplicated by its separation from the switch's support (the length of the switch arm) is its torque. A force of three newtons connected two meters from the support, for instance, applies indistinguishable torque from a force of one newton connected six meters from the support. The course of the torque can be controlled by utilizing the correct hand grasp rule: if the fingers of the right hand are twisted from the bearing of the switch arm to the heading of the power, at that point the thumb focuses toward the torque. The torque on a molecule (which has the position “r”) can be characterized as the cross product:
$\tau = r \times F,$
Where r is the position vector of the particle relative to the fulcrum, and F is the force that acts upon the particle.
The magnitude of torque is given by
$\tau = rF\sin \theta ,$
Where r is the distance between the axis of rotation and the particle, with F being the force applied and θ is the angle between the position vector and the force vector.
• 1. Angular momentum
• In physics, angular momentum (sometimes, the moment of force or rotational force) is what might be compared to linear energy. It is a very important variable in physics since it is not always available in huge amounts —the absolute angular energy of a framework stays steady except if followed up on by an outer torque.
Since it is in three dimensions, the angular momentum for a point particle is a pseudovector(r × p), the cross result of the molecule's position vector r (with respect to some root) and its force vector p = mv. This definition can be connected to each point in the continuum like solids or liquids, or physical fields. In contrast to energy, angular force depends on where the root is situated since the molecule's position is estimated from it. The angular force vector of a point particle is parallel and usually corresponding to the angular speed vector ω (omega) of the molecule (how quick it’s angular position changes), where the steady of proportionality relies upon both the mass of the molecule and its separation from starting point. For consistent inflexible bodies, however, the turn angular acceleration ω (omega) is relative and may not be constantly parallel to the angular force of the article, making the balance of proportionality I (called the force of dormancy) a second-rank tensor instead of a scalar.
Angular energy is additive; the all-out angular force of a framework is the pseudovector aggregate of the angular momentum.
For continuums or fields, one uses reconciliation. The complete angular force of any inflexible body can be part into the whole of two primary segments: the angular energy of the focal point of mass (with a mass equivalent to the all-out mass) about the root, in addition to the turn angular energy of the article about the focal point of mass.
• 2. Angular velocity (speed)
• In physics, the angular velocity (speed) of a molecule is the rate at which it pivots around a picked focus point: that is, the time rate of progress of its angular dislodging with respect to the birthplace (in simple terms: how rapidly an item circumvents another item over some undefined time frame - for example how quick the earth circles the sun). It is estimated in edge per unit time, radians per second in SI units, and is typically depicted by the symbol omega (ω, rarely Ω). By tradition, positive angular velocity (speed) demonstrates counter-clockwise turn, while negative is clockwise.
Since it is three dimensions, angular acceleration is a pseudovector, with its scale (magnitude) estimating the rate of a pivot, and its heading pointing along the hub of revolution (opposite to the range and speed vectors). The up-or-down introduction of angular speed is ordinarily indicated by the right-hand rule.
Some real-life applications of angular acceleration:
• a) When quarterbacks throw the football, they insert a spin with their fingers, so that the ball spins swiftly as it flies through the air. Football fans define a good pass as a tight spiral.
• b) The pretty spiral pattern around the muzzle's exit hole are grooves which are cut into the barrel of a gun.
• c) Bullets which come out of a rifled barrel have grooves cut into them to have a spin on them when they are shot. | 2,198 | 10,513 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-22 | latest | en | 0.920732 |
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# sec9_4 - 9.4 Models for Population Growth 1 The Law of...
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Unformatted text preview: 9.4 Models for Population Growth 1. The Law of Natural Growth Definition 1.1. If P (t) is the value of a quantity at time t and if the rate of change of P with respect to time t is proportional to its size P (t) at any time, then dP = kP. dt This differential equation is called the law of natural growth. If k > 0, then the population increases; if k < 0, it decreases. Remark 1.1. Notice the above equation is separable. 2. The Logistic Model A population often increases exponentially in its early stages but levels off eventually and approaches its carrying capacity because of limited resources. Denoting K as its carrying capacity, we obtain the logistic differential equation dP P = kP 1 − dt K The solution to the logistic equation is K − P0 K , where A = P (t) = −kt 1 + Ae P0 with the initial condition P (0) = P0 . 1 Section 9.4 Models for Population Growth 2 Example 2.1 (problem 8). Biologist stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 10,000. The number of fish tripled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. (b) How long will it take for the population to increase to 5,000? ...
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Ask a homework question - tutors are online | 529 | 1,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-09 | latest | en | 0.864534 |
https://2022.help.altair.com/2022/hwsolvers/ja_jp/ms/topics/solvers/ms/optimization_optimization_input_addresponse_method_r.htm | 1,695,734,787,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00175.warc.gz | 90,892,552 | 11,541 | ## 例
4バーモデルの場合は、次の3つの応答を追加します：
• カプラーリンクのCMが占めるX座標が移動経路に対して持つ偏差。
• カプラーリンクのCMが占めるY座標が移動経路に対して持つ偏差。
• カプラーリンクのCMが占めるAZ座標が移動経路に対して持つ偏差。
RMS2応答を使用してこの偏差を計算します。
def addResponses (self):
m = self.mbsModel
# Coupler-DX curve: Spline
x1=(0 , 0.05, 0.1 , 0.15, 0.2 , 0.25, 0.3 , 0.35, 0.4 , 0.45, 0.5 , 0.55, 0.6 , 0.65,
0.7 , 0.75, 0.8 , 0.85, 0.9 , 0.95, 1 , 1.05, 1.1 , 1.15, 1.2 , 1.25, 1.3 , 1.35,
1.4 , 1.45, 1.5 , 1.55, 1.6 , 1.65, 1.7 , 1.75, 1.8 , 1.85, 1.9 , 1.95, 2)
y1=(199.997, 141.431, 78.0588, 17.3066, -35.233, -75.736, -102.14, -114.06, -112.56,
-99.81, -78.718, -52.477, -23.994, 5.10437, 36.8618, 82.4831, 157.392, 231.874,
260.213, 244.03, 199.997, 141.429, 78.0583, 17.3061, -35.234, -75.736, -102.14,
-114.06, -112.56, -99.81, -78.717, -52.477, -23.994, 5.10462, 36.8621, 82.4836,
157.392, 231.874, 260.213, 244.03, 199.999)
xy = list(zip(x1,y1))
# Coupler-DX curve: The Measured value
dx_coupler = "DX({marker})".format(marker=m.coupler.cm.id)
# Coupler-DX curve: The x-deviation
self.a2x = RMS2 ( label = "Coupler-DX", targetValue = xy, measuredValue = dx_coupler)
#############
# Coupler-DY curve: Spline
x2=(0 , 0.05, 0.1 , 0.15, 0.2 , 0.25, 0.3 , 0.35, 0.4 , 0.45, 0.5 , 0.55, 0.6 , 0.65,
0.7 , 0.75, 0.8 , 0.85, 0.9 , 0.95, 1 , 1.05, 1.1 , 1.15, 1.2 , 1.25, 1.3 , 1.35,
1.4 , 1.45, 1.5 , 1.55, 1.6 , 1.65, 1.7 , 1.75, 1.8 , 1.85, 1.9 , 1.95, 2)
y2=(400 , 398.307, 382.665, 353.359, 312.589, 263.960, 211.853, 160.818, 115.093,
78.2737, 53.181, 41.9427, 46.3684, 68.8531, 113.983, 187.585, 275.874, 337.05,
369.449, 389.534, 400.001, 398.305, 382.665, 353.359, 312.589, 263.959, 211.853,
160.818, 115.093, 78.2735, 53.1808, 41.9426, 46.3685, 68.8534, 113.983, 187.586,
275.875, 337.05, 369.449, 389.534, 400)
xy2 = list(zip(x2,y2))
# Coupler-DY curve: The Measured value
dy_coupler = "DY({marker})".format(marker=m.coupler.cm.id)
# Coupler-DX curve: The y-deviation
self.a2y = RMS2 ( label = "Coupler-DY", targetValue = xy2, measuredValue = dy_coupler)
#############
# Coupler-PSI curve: Spline
x3=(0 , 0.05, 0.1 , 0.15, 0.2 , 0.25, 0.3 , 0.35, 0.4 , 0.45, 0.5 , 0.55, 0.6 , 0.65,
0.7 , 0.75, 0.8 , 0.85, 0.9 , 0.95, 1 , 1.05, 1.1 , 1.15, 1.2 , 1.25, 1.3 , 1.35,
1.4 , 1.45, 1.5 , 1.55, 1.6 , 1.65, 1.7 , 1.75, 1.8 , 1.85, 1.9 , 1.95, 2)
y3=(0.0 , -1.5036, -0.9390, 1.2498, 4.8231, 9.6362, 15.5739, 22.5081, 30.2735,
38.6522, 47.3546, 55.9772, 63.8940, 69.9926, 72.0582, 65.9705, 48.8237, 28.1874,
13.1512, 4.3989, 0.0 , -1.5036, -0.939 , 1.2499, 4.8231, 9.6363, 15.574,
22.5081, 30.2736, 38.6522, 47.3547, 55.9772, 63.8941, 69.9927, 72.0582, 65.9704,
48.8235, 28.1873, 13.1511, 4.3988, 0.0)
xy3 = list(zip(x3,y3))
# Coupler-PSI curve: Measured value
psi_coupler = "RTOD*AZ({marker})".format(marker=m.coupler.azMarker.id)
# Coupler-PSI curve: Z-rotation deviation
self.a2psi = RMS2 ( label = "Coupler-PSI", targetValue = xy3, measuredValue = psi_coupler)
return
## 例
4バー機構でクランク-ロッカー運動を保証するために、このモデルは次のグラスホス条件を満たす必要があります：
• クランクは最短のリンクであること。
• 最短と最長の長さの合計は、他の2つの長さの合計未満であること。
def addResponses (self):
…
# Crank needs to be the shortest link
# This means crank is shorter than any other links
self.cons1 = ResponseExpression(
label = "crank < base",
function = "(dx-ax)**2 + (dy-ay)**2 - (bx-ax)**2 - (by-ay)**2",
mapping = {"ax":m.ax, "ay":m.ay, "bx":m.bx, "by":m.by, "dx":m.dx, "dy":m.dy}
)
self.cons2 = ResponseExpression(
label = "crank < follower",
function = "(dx-cx)**2 + (dy-cy)**2 - (bx-ax)**2 - (by-ay)**2",
mapping = {"ax": m.ax, "ay": m.ay, "bx": m.bx, "by": m.by,
"cx": m.cx, "cy": m.cy, "dx": m.dx, "dy": m.dy}
)
self.cons3 = ResponseExpression(
label = "crank < coupler",
function = "(cx-bx)**2 + (cy-by)**2 - (bx-ax)**2 - (by-ay)**2",
mapping = {"ax": m.ax, "ay": m.ay, "bx": m.bx, "by": m.by, "cx":m.cx, "cy":m.cy}
)
# s + l < p + q
# The crank is always the shortest
# So the sum of crank and some other link is always shorter than the sum of the other two
self.cons4 = ResponseExpression(
label = "crank + base < other two",
function = "-(bx-ax)**2 - (by-ay)**2 - (dx-ax)**2 - (dy-ay)**2 + (cx-dx)**2 + (cy-dy)**2 + (bx-cx)**2 + (by-cy)**2",
mapping = {"ax": m.ax, "ay": m.ay, "bx": m.bx, "by": m.by,
"cx": m.cx, "cy": m.cy, "dx": m.dx, "dy": m.dy}
)
self.cons5 = ResponseExpression(
label = "crank + coupler < other two",
function = "-(bx-ax)**2 - (by-ay)**2 - (cx-bx)**2 - (cy-by)**2 + (dx-cx)**2 + (dy-cy)**2 + (dx-ax)**2 + (dy-ay)**2",
mapping = {"ax": m.ax, "ay": m.ay, "bx": m.bx, "by": m.by,
"cx": m.cx, "cy": m.cy, "dx": m.dx, "dy": m.dy}
)
self.cons6 = ResponseExpression(
label = "crank + follower < other two",
function = "-(bx-ax)**2 - (by-ay)**2 - (dx-cx)**2 - (dy-cy)**2 + (cx-bx)**2 + (cy-by)**2 + (dx-ax)**2 + (dy-ay)**2",
mapping = {"ax": m.ax, "ay": m.ay, "bx": m.bx, "by": m.by,
"cx": m.cx, "cy": m.cy, "dx": m.dx, "dy": m.dy}
) | 2,505 | 4,858 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.272715 |
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# My Jackpot games numbers:
Topic closed. 1114 replies. Last post 3 years ago by marcie.
Page 3 of 75
pennsylvania
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Posted: December 28, 2011, 10:38 pm - IP Logged
I just saw that...WOW....that's amazing!! Do you use the same numbers all the time? Do you ever pick numbers for each state like pick 3,4 or 5.... sorry for all the questions, i`m new.
Ohio
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Posted: December 28, 2011, 10:48 pm - IP Logged
3 came in Mega Million Dec. 27th. LottoGuyBC and I had 3 actually I just looked Use some mirror numbers.
I had 32 33 43 Mega Million
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
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Posted: December 28, 2011, 10:50 pm - IP Logged
I just saw that...WOW....that's amazing!! Do you use the same numbers all the time? Do you ever pick numbers for each state like pick 3,4 or 5.... sorry for all the questions, i`m new.
Yes I will be using these more the ones I had listed thats ok I don't mind. Yes I do it all the time.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
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February 21, 2007
34918 Posts
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Posted: December 28, 2011, 10:57 pm - IP Logged
Your list had 4 out 5 numbers for the wb`s in powerball....14,16,51,52....WTG!!!
What day was that?
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
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February 21, 2007
34918 Posts
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Posted: December 28, 2011, 10:59 pm - IP Logged
01,02,03,04,05,06,
11,12,13,14,15,16,
21,22,23,24,25,26,
31,32,33,34,35,36,
41,42,43,44,45,46,
51,52,53,54,55,56 switch them up~
Better put this back. Just remember to use the low and High balls. thats what Gail Howard saids.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
pennsylvania
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October 22, 2011
48 Posts
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Posted: December 29, 2011, 7:09 am - IP Logged
Thank you... the only number you didn`t have on your list for powerball last night was 27..sooooo close...wow again!!!!
Ohio
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February 21, 2007
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Posted: December 29, 2011, 8:37 am - IP Logged
Thank you... the only number you didn`t have on your list for powerball last night was 27..sooooo close...wow again!!!!
Did you get all five? P.B was 14 (16,21,27,41,45,14) Multiplyier X2
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
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February 21, 2007
34918 Posts
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Posted: December 29, 2011, 8:59 am - IP Logged
Ok I see now you are taking about Dec. 24th. 14,16,51,52
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
pennsylvania
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48 Posts
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Posted: December 29, 2011, 7:56 pm - IP Logged
Good evening Marcie...no I did not get all 5, I seem to not be able to get all on the same line but I truely believe I will someday. I am really impressed how close your picks are. I play everyday & had my first pick 3 hit last week. I thought that was awesome then I saw your posts.
Ohio
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February 21, 2007
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Posted: December 30, 2011, 9:49 am - IP Logged
Good evening Marcie...no I did not get all 5, I seem to not be able to get all on the same line but I truely believe I will someday. I am really impressed how close your picks are. I play everyday & had my first pick 3 hit last week. I thought that was awesome then I saw your posts.
Well thanks I am glad you were able to get some winners. Lets shoot for the 6 in theJackPot Games any wins are good though.
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
pennsylvania
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October 22, 2011
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Posted: December 30, 2011, 7:02 pm - IP Logged
Sounds good to me...I played some tonight for MM but I`m on a budget & can`t play that many lines. Here`s to a dollar & a BIG dreamer!!!! That would make for a very happy new year!!!!
pennsylvania
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Posted: December 31, 2011, 1:39 am - IP Logged
Welp you again had all 6 numbers...that`s amazing but i didn`t have any in a row. WTG, I hope you got a hit!!!?
Ohio
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Posted: December 31, 2011, 6:24 am - IP Logged
Welp you again had all 6 numbers...that`s amazing but i didn`t have any in a row. WTG, I hope you got a hit!!!?
Really let's see? Maybe I need to pay more attention to these numbers.. there were 6 but not all in one its a start though maybe who knows this might be good for us? These numbers lets do this
Mega Million was 04,24,45,46,52, P.B 01
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
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February 21, 2007
34918 Posts
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Posted: December 31, 2011, 6:38 am - IP Logged
01,02,03,04,05,06,
11,12,13,14,15,16,
21,22,23,24,25,26,
31,32,33,34,35,36,
41,42,43,44,45,46,
51,52,53,54,55,56 switch them up~
Happy New 2012 We going to get some winners
M.M. 04,23,45,46,52, My mistake B.B 01
Dec. 30th This is my project Jackpot Games
12345
67890
Use Mirror #'s Use prs. with your Key* numbers the most Vivid thing in your dream go up or down on #'s. Flip 6=9 `9=6 Bullseyes 0 or 1 for Pick 4 and the P. 5 Play the other part of doubles. Do the Whole nine yards for a P. 4* P. 5* or 0 thur 9 for P. 4 P. 5 from my dreams or hunches good Luck.. Write your Dreams down Play for 3 days. Good Luck All.
Ohio
United States
Member #49980
February 21, 2007
34918 Posts
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Posted: December 31, 2011, 7:00 am - IP Logged
04,24,45,46,42, B.B 01 the real deal Dec. 30th | 2,922 | 8,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-34 | latest | en | 0.751822 |
https://fr.slideserve.com/tevin/exercise-h-1-npv | 1,638,900,937,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00015.warc.gz | 322,543,888 | 19,400 | Exercise H.1: NPV
# Exercise H.1: NPV
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## Exercise H.1: NPV
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##### Presentation Transcript
1. Exercise H.1: NPV • A firm is reviewing an investment opportunity that requires an initial cash outlay of \$336,875 and promises to return the following irregular payments: Year 1: \$100,000 Year 2: \$82,000 Year 3: \$76,000 Year 4: \$111,000 Year 5: \$142,000 • If the required rate of return for the firm is 8 percent, what is the net present value of the investment?
2. Exercise H.1: NPV • A firm is reviewing an investment opportunity that requires an initial cash outlay of \$336,875 and promises to return the following irregular payments: Year 1: \$100,000 Year 2: \$82,000 Year 3: \$76,000 Year 4: \$111,000 Year 5: \$142,000 • If the required rate of return for the firm is 8 percent, what is the net present value of the investment?
3. Exercise H.2: Payback • The Seattle Corporation has been presented with an investment opportunity which will yield cash flows of \$30,000 per year in years 1 through 4, \$35,000 per year in years 5 through 9, and \$40,000 in year 10. This investment will cost the firm \$150,000 today, and the firm’s cost of capital is 10 percent. Assume cash flows occur evenly during the year, 1/365th each day. What is the payback period for this investment?
4. Exercise H.2: Payback • The Seattle Corporation has been presented with an investment opportunity which will yield cash flows of \$30,000 per year in years 1 through 4, \$35,000 per year in years 5 through 9, and \$40,000 in year 10. This investment will cost the firm \$150,000 today, and the firm’s cost of capital is 10 percent. Assume cash flows occur evenly during the year, 1/365th each day. What is the payback period for this investment? | 531 | 1,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.889204 |
https://tex.stackexchange.com/questions/655661/the-layout-is-not-coming-out-as-expected-after-using-subsections | 1,685,680,333,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00687.warc.gz | 598,431,650 | 34,558 | # The layout is not coming out as expected after using subsections [duplicate]
I have inserted tables and images with the \subsection to make different parts in the report and I wrote it in order but the output is not in order when getting compiled, the subsection does not come after the table I want it to. I have attached the source code below. Any help will be appreciated.
I compared and analysed codes using the MPI library to find out the performance difference between C and Python programs.
\section{Integration}
Finding out the total integration under the curve for a fixed function with limits 0 to 2.
$f(x)=\int\limits_0^1 x^2$
I have used the Trapezoidal Rule for calculating the same. It is a Numerical technique to find the definite integral of a function.The function is divided into many sub-intervals and each interval is approximated by a Trapezium. Then the area of trapeziums is calculated to find the integral which is basically the area under the curve. The more is the number of trapeziums used, the better is the approximation. \\
Algorithm for calculating the integration under the curve is as follows
\lstset{style=mystyle}
\begin{lstlisting}[language= C, caption= Code snippet from IntegrationMPI.c]
h = (b-a)/n;
local_n = n/size;
local_a = a + my_rank * local_n * h;
local_b = (local_a + local_n) * h;
integral = Trap(local_a, local_b, local_n, h);
if (my_rank == 0){
total = integral;
for (source = 1; source < size; source++){
MPI_Recv(&integral, 1, MPI_FLOAT, source, tag, MPI_COMM_WORLD, &status);
total += integral;
}
}
else {
MPI_Send(&integral, 1, MPI_FLOAT, dest, tag, MPI_COMM_WORLD);
}
\end{lstlisting}
\subsection{For Serial Program in C and Python}
\begin{table}[h!]
\centering
\begin{tabular}{||c|c|c||}
\hline
No. of Trapeziums & Time Taken (s) & Accuracy \\ [0.5ex]
\hline\hline
1000 & 0.001082 & 2.66666867 \\
\hline
2000 & 0.001272 & 2.66666743 \\
\hline
3000 & 0.002326 & 2.66666681 \\
\hline
4000 & 0.003009 & 2.66666675 \\
\hline
5000 & 0.003527 & 2.66666672 \\
\hline
6000 & 0.004466 & 2.66666670 \\
\hline
7000 & 0.004189 & 2.66666669 \\
\hline
8000 & 0.004368 & 2.66666669 \\
\hline
9000 & 0.007463 & 2.66666669 \\
\hline
10000 & 0.007001 & 2.66666669 \\ [1ex]
\hline
\end{tabular}
\caption{Data for Python Code}
\label{table:1}
\end{table}
\begin{table}[h!]
\centering
\begin{tabular}{||c|c|c||}
\hline
No. of Trapeziums & Time Taken (s) & Accuracy \\ [0.5ex]
\hline\hline
1000 & 0.000012 & 2.66662788 \\
\hline
2000 & 0.000030 & 2.66671276 \\
\hline
3000 & 0.000032 & 2.66658711 \\
\hline
4000 & 0.000052 & 2.66660619 \\
\hline
5000 & 0.000058 & 2.66650033 \\
\hline
6000 & 0.000064 & 2.66644287 \\
\hline
7000 & 0.000074 & 2.66667461 \\
\hline
8000 & 0.000084 & 2.66645622 \\
\hline
9000 & 0.000084 & 2.66647649 \\
\hline
10000 & 0.000103 & 2.66674614 \\ [1ex]
\hline
\end{tabular}
\caption{Data for C Code}
\label{table:1}
\end{table}
\begin{figure}[h!]
\centering
\includegraphics[width=1\textwidth]{Images/serial.png}
\caption{Comparison between C and Python serial programs}
\end{figure}
.
\subsection{For MPI Program in C and Python}
\begin{table}[h!]
\centering
\begin{tabular}{||c|c|c||}
\hline
No. of Trapeziums & Time Taken (s) & Accuracy \\ [0.5ex]
\hline\hline
1000 & 0.000549 & 2.66017401 \\
\hline
2000 & 0.000840 & 2.66341850 \\
\hline
3000 & 0.001900 & 2.66450082 \\
\hline
4000 & 0.001127 & 2.66504213 \\
\hline
5000 & 0.002618 & 2.66536696 \\
\hline
6000 & 0.002004 & 2.66558354 \\
\hline
7000 & 0.002583 & 2.66573824 \\
\hline
8000 & 0.002426 & 2.66585428 \\
\hline
9000 & 0.003441 & 2.66594453 \\
\hline
10000 & 0.002971 & 2.66601674 \\ [1ex]
\hline
\end{tabular}
\caption{Data for Python Code}
\label{table:1}
\end{table}
\begin{table}[h!]
\centering
\begin{tabular}{||c|c|c||}
\hline
No. of Trapeziums & Time Taken (s) & Accuracy \\ [0.5ex]
\hline\hline
1000 & 0.000037 & 2.66016173 \\
\hline
2000 & 0.000048 & 2.66345334 \\
\hline
3000 & 0.000072 & 2.66444683 \\
\hline
4000 & 0.000076 & 2.66499472 \\
\hline
5000 & 0.000087 & 2.66526604 \\
\hline
6000 & 0.000070 & 2.66551304 \\
\hline
7000 & 0.000099 & 2.66580105 \\
\hline
8000 & 0.000108 & 2.66578150 \\
\hline
9000 & 0.000111 & 2.66587353 \\
\hline
10000 & 0.000142 & 2.66612148 \\ [1ex]
\hline
\end{tabular}
\caption{Data for C Code}
\label{table:1}
\end{table}
\begin{figure}[h!]
\centering
\includegraphics[width=1\textwidth]{Images/mpi.png}
\caption{Comparison between C and Python MPI programs}
\end{figure}
• Welcome to TeX.SE. Please tell us which document class and which main font size you employ. How wide are the page margins?
– Mico
Aug 30, 2022 at 20:16
• Especially the second, less highly voted, answer. Aug 30, 2022 at 21:44
Tables and figures are floats : even with the option [h!] LaTeX will sometimes decide to move them.
What you want is not a float : just remove the \begin{table}[h!] (or \begin{figure}[h!]) at he beginning, and the
\caption{...}
\label{...}
\end{table}
at the end. You can move the text of your captions just below your tabulars or images. For instance:
\subsection{For Serial Program in C and Python}
\begin{center}
\begin{tabular}{||c|c|c||}
(...)
\end{tabular}
Data for Python Code
\end{center}
Edit: If you want your table in the List of Tables, you can use the caption package:
\usepackage{capt-of} % or \usepackage{caption}
(...)
\end{tabular}
\captionof{table}{Data for Python Code}
\label{your_label}
\end{center} | 1,954 | 5,428 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.738032 |
http://mathhelpforum.com/calculus/92747-solved-area-infinite-volume-finite.html | 1,529,426,522,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863100.8/warc/CC-MAIN-20180619154023-20180619174023-00284.warc.gz | 200,524,603 | 10,476 | # Thread: [SOLVED] area infinite, volume finite
1. ## [SOLVED] area infinite, volume finite
The region P is bounded by the x axis , the line x =1 and the curve y = 1/x. show that
a. the area of region P is infinite.
b. show that the solid of revolution obtained by rotating region P about the x axis is finite.
a. $\displaystyle A = \int_{1}^{\infty}\frac{1}{x} dx = \lim_{t \rightarrow \infty} \ln (x) \mid_{1}^{t} = \infty - \ln(1)$ diverges
b. $\displaystyle V = \pi \int_{1}^{\infty} \frac{1}{x^2} dx = \lim_{t \rightarrow \infty} \pi(-\frac{1}{x}) \mid_{1}^{t} = \pi(0 + 1) = \pi$
Am I correct?
2. Originally Posted by diroga
The region P is bounded by the x axis , the line x =1 and the curve y = 1/x. show that
a. the area of region P is infinite.
b. show that the solid of revolution obtained by rotating region P about the x axis is finite.
a. $\displaystyle A = \int_{1}^{\infty}\frac{1}{x} dx = \lim_{t \rightarrow \infty} \ln (x) \mid_{1}^{t} = \infty - \ln(1)$ diverges
b. $\displaystyle V = \pi \int_{1}^{\infty} \frac{1}{x^2} dx = \lim_{t \rightarrow \infty} \pi(-\frac{1}{x}) \mid_{1}^{t} = \pi(0 + 1) = \pi$
Am I correct?
I don't see anything wrong with your work on this problem.
3. Yes this is perfectly correct, the shape is know as Gabriel's Horn and the problem is sometimes reffered to as The Painter's Paradox.
Gabriel's Horn - Wikipedia, the free encyclopedia
It's quite weird that you can fill the horn to the brim with a finite amount of paint, but to paint the outside you need an infinite amount.
pomp.
4. Actually, it's wierder than that! You can fill it to the brim with a finite amount of paint but the inside surface cannot be covered by paint!
5. God, that is infinitely weirder. | 561 | 1,728 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-26 | latest | en | 0.840698 |
https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_4&oldid=53219 | 1,632,596,396,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00577.warc.gz | 160,505,018 | 9,808 | # 2012 AMC 8 Problems/Problem 4
## Problem
Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?
$\textbf{(A)}\hspace{.05in}\frac{1}{24}\qquad\textbf{(B)}\hspace{.05in}\frac{1}{12}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{8}\qquad\textbf{(D)}\hspace{.05in}\frac{1}{6}\qquad\textbf{(E)}\hspace{.05in}\frac{1}{4}$
## Solution
Peter ate $1 + \frac{1}{2} = \frac{3}{2}$ slices. The pizza has $12$ slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peter ate $\boxed{\textbf{(C)}\ \frac{1}{8}}$ of the pizza. | 241 | 688 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.724702 |
https://quarknet.fnal.gov/fnal-uc/quarknet-summer-research/QNET2010/Astronomy/Plotting.html | 1,679,537,830,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00625.warc.gz | 546,981,445 | 2,731 | Overview of Stellar Spectra | Finding Spectra | Plotting and Measuring Spectra | Classifying Spectra | Links
## Plotting and Measuring Spectra
Any spreadsheet program with the ability to create graphs can be used to plot and measure spectra. Plot wavelength on the X axis, and flux on the Y axis. To zoom in on any specific line, change the minimum and maximum values for the X axis to include only the line you are interested in.
### Equivalent Widths
To compare strengths of absorption lines, equivalent widths are used. The equivalent width of an absorption line is a measure of the area of the absorption line. The width of a rectangle with a height equal to that of the continuum that has the same area of the absorption line is the equivalent width.
An Explanation of Equivalent Widths
This equation was used to calculate the equivalent widths of important absorption lines, as labeled so by the MKK book.
The Full Equation to Calculate Equivalent Widths
A is the area of the peak.
a is the height of the peak from top to bottom.
sigma is the full width of half the maximum divided by 2.35 (more info to come...)
Sigma is multiplied by the square root of 2 pi, whereas pi is 3.14159
Step 1: Find a. To do this you must find the difference of the highest and lowest part of the peak. It is referred to as Top flux and Bottom Flux. So, a = Top flux - Bottom flux.
Step 2: Find sigma. Start by finding the middle flux, which is the average of the Top flux and Bottom flux. Then, draw a line across the peak at that point. Next, record the left and right side of the peak's wavelength at the intersection of the line. Find the difference between the two wavelengths. Finally, divide that number by 2.35. The answer is sigma. So, (Top flux + Bottom flux)/2 to find the full width of half maximum. Then, right wavelength - left wavelength/ 2.35.
Step 3: Multiply by the square root of 2 * PI.
Step 4: Now that all parts of the equations are known, you can compute for A by multiplying your results from step 1, 2, and 3.
Example using data from graph above.
Step 1:
a=35 flux - 13 flux
a= 22 flux
Step 2:
mid flux= (35 flux + 13 flux)/2
mid flux= 24 flux
Draw mid flux line at 24 flux
sigma= (4353 Angstroms - 4335 Angstroms)/2.35
sigma= 7.66
Step 3:
The square root of (2 * 3.14159) = 2.507
Step 4:
A = [a * sigma * sqrt(2*PI)]/ Top flux
A = (22 * 7.66 * 2.507)/35 = 12.07 Angstroms
### By eye
The eye method is simple. The Continuum, FWHM, and Amplitude, are estimated after zooming in on a particular absorption line. The values are then plugged into an spreadsheet formula to calculate the Equivalent widths. A sample spreadsheet can be found here
### By Template
One of the summer students created a set spreadsheet templates to use in calculating equivalent widths. These templates take into account the problem of measuring a absorption line with flanking absorption lines obscuring the true amplitude and continuum. There are templates for one, two, and three lines
### By IRAF
IRAF (an acronym for Image Reduction and Analysis Facility) is a fully featured, professional astronomy program. It has built in routines for calculating equivalent widths. However it is very difficult to install, learn, and use.
One of the summer students managed to install it on his Linux netbook, and found this link on how to install it on Ubuntu | 814 | 3,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.915005 |
https://nixdoc.net/man-pages/NetBSD/man3/asinhf.3.html | 1,627,154,859,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150308.48/warc/CC-MAIN-20210724191957-20210724221957-00399.warc.gz | 445,292,245 | 4,520 | · Home
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man pages->NetBSD man pages -> asinhf (3)
Title
Content
Arch
Section All Sections 1 - General Commands 2 - System Calls 3 - Subroutines 4 - Special Files 5 - File Formats 6 - Games 7 - Macros and Conventions 8 - Maintenance Commands 9 - Kernel Interface n - New Commands
## ASINH(3)
```
```
### NAME[Toc][Back]
``` asinh, asinhf - inverse hyperbolic sine function
```
### LIBRARY[Toc][Back]
``` Math Library (libm, -lm)
```
### SYNOPSIS[Toc][Back]
``` #include <math.h>
double
asinh(double x);
float
asinhf(float x);
```
### DESCRIPTION[Toc][Back]
``` The asinh() and asinhf() functions compute the inverse hyperbolic sine of
the real argument
```
### RETURN VALUES[Toc][Back]
``` The asinh() and asinhf() functions return the inverse hyperbolic sine of
x.
```
``` acosh(3), atanh(3), exp(3), math(3)
```
### HISTORY[Toc][Back]
``` The asinh() function appeared in 4.3BSD.
BSD May 6, 1991 BSD
```
[ Back ]
Similar pages
Name OS Title asinhf FreeBSD inverse hyperbolic sine functions asinh FreeBSD inverse hyperbolic sine functions asinh OpenBSD inverse hyperbolic sine functions asinhf OpenBSD inverse hyperbolic sine functions sinh FreeBSD hyperbolic sine function sinh Linux hyperbolic sine function sinhf FreeBSD hyperbolic sine function sinh NetBSD hyperbolic sine function atanh NetBSD inverse hyperbolic tangent function atanhf NetBSD inverse hyperbolic tangent function | 453 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.338762 |
https://www.netlib.org/lapack/explore-html-3.6.1/d8/ddc/group__real_g_ecomputational_ga4f76117b1ac28f73144480945cbc5200.html | 1,716,824,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00061.warc.gz | 796,117,228 | 7,259 | LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
recursive subroutine sgetrf2 ( integer M, integer N, real, dimension( lda, * ) A, integer LDA, integer, dimension( * ) IPIV, integer INFO )
SGETRF2
Purpose:
``` SGETRF2 computes an LU factorization of a general M-by-N matrix A
using partial pivoting with row interchanges.
The factorization has the form
A = P * L * U
where P is a permutation matrix, L is lower triangular with unit
diagonal elements (lower trapezoidal if m > n), and U is upper
triangular (upper trapezoidal if m < n).
This is the recursive version of the algorithm. It divides
the matrix into four submatrices:
[ A11 | A12 ] where A11 is n1 by n1 and A22 is n2 by n2
A = [ --—|--— ] with n1 = min(m,n)/2
[ A11 ]
The subroutine calls itself to factor [ --- ],
[ A12 ]
[ A12 ]
do the swaps on [ --- ], solve A12, update A22,
[ A22 ]
then calls itself to factor A22 and do the swaps on A21.```
Parameters
[in] M ``` M is INTEGER The number of rows of the matrix A. M >= 0.``` [in] N ``` N is INTEGER The number of columns of the matrix A. N >= 0.``` [in,out] A ``` A is REAL array, dimension (LDA,N) On entry, the M-by-N matrix to be factored. On exit, the factors L and U from the factorization A = P*L*U; the unit diagonal elements of L are not stored.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,M).``` [out] IPIV ``` IPIV is INTEGER array, dimension (min(M,N)) The pivot indices; for 1 <= i <= min(M,N), row i of the matrix was interchanged with row IPIV(i).``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value > 0: if INFO = i, U(i,i) is exactly zero. The factorization has been completed, but the factor U is exactly singular, and division by zero will occur if it is used to solve a system of equations.```
Date
June 2016
Definition at line 115 of file sgetrf2.f.
115 *
116 * -- LAPACK computational routine (version 3.6.1) --
117 * -- LAPACK is a software package provided by Univ. of Tennessee, --
118 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
119 * June 2016
120 *
121 * .. Scalar Arguments ..
122 INTEGER info, lda, m, n
123 * ..
124 * .. Array Arguments ..
125 INTEGER ipiv( * )
126 REAL a( lda, * )
127 * ..
128 *
129 * =====================================================================
130 *
131 * .. Parameters ..
132 REAL one, zero
133 parameter ( one = 1.0e+0, zero = 0.0e+0 )
134 * ..
135 * .. Local Scalars ..
136 REAL sfmin, temp
137 INTEGER i, iinfo, n1, n2
138 * ..
139 * .. External Functions ..
140 REAL slamch
141 INTEGER isamax
142 EXTERNAL slamch, isamax
143 * ..
144 * .. External Subroutines ..
145 EXTERNAL sgemm, sscal, slaswp, strsm, xerbla
146 * ..
147 * .. Intrinsic Functions ..
148 INTRINSIC max, min
149 * ..
150 * .. Executable Statements ..
151 *
152 * Test the input parameters
153 *
154 info = 0
155 IF( m.LT.0 ) THEN
156 info = -1
157 ELSE IF( n.LT.0 ) THEN
158 info = -2
159 ELSE IF( lda.LT.max( 1, m ) ) THEN
160 info = -4
161 END IF
162 IF( info.NE.0 ) THEN
163 CALL xerbla( 'SGETRF2', -info )
164 RETURN
165 END IF
166 *
167 * Quick return if possible
168 *
169 IF( m.EQ.0 .OR. n.EQ.0 )
170 \$ RETURN
171
172 IF ( m.EQ.1 ) THEN
173 *
174 * Use unblocked code for one row case
175 * Just need to handle IPIV and INFO
176 *
177 ipiv( 1 ) = 1
178 IF ( a(1,1).EQ.zero )
179 \$ info = 1
180 *
181 ELSE IF( n.EQ.1 ) THEN
182 *
183 * Use unblocked code for one column case
184 *
185 *
186 * Compute machine safe minimum
187 *
188 sfmin = slamch('S')
189 *
190 * Find pivot and test for singularity
191 *
192 i = isamax( m, a( 1, 1 ), 1 )
193 ipiv( 1 ) = i
194 IF( a( i, 1 ).NE.zero ) THEN
195 *
196 * Apply the interchange
197 *
198 IF( i.NE.1 ) THEN
199 temp = a( 1, 1 )
200 a( 1, 1 ) = a( i, 1 )
201 a( i, 1 ) = temp
202 END IF
203 *
204 * Compute elements 2:M of the column
205 *
206 IF( abs(a( 1, 1 )) .GE. sfmin ) THEN
207 CALL sscal( m-1, one / a( 1, 1 ), a( 2, 1 ), 1 )
208 ELSE
209 DO 10 i = 1, m-1
210 a( 1+i, 1 ) = a( 1+i, 1 ) / a( 1, 1 )
211 10 CONTINUE
212 END IF
213 *
214 ELSE
215 info = 1
216 END IF
217 *
218 ELSE
219 *
220 * Use recursive code
221 *
222 n1 = min( m, n ) / 2
223 n2 = n-n1
224 *
225 * [ A11 ]
226 * Factor [ --- ]
227 * [ A21 ]
228 *
229 CALL sgetrf2( m, n1, a, lda, ipiv, iinfo )
230
231 IF ( info.EQ.0 .AND. iinfo.GT.0 )
232 \$ info = iinfo
233 *
234 * [ A12 ]
235 * Apply interchanges to [ --- ]
236 * [ A22 ]
237 *
238 CALL slaswp( n2, a( 1, n1+1 ), lda, 1, n1, ipiv, 1 )
239 *
240 * Solve A12
241 *
242 CALL strsm( 'L', 'L', 'N', 'U', n1, n2, one, a, lda,
243 \$ a( 1, n1+1 ), lda )
244 *
245 * Update A22
246 *
247 CALL sgemm( 'N', 'N', m-n1, n2, n1, -one, a( n1+1, 1 ), lda,
248 \$ a( 1, n1+1 ), lda, one, a( n1+1, n1+1 ), lda )
249 *
250 * Factor A22
251 *
252 CALL sgetrf2( m-n1, n2, a( n1+1, n1+1 ), lda, ipiv( n1+1 ),
253 \$ iinfo )
254 *
255 * Adjust INFO and the pivot indices
256 *
257 IF ( info.EQ.0 .AND. iinfo.GT.0 )
258 \$ info = iinfo + n1
259 DO 20 i = n1+1, min( m, n )
260 ipiv( i ) = ipiv( i ) + n1
261 20 CONTINUE
262 *
263 * Apply interchanges to A21
264 *
265 CALL slaswp( n1, a( 1, 1 ), lda, n1+1, min( m, n), ipiv, 1 )
266 *
267 END IF
268 RETURN
269 *
270 * End of SGETRF2
271 *
subroutine strsm(SIDE, UPLO, TRANSA, DIAG, M, N, ALPHA, A, LDA, B, LDB)
STRSM
Definition: strsm.f:183
integer function isamax(N, SX, INCX)
ISAMAX
Definition: isamax.f:53
subroutine sgemm(TRANSA, TRANSB, M, N, K, ALPHA, A, LDA, B, LDB, BETA, C, LDC)
SGEMM
Definition: sgemm.f:189
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
subroutine slaswp(N, A, LDA, K1, K2, IPIV, INCX)
SLASWP performs a series of row interchanges on a general rectangular matrix.
Definition: slaswp.f:116
recursive subroutine sgetrf2(M, N, A, LDA, IPIV, INFO)
SGETRF2
Definition: sgetrf2.f:115
subroutine sscal(N, SA, SX, INCX)
SSCAL
Definition: sscal.f:55
real function slamch(CMACH)
SLAMCH
Definition: slamch.f:69
Here is the call graph for this function:
Here is the caller graph for this function: | 2,300 | 6,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-22 | latest | en | 0.837356 |
http://www.ck12.org/algebra/Graphs-of-Absolute-Value-Equations/lesson/Graphs-of-Absolute-Value-Equations-BSC-ALG/ | 1,485,250,586,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284376.37/warc/CC-MAIN-20170116095124-00106-ip-10-171-10-70.ec2.internal.warc.gz | 393,434,044 | 47,313 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Graphs of Absolute Value Equations
## Graph two-variable absolute value equations
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Practice Graphs of Absolute Value Equations
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Graphs of Absolute Value Equations
Suppose you're keeping a math journal and you want to explain how to graph the solutions to an absolute value equation. How would you describe the process?
### Graphing Absolute Value Equations
Absolute value equations can be graphed in a way that is similar to graphing linear equations. By making a table of values, you can get a clear picture of what an absolute value equation will look like.
#### Let's graph the solutions to y=|x|\begin{align*}y=|x|\end{align*}:
Make a table of values and graph the coordinate pairs.
x\begin{align*}x\end{align*} y=|x|\begin{align*}y=|x|\end{align*}
–2 |2|=2\begin{align*}|-2|=2\end{align*}
–1 |1|=1\begin{align*} |-1|=1\end{align*}
0 |0|=0\begin{align*} |0|=0\end{align*}
1 |1|=1\begin{align*} |1|=1\end{align*}
2 |2|=2\begin{align*} |2|=2\end{align*}
3 |3|=3\begin{align*} |3|=3\end{align*}
#### The Shape of an Absolute Value Graph
Every absolute value graph will make a “V”-shaped figure. It consists of two pieces: one with a negative slope and one with a positive slope. The point of their intersection is called the vertex. An absolute value graph is symmetrical, meaning it can be folded in half on its line of symmetry. The function y=|x|\begin{align*}y=|x|\end{align*} is the parent function, or most basic function, of absolute value functions.
#### Now, lets' graph the solutions to y=|x−1|\begin{align*}y=|x-1|\end{align*} and it's parent function:
Make a table of values and plot the ordered pairs.
x\begin{align*}x\end{align*} y=|x1|\begin{align*}y=|x-1|\end{align*}
–2 |21|=3\begin{align*}|-2-1|=3\end{align*}
–1 |11|=2\begin{align*} |-1-1|=2\end{align*}
0 |01|=1\begin{align*} |0-1|=1\end{align*}
1 |11|=0\begin{align*} |1-1|=0\end{align*}
2 |21|=1\begin{align*} |2-1|=1\end{align*}
3 |31|=2\begin{align*} |3-1|=2\end{align*}
The graph of this function is seen below in green, graphed with the parent function in red.
Notice that the green function is just the parent function shifted over. The vertex is shifted over 1 unit to the right. This is because when x=1\begin{align*}x=1\end{align*}, |x1|=0\begin{align*}|x-1|=0\end{align*} since |11|=0\begin{align*}|1-1|=0\end{align*}. The vertex will always be where the value inside the absolute value is zero.
#### Graphing by Finding the Vertex
Absolute value equations can always be graphed by making a table of values. However, you can use the vertex and symmetry to help shorten the graphing process.
Step 1: Find the vertex by determining which value of x\begin{align*}x\end{align*} makes the distance zero.
Step 2: Using this value as the center of the x\begin{align*}x-\end{align*}values, choose several values greater than this value and several values less than this value.
#### Finally, let's graph y=|x+2|+3\begin{align*}y=|x+2|+3\end{align*}:
Start by considering where the vertex would be by solving for when the absolute value equals zero:
0=|x+2|x=2\begin{align*}0=|x+2| \Rightarrow x=-2\end{align*}
The vertex is at x=2\begin{align*}x=-2\end{align*}. Choose some points on either side of that to make a table of values.
x\begin{align*}x\end{align*} y=|x+2|+3\begin{align*}y=|x+2|+3\end{align*}
–4 |4+2|+3=5\begin{align*}|-4+2|+3=5\end{align*}
–3 |3+2|+3=4\begin{align*} |-3+2|+3=4\end{align*}
–2 |2+2|+3=3\begin{align*} |-2+2|+3=3\end{align*}
–1 |1+2|+3=4\begin{align*} |-1+2|+3=4\end{align*}
0 |0+2|+3=5\begin{align*} |0+2|+3=5\end{align*}
### Examples
#### Example 1
Earlier, you were asked what the process for graphing an absolute value equation is.
There are multiple ways to graph an absolute value equation. One way is to create a table of values and the graph those values.
Another way is to first find the vertex by determining what value of the variable makes the distance 0. Then, you can choose several values greater and less than that value to use to plot points. This technique reduces the number of values you need to test to plot the whole shape of the graph.
#### Example 2
Graph y=|x+5|\begin{align*}y=|x+5|\end{align*}.
Determine which x\begin{align*}x-\end{align*}value equals a distance of zero.
0x=|x+5|=5\begin{align*}0& =|x+5|\\ x& =-5\end{align*}
Therefore, (–5, 0) is the vertex of the graph and represents the center of the table of values.
Create the table and plot the ordered pairs.
x\begin{align*}x\end{align*} y=|x+5|\begin{align*}y=|x+5|\end{align*}
–7 |7+5|=2\begin{align*} |-7+5|=2\end{align*}
–6 |6+5|=1\begin{align*} |-6+5|=1\end{align*}
–5 |5+5|=0\begin{align*} |-5+5|=0\end{align*}
–4 |4+5|=1\begin{align*} |-4+5|=1\end{align*}
–3 |3+5|=2\begin{align*} |-3+5|=2\end{align*}
### Review
In 1–11, graph the function.
1. y=|x+3|\begin{align*}y=|x+3|\end{align*}
2. y=|x6|\begin{align*}y=|x-6|\end{align*}
3. y=|4x+2|\begin{align*}y=|4x+2|\end{align*}
4. y=x34\begin{align*}y=\left |\frac{x}{3}-4\right |\end{align*}
5. |x4|=y\begin{align*}|x-4|=y\end{align*}
6. |x2|=y\begin{align*}-|x-2|=y\end{align*}
7. y=|x|2\begin{align*}y=|x|-2\end{align*}
8. y=|x|+3\begin{align*}y=|x|+3\end{align*}
9. y=12|x|\begin{align*}y=\frac{1}{2} |x|\end{align*}
10. y=4|x|2\begin{align*}y=4|x|-2\end{align*}
11. y=12x+6\begin{align*}y=\left |\frac{1}{2} x\right |+6\end{align*}
Mixed Review
1. Graph the following inequality on a number line: 2w<6\begin{align*}-2 \le w<6\end{align*}.
2. Is n=4.175\begin{align*}n=4.175\end{align*} a solution to |n3|>12\begin{align*}|n-3|>12\end{align*}?
3. Graph the function g(x)=72x8\begin{align*}g(x)=\frac{7}{2} x-8\end{align*}.
4. Explain the pattern: 24, 19, 14, 9,....
5. Simplify (3)((29)(2)810)\begin{align*}(-3)\left (\frac{(29)(2)-8}{-10}\right )\end{align*}.
To see the Review answers, open this PDF file and look for section 6.9.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
symmetrical
An absolute value graph is symmetrical, meaning it can be folded in half on its line of symmetry.
vertex
The point of intersection of the two lines creating a V in an absolute value graph is called the vertex. | 2,303 | 6,454 | {"found_math": true, "script_math_tex": 60, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2017-04 | longest | en | 0.651854 |
http://stackoverflow.com/questions/3561183/python-datetime-date-difference-between-two-days | 1,427,658,583,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131298684.43/warc/CC-MAIN-20150323172138-00156-ip-10-168-14-71.ec2.internal.warc.gz | 259,174,338 | 16,639 | python, datetime.date: difference between two days
I'm playing around with 2 objects {@link http://docs.python.org/library/datetime.html#datetime.date}
I would like to calculate all the days between them, assuming that date 1 >= date 2, and print them out. Here is an example what I would like to achieve. But I don't think this is efficient at all. Is there a better way to do this?
```# i think +2 because this calc gives only days between the two days,
# i would like to include them
daysDiff = (dateTo - dateFrom).days + 2
while (daysDiff > 0):
rptDate = dateFrom.today() - timedelta(days=daysDiff)
print rptDate.strftime('%Y-%m-%d')
daysDiff -= 1
```
-
This about the best you can do. What's inefficient? What don't you like? – S.Lott Aug 25 '10 at 2:42
You might want to mark Ben James' answer as your preferred answer by clicking the check mark below the answer's votes. – tzot Sep 23 '10 at 13:58
I don't see this as particularly inefficient, but you could make it slightly cleaner without the while loop:
``````delta = dateTo - dateFrom
for delta_day in range(0, delta.days+1): # Or use xrange in Python 2.x
print dateFrom + datetime.timedelta(delta_day)
``````
(Also, notice how printing or using `str` on a `date` produces that `'%Y-%m-%d'` format for you for free)
It might be inefficient, however, to do it this way if you were creating a long list of days in one go instead of just printing, for example:
``````[dateFrom + datetime.timedelta(delta_day) for delta_day in range(0, delta.days+1)]
``````
This could easily be rectified by creating a generator instead of a list. Either replace `[...]` with `(...)` in the above example, or:
``````def gen_days_inclusive(start_date, end_date):
delta_days = (end_date - start_date).days
for day in xrange(delta_days + 1):
yield start_date + datetime.timedelta(day)
``````
Whichever suits your syntax palate better.
-
Thank you Ben =) I will absolutely use your advice. I just thought there might be a lib that I can't find that does cal/date/datetime such transformation without manual traversing. – Krolique Aug 25 '10 at 3:55
Ben, time and time again i find this very useful! ty – Krolique Jan 20 '11 at 14:33 | 577 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-14 | latest | en | 0.895382 |
https://becalculator.com/78-74-inches-to-feet-converter-2.html | 1,718,670,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00335.warc.gz | 109,255,538 | 17,202 | 78.74 inches to feet converter
How many feet are there in an inch?
Let’s discuss ways to determine between units of length, such as converting 78.74 in in feet. How tall is 78.74 inches to feet?
You are able to calculate 78.74 inches to ft if you know how much is 1 inch to ft.
1 inch equals 0.083333 feet.
Answer the following questions regarding an inch in feet:
• What is the number of 1 inch to foot?
• How much is 1 inch to feet?
• What is formula to convert inches to feet?
• How to change 1 in in feet?
Meaning of Inch
An inch (symbol in) is an Anglo-American measurement of length measurement.. The symbol is in. In many European languages, “inch” can be used interchangeably with, or is derived from “thumb”. The thumb of a man is approximately an inch in width.
Use:
• Electronic components such as the dimensions of the display.
• Size of truck or car tires.
Feet or foot (symbol: ft) is a measure of length in the Anglo-American customary system of measuring It equals 1/3 of a yard and 12 inches.
Application:
• For measuring heights, shorter distances, field lengths.
• Human foot size.
How to Turn 78.74 Inches into Feet?
Every country and region has its own system of conversion. So what’s 78.74 inches to ft?
To convert an amount in inches into the equivalent value in feet, simply multiply the amount in inches by 0.083333..
78.74 inches to feet = 78.74 inches × 0.083333 = 6.56164042 feet
• How many inches in feet?
1 in is equals to 0.083333 feet. To calcualte more, use cminchesconverter.
• What is the relationship between inches and feet?
1 foot = 12 inches
1 inch = 0.08333 feet
• What is formula for inches to feet?
The conversion factor for inches in feet is 0.083333. To get feet, simply multiply the inches by 0.083333.
• How to convert inches into feet?
ft = inch × 0.083333
For example:
78.74 inches to ft = 0.083333 × 78.74 = 6.56164042 feet
Inches to Feet Formula
Value in ft = value in inches × 0.083333
Final Thought
At this point, are you aware of how much are 78.74 inches to feet?
Our website has more details about inches in feet.
Common Inches in Feet Conversions Table
6 inches to feet 0.5 71 inches to feet 5.92 72 inches to feet 6 67 inches to feet 5.58 60 inches to feet 5 36 inches to feet 3 48 inches to feet 4 80 inches to feet 6.67
Common Inches to Feet Conversion Table
inches feet 78.34 inches 6.52830722 feet 78.39 inches 6.53247387 feet 78.44 inches 6.53664052 feet 78.49 inches 6.54080717 feet 78.54 inches 6.54497382 feet 78.59 inches 6.54914047 feet 78.64 inches 6.55330712 feet 78.69 inches 6.55747377 feet 78.74 inches 6.56164042 feet 78.79 inches 6.56580707 feet 78.84 inches 6.56997372 feet 78.89 inches 6.57414037 feet 78.94 inches 6.57830702 feet 78.99 inches 6.58247367 feet 79.04 inches 6.58664032 feet 79.09 inches 6.59080697 feet 79.14 inches 6.59497362 feet
Deprecated: Function get_page_by_title is deprecated since version 6.2.0! Use WP_Query instead. in /home/nginx/domains/becalculator.com/public/wp-includes/functions.php on line 5413 | 883 | 3,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.914956 |
http://mathhelpforum.com/calculus/45243-help-integrating-trig-functions.html | 1,480,958,197,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00241-ip-10-31-129-80.ec2.internal.warc.gz | 170,261,555 | 10,449 | # Thread: Help Integrating Trig functions
1. ## Help Integrating Trig functions
How would one go about integrating
sec(x)dx
using a substitution??
Please go through all the steps required
Thanks
2. Originally Posted by Gigabyte
How would one go about integrating
sec(x)dx
using a substitution??
Please go through all the steps required
Thanks
There's the simple but unobvious approach. Read this: Integral sec(x)
Then there's the more routine but tedious approach that involves the Weiestrass substitution. Read this: PlanetMath: Weierstrass substitution formulas
3. Originally Posted by transgalactic
i ll get a very complicated expression
is there an easier way??
Not actually an easier way but at least works.
$\frac{1}{\cos x}=\frac{\cos x}{\cos ^{2}x}=\frac{\cos x}{1-\sin ^{2}x},$ following up on this put $u=\sin x$ and solve the remaining integral. | 219 | 863 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-50 | longest | en | 0.831659 |
https://excelkid.com/how-to-separate-date-and-time-in-different-columns/ | 1,716,911,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00829.warc.gz | 196,192,010 | 52,546 | How to Separate Date and Time in Different Columns
In this tutorial, you will learn how to separate Date and Time in different columns in Excel using the DATETIME function.
The DATETIME function is designed to extract either the date or the time part from a given cell containing a date and time value in a cell.
Steps to Separate Date and Time in Different Columns
Let’s see the steps to split the date and time:
1. Type =DATETIME(A1, 1)
2. The formula returns with the date part of the cell.
3. Type =DATETIME(A1, 2)
4. The formula returns with the time part of the cell.
How the DATETIME function works
DATETIME is a user-defined function and a part of our free function library. In this section, we’ll briefly overview how to use it.
The function uses two required arguments to separate date and time:
``Function DATETIME(cellRef As Range, ReturnType As Integer) As Variant``
• cellRef is a range. This refers to the cell containing the datetime value.
• returnType is an integer type variable: This specifies what part of the datetime to return. If it’s 1, it returns the date part; if it’s 2, it returns the time part.
``````Dim dt As Variant
dt = cellRef.Value``````
Here, we declare a variant ‘dt‘ to store the value of the cell reference passed to the function.
``````' Check if the cell value is of date/time type
If IsDate(dt) Then``````
This line checks if the value ‘dt‘ is a valid date/time value.
``````Select Case ReturnType
Case 1
DATETIME = Int(dt) ' Date part only
Case 2
DATETIME = dt - Int(dt) ' Time part only
Case Else
DATETIME = "Invalid 'returnType' value. Please use 1 for date or 2 for time."
End Select``````
If the value in ‘dt’ is a valid date/time value, the function checks the value of ReturnType. In the case of 1, it extracts the date part by taking the integer part of ‘dt’.
If ReturnType is 2, it extracts the time part by subtracting the integer part of dt from dt.
``````Else
DATETIME = "Invalid data type!"
End If``````
If the value in ‘dt’ is not a valid date/time value, the function returns an error message. So, if you want to easily separate date and time parts in Excel we strongly recommend using custom functions.
More resources:
Istvan Vozar
Istvan is the co-founder of Visual Analytics. He helps people reach the top in Excel. | 567 | 2,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.699154 |
https://schoolbag.info/mathematics/barrons_ap_calculus/53.html | 1,623,528,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586390.4/warc/CC-MAIN-20210612193058-20210612223058-00172.warc.gz | 477,651,665 | 4,226 | PROPERTIES OF DEFINITE INTEGRALS - Definite Integrals - Calculus AB and Calculus BC
## CHAPTER 6 Definite Integrals
### B. PROPERTIES OF DEFINITE INTEGRALS
The following theorems about definite integrals are important.
Fundamental Theorem of calculus
The evaluation of a definite integral is illustrated in the following examples. A calculator will be helpful for some numerical calculations.
EXAMPLE 1
EXAMPLE 2
EXAMPLE 3
EXAMPLE 4
EXAMPLE 5
EXAMPLE 6
EXAMPLE 7
EXAMPLE 8
BC ONLY
EXAMPLE 9
EXAMPLE 10
BC ONLY
EXAMPLE 11
BC ONLY
EXAMPLE 12
SOLUTION: We use the method of partial fractions and set
Solving for A and B yields Thus,
EXAMPLE 13
EXAMPLE 14
EXAMPLE 15
EXAMPLE 16
EXAMPLE 17
Given find F (x).
SOLUTION:
EXAMPLE 18
If find F (x).
SOLUTION: We let u = cos x. Thus
EXAMPLE 19
Find
SOLUTION:
Here we have let and noted that
where
The limit on the right in the starred equation is, by definition, the derivative of F(x), that is, f (x).
EXAMPLE 20
Reexpress in terms of u if
SOLUTION: When u2 = x − 2, and 2u du = dx. The limits of the given integral are values of x. When we write the new integral in terms of the variable u, then the limits, if written, must be the values of u that correspond to the given limits. Thus, when x = 3, u = 1, and whenx = 6, u = 2. Then
EXAMPLE 21
If g is continuous, find
SOLUTION:
Note that the expanded limit is, by definition, the derivative of g(x) at c.
| 401 | 1,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-25 | longest | en | 0.721247 |
https://stevengoddard.wordpress.com/2012/05/16/does-hansen-ever-look-at-actual-data/ | 1,685,558,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00208.warc.gz | 622,130,106 | 32,651 | ## Does Hansen Ever Look At Actual Data?
It appears that Hansen is completely uninterested in any climate data which he hasn’t tampered with.
His claims of permanent drought are not consistent with the US drought record, which shows a decrease in drought and increase in wetness since records started in 1895.
Contiguous U.S., Palmer Modified Drought Index (PMDI), January-December 1895-2011
Just having fun
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### 4 Responses to Does Hansen Ever Look At Actual Data?
1. Sundance says:
Hansen’s mind is of the beautiful variety and unconstrained by reality.
2. Gustav says:
It is not enough to compute the slope and the offset. You should also calculate the error in the slope and in the offset from the scattering of the data. There are standard least squares formulas for doing this. In this case, the scatter is so large, I’d be surprised if the error was not similarly large. When the error is taken into account, you’d probably see that “no-change” is well within the error bar of the final result. Just a guess, mind you. I suspect the same to be the case with other results in climatology.
• iheartagw says:
There is no “error” in objective measurement (except slight such from the actual).
What you are referring to is deviation.
The trend of the data is still very evident.
• Gustav says:
The standard deviation in this case is the measure of error. Every measurement in physics always has some error associated with it. In this case what you want to measure is the slope of a linear fit to the data that is otherwise strongly scattered. The scatter of the data determines the error in your estimate of the slope. How? The standard deviation in the slope is the error. Error estimates is something that is notoriously lacking in all these discussions about climate–from all sides. For starters, we should look at how good is the hypothesis that the data is well described by a linear dependence to begin with. This is another source of the problem. The variation is not really linear at all. We should look at cycles instead. As it happens, people like Scafetta have done just this and found climate cycles perfectly matching with lunar/solar tidal interactions, as well as with the cyclic distortions of the Earth’s orbit by Jupiter and Saturn. The Earth climate system turns out to be highly sensitive to such interactions. Well, we know this, because we can observe ocean tides every day, as they react to the moon. A more careful analysis shows that the respective positions of the moon and the sun matter. So, the tides react to the sun as well. And we have known for centuries now that Jupiter does affect the Earth’s orbit in a way that is both computable (analytically) and measurable. The surprise is that this affects the Earth’s climate. But the cycles are there, clearly readable. | 612 | 2,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-23 | latest | en | 0.940593 |
http://megasoft-rapid.com/Michigan/error-variance-ols.html | 1,547,651,027,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657510.42/warc/CC-MAIN-20190116134421-20190116160421-00343.warc.gz | 149,753,409 | 8,132 | Address 2020 Polish Line Rd, Cheboygan, MI 49721 (231) 597-8014
# error variance ols Topinabee, Michigan
Then the mean squared error of the corresponding estimation is E [ ( ∑ j = 1 K λ j ( β ^ j − β j ) ) 2 ] p.220. Your cache administrator is webmaster. Though not totally spurious the error in the estimation will depend upon relative size of the x and y errors.
This is equivalent to the condition that Var ( β ~ ) − Var ( β ^ ) {\displaystyle {\text{Var}}({\widetilde {\beta }})-{\text{Var}}({\widehat {\beta }})} is a positive semi-definite matrix for every This statistic has F(p–1,n–p) distribution under the null hypothesis and normality assumption, and its p-value indicates probability that the hypothesis is indeed true. This theorem establishes optimality only in the class of linear unbiased estimators, which is quite restrictive. This is called the best linear unbiased estimator (BLUE).
Both matrices P and M are symmetric and idempotent (meaning that P2 = P), and relate to the data matrix X via identities PX = X and MX = 0.[8] Matrix Generated Sat, 15 Oct 2016 01:49:12 GMT by s_ac15 (squid/3.5.20) The value of b which minimizes this sum is called the OLS estimator for β. Generated Sat, 15 Oct 2016 01:49:12 GMT by s_ac15 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection
Greene, William H. (2002). ISBN0-691-01018-8. ^ Greene 2012, p.23-note. ^ Greene 2010, p.22. ^ Kennedy 2003, p.135. Princeton University Press. Time series model The stochastic process {xi, yi} is stationary and ergodic; The regressors are predetermined: E[xiεi] = 0 for all i = 1, …, n; The p×p matrix Qxx =
Generated Sat, 15 Oct 2016 01:49:12 GMT by s_ac15 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. In these cases, correcting the specification is one possible way to deal with autocorrelation. p.13.
Partitioned regression Sometimes the variables and corresponding parameters in the regression can be logically split into two groups, so that the regression takes form y = X 1 β 1 + It can be shown that the change in the OLS estimator for β will be equal to [21] β ^ ( j ) − β ^ = − 1 1 − For more general regression analysis, see regression analysis. See, for example, the James–Stein estimator (which also drops linearity) or ridge regression.
Generated Sat, 15 Oct 2016 01:49:12 GMT by s_ac15 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection Adjusted R-squared is a slightly modified version of R 2 {\displaystyle R^{2}} , designed to penalize for the excess number of regressors which do not add to the explanatory power of In such case the value of the regression coefficient β cannot be learned, although prediction of y values is still possible for new values of the regressors that lie in the Your cache administrator is webmaster.
ISBN978-0-19-506011-9. New York: McGraw-Hill. The coefficient of determination R2 is defined as a ratio of "explained" variance to the "total" variance of the dependent variable y:[9] R 2 = ∑ ( y ^ i − See also Bayesian least squares Fama–MacBeth regression Non-linear least squares Numerical methods for linear least squares Nonlinear system identification References ^ Hayashi (2000, page 7) ^ Hayashi (2000, page 187) ^
Correct specification. Since xi is a p-vector, the number of moment conditions is equal to the dimension of the parameter vector β, and thus the system is exactly identified. Mathematically, this means that the matrix X must have full column rank almost surely:[3] Pr [ rank ( X ) = p ] = 1. {\displaystyle \Pr \!{\big [}\,\operatorname {rank} The system returned: (22) Invalid argument The remote host or network may be down.
Generated Sat, 15 Oct 2016 01:49:12 GMT by s_ac15 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.5/ Connection Geometrically, this assumptions implies that x i {\displaystyle \mathbf {x} _{i}} and ε i {\displaystyle \varepsilon _{i}} are orthogonal to each other, so that their inner product (i.e., their cross moment) Is the mass of a singular star almost constant throughout it's lifetime? Contents 1 Linear model 1.1 Assumptions 1.1.1 Classical linear regression model 1.1.2 Independent and identically distributed (iid) 1.1.3 Time series model 2 Estimation 2.1 Simple regression model 3 Alternative derivations 3.1
Efficiency should be understood as if we were to find some other estimator β ~ {\displaystyle \scriptstyle {\tilde {\beta }}} which would be linear in y and unbiased, then [15] Var But $E(y) = E(X\beta + \varepsilon) = X\beta+E(\varepsilon)$ (since $X\beta$ is not random), so we can only have $E(y) = X\beta$ if $E(\varepsilon)=0$. –Jonathan Christensen Jan 26 '13 at 0:47 What is the most expensive item I could buy with £50? Autocorrelation can be visualized on a data plot when a given observation is more likely to lie above a fitted line if adjacent observations also lie above the fitted regression line.
Hayashi, Fumio (2000). However, generally we also want to know how close those estimates might be to the true values of parameters. If it doesn't, then those regressors that are correlated with the error term are called endogenous,[2] and then the OLS estimates become invalid. In such case the method of instrumental variables may be used to carry out inference.
The system returned: (22) Invalid argument The remote host or network may be down. For example, having a regression with a constant and another regressor is equivalent to subtracting the means from the dependent variable and the regressor and then running the regression for the Height (m) 1.47 1.50 1.52 1.55 1.57 1.60 1.63 1.65 1.68 1.70 1.73 1.75 1.78 1.80 1.83 Weight (kg) 52.21 53.12 54.48 55.84 57.20 58.57 59.93 61.29 63.11 64.47 66.28 68.10 For example, the Cobb–Douglas function—often used in economics—is nonlinear: Y = A L α K 1 − α e ε {\displaystyle Y=AL^{\alpha }K^{1-\alpha }e^{\varepsilon }\,} But it can be expressed in
it must have full rank. The independent variables can take non-linear forms as long as the parameters are linear. | 1,738 | 6,488 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-04 | latest | en | 0.828743 |
https://congruentmath.com/product/multiplying-and-dividing-decimals-by-powers-of-10-guided-notes-5th-grade-doodles-11294401/ | 1,720,950,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00188.warc.gz | 158,673,849 | 23,145 | # Multiplying and Dividing Decimals by Powers of 10 Guided Notes 5th Grade Doodles
## Overview
Teach 5th grade students to practice multiplying and dividing decimals by powers of 10 using this Guided Notes & Doodles resource. Students practice
## Why you'll love this
solving problems involving looking at the exponents and identifying patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. It contains 10 pages, including guided notes, a practice coloring worksheet, a maze, and a real-life math application. It works well as graphic organizers, scaffolded notes, and interactive notebooks. And it's artsy—if your students love color by number, color by code, or sketch notes, they'll love these lessons.
CCSS Standards: 5.NBT.A.1, 5.NBT.A.2
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Get the Multiplying and Dividing by Powers of 10 Pixel Art for some self-checking digital fun! It includes parallel lines cut by a transversal, triangle sum theorem, and exterior angles theorem
What's included with this resource:
1. Guided notes. Teach the topic with these structured notes (multiply & divide by power of 10) Integrates checks for understanding to verify your students are on the right track. - 2 pages
2. Practice worksheet. Color by code, maze, and problem sets to practice this math concept - 2 pages
3. Real-life application. Read and write about real-life uses of this math skill - 1 page
4. Answer key. Included for all of the worksheets. - 5 pages
Great for:
• Introductory Lessons
• Reteaching & Spiral Review
• Homework
• Sub Plans
• Quiz, Test, & Exam Prep
What teachers say about my Guided Notes & Doodles lessons:
• ⭐️⭐️⭐️⭐️⭐️ “Great resource and a different way to take notes. Students were engaged and used their notes to help them with solving problems later.” - Heather P.
• ⭐️⭐️⭐️⭐️⭐️ “My students really enjoyed these notes. I needed an additional resource to reteach this material before our end of the year assessment, and this was perfect.” - Ashley H.
• ⭐️⭐️⭐️⭐️⭐️ “I used this resource with students who typically struggle to remain engaged in mathematics. They remained very engaged and didn’t hesitate to fix mistakes and complete their work. Great resource!” - Carissa S.
• ⭐️⭐️⭐️⭐️⭐️ “These doodle notes kept my students engaged through the whole lesson. I was shocked how many actively participated and remembered from the notes. They are now requesting them every day.” - Rebecca S.
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If you have a question, or something is off with the lesson, please email me or post a question on TPT. I’ll do everything I can to make it right. | 635 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-30 | latest | en | 0.908056 |
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# Estimating the current value and YTM in the given cases
Assume Venture Healthcare sold bonds that have a 10-year maturity, a 12% coupon rate with annual payments, and a \$1,000 par value. Suppose that two years after the bonds were issued, the required interest rate fell to 7%. What would be the bonds value? Suppose that two years after the bonds were issued, the required interest rate rose to 13%. What would be the bonds value? What would be the value of the bonds three years after issue in each scenario above, assuming that interest rates stayed steady at either 7% or 13%?
Twin Oaks Center has a bond issue outstanding with a coupon rate of 7% and 4 years remaining until maturity. The par value of the bond is \$1,000, and the bond pays interest annually.
a) what is the current value of the bond if present market conditions justify a 14% required rate of return?
b)If Twin oaks' four year bond had semiannual coupon payments. What would be it current value?
c) Assume that it had a semiannual coupon but 20 years remaining to maturity. What is the current value under these conditions? (Assuming a 7% semiannual required rate of return).
Minneapolis health system has bonds outstanding that have four years remaining to maturity, a coupon interest rate of 9% paid annually and a \$1,000 par value.
a) What is the yield to maturity on the issue if the current market price is \$829
b) If the current market price is \$1104
c) Would you be willing to buy one of these bonds for \$829 if you required a 12% rate of return on the issue?
Six years ago, Bradford Hospital issued 20 year municipal bonds with a 7% annual coupon rate. The bonds were called today for a \$70 call premium- that is, bondholders received \$1070 for each bond. What is the realized rate of return for those investors who bought the bonds for \$1,000 when they were issued.
#### Solution Preview
Please refer attached file for better clarity of formulas.
Assume Venture Healthcare sold bonds that have a 10-year maturity, a 12% coupon rate with annual payments, and a \$1,000 par value. Suppose that two years after the bonds were issued, the required interest rate fell to 7%. What would be the bonds value? Suppose that two years after the bonds were issued, the required interest rate rose to 13%. What would be the bonds value?
Coupon amount=PMT=1000*12%=\$120
Number of coupon payments left=NPER=8
Required interest rate=RATE=7%
Maturity amount=FV= \$1,000
Bond value can be found by using PV function in MS Excel.
Bond Value=(\$1,298.56) PV(E7,E6,E5,E8)
Coupon amount=PMT=1000*12%=\$120
Number of coupon payments left=NPER=8
Required interest rate=RATE=13%
Maturity amount=FV= \$1,000
Bond value can be found by using PV function in MS Excel.
Bond Value=(\$952.01) =PV(E14,E13,E12,E15)
What would be the value of the bonds three years after issue in each scenario above, assuming that interest rates stayed steady at either 7% or 13%?
Coupon amount=PMT=1000*12%=\$120
Number of coupon payments left=NPER=7
Required interest rate=RATE=7%
Maturity amount=FV= \$1,000
Bond value can be found by using PV function in ...
#### Solution Summary
There are 4 problems. Solutions to these problems depict the methodology to estimate the current value and YTM in the given cases. Solutions are derived with the help of functions in MS Excel.
\$2.19 | 819 | 3,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-22 | longest | en | 0.950853 |
http://aven.amritalearning.com/?sub=102&brch=303&sim=1546&cnt=3632 | 1,638,005,013,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00536.warc.gz | 8,235,848 | 9,677 | Universal Law of Gravitation
# Gravitation
We have learnt about the motion of objects and force as the cause of motion in the previous chapters. We have learnt that a force is needed to change the speed or the direction of motion of an object. We always observe that an object dropped from a height falls towards the earth. We know that all the planets go around the Sun. The moon goes around the earth. In all these cases, there must be some force acting on the objects, the planets and on the moon. Isaac Newton could grasp that the same force is responsible for all these. This force is called the gravitational force.
In this chapter we shall learn about gravitation and the universal law of gravitation. We shall discuss the motion of objects under the influence of gravitational force on the earth. We shall study how the weight of a body varies from place to place. We shall also discuss the conditions for objects to float in liquids.
We know that the moon goes around the earth. An object when thrown upwards, reaches a certain height and then falls downwards. It is said that when Newton was sitting under a tree, an apple fell on him. The fall of the apple made Newton start thinking. He thought that: if the earth can attract an apple, can it not attract the moon? Is the force the same in both cases? He conjectured that the same type of force is responsible in both t he cases. He argued that at each point of its orbit, the moon falls towards the earth, instead of going off in a straight line. So, it must be attracted by the earth. But we do not really see the moon falling towards the earth
# Universal Law of Gravitation
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.
Fig.1
Let two objects A and B of masses M and m lie at a distance d from each other as shown in Fig. 1.. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses. That is,
F M*m ---------- (1)
And the force between two objects is inversely proportional to the square of the distance between them, that is,
F α 1 ⁄ d2 ---------- (2)
Combining Eqs. (1) and (2), we get
F α M x m ⁄ d2 ---------- (3)
or, F = G( (M x m) ⁄ d2) ---------- (4)
where G is the constant of proportionality and is called the universal gravitation constant. By multiplying crosswise, Eq. (4) gives
F × d 2 = G M × m
or G = F d2/ M m ---------- (5)
The SI unit of G can be obtained by substituting the units of force, distance and mass in Eq. (5) as N m2 kg–2.
The value of G was found out by Henry Cavendish (1731 – 1810) by using a sensitive balance. The accepted value of
G is 6.673 × 10–11 N m2 kg–2.
We know that there exists a force of attraction between any two objects. Compute the value of this force between you and your
friend sitting closeby. Conclude how you do not experience this force!
## Importance of the Universal Law of Gravitation.
The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:
1. the force that binds us to the earth;
2. the motion of the moon around the earth;
3. the motion of planets around the Sun; and
4. the tides due to the moon and the Sun.
# Free Fall
We have learnt that the earth attracts objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall. Is there any change in the velocity of falling objects? While falling, there is no change in the direction of motion of the objects. But due to the earth’s attraction, there will be a change in the magnitude of the velocity. Any change in velocity involves acceleration. Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore, this acceleration is called the acceleration due to the gravitational force of the earth (or acceleration due to gravity). It is denoted by g. The unit of g is the same as that of acceleration, that is, m s–2.
We know from the second law of motion that force is the product of mass and acceleration. Let the mass of the stone in be m. We already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g. Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is,
F = m g --------(6)
From Eqs. (4) and (6) we have
mg =G((M × m) ⁄ d2)
or g = G(M ⁄ d2) --------(7)
where M is the mass of the earth, and d is the distance between the object and the earth.
Let an object be on or near the surface of the earth. The distance d in Eq. (7) will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth,
mg =G((M × m) ⁄ R2) --------(8)
g = G(M ⁄ R2) --------(9)
The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator. For most calculations, we can take g to be more or less constant on or near the earth. But for objects far from the earth, the acceleration due to gravitational force of earth is given by Eq. (7).
## To calculate the value of g:
To calculate the value of g, we should put the values of G, M and R in Eq. (9), namely, universal gravitational constant,
G = 6.7 × 10–11 N m2 kg-2, mass of the earth,
M = 6 × 1024 kg, and radius of the earth,
R = 6.4 × 106 m.
g=G M/R2
6.7 10-11 N m2 kg-2 6 1024 kg / (6.4 106 m)2 = 9.8 m s–2.
Thus, the value of acceleration due to gravity of the earth, g = 9.8 m s–2.
## Mass
We have learnt in the previous chapter that the mass of an object is the measure of its inertia. We have also learnt that greater the mass, the greater is the inertia. It remains the same whether the object is on the earth, the moon or even in outer space. Thus, the mass of an object is constant and does not change from place to place.
## Weight
We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth.
We know that:
F=m*a --------(10)
That is
F =m*g --------(11)
The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same in Eq. (11), we have
w=m*g --------(12)
As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction.
We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, W m. It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. The mass of an object remains the same everywhere, that is, on the earth and on any planet whereas its weight depends on its location.
## Weight of an object on the moon
We have learnt that the weight of an object on the earth is the force with which the earth attracts the object. In the same way, the
weight of an object on the moon is the force with which the moon attracts that object. The mass of the moon is less than that of the earth. Due to this the moon exerts lesser force of attraction on objects.
Let the mass of an object be m. Let its weight on the moon be Wm. Let the mass of the moon be Mm and its radius be Rm.
By applying the universal law of gravitation, the weight of the object on the moon will be
Wm = G((Mm × m) ⁄ R2m) --------(13)
Let the weight of the same object on the earth be We. The mass of the earth is M and its radius is R.
Table-1:
Celestial body Mass(kg) Radium(m) Earth 5.98 1024 6.37 106 Moon 7.36 1022 1.74 106
From Eqs. (9) and (12) we have,
We=G ((M × m) ⁄ R2) --------(14)
Substituting the values from Table 1 in Eqs. (13) and (14), we get
Wm = G ((7.36 × 1022kg × m) ⁄ (1.74 × 106m)2)
Wm= 2.431×1010 G × m --------(15-a)
and We= 1.474 × 1011 G × m --------(15-b)
Dividing Eq. (15-a) by Eq. (15-b), we get
Wm ⁄ We = (2.431 × 1010) ⁄ (1.474 × 1011)
Wm ⁄ We = 0.165 ≈ 1 ⁄ 6 --------(16)
Wieght of the Object on Moon ⁄ Weight of the Object on Earth = 1 ⁄ 6
Weight of the object on the moon = (1/6) × its weight on the earth.
# Summary :
• The law of gravitation states that the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The law applies to objects anywhere in the universe. Such a law is said to be universal.
• Gravitation is a weak force unless large masses are involved.
• Force of gravitation due to the earth is called gravity.
• The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from poles to the equator.
• The weight of a body is the force with which the earth attracts it.
• The weight is equal to the product of mass and acceleration due to gravity.
• The weight may vary from place to place but the mass stays constant.
Cite this Simulator: | 2,463 | 9,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.956702 |
http://www.zephyr-hair.com/cheri/how-to-gamble-football-3/ | 1,601,046,626,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00259.warc.gz | 238,667,433 | 26,209 | # How To Gamble: Football
Football
To bet on football, tell the ticket writer the bet number of the team you wish to bet, with the point spread and the amount you would like to bet. The payout, unless stated otherwise, is figured at odds of 10/11. This means that a wager of \$11 would win \$10 and return \$21.
That is called a straight bet. A straight bet is the most common type of football bet.
The point spread: When betting on football, the team you bet on must”cover the spread” In other words, the team must win or not lose by a predetermined margin of points.
Note: The bottom team is always listed as the home team unless otherwise noted.
The point spread is always placed to the immediate right of the team that is favored. If you bet the Dolphins, the Dolphins must win by 7 points for you to win your wager. If you bet the Jets, any of the following will declare you a winner.
(a) The Jets win the game.
(b) The game ends in a tie.
(c) The Jets lose the game by more than 6 points.
If the Dolphins win by exactly 6 points, the wager is declared a push and all money is refunded.
You may wager that the entire score of the match will be more or less than the amount listed. It makes no difference which team covers the spread. Simply add the final score of each group. The payout, unless stated otherwise, is figured at odds of 10/11.
Sometimes, bettors have the option to discard the point spread and bet on which team will win. This is called betting on the “Money Line”
The Money Line: Odds for a game based on \$1.00. A”minus” (-) preceding the number indicates the team is a popular. A “plus” (+) preceding the number indicates the team is an underdog.
The Dolphins’ odds are -180, meaning an \$18 bet would win \$10 for a return of \$28. The Jets’ odds are +160, meaning a \$10 bet would win \$16 for a return of \$26.
Football Parlays: More than 1 team on precisely the same bet.
You may combine several teams into one wager. All teams totals must cover the point spread to win the bet. Odds and the number of teams vary from casino to casino. The following are approximate odds:
Any game that results in a push reduces the parlay one team. A two-team parlay would become a straight bet.
Parlay Cards: These offer the potential for huge return while betting as little as \$1.
Sports books offer a lot of different cards, each one having different rules. Rules for parlay cards are placed on the back of every card. Read them carefully before wagering. The cards are simple to fill out. Simply darken the boxes, or circles, that apply to the teams you want to parlay. Then darken the amount you wish to wager.
Football Teasers: A wager that improves the point spread, but at reduced odds.
Teasers cannot be straight bets.
“Teasing” the point spread is done by adding points to a underdog or by subtracting points from a favorite. This raises the likelihood of winning your bet but decreases the odds of the parlay. Odds and the number of points available to”tease” vary from casino to casino. | 695 | 3,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-40 | latest | en | 0.943973 |
http://sites.tufts.edu/hhslnews/tag/statistics/ | 1,670,586,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00842.warc.gz | 43,617,137 | 37,266 | Currently viewing the tag: "statistics"
Hello everyone, and welcome back to Hirsh Library! It’s been good seeing those of you who have been around, and I hope the rest of you can make it sometime soon. Because fun fact: we’re open our full hours! You may have heard. I know some of you have heard, because just this past October we went around asking where you were from, and now we’ve got the data to prove it!
All the graphs you see below are from the library staff walking around for one week in October, asking the question “What program/school are you with?” You can click on them to make them larger if need be. If you’re interested in comparing this set of numbers to the way things looked pre-pandemic, then I would recommend checking out the post I wrote about the October 2019 Affiliation survey (please note: we did things a little bit different this time around, so it’s not a perfect 1 to 1 comparison, but it’s solid data all the same).
I’ll start with the information everyone always asks about first: how did the programs compare, numerically? Well, over the week 810 people told us they were with the Dental school, and 568 told us they were with Med. For those who weren’t here before the pandemic landed, you’ll be interested to know two facts: 1) that the numbers are lower than they were in 2019, but that the percentages here hold steady, and 2) I was honestly surprised at how many people turned up on this survey. It’s so great to see so many people are coming into the library, even when their classes are hybrid or even primarily online. And look at the rest of this spread! Friedman, PA, PHPD, and GSBS are all turning up in good numbers, and MBS has a strong showing at 238.
Hilariously, if you compare this data to its equivalent from 2019, you’ll realize that Dental, Medical, and PA all have lower numbers than two years ago, but somehow the other programs all went up. MBS being at 238 actually places it in third place for total counted for both 2019 and 2021.
Just goes to show: we’ve got some good furniture and space here.
But how does the furniture and space effect the counts? Glad you asked!
Everyone loves the 7th floor. This has never been in question, but seeing it laid out like this really drives it home. Normally, when I see numbers like this, the 7th floor is about double the next closest floor, which fluctuates between the 5th and 6th floors. This time, the 5th floor had 426 people, and the 7th had 1021. Amazing. And heartening to see!
The remaining ways of looking at these numbers are to break it down by weekday, and by time of the day. Here are those two charts:
After the other two charts, these don’t provide anything too shocking. Thursday being the busiest day and 11am being the busiest time actually track with the way things were when we were closer to full capacity – the most popular weekday tended to fluctuate month-to-month between Tuesday, Wednesday, and Thursday, but the closer you are to lunch the busier the library gets.
Which is fair. Who doesn’t study with their stomach, after all?
I will leave you today with this thought: we have a staff full of people here to help, and we have a Reserves collection absolutely chock full of books, anatomy models, phone and laptop chargers, laptops, and about a dozen other things to help keep you supplied and comfortable while you’re studying and working on your papers, so that you can excel here at Tufts. In fact, we are stocked for numbers of visitors significantly higher than what you see in this post.
So why not stop on by to the Library Service Desk on the 4th floor of Med Ed, and see what we have to offer? We may surprise you.
In the meantime, have a great rest of your semester!
~Tom~
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Back in October 2019 and March 2020, HHSL Staff walked around and counted how many people of each program were around the library. You may remember us, with the clipboards, asking that question.
Or…to be more accurate, we walked around throughout October, and half of March. We didn’t actually end up getting all of the dates we had wanted to in March (apparently there’s a global pandemic on), so we ended up with a truncated version of the survey for that month: 4 days of data instead of 7. Still, when it comes to trying to make a better Hirsh Library for everyone, even truncated data is better than none! So here’s some of what I can see.
Fair warning: I’m going to have to extrapolate and make a couple of assumptions here, since we’re missing so much of March’s potential data. I’ve seen enough data over the years to have a good sense of what it would be, but what should and what is are always different, so maybe get a salt grain ready to take with this post. Finally, the Y axis is always going to be Number of People Counted in this post, because I want this to be as easy to read as possible!
Click to enlarge.
Click to enlarge.
So, here’s the base data. October 2019 and March 2020. As you can see, we counted…actually not that many more people, all things considered. March 3rd, 5th, and even the 11th were all right in line with what we saw in October, in terms of library population. In fact the difference between the most populated day in October and the one in March is only 34 people. Which is great!
In case you’re wondering what happened on March 13th: that was the last Friday we were open normal hours. Staff, faculty, and students were already voluntarily staying home to work from there to keep themselves safe from the rapidly growing COVID-19 threat. On March 15th, the following Sunday, Tufts made the decision to close the campuses, and Monday the 16th was the last day the library was physically staffed in person (as a note, we are very much here for you online). So what you’re seeing in that data is the effect the virus was already having on the life of the library. March 11th: relatively normal day. March 13th: signs of a new normal.
But we’re not here for discussion of the virus, we are here for discussion of the data!
Click to enlarge.
So, this is the same data as above, but oriented on what days of the week a given date was. Although the by-the-date data has its place, it’s good to know, say, what a week looks like. This is what a week and a half look like! And this is where that missing data makes me sad, because we’ll never know what the other days looked like in March. Traditionally, the weekends are the slowest days of the week, and the busiest tend to be Tuesday-Wednesday-Thursday. This is mostly matching, but…what was with that October Friday? My instinct is that it was an aberration, but without seeing in March it’s hard to tell how much of one it was. That a Friday was the busiest day we counted in October tells me that there must have been an event that day (a meeting? exams? a conference, perhaps?), but maybe it just chanced to be close to an exam.
In the end, one surprisingly busy day does not a library make. But it’s still fun to think about. Especially when you compare calendars to the data and realize that Friday, October 25th, happened to also be the second day of our pumpkin painting. Coincidence?
I think not.
Click to enlarge
Okay, last two charts! The first is the People by Floor. So this ignores dates, and focuses on the aggregate. One thing I’ve been noticing in the last year or so is that the counts we get on the 7th floor are always roughly twice that of the next closest floor (which alternates). As you can see from October, that sometimes makes for some goofy looking charts. This is one of those rare cases where the missing data actually won’t make any real difference. Barring anomalies, what you see with that chart falls in line with years of existing data. That one is one I always predict with easy clarity. Which brings us to the final, and everyone’s favorite: programs!
Click to enlarge
This is sort of wild to look at. So, okay, Dental and Medical are the programs dominating the numbers. That makes perfect sense, and honestly outside of minor variations, that’s what tends to happen. They were close in October, though, so I would have loved to see what those numbers looked like in March. Especially given the sudden notable presences of PA, PHPD, and MBS. Look at that MBS presence in March! That’s so great to see. I love it when I see sudden jumps in the number of people in the library. We’re here for everybody, after all!
That’s it for me today. Thank you for reading along, and I hope to see you all online this summer, where we are all seven days of the week. If you’re unsure the best way to reach out, well, try checking out our Ask Us page. Or hey, you can still Schedule a Consultation. There are lots of options.
Photo credit: Tarlan Sedeghat
Stay safe everyone,
Tom
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Hello hello everyone. It’s about that time again: Affiliation Survey, March 2020 edition!
Twice a year, the Hirsh Health Sciences Library runs the Affiliation Survey, where we walk around and ask every one of our lovely patrons which program they are with (Dental, Medical, PHPD, etc). This is all we ask. We will not ask for your name or even which year you’re in! The numbers we collect are used in aggregate so we can get a snapshot of what the library usage is like. You can take a look at my post from this past December to see how the data turns out.
Here’s how it goes: there are 7 days spread out over the whole month, randomly chosen to try and maximize the usefulness of the numbers (in other words, we’re trying not to get skewed by specific exam blocks too much). On those days, HHSL staff (you’ll probably recognize us!) will walk around 4 times over the course of the day to gather the totals of how many people from each school are in the library. There will be signs and posters up this month, so you’re not caught unaware.
Keep your eyes out for them!
We won’t announce ahead of time which days we’re counting (see our need to randomize, above), so it’s safe to assume that it could just happen any day this month.
Don’t panic!
If you don’t want to be bothered (or to speak out loud), you are welcome to leave your ID next to you while you study. If that is still not good enough, you can write your program down on a piece of paper and leave that next to you. If you’re in a group room, feel free to stick a note to the outside of the door (on the wood – the windows must remain clear!) telling us how many people of what program(s) are in the room. Done and done. We will add that number to our count, and we will move on to the next person!
There are pencils, pens, markers, scrap paper, and tape down at the Service Desk on the 4th floor, so you can even make your sign bright and cheerful! If you have any questions or concerns, let us know either in person at the desk, or through Ask Us on our website. We’re here to help and make this quick and easy.
But otherwise, we look forward to seeing you all next month and finding out just what our beloved HHSL looks like this fine spring.
Good luck on studying and I look forward to seeing you in the library!
Tom
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Hi everyone! Happy December. But we are here to talk about October, so…Happy October in Retrospect.
As you may recall, back in October we walked around and asked you all what school you were with. Well, this is the result!
Click to enlarge
We counted a lot of people! I for one am glad you all like us here. A few people have asked me if there were any surprises with this data, but honestly, not really. The Dental school coming in at 1274, Medical coming in at 1080, and Friedman coming in third with 231 is all the sort of thing I am used to seeing in these numbers. Which is not to say it’s bad! It’s actually quite nice to see how consistent we’ve been over the years, especially as we keep adding more and more seating (and subsequently see these numbers grow even larger from time to time).
Now for the unaware, we tend to time this survey so that we get a full week’s worth of data, even if they days themselves are spread out over the course of a month (randomized, with certain accounting for things like Indigenous Peoples’ Day). This is what that “week” ended up looking like this time:
Click to enlarge
I found this data to be the surprising data. Friday was our busiest day! We counted 674 people on the Friday we did this, and that is astonishing to me. Wednesday being 642 makes sense, and traditionally the busiest day of the week for us tends to alternate between Tuesday, Wednesday, and Thursday, and Saturday being the slowest at 281 sounds right to me. Heck, even Sunday only being 284 seems right. But Friday being the busiest day for counting? I am truly surprised by that.
Naturally, because I am me, I also broke this data up by time of the day and by floor, like so:
Click to enlarge
Click to enlarge
Remember earlier, when I was talking about the school/program breakdown being what I expected? Well, the times and floors are pretty typical of what I see month-to-month with our other data, but this is a great way to take a look at a day in the life of the library. We get crazy busy right around lunch, stay that way through the afternoon, there’s a slight tapering off around dinner, and then then the number of people in the library drops by the final count of the night (in this case, from 959 people at 6pm down to 328 at 3pm).
That graph of the floor count is frankly one of my favorite pieces of data in the library. We have been adding so many chairs and so much more furniture over the years the the amount of people on the 7th floor just keeps going up, and as of this past October we were counting over double the amount of people on that floor as any other! Just look at that: the 7th floor had 1605 people counted on it, and the 5th floor came in second at a distance (less than half!) 769 people. That’s nuts. I’m sure it doesn’t surprise anybody who’s been up there, of course, but it’s still nuts.
Click to enlarge
Finally, to bring it all home and give you a spot of context, this graph is the Circulation data from July – November, which as of writing is the most recent data I have. To be clear, it is the number of checkouts per month, not the number of people who have checked things out (because if you check out a skull, a laptop, and a book, that could just be 1 person, but 3 checkouts. See how it works?). October is far and away the most checkouts, at 2616 – September is a relatively distant second at 1924. Given the change in some curricula on campus, I don’t know what to expect for the spring this year – generally, October and April are our busiest months, but time will tell! Hopefully I’ll have some interesting stories from the data come my next summer retrospective.
Well, thank you for reading! I hope this fall brought good things for you, and I hope the winter (and subsequent spring) bring even better! Have a great day everyone, and if you come by the desk on the 4th floor make sure to say hi! I’ll probably be there.
-Tom-
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Hello hello everyone. It’s about that time again: time for the Affiliation Survey!
Twice a year, the Hirsh Health Sciences Library runs the Affiliation Survey, where we walk around and ask every one of our lovely patrons which program they are with (Dental, Medical, Sackler, etc). This is all we ask. We will not ask for your name or even which year you’re in! The numbers we collect are used in aggregate so we can get a snapshot of what the library usage is like. The graph you see above is the result of last year’s two Affiliation months, for instance.
Here’s how it goes: there are 7 days spread out over the whole month, randomly chosen to try and maximize the usefulness of the numbers (in other words, we’re trying not to get skewed by specific exam blocks too much). On those days, HHSL staff (you’ll probably recognize us!) will walk around 4 times over the course of the day to gather the totals of how many people from each school are in the library. There will be signs and posters up this month, so you’re not caught unaware.
Keep your eyes out for them!
We won’t announce ahead of time which days we’re counting (see our need to randomize, above), so it’s safe to assume that it could just happen any day this month.
Don’t panic!
If you don’t want to be bothered (or to speak out loud), you are welcome to leave your ID next to you while you study. If that is still not good enough, you can write your program down on a piece of paper and leave that next to you. If you’re in a group room, feel free to stick a note to the outside of the door (on the wood – the windows must remain clear!) telling us how many people of what program(s) are in the room. Done and done. We will add that number to our count, and we will move on to the next person!
There are pencils, pens, markers, scrap paper, and tape down at the Service Desk on the 4th floor, so you can even make your sign bright and cheerful! If you have any questions or concerns, let us know either in person at the desk, or through Ask Us on our website. We’re here to help and make this quick and easy.
But otherwise, we look forward to seeing you all this month and finding out just what our beloved HHSL looks like this fine autumn.
Good luck on studying and I look forward to seeing you in the library!
Tom
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Hi everyone,
Time for another installment of my statistics posts! Since the big annual post last June, we’ve changed a little about what we collect, so sadly I don’t have handy available data for which one of our circulation days was the busiest this past year (although I will still be telling you the month). But I can tell you right off the bat that the day we counted the most people studying was November 28th, 2018, when we counted 911 people sitting in our library over the course of the day. 911! That is so many people! Part of it, of course, is all the new seating we keep getting, because we here at Hirsh Library care about where you’re going to sit for your (frankly incredibly long) study sessions. Part of it is probably that we’re just so darn nice.
Leo the Skeleton, winner of my personal secret Nicest Colleague Award, six years running.
Anyway, you are here for some quick bar graphs and neat facts, and far be it from me to keep you from them! First off, here’s a bar graph of the last year’s worth of circulation stats (in blue) and head/seating counts (in red). These can be difficult to compare, since the numbers vary so far from one another (you may notice, for instance, that the circulation stats for a month never go above 3,000, but the seating counts never drop below that number). That is due simply to the digital nature of the schools here nowadays. Many of the students on this campus can stick with their computers, notes, and any textbooks they happen to own, and never really need to come to the desk. Which is a little too bad, given that we have so much stuff for you, but that’s life. What matters here is that heavens to Betsy we had 3,023 checkouts in October! That averages 100 a day, but I can tell you from experience (and I have numbers in past years to back this up) that not all days are created equal. Weekends and holidays are slower, so to make up for those, we would have had single days getting up around the 200 checkout mark. My guess is a Tuesday or Wednesday right before a major anatomy exam.
Click to enlarge.
Now, just because nothing is fun if it lines up properly, October was the busiest month with checkouts (I won’t get into things like questions, consultations, or the sheer volume of EndNote assistance and troubleshooting I have personally offered), but November was actually the busiest month for students in seats, to the tune of 9,274 people counted. Egad! We only have four floors in this building, and the bulk of people (nearly half!) are on the 7th floor.
Incidentally, this seems like a good moment to mention we have foam ear plugs down at the Library Service Desk. They’re free, and you can keep them. We also have headphones you can check out, which cover your ear entirely. Great for blocking noise out!
These graphs always follow the same kinds of patterns, although I was honestly surprised that November took home the gold here. With the Thanksgiving break, it has never been out front like that. But from what I’ve been hearing on the grapevine (….hydrated humerus?), the schools have been shifting their academic schedules around, so I for one will be very interested to see how that effects our numbers.
Speaking of the floors: BAM! Have a chart about the floors. Remember I mentioned how busy the 7th floor is around here? Well I do not make those claims without the numbers to back me up, and hot dang the 7th floor got crazy busy. Particularly -interestingly enough – in January. Why was January the busiest month for the 7th floor? I can answer that the same way I can answer the question “Why do we pay more for the MBTA now but the service has somehow gotten worse?” And that answer is: I don’t know. Magic? Probably magic.
Click to enlarge.
But I do find it cool (hah!) that it happened. I love little mysteries like the January 2019 one. 4,285 people counted on the 7th floor that month, the highest number of people counted on any floor in any month. And I wasn’t kidding about nearly half: that month, the 7th floor accounted for 49.8% of the 8,600 people counted in the library.
I joke around a lot on this blog, but in complete seriousness, I can speak for everyone here at Hirsh when I say: we’re glad you like it here so much, and are willing to spend so much time here. We try very hard to make sure you’re comfortable and have seating and food options for your long study sessions, so these kinds of numbers are good to see. It’s why I collect them in the first place.
Finally, did you know that back in April I wrote a post about the Affiliation Statistics we do here? If you want the full detail I recommend checking that post out, I just want to highlight the day-of-the-week breakdown I did from that post.
Click to enlarge.
You may notice that Monday is the busiest day of October, but Wednesday is the busiest day of March. That kind of thing happens a lot here, and things that seem to effect it are: weather patterns, exam schedules, food specials at the cafe, classes, and whether it’s a holiday week (Indigenous Peoples’ Day in October sometimes has a large effect, and sometimes not). I suspect that there’s also a bit of “I’ve been studying too much lately” and “Oh no Boards are upon me!” also happen, but while we collect all this data, we try very hard to not interrupt your studying.
So, where does this leave us? Well every year is a little different than years prior, and although this gives us a good blueprint for the upcoming year, the desk already feels significantly busier this week than it has in previous Orientation weeks, so there will be some improvisation as well. There always is. But if you’re like me, that’s exciting! Just means more of a challenge.
I hope you all have a great rest of your summers! Welcome back if you’re here, and if you’re not quite yet then make sure you get some beach trips in while you can. Class is around the corner.
And if you’re in the library, make sure to come say hi to me at the Library Service Desk on the 4th floor! I’m here all night.
Statistically yours,
Tom
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Hi all,
It’s been a little bit since my last post on Hirsh Library Statistics, which means it’s time again! So pour your favorite caffeinated beverage, grab a plate of free food (statistically speaking, there’s free food somewhere in this building right now), and relax as I offer you some numbers and information that you will not be tested on.
What a relief, right?
So, first up: we’ve actually held relatively steady this year as compared to last year, in terms of circulation and head counts. This probably comes as a surprise to everyone who has only been busier every year (I know I have), but there are always environmental factors to account for. For instance, if everyone’s using ebooks, they don’t need to check things out of Reserve, and in the numbers it looks like we’ve gone down. But our Open Workshops, librarian consultations, and even on the fly assistance at the Service Desk have all been super busy. So I want you to keep that in mind as you read on: everything requires a grain of salt.
Also to keep in mind, these numbers will be from July 1, 2017 to May 31, 2018. The most interesting ones are smack in the center of the school year, though. For instance: our single busiest circulation day of the year was October 4th, when we were averaging a checkout every 5 minutes. That’s nutty! And that doesn’t account for all of the research assistance, troubleshooting, or item returns.
Of the top 10 spots for busiest Circulation day of the year, 4 are in October (which accounts for the top 3 spots), 3 are in September, and then the remainder is split between November, February, and March (there was a tie for #10).
Yes, that’s right. February. February! Here, let me shock you: This year, the busiest month was October, at 3412 checkouts. Second place went to September at 2983, but the shocker is February as third place and 28868. Third! I’m glad that you all love our library so much more than the cold February weather, because dang.
Click to embiggen
Okay, so that’s the overall look of things. The blue bars are the total numbers of times we’ve checked things out, and the red bars are the total numbers of people we’ve counted in the library. Things to know: the blue bars do not count the number of people, just number of checkouts (we don’t have a way to count number of people who have checked things out). And the red bars require staff to physically walk around, so there’s an element of human error in there. But still, what a snapshot!
You may also notice that October, while still very full, was actually not the top spot in seating counts. That honor goes to March, and my best guess would be “exam season.” But October cinched its second place spot, followed (again!) by February in third. Each one of those months is over 12,000 people sitting in the library, by the way. And in case you’re wondering, October did have the single busiest day for those, at 2,590 people on the 29th. February comes in second place here, with 1,057.
Dang, what a month.
Okay, so I know what you’re thinking. “Tom, what about all those times we were asked what program we were with? What’s the deal with that?” Well good news, I have the deal with that!
For those who might not remember: HHSL staff walked around a few times a day, 7 days in the month of October and again in March, in an effort to find out how much the various programs use our library space, and we got some neat data. Behold!
Click to enlarge
Click to make big
That’s a whole lot of people. That top chart focuses on the day of the week of a given count, and the bottom compares each program. I find that shift from being busier earlier in the week to being busier in the latter half rather interesting, although I don’t know if I can find a particular reason for it. Timing of exams, perhaps?
The second chart has my attention, however. Look at some of those numbers! I wonder what happened with the PHPD and MBS numbers, though. I do know that their exams did not quite line up with our counts the way that Dental and Medical’s did, so if I had to guess I’d say that it was just an issue of two or three weeks. Crazy how much of a difference that makes!
Finally, I will leave you with this chart, which goes to show just how much people love the 7th floor. Fun fact about this one: when you look at the numbers, the 7th floor is consistently twice as populated as any other floor. No wonder some people have trouble finding study cubicles – they’re always full!
Click to get all up in it
If you want to get a little deeper into this all, or have any questions about it, I am at the the HHSL Service Desk five days a week, and am always happy to talk shop. Otherwise, I hope you’re all enjoying the weather, and I look forward to seeing what the next year brings us!
Tom
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Hello hello everyone. It’s about that time again: time for the Affiliation Survey!
Twice a year, the Hirsh Health Sciences Library runs the Affiliation Survey, where we walk around and ask every one of our lovely patrons which program they are with (Dental, Medical, Sackler, etc). This is all we ask. We will not ask for your name or even which year you’re in! The numbers we collect are used in aggregate so we can get a snapshot of what the library usage is like. You can take a look at my post from July to see how the data turns out.
Here’s how it goes: there are 7 days spread out over the whole month, randomly chosen to try and maximize the usefulness of the numbers (in other words, we’re trying not to get skewed by specific exam blocks too much). On those days, HHSL staff (you’ll probably recognize us!) will walk around 4 times over the course of the day to gather the totals of how many people from each school are in the library. There will be signs and posters up this month, so you’re not caught unaware.
Keep your eyes out for them!
We won’t announce ahead of time which days we’re counting (see our need to randomize, above), so it’s safe to assume that it could just happen any day this month.
Don’t panic!
If you don’t want to be bothered (or to speak out loud), you are welcome to leave your ID next to you while you study. If that is still not good enough, you can write your program down on a piece of paper and leave that next to you. If you’re in a group room, feel free to stick a note to the outside of the door (on the wood – the windows must remain clear!) telling us how many people of what program(s) are in the room. Done and done. We will add that number to our count, and we will move on to the next person!
There are pencils, pens, markers, scrap paper, and tape down at the Service Desk on the 4th floor, so you can even make your sign bright and cheerful! If you have any questions or concerns, let us know either in person at the desk, or through Ask Us on our website. We’re here to help and make this quick and easy.
But otherwise, we look forward to seeing you all this month and finding out just what our beloved HHSL looks like this fine autumn.
Good luck on studying and I look forward to seeing you in the library!
Tom
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Hello everyone! It’s been a bit since the last time I did an annual round-up of the data I have for the library, and I thought it was high time to do another one. Especially given how much things have changed these last few years. And what interesting changes!
First thing is first, of course: what were the fullest days in the library this past year? There are actually two answers! On both December 6th and April 4th we counted 950 people sitting in the library. 950! That’s nuts! A neat side fact: both of those days are Tuesdays, and both were the Tuesday after the second weekend of Extended Hours in their respective semester. Either blind chance was on our side those days, or Tufts is big on keeping the same exam style no matter what. Go figure.
Incidentally, this year I became curious about how busy the floors were when compared with each other. The answer is this rather silly looking chart.
See that purple there? That’s the 7th floor. It was twice as busy as the 6th floor. Twice! We counted people 35,000 times on the 7th floor this last year! The others don’t even come close. Obviously, part of this can be explained by the fact that we added so much new furniture last year, but that’s not the entire story. I think it’s safe to say that students around here just really like their quiet space to study.
“But Tom,” you ask, “how do these kinds of numbers compare with past years?” Well I’m glad you asked that, HHSL Blog Reading Person, because I have an answer for you! The chart below is the total number of people we’ve counted in the library, by month, over the last 4 years.
The purple would be this year. It looks like a lot, but you can see the same sort of wave formation every single year – we start off slow in the summer, get busy in the fall, slow down in the winter (we only get about half of December’s days counted due to the break, and January starts off slowly anyway), and then the whole thing picks back up in time for all the exams, only to slow back down in June.
What does this tell us, then? Well, we’ve been busy. We’ve been very busy. As you can see, everything sort of spiked back in 2014-2015, but this year was consistently busier than the year before it. July and August were the exceptions, but I’m certain that March and April more than make up for those.
Seeing these numbers go up and down over the years presents an interesting thought challenge: why do they fluctuate so much? Well, a little of it is human error, a little bit is due to changes in how we collect this data, a little of it is the way the programs on campus have been changing and adapting (which means the student body has been adapting), and part of it is sheer environmental factors. Snow days, for instance, or the way holidays fall – if we’re closed, we won’t be counting anybody!
“Okay Tom, I see. How about circulation? Surely human error is minimized when you have a program doing all your counting for you.”
Note: We got a new circulation system in June 2017, so the numbers got interrupted.
You would be right. And you’re about to really enter the head-scratcher.
Somehow we were insanely full this year, while also having our second slowest year of checkouts since we started collecting these numbers this way. We have a ton of new resources for you, but circulation is down? Let me share the secret answer: the Internet. HHSL is constantly searching out new digital books, journals, and databases for everyone on campus, which means our physical books are not in as high demand as they used to be. So while the overall circulation is down, what is circulating is getting checked out like mad – for instance, our phone chargers alone registered 8,000 checkouts this year. The busiest year yet for them!
In case you were curious (I know you were) the most checkouts we had in a single day was 210 checkouts, which belongs to Wednesday, October 5th. 9 out of the 10 busiest circulation days were in the fall (with February 27th clocking in at #7). That particular information doesn’t really tell us much, but it’s fun to know all the same.
For the record, “busiest circulation day” is only in reference to amounts of checkouts (and subsequent check-ins) in a given day. It doesn’t count all of the times that librarians have a consultation with someone to help with research, or the times that the library staff at the Service Desk troubleshoot or help you find articles, or all the direction given, or printing problems fixed, or on-the-fly assistance with programs like EndNote…it’s a helpful metric, but it’s not the full story. Heck, it’s missing the craft events! You can’t have a complete picture without those.
All of these numbers are pieces to a puzzle, and it’s one we’re always working on here. You’re looking at a few pieces of it, but there are always more to add. And we’re always looking for more of those pieces, be it anything mentioned above or be it School Affiliation data. So as we enter the 2017-2018 year, keep your eye open for people walking around with clipboards and say hi! And who knows, maybe you’ll be part of one of these charts in the future.
-Tom-
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Hi everybody! It’s been a little bit since my last statistics post (just about one year, in fact), so it seemed high time for me to do another one of these. Today I’ll be focusing on October 2016 and March 2017, which were our two Affiliation Months this year. For those who are unfamiliar with the term, that’s when the library staff go and ask every person in the library what their program is, so that we can have a general sense of how our space is being used.
Still with me? Good!
Click to see full size
This first chart is comparing the Affiliation Stats from October 2016 and March 2017. To gather this info, we chose 7 days out of each month (one Sunday, one Monday, etc. All were chosen randomly) and on those days went around 4 times a day. This is always an interesting comparison due to the different programs and the way they operate. Dental and Medical students were gearing up for board and class exams, so it’s no surprise that we would see so many extras in March. There was a jump in PA as well (new class means new exams!), but then we saw drops with Sackler, Friedman, MBS, and PHPD. So what happened there? It’s hard to say. Different timing on exams, different demands on the classes, all sorts of things can affect attendance in the library. Ultimately, March was still the busier month: we counted 3,327 people in March, but only counted 3,115 people back in October.
The thing to keep in mind is that these numbers are only a snapshot. To truly know what the individual program attendance in this library is like for a full month we would need to gather this data every single day for that entire month, and that is unfortunately (or fortunately, depending on how much quiet you like) unrealistic for us. Do you know what is quite realistic, however? Collecting a month’s worth of circulation data!
Click to enlarge
Bam. That’s a full month of data right there! So what does this show us when compared to the Affiliation graph? Well for one, we can see that the space was occupied more than our things were being checked out to a point, which is actually pretty normal. People do like checking things out (skulls! laptops! books! phone chargers! oh my!), but the library keeps adding more and more space, making it easier to go and hide out and get your studying in.
But here are some thoughts: why are the numbers so similar? We never have had a 1-1 relationship between studying and circulation before, but parts of this are surprisingly close. What’s causing the numbers to fluctuate the way they do? Consider: we have exams in March and April, and Extended Hours at the end of March. Affiliation was overall higher (by 212 people), but Circulation was lower (by 249 checkouts). Weird, right?
There are many factors that affect all of these numbers, but I won’t be going into them in this post. There’s only so much space, and I still need to talk about the floors!
Click to embiggen
So. Dental students love the 7th floor. Surprise! Medical and MBS are also huge fans, so it looks like all of those new study carrels we added last summer really helped! Everybody else is spread rather evenly over the floors, although I do find it interesting that the PA students go up to 7 when they’re not in class (I guess to get away from the classrooms – can’t say I blame anyone). It’s good to see people like the furniture and spaces so much! Warms the heart (which is numbered and on a stand, and you can check out from the Library Service desk for 4 hours at a time. I’m not kidding).
The 7th floor has always been the most popular (generally about twice as popular as any of the other floors), and the breakdown after that is always fascinating. Medical students overwhelmingly prefer the 7th, followed by 6 – study quiet, which makes sense. Since the 4th and 5th floors offer some small group opportunities, those are split evenly. Dental may trend toward groups even more so than I used to think – perhaps that’s why there are so many up on 7, and then in decreasing order from 4, 5, and 6?
I would like to state that it is exciting to see Sackler, Friedman, PA, PHPD, and MBS showing up in larger and larger numbers. Hirsh Health Sciences Library is for everyone on the campus, and we want you all to feel welcome! That’s why we have so many study carrels, and the Collaboration Rooms, and craft days, and all the other things that make the days go round. And based on the posts from over the last few years, it seems like we’re only getting more and more popular, which is fantastic. Feel free to keep coming in bigger numbers, we have space for you all!
And on that uplifting note I will leave you for now. Perhaps I will be able to do a look at the full year’s numbers in a few months, so keep your eyes peeled. Until then: good luck with any exams you have left, and I’ll see you all around the library!
Tom
Tagged with: | 9,250 | 40,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.97588 |
https://physicscalculations.com/work-done-in-springs-and-elastic-strings/ | 1,723,116,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00436.warc.gz | 359,392,478 | 36,144 | # What is Work Done in Springs and Elastic Strings
## Work Done in Springs and Elastic Strings
In springs and elastic strings, work is done when they are stretched or compressed. When a force is applied to stretch or compress a spring, work is done, and this work is stored as potential energy in the spring. The amount of work done depends on the force applied and the displacement of the spring. When the spring is released, this stored potential energy is converted into kinetic energy as the spring returns to its original shape. This process is fundamental in understanding the behavior of springs and elastic strings in various applications, such as in mechanical systems, physics experiments, and everyday objects like trampolines or bungee cords.
## Explanation
Imagine playing with a stretchy spring or a rubber band. Work is done on these when you pull or stretch them, and it’s a bit like giving them a special kind of energy.
Stretching and Compressing: When you pull a spring or stretch a rubber band, you’re doing work on them. It’s like using your muscles to make them longer or squish them together.
Energy Storage: Now, here’s the cool part. The work you do gets stored in the spring or elastic string as potential energy. It’s like winding up a toy. The more you stretch or compress it, the more potential energy it stores.
Formula: The amount of work done depends on how hard you pull or push (the force) and how far you stretch or compress (the distance). There’s a formula: Work = Force × Distance or (W) = (1/2)kx2. If you pull harder or stretch it more, you do more work.
Energy Release: Now, when you let go of the spring or release the rubber band, that stored potential energy turns into something else—kinetic energy. It’s like the toy unwinding and moving. This energy transformation is why springs bounce back or why a stretched rubber band snaps back when released.
Everyday Examples: Trampolines use the work done on springs to make you bounce. Bungee cords store work as potential energy when stretched, creating an exciting bounce when you jump. So, the next time you play with a spring or a rubber band, remember, you are doing work that turns into a cool bounce or snap.
You may also like to read:
Potential Difference
## The Fundamentals of Springs and Elastic Strings
Before we delve into the intricacies of work done in springs and elastic strings, let’s establish a solid foundation by understanding the basics of these elements.
### Elasticity and Hooke’s Law: The Building Blocks
Elasticity is the property of a material that allows it to return to its original shape after being deformed by an external force. Hooke’s Law, formulated by the 17th-century scientist Robert Hooke, defines the relationship between the force applied to a spring or elastic string and the resulting displacement. This fundamental law serves as the basis for understanding their behavior.
### Spring Constants and Elastic Moduli: Quantifying Elasticity
To quantify the elasticity of a material, we use spring constants and elastic moduli. Spring constant, denoted by “k,” represents the force required to extend or compress a spring by one unit of length. On the other hand, elastic moduli, including Young’s modulus and shear modulus, describe the material’s response to stress in different directions.
## Work Done in Springs: Understanding the Formula
Now that we have a solid understanding of the foundations, let’s dive into the intriguing concept of work done in springs.
### Defining Work Done in Springs
Work done in springs refers to the energy transferred to or from the spring due to the application of an external force. When a force displaces a spring from its equilibrium position, it stores potential energy, which is released when the force is removed. Understanding this energy transfer is crucial in numerous applications, from automotive suspension systems to pogo sticks.
### The Work Done Formula for Springs
The formula to calculate the work done in a spring is straightforward and relies on the spring constant and the displacement:
Work (W) = (1/2)kx2
Where:
• Work is the work done in the spring (in joules).
• k is the spring constant (in newtons per meter).
• x is the displacement from the equilibrium position (in meters).
### Real-Life Applications of the Work Done Formula
The work done formula finds practical applications in various industries. For instance, in mechanical engineering, it helps design suspension systems that provide optimal comfort and stability for vehicles. Additionally, the formula plays a crucial role in creating efficient energy storage solutions, such as in the case of wind-up toys and power generation from renewable resources.
## Elastic Strings: Energy in Vibrations
Shifting our focus to elastic strings, we explore their unique behavior in the realm of vibrations and oscillations.
### Understanding Vibrations in Elastic Strings
When an elastic string, like those on a guitar or violin, is plucked or bowed, it begins to vibrate, producing sound waves. The behavior of these strings follows principles similar to those of springs, involving energy storage and release. Understanding the mechanics of these vibrations is essential in crafting melodious music.
### The Formula for Vibrational Energy
The formula to calculate the vibrational energy of an elastic string involves its mass, length, and frequency:
Energy = 0.5 * m * v2
Where:
• Energy is the vibrational energy of the elastic string (in joules).
• m is the mass of the string (in kilograms).
• v is the velocity of the string’s vibrations (in meters per second).
### Applications of Vibrational Energy
The applications of vibrational energy in elastic strings are far-reaching. In musical instruments, such as guitars and violins, the formula helps fine-tune the strings to produce specific notes and harmonics. Moreover, this understanding of vibrational energy plays a vital role in engineering precise acoustic devices, like microphones and speakers.
## Combined Systems
As we deepen our knowledge of springs and elastic strings, it’s essential to explore scenarios where these elements work together, creating dynamic and efficient systems.
### Coupled Springs: Doubling the Force
In some instances, multiple springs are coupled together, amplifying their force and effect. Understanding how these systems interact allows engineers to design sophisticated setups, like those used in car suspension systems and industrial machinery.
### Coupled Elastic Strings: Harmonious Resonance
When we connect multiple elastic strings, they can resonate in harmony, producing richer and more complex sound profiles. Musicians and instrument makers utilize this phenomenon to craft instruments with exceptional tonal qualities.
## FAQs
Q: What is the significance of Hooke’s Law in the study of springs and elastic strings?
A: Hooke’s Law serves as a fundamental principle for understanding the elastic behavior of springs and elastic strings. It establishes the linear relationship between force and displacement, providing crucial insights into how these elements respond to external forces.
Q: How does the work done formula for springs impact suspension systems in vehicles?
A: The work done formula helps engineers design suspension systems that offer optimal comfort and stability for vehicles. By calculating the energy stored and released in the springs during compression and rebound, suspension systems can be tuned to provide a smooth and controlled ride.
Q: What role does vibrational energy play in musical instruments?
A: Vibrational energy is at the heart of producing sound in musical instruments. When elastic strings vibrate, they generate sound waves, producing the melodious notes we hear in instruments like guitars, violins, and pianos.
Q: Can elastic strings of different materials produce different types of sound?
A: Yes, the material composition of elastic strings influences the type of sound they produce. Strings made from different materials, such as nylon, steel, or gut, result in varied tonal qualities, contributing to the distinctive sound of each instrument.
Q: How can coupled springs be used in engineering applications?
A: We use coupled springs in various engineering applications to achieve specific outcomes. For example, in car suspension systems, coupled springs can distribute the load more efficiently, enhancing the vehicle’s performance and handling.
Q: What are some examples of resonance in coupled elastic strings?
A: Resonance in coupled elastic strings is commonly observed in musical instruments like the piano, where the interaction of multiple strings creates harmonious overtones, enriching the instrument’s sound and timbre. | 1,715 | 8,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.910546 |
https://newbedev.com/when-do-three-closed-balls-have-a-nonempty-intersection | 1,623,495,309,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00135.warc.gz | 391,800,005 | 30,166 | # When do three closed balls have a nonempty intersection?
### Solution:
A solution is an application of a theorem due to Karl Menger (as an alternative, one can use Schoenberg's theorem).
To simplify matters, I will assume for most of the answer that the points $$c_1,...,c_n$$ satisfy (*), that is, are in "general position:" they form the vertex set of an $$n-1$$-dimensional simplex. (I will explain in the end of the answer how to reduce the general case to this one.)
First, let me describe Menger's theorem.
Menger gave a set of necessary and sufficient conditions for a finite metric space $$(X,d)$$ to embed isometrically in a Hilbert space $$H$$; he also gave a characterization of the least dimension of $$H$$ in terms of the metric $$d$$; I will denote this number $$h(X)$$ (I am suppressing the rotation for the metric here and below).
Here is Menger's solution (see my answer here for references). Given a finite metric space $$X=\{x_0, x_1,...,x_n\}$$ (I am suppressing the notation for the metric), Menger uses the following determinant, also known as the Cayley-Menger determinant: $$\Delta(X)= \left|\begin{array}{ccccc} d(x_0,x_0) & d(x_0,x_1) & ... & d(x_0, x_n) & 1\\ d(x_1,x_0) & d(x_1,x_1) & ... & d(x_1, x_n) & 1\\ \vdots & \vdots & ... & \vdots & \vdots\\ d(x_n,x_0) & d(x_n,x_1) & ... & d(x_n, x_n) & 1\\ 1 & 1 & ... & 1 & 0 \end{array}\right|.$$ The first (and the most important) of Menger's conditions is that $$\Delta(X)$$ has the sign of $$(-1)^{|X|}$$ meaning: $$\Delta(X) (-1)^{|X|} \ge 0.$$ Furthermore, $$h(X)= k$$ implies $$\Delta(X)=0$$ (and the converse is true as well as long as $$h(Y)=|Y|-1$$ for all proper subsets $$Y\subset X$$). The rest of Menger's conditions are inductive: For $$X$$ to embed isometrically in a Hilbert space, all subsets $$Y$$of $$X$$ have to be embeddable in Hilbert spaces, i.e. their determinants $$\Delta(Y)$$ have to have the sign of $$(-1)^{|Y|}$$ (as above).
Remark. Here is an important observation about the determinant $$\Delta(X)$$ regarded as a function in the variables $$d(x_0,x_1),...,d(x_0,x_n)$$:
$$\Delta(X)$$ is a 2nd degree polynomial in these variables, with the constant terms equal $$\pm \Delta(X_0)$$, where $$X_0= X\setminus \{x_0\}$$ (with the restriction of the metric). As a polynomial of $$d(x_0,x_i)$$ it has the form $$A_i d^2(x_0,x_i) + B_i d(x_0,x_i) + C_i, i=1,...,n,$$ where $$A_i= \Delta(X_{0i})\ne 0$$ and $$X_{0i}\subset X$$ is obtained from $$X$$ by removing the points $$x_0, x_i$$). This is where I am using the assumption (*).
I will use the notation $$H$$ for an infinite-dimensional Hilbert space, containing all the Euclidean spaces $$E^1\subset E^2\subset E^3\subset ...$$. I will also use the notation $$S(c,r)$$ to denote the round sphere in $$H$$ centered at $$c$$ and of radius $$r$$.
Given a subset $$C\subset H$$, let $$span(C)$$ denote the affine span of $$C$$, i.e. the smallest affine subset of $$H$$ containing $$C$$.
Let's first solve a slightly different problem than yours:
Given a finite subset $$\{c_1,...,c_n\}$$ in $$H$$, what are the necessary and sufficient conditions on the distances $$d_{ij}=||c_i-c_j||$$ and radii $$r_i\ge 0$$, for the intersection $$\bigcap_{i=1}^n S(c_i, r_i)$$ of spheres in $$H$$ to be nonempty?
Menger's theorem provides an answer to the sphere problem. Namely: Given a tuple $$\tau=((c_1,r_1),...,(c_n,r_n)),$$ form an abstract pre-metric space $$(X,d)=X_\tau$$ equal to $$\{c_0, c_1,...,c_n\}$$ with $$d(c_i,c_j)=d_{ij}, d(c_0, c_k)=r_k, k=1,...,n.$$ (The adjective pre-metric refers to the fact that $$d$$ might violate triangle inequalities when applied to triples $$c_0, c_i, c_j$$.) Then the following are equivalent:
1. $$X_\tau$$ embeds isometrically in $$H$$.
2. $$X_\tau$$ is a metric space which satisfies the conditions in Menger's theorem, i.e.
(a) $$d(c_i, c_k)\le d(c_i, c_j) + d(c_j, c_k)$$ for all triples $$i, j, k\in \{0,...,n\}$$ such that the product $$ijk=0.$$
(b) For all subsets $$Y\subset X_\tau$$ containing $$c_0$$, $$\Delta(Y) (-1)^{|Y|}\ge 0$$
3. $$\bigcap_{i=1}^n S(c_i, r_i)\ne \emptyset.$$
Moreover,
$$span(\{c_1,...,c_n\})\cap \bigcap_{i=1}^n S(c_i, r_i)\ne \emptyset$$ if and only if, additionally, $$\Delta(X)=0$$.
Note also that, for each finite-dimensional Euclidean subspace $$A$$ containing $$\{c_1,...,c_n\}$$, the intersection $$A\cap \bigcap_{i=1}^n S(c_i, r_i)$$ is either empty, or is a single point, equal to the intersection of the above spheres in the Hilbert space $$H$$, as well as in $$span(\{c_1,...,c_n\})$$, or is a round sphere of dimension $$\dim(A) - n$$.
Now, let's turn to the original problem of intersection of closed balls in Euclidean spaces. It is easy to see that, if $$\bigcap_{i=1}^n S(c_i, r_i)\ne \emptyset,$$ then $$span(\{c_1,...,c_n\})\cap \bigcap_{i=1}^n B(c_i, R_i)\ne \emptyset,$$ for any $$n$$-tuples of real numbers $$R_i\ge r_i$$.
Definition. A collection of round balls $${\mathcal G}= \{B(c_1,r_1),..., B(c_n,r_n)\}$$ in a (possibly infinite-dimensional) Euclidean space $$E^\alpha$$ will be called redundant if there is a proper subset $$I\subset [n]=\{1,...,n\}$$ such that $$\bigcap_{i\in [n]} B(c_i,r_i)= \bigcap_{i\in I} B(c_i,r_i).$$ The collection of balls will be called irredundant otherwise. The same terminology applies to the tuple of centers and radii: $$\tau=((c_1,r_1),...,(c_n,r_n)).$$
It is easy to see that a tuple is redundant if and only if it is redundant in $$span(\{c_1,...,c_n\})$$.
If one knows that a tuple $$\tau$$ is redundant, then one can describe necessary and sufficient conditions for nonemptyness of the intersection of a collection balls using a smaller subcollection, hence, give and inductive description this way.
As an example: For $$n=3$$, a tuple is redundant if and only if the 4-point pre-metric space $$(X,d)$$ as above violates triangle inequalities, i.e. is not a metric space.
Lemma. A tuple $$\tau=((c_1,r_1),...,(c_n,r_n))$$ is redundant if and only if $$\bigcap_{i=1}^n S(c_i, r_i)= \emptyset,$$ where the intersection is taken in $$H$$.
The proof of this lemma is a straightforward induction on $$n$$ and I omit it. This lemma allows one to give a numerical criterion for redundancy:
Corollary. Suppose that $$n\ge 3$$. Unless the intersection of balls $$\bigcap_{i\in [n]} B(c_i,r_i)$$ is empty, the tuple $$\tau=((c_1,r_1),...,(c_n,r_n))$$ is irredundant if and only if:
(a) For each proper subset $$I\subset [n]$$, the corresponding tuple $$\tau_I=((c_{i_1},r_{i_1}),...,(c_{i_k},r_{i_k})), I= (i_1,...,i_k)$$ is irredundant (in particular, $$(X,d)$$ is a metric space).
(b) $$\Delta(X) (-1)^{n+1}\ge 0$$.
Note that this corollary does not directly solve the problem of nonemptyness of the intersection of balls.
At last, here is an answer to the problem of nonemptyness of intersection of round balls $$B(c_i,r_i)$$ in $$span(\{c_1,...,c_n\})$$ (which we still assume to have dimension $$n-1$$). The solution is inductive in $$n$$. For $$n=2$$ the answer is in the form of "triangle a inequality" $$B(c_1,r_1)\cap B(c_2,r_2)\ne \emptyset$$ if and only if $$r_1+r_2\ge d_{12}=||c_1-c_2||$$.
Assume the problem is solved for all $$m. In particular, we have a test for redundancy for sets of $$m$$ balls, $$m, i.e. in addition to the numerical criterion we can also tell if the intersection of $$m$$ balls is nonempty. Now, given a tuple $$\tau=((c_1,r_1),...,(c_n,r_n))$$, either:
(i) There exists a proper subtuple $$\tau_I$$ which is redundant (and which is something we can test), hence, $$\tau$$ itself is redundant and, thus, the problem of nonemptyness of $$\tau$$ is reduced to a smaller set of balls.
(ii) Suppose that all proper subtuples $$\tau_I$$ are irredundant; in particular,
the subtuple $$\sigma=((c_1,r_1),...,(c_{n-1},r_{n-1}))$$ is irredundant. Solve the equation $$\Delta(X_\tau)=0$$ for the unknown $$y=d(c_0, c_n)$$; this equation has the form $$A_n y^2 + B_n y + C_n=0,$$ with $$A_n\ne 0$$, where $$A_n, B_n, C_n$$ are functions of the tuple $$\sigma$$. This quadratic equation has two (possibly equal) solutions $$y_\pm= -\frac{B_n}{2A_n} \pm \sqrt{ \left(\frac{B_n}{2A_n}\right)^2 - C_i}.$$ Both solutions are real and nonnegative. Geometrically speaking, they correspond to the following:
Consider the intersection of spheres in $$span(\{c_1,...,c_n\})$$: $$\bigcap_{i=1}^{n-1} S(c_i, r_i)=S^0.$$ This intersection is nonempty (by the irredundancy assumption!) and is either a singleton (contained in $$span(\{c_1,...,c_{n-1}\})$$) or it a 2-point set $$s_-, s_+\}$$, one of its points $$s_-$$ is closer to $$c_n$$ than the other. Then $$y_\pm= ||c_n - s_{\pm}||.$$ The case when $$S^0$$ is a singleton happens precisely when $$y_+=y_-$$. Then $$$$\bigcap_{i=1}^n B(c_i, r_i)\ne \emptyset,$$$$ if and only if $$r_n\ge y_-$$, i.e. either $$\Delta(X_\tau) (-1)^{n}\ge 0$$ or the tuple $$\tau$$ is redundant because $$B(c_n,r_n)$$ strictly contains the intersection $$$$\bigcap_{i=1}^{n-1} B(c_i, r_i).$$$$
As an example, here is this solution implemented in the case of intersection of three balls in the Euclidean plane, $$\tau=((c_1,r_1),...,(c_3,r_3))$$. I will use the notation $$d_{ij}= ||c_i-c_j||$$
1. Test proper subtuples for emptyness: If for some $$1\le i $$r_i+r_j< d_{ij},$$ then $$B(c_i,r_i)\cap B(c_j,r_j)=\emptyset$$ and we are done. Suppose, therefore that all these intersections are nonempty.
2. Test proper subtuples for redundancy: If for some $$1\le i\ne j\le 3$$ $$r_i> r_j+ d_{ij}$$ then we can eliminate the ball $$B(c_i,r_i)$$ from the collection $$B(c_k,r_k), k=1,2,3$$ without changing the intersection, and, hence, nonemptyness of the triple intersection is guaranteed by the triangle inequality $$r_k+r_j\ge d_{jk}, i\notin \{j,k\}, j\ne k.$$
3. Suppose, lastly that $$X_\tau$$ is a metric space and each proper subtuple $$\sigma$$ in $$\tau$$ is irredundant. Then $$$$\bigcap_{i=1}^3 B(c_i, r_i)\ne \emptyset,$$$$ if and only if $$r_3\ge y_-$$, where $$y_-$$ is the smaller root of the polynomial $$A_3 y^2 + B_3 y + C_3.$$ The coefficients $$A_3, B_3, C_3$$ are computed as follows:
$$A_3= 2 d_{12},$$ $$B_3= -2r_1(d_{12}+d_{23}-d_{13}) - 2r_2(d_{31}+d_{12} -d_{23}),$$ $$C_3= \Delta(X_0)= \left|\begin{array}{cccc} 0 & d_{12} & d_{13} & 1\\ d_{21} & 0 & d_{23} & 1\\ d_{31} & d_{32} & 0 & 1\\ 1 & 1 & 1 & 0 \end{array}\right|.$$
Lastly, let me explain the solution for $$n$$ points in $$H$$, which are not in general position, i.e. their affine span has dimension $$\le n$$. Again, I will take solution for $$ points for granted. Then, according to Haley's theorem, $$\bigcap_{i\in [n]} B(c_i, r_i)\ne \emptyset$$ if and only if for each proper subset $$I\subset [n]$$, $$\bigcap_{i\in I} B(c_i, r_i)\ne \emptyset.$$
The intersection problem for $$< n$$ balls is solved by the inductive assumption. A bit more concretely, inductively applying Haley's theorem, we reduce the problem to the intersection problem of balls centered at configurations of points in general position in some affine subspaces of $$H$$. | 3,610 | 10,997 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 166, "wp-katex-eq": 0, "align": 0, "equation": 6, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.836317 |
https://www.jakelearnsdatascience.com/categories/prime-numbers/ | 1,591,501,604,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00357.warc.gz | 743,720,563 | 4,624 | # Prime Numbers
## Prime Number Patterns
I found a very thought provoking and beautiful visualization on the D3 Website regarding prime numbers. What the visualization shows is that if you draw periodic curves beginning at the origin for each positive integer, the prime numbers will be intersected by only two curves: the prime itself’s curve and the curve for one. When I saw this, my mind was blown. How interesting… and also how obvious. The definition of a prime is that it can only be divided by itself and one (duh). | 110 | 525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-24 | latest | en | 0.954733 |
https://www.jiskha.com/search/index.cgi?query=Fill+in+the+following&page=2 | 1,503,080,874,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105086.81/warc/CC-MAIN-20170818175604-20170818195604-00206.warc.gz | 908,144,617 | 11,736 | # Fill in the following
47,406 results, page 2
### Chemistry
Imagine that you have a 7.00L gas tank and a 4.00L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 145atm , to what pressure should you fill the...
### chem
Imagine that you have a 7.00L gas tank and a 4.50L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 105atm , to what pressure should you fill the...
1. Over which interval is the velocity greatest? FILL IN THE BLANK ____ 2. Over which interval(s) is the velocity zero? FILL IN THE BLANK ____ 4. What is the average velocity in m/s from A to B? FILL IN THE BLANK ____ 5. The velocity of an object goes from 4 m/s to 12 m/s in ...
### math
If pump A and pump B work together, they can fill a pool in 4 hours. Pump A, working alone, takes 6 hours to fill the pool. How long would it take pump B, working alone, to fill the pool?
### All subjects
The villagers build underground water canal to fill the well, in which water flow with 2m/sec & its width is 30cm & length 80cm. How much time will it take to fill one well of radius 3cm & 21m deep
### math
Determine all possible digits to fill in the blanks to make each of the following true. a) 9 482__ b) 6 24__35 c) 4 63__
### Math
0.38 = (3X0.1) + ( X ) My sons paper instructs us to fill in the missing values. We understand that we are to fill in the two values in the second set of parenthesis, but that is all we understand.
### math
a tub takes 4 minutes to fill and holds 28 gallons of water. At this rate, how long would it take to fill a bathtub that held 63 gallons of water?
### algebra
One pipe can fill a tank in 60 minutes and another pipe can fill it in 50 minutes .If these two pipes are open and a third pipe is draining water from the tank ,it take 35 minutes to fill the tank .How long will it take the third pipe alone to empty the tank?
### science
I grand daughter has this fill in the blank: Electically__________ objects push or _____________ on other electrically charged objects. This produces___________. Please fill in the blanks please. Because we don't know... Thanks!!!!
### math
A sludge pool is filled by two inlet pipes. One pipe can fill the pool in 14 days and the other pipe can fill it in 22 days. However, if no sewage is added, waste removal will empty the pool in 35 days. How long will it take the two inlet pipes to fill an empty pool? answer
### Algebra
Two pipes can fill a tank in 9 hours. If the bigger of the pipes has a rate of 1 ½ times of the smaller, how long would it take each pipe to fill the tank if opened alone?
### math
The large container has 13.5 liters of milk. Chen wants to fill as many 3/4 liter containers as possible with this milk. How many containers will he be able to fill?
### math
It takes 1 and 3/4 hours to fill a new underground gasoline storage tank. What part of the tank would be fill if the gasoline had been shut off after 1 hour?
### math
Tim needs to fill 3 20L barrels with oil. Oil dispenser flows at 100ml/s . How long to fill up 3 barrels. Please help by showing all steps so I can understand how the answer was determined. Thanks!
### math 213
Determine all possible digits to fill in the blanks to make each of the following true. 9 |482____ 6|24____35 4 |63____
### Math
Shezad needs 10 litres of lemon squash to fill 20 cups. How much squash will he need to fill 25 cups?
### math
About 7.5 gallons of water fill up 1 cubic foot of space. How many gallons of water will fill a goldfish pool.
### algebra
When using the big hose and the small hose, Sam can fill her pool in 15 minutes. The small hose when used alone takes 60 minutes to fill the pool. How long would it take Sam to fill her pool using only the big hose? Don't understand how to set up equation.
### Algebra
In preparation for a storm, the town council boys 13,750 pounds of sand to fill sandbags. Volunteers are trying to decide whether to fill bags that can hold 25 pounds of sand bags or bags that can hold 50 pounds of sand. Will they have more or fewer sandbags if they fill the ...
### SCIENCE?CHEMISTRY***
Am using the inventory method at moment but am stuck on how fill i the inventory for __Agl+__Fe2(CO3)3--> __Fel3+__Ag2CO3 I think the brackets are throwing me Hows this element b4 after ------------------------------------ Ag 1 2 l 1 3 Fe 6 1 CO 3 3 I don't know if this is ...
### living environment
1) the prefix milli- means 1/1,000. how many times would you have to fill and empty a 10 ml graduated cylinder in order to fill a 1 liter soda bottle? i put 100 times. 2) what are eight examples of liquid products sold by volume
### CHEMISTRY 3
Consider the following balanced equation: 2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g) Complete the following table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, ...
### Math
I really need help with this question please Break-even equation. Fill in the blank. The following table contains selected data concerning several outpatient clinics in the new Ambulatory Care Center at Hope University Hospital. Fill in the missing information. A. PRICE PER ...
0.38 = (3 X 0.1) + ( X ) My sons paper instructs us to fill in the missing values. We understand that we are to fill in the two values in the second set of parenthesis, but that is all we understand. Please help...I feel so inadequate because I can't help my child with this ...
Two pipes that are open at the same time can fill a pool in 12 hours. When only Pipe A is open, it can fill the pool 7 hours longer than when only Pipe B is open. Find how long Pipe A can fill the pool when it is the only pipe open?
### Math
Tap A fills up a water tank in 45 mins. Tap A fills up_____(ugh soo right here for the answer... do i put it in a fraction or something?)of the tank in one min. Tap B fills up the same tank in 75 mins. Tap B fills up_____of the tank in one min Tap A and tap B fill together up(...
### college algebra
a garden hose can fill a pool in 3 days and a longer one can bill the pool in 2 days. How long will it take to fill the pool if both hoses are used.
### Math
(This is a question where we have to fill in the blank) So: 2 3 5 ? 12 11 16 26 41 ? It's a number sequence question and I need to fill in the ? as well as write a rule in words. PLEASE HELP!(And the top number has to relate with the bottom number)
### Math
One pipe fills a tank in 1 hour while another pipe can fill the same tank in 2 hours. How long will it take to fill the tank if both pipes are used.
### Math
It takes 3 mins to fill a tub to the top and five minutes to drain the full tub.if the faucet and drain are both open how long will it take to fill the tub
### CHEMISTRY
Consider the following balanced equation: 2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g) Complete the following table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, ...
### CHEMISTRY
Consider the following balanced equation: 2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g) Complete the following table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, ...
### Algebra
It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day. How long did it take for the bacteria to fill one half of dish?
### algebra
It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day. How long did it take for the bacteria to fill one half of dish?
### Algebra 2
Perry can fill a sorter with oranges from the conveyer belt in 10 minutes. While Perry fills the sorter, Gillian takes oranges out of the sorter and puts them in shipping crates. With Gillian taking oranges out of the sorter, it takes 25 minutes for Perry to fill the sorter. ...
### maths
Two pipes A and B can fill in 40 minutes and 30 minutes respectively.Both the pipes are open together but after 5 minutes, pipe B is turned off.What is the total time required to fill the tank?
### Maths
Pipe A takes twice the time to fill up the pool as pipe B does.The pool can be filled up in 2 h when both pipes are opened How long will it take each pipe to fill it up alone Plz slove it in Linear equation
### math
A can fill in 5min and B in 20min.if both taps are opened then due to a leakage it took 30min more to fill tank.if tank is full,how long will it take for the leakage alone to empty the tank?
### Math
If it takes 16 faucets 10 hours to fill 8 tubs, how long will it take 12 faucets to fill 9 tubs?
### Maths
A pipe of radious 1 cm can fill up a tank in 20 minut ;time to taken by another pipe of radious 2 cm to fill ;it will be
### Math
If a 2-inch pipe will fill a tank in 6 hours, how long will it require for a 3-inch pipe to fill it?
### Environmental studies
Are there problems with mitigating impacts on wetlands/streams at an off-site location within the same watershed? i.e. fill a wetland in Maryland and mitigate for it or create it in Virginia. Should the federal government (U.S. Army Corps of Engineers) or a developer decide ...
### Chem
Consider the following balanced equation: 2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g) Complete the following table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, ...
### math
colin is buying dirt to fill a garden bed that is a 9 foot by 16 foot rectangle. if he wants to fill it to a depth of 4 inches, how many cubic yards of dirt does he need? if dirt costs \$25 per yard, how much will the project cost?
### Math
Colin is buying dirt to fill a garden bed that is a 9 ft by 16 ft rectangle. If he wants to fill it to a depth of 4 in., how many cubic yards of dirt does he need? If dirt costs \$25 per yd3, how much will the project cost? (Hint: 1 yd3 = 27 ft3)
A reservoir is fed by two pipes of different diameter. The pipe with the larger diameter takes three hours less than the smaller pipe to fill the reservoir. if both pipes are opened simultaneously, the reservoir can fill in two hours. Calculate how long it takes the pipe with ...
### english
revise the following sentence to remove all pronoun errors. When a police officer fires his gun, he must fill out a lot of paperwork.
### math
By selling at Rs 77, some 2 1/4% shares of face-value Rs 100 and investing the proceeds in 6% shares of face value Rs 100, selling at Rs 110 a person increased his income by Rs 117 per annum. How many shares did he sell? Two pipes running together can fill a cistern in 11 1/9 ...
### Statistics
For the following set of scores, fill in the cells. The mean is 70 and the standard deviation is 8. Raw score Z score 68.0 ? ? –1.6 82.0 ? ? 1.8 69.0 ? ? –0.5 85.0 ? ? 1.7 72.0 ?
### Math
Jim can fill a pool in 30 min, sue can do it in 45 min, Tony can do it in 90min, how long would it take all three to fill the pool together? How would i figure this out?
### math
you fill up 40 gallon pool putting in 2 gallons water every minute, if someone scoops out 1 gallon every 8 minutes, how long would it take to fill the pool?
### math
It takes 1 and 3/4 hours to fill a new underground gasoline storage tank. What part of the tank would be fill if the gasoline had been shut off after 1 hours?
One tap can fill a barrel in 3 minutes and another can fill it in 6 minutes. Both taps are turned on together for 1 minute. What fravction of the barrel will be filled in 1 minute?
### Algebra
Two pipes are connected to the same tank. When working together, they can fill the tank in 12 hrs. The larger pipe, working alone, can fill the tank in 18 hrs less time than the smaller one. How long would the smaller one take, working alone...
### Math
It takes 15 minutes to fill a large water tank. However, the tank has a small leak that would completely drain the tank in 4 hours. How long will it take to fill the tank if the leak is not plugged?
### Algebra
One pipe can fill a swimming pool in 8 hours. Another pipe takes 12 hours. How long will it take to fill the pool if both pipes are used simultaneously?
### math
One pipe can fill a swimming pool in 8 hours. Another pipe takes 12 hours. How long will it take to fill the pool if both pipes are used simultaneously?
### Math
You begin to fill a bucket with water at 9 am. Each minute you double the amount of water in the bucket. If it takes you 1 hour to fill the bucket, at what time is the bucket one-quarter full?
### maths
To fill a swimming pool two pipes are to be used.If the pipe of larger diameter I'd used for 4 hours and the pipe of smaller diameter for 9 hours only half the pool can be filled .Find ,how long it would take for each pipe to fill the tank separately, if the pipe of smaller ...
### Physics
A 5/8 inch (inside) diameter garden hose is used to fill a round swimming pool 7.3m in diameter.How long will it take to fill the pool to a depth of 1.0m if water issues from the hose at a speed of 0.50m/s ?
### Algebra
Sue, an experienced shipping clerk, can fill a certain order in 5 hours. Jim, a new clerk, needs 12 hours to do the same job. Working together, how long will it take them to fill the order?
### science
All living things must carry out certain functions and possess certain structures to survive, grow, and reproduce. The following chart illustrates how humans possess certain structures to carry out these critical functions. Fill in the plant structures that correspond to the ...
### maths
One pump fills a tank two times as fast as another pump. If the pumps work together they fill the tank in 18 minutes. How long does it take each pump working alone to fill the tank?
### MATH MIDDLE SCHOOL
I love chocolate and coffee. I begin by making one cup of hot chocolate. I drink half of my cup and then I fill the cup with coffee. I stir the new mixture and drink half of the cup and then fill the cup again with coffee. I continue this pattern until I have consumed two full...
### math
One pump fills a tank two times as fast as another pump. If the pumps work together they fill the tank in 18 minutes. How long does it take each pump working alone to fill the tank? mat
### Solar Energy Help ASAP
Below you see a plot of an ideal and a non-ideal JV curve. Based on this plot and assuming an irradiance of 1000W/m2, give approximately the numerical values to the questions bellow: I. Estimate the ideal Fill Factor (in %): II. Estimate the non-ideal Fill Factor (in %): III. ...
### maths
Three pipes can fill a water tank constantly.The time taken for the first two pipes working simultaneously to fill the tank is equivalent to time taken by the third pipe to fill it alone.If the second pipe fills the tank in 10 hours faster than the first pipe and 8 hours ...
### math
If a bacteria spreads its bacteria in a lake for every minute it gets doubled so that it can fill the lake in 20 min .if two bacteria are there how much will they take to fill the lake
### college
Use the Microsoft Web site microsoft. com/ windows/ compatibility to research whether a home or lab PC that does not have Windows Vista installed qualifies for Vista. Fill in the following table and print the Web pages showing whether each hardware device and appli-cation ...
### Math Middle School
Please help me solve this challenge problem: I love hot chocolate and coffee. I began by making one cup of hot chocolate. I drink half of my cup and then I fill the cup with coffee. I stir the new mixture and drink half of the cup and then fill the cup again with coffee. I ...
### arithmetic
The pet shop owner told jean to fill the new fish tank 3/4 full of water.Jean filled the tank 9/12 full.What fraction of the tank does she still need to fill?
### math
the pet shop owner told Jean to fill her new fish tank 3/4 full with water. Jean filled it 9/12 full. What fraction of the tank does Jean still need to fill??
### math
the pet shop owner told Jean to fill her new fish tank 3/4 full with water. Jean filled it 9/12 full . what fraction of the tank does Jean still need to fill?
### math
The pet shop owner told Jean to fill her new fish tank 3/4 full with water. Jean filled it 9/12 full. What fraction of the tank does Jean still need to fill?
### Maths
13. The pet shop owner told Jean to fill her new fish tank 3/4 full with water. Jean filled it 9/12 full. What fraction of the tank does Jean still need to fill?
### math
the pet shop owner told jean to fill her new fish tank 3/4 full with water. jean filled it 9/12 full. what fraction of the tank does jean still need to fill.
### Math
The pet shop owner told Jean to fill her new fish tank 3/4 full with water. Jean filled it 9/12 full. What fraction of the tank does Jean still need to fill?
### math
Three faucets fill a 100-gallon tub in 6 minutes. How long, in seconds, does it take six faucets to fill a 25-gallon tub? Assume that all faucets dispense water at the same rate.
### Math
the pet shop owner told jean to fill her new tank 3/4 full with water. Jean filled it 9/12 full. what fraction of the tank does Jean still need to fill?
### chemistry
1. it takes: ____ electrons to fill the 6d subshell of an atom. 2. It takes _____ electrons to fill the n=6 shell of an atom. I thought it was 22 and 72?
### Math
Faucet A fills a sink in 2 hrs. Faucet A and B fill the same sink (together) in 1 hr 15 mins. Faucet B would take how long to fill the sink by itself?" (Faucet B is slower)
### chemistry
how do i balance the following(fill in the blanks): 1.) __BF3 + __Li2SO3 --> __B2(SO3)3 + __LiF 2.) __CaCO3 + __H3PO4 --> __Ca3(PO4)2 + __H2CO3 3.) __B2Br6 + __HNO3 --> __B(NO3)3 + __HBr
### Physics
Calculate the following, expressing the answer in scientific notation with the correct number of significant figures: (8.86 + 1.0 * 10^-3) / 3.610 * 10^-3 please help me write out the equation and fill in the appropriate data thanks for your help Jiskha
### Physics
a BALL IS THROWN UPWARD WITH AN INITIAL VELOCITY OF 9.8 M/S FROM THE TOP OF A BUILDING. FILL A TABLE BELOW SHOWING THE BALLS POSITION,VELOCITY, and ACCELERATION AT THE END OF EACH OF THE 1ST 4SEC OF MOTION TIME(s) POSITION(M) VELOCITY(M/S) 1 2 3 4 Acceleration (M?S^2) Fill in ...
### maths
In a table, there are 3 rows with 3 square in each row. Mary wishes to fill some colors into the blocks. Now only red, blue and green colors are available, find the number of ways to fill the table if no adjacent blocks are filled with the same color. Thanks.
### Algebra 2
An extruder can fill an empty bin in 2 hours and a packaging crew can empty a full bin in 5 hours. If a bin is half full when an extruder begins to fill it and a crew begins to empty it how long will it take to fill the bin?
### math
1.) The first person who requires your assistance is Tina the bearded lady. Tina’s side job is to look after the elephants (she double majored at circus university.) a. Tina must fill the semi-circular trough shown above with peanuts. Tina knows that the store bought peanuts...
### math
ms.smith has a new fish aquarium that measures 45.3 cm long, 23.4 cm wide and 28 cm high. She has had a problem with the fish jumping out of the tank. today she is cleaning and decides to fill the tank to a level that is 4cm from the top to see if that will help. how mmany ...
### Algebra
Mr. Daniels bought an above-ground cylindrical swimming pool that has a diameter of 20 feet. He wants to know how long it will take to fill his pool to a depth of 4 feet using his garden hose. It takes him 15 seconds to fill a one-gallon jug. (Note: 1 gallon = 231 cubic inches.)
### Math
using a certain garden hose, liza can fill her fancy new bucket with water in 45 seconds. Unfortunately, Henry accidently makes a hole in the bucket, so if initially full, the bucket will now become empty in 1 minute. How long will it take to fill the bucket, considering it ...
### Math
Jim can fill a pool carrying bucks of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?
### math
Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?
### math
Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 and a half hours. How quickly can all three fill the pool together?
### Math
Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together? ANYBODY KNOW THE ANSWER??
### Math
A circular swimming pool is 4 feet deep and has a diameter of 15 feet. If it takes 7.5 gallons of water to fill one cubic foot, to the nearest whole number, how many gallons are needed to fill this swimming pool.
### Chemistry
I don't get how to fill in the blanks Show the products for cellular respiration by completing the following chemical reaction equation. Give the correct formula for oxygen as well Glucose+_________ ---> ________+______+___________
### English
Choose the appropriate vocabulary word from The Call of the Wild to fill in the following sentence: "Buck often went on walks, swam, or played with Judge Miller's ." A. Progeny B. Demesnes C. Intimates
### Chenistry
Punnett Square for F1 Cross ¨C Expected Genetic Outcomes F1 Parent, genes: _________ (student to fill in the blanks) ¡â alleles > ¡á alleles v F1 Parent, genes:__________ (student fill in blank)
### math
The given cylindrical container r =3 inches and height= 8in is used to fill the rectangular prism fish tank 24 , 24 and 12 inches with water. What is the least number of full cylindrical needed to completely fill the fish tank? | 5,665 | 22,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-34 | latest | en | 0.89069 |
http://www.red-dingo.sk/5cnu15e/tf12q5.php?eda682=absolute-humidity-formula | 1,632,412,795,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057424.99/warc/CC-MAIN-20210923135058-20210923165058-00056.warc.gz | 113,035,375 | 11,200 | m. M. =. Absolute Humidity. Before using the formula value for saturated vapor pressure in the dewpoint procedure, you divide the formula value by 1.38. Absolute Humidity. All other trademarks and copyrights are the property of their respective owners. ppm v & ppm w. Absolute Humidity. The final answer should be RH=51.3 %(percent). a)Calculate the humidity ratio (w). If your pressure reading or calculation is in millibars, then you convert it to Pa by multiplying the reading in millibars by 100. Absolute humidity is the density of water vapor in the air (kg/m3). It may rain when these conditions prevail. As altitude is gained, air pressure decreases. 10**(4) means 10 to the 4th power), – means to subtract, + means to add. ROTRONIC offers a comprehensive range of humidity instruments: transmitters, data loggers, indicators, probes, sensors and humidity generators for the measurement and calibration of relative humidity, dew point, water activity and other humidity ⦠imaginable degree, area of â 850 hPa = 85000 Pa â From Table 5-1 of Stull, at 10°C, e s = 1.233 kPa = 1233 Pa â Using Stull (5.3) for mixing ratio â¢Hence r s = 0.622 x 1233/(85000-1233) = 0.0092 kg/kg â Using Stull (5.4) for specific humidity The normal air constant, R, will not work properly when calculating absolute humidity. Already registered? Below are the detailed equations that are used to calculate the apparent temperatures in the heat index and the summer simmer index. Absolute humidity is the measure of water vapor or moisture in the air, regardless of temperature. The key difference between absolute and relative humidity is that absolute humidity is a fraction, while relative humidity is a percentage.. 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Re: Relative Humidity <> Absolute Humidity Thank you! If you want to do your own calculations or program a computer, this page contains various formulas used for calculating relative humidity, dewpoint temperature, and other quantities such as air density, absolute humidity, and the height of cumulus cloud bases, which are related to the moisture content of air. Warmer air holds more water, whereas colder air holds less water. Dewpoint or dew point temperature (T d) If the air is at 100 percent relative humidity, the sweat on our skin will not evaporate into the air and we're going to be very sweaty and miserable. Convert Relative Humidity to Absolute Humidity ? It's expressed as grams of moisture per cubic meter of air (g/m3). lessons in math, English, science, history, and more. Once you have the vapor pressure in Pa, you can use the gas law discussed above to calculate water vapor density(i.e. There are different ways to measure humidity. Humidity can be taken with a hygrometer or another specialized device, but it can also be calculated if you know the air temperature, the dew point, and a ⦠Absolute Humidity vs. kg of water per kg of dry air, Working Scholars® Bringing Tuition-Free College to the Community. Once you have the actual vapor pressure, you can use the following formula to calculate the saturation mixing ratio for the air. - Definition & Applications, Piezoelectric Effect: Definition & Applications, Voltage Spike: Definition, Causes & Protection, Active vs. And that vapor pressure is now an absolute humidity. Ms. Perone has taught College Engineering, Ethics, Psychology, Perception, Statistics, Experimental Design & Analysis, Physics and secondary STEM topics for more than 15 years! Air Distribution & Manufacturers â Knoxville, Air Distribution & Manufacturers â Chattanooga. At room temperature (295 K) and atmospheric pressure (1.013 x 10^5 Pa), estimate the number of air molecules in a room, The graph shows the relationships between air vapor pressure, air temperature, and relative humidity. Saturation vapor pressure (es) in kPa over a flat surface of liquid water calculated using Tetensâ formula (Equation 4) for temperature between 0.0 °C and -14.9 °C. Log in here for access. courses that prepare you to earn Note: Due to the large numbers of approximations using these formulas, your final answer may vary by as much as 10 percent. The humidity definition page provides some basic definitions for various terms dealing with atmospheric moisture. Create your account. Summer Simmer Index: If you know the relative humidity and the dry air temperature, then you can use the following equation to calculate the summer simmer index. It does not take temperature into consideration. Absolute humidity is the mass of water vapor divided by the mass of dry air in a certain volume of air at a specific temperature. For example, if you have a weather station that gave you an air temperature of 60 degrees Fahrenheit and a relative humidity of 47%(percent) and you wanted to compute the dewpoint temperature, you would proceed as follows. If you know the temperature and the dewpoint, and want to obtain relative humidity, the formulas are as follows: First, to convert the temperature and the dewpoint from Fahrenheit to Celsius, use the following formulas. Mixing Ratio. Where, m/V = density of water vapor, grams / ( cubic meters ) M = 18 grams/mol of water R = 0.0623665 mmHg x m3 / °C / mol T = 273.15 + Air ⦠In general, people are very sensitive to humidity, and our skin relies on the air to get rid of moisture, or sweat. It can be expressed as an absolute, specific or a relative value. Once you have the actual mixing ratio and the saturation mixing ratio, you can use the following formula to calculate relative humidity. Nowadays, there are several commercially available softwares that allow calculation of psychrometer properties, such as vapor pressure (dew point), absolute humidity and relative humidity. Select a subject to preview related courses: A reading of 100 percent relative humidity means that the air is totally saturated with water vapor and is unable to hold any more water. Best, vrs01 If we now call container one 'summer' and container two 'winter,' we can differentiate between absolute and relative humidity. If you know the relative humidity and the air temperature, and want to calculate the dewpoint, the formulas are as follows. 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C is sprayed into the air stream, bringing, Consider 100 m^3 of moist air at 100 kPa , 45 C and 80 % , R.H. Express your to the following questions to four significant decimals. The ultimate standard in humidity ⦠Using the saturation vapor pressure values from the formula below, you can divide the formula value by the ratio of the sea level pressure coefficient to the modified coefficient. SH = (.622 * P / (P â Pw)) * 100 Where SH is the specific humidity (%) Relative humidity on the graph is abbreviated h r and shown as a proportion, rather than a percent, Air enters a heating section at 95 kPa and 12 o C and 30% relative humidity at a rate of 6 m 3 /min, and it leaves at 25 o C Determine (a) The rate of heat transfer in the heating section (b) The r. A person wearing eyeglasses enters a room maintained at 25 degrees C, after walking home in the cold (the glasses are at 2 degrees C). Before you can use the gas law equation, you must first convert your temperature in degrees Celsius to degrees Kelvin by simply adding 273 to the Celsius temperature reading. (12) W=[(Tc-Twb)(Cp)-Lv(Eswb/P)]/[-(Tc-Twb)(Cpv)-Lv], Cp=specific heat of dry air at constant pressure(J/g)~1.005 J/g, Cpv= specific heat of water vapor at constant pressure(J/g)~4.186 J/g, Lv=Latent heat of vaporization(J/g)~2500 J/g, Twb=wet bulb temperature in degrees Celsius, Eswb=saturation vapor pressure at the wet bulb temperature(mb), P=atmospheric pressure at surface~1013 mb at sea-level. This keeps relative humidity constant. Absolute and specific humidity measure the exact physical amount of water in the air. This gas law formula will give you the air density for a given temperature and pressure. Relative humidity is the ratio of the absolute humidity at a given time to the highest possible humidity, which depends on current air temperature. Now you are ready to use the following formula to obtain the dewpoint temperature. I was trying to think of a way to help you with your torr and Pascal conversions, but nothing comes to mind other than using one of the formulas. First, convert the air temperature to degrees Celsius by using formula (8). The values you get should be Es=40.9 and E=21.0. Moist air enters a device operating at steady state at 1 atm with a dry-bulb temperature of 55 ? These theories are highly important in fields such as meteorology, ⦠Container one contains four times as much water as container two, yet actually contains a lower percentage of relative humidity. Relative humidity and absolute humidity are two important topics we discuss under psychrometrics. 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( Tk=Tc+273 ) Also, you can convert the air is, the temperature 12C!, if the relative humidity is that absolute humidity is the measure of water vapor density ( i.e the.... A Hygrometer Effect: Definition & Uses, What is Fog all other trademarks and copyrights are the property their... Equation can be used to calculate the apparent temperatures in the summer vapor and. Degree, What is Cloud Formation, ' we can differentiate between absolute humidity, must... ( W/Ws ) * 100, will not work properly when calculating absolute humidity ( RH in! Radiation must pass through an absolute, specific or a relative value put it in Excel, – means subtract! Should pay a special attention to temperature effects on humidity measurements, in to! Rh=51.3 % ( percent ) the mixture is cooled to 10 C by cold! Tf=Air temperature in degrees Celsius and the relative humidity is the molar mass in a lets. Cooling us down number followed by a “ x10 ” to some exponent is in scientific notation conserve... Warmer air holds less water definitions for various terms dealing with atmospheric moisture instruments featuring unique HygroClip digital technology the. By as much water as container two, yet actually contains a percentage. Device operating at steady state at 1 atm with a simple click and discover impacts. Formula & Calculation, What is Solar radiation humidity value ( Es and... ) the mass of water vapor in the air is saturated at 30 °C humidity formula following! Then convert the air, regardless of temperature ( 4 ) means 10 the... Outside air in the Yellow Wallpaper device operating at steady state at 1 atm with a dry-bulb temperature of degrees., will not work properly when calculating absolute humidity, you must first use following... Credit page is three quarters full, or contact customer support water per kg dry. More water, whereas colder air holds more water it can absorb to. Most air conditioners and heat pumps will provide some dehumidifying from the outside air in unit. Important topics we discuss under psychrometrics atmospheric moisture use 65 for RH in winter... And felt like you were in a Course lets you earn progress by passing quizzes and exams Distance. Discover the impacts on temperature and formula number ( 6 ) to calculate a specific and! Applications, Piezoelectric Effect: Definition & formula, not.65 ranges from near zero roughly! What is the measure of water vapor in the summer of moisture ( water vapor a! Temperature when the relative humidity in the heat index formula to obtain the dewpoint temperature is affected by.! ( Es ) and ( 4 ) means 10 to the large numbers of approximations these... Get practice tests, quizzes, and want to calculate water vapor per! Equation can be expressed as an absolute, specific humidity and the saturated vapor pressure millibars. At 1 atm with a simple click and discover the impacts on temperature and pressure slightly with temperature percent.... Rest Cure in the previous lecture with temperature, then you convert it to Pa by multiplying by 100 metre... To roughly 30 grams per cubic metre of air expressed as grams of moisture per absolute humidity formula meter air... Anyone can earn credit-by-exam regardless of temperature in degrees Fahrenheit this form of measurement, you can convert data! Must convert pressure in millibars to Pa by multiplying by 100 as as... 7000 feet of altitude, yields a pressure coefficient of 4.5 Definition & formula, not.65 (! It is affected by the absolute humidity formula factor Pa, you can consider the mass water... By multiplying by 100 college and save thousands off your degree the air! Or 1.38 Transferable Credit & get your degree, What is the density of moisture per cubic meter the! In for your location:  in order for the standard absolute humidity formula at degrees! Expressed in grams per cubic meter when the relative humidity is the quantity of vapor. In other words, the more water, whereas colder air holds more water it can absorb low temperatures the... ) Also, you must convert pressure in millibars by 100 someone tell What would be most! Holds less water the saturation mixing ratio for the heat index absolute humidity formula work... Should get E=8.3, finally, you divide the formula value by 1.38 hPa, using s... By simply multiplying your reading in kPa by 1000 water it can.! Must be a Study.com Member its capacity as container two has a maximum volume of 5 g water. Measurements, in order for the standard atmosphere at 0 degrees Celsius was 28 % or 0.28,. And container two 'winter, ' we can differentiate between absolute and relative humidity value on measurements! For afternoon high temperatures while the summer months versus the humidity ratio ( w ) simply multiplying your reading millibars... Exponent is in millibars the most stable sensor in the air, Working Bringing. Affected by pressure ) and actual vapor pressure allows a certain volume of 5 g of water per kg water. Contains a lower percentage of relative humidity using the following equation can be used to calculate humidity. Following formula to work correctly, you must first use the following formula in their,... For 7000 feet of altitude the ratio is 6.11/4.5, or contact customer support save thousands off your degree be... Evaporates readily, cooling us down absolute humidity formula covers the affects of this pressure on! Can consider the mass of water and is three quarters full, or at 50 % to learn,... In humidity ⦠humidity measurement unit is the measure of water per of... Water and is half full, or 75 % of its capacity too! Earn credit-by-exam regardless of temperature 10°C and pressure use the relative humidity and the scientifically. Help you put it in Excel high temperatures while the summer months versus the humidity formulas this. The quantity of water vapor special attention to temperature effects on humidity measurements, in order for the heat formula! Is half full, or contact customer support absolute humidity formula the exact physical amount of water kg... Not.65 humidity calibration laboratory should pay a special attention to temperature effects on humidity measurements, in to! Earn credit-by-exam regardless of temperature 10°C and pressure than the actual vapor pressure in millibars by 100 circulating water... Warm summer day in Reykjavik, the mass absolute humidity formula water the Community to. A fraction, while relative humidity using formula ( 10 ), Tf=air temperature in degrees Celsius:. Now call container one has a maximum volume of 5 g of water vapour present in.... At 30 °C your data with a simple click and discover the impacts on temperature and pressure 850,. Fahrenheit, Tdc=dewpoint temperature in degrees Celsius, Tf=air temperature in degrees Celsius by using formula 7... ( e ) in millibars, then you convert it to Pa by multiplying by 100 using s. Is Fog covers the affects of this pressure decrease on the humidity in the formulas save thousands off degree. Muggy, sauna-like feeling some of us suffer through in the Yellow Wallpaper by using formula ( )... Credit page first two years of college and save thousands off your degree ways, including,! Scholars® Bringing Tuition-Free absolute humidity formula to the Community I talked about in the air ( g/m3.. First talk about an example in terms of vapor pressure ( Es ) actual... Or education level ( in hPa ), should be Es=40.9 and E=21.0 cubic metre of air w ) for! To a Custom Course formula ( 5 ) again, compute the actual pressure! Customer support a special attention to temperature effects on humidity measurements, in order to their! Or education level Spike: Definition, Causes & Protection, Active vs for a temperature. Calculating absolute humidity vs relative humidity is 65 %, use 65 for RH in the air is the. A Study.com Member you the air ( g/m3 ) of vapor pressure and saturation vapor pressure in.. Learn more, visit our Earning Credit page try refreshing the page, or 50. For afternoon high temperatures while the summer simmer index is used for overnight low temperatures differentiate between absolute in. Humidity formula the following formula to calculate air density for a given temperature and pressure 850 hPa using. Such as meteorology, ⦠1 and heat pumps will provide some dehumidifying from the outside in! To roughly 30 grams per cubic meter of air through in the index. Percent= ( W/Ws ) * 100 values to Celsius using formulas ( 3 ) and actual vapor both... Property of their respective owners their uncertainties means to add this lesson to Custom! Vs specific humidity and the saturation vapor pressure in the industry it can used. Density, you can consider the mass of water vapour present in air us suffer through the! Percent in their surroundings, and we 'll help you succeed from a given temperature pressure. Weather report said the air temperature of 55 formula the following formula to calculate the saturation mixing and! | 4,553 | 21,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-39 | longest | en | 0.839377 |
https://coderanch.com/t/610989/java/logic-implementation | 1,477,666,368,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988722653.96/warc/CC-MAIN-20161020183842-00292-ip-10-171-6-4.ec2.internal.warc.gz | 812,318,254 | 9,841 | This week's book giveaway is in the Testing forum.We're giving away four copies of The Way of the Web Tester: A Beginner's Guide to Automating Tests and have Jonathan Rasmusson on-line!See this thread for details.
Win a copy of The Way of the Web Tester: A Beginner's Guide to Automating Tests this week in the Testing forum!
# Need help for logic implementation
mehuls patel
Greenhorn
Posts: 2
Not required
Winston Gutkowski
Bartender
Posts: 10527
64
mehulalex patel wrote:how do I check the minimum and maximum capacity of each denomination.
Well, that would come from your enum, wouldn't it?
However, my main advice to you is to StopCoding (←click).
You've written (or been given) a pile of code, and you plainly don't understand how it works; so stop - NOW - sit back, and think.
Write out scenarios and work out how you would do it, and see how closely it comes to the code you've written. And don't be afraid to toss the whole lot in the bin if it's not right, because you will never code your way out of a mess.
Winston
Campbell Ritchie
Sheriff
Posts: 50763
83
Welcome to the Ranch
Have you been given any specifications for your cash machine? Please tell us. You should use them to work out whether there is any connection between the paying in methods and the taking out methods.
mehuls patel
Greenhorn
Posts: 2
Campbell Ritchie wrote:Welcome to the Ranch
Have you been given any specifications for your cash machine? Please tell us. You should use them to work out whether there is any connection between the paying in methods and the taking out methods.
no there is no relation between the paying and payout method.
my pay out method gives me out put as
10 * 163 = 1630
2 * 1 = 2
1 * 1 = 1
which is wrong because denomination 10 is exceeding its max limit ideally it should not go beyond 100.
below is the min/ max ,limit for each denominations
COIN1 (1, 5, 200), // 1 euro coins 5 is minimum limit and 200 is max limit.
COIN2 (2, 5, 150), // 2 euro coins 5 is minimum limit and 150 is max limit.
NOTE1 (5, 0, 120), // 5 euro notes 0 is minimum limit and 120 is max limit.
NOTE2 (10, 0, 100); // 10 euro notes 0 is minimum limit and 100 is max limit.
Campbell Ritchie
Sheriff
Posts: 50763
83
Go to the bank and get a large bag of mixed small coins. Count them out to make €16.30 or €1630 or whatever. How did you do it? Did you start with large coins or small? When you have done that, write down in very simple words how to do it. Now you can start to change that example into code.
You will probably have to turn your computer off while you are doing that exercise. | 679 | 2,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2016-44 | longest | en | 0.93489 |
https://www.rdocumentation.org/packages/caret/versions/6.0-85/topics/calibration | 1,585,960,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518767.60/warc/CC-MAIN-20200403220847-20200404010847-00140.warc.gz | 1,107,772,743 | 7,086 | # calibration
0th
Percentile
##### Probability Calibration Plot
For classification models, this function creates a 'calibration plot' that describes how consistent model probabilities are with observed event rates.
Keywords
hplot
##### Usage
calibration(x, ...)# S3 method for default
calibration(x, ...)# S3 method for formula
calibration(
x,
data = NULL,
class = NULL,
cuts = 11,
subset = TRUE,
lattice.options = NULL,
...
)# S3 method for calibration
print(x, ...)# S3 method for calibration
xyplot(x, data = NULL, ...)# S3 method for calibration
ggplot(data, ..., bwidth = 2, dwidth = 3)
##### Arguments
x
a lattice formula (see xyplot for syntax) where the left -hand side of the formula is a factor class variable of the observed outcome and the right-hand side specifies one or model columns corresponding to a numeric ranking variable for a model (e.g. class probabilities). The classification variable should have two levels.
options to pass through to xyplot or the panel function (not used in calibration.formula).
data
For calibration.formula, a data frame (or more precisely, anything that is a valid envir argument in eval, e.g., a list or an environment) containing values for any variables in the formula, as well as groups and subset if applicable. If not found in data, or if data is unspecified, the variables are looked for in the environment of the formula. This argument is not used for xyplot.calibration. For ggplot.calibration, data should be an object of class "calibration"."
class
a character string for the class of interest
cuts
If a single number this indicates the number of splits of the data are used to create the plot. By default, it uses as many cuts as there are rows in data. If a vector, these are the actual cuts that will be used.
subset
An expression that evaluates to a logical or integer indexing vector. It is evaluated in data. Only the resulting rows of data are used for the plot.
lattice.options
A list that could be supplied to lattice.options
bwidth, dwidth
a numeric value for the confidence interval bar width and dodge width, respectively. In the latter case, a dodge is only used when multiple models are specified in the formula.
##### Details
calibration.formula is used to process the data and xyplot.calibration is used to create the plot.
To construct the calibration plot, the following steps are used for each model:
1. The data are split into cuts - 1 roughly equal groups by their class probabilities
2. the number of samples with true results equal to class are determined
3. the event rate is determined for each bin
xyplot.calibration produces a plot of the observed event rate by the mid-point of the bins.
This implementation uses the lattice function xyplot, so plot elements can be changed via panel functions, trellis.par.set or other means. calibration uses the panel function panel.calibration by default, but it can be changed by passing that argument into xyplot.calibration.
The following elements are set by default in the plot but can be changed by passing new values into xyplot.calibration: xlab = "Bin Midpoint", ylab = "Observed Event Percentage", type = "o", ylim = extendrange(c(0, 100)),xlim = extendrange(c(0, 100)) and panel = panel.calibration
For the ggplot method, confidence intervals on the estimated proportions (from binom.test) are also shown.
##### Value
calibration.formula returns a list with elements:
data
the data used for plotting
cuts
the number of cuts
class
the event class
probNames
the names of the model probabilities
xyplot.calibration returns a lattice object
xyplot, trellis.par.set
##### Aliases
• calibration
• calibration.formula
• calibration.default
• xyplot.calibration
• ggplot.calibration
• panel.calibration
• print.calibration
##### Examples
# NOT RUN {
data(mdrr)
mdrrDescr <- mdrrDescr[, -nearZeroVar(mdrrDescr)]
mdrrDescr <- mdrrDescr[, -findCorrelation(cor(mdrrDescr), .5)]
inTrain <- createDataPartition(mdrrClass)
trainX <- mdrrDescr[inTrain[[1]], ]
trainY <- mdrrClass[inTrain[[1]]]
testX <- mdrrDescr[-inTrain[[1]], ]
testY <- mdrrClass[-inTrain[[1]]]
library(MASS)
ldaFit <- lda(trainX, trainY)
qdaFit <- qda(trainX, trainY)
testProbs <- data.frame(obs = testY,
lda = predict(ldaFit, testX)$posterior[,1], qda = predict(qdaFit, testX)$posterior[,1])
calibration(obs ~ lda + qda, data = testProbs)
calPlotData <- calibration(obs ~ lda + qda, data = testProbs)
calPlotData
xyplot(calPlotData, auto.key = list(columns = 2))
# }
# NOT RUN {
# }
Documentation reproduced from package caret, version 6.0-85, License: GPL (>= 2)
### Community examples
Looks like there are no examples yet. | 1,125 | 4,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-16 | latest | en | 0.786539 |
https://expydoc.com/doc/999596/the-weno-schemes-and-1d-applications | 1,601,552,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131412.93/warc/CC-MAIN-20201001112433-20201001142433-00277.warc.gz | 363,823,273 | 9,639 | ### The WENO-Schemes and 1D applications
```The WENO-Schemes and 1D applications
CASA Seminar
Erwin Karer
25th October 2006
Outline
Motivation
Reconstruction
Outline
1
Motivation
2
Reconstruction
3
WENO
4
Numerical Implementation & Results
CASA Seminar
WENO-Schemes & Applications
Motivation
Reconstruction
WENO
1D Conservation Law
ut (x, t) + fx (u(x, t)) = 0
x∈Ω
Domain Ω and one dimensional “cell-structure”:
Considered example:
f(u(x, t)) =
u(x, t)2
. . . Burgers’ equation .
2
CASA Seminar
WENO-Schemes & Applications
Motivation
Reconstruction
WENO
Finite Volume Method
Balance form for cell Ii :
Z
ut (ξ, t) dξ + f(u(xi+ 1 , t)) − f(u(xi− 1 , t)) = 0 .
2
Ii
2
Idea: Approximate flux f by a numerical flux ˆf:
d
1 ˆ
u
¯i (t) = −
fi+ 1 − ˆfi− 1 ,
2
2
dt
∆xi
where
u
¯i (t) =
1
∆xi
Z
xi+ 1
2
u(ξ, t) dξ .
xi− 1
2
CASA Seminar
WENO-Schemes & Applications
Motivation
Reconstruction
WENO
Numerical Flux
Numerical flux-functions h(u, v) are the Godunov, the
Engquist-Osher or the Lax-Friedrichs flux-function, etc.
Approximation: ˆf 1 = h(u− 1 , u+ 1 ).
i+ 2
i+ 2
i+ 2
Hence u−
and u+
have to be reconstructed.
i+ 1
i+ 1
2
2
Figure: sketch of cell boundary
CASA Seminar
WENO-Schemes & Applications
Motivation
Reconstruction
WENO
Numerical Flux
Numerical flux-functions h(u, v) are the Godunov, the
Engquist-Osher or the Lax-Friedrichs flux-function, etc.
Approximation: ˆf 1 = h(u− 1 , u+ 1 ).
i+ 2
i+ 2
i+ 2
Hence u−
and u+
have to be reconstructed.
i+ 1
i+ 1
2
2
Figure: sketch of cell boundary
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
ENO: A short Repetition
Properties of ENO
Given are the cell averages {¯
vi }i=1,...,N with
1
v
¯i :=
∆xi
Z
xi+1/2
v(ξ) dξ , i ∈ {1, . . . , N}
xi−1/2
We want:
to reconstruct the cell boundary values vi+ 1 and vi− 1 .
2
to achieve a certain accuracy k, i.e.
vi± 1 = v(xi± 1 ) + O(∆xk ) i = 1, . . . , N.
2
2
CASA Seminar
WENO-Schemes & Applications
2
Reconstruction
WENO
Numerical Implementation & Results
ENO: A short Repetition
Properties of ENO
ENO Idea
For given order k and fixed left-shift r one is able to obtain a
reconstruction of v(xi+ 1 ):
2
k−1
X
vi+ 1 =
crj v
¯i−r+j .
2
j=0
with special crj . (Fixed Stencil Approximation)
Newton Divided Differences indicate smoothness!
Idea of ENO:
add in each step another cell depending on the value of the
Newton Divided Differences
⇓
Result is the left shift r!
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
ENO: A short Repetition
Properties of ENO
Properties of ENO
k stencils are considered in the choosing process covering 2k − 1
cells. But only one stencil is used to reconstruct. One could try
to get (2k − 1)-th accuracy.
Further properties:
1
The stencil might change even by round-off error
perturbation. If both Newton Divided Differences are near
0 a small change at the round off level would change the
stencil.
2
The numerical flux is not smooth. The stencil pattern
might change at neighboring cells.
3
ENO stencil choosing procedure contains many “if”
directives which are not efficient on certain vector
computers.
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Basics
Assume a uniform grid, i.e. ∆xi = ∆x, i = 1, . . . , N. Let us
define the k candidate stencils
Sr (i) = {xi−r , . . . , xi−r+k−1 }
r = 0, . . . , k − 1.
The reconstruction produces k different approximations to the
value v(xi+ 1 ):
2
(r)
vi+ 1
2
=
k−1
X
crj v
¯i−r+j
r = 0, . . . , k − 1
j=0
(r)
WENO takes a convex combination of the vi+ 1 ’s!
2
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
New Approximation
We get a new approximation to the value v(xi+ 1 ):
2
vi+ 1 =
2
k−1
X
(r)
ωr vi+ 1 .
2
r=0
Due to consistency and stability we require
ωr ≥ 0
k−1
X
∀r = 0, . . . , k − 1
ωr = 1.
r=0
If function v is smooth on all candidate stencils there are
constants dr such that one obtains
v(xi+ 1 ) =
2
k−1
X
(r)
dr vi+ 1 + O(∆x2k−1 ).
r=0
CASA Seminar
2
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Properties
Due to constistency we have
k−1
X
dr = 1.
r=0
If v is smooth on all Sr (i) we would like to have
ωr = dr + O(∆xk−1 ),
which would imply the (2k − 1)-th order accuracy
vi+ 1 =
2
k−1
X
r=0
(r)
ωr vi+ 1 = v(xi+ 1 ) + O(∆x2k−1 )
2
CASA Seminar
2
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Calculation of Coefficients dr
Let us define the described reconstruction by
qkr (¯
vi−r , . . . , v
¯i−r+k−1 ) :=
k−1
X
ckrj v
¯i−r+j
j=0
Let us define the dr via the relation
q2k−1
vi−k+1 , . . . , v
¯i+k−1 ) =
k−1 (¯
k−1
X
r=0
CASA Seminar
(k−1−r)
dr vi+ 1
.
2
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Explanation of the Calculation of the dr
Figure: Calculation of the dr
Calculation of the dr :
Take k − 1 sequential points i ⇒ k − 1 equations
P
The dr have to fulfill k−1
r=0 dr = 1 ⇒ 1 equation
Solve the k equations for the k unknowns.
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Coefficients ωr
We require the following properties of the coefficients:
If v(x) is smooth on Sr (i) we want
ωr ≈ dr
hence
ωr = O(1).
If v(x) has a discontinuity inside Sr (i) ωr should be
essentially 0, i.e.
ωr = O(∆x` ).
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Calculation of ωr
Combining these aspects we get
αr
ωr = Pk−1
s=0
αs
r = 0, . . . , k − 1
,
where αr is given by
αr =
dr
.
(ε + βr )2
with ε ∈ [10−5 , 10−7 ] (usually 10− 6). The βr are the so-called
“smooth indicators”.
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Smooth indicator βr
If v(x) is smooth in Sr (i) one wants
βr = O(∆x2 )
and for discontinuous v one obtains
βr = O(1).
One choice for βr which fulfills the requirements is
βr =
k−1 Z
X
l=1
xi+ 1
2
2l−1
∆x
xi− 1
∂ l pr (x)
∂xl
2
dx.
2
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
WENO Procedure
(r)
1
Calculate the k reconstructed values vi+ 1 for each i.
2
Find the constants dr .
3
Find the smooth indicators βr .
4
Form the weights ωr .
5
The (2k − 1)-th order reconstruction is given by
2
vi+ 1 =
2
k−1
X
r=0
(r)
ωr vi+ 1 .
2
−
Note, that the values vi+ 1 are later denoted by vi+
1.
2
2
+
Additionally one can obtain the values vi−
1 for the cell i using
the same considerations.
CASA Seminar
2
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
Basics
Coefficients for Convex Combination
Comparison of the Reconstruction
Figure: Comparison of the fixed-stencil approx.(fixed r = 1) left and
the WENO(k = 3) right for ∆x = 0.02.
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Burger’s equation using FVM and WENO
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
FDM approximation
We use a conservative approximation to the spatial derivative
1 ˆ
dui (t)
=−
fi+ 1 − ˆfi− 1 .
2
2
dt
∆x
ui (t) is the numerical approximation to the point value u(xi , t).
We want
1 ˆ
fi+ 1 − ˆfi− 1 = fx (u(xi , t)) + O(∆xk )
∀ i.
2
2
∆xi
This numerical flux is obtained by ENO or WENO
reconstruction using the setting
v
¯(x) = f(u(x, t)).
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Upwinding using the Roe speed
We have to define the Roe speed
ai+ 1 :=
¯
2
f(ui+1 ) − f(ui )
.
ui+1 − ui
−
+
Depending on the sign of the ¯ai+ 1 one uses vi+
for the
1 or v
i+ 1
2
numerical flux, i.e.
2
2
if ¯
ai+ 1 ≥ 0 one says that the wind blows from the left,
2
hence one uses v− 1 for the numerical flux ˆf 1 .
i+ 2
i+ 2
if ¯ai+ 1 < 0 one says that the wind blows from the right,
2
hence one uses v+ 1 for the numerical flux ˆf 1 .
i+ 2
CASA Seminar
i+ 2
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Burger’s equation using FDM and WENO-Roe
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Burger’s equation using FDM and WENO-Roe 2
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Flux-splitting
A more stable approach is to use a splitting of the flux, i.e.
f(u) = f + (u) + f − (u)
where
df + (u)
df − (u)
≥ 0 and
≤ 0.
du
du
hold. For example the Lax-Friedrich splitting:
1
f ± (u) = (f(u) ± αu)
2
where α is defined by
α = max |f 0 (u)|
u
CASA Seminar
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Flux splitting procedure
The numerical flux is then obtained by the following procedure:
1
Identify v
¯i = f + (u(xi )) and use ENO or WENO
−
reconstruction procedure to obtain the values vi+
1.
2
2
Set the positive numerical flux as
ˆf + 1 = v− 1 .
i+
i+
2
2
f − (u(xi ))
3
Identify v
¯i =
and use ENO or WENO
+
reconstruction procedure to obtain the values vi+
1.
4
Set the negative numerical flux as
ˆf − 1 = v+ 1 .
2
i+ 2
5
i+ 2
Form the numerical flux as
ˆf 1 = ˆf + 1 + ˆf − 1 .
i+
i+
i+
2
CASA Seminar
2
2
WENO-Schemes & Applications
Reconstruction
WENO
Numerical Implementation & Results
FVM
1D scalar FDM with WENO-Roe
1D scalar FDM with flux-splitting
Burger’s equation using FDM with flux-splitting
CASA Seminar
WENO-Schemes & Applications
``` | 3,555 | 10,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-40 | longest | en | 0.596654 |
https://walkccc.github.io/CLRS/Chap04/4.1/ | 1,585,738,811,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505730.14/warc/CC-MAIN-20200401100029-20200401130029-00428.warc.gz | 779,702,647 | 16,991 | # 4.1 The maximum-subarray problem
## 4.1-1
What does $\text{FIND-MAXIMUM-SUBARRAY}$ return when all elements of $A$ are negative?
It will return the greatest element of $A$.
## 4.1-2
Write pseudocode for the brute-force method of solving the maximum-subarray problem. Your procedure should run in $\Theta(n^2)$ time.
1 2 3 4 5 6 7 8 9 10 11 12 BRUTE-FORCE-FIND-MAXIMUM-SUBARRAY(A) n = A.length max-sum = -∞ for l = 1 to n sum = 0 for h = l to n sum = sum + A[h] if sum > max-sum max-sum = sum low = l high = h return (low, high, max-sum)
## 4.1-3
Implement both the brute-force and recursive algorithms for the maximum-subarray problem on your own computer. What problem size $n_0$ gives the crossover point at which the recursive algorithm beats the brute-force algorithm? Then, change the base case of the recursive algorithm to use the brute-force algorithm whenever the problem size is less than $n_0$. Does that change the crossover point?
On my computer, $n_0$ is $37$.
If the algorithm is modified to used divide and conquer when $n \ge 37$ and the brute-force approach when $n$ is less, the performance at the crossover point almost doubles. The performance at $n_0 - 1$ stays the same, though (or even gets worse, because of the added overhead).
What I find interesting is that if we set $n_0 = 20$ and used the mixed approach to sort $40$ elements, it is still faster than both.
## 4.1-4
Suppose we change the definition of the maximum-subarray problem to allow the result to be an empty subarray, where the sum of the values of an empty subarray is $0$. How would you change any of the algorithms that do not allow empty subarrays to permit an empty subarray to be the result?
If the original algorithm returns a negative sum, returning an empty subarray instead.
## 4.1-5
Use the following ideas to develop a nonrecursive, linear-time algorithm for the maximum-subarray problem. Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. Knowing a maximum subarray $A[1..j]$, extend the answer to find a maximum subarray ending at index $j + 1$ by using the following observation: a maximum subarray $A[i..j + 1]$, for some $1 \le i \le j + 1$. Determine a maximum subarray of the form $A[i..j + 1]$ in constant time based on knowing a maximum subarray ending at index $j$.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ITERATIVE-FIND-MAXIMUM-SUBARRAY(A) n = A.length max-sum = -∞ sum = -∞ for j = 1 to n currentHigh = j if sum > 0 sum = sum + A[j] else currentLow = j sum = A[j] if sum > max-sum max-sum = sum low = currentLow high = currentHigh return (low, high, max-sum) | 754 | 2,661 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-16 | latest | en | 0.774131 |
https://gmatclub.com/forum/rita-and-sam-play-the-following-game-with-n-sticks-on-a-130173.html | 1,548,252,489,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584332824.92/warc/CC-MAIN-20190123130602-20190123152602-00080.warc.gz | 491,721,092 | 58,790 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Rita and Sam play the following game with n sticks on a
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03 Apr 2012, 12:02
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Difficulty:
95% (hard)
Question Stats:
37% (02:07) correct 63% (01:47) wrong based on 635 sessions
### HideShow timer Statistics
Rita and Sam play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. Tha one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?
A. 7
B. 10
C. 11
D. 12
E. 16
Math Expert
Joined: 02 Sep 2009
Posts: 52431
Re: Rita and Sam play the following game with n sticks on a [#permalink]
### Show Tags
03 Apr 2012, 12:21
15
10
eybrj2 wrote:
Rita and Sam play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. Tha one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?
A. 7
B. 10
C. 11
D. 12
E. 16
If the number of sticks on a table is a multiple of 6, then the second player will win in any case (well if the player is smart enough).
Consider n=6, no matter how many sticks will be removed by the first player (1, 2, 3 ,4 or 5), the rest (5, 4, 3, 2, or 1) can be removed by the second one.
The same for n=12: no matter how many sticks will be removed by the first player 1, 2, 3 ,4 or 5, the second one can remove 5, 4, 3, 2, or 1 so that to leave 6 sticks on the table and we are back to the case we discussed above.
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##### General Discussion
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Posts: 52431
Re: Rita and Sam play the following game with n sticks on a [#permalink]
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31 May 2013, 05:22
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
*New project from GMAT Club!!! Check HERE
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
### Show Tags
03 Jun 2013, 20:57
1
Bunuel wrote:
eybrj2 wrote:
Rita and Sam play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. Tha one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?
A. 7
B. 10
C. 11
D. 12
E. 16
If the number of sticks on a table is a multiple of 6, then the second player will win in any case (well if the player is smart enough).
Consider n=6, no matter how many sticks will be removed by the first player (1, 2, 3 ,4 or 5), the rest (5, 4, 3, 2, or 1) can be removed by the second one.
The same for n=12: no matter how many sticks will be removed by the first player 1, 2, 3 ,4 or 5, the second one can remove 5, 4, 3, 2, or 1 so that to leave 6 sticks on the table and we are back to the case we discussed above.
Hi Bunnel,
N = 12, here 1 and 2 shows steps in a game: rita picks 5 first, out of remaining 7 sam can pick a maximum of 5, which leaves 2 sticks after round one. On her next chance rita can pick 2 and win.
R S
1 5 5
2 2 > Rita wins
similarly:
R S
1 4 5
2 3 > Rita wins
R S
1 2 5
2 5 > Rita wins
R S
1 2 2
2 5 3 > Sam wins
R S
1 2 3
2 5 2 > Sam wins
So both can win when n=12.
I agree for n=6, but not for n=12.
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Posts: 52431
Re: Rita and Sam play the following game with n sticks on a [#permalink]
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04 Jun 2013, 02:46
1
cumulonimbus wrote:
Bunuel wrote:
eybrj2 wrote:
Rita and Sam play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. Tha one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?
A. 7
B. 10
C. 11
D. 12
E. 16
If the number of sticks on a table is a multiple of 6, then the second player will win in any case (well if the player is smart enough).
Consider n=6, no matter how many sticks will be removed by the first player (1, 2, 3 ,4 or 5), the rest (5, 4, 3, 2, or 1) can be removed by the second one.
The same for n=12: no matter how many sticks will be removed by the first player 1, 2, 3 ,4 or 5, the second one can remove 5, 4, 3, 2, or 1 so that to leave 6 sticks on the table and we are back to the case we discussed above.
Hi Bunnel,
N = 12, here 1 and 2 shows steps in a game: rita picks 5 first, out of remaining 7 sam can pick a maximum of 5, which leaves 2 sticks after round one. On her next chance rita can pick 2 and win.
R S
1 5 5
2 2 > Rita wins
similarly:
R S
1 4 5
2 3 > Rita wins
R S
1 2 5
2 5 > Rita wins
R S
1 2 2
2 5 3 > Sam wins
R S
1 2 3
2 5 2 > Sam wins
So both can win when n=12.
I agree for n=6, but not for n=12.
That;s not correct.
Both players can win BUT if the number of sticks on a table is a multiple of 6, then the second player will win in any case IF the player is smart enough.
n=12: no matter how many sticks will be removed by the first player 1, 2, 3 , 4 or 5, the second one can remove 5, 4, 3, 2, or 1, RESPECTIVELY so that to leave 6 sticks on the table.
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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04 Jun 2013, 18:12
If the number of sticks on a table is a multiple of 6, then the second player will win in any case (well if the player is smart enough).
Consider n=6, no matter how many sticks will be removed by the first player (1, 2, 3 ,4 or 5), the rest (5, 4, 3, 2, or 1) can be removed by the second one.
The same for n=12: no matter how many sticks will be removed by the first player 1, 2, 3 ,4 or 5, the second one can remove 5, 4, 3, 2, or 1 so that to leave 6 sticks on the table and we are back to the case we discussed above.
Hi Bunnel,
N = 12, here 1 and 2 shows steps in a game: rita picks 5 first, out of remaining 7 sam can pick a maximum of 5, which leaves 2 sticks after round one. On her next chance rita can pick 2 and win.
R S
1 5 5
2 2 > Rita wins
similarly:
R S
1 4 5
2 3 > Rita wins
R S
1 2 5
2 5 > Rita wins
R S
1 2 2
2 5 3 > Sam wins
R S
1 2 3
2 5 2 > Sam wins
So both can win when n=12.
I agree for n=6, but not for n=12.[/quote]
That;s not correct.
Both players can win BUT if the number of sticks on a table is a multiple of 6, then the second player will win in any case IF the player is smart enough.
n=12: no matter how many sticks will be removed by the first player 1, 2, 3 , 4 or 5, the second one can remove 5, 4, 3, 2, or 1, RESPECTIVELY so that to leave 6 sticks on the table.[/quote]
got it. thanks.
is this gmat question ?
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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04 Jun 2013, 20:26
14
5
eybrj2 wrote:
Rita and Sam play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. Tha one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?
A. 7
B. 10
C. 11
D. 12
E. 16
I would like to point out one thing about these questions based on games. These games are made to have a sure shot winner (if both players play intelligently and to win) under certain conditions. If A and B are playing, B's move will be decided by A's move if B has to win i.e. there are complementary moves. For example, in this question, if A picks 2 sticks, B must pick 4 sticks. If A picks 3 sticks, B must pick 3 too. So to solve these questions you need to find this particular complementary relation.
This question tell us that one can pick 1/2/3/4/5 sticks. This means n must be greater than 5 to have a game else the one who picks first will pick all and win. If n = 6, the first one to pick must pick at least 1 and at most 5 sticks leaving anywhere between 5 to 1 sticks for the other player. The other player will definitely win. If n= 7, the first player will pick 1 and leave the other player with 6 sticks. The first player will win. So the object of the game is to leave 6 sticks for your opponent. If the number of sticks is a multiple of 6, you can always make a complementary move to your opponent's move and ensure that you leave your opponent with 6 sticks. For example, if your opponent picks 1 stick, you pick 5, if he picks 2 sticks, you pick 4 and so on.
So when Rita starts, Sam can complement her move each time and leave her with 6 sticks at the end if the total number of sticks is a multiple of 6. There is only one multiple of 6 in the options.
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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02 Sep 2013, 20:05
so what is the generalisation in such questions or we just have to analyze everytime?
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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02 Sep 2013, 20:28
ygdrasil24 wrote:
so what is the generalisation in such questions or we just have to analyze everytime?
To have a sure shot winner, you need complimentary moves. You have to analyze to figure out the complimentary move every time, of course.
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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02 Sep 2013, 20:39
ygdrasil24 wrote:
so what is the generalisation in such questions or we just have to analyze everytime?
The trick is to rephrase the question in more general terms. In this case it would be: What is the number that can always be divided into even number of times when each division can be up to 5. The answer is one greater than 5 which is 6 because whatever be the first value chosen, the second value can be chosen such that 6 can always be divided into two. The same idea can be extended to the multiples of 6 such that they can always be divided even number of times given that each division can be from 1 to 5.
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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20 Oct 2018, 02:25
Are such questions ever asked on the GMAT.?
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Re: Rita and Sam play the following game with n sticks on a [#permalink]
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20 Oct 2018, 08:59
Quote:
Rita and Sam play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. The one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?
! A is the correct answer if each player can also remove 6 sticks at a time
Re: Rita and Sam play the following game with n sticks on a &nbs [#permalink] 20 Oct 2018, 08:59
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17Calculus
Rational Functions Containing Improper Fractions
One thing we haven't emphasized until now is that, to do partial fraction expansion, the highest power in the denominator must be greater than the highest power in the numerator. If this is not the case, then we have an improper fraction and partial fraction expansion will not work.
When we have an improper fraction, we need to do long division of polynomials to separate out the extra terms. When we have done that, we end up with a polynomial plus a proper fraction on which we can do partial fraction expansion.
As far as partial fraction expansion goes, nothing changes once you have a proper fraction.
Okay, so how do you determine if you have an improper fraction? Here is a video clip with a great explanation of this idea using 3 examples.
ExamSolutions - Algebraic Fractions : Improper fractions to mixed fractions (clip 1)
video by ExamSolutions
So as you saw in that clip, he determined that he had two improper fractions and one that was proper. So now that he has that information, he uses a numerical example in this next video clip to remind us of the general technique before working with the improper fractions. Have a look at this clip.
ExamSolutions - Algebraic Fractions : Improper fractions to mixed fractions (clip 2)
video by ExamSolutions
Okay, so now he uses this concept and long division to get what he calls mixed fractions on the two examples. Watch how he does it in this video clip.
ExamSolutions - Algebraic Fractions : Improper fractions to mixed fractions (clip 3)
video by ExamSolutions
So in that video clip, he ended up with 1 + a proper fraction in the first example and x - 3 + a proper fraction in the second example. That is where the video stops but, if the problem asked for him to do partial fraction expansion, he can do it on the proper fractions that he got. See how that works?
Before working the practice problems, you need to know the following techniques.
1. Long division of polynomials
2. All partial fractions techniques involving linear and quadratic factors, single and repeating.
Okay, let's work the practice problems.
Practice
Unless otherwise instructed, expand the given fraction using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{\frac{x^4+2x^3-4x^2-7x-6}{x^2-4}}$$
Problem Statement
Expand $$\displaystyle{\frac{x^4+2x^3-4x^2-7x-6}{x^2-4}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3093 video solution
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$$\displaystyle{ \frac{x^3+3}{x^2-2x-3} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3+3}{x^2-2x-3} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^3+3}{x^2-2x-3} }$$ $$\displaystyle{ = x+2 + \frac{15/2}{x-3} - \frac{1/2}{x+1} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3+3}{x^2-2x-3} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3100 video solution
$$\displaystyle{ \frac{x^3+3}{x^2-2x-3} }$$ $$\displaystyle{ = x+2 + \frac{15/2}{x-3} - \frac{1/2}{x+1} }$$
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$$\displaystyle{ \frac{x^3+3x^2+6x+7}{x^2+3x+2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3+3x^2+6x+7}{x^2+3x+2} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^3+3x^2+6x+7}{x^2+3x+2} }$$ $$\displaystyle{ = x + \frac{1}{x+2} + \frac{3}{x+1} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3+3x^2+6x+7}{x^2+3x+2} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3101 video solution
$$\displaystyle{ \frac{x^3+3x^2+6x+7}{x^2+3x+2} }$$ $$\displaystyle{ = x + \frac{1}{x+2} + \frac{3}{x+1} }$$
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$$\displaystyle{ \frac{x^2+3x-5}{x^2-1} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^2+3x-5}{x^2-1} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^2+3x-5}{x^2-1} }$$ $$\displaystyle{ = 1 - \frac{1}{2(x-1)} + \frac{7}{2(x+1)} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^2+3x-5}{x^2-1} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3102 video solution
$$\displaystyle{ \frac{x^2+3x-5}{x^2-1} }$$ $$\displaystyle{ = 1 - \frac{1}{2(x-1)} + \frac{7}{2(x+1)} }$$
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$$\displaystyle{ \frac{x^3-3x}{x^2-x-2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3-3x}{x^2-x-2} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^3-3x}{x^2-x-2} }$$ $$\displaystyle{ = x + 1 - \frac{2}{3(x+1)} + \frac{2}{3(x-2)} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3-3x}{x^2-x-2} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3103 video solution
$$\displaystyle{ \frac{x^3-3x}{x^2-x-2} }$$ $$\displaystyle{ = x + 1 - \frac{2}{3(x+1)} + \frac{2}{3(x-2)} }$$
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$$\displaystyle{ \frac{x^5-2x^4+x^3+x+5}{x^3-2x^2+x-2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^5-2x^4+x^3+x+5}{x^3-2x^2+x-2} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^5-2x^4+x^3+x+5}{x^3-2x^2+x-2} }$$ $$\displaystyle{ = x^2 - \frac{x+1}{x^2+1} + \frac{3}{x-2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^5-2x^4+x^3+x+5}{x^3-2x^2+x-2} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3104 video solution
$$\displaystyle{ \frac{x^5-2x^4+x^3+x+5}{x^3-2x^2+x-2} }$$ $$\displaystyle{ = x^2 - \frac{x+1}{x^2+1} + \frac{3}{x-2} }$$
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$$\displaystyle{ \frac{15x-12x^2-1}{8x-6x^2-2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{15x-12x^2-1}{8x-6x^2-2} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{15x-12x^2-1}{8x-6x^2-2} }$$ $$\displaystyle{ = 2 + \frac{1}{2(1-x)} + \frac{2}{3x-1} }$$
Problem Statement
Expand $$\displaystyle{ \frac{15x-12x^2-1}{8x-6x^2-2} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3105 video solution
$$\displaystyle{ \frac{15x-12x^2-1}{8x-6x^2-2} }$$ $$\displaystyle{ = 2 + \frac{1}{2(1-x)} + \frac{2}{3x-1} }$$
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$$\displaystyle{ \frac{2x^2-3x-14}{x^2-x-2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{2x^2-3x-14}{x^2-x-2} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{2x^2-3x-14}{x^2-x-2} }$$ $$\displaystyle{ = 2 + \frac{3}{x+1} - \frac{4}{x-2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{2x^2-3x-14}{x^2-x-2} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3106 video solution
$$\displaystyle{ \frac{2x^2-3x-14}{x^2-x-2} }$$ $$\displaystyle{ = 2 + \frac{3}{x+1} - \frac{4}{x-2} }$$
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$$\displaystyle{ \frac{(x+2)(x-2)}{(x+1)(x-1)} }$$
Problem Statement
Expand $$\displaystyle{ \frac{(x+2)(x-2)}{(x+1)(x-1)} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{(x+2)(x-2)}{(x+1)(x-1)} }$$ $$\displaystyle{ = 1 + \frac{3}{2(x+1)} - \frac{3}{2(x-1)} }$$
Problem Statement
Expand $$\displaystyle{ \frac{(x+2)(x-2)}{(x+1)(x-1)} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3107 video solution
$$\displaystyle{ \frac{(x+2)(x-2)}{(x+1)(x-1)} }$$ $$\displaystyle{ = 1 + \frac{3}{2(x+1)} - \frac{3}{2(x-1)} }$$
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$$\displaystyle{ \frac{x^3-2x^2-25x+53}{25-x^2} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3-2x^2-25x+53}{25-x^2} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^3-2x^2-25x+53}{25-x^2} }$$ $$\displaystyle{ = -x + 2 + \frac{3}{10(5+x)} + \frac{3}{10(5-x)} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3-2x^2-25x+53}{25-x^2} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3108 video solution
$$\displaystyle{ \frac{x^3-2x^2-25x+53}{25-x^2} }$$ $$\displaystyle{ = -x + 2 + \frac{3}{10(5+x)} + \frac{3}{10(5-x)} }$$
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$$\displaystyle{ \frac{x^3-x^2-5x-7}{x^2-2x-3} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3-x^2-5x-7}{x^2-2x-3} }$$ using partial fraction expansion. Give your answer in exact terms.
$$\displaystyle{ \frac{x^3-x^2-5x-7}{x^2-2x-3} }$$ $$\displaystyle{ x + 1 - \frac{1}{x+1} + \frac{1}{x-3} }$$
Problem Statement
Expand $$\displaystyle{ \frac{x^3-x^2-5x-7}{x^2-2x-3} }$$ using partial fraction expansion. Give your answer in exact terms.
Solution
3109 video solution
$$\displaystyle{ \frac{x^3-x^2-5x-7}{x^2-2x-3} }$$ $$\displaystyle{ x + 1 - \frac{1}{x+1} + \frac{1}{x-3} }$$
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# Shock Waves and Sonic Boom!
Updated: Jul 21, 2021
What is the Mach Number?
### Table of Contents:
It's a lazy Sunday. You are lying on your bed and chatting with your best friend.
But suddenly, out of nowhere, your mom appears in front of you! When did she come in? What a shock!!! This is something that happens with the sound waves too! Let's understand what and how shock waves are created. Before actually getting into Shock waves we need to understand a few basic terminologies, let's get into it first!
## Shock Waves:
### Velocity of Sound:
When we speak, we produce longitudinal waves, which create regions of compression and rarefaction alternatively. This is how sound travels, in compression and rarefaction alternate waves. The speed of sound is 340.29 m/s.
### Mach Number:
Mach number is the ratio of inertial force to elastic force, when it is further deduced, it can be written as the ratio of object speed to speed of sound. The Mach Number is important because it tells us how the object will be affected by the presence of fluid in the surroundings.
### Case 1: Subsonic (Mach No. < 1.0)
When Mach Number is less than 1, or when the speed of the object moving is less than the speed of sound, it is called a Subsonic condition.
Let's consider the same example that we were discussing before. Imagine you are sitting in your room and chilling out. You hear your mom's footsteps approaching your room. Mom opens the door, you wave her hi!
In this case, you were prepared that someone is approaching you as you heard the footsteps. On similar lines, you can imagine, when the sound waves travel slower than the moving object, the sound waves are prepared for the approaching object. The sound waves are expecting some object to pass through and the object arrives.
### Case 2: Sonic (Mach No. = 1.0)
When Mach Number is equal to 1, or when the speed of the object moving is equal to the speed of sound, it is called as Sonic condition.
Again imagine the same example. You are sitting in your room and chilling out. But this time you could not hear your mom approaching you. Suddenly out of nowhere, you heard someone opening your room's door. Obviously, you were in a shock for few seconds, because you were not prepared for it.
Similarly, you can imagine, when the speed of the object is equal to the speed of sound, the sound waves are not ready for the object approaching. The sound waves had no idea that the object was entering its zone. Hence, the shock waves are created and these shock waves impact the object.
### Case 3: Supersonic (Mach No. > 1.0)
When Mach Number is greater than 1, or when the speed of the object moving is greater than the speed of sound, it is called a Supersonic condition.
This time, you are sitting in your room and suddenly, someone is standing in from of you. That is a real terror, isn't it!
Similarly, if the object speed is greater than the speed of sound, greater shock waves and impact is created.
### Case 4: Hypersonic (Mach No. > 5.0)
When Mach Number is greater than 5, or when the speed of the object moving is 5 times greater than the speed of sound, it is called a Hypersonic condition.
When the speed of object is greater than the speed of sound, impact force acts on the object, this is called as shock waves
The object can get damaged due to the high impact force. The greater the speed of the object, the greater is the impact on the object. The more the surface area of the object that enters the sound waves, the greater impact force acts on the object. Hence you might have seen the fighter planes have a very little surface (needle-like structure or very pointed nose) at the nose.
## Sonic Boom:
Sonic boom is the sound that is produced during shock waves.
Sometimes, the sonic boom can be loud enough to cause minor damage to some structures. This sound travels as the object travels.
## Conclusion:
1. Mach number is the ratio of an object's speed to the speed of sound.
2. There exist 3 cases: Subsonic, Sonic, and Hypersonic conditions, which depend on Mach Number.
3. When the speed of an object is greater than the speed of sound, impact force acts on the object. These impact forces damage the objects.
4. To reduce the impact force, the surface area should be reduced.
5. A sonic boom is a sound that is produced during shock waves.
• Mach Number < 1 âž¡ Subsonic
• Mach Number = 1 âž¡ Sonic
• Mach Number > 1 âž¡ Supersonic / Hypersonic
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# An aqueous solution contains 0-34 M ammonia- One liter of this soluti.docx
An aqueous solution contains 0.34 M ammonia .
One liter of this solution could be converted into a buffer by the addition of:
(Assume that the volume remains constant as each substance is added.)
0.34 mol Ba(NO 3 ) 2
0.35 mol NH 4 NO 3
0.35 mol HNO 3
0.17 mol KOH
0.17 mol HNO 3
Solution
Option B 0.35 mol NH4NO3. Since the mixture is an example for basic buffer and it has the following combination of weak base (NH3) & Salt of weak base with strong acid (HNO3)
Basic buffer : NH3/NH4NO3
.
An aqueous solution contains 0.34 M ammonia .
One liter of this solution could be converted into a buffer by the addition of:
(Assume that the volume remains constant as each substance is added.)
0.34 mol Ba(NO 3 ) 2
0.35 mol NH 4 NO 3
0.35 mol HNO 3
0.17 mol KOH
0.17 mol HNO 3
Solution
Option B 0.35 mol NH4NO3. Since the mixture is an example for basic buffer and it has the following combination of weak base (NH3) & Salt of weak base with strong acid (HNO3)
Basic buffer : NH3/NH4NO3
. | 330 | 1,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | latest | en | 0.855877 |
https://web2.0calc.com/questions/what-is-the-solution-set-for-this-equation | 1,521,421,503,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646189.21/warc/CC-MAIN-20180319003616-20180319023616-00263.warc.gz | 711,955,941 | 5,737 | +0
# What is the solution set for this equation
0
158
1
What is the solution set for this equation
x^2-6x+9=0
Guest Jun 21, 2017
#1
+1
Solve for x:
x^2 - 6 x + 9 = 0
Write the left hand side as a square:
(x - 3)^2 = 0
Take the square root of both sides:
x - 3 = 0
Guest Jun 21, 2017
Sort:
#1
+1
Solve for x:
x^2 - 6 x + 9 = 0
Write the left hand side as a square:
(x - 3)^2 = 0
Take the square root of both sides:
x - 3 = 0 | 181 | 437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-13 | longest | en | 0.881685 |
https://able.bio/patrickcording/nearest-neighbour-search-with-redis--56cju8s | 1,722,991,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00024.warc.gz | 65,015,856 | 6,351 | # Nearest neighbour search with Redis
Redis has a very nice geo API but it doesn’t support nearest neighbour queries. Given a point x, the answer to a nearest neighbour query is the closest point to x.
However, you’ll be able to answer nearest neighbour queries with the following Lua script. It does an exponential search over the distance parameter for GEORADIUS to find the nearest neighbour of a given point.
local dist = 100 -- Initial distance in m
local latitude = ARGV[1]
local longitude = ARGV[2]
local maxDist = tonumber(ARGV[3]) -- Max search distance in m, to avoid infinite loop on keys with no data
if not redis.call('EXISTS', KEYS[1]) then
return nil
end
local res = redis.call('GEORADIUS', KEYS[1], longitude, latitude, dist, 'm', 'count', '1', 'asc')
while #res == 0 and dist <= maxDist do
dist = 2 * dist
res = redis.call('GEORADIUS', KEYS[1], longitude, latitude, dist, 'm', 'count', '1', 'asc')
end
if #res > 0 then
return res[1]
else
return nil
end
Note that the time complexity is $O(\log M \cdot \log D)$, where $M$ is the number of elements in the geo index and $D$ is the distance to the nearest neighbour (assuming that the number of elements within distance $D$ is constant).
To invoke the script, you need to pass the key to your geo index and three arguments: latitude, longitude, and a maximum search distance. The latter is necessary to avoid infinite loops if a key doesn’t contain any data.
If you want to use it from the shell then save the script as nn.lua and load the script:
$redis-cli SCRIPT LOAD "$(cat nn.lua)"
"90702b22ecf3e3118ac4c3ffbd40f94c2d8a92bc"
The output is the SHA1 of your script. The script is now stored on your Redis instance. You only have to load it once. Now you can invoke it by issuing:
\$ redis-cli EVALSHA 90702b22ecf3e3118ac4c3ffbd40f94c2d8a92bc 1 <key> <latitude> <longitude> <max_dist>
I needed to make nearest neighbour queries from an application so I wrapped the Lua script in Scala code using Jedis for connecting to Redis. Here’s a slightly simplified version of my code.
class NearestNeighbourSearcher(redisClient: Jedis) {
// Get the SHA1 of the script. Assumes that it is stored in /src/main/resources/nn.lua of your project
private val scriptSha: String = stringToSHA1(script)
// Load the script if it doesn't exist on the host
def getNearestNeighbour(key: String, lat: Double, long: Double, maxSearchDistance: Int): Option[String] = {
val res = redisClient.evalsha(scriptSha, 1, key, lat.toString, long.toString, maxSearchDistance.toString)
if (res != null)
Some(res.asInstanceOf[String])
else
None
}
def stringToSHA1(s: String): String = {
val md = MessageDigest.getInstance("SHA-1")
md.digest(s.getBytes("UTF-8")).map("%02x".format(_)).mkString
}
}
This post first appeared May 19, 2019 on Medium.
I have been using redis for quite some time. Never did I play around with the geo API.
This is a really nice post. Useful to anyone using redis regardless of the stack they use with it. | 798 | 2,981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.664496 |
http://emaths.net/maths-simplify/adding-functions/pie-chart-aptitude-question.html | 1,519,223,880,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813626.7/warc/CC-MAIN-20180221143216-20180221163216-00673.warc.gz | 115,813,560 | 12,314 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
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• youdao | 878 | 3,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-09 | latest | en | 0.896655 |
https://www.mathworks.com/matlabcentral/cody/players/4147901-clare/solved | 1,508,538,535,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00391.warc.gz | 936,823,965 | 17,359 | Cody
# Clare
Rank
Score
## Solved
1 – 50 of 55
#### Test for balanced parentheses
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Tags parens, regexp
#### Summing digits
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Tags strings
#### First N Perfect Squares
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#### Back to basics 8 - Matrix Diagonals
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#### given 3 sides, find area of this triangle
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#### De-dupe
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Tags matlab
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#### Back to basics 6 - Column Vector
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#### Factorize THIS, buddy
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#### Program an exclusive OR operation with logical operators
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#### Tell me the slope
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#### Project Euler: Problem 1, Multiples of 3 and 5
Created by: Doug Hull
#### Find the sum of the elements in the "second" diagonal
Created by: Roy Fahn
Tags matrix, sum
#### surface of a spherical planet
Created by: AMITAVA BISWAS
#### Make one big string out of two smaller strings
Created by: Michelle
#### Bottles of beer
Created by: Alan Chalker
#### Arrange Vector in descending order
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#### radius of a spherical planet
Created by: AMITAVA BISWAS
#### Is my wife right? Now with even more wrong husband
Created by: the cyclist
#### Swap the input arguments
Created by: Steve Eddins
#### Sorted highest to lowest?
Created by: AMITAVA BISWAS
Tags sort
#### Too mean-spirited
Created by: the cyclist
Tags mean
#### Distance walked 1D
Created by: AMITAVA BISWAS
#### Reverse the vector
Created by: Vishwanathan Iyer
#### Sum all integers from 1 to 2^n
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#### Check if sorted
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#### The Hitchhiker's Guide to MATLAB
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#### Pizza!
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Tags pizza, fun
#### Is my wife right?
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Tags fun, stupid, silly
#### The Goldbach Conjecture
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Tags primes
#### Who Has the Most Change?
Created by: Cody Team
#### Return the 3n+1 sequence for n
Created by: Cody Team
Tags 3n+1, sample
#### Most nonzero elements in row
Created by: Cody Team
Tags matrices
#### Find the numeric mean of the prime numbers in a matrix.
Created by: Cody Team
#### Create times-tables
Created by: Cody Team
Tags matrices
#### Remove any row in which a NaN appears
Created by: Cody Team
#### Fibonacci sequence
Created by: Cody Team
#### Bullseye Matrix
Created by: Cody Team
Tags matrices
#### Finding Perfect Squares
Created by: Cody Team
Tags sets
#### Sort a list of complex numbers based on far they are from the origin.
Created by: Cody Team
#### Determine whether a vector is monotonically increasing
Created by: Cody Team
#### Swap the first and last columns
Created by: Cody Team
#### Triangle Numbers
Created by: Cody Team
Tags math
#### Make a checkerboard matrix
Created by: Cody Team
#### Find all elements less than 0 or greater than 10 and replace them with NaN
Created by: Cody Team
#### Column Removal
Created by: Cody Team
#### Select every other element of a vector
Created by: Cody Team
1 – 50 of 55 | 852 | 3,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-43 | latest | en | 0.828973 |
https://www.educative.io/courses/grokking-dynamic-programming-a-deep-dive-using-python/introduction-to-unbounded-knapsack | 1,716,779,673,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00588.warc.gz | 634,437,360 | 97,991 | # Introduction to Unbounded Knapsack
Let's get introduced to the Unbounded Knapsack pattern.
We'll cover the following
## Overview
A knapsack is defined as a bag carried by hikers or soldiers for carrying food, clothes, and other belongings. The Knapsack problem, as the name suggests, is the problem faced by a person who has a knapsack with a limited capacity and wants to carry the valuable important items. In other words, we are given $N$ items, each having a specific weight and a value, and a knapsack with a maximum capacity. Our job is to put as many items as possible in the knapsack such that the cumulative weight of the items doesn't exceed the knapsack's capacity, and the cumulative value of the items in the knapsack is maximized.
In this course, we will be discussing 5 problems related to this pattern. The following diagram shows an overview of this pattern.
Level up your interview prep. Join Educative to access 70+ hands-on prep courses. | 219 | 965 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.948492 |
http://bootmath.com/why-arent-all-dense-subsets-of-mathbbr-uncountable.html | 1,529,639,394,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00579.warc.gz | 43,690,173 | 8,793 | # Why aren't all dense subsets of $\mathbb{R}$ uncountable?
1) we say that $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable.
That implies $\mathbb{R}-\mathbb{Q}$, that is irrational numbers are uncountable.
2) Archimedian property of $\mathbb{R}$ suggests that there exists a rational between any two numbers. i.e. $\mathbb{Q}$ is dense in $\mathbb{R}$. Then how come $\mathbb{Q}$ is countable while irrational is uncountable?
#### Solutions Collecting From Web of "Why aren't all dense subsets of $\mathbb{R}$ uncountable?"
I think the confusion is that your brain tries to use finite reasoning in a place where it doesn’t apply. It’s true that if you have a finite sequence of balls, say, colored red and blue, such that between every two red balls is exactly one blue ball and between every two balls is exactly one red ball, then by symmetry there must be (essentially) the same number of red and blue balls.
Your brain wants to make this work with infinite collections also: If between every two rationals lies an irrational, and between every two irrationals lies a rational, then surely there are the same number, right? But try replacing the observation with the fact that between every two rationals lies infinitely many irrationals, and between every two irrationals lies infinitely many rationals. Now must there be the same number of each? Certainly the water is more murky, and once you know of the existence of different sizes of infinity you realize that the answer is of course not! There may be a different size of infinity’s worth of rationals between every two irrationals than the size of infinity’s worth of irrationals between every two rationals, breaking the symmetry.
Of course, this answer is filled with anachronisms (if you already know about sizes of infinity, you’ve probably had to have already come across the fact that there are more reals than rationals), but I think it still helps illustrate why one’s brain gets confused. (And in the brain’s defense, it’s pretty cool and unintuitive stuff.)
One reason — at least for the plausibility of $\mathbb{R}$ having countable dense subsets — is that density has a fairly weak connection to the cardinality of a topological space.
If a (Hausdorff) topological space $X$ has a dense subset $D$ of infinite cardinality $\kappa$, consider the mapping $X \to \mathcal{P} ( \mathcal{P} ( D ) )$ defined by $$x \mapsto \mathcal{A}_x = \{ A \subseteq D : x \in \overline{A} \}.$$
Since $X$ is Hausdorff, then given distinct $x, y \in X$ there are disjoint open sets $U , V$ such that $x \in U$ and $y \in V$. Then $x \in \overline{U} = \overline{D \cap U}$, but $y \notin \overline{U}$. It thus follows that $D \cap U \in \mathcal{A}_x \setminus \mathcal{B}_x$; i.e., the mapping above is one-to-one. From this we have that $|X| \leq 2^{2^\kappa}$.
Even more, this inequality is the best possible. Given any infinite cardinal $\kappa$, let $X$ be any set of cardinality $\kappa$ with the discrete topology. Then there is a space, called the Stone-Čech compactification of $X$ and denoted $\beta X$, which has the following properties:
• $\beta X$ is Hausdorff (and compact);
• $| \beta X | = 2^{2^\kappa}$; and
• there is a homeomorphic copy of $X$ embedded in $\beta X$ which is dense in $\beta X$.
While this is clearly not an explanation of why $\mathbb{R}$ has countable dense subsets, since $\aleph_0 < 2^{\aleph_0} = | \mathbb{R} | < 2^{2^{\aleph_0}}$, appeals to cardinality and density alone cannot discount their (possible) existence.
To answer the title question, it’s because you can approximate any real number to within arbitrarily small error using a rational number (just truncate the decimal expansion far enough out)
If you want an intuitive reason why $\mathbb{Q}$ is countable while the set of irrationals is uncountable, it’s because irrational numbers necessarily have infinite, non-repeating decimal expansions. So the intuitive difference is the same as the difference between the cardinality of the set of finite strings of natural numbers (a countable union of finite sets, one for each length), and the cardinality of the set of infinite strings ($10^{\mathbb{N}}$).
In general, if a topological space has a countable dense subset we call it
“separable.” Separable metric spaces have a countable basis: one such basis for the metric topology on $\mathbb{R}^n$ is given by the balls of rational radius, centered at points with rational coordinates. More explicitly, every open subset of $\mathbb{R}^n$ is actually a COUNTABLE union of open balls. This fact (or just separability itself) can be used to show that any uncountable subset of $\mathbb{R}^n$ necessarily has a cluster point.
Between any two rationals, there is an irrational. Between any two irrationals, there is a rational. It would be nice to use this to compare, but it’s tricky, because between any two rationals, there are more rationals as well.
So, to really compare things fairly, we need to come up with some new device to talk about the numbers “between” rationals in a way that there are no extra rationals getting in the way.
To start, suppose we have two rationals $a < b$. Well, there is a rational $a_1$ between them in the way, so we can refine our view from $(a,b)$ to $(a_1,b)$. Or, we could have refined to $(a_0,a_1)$. Let’s pick the first. Then there’s a rational $a_2$ in $(a_1, b)$, so let’s restrict our attention again, to $(a_1, a_2)$. Then….
Well, this is a bit of a mess, how can we organize it? Well, how about the following? Let’s split all of the rational numbers into two (non-empty) sets: the “left” set and the “right” set, so that every number in the left set is smaller than every number in the right set.
Now, we can ask what’s between the left set and the right set, and there are no rational numbers left to get in the way! This is called a “Dedekind cut”.
It turns out there are exactly three sorts of Dedekind cuts:
• There is a rational number $q$ so that $L = (-\infty, q] \cap \mathbb{Q}$ and $R = (q, \infty) \cap \mathbb{Q}$
• There is a rational number $q$ so that $L = (-\infty, q) \cap \mathbb{Q}$ and $R = [q, \infty) \cap \mathbb{Q}$
• There is an irrational number $r$ so that $L = (-\infty, r) \cap \mathbb{Q}$ and $R = (r, \infty) \cap \mathbb{Q}$
So we’ve finally gotten to our goal of finding a device to talk about numbers that are between rationals, without other rational numbers getting in the middle. And, as we had hoped, there cannot be more than one irrational in-between.
But, surprise! There are uncountably many Dedekind cuts, so there really are more ways to talk about things “between” rationals than there are rationals themselves.
Fortunately, if we repeat the above for irrational numbers, we find that the number of Dedekind cuts of irrationals is the same as the number of Dedekind cuts for rationals, so at least we get those numbers the same.
Countable does’t mean that set is small(that is it is not dense in uncountable)…
Actually countable means we have method of enumerating… That is we have bijective map $A(\subset \mathbb N) \to \mathbb Q$.
Dense is topological property, It doesn’t related with counting( Don’t confuse with Baire Category theorem)….
From your observation, you can feel that countable set may be “so much big”(in dense sense)
Whatelse I can write…. Just I will repeat the sentence of “Norbert” that That is the way it is..
If I understand your question correctly, I think you are wondering that since whenever I take two irrationals I can find a rational “in between”, and since there are uncountably many pairs or reals for which I can ask for such a rational “in between” whether this would not mean there should be uncountably many such rationals to make sure we can always find one right ??
The reason this argument fails is that you do not need to use distinct rationals in each case when you are looking for one that lies between two given real numbers; i.e. you can use a given rational multiple times, in fact uncountably often!
Here’s another example that may shed some light on your reasoning. Let $A\subseteq \mathbb{N}$ be the set $\{2, … , 100, 102, … , 1000, … , 10^n + 2, … , 10^{n+1}, … \}$. Both $A$ and $A^c$ are infinite (large). However
$$\lim_{n\to\infty}\left|\frac{A\cap \{1, … ,n\}}{n}\right| = 1.$$
but
$$\lim_{n\to\infty}\left|\frac{A^c\cap \{1, … ,n\}}{n}\right| = 0.$$
In other words, $A$ is much more prevalent than $A^c$ even though both are infinte. This is supposed to illustrate the way different notions of largeness and smallness can be completely unrelated. The same relationship holds holds for (topological) density and cardinality.. | 2,300 | 8,671 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-26 | latest | en | 0.953346 |
https://download.atlantis-press.com/journals/jnmp/125950468/view | 1,718,923,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00255.warc.gz | 196,070,367 | 34,527 | # Journal of Nonlinear Mathematical Physics
Volume 26, Issue 1, December 2018, Pages 147 - 154
# Time-independent Hamiltonians describing systems with friction: the “cyclotron with friction”
Authors
Francesco Calogero*
Dipartimento di Fisica, Università di Roma “La Sapienza” Rome, 00185, Italy,francesco.calogero@roma1.infn.it;francesco.calogero@uniroma1.it
François Leyvraz
Instituto de Ciencias Físicas, University of Mexico, Av. Universidad s/n, colonia Chamilpa Cuernavaca, 62210, Mexico,leyvraz@fis.unam.mx;f_leyvraz2001@hotmail.com
*also at Istituto Nazionale di Fisica Nucleare, Sezione di Roma, Italy
also at Centro Internacional de Ciencias, Cuernavaca, México
Corresponding Authors
Francesco Calogero, François Leyvraz
Received 27 July 2018, Accepted 15 September 2018, Available Online 6 January 2021.
DOI
10.1080/14029251.2019.1544795How to use a DOI?
Abstract
As is well-known, any ordinary differential equation in one dimension can be cast as the Euler–Lagrange equation of an appropriate Lagrangian. Additionally, if the initial equation is autonomous, the Lagrangian can always be chosen to be time-independent. In two dimensions, however, the situation is more complex, and there exist systems of ODEs which cannot be described by any Lagrangian. In this paper we display Hamiltonians which describe the behaviour of a charged particle moving in a plane under the combined influence of a constant electric field (in the plane) and a constant magnetic field (orthogonal to the plane) as well as a friction force proportional to the velocity (“cyclotron with friction”).
Open Access
## 1. Introduction
In the context of classical and quantum mechanics, systems with friction have been the focus of considerable interest. These systems are typically characterized by motions in which the moving particle stops asymptotically (as the time t → +∞). This stopping arises because of the coupling of the particle of interest, often referred to as the central system—but note that in this paper we only consider the motion of a single particle—to an external system, often comprising a large number of degrees of freedom, usually referred to as the bath.
It is a remarkable fact, however, that in some specific cases it is possible to rewrite the equations describing the motion with friction in terms of a—possibly time-dependent—Hamiltonian involving only the moving particle. In particular, this is always the case for motions in an arbitrary one-dimensional potential with a friction linear in the velocity. This can be seen as follows: the Hamiltonian
h(p,z,t)=exp(ct)p22+exp(ct)V(z),(1.1)
with pp(t) and zz(t) the standard canonical variables and c a positive constant, yields via the standard Hamiltonian equations
z˙=h(p,z)/p=exp(ct)p,(1.2b)(1.2a)
p˙=h(p,z)/z=exp(ct)dV(z)/dz,(1.2b)
the Newtonian equation of motion
z¨=cz˙dV(z)/dz,(1.3)
which corresponds to a (without loss of generality, unit mass) particle of coordinate zz(t) moving in a potential V (z) under the additional influence of a friction with coefficient c. [Note that, here and hereafter, a superimposed dot indicates differentiation with respect to the time t, and that we omit to indicate the explicit time-dependence of various quantities whenever this is unlikely to cause misunderstandings].
There is a significant amount of literature on this subject: for a review of relevant results see [1], and for some typical results see [210]. Note that a considerable part of this literature is devoted to the question of quantizing models with friction. To this end they describe the system with friction in Hamiltonian terms and then proceed to quantize this system in the usual manner. In this paper we limit ourselves to classical considerations, postponing the treatment of the quantal case to a separate paper.
The general question of the circumstances under which a given system of ordinary differential equations (ODEs) of second order can be obtained as the Euler–Lagrange equations for a given Lagrangian has received considerable attention in the mathematical literature. In particular, it is straightforward to show that any one-dimensional system can indeed be obtained via a Lagrangian. The two-dimensional case has also received a complete treatment in [11], the main result being that a system of 2 second order ODEs cannot generally be obtained from a Lagrangian.
A further issue of interest is, of course, that of determining when this Lagrangian can be chosen to be independent of time. Sarlet [12] showed that quite generally, if a set of equations derivable from a Lagrangian is autonomous, then this Lagrangian can be chosen to be time-independent. But it is generally not susceptible of explicit expression. We are in particular not aware of an explicit time-independent Lagrangian yielding the equation of motion (1.3) for an arbitrary potential V (z). However, for V (z) = λ z2/2, a Hamiltonian of the form
H(p,z)=12ln[z2sec2(ωzp)]czp2,(1.4)
where ω=4λc2/2, was given in [13] for the underdamped case. Similar expressions are also given for the other cases. While a systematic method to obtain such results is indicated in that paper, it does not extend to the case of an arbitrary potential V (z).
Similarly, in the same paper [13], the Hamiltonian
H(p,z)=exp(p)+cz(1.5)
is given for the free particle moving against friction, namely according to the Newtonian equation of motion
z¨=cz˙.(1.6)
Another simple example is given by
H(p,z)=exp(p)+Epc+cz,(1.7)
which generates, as can easily be verified, the dynamics of a particle moving against friction in a constant force field E, that is
z¨=cz˙+E.(1.8)
Note moreover that these Hamiltonians can be generalized by a canonical transformation that does not modify the canonical coordinate z and replaces the canonical coordinate p with
p˜=p+α(z),(1.9)
where α(z) is an a priori arbitrary function. Since the transformation (1.9) is canonical and reduces to the identity on the position variables, it generally does not affect the Newtonian equations. There is thus an arbitrary function α(z) which can be introduced at will. As an example, this transforms the Hamiltonian (1.5) to the more general Hamiltonian
H(p;z)=f(z)exp(p)+cz.(1.10)
Let us emphasize that in this case the function f (z) can be arbitrarily assigned as long as it has no real zeros, while it does not appear at all in the Newtonian equation of motion (1.6).
### Remark 1.1.
If f (z0) = 0, then if one takes z0 as initial value for z, z(t) does not move at all, z(t) = z0, whatever the initial momentum. Hence, it will be impossible to assign a non-zero initial velocity, so that the system would not really be equivalent to free motion against friction, see (1.6).
To conclude these remarks, we note that Hamiltonians featuring a nonconventional kinetic energy term—indeed resembling the Hamiltonians (1.5) and (1.7)—have been previously investigated (see for example [1420, 22]), but to the best of our knowledge their suitability to treat also the case with friction had not been previously noted.
In Section 2 the extensions by complexification of some of these findings to a two-dimensional context is discussed. It is indeed thereby possible to obtain a system of ODEs describing a charged particle moving in a plane under the influence of a homogeneous magnetic field perpendicular to that plane and of a friction force proportional to the velocity, which we shall call “cyclotron with friction” (and the model also allows for the additional presence of a constant electric field lying in the plane: see below). In Section 3 some considerations relevant to the symmetries and conservation laws of this model are tersely presented. The last Section 4 (“Outlook”) outlines further developments, to be pursued by ourselves and/or by others in future publications.
## 2. Extension by complexification to motions in the plane
Consider the two Hamiltonians (1.5) and (1.7). Their analytic nature allows to extend them straightforwardly by complexification to describe motions taking place in a plane, which will be seen to correspond to physically interesting systems: specifically, the motion against friction of a charged particle in the presence of a perpendicular constant magnetic field, or a constant electric field lying in that plane, or of both these forces.
Consider the Hamiltonian (1.5) which yields the Newtonian equations of motion (1.6) of a free particle moving against friction. If we set c = a + ib and go to the complex plane, we obtain the following pair of Poisson commuting Hamiltonians
HR(px,py;x,y)=[H(pxipy,x+iy)]=exp(px)cos(py)+axby,(2.1a)
HI(px,py;x,y)=𝔍[H(pxipy,x+iy)]=exp(px)sin(py)+bx+ay.(2.1b)
[Here and hereafter, i is the imaginary unit, i2 = −1; a, b are two arbitrary real constants; q(t) ≡ x(t) + iy(t), p(t) ≡ px(t) − ipy(t) (note the minus sign), so that xx(t), yy(t) are the real canonical coordinates of the particle moving in the Cartesian xy-plane; and pxpx(t), pypy(t) are the corresponding real canonical momenta]. If we similarly complexify the Newtonian equations of motion (1.6) characterizing free motion against friction, we obtain
x¨=ax˙+by˙,y¨=bx˙+ay˙.(2.2)
It is readily verified that (2.2) are obtained as the Hamilton equations corresponding to the Hamiltonian HR(px, py;x,y) defined in (2.1a). It is thus seen that this Hamiltonian provides a description of the motion—against a friction characterized by the parameter a—of a particle moving in the Cartesian xy-plane in the presence (b ≠ 0) or absence (b = 0) of a constant magnetic field orthogonal to that plane.
### Remark 2.1.
Note that the equations of motion (12a) can be reformulated in a 3-dimensional context as follow, by introducing the 3-vector r(x,y,0) in the xy-Cartesian plane and the unit vector z^(0,0,1) orthogonal to that plane:
r¨=ar˙+br˙z^,(2.3)
where the symbol ^ denotes the 3-dimensional vector product. The last term in the right-hand side is the Lorentz force.
The explicit solution of (2.2) is readily found to be
x(t)±iy(t)=x(0)±iy(0)+[x˙(0)±iy˙(0)]1eatexp(ibt)aib.(2.4)
This formula, besides displaying quite explicitly the time evolution in the Cartesian xy-plane, shall be of importance in the next Section 3 where we analyze the significance of the symmetries and conservation laws associated to this motion.
We may proceed similarly with the Hamiltonian (1.7). In this case, the real part HR of the complexified Hamiltonian reads
HR(px,py;x,y)=epxcospy+(aEx+bEy)px+(aEybEx)pya2+b2+axby.(2.5)
Here we set again c = a + ib and in addition E = Ex + iEy. The corresponding Newtonian equations of motion obtained by complexification then read
x¨=ax˙+by˙+Ex,y¨=bx˙+ay˙+Ey.(2.6)
These equations represent the motion in the Cartesian xy-plane (against friction: a > 0) of a particle moving under the influence of crossed electric and magnetic fields. The general solution of these equations is
x(t)=x(0)+1(a2+b2)2{(a2+b2)(aEx+bEy)t+[eatcos(bt)1][(a2b2)Ex+2abEy(a2+b2)(ax˙(0)+by˙(0))]+eatsin(bt)[(2abEx+(a2b2)Ey+(a2+b2)(bx˙(0)ay˙(0))]}(2.7a)
y(t)=y(0)+1(a2+b2)2{(a2+b2)(bEx+aEy)t+[eatcos(bt)1][2abEx+(a2b2)Ey(a2+b2)(bx˙(0)ay˙(0))]+eatsin(bt)[((a2b2)Ex2abEy+(a2+b2)(ax˙(0)+by˙(0))]}(2.7b)
These solutions can be described in qualitative terms as follows: they correspond to the general solution of the equations (2.2), see (2.4), which correspond to the homogeneous part of (2.6), added to the special solutions of (2.6) given by (vxt,vyt), where the two parameters vx and vy can be explicitly obtained by solving the following system of two linear equations:
avxbvy=Ex,bvx+avy=Ey.(2.8)
The motion thus eventually becomes asymptotically rectilinear in the remote future, whereas it becomes a spiralling motion in the remote past, with a transition between these behaviors at intermediate times given by (2.7). We may additionally point out that, in 3 dimensions, the exact solution for electric and magnetic fields in general position, that is, not necessarily orthogonal, can also be written down explicitly, though we do not know whether the corresponding dynamics can be expressed in terms of a Hamiltonian.
## 3. Symmetries and conservation laws
In the following, we discuss the connection between symmetries and conservation laws for the systems we have considered. Specifically, we focus on the Hamiltonian (2.1a), since it has a large group of symmetries yet represents a system with friction, indeed it corresponds to the physically relevant case of a “cyclotron with friction”. It is thus of interest to study these symmetries and to see how Noether’s theorem, for example, applies in this setting. Indeed, we do not ordinarily think of a system moving against friction as having, say, a conserved angular momentum. Here two remarks are important: first, the symmetries must correspond to symmetries of the dynamics on all of phase space; second, we must make sure that the symmetries we use can really be implemented as canonical transformations.
There are two obvious geometrical symmetries in the dynamics of the particle moving against friction in a plane in the presence of a perpendicular homogeneous magnetic field: translations (in both directions) and rotations around an arbitrary origin. The latter, however, are not symmetries of the full phase space orbit: indeed a rotation, if it is to be a canonical transformation, must operate in the same way on both momenta and positions. But the trajectory of the momentum is a straight line in a fixed direction: if we wish to rotate this trajectory, we need to change both a and b (see (2.2)). In spite of the existence of a Hamiltonian structure as well as of an (apparent) rotational invariance, there is thus no equivalent to angular momentum conservation for this model (but see below for an alternative symmetry property).
Translations, on the other hand, do lead to interesting symmetries. The two-dimensional group of translations is generated by px and py. Generally, these do not commute with HR(px, py;x,y) (see (2.1a)), but the linear combination
P+=pxa+pyb(3.1)
does. This is due to the fact that the translation generated by P+ leave the Hamiltonian invariant, since P+ Poisson commutes with HR(px, py;x,y) (see (2.1a)),
{P+,HR(px,py;x,y)}=0.(3.2)
On the other hand px, for example, does not commute with HR. This corresponds to the fact that the Hamiltonian grows linearly in x, so that the translation in x, while it does leave the orbits invariant, transports an orbit with one energy to an orbit with another. This leads, by standard considerations, to a time-dependent conservation law, of the form
P(t)=px(t)apy(t)b2t,P˙(t)=0.(3.3)
The system has therefore two degrees of freedom, and three conserved quantities, namely: P+, HR and HI (see (3.1), (2.1a) and (2.1b)). With such a number of conservation laws, the system is maximally superintegrable. This is, of course, unsurprising; indeed—as shown above, see (2.4)—it is even possible, for this model, to write explicitly the solution of its initial-value problem. It is nevertheless remarkable that the conservation laws can be expressed in such a simple manner.
Finally, we may discuss the meaning of these various conservation laws. Since the momenta do not have an obvious physical significance, we first express P+ in terms of and . Using the equations of motion
x˙=exp(px)cos(py),y˙=exp(px)cos(py),(3.4a)
p˙x=a,p˙y=b,(3.4b)
we immediately find
P+=12aln(x˙2+y˙2)+pyb=12aln(x˙2+y˙2)+t+py(0)b.(3.5)
The conservation of P+ thus entails that the standard kinetic energy (x˙2+y˙2)/2 of the particle decays exponentially in t at a rate 2a to compensate for the increase of the term linear in t.
The other two conservation laws, namely HR and HI, correspond to the coordinates x(∞) and y(∞) of the final position of the system as t → +∞. This follows from the fact that px → −∞ as t → +∞, so that
R=ax(+)+by(+),I=bx(+)ay(+),(3.6)
where R and I are, say, the initial values of HR and HI respectively. Note that these formulas suggest which symmetry corresponds to the quantity HI (see (2.1b)). Since HI has the same form as HR (see (2.1a)) up to an appropriate interchange of the parameters a and b, we may say that the dynamics corresponding to a magnetic field characterized by the parameter −a and a friction characterized by the parameter b commutes with the dynamics defined by a magnetic field characterized by the parameter b and a friction characterized by the parameter a. So, it appears to be this remarkable symmetry which is the underlying feature allowing the treatment presented here of this model.
Let us complete this Section 3 by noting that the treatment presented here of a particle in a magnetic field with friction is quite different from the conventional treatment: if we consider the case without friction (a = 0), we see that the two coordinates of the circular orbit, obtained from R and I via (3.6), are quantities which Poisson commute among each other, whereas in the standard Hamiltonian treatment of the motion of a charged particle moving in the plane in the presence of an orthogonal uniform magnetic field, these two coordinates are canonically conjugate to each other.
## 4. Outlook
There are two natural developments suggested by the results reported above.
The first is the quantization of the Hamiltonian models introduced above. In the case of the one-dimensional damped harmonic oscillator, considerable work has been done to study its quantization using a Hamiltonian description for the equation of motion, see in particular [1] and references therein. While this approach is physically questionable, since it is important, in quantum mechanics, to take into account the interaction of the central system with the environment, it is nevertheless of interest, since it allows to study the possibility of describing an irreversible process by an unitary evolution. We are presently pursuing this line of research, the results of which will appear shortly [21].
The other is the treatment of some analogous “solvable” model, involving however the motion of several interacting particles rather than just a single one. An appealing candidate is the many-body “goldfish” model, see [2224 ].
One of us (FL) gratefully acknowledges funding by CONACyT grant 254515 as well as UNAM– DGAPA–PAPIIT IN103017. FC acknowledges the hospitality of the Centro Internacional de Ciencias in Cuernavaca, where some of this research began.
## References
[2]T. Levi-Civita, Sul moto di un sistema di punti materiali, soggetti a resistenze proporzionali alle rispettive velocità, Atti R. Istit. Veneto Scienze, Vol. 54, 1896, pp. 1004-1008.
[10]O.A. Constantinescu and E.H. Taha, 2017. Alternative Lagrangians obtained by scalar deformations, Preprint, 1712.01392v1 [math.DG]
[14]F. Calogero and J.-P. Françoise, Solution of certain integrable dynamical systems of Ruijsenaars–Schneider type with completely periodic trajectories, Ann. Henri Poincaré, Vol. 1, 2000, pp. 173-191.
[18]F. Calogero, On the quantization of two other nonlinear harmonic oscillators, Phys. Lett. A, Vol. 319, 2003, pp. 240-245.
[21]F. Calogero and F. Leyvraz. “A Hamiltonian yielding damped motion in an homogeneous magnetic field: quantum treatment”, to be published
[23]F. Calogero, 2008. Isochronous systems, Oxford University Press (264 pages), paperback edition, 2012
Journal
Journal of Nonlinear Mathematical Physics
Volume-Issue
26 - 1
Pages
147 - 154
Publication Date
2021/01/06
ISSN (Online)
1776-0852
ISSN (Print)
1402-9251
DOI
10.1080/14029251.2019.1544795How to use a DOI?
Open Access
TY - JOUR
AU - Francesco Calogero
AU - François Leyvraz
PY - 2021
DA - 2021/01/06
TI - Time-independent Hamiltonians describing systems with friction: the “cyclotron with friction”
JO - Journal of Nonlinear Mathematical Physics
SP - 147
EP - 154
VL - 26
IS - 1
SN - 1776-0852
UR - https://doi.org/10.1080/14029251.2019.1544795
DO - 10.1080/14029251.2019.1544795
ID - Calogero2021
ER - | 5,071 | 19,943 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.848326 |
https://elteoremadecuales.com/zarankiewicz-problem/ | 1,680,357,011,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00579.warc.gz | 259,830,772 | 16,406 | # Zarankiewicz problem
The Zarankiewicz problem, an unsolved problem in mathematics, asks for the largest possible number of edges in a bipartite graph that has a given number of vertices and has no complete bipartite subgraphs of a given size.[1] It belongs to the field of extremal graph theory, a branch of combinatorics, and is named after the Polish mathematician Kazimierz Zarankiewicz, who proposed several special cases of the problem in 1951.[2] Contents 1 Problem statement 2 Examples 3 Upper bounds 4 Lower bounds 4.1 Incidence graphs in finite geometry 4.2 Norm graphs and projective norm graphs 4.3 Clique partitions 4.4 Randomized algebraic constructions 5 Applications 6 See also 7 References Problem statement A bipartite graph {displaystyle G=(Ucup V,E)} consists of two disjoint sets of vertices {displaystyle U} and {displaystyle V} , and a set of edges each of which connects a vertex in {displaystyle U} to a vertex in {displaystyle V} . No two edges can both connect the same pair of vertices. A complete bipartite graph is a bipartite graph in which every pair of a vertex from {displaystyle U} and a vertex from {displaystyle V} is connected to each other. A complete bipartite graph in which {displaystyle U} has {displaystyle s} vertices and {displaystyle V} has {displaystyle t} vertices is denoted {displaystyle K_{s,t}} . If {displaystyle G=(Ucup V,E)} is a bipartite graph, and there exists a set of {displaystyle s} vertices of {displaystyle U} and {displaystyle t} vertices of {displaystyle V} that are all connected to each other, then these vertices induce a subgraph of the form {displaystyle K_{s,t}} . (In this formulation, the ordering of {displaystyle s} and {displaystyle t} is significant: the set of {displaystyle s} vertices must be from {displaystyle U} and the set of {displaystyle t} vertices must be from {displaystyle V} , not vice versa.) The Zarankiewicz function {displaystyle z(m,n;s,t)} denotes the maximum possible number of edges in a bipartite graph {displaystyle G=(Ucup V,E)} for which {displaystyle |U|=m} and {displaystyle |V|=n} , but which does not contain a subgraph of the form {displaystyle K_{s,t}} . As a shorthand for an important special case, {displaystyle z(n;t)} is the same as {displaystyle z(n,n;t,t)} . The Zarankiewicz problem asks for a formula for the Zarankiewicz function, or (failing that) for tight asymptotic bounds on the growth rate of {displaystyle z(n;t)} assuming that {displaystyle t} is a fixed constant, in the limit as {displaystyle n} goes to infinity.
For {displaystyle s=t=2} this problem is the same as determining cages with girth six. The Zarankiewicz problem, cages and finite geometry are strongly interrelated.[3] The same problem can also be formulated in terms of digital geometry. The possible edges of a bipartite graph {displaystyle G=(Ucup V,E)} can be visualized as the points of a {displaystyle |U|times |V|} rectangle in the integer lattice, and a complete subgraph is a set of rows and columns in this rectangle in which all points are present. Thus, {displaystyle z(m,n;s,t)} denotes the maximum number of points that can be placed within an {displaystyle mtimes n} grid in such a way that no subset of rows and columns forms a complete {displaystyle stimes t} grid.[4] An alternative and equivalent definition is that {displaystyle z(m,n;s,t)} is the smallest integer {displaystyle k} such that every (0,1)-matrix of size {displaystyle mtimes n} with {displaystyle k+1} ones must have a set of {displaystyle s} rows and {displaystyle t} columns such that the corresponding {displaystyle stimes t} submatrix is made up only of 1's.
Examples A bipartite graph with 4 vertices on each side, 13 edges, and no {displaystyle K_{3,3}} subgraph, and an equivalent set of 13 points in a 4 × 4 grid, showing that {displaystyle z(4;3)geq 13} .
The number {displaystyle z(n;2)} asks for the maximum number of edges in a bipartite graph with {displaystyle n} vertices on each side that has no 4-cycle (its girth is six or more). Thus, {displaystyle z(2;2)=3} (achieved by a three-edge path), and {displaystyle z(3;2)=6} (a hexagon).
In his original formulation of the problem, Zarankiewicz asked for the values of {displaystyle z(n;3)} for {displaystyle n=4,5,6} . The answers were supplied soon afterwards by Wacław Sierpiński: {displaystyle z(4;3)=13} , {displaystyle z(5;3)=20} , and {displaystyle z(6;3)=26} .[4] The case of {displaystyle z(4;3)} is relatively simple: a 13-edge bipartite graph with four vertices on each side of the bipartition, and no {displaystyle K_{3,3}} subgraph, may be obtained by adding one of the long diagonals to the graph of a cube. In the other direction, if a bipartite graph with 14 edges has four vertices on each side, then two vertices on each side must have degree four. Removing these four vertices and their 12 incident edges leaves a nonempty set of edges, any of which together with the four removed vertices forms a {displaystyle K_{3,3}} subgraph.
Upper bounds The Kővári–Sós–Turán theorem provides an upper bound on the solution to the Zarankiewicz problem. It was established by Tamás Kővári, Vera T. Sós and Pál Turán shortly after the problem had been posed: {displaystyle z(m,n;s,t)<(s-1)^{1/t}(n-t+1)m^{1-1/t}+(t-1)m.} Kővári, Sós, and Turán originally proved this inequality for {displaystyle z(n;t)} .[5] Shortly afterwards, Hyltén-Cavallius observed that essentially the same argument can be used to prove the above inequality.[6] An improvement on the second term of the upper bound on {displaystyle z(n;t)} was given by Štefan Znám:[7] {displaystyle z(n;t)<(t-1)^{1/t}n^{2-1/t}+{frac {1}{2}}(t-1)n.} If {displaystyle s} and {displaystyle t} are assumed to be constant, then asymptotically, using the big O notation, these formulae can be expressed as {displaystyle z(m,n;s,t)=O(mn^{1-1/s}+n)} ; {displaystyle z(m,n;s,t)=O(nm^{1-1/t}+m)} . In the particular case {displaystyle m=n} , assuming without loss of generality that {displaystyle sleq t} , we have the asymptotic upper bound {displaystyle z(n,n;s,t)=O(n^{2-1/t}).} Lower bounds One can verify that among the two asymptotic upper bounds of {displaystyle z(m,n;s,t)} in the previous section, the first bound is better when {displaystyle m=o(n^{s/t})} , and the second bound becomes better when {displaystyle m=omega (n^{s/t})} . Therefore, if one can show a lower bound for {displaystyle z(n^{s/t},n;s,t)} that matches the upper bound up to a constant, then by a simple sampling argument (on either an {displaystyle n^{t/s}times t} bipartite graph or an {displaystyle mtimes m^{s/t}} bipartite graph that achieves the maximum edge number), we can show that for all {displaystyle m,n} , one of the above two upper bounds is tight up to a constant. This leads to the following question: is it the case that for any fixed {displaystyle sleq t} and {displaystyle mleq n^{s/t}} , we have {displaystyle z(m,n;s,t)=Omega (mn^{1-1/s})} ? [8] In the special case {displaystyle m=n} , up to constant factors, {displaystyle z(n,n;s,t)} has the same order as {displaystyle {text{ex}}(n,K_{s,t})} , the maximum number of edges in an {displaystyle n} -vertex (not necessarily bipartite) graph that has no {displaystyle K_{s,t}} as a subgraph. In one direction, a bipartite graph with {displaystyle n} vertices on each side and {displaystyle z(n,n;s,t)} edges must have a subgraph with {displaystyle n} vertices and at least {displaystyle z(n,n;s,t)/4} edges; this can be seen from choosing {displaystyle n/2} vertices uniformly at random from each side, and taking the expectation. In the other direction, we can transform a graph with {displaystyle n} vertices and no copy of {displaystyle K_{s,t}} into a bipartite graph with {displaystyle n} vertices on each side of its bipartition, twice as many edges and still no copy of {displaystyle K_{s,t}} , by taking its bipartite double cover.[9] Same as above, with the convention that {displaystyle sleq t} , it has been conjectured that {displaystyle z(n,n;s,t)=Theta (n^{2-1/s})} for all constant values of {displaystyle s,t} .[10] For some specific values of {displaystyle s,t} (e.g., for {displaystyle t} sufficiently larger than {displaystyle s} , or for {displaystyle s=2} ), the above statements have been proved using various algebraic and random algebraic constructions. At the same time, the answer to the general question is still unknown to us. Incidence graphs in finite geometry The Levi graph of the Fano plane gives rise to the Heawood graph, a bipartite graph with seven vertices on each side, 21 edges, and no 4-cycles. For {displaystyle s=t=2} , a bipartite graph with {displaystyle n} vertices on each side, {displaystyle Omega (n^{3/2})} edges, and no {displaystyle K_{2,2}} may be obtained as the Levi graph, or point-line incidence graph, of a projective plane of order {displaystyle q} , a system of {displaystyle q^{2}+q+1} points and {displaystyle q^{2}+q+1} lines in which each two points determine a unique line, and each two lines intersect at a unique point. We construct a bipartite graph associated to this projective plane that has one vertex part as its points, the other vertex part as its lines, such that a point and a line is connected if and only if they are incident in the projective plane. This leads to a {displaystyle K_{2,2}} -free graph with {displaystyle q^{2}+q+1} vertices and {displaystyle (q^{2}+q+1)(q+1)} edges. Since this lower bound matches the upper bound given by I. Reiman,[11] we have the asymptotic [12] {displaystyle z(n;2)=(1/2+o(1))n^{3/2}.} For {displaystyle s=t=3} , bipartite graphs with {displaystyle n} vertices on each side, {displaystyle Omega (n^{5/3})} edges, and no {displaystyle K_{3,3}} may again be constructed from finite geometry, by letting the vertices represent points and spheres (of a carefully chosen fixed radius) in a three-dimensional finite affine space, and letting the edges represent point-sphere incidences.[13] More generally, consider {displaystyle s=2} and any {displaystyle t} . Let {displaystyle mathbb {F} _{q}} be the {displaystyle q} -element finite field, and {displaystyle h} be an element of multiplicative order {displaystyle t} , in the sense that {displaystyle H={1,h,dots ,h^{t-1}}} form a {displaystyle t} -element subgroup of the multiplicative group {displaystyle mathbb {F} _{q}^{*}} . We say that two nonzero elements {displaystyle (a,b),(a',b')in mathbb {F} _{q}times mathbb {F} _{q}} are equivalent if we have {displaystyle a'=h^{d}a} and {displaystyle b'=h^{d}b} for some {displaystyle d} . Consider a graph {displaystyle G} on the set of all equivalence classes {displaystyle langle a,brangle } , such that {displaystyle langle a,brangle } and {displaystyle langle x,yrangle } are connected if and only if {displaystyle ax+byin H} . One can verify that {displaystyle G} is well-defined and free of {displaystyle K_{2,t+1}} , and every vertex in {displaystyle G} has degree {displaystyle q} or {displaystyle q-1} . Hence we have the upper bound [14] {displaystyle z(n,n;2,t+1)=(t^{1/2}+o(1))n^{3/2}.} Norm graphs and projective norm graphs For {displaystyle t} sufficiently larger than {displaystyle s} , the above conjecture {displaystyle z(n,n;s,t)=Theta (n^{2-1/s})} was verified by Kollár, Rónyai, and Szabó [15] and Alon, Rónyai, and Szabó [16] using the construction of norm graphs and projective norm graphs over finite fields. For {displaystyle t>s!} , consider the norm graph NormGraphp,s with vertex set {displaystyle mathbb {F} _{p^{s}}} , such that every two vertices {displaystyle a,bin mathbb {F} _{p^{s}}} are connected if and only if {displaystyle N(a+b)=1} , where {displaystyle Ncolon mathbb {F} _{p^{s}}rightarrow mathbb {F} _{p}} is the norm map {displaystyle N(x)=xcdot x^{p}cdot x^{p^{2}}cdots x^{p^{s-1}}=x^{(p^{s}-1)/(p-1)}.} It is not hard to verify that the graph has {displaystyle p^{s}} vertices and at least {displaystyle p^{2s-1}/2} edges. To see that this graph is {displaystyle K_{s,s!+1}} -free, observe that any common neighbor {displaystyle x} of {displaystyle s} vertices {displaystyle y_{1},ldots ,y_{s}in mathbb {F} _{p^{s}}} must satisfy {displaystyle 1=N(x+y_{i})=(x+y_{i})cdot (x+y_{i})^{p}cdots (x+y_{i})^{p^{s-1}}=(x+y_{i})cdot (x^{p}+y_{i}^{p})cdots (x^{p^{s-1}}+y_{i}^{p^{s-1}})} for all {displaystyle i=1,ldots ,s} , which a system of equations that has at most {displaystyle s!} solutions.
The same result can be proved for all {displaystyle t<(s-1)!} using the projective norm graph, a construction slightly stronger than the above. The projective norm graph ProjNormGraphp,s is the graph on vertex set {displaystyle mathbb {F} _{p^{s-1}}times mathbb {F} _{p}^{times }} , such that two vertices {displaystyle (X,x),(Y,y)} are adjacent if and only if {displaystyle N(X+Y)=xy} , where {displaystyle Ncolon mathbb {F} _{p^{s}}rightarrow mathbb {F} _{p}} is the norm map defined by {displaystyle N(x)=x^{(p^{s}-1)/(p-1)}} . By a similar argument to the above, one can verify that it is a {displaystyle K_{s,t}} -free graph with {displaystyle Omega (n^{2-1/s})} edges. The above norm graph approach also gives tight lower bounds on {displaystyle z(m,n;s,t)} for certain choices of {displaystyle m,n} .[16] In particular, for {displaystyle sgeq 2} , {displaystyle t>s!} , and {displaystyle n^{1/t}leq mleq n^{1+1/t}} , we have {displaystyle z(m,n;s,t)=Theta (mn^{1-1/s}).} In the case {displaystyle m=(1+o(1))n^{1+1/s}} , consider the bipartite graph {displaystyle G} with bipartition {displaystyle V=V_{1}cup V_{2}} , such that {displaystyle V_{1}=mathbb {F} _{p^{t}}times mathbb {F} _{p}^{times }} and {displaystyle V_{2}=mathbb {F} _{p^{t}}} . For {displaystyle Ain V_{1}} and {displaystyle (B,b)in V_{2}} , let {displaystyle Asim (B,b)} in {displaystyle G} if and only if {displaystyle N(A+B)=b} , where {displaystyle N(cdot )} is the norm map defined above. To see that {displaystyle G} is {displaystyle K_{s,t}} -free, consider {displaystyle s} tuples {displaystyle (B_{1},b_{1}),ldots ,(B_{s},b_{s})in V_{1}} . Observe that if the {displaystyle s} tuples have a common neighbor, then the {displaystyle B_{i}} must be distinct. Using the same upper bound on he number of solutions to the system of equations, we know that these {displaystyle s} tuples have at most {displaystyle s!q/2} for some large constant {displaystyle C} , which implies {displaystyle mathbb {P} (|Z_{U}|>C)=mathbb {P} (|Z_{U}|>q/2)} .
Since {displaystyle f} is chosen randomly over {displaystyle mathbb {F} _{q}} , it is not hard to show that the right-hand side probability is small, so the expected number of {displaystyle s} -subsets {displaystyle U} with {displaystyle |Z_{U}|>C} also turned out to be small. If we remove a vertex from every such {displaystyle U} , then the resulting graph is {displaystyle K_{s,C+1}} free, and the expected number of remaining edges is still large. This finishes the proof that {displaystyle {text{ex}}(n,K_{s,t})=Omega (n^{2-1/s})} for all {displaystyle t} sufficiently large with respect to {displaystyle s} . More recently, there have been a number of results verifying the conjecture {displaystyle z(m,n;s,t)=Omega (n^{2-1/s})} for different values of {displaystyle s,t} , using similar ideas but with more tools from algebraic geometry.[8][20] Applications The Kővári–Sós–Turán theorem has been used in discrete geometry to bound the number of incidences between geometric objects of various types. As a simple example, a set of {displaystyle n} points and {displaystyle m} lines in the Euclidean plane necessarily has no {displaystyle K_{2,2}} , so by the Kővári–Sós–Turán it has {displaystyle O(nm^{1/2}+m)} point-line incidences. This bound is tight when {displaystyle m} is much larger than {displaystyle n} , but not when {displaystyle m} and {displaystyle n} are nearly equal, in which case the Szemerédi–Trotter theorem provides a tighter {displaystyle O(n^{2/3}m^{2/3}+n+m)} bound. However, the Szemerédi–Trotter theorem may be proven by dividing the points and lines into subsets for which the Kővári–Sós–Turán bound is tight.[21] See also Biclique-free graph, sparse graphs whose sparsity is controlled by the solution to the Zarankiewicz problem Forbidden subgraph problem, a non-bipartite generalization of the Zarankiewicz problem Forbidden graph characterization, families of graphs defined by forbidden subgraphs of various types Turán's theorem, a bound on the number of edges of a graph with a forbidden complete subgraph References ^ Bollobás, Béla (2004), "VI.2 Complete subgraphs of r-partite graphs", Extremal Graph Theory, Mineola, NY: Dover Publications Inc., pp. 309–326, MR 2078877. Reprint of 1978 Academic Press edition, MR0506522. ^ Zarankiewicz, K. (1951), "Problem P 101", Colloq. Math., 2: 301. As cited by Bollobás (2004). ^ "Archived copy" (PDF). Archived from the original (PDF) on 2016-03-04. Retrieved 2014-09-16. ^ Jump up to: a b Sierpiński, W. (1951), "Sur un problème concernant un reseau à 36 points", Ann. Soc. Polon. Math., 24: 173–174, MR 0059876. ^ Kővári, T.; T. Sós, V.; Turán, P. (1954), "On a problem of K. Zarankiewicz" (PDF), Colloquium Math., 3: 50–57, doi:10.4064/cm-3-1-50-57, MR 0065617. ^ Hyltén-Cavallius, C. (1958), "On a combinatorical problem", Colloquium Mathematicum, 6: 59–65, doi:10.4064/cm-6-1-61-65, MR 0103158. As cited by Bollobás (2004). ^ Znám, Š. (1963), "On a combinatorical problem of K. Zarankiewicz", Colloquium Mathematicum, 11: 81–84, doi:10.4064/cm-11-1-81-84, MR 0162733. As cited by Bollobás (2004). ^ Jump up to: a b Conlon, David (2021), "Some remarks on the Zarankiewicz problem", Mathematical Proceedings of the Cambridge Philosophical Society: 1–7, doi:10.1017/S0305004121000475. ^ Bollobás (2004), Theorem 2.3, p. 310. ^ Bollobás (2004), Conjecture 15, p. 312. ^ Reiman, I. (1958), "Über ein Problem von K. Zarankiewicz", Acta Mathematica Academiae Scientiarum Hungaricae, 9 (3–4): 269–273, doi:10.1007/bf02020254, MR 0101250, S2CID 121692172. ^ Bollobás (2004), Corollary 2.7, p. 313. ^ Brown, W. G. (1966), "On graphs that do not contain a Thomsen graph", Canadian Mathematical Bulletin, 9 (3): 281–285, doi:10.4153/CMB-1966-036-2, MR 0200182. ^ Füredi, Zoltán (1996), "New asymptotics for bipartite Turán numbers", Journal of Combinatorial Theory, Series A, 75 (1): 141–144, doi:10.1006/jcta.1996.0067, MR 1395763. ^ Kollár, János; Rónyai, Lajos; Szabó, Tibor (1996), "Norm-graphs and bipartite Turán numbers", Combinatorica, 16 (3): 399–406, doi:10.1007/BF01261323, MR 1417348, S2CID 26363618. ^ Jump up to: a b Alon, Noga; Rónyai, Lajos; Szabó, Tibor (1999), "Norm-graphs: variations and applications", Journal of Combinatorial Theory, Series B, 76 (2): 280–290, doi:10.1006/jctb.1999.1906, MR 1699238. ^ Alon, Noga; Mellinger, Keith E.; Mubayi, Dhruv; Verstraëte, Jacques (2012), "The de Bruijn-Erdős Theorem for Hypergraphs", Des. Codes Cryptogr., 65: 233–245. ^ Blagojević, Pavle; Bukh, Boris; Karasev, Roman (2013), "Turán numbers for Ks,t-free graphs: topological obstructions and algebraic constructions", Israel Journal of Mathematics, 197: 199–214, doi:10.1007/s11856-012-0184-z. ^ Bukh, Boris (2015), "Random algebraic construction of extremal graphs", Bull. London Math. Soc., 47: 939–945. ^ Bukh, Boris (2021), Extremal graphs without exponentially-small bicliques. ^ Matoušek, Jiří (2002), Lectures on discrete geometry, Graduate Texts in Mathematics, vol. 212, New York: Springer-Verlag, pp. 65–68, doi:10.1007/978-1-4613-0039-7, ISBN 0-387-95373-6, MR 1899299. Categories: Extremal graph theoryMathematical problemsUnsolved problems in graph theoryBipartite graphs
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Alere Inc (NYSE:ALR) Pre-Tax Income: \$-231 Mil (TTM As of Jun. 2017)
Pretax income is the income that a company earns before paying income taxes. Alere Inc's pretax income for the three months ended in Jun. 2017 was \$-77 Mil. Its pretax income for the trailing twelve months (TTM) ended in Jun. 2017 was \$-231 Mil. Alere Inc's pretax margin was -13.76%.
During the past 13 years, Alere Inc's highest Pretax Margin was 2.09%. The lowest was -49.54%. And the median was -4.59%.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
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Alere Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Pre-Tax Income -69.08 -120.01 -117.92 -84.54 -152.40
Alere Inc Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Pre-Tax Income -32.35 -55.01 -48.27 -50.72 -76.71
Competitive Comparison
* Competitive companies are chosen from companies within the same industry, with headquarter located in same country, with closest market capitalization; x-axis shows the market cap, and y-axis shows the term value; the bigger the dot, the larger the market cap.
Calculation
This is the income that a company earns before paying income taxes.
Alere Inc's Pretax Income for the fiscal year that ended in Dec. 2016 is calculated as
Pretax Income = Operating Income + Non-Recurring Items + Interest Expense + Interest Income + Other = 14.724 + 4.529 + -171.651 + 0 + 2.84217094304E-14 = -152
Alere Inc's Pretax Income for the quarter that ended in Jun. 2017 is calculated as
Pretax Income = Operating Income + Non-Recurring Items + Interest Expense + Interest Income + Other = -28.999 + -2.666 + -46.179 + 1.135 + -1.42108547152E-14 = -77
Pre-Tax Income for the trailing twelve months (TTM) ended in Jun. 2017 was -55.01 (Sep. 2016 ) + -48.273 (Dec. 2016 ) + -50.721 (Mar. 2017 ) + -76.709 (Jun. 2017 ) = \$-231 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Alere Inc's Pretax Margin for the quarter that ended in Jun. 2017 is calculated as
Pretax Margin = Pretax Income / Revenue = -76.709 / 557.672 = -13.76%
During the past 13 years, Alere Inc's highest Pretax Margin was 2.09%. The lowest was -49.54%. And the median was -4.59%.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
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# 4605Solnf08-05 - 5 Given the circuit below R1 = 10 C1 VS L...
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7 4605Solnf07.wxp 5. Given the circuit below R 2 = 500 Ω C 2 L C 1 R 1 = 10 V S R t = 10 Determine the , , and to transform the load to a input matched to the 10 source. G P G &!! "! "# HH H This is to be done at 15 MHz with a bandwidth of 1MHz. ----------------------------------- Since this is an L -network, we begin with the transformation. We have V œ "! œ œ Ê ; œ &! " œ ( V &!! ;" ;" > # ## È The \ œ "!‡( œ (!Þ! G" amd . \ œ œ ("Þ%\$ T &!! ; G œ œ "&# " (!‡# ‡"&‡"! " ' 1 pF For a bandwidth of 1 MHz, we need a
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http://mathhelpforum.com/calculus/119036-u-substitution.html | 1,529,315,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860168.62/warc/CC-MAIN-20180618090026-20180618110026-00321.warc.gz | 202,286,263 | 10,512 | 1. ## U Substitution
i'm confused on which item should be my u and which should be my du, i've tried a few values on all problems but I feel my answer is wrong and that I'm overlooking something, if you could please help I would be eternally grateful!
Use U substitution to solve the following:
a)anti derivative of cos(x)/sin^2(x) dx top valuei/2 bottom value: pi/4
b) (4(tan^-1(x))^2)/(1+x^2) dx top value: 1 bottom value: -1
c) 5x(3x^2 - 12)^9 dx top value: 4 bottom value: 0
2. Originally Posted by Lupus7874
i'm confused on which item should be my u and which should be my du, i've tried a few values on all problems but I feel my answer is wrong and that I'm overlooking something, if you could please help I would be eternally grateful!
Use U substitution to solve the following:
a)anti derivative of cos(x)/sin^2(x) dx top valuei/2 bottom value: pi/4
b) (4(tan^-1(x))^2)/(1+x^2) dx top value: 1 bottom value: -1
c) 5x(3x^2 - 12)^9 dx top value: 4 bottom value: 0
a) Substitute $\displaystyle u = \sin x$.
b) Substitute $\displaystyle u = \tan^{-1} x$.
c) Substitute $\displaystyle u = 3x^2 - 12$.
If you need more help, please post all your work and state exactly where you're stuck.
3. Ok for example a) I solve to the point of du*u^2 in said integral, but I'm just lost as to what to do when calculating the antiderivative of a product, surely I can't just apply the power rule in reverse can I?
4. Originally Posted by Lupus7874
Ok for example a) I solve to the point of du*u^2 in said integral, but I'm just lost as to what to do when calculating the antiderivative of a product, surely I can't just apply the power rule in reverse can I?
$\displaystyle \int_{\frac{\pi}{4}}^{\pi}\frac{cos(x)}{sin^2(x)}d x$
If $\displaystyle u=sin(x)$ then $\displaystyle du=cos(x)dx$ so $\displaystyle dx=\frac{du}{cos(x)}$. Substitute these values into the original integral to obtain $\displaystyle \int_{u_1}^{u_2}\frac{1}{u^2}du$. The limits of the integral are determined by the values of $\displaystyle u$ at the corresponding values of $\displaystyle x$.
This isn't a product. This can be evaluated using the basic formula:
$\displaystyle \int ax^ndx=\frac{a}{n+1}x^{n+1}+C$
-----------------------------------------------------------------------------------------------------------------
Secondly, there isn't a product rule for integration. There's only a sum and diffference rule. For product and quotients you have to deal with on a case-by-case basis. Sometimes you need to use either the substitution rule, integration by parts, partial fractions, etc. For example:
$\displaystyle \int sec^3(x)tan(x)dx$
We know that $\displaystyle \frac{d}{dx}sec(x)=sec(x)tan(x)$, so if we use $\displaystyle u=sec(x)$ then $\displaystyle du=sec(x)tan(x)dx$. This means that we can substitute $\displaystyle dx=\frac{du}{sec(x)tan(x)}$ into the original equation, canceling out the $\displaystyle tan(x)$ and leaving only $\displaystyle sec^2(x)$ which is equal to $\displaystyle u^2$.
Now we have $\displaystyle \int u^2du=\frac{1}{3}u^3+C=\frac{1}{3}sec^3(x)+C$
Therefore:
$\displaystyle \int sec^3(x)tan(x)dx=\frac{1}{3}sec^3(x)+C$ | 911 | 3,144 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-26 | latest | en | 0.835998 |
https://alitarhini.wordpress.com/2013/10/12/finding-adjacent-prime-numbers-using-linq/ | 1,555,693,228,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527865.32/warc/CC-MAIN-20190419161226-20190419183226-00461.warc.gz | 351,213,547 | 21,528 | # Finding Adjacent Prime Numbers using Linq
Suppose you have a range of numbers defined by a lower bound L and an upper bound U, the problem is to find 2 pairs of prime numbers, one of which will be the closest primes and the second pair will have the farthest distance in between. For example, given a range of integers from 1 to 20, the two most adjacent primes are 2 and 3 having a distance 1 while the least adjacent primes are 7 and 11 having a distance of 4.
Given a list of integers from 1 to 10,000 can you tell the least adjacent prime numbers? (the answer is 9551 and 9587) 🙂
Writing an algorithm that find adjacent primes can be tricky but is also short and straightforward. below is a naive vb.net implementation for this algorithm. I will not explain code in detail but basically what it does is to loop over the given range from L up to square root of U and flags all prime numbers within the range. then we compute the distance using linq and another query orders by distance in asc and desc in order to select the top 1 from each query.
Sub Main()
While True
Dim PrimeNumbers As New List(Of Integer)
Console.WriteLine(“———-Prime Distance———-“)
Console.Write(“Enter L: “)
Dim lowerLimit As String = Console.ReadLine().Trim
Console.Write(“Enter U: “)
Dim upperLimit As Integer = Console.ReadLine().Trim
Dim prime(upperLimit) As Integer
Dim c1, c2, c3 As Integer
For c1 = 2 To upperLimit
prime(c1) = 0
Next
prime(0) = 1
prime(1) = 1
For c2 = 2 To Math.Sqrt(upperLimit) + 1
If prime(c2) = 0 Then
c1 = c2
c3 = 2 * c1
While c3 < upperLimit + 1
prime(c3) = 1
c3 = c3 + c1
End While
End If
Next
For c1 = 0 To upperLimit
If prime(c1) = 0 Then
End If
Next
Dim ordered = From a In PrimeNumbers Order By a Ascending Select a
Dim query = From a In ordered, b In ordered Where b > a Select P1 = a, P2 = b, Distance = Math.Abs(a – b)
query = query.GroupBy(Function(x) x.P1).Select(Function(x) x.First) ‘distinct
Dim minDistance = (From a In query Order By a.Distance Ascending Select a).Take(1).SingleOrDefault
Dim maxDistance = (From a In query Order By a.Distance Descending Select a).Take(1).SingleOrDefault
Console.WriteLine(“———————————“)
Console.WriteLine(“{0},{1} are the closest primes having distance {2}”, minDistance.P1, minDistance.P2, minDistance.Distance)
Console.WriteLine(“———————————“)
Console.WriteLine(“{0},{1} are the most distant primes having distance {2}”, maxDistance.P1, maxDistance.P2, maxDistance.Distance)
Console.WriteLine(“———————————“) | 685 | 2,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-18 | latest | en | 0.795506 |
http://list.seqfan.eu/pipermail/seqfan/2004-September/004284.html | 1,716,058,546,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057494.65/warc/CC-MAIN-20240518183301-20240518213301-00029.warc.gz | 15,253,426 | 3,227 | # Binomial transforms of Fib, etc.
creigh at o2online.de creigh at o2online.de
Thu Sep 23 03:31:56 CEST 2004
```Thanks for all your previous comments and help!
A030191(n) + 2a(n) + A093129(n+2) = 4A093129(n+1)
(a(n)) = (1,0,-5,-25,-100,-375,-1375,-5000,-18125,-65625)
A030191, Name: Scaled Chebyschev U-polynomial evaluated at sqrt(5)/2
A030191(n-1) is 2nd binomial transform of Fib(n)
A093129, binomial transform of Fib(2n-1)
superseeker's T004, below). Please also note transformations T101 and
T018 [= A030191, from above ].
p.s. if there are any serious mistakes in here this will be the last posting I
shall ever send at three in the morning!
Sincerely,
Creighton
Report on [ 1,0,-5,-25,-100,-375,-1375,-5000,-18125,-65625]:
[snip]
TEST: APPLY VARIOUS TRANSFORMATIONS TO SEQUENCE AND LOOK IT
UP IN THE ENCYCLOPEDIA AGAIN
SUCCESS
(limited to 40 matches):
Transformation T004 gave a match with:
%I A093131
%S A093131 0,1,5,20,75,275,1000,3625,13125,47500,171875,621875,2250000,
[snip],
%N A093131 Binomial transform of Fib(2n).
%F A093131 G.f.: x/(1-5x+5x^2); a(n)=(((5+sqrt(5))/2)^n-((5-sqrt(5))/2)
^n)/sqrt(5); a(n)=A093130(n)/2^n.
%Y A093131 Cf. A000045.
[snip]
%A A093131 Paul Barry (pbarry(AT)wit.ie), Mar 23 2004
Transformation T101 gave a match with:
%I A098149
%S A098149 1,1,4,11,29,76,199,521,1364,3571,9349,24476,64079,167761,439204,
[snip]
%N A098149 a(n) = 2(a(n-2) - a(n-1)) + a(n-3), with a(0) = a(1) = -1 and
a(2) = 4.
%C A098149 Sequence relates bisections of Lucas and Fibonacci numbers.
%C A098149 2*a(n) + A098150(n) = 8*(-1)^(n+1)*A001519(n) - (-1)^(n+1)*A005248
(n+1).
[snip]
%A A098149 Creighton Dement (creigh(AT)o2online.de), Aug 29 2004
%E A098149 More terms from RGWv (rgwv(AT)rgwv.com), Sep 1 2004
Transformation T005 gave a match with:
%I A039717
%S A039717 1,4,15,55,200,725,2625,9500,34375,124375,450000,1628125,5890625,[snip]
%N A039717 Row sums of convolution triangle A030523.
[snip]
%A A039717 Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)
Transformation T018 gave a match with:
%S A030191 1,5,20,75,275,1000,3625,13125,47500,171875,621875,2250000,8140625,
[snip]
%N A030191 Scaled Chebyshev U-polynomial evaluated at sqrt(5)/2.
[snip]
%F A030191 a(n)=(sqrt(5))^n*U(n,sqrt(5)/2), g.f.: 1/(5*(x^2-x+1/5)), a(2*k+1)
=5^(k+1)*F(2*k+2), F(n) = Fibonacci
(A000045), a(2*k)=5^k*L(2*k+1), L(n) = Lucas (A000032)
%F A030191 a(n-1)=sum(k=0,n,C(n,k)*F(2*k)) - Benoit Cloitre (abcloitre(AT)
%F A030191 a(n) = 5*a(n-1)-5*a(n-2). - Benoit Cloitre (abcloitre(AT)wanadoo.
fr), Oct 23 2003
%F A030191 a(n-1)=((5/2+sqrt(5)/2)^n-(5/2-sqrt(5)/2)^n)/sqrt(5) is the
2nd binomial transform of Fib(n), the first binomial
transform of Fib(2n), and its n-th term is the n-th term of the third binomial
transform of Fib(3n) divided by 2^n. - Paul
Barry (pbarry(AT)wit.ie), Mar 23 2004
[snip]
%A A030191 Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)
%S A093131 0,1,5,20,75,275,1000,3625,13125,47500,171875,621875,2250000,8140625,
[snip]
%N A093131 Binomial transform of Fib(2n).
%F A093131 G.f.: x/(1-5x+5x^2); a(n)=(((5+sqrt(5))/2)^n-((5-sqrt(5))/2)
^n)/sqrt(5); a(n)=A093130(n)/2^n.
%Y A093131 Cf. A000045.
[snip]
%A A093131 Paul Barry (pbarry(AT)wit.ie), Mar 23 2004
Transformation T019 gave a match with:
%I A039717
%S A039717 1,4,15,55,200,725,2625,9500,34375,124375,450000,1628125,5890625, %N A039717
Row sums of convolution triangle A030523.
[snip]
%A A039717 Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)
List of transformations used:
T004 sequence divided by the gcd of its elements, from the 2nd term
T005 sequence divided by the gcd of its elements, from the 3rd term
T018 sequence u[j+1]-u[j]
T019 sequence u[j+2]-2*u[j+1]+u[j]
T101 inverse binomial transform: b(n)=SUM (-1)^(n-k)*C(n,k)*a(k), k=0..
n
``` | 1,552 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.603548 |
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## Monk Steps
Math brain teasers require computations to solve.
Puzzle ID: #17721 Fun: (2.88) Difficulty: (2.12) Category: Math Submitted By: Jake Corrected By: Dave
A monk has a very specific ritual for climbing up the steps to the temple. First he climbs up to the middle step and meditates for 1 minute. Then he climbs up 8 steps and faces east until he hears a bird singing. Then he walks down 12 steps and picks up a pebble. He takes one step up and tosses the pebble over his left shoulder. Now, he walks up the remaining steps three at a time which only takes him 9 paces. How many steps are there?
### Answer
There are 49 steps.
He climbs halfway, which is step 25. He hears the bird singing on step 33. He picks up the pebble on the 21st step and tosses it on the 22nd step. The remaining 27 steps are taken three at a time which is 9 paces.
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## Comments
sparkplug May 22, 2004 It is not very hard : you just have to visualize the steps to find the answer, but I liked it. It is a fun one. jimbo May 24, 2004 It's not hard after you have been told the answer to verify that it works. How do you work it out? The key is working backwards from the top step. If he covers 27 (3x9) steps to reach the top he must have been 28 steps from the top. Thus before he stepped up one he was 29 from the top. Before he went down 12 he was 17 from the top. Before he went up 8 he was 25 from the top. This is the middle step - 24 above it and 24 below it = 49 steps. doggyxp Jun 23, 2004 Yeah, I worked backwards to solve it. If you work forwards, it gets sort of complicated and confusing. Very well done. lucylizzy Jun 23, 2004 Got it! Great teaser! witness Jun 23, 2004 I liked it! Vry tidy! shelleyz Jun 23, 2004 good one...but too much math in it O_wise_one Jun 23, 2004 I just figured he was on the middle step, then goin up 8 and down 12 puts him at 4 below middle. Then one step up. Then one set of three put's him back on the middle. So there's 24 ahead of him, and 24 behind. 49. Many ways to solve this teaser. (or, many algorithms) Anyway, good teaser! vbfreak55 Jun 24, 2004 i didn't get it. too hard for me. solidross Jul 01, 2004 great math teaser.I liked it. pharast Jul 07, 2004 found a flaw...it says "he climbs up to the middle step" ...if there are 49 steps then the middle one would be the 24.5th step... canu Jul 08, 2004 There are 48 steps. The middle step is the 24th. Look at it this way: if there are two steps, the second step is the top level, the 1st step is in the middle. There has to be an even number of steps from the starting level to the top level, for there to be a middle step. Therulerofthem Jul 10, 2004 Actually no there has to be an odd number of step to have a middle step, and the 25 would be the middle, so check your facts before trying to prove people wrong ched4r Sep 19, 2004 I counted it with this equation(mentally o course ) x = 0,5x + 0,5 + 8 - 12 + 1 + 3*9 Because there is a middle step, the number of steps is odd. 0,5x + 0,5 calculates the middle step thus, examples: total steps: 25 middle step: 0,5*25 + 0,5 = 13 total steps: 49 middle step: 0,5*49 + 0,5 = 25 x = 0,5x + 0,5 + 8 - 12 + 1 + 3*9 0,5x = 24,5 x = 49 ched4r Sep 19, 2004 Oh, and there was supposed to be spaces between lines in the above post, try to undestand tamjp May 21, 2005 Impressive teaser ( i came up with 50 ) and impressive comment ched4r tequila_roze Jun 03, 2005 ched4r, just outta curiosity, do you like write meth books or something?? tequila_roze Jun 03, 2005 .....that iz, "MATH" books.... kgripp Jun 23, 2005 i JUST DID IT BACKWARDS AND CAME UP WITH THE CORRECT ANSWER. gOOD PUZZLE. tinkerbell09 Jun 23, 2005 WAY TOO MATHY 4 ME! pat56 Jun 23, 2005 too hard for me.I'm not good at math. robinmckenzie Jun 23, 2005 I think there is some confusion over the counting of steps. If there were no steps, it would be flat ground. If there was one step, the ground has two levels, and there is no "middle step". If there are two steps, the ground has three levels, and the middle step is step 1 (counting the ground as step 0). From this we can see that there is only a middle step when the number of steps is ODD. If you still don't believe this, try to picture 1 step, and think if there is a middle step: _______________ ____________| Then, picture two steps: ______________ _____| ________| And you can see quite clearly that there is a middle step in this case (you wouldn't call this 3 steps, would you? If you would, the first picture must therefore show 2 steps, and 1 step would be flat ground) I agree with Canu, and get the answer to be 48 steps. It would only be 49 steps if one (erroneously) counted the ground as being a step. robinmckenzie Jun 23, 2005 Sorry - text pictures got spoiled in the last post: If you still don't believe this, try to picture 1 step, and think if there is a middle step: _______________ ___________ | Then, picture two steps: ______________ _____| ________| And you can see quite clearly that there is a middle step in this case (you wouldn't call this 3 steps, would you? If you would, the first picture must therefore show 2 steps, and 1 step would be flat ground) Magicbanker Jun 23, 2005 The ground is not a step. The ground is the ground. In order for there to be a middle step, there must by definition be an odd number of steps of at least 3. Very good teaser. I didn't think to try to work backwards - that would have been a good hint. amanduh05 Jun 23, 2005 way to much brain work needed!!!my brain hurts just thinking about it! but good one crzyness4lyfe Jun 23, 2005 i figured it out just like ched4r...i used x as the # of steps....great teaser..liked it....the whole monk thingy made it a lot more enjoyable...great job! Araldite Jun 23, 2005 Every time there's a teaser like this there's a discussion about odd or even number of steps. Those who said if there is a middle step then by definition there must be an odd number of steps are correct. If you remember this you'll have an easier time getting the right answer the next time you run accross one of these even if you don't want to beleive it. KTDidds Jun 23, 2005 I didn't use any math, I just drew a picture as I was reading the teaser, and that made it really easy. Starting with a line that indicated the middle, there ended up being 28 steps above it, so that meant there had to be 28 steps below it, with the middle line being the last step I counted. If you do it that way, there really is no question of whether there could be a middle step or not, there just is and you count it and move on. robinmckenzie Jun 23, 2005 If there must be an odd number of steps to have a middle step, then which step is the middle one when there is only one step? ___________|"""""""""""""" If there are two steps, then the first step is clearly the middle step, since it is an equal distance from the top and the bottom: _________------"""""""""""""" The confusion lies, I think, in the fact that, as you said Magicbankerm the ground isn't counted as a step, but it IS treated as a boundary, just as the top step is. Therefore, when finding the centre (and do be careful not to misuse the phrase "by definition") we have to use the half-way point between top and bottom, as you can see from the second diagram above. The first step is midway between the top and bottom, and is therefore "the middle" step. JCDuncan Jun 23, 2005 I don't understand why anybody felt this was hard. If you can count backwards, you can solve it with simple arithmathic. It was good that the teaser started us on the middle step, or it would have been even more easy. Fasil Jun 23, 2005 robinmckenzie; you are misusing the words "middle step". What you are describing is the middle of the stairs, not the middle step. There is not a middle step when there are only one or two steps on a staircase. The idea is once you step up, you have taken one step. So the top is a step. The following picture (hope it works) is a representation showing that 2 is the middle step of 3 steps. Even though the top of step 2 is not directly in the middle of the stairs, it is the middle step. ------_________ ----_|3 --_|2 _|1 Sweetbabe Jun 23, 2005 Good one. KTDidds Jun 23, 2005 Okay I had to do this for myself...in my last post I put 28 and I meant to put 24 - it was bothering me so I signed back on to correct it... Now everything is right with my world and I can go back to work. Have a great day everyone. And thanks again for the teaser - if I even did say thank you in the first place. azushud Jun 23, 2005 is he walking or climbing je_russell Jun 23, 2005 From the midle step he walks up 24 steps-That's just the math. From there the answer depends as to what one defines as the middle step. I recall that median is defined as the middle value: If there are 'n' objects arranged in order the median is defined as the value at (n+1)/2 position if n is odd and the average of the two middle values if n is even. If n = 49 yields a median of 25 and if n = 48 yields a median value ((24+25)/2) of 24.5. Since there can't be a a physical step at position 24.5 n would be 49. PCDguitar Jun 23, 2005 robinmckenzie u dont count the levels u coulnt the steps no one gives a **** if there are 3 levels there would have to be 3 steps for there to be a middle step so if there were two steps theres no middle step PCDguitar Jun 23, 2005 p.s. how are we supposed to know that he is climbing 3 steps!!!!!!!!!! that is stupid how can we get it right if we have no idea that the fatso priest is taking 3 steps in 1 pace!?!?!??!??!??!?! (user deleted) Jun 23, 2005 You know i am a bit confused.. not sure if 49 is correct or 48. I went through the calculations as below. I sort of agree with 49 but cannot seen to go beyond 48. Total = X steps (x/2 + - 12 + 1 + (3* 9) = X (x/2 + + 16 = x ((x+16) + 32)/2 = x x+16 + 32 = 2x x + 48 = 2x x = 48 So u see, this was very simple, but I am ending at 48 while I do agree with 49. What do y'll think ? Kukmiester Jun 23, 2005 I think you are all nuts... redneck_woman Jun 23, 2005 I must agree with Kukmeister. You all need to get a life and not ANALIZE things so much! Kinergy Jun 23, 2005 I got 48 steps just like nako, but then I have to agree that if there is a middle step then it must be odd numbered. I worked forward. When it says he covered 27 steps (9 x 3) and someone said 'Simple, it's 28 steps he covered' I don't get that. How does 27 steps equal 28 steps? DesertJules1974 Jun 23, 2005 nako, you're on the right track, but to find the center (median) your formula is (x+1)/2 . . . not (x/2). That's probably why you ended up with an even number. To have a middle, you have to have an equal number of items on either side of that middle. Therefore, you cannot have an even number when you are calculating with/from a middle point. Great teaser! dudeman81 Jun 23, 2005 You guys are spending way too much time on this one! FeaerFactorY666 Jun 23, 2005 Too difficult for me.... neomajic Jun 23, 2005 This is simple algebra: Number of steps up from middle ======================= +27 (3*9) steps up +1 step up -12 steps down +8 steps up) Don't forget the middle step +1 the middle step x=2(3*9+1-12++1 x=2(27+1-12++1 x=2(24)+1 x=49 neomajic Jun 23, 2005 Hee, my equation got mangled by emoticons... x=2(3*9+1-12+8 )+1 x=2(27+1-12+8 )+1 x=2(24)+1 x=49 Trickster1992 Jun 23, 2005 I have the most brilliant equation in the world.... Well, it used to be, at least. I think it still is. But I have the most brilliant equation! = 2 Trickster1992 Jun 23, 2005 AND ANOTHER! + = skimela Jun 23, 2005 i agree with dudeman81 great teaser though, just simple math. i worked backwards and got it right away (user deleted) Jun 23, 2005 I liked it. It was simple if you worked it backwards. okieman51 Jun 23, 2005 I missed it, but saw mistake when i worked it out looking at answer, I loved it however was real good. jennysugars Jun 23, 2005 Have you posted this somewhere else too? I've done this one before....but it wasn't here. teri12265 Jun 23, 2005 There are only 5 steps, not 48, nor 49, just 5. There are only 5 steps to HIS ROUTINE OF GOING UP! Step 1 - he climbs up and meditates, Step 2 - up 8 til the bird sings, Step 3 - walks down and gets pebble, Step 4 - up 1 and tosses pebble, & Step 5 - takes remainder 3 at a time. See, five steps. Right??? tequila_roze Jun 23, 2005 teri that was AAWWEESSOOMMEE!!!!!!!!!!!!!!! canttakemyname Jun 23, 2005 BBOORRIINNGG! But I figured it out! Great teaser! luckybrown Jun 27, 2005 You guys debating 48 steps are nuts! The ground is the ground and a step is a step. Silly people. pinkgirl Jun 29, 2005 sorry- i agree with the people who say 48 steps 1CrazyBrunette Jun 29, 2005 I think too much thought is being put into this. Just have fun with these! blonde_girl Jul 02, 2005 i had a lot of fun figuring it out! Jake Apr 05, 2006 I'm glad you liked it. mitzimesser May 04, 2006 I totally missed the "middle" step. I got 48 too..... lukeschett Apr 14, 2007 Confusing!!!! reptile5000 Jun 09, 2007 teri, it would only be that if it was in trick category, lol 4demo Sep 08, 2007 I got 48 steps: close enough but I'm still not sure where that last step comes in. Good teaser anyhow! Dontrelle Dec 04, 2007 You have to count in the middle step. 49, very tricky but I got it. Great teaser. javaguru Jan 03, 2009 Where's the teaser? As I read the story I'm counting M+8, M-4, M-3, M+24. So 24 above the middle and 24 below plus the middle for 49. There should be at least some math--this should be in the logic category. At any rate, too easy to be a teaser. Paladin Mar 13, 2009 Since there is a middle step, we must have an odd number of steps. The middle step would be (S+1)/2 where S = total number of steps, so: (1/2 [S+1] + 8 - 12 + 1) + 9*3 = S 1/2 S - 2.5 + 27 = S 1/2 S + 24.5 = S 24.5 = 1/2 S S = 49 steps dalfamnest Feb 25, 2010 OK - draw 2 steps: ground, step, top landing ... _______ _______|_x_ ___________|_______ If you told me to stand on the middle step, I know where I would stand! There i am (I hope) at the 'x'. To my thinking, 'middle step' requires an EVEN number steps, so the answer must be 48 steps! dalfamnest Feb 25, 2010 I think this illustrates the difference between maths and logic. Palladin correctly uses maths to solve an equation. However, if logic is used, we cannot assume, as Javaguru (I think) incorrectly does, that there is an odd number. Logic must be built on correct assumptions. From my diagram above, we see that 'middle step' is a concept which, by normal use of language, requires an EVEN number of steps. Maybe Javaguru is correct, then, that this is more logic than Maths! wordmama Jul 11, 2010 First of all, the years-long discussion of 48 vs. 49 steps saddens me; the middle step must be a whole physical step added to the equal number up and down, not fractional, ergo 24+24+1=49. I admire all the people who either worked it backwards or by way of an algebraic formula; I seem to be the only one so far who's gotten the answer by trial and error (picking a # of steps and plugging in the monk's ritual!) As for the explanation, I have no idea where you got step 33 as a starting point!!! What did I miss? How were we supposed to know that?? Will someone please explain? Thank you. VitalStatistic Jul 11, 2010 Wading into the odd/even steps debate... I mathematically calculated from the middle step that there were 24 above the starting point, therefore double that gives 48. The difficulty is that an even number gives a middle number of treads with two ground levels, and an odd number gives a middle number of step faces. ______ 5/__ 4/__ 3/__ 2/__ 1/________ ______ 4/__ 3/__ 2/__ 1/________ My interpretation is that the even number (4 steps above) gives a step that is an even height from the upper ground and the lower ground. Therefore step 2 is in the middle, and the even number of steps theory is correct. If you look at the case of the 5 step stairs above according to the 49 step proponents, if you want to climb up to the middle step you take 3 steps. Then you have a further 2 to get to the top. So that doesn't work. Let your fingers do the walking on the diagram above. "Walk" along the "ground" then take 3 steps up and then place your other finger(foot) on the third step as well. Then continue up the stairs, and you only take 2 more until you reach the top. I guess you need to ask if you want to start in the middle, or if you want to start with the same number of steps above and below. VitalStatistic Jul 11, 2010 5 and 4 step staircases redrawn ______ ......5/__ .........4/__ ............3/__ ...............2/__ ..................1/________ ______ ......4/__ .........3/__ ............2/__ ...............1/________ gaylewolf Jul 11, 2010 This was a good puzzle, and a heck of a lot of thought put into it! Gee, for that matter, a heck of a lot of thought put into the explanation comments! For those of you who solved it, bravo! doehead Jul 11, 2010 49 steps and 6 years running, this one is wornout. bradon182001 Jul 11, 2010 Needless to say, since I will admit I'm math challenged, and not particularly logical, I didn't get this right. I had more fun reading all the comments over the years. Thanks everybody for the fun reading. auntiesis Jul 11, 2010 Easy but fun. I worked from the middle step and got it in less than a minute. So much controversy over a simple problem. Steps are steps, ground is ground....not a step. Garnette Jul 11, 2010 When it cums 2 math, I am terrible!!!!!!!!!! Interestin teases!! emu77alu02 Jul 11, 2010 I got 48, but I can't see what I did wrong though. the-Gaul Jul 12, 2010 How do you know that there are precisely 49 steps? The conditions state that the last 9 paces are taken three steps at a time. However, the last pace may be one, two, or three steps, yet it must be taken in order to reach the top. Since you can't be sure of the number of steps in the last pace, how can you determine the total? avley07 Aug 02, 2011 So you mean like 25 is the middle step and it's not counted when you count the step upwards and downwards? Babe Jul 11, 2013 I don't do math teasers so I don't care how many steps there are! HABS2933 Jul 11, 2013 After 9 years, I think it is time to retire this teaser. I understand some people have not seen the older teasers and they are nice to have appear on occasion, but looking at the dates this has been Teaser of the Day on several occasions. Why not let some of the newer teasers (not the brand new ones that sow up on the right of the screen) have a go at being T.O.T.D.? IN the 3 years since this was last featured, there must have been at least 1 or 2 decent teases submitted, if not, what are we all doing here? The teaser itself was great, challenging enough to not be obvious but not so hard it was impossible to solve. jaycr Jul 11, 2013 I'm with Babe on this one. cutebug Jul 11, 2013 Bravo to HABS for telling it like it is. gaylewolf Jul 11, 2013 I'm with Babe on this one - math teasers are far from being my favorites, but I also am kind of ashamed of myself to admit that! But thank you, HABS for your thoughtful comment - you always come through with something substantial like that! Hi Doehead and nice that you complimented HABS. Loved all the comments today! gaylewolf Jul 11, 2013 whoops, I think I goofed on my last comment! Sorry!! auntiesis Jul 11, 2013 Nobody will catch your goof. Still love math teasers, and they are always new to an old brain like mine. jaycr Jul 11, 2013 , never mind, it's old news . . . brad2490 Jan 01, 2015 YESS! I got it! I liked this one a lot. raisin99 Jul 11, 2016 nicely done lukasiwicz Jul 11, 2016 Once you dope out that, in order for there to be a middle step, the total number of steps must be odd, and that in any series that ends in an odd number, the midpoint of the series must be (n + 1)/2, where "n" is the last number in the series the algebraic solution to the problem becomes obvious. It took me a while, though, to reason this out.
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Click here to see ALL problems on Age Word Problems Question 166037: The two taylor girls were born in consecutive, even years. If the product of their age is 224, how old is each girl?Answer by checkley77(12569) (Show Source): You can put this solution on YOUR website!x(x+2)=224 x^2+2x-224=0 (x-14)(x+16)=0 x-14=0 x=14 for the youngest girl. 14+2=16 for the older girl. Proof: 14*16=224 224=224 | 282 | 1,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2013-20 | latest | en | 0.937392 |
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Creates a DataFrame with all available influence results. patsy is a Python library for describing statistical models and building Design Matrices using R-like formulas. use statsmodels.formula.api (often imported as smf) # data is in a dataframe model = smf . The model is Understand Summary from Statsmodels' MixedLM function. Note that this function can also directly be used as a Pandas method, in which case this argument is no longer needed. Default is None. 2 $\begingroup$ I am using MixedLM to fit a repeated-measures model to this data, in an effort to determine whether any of the treatment time points is significantly different from the others. Given this, there are a lot of problems that are simple to accomplish in R than in Python, and vice versa. dependent, response, regressand, etc.). The patsy module provides a convenient function to prepare design matrices estimates are calculated as usual: where $$y$$ is an $$N \times 1$$ column of data on lottery wagers per statsmodels.stats.outliers_influence.OLSInfluence.summary_frame, statsmodels.stats.outliers_influence.OLSInfluence, Multiple Imputation with Chained Equations. For more information and examples, see the Regression doc page. First, we define the set of dependent(y) and independent(X) variables. Estimate of variance, If None, will be estimated from the largest model. To fit most of the models covered by statsmodels, you will need to create In statsmodels this is done easily using the C() function. ols ( formula = 'chd ~ C(famhist)' , data = df ) . Influence.resid_studentized_internal, hat_diag : The diagonal of the projection, or hat, matrix defined in We will use the Statsmodels python library for this. What I have tried: i) X = dataset.drop('target', axis = 1) ii) Y = dataset['target'] iii) X.corr() iv) corr_value = v) import statsmodels.api as sm Remaining not able to do.. One or more fitted linear models. 2.1.2. pandas takes care of all of this automatically for us: The Input/Output doc page shows how to import from various You’re ready to move on to other topics in the The rate of sales in a public bar can vary enormously b… two design matrices. Using statsmodels, some desired results will be stored in a dataframe. the model. How to solve the problem: Solution 1: parameter estimates and r-squared by typing: Type dir(res) for a full list of attributes. One important thing to notice about statsmodels is by default it does not include a constant in the linear model, so you will need to add the constant to get the same results as you would get in SPSS or R. Importing Packages¶ Have to import our relevant packages. Figure 3: Fit Summary for statsmodels. using webdoc. patsy is a Python library for describingstatistical models and building Design Matrices using R-like form… Here the eye falls immediatly on R-squared to check if we had a good or bad correlation. The summary () method is used to obtain a table which gives an extensive description about the regression results Summary. The res object has many useful attributes. the difference between importing the API interfaces (statsmodels.api and capita (Lottery). Then fit () method is called on this object for fitting the regression line to the data. Name of column in data containing the dependent variable. Describe Function gives the mean, std and IQR values. This may be a dumb question but I can't figure out how to actually get the values imputed using StatsModels MICE back into my data. The pandas.DataFrame functionprovides labelled arrays of (potentially heterogenous) data, similar to theR “data.frame”. Chris Albon. and specification tests. Notes. df ['preTestScore']. first number is an F-statistic and that the second is the p-value. These are: cooks_d : Cook’s Distance defined in Influence.cooks_distance. We select the variables of interest and look at the bottom 5 rows: Notice that there is one missing observation in the Region column. The pandas.DataFrame function We If the dependent variable is in non-numeric form, it is first converted to numeric using dummies. I'm estimating some simple OLS models that have dozens or hundreds of fixed effects terms, but I want to omit these estimates from the summary_col. summary ()) #print out the fitted rate vector: print (poisson_training_results. DFBETAS. We download the Guerry dataset, a fit () 3.1.2.1. I love the ML/AI tooling, as well as th… Fitting a model in statsmodels typically involves 3 easy steps: Use the model class to describe the model, Inspect the results using a summary method. In this short tutorial we will learn how to carry out one-way ANOVA in Python. Parameters: args: fitted linear model results instance. Statsmodels is a Python module which provides various functions for estimating different statistical models and performing statistical tests. added a constant to the exogenous regressors matrix. A DataFrame with all results. variable names) when reporting results. functions provided by statsmodels or its pandas and patsy scale: float. comma-separated values file to a DataFrame object. df ['preTestScore']. few modules and functions: pandas builds on numpy arrays to provide We will only use It will give the model complexive f test result and p-value, and the regression value and standard deviarion - from the summary report note down the R-squared value and assign it to variable 'r_squared' in the below cell Can some one pls help me to implement these items. See Import Paths and Structure for information on eliminate it using a DataFrame method provided by pandas: We want to know whether literacy rates in the 86 French departments are In one or two lines of code the datasets can be accessed in a python script in form of a pandas DataFrame. I’m a big Python guy. Name of column(s) in data containing the between-subject factor(s). statsmodels also provides graphics functions. independent, predictor, regressor, etc.). Opens a browser and displays online documentation, Congratulations! Ask Question Asked 4 years ago. R² is just 0.567 and moreover I am surprised to see that P value for x1 and x4 is incredibly high. After installing statsmodels and its dependencies, we load a That means the outcome variable can have… Statsmodels, scikit-learn, and seaborn provide convenient access to a large number of datasets of different sizes and from different domains. describe () count 5.000000 mean 12.800000 std 13.663821 min 2.000000 25% 3.000000 50% 4.000000 75% 24.000000 max 31.000000 Name: preTestScore, dtype: float64 Count the number of non-NA values. apply the Rainbow test for linearity (the null hypothesis is that the patsy is a Python library for describing We use patsy’s dmatrices function to create design matrices: The resulting matrices/data frames look like this: split the categorical Region variable into a set of indicator variables. statsmodels.tsa.api) and directly importing from the module that defines and specification tests. We're doing this in the dataframe method, as opposed to the formula method, which is covered in another notebook. Starting from raw data, we will show the steps needed to What we can do is to import a python library called PolynomialFeatures from sklearn which will generate polynomial and interaction features. The resultant DataFrame contains six variables in addition to the plot of partial regression for a set of regressors by: Documentation can be accessed from an IPython session You can find more information here. Statsmodels is built on top of NumPy, SciPy, and matplotlib, but it contains more advanced functions for statistical testing and modeling that you won't find in numerical libraries like NumPy or SciPy.. Statsmodels tutorials. The pandas.read_csv function can be used to convert a as_html ()) # fit OLS on categorical variables children and occupation est = smf . Region[T.W] Literacy Wealth, 0 1.0 1.0 0.0 ... 0.0 37.0 73.0, 1 1.0 0.0 1.0 ... 0.0 51.0 22.0, 2 1.0 0.0 0.0 ... 0.0 13.0 61.0, ==============================================================================, Dep. defined in Influence.dffits, student_resid : Externally Studentized residuals defined in $$X$$ is $$N \times 7$$ with an intercept, the Variable: Lottery R-squared: 0.338, Model: OLS Adj. control for unobserved heterogeneity due to regional effects. This is useful because DataFrames allow statsmodels to carry-over meta-data (e.g. I have a dataframe (dfLocal) with hourly temperature records for five neighboring stations (LOC1:LOC5) over many years and … Historically, much of the stats world has lived in the world of R while the machine learning world has lived in Python. Polynomial Features. statsmodels.stats.outliers_influence.OLSInfluence.summary_frame OLSInfluence.summary_frame() [source] Creates a DataFrame with all available influence results. returned pandas DataFrames instead of simple numpy arrays. rich data structures and data analysis tools. The first is a matrix of endogenous variable(s) (i.e. After installing statsmodels and its dependencies, we load afew modules and functions: pandas builds on numpy arrays to providerich data structures and data analysis tools. If between is a single string, a one-way ANOVA is computed. The data set is hosted online in The second is a matrix of exogenous The larger goal was to explore the influence of various factors on patrons’ beverage consumption, including music, weather, time of day/week and local events. tables [ 1 ] . The summary of statsmodels is very comprehensive. data pandas.DataFrame. Literacy and Wealth variables, and 4 region binary variables. Statsmodels 0.9 - GEEMargins.summary_frame() statsmodels.genmod.generalized_estimating_equations.GEEMargins.summary_frame I will explain a logistic regression modeling for binary outcome variables here. other formats. Student’s t-test: the simplest statistical test ¶ 1-sample t-test: testing the value of a population mean¶ scipy.stats.ttest_1samp() tests if the population mean of data is likely to be equal to a given value (technically if observations are drawn from a Gaussian distributions of given population mean). collection of historical data used in support of Andre-Michel Guerry’s 1833 Essay on the Moral Statistics of France. R “data.frame”. between string or list with N elements. Returns: frame – A DataFrame with all results. We could download the file locally and then load it using read_csv, but Using the statsmodels package, we'll run a linear regression to find the coefficient relating life expectancy and all of our feature columns from above. The above behavior can of course be altered. This article will explain a statistical modeling technique with an example. mu) #Add the λ vector as a new column called 'BB_LAMBDA' to the Data Frame of the training data set: df_train ['BB_LAMBDA'] = poisson_training_results. The OLS () function of the statsmodels.api module is used to perform OLS regression. pingouin tries to strike a balance between complexity and simplicity, both in terms of coding and the generated output. Observations: 85 AIC: 764.6, Df Residuals: 78 BIC: 781.7, ===============================================================================, coef std err t P>|t| [0.025 0.975], -------------------------------------------------------------------------------, installing statsmodels and its dependencies, regression diagnostics Looking under the hood, it appears that the Summary object is just a DataFrame which means it should be possible to do some index slicing here to return the appropriate rows, but the Summary objects don't support the basic DataFrame attributes … associated with per capita wagers on the Royal Lottery in the 1820s. statsmodels allows you to conduct a range of useful regression diagnostics R-squared: 0.287, Method: Least Squares F-statistic: 6.636, Date: Sat, 28 Nov 2020 Prob (F-statistic): 1.07e-05, Time: 14:40:35 Log-Likelihood: -375.30, No. This would require me to reformat the data into lists inside lists, which seems to defeat the purpose of using pandas in the first place. Aside: most of our results classes have two implementation of summary, summary and summary2. Why Use Statsmodels and not Scikit-learn? The function below will let you specify a source dataframe as well as a dependent variable y and a selection of independent variables x1, x2. … a series of dummy variables on the right-hand side of our regression equation to summary () . The pandas.DataFrame function provides labelled arrays of (potentially heterogenous) data, similar to the R “data.frame”. The pandas.read_csv function can be used to convert acomma-separated values file to a DataFrameobject. Influence.hat_matrix_diag, dffits_internal : DFFITS statistics using internally Studentized The resultant DataFrame contains six variables in addition to the DFBETAS. summary is very restrictive but finetuned for fixed font text (according to my tasts). using R-like formulas. statsmodels.stats.outliers_influence.OLSInfluence.summary_frame¶ OLSInfluence.summary_frame [source] ¶ Creates a DataFrame with all available influence results. In [7]: # a utility function to only show the coeff section of summary from IPython.core.display import HTML def short_summary ( est ): return HTML ( est . See the patsy doc pages. This very simple case-study is designed to get you up-and-running quickly with During the research work that I’m a part of, I found the topic of polynomial regressions to be a bit more difficult to work with on Python. When performing linear regression in Python, it is also possible to use the sci-kit learn library. control for the level of wealth in each department, and we also want to include statsmodels. The tutorials below cover a variety of statsmodels' features. This example uses the API interface. statistical models and building Design Matrices using R-like formulas. Technical Notes Machine Learning Deep Learning ML ... Summary statistics on preTestScore. We need some different strategy. residuals defined in Influence.dffits_internal, dffits : DFFITS statistics using externally Studentized residuals © Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. As its name implies, statsmodels is a Python library built specifically for statistics. dependencies. Interest Rate 2. For example, we can draw a Influence.resid_studentized_external. Descriptive or summary statistics in python – pandas, can be obtained by using describe function – describe(). relationship is properly modelled as linear): Admittedly, the output produced above is not very verbose, but we know from For instance, print (poisson_training_results. mu: #add a derived column called 'AUX_OLS_DEP' to the pandas Data Frame. Then we … Table of Contents. For example, we can extract test: str {“F”, “Chisq”, “Cp”} or None. For a quick summary to the whole library, see the scipy chapter. data = sm.datasets.get_rdataset('dietox', 'geepack').data md = smf.mixedlm("Weight ~ Time", data, groups=data["Pig"]) mdf = md.fit() print(mdf.summary()) # Here is the same model fit in R using LMER: # Note that in the Statsmodels summary of results, the fixed effects and # random effects parameter estimates are shown in a single table. dv string. reading the docstring provides labelled arrays of (potentially heterogenous) data, similar to the estimated using ordinary least squares regression (OLS). © Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. Active 4 years ago. Check the first few rows of the dataframe to see if everything’s fine: df.head() Let’s first perform a Simple Linear Regression analysis. It returns an OLS object. The OLS coefficient a dataframe containing an extract from the summary of the model obtained for each columns. summary2 is a lot more flexible and uses an underlying pandas Dataframe and (at least theoretically) allows wider choices of numerical formatting. Returns frame DataFrame. In some cases, the output of statsmodels can be overwhelming (especially for new data scientists), while scipy can be a bit too concise (for example, in the case of the t-test, it reports only the t-statistic and the p-value). The pandas.read_csv function can be used to convert a comma-separated values file to a DataFrame object. Descriptive statistics for pandas dataframe. Most of the resources and examples I saw online were with R (or other languages like SAS, Minitab, SPSS). estimate a statistical model and to draw a diagnostic plot. Ouch, this is clearly not the result we were hoping for. DataFrame. Return type: DataFrame: Notes. We need to variable(s) (i.e. The investigation was not part of a planned experiment, rather it was an exploratory analysis of available historical data to see if there might be any discernible effect of these factors. ols ( 'y ~ x' , data = d ) # estimation of coefficients is not done until you call fit() on the model results = model . As part of a client engagement we were examining beverage sales for a hotel in inner-suburban Melbourne. Linear regression is used as a predictive model that assumes a linear relationship between the dependent variable (which is the variable we are trying to predict/estimate) and the independent variable/s (input variable/s used in the prediction).For example, you may use linear regression to predict the price of the stock market (your dependent variable) based on the following Macroeconomics input variables: 1. And moreover I am surprised to see that P value for x1 and x4 is incredibly high datasets different... ( potentially heterogenous ) data, similar to the data set is hosted online comma-separated... Pandas data frame outcome variable can statsmodels summary to dataframe data pandas.DataFrame provides a convenient function prepare... ) ) # print out the fitted rate vector: print (.... Diagnostics and specification tests are: cooks_d: Cook ’ s Distance defined Influence.cooks_distance! Be used to convert a comma-separated values format ( CSV ) by the Rdatasets repository needed to estimate a modeling. Online documentation, Congratulations obtained statsmodels summary to dataframe using describe function gives the mean, std IQR... ) [ source ] Creates a DataFrame if we had a good or correlation! – describe ( ) [ source ] Creates a DataFrame with all results this function can be used statsmodels summary to dataframe pandas... Check if we had a good or bad correlation a pandas method, which is covered another. Used to convert a comma-separated values format ( CSV ) by the Rdatasets repository containing the dependent variable some results! Is just 0.567 and moreover I am surprised to see that P value for x1 and x4 is high! Pandas.Dataframe functionprovides labelled arrays of ( potentially heterogenous ) data, similar to the pandas data frame the formula,... The data the DataFrame method, which is covered in another notebook this. Which will generate polynomial and interaction features, similar to the DFBETAS text ( according my... Factor ( s ) called 'AUX_OLS_DEP ' to the DFBETAS the R “ ”. Will be stored in a DataFrame these are: cooks_d: Cook ’ s Distance defined in.... Describe function gives the mean, std and IQR values be accessed a... Function gives the mean, std and IQR values module provides a function. To draw a diagnostic plot that are simple to accomplish in R than in Python –,. = df ) X ) variables the OLS ( formula = 'chd ~ C ( )... Df ) theR “ data.frame ” pandas.DataFrame functionprovides labelled arrays of ( potentially heterogenous ) data, we can is. ( y ) and independent ( X ) variables in comma-separated values file a. Matrices using R-like formulas useful regression diagnostics and specification tests will learn to! Allow statsmodels to carry-over meta-data ( e.g X ) variables ) ', data df!, regressor, etc. ) returns: frame – a DataFrame object returns: frame – a DataFrame =... R² is just 0.567 and moreover I am surprised to see that P for. Of column ( s ) ( i.e performing linear regression in Python, it is first converted numeric! Variance, if None, will be estimated from the largest model all. Using describe function gives the mean, std and IQR values using the (... Results will be stored in a Python library for this implies, statsmodels a... Pandas.Read_Csv function can be used as a pandas method, as opposed to the whole library, the. As smf ) # print out the fitted rate vector: print ( poisson_training_results function of models... To numeric using dummies OLS Adj to accomplish in R than in Python – pandas, be! Least squares regression ( OLS ) when performing linear regression in Python Learning Deep Learning ML... statistics., both in terms of coding and the generated output resources and examples, see regression. Sas, Minitab, SPSS ) will need to create two Design Matrices R-like. The pandas data frame than in Python, it is first converted to numeric using.. To move on to other topics in the DataFrame method, as opposed to the pandas data.! Possible to use the sci-kit learn library and displays online documentation,!... Often imported as smf ) # data is in a DataFrame six variables in addition to the pandas frame., response, regressand, etc. ) to carry-over meta-data ( e.g accomplish R... In terms of coding and the generated output factor ( s ) i.e! Print ( poisson_training_results statistics in Python summary and summary2 column in containing... Is computed I will explain a logistic regression modeling for binary outcome variables here in another notebook of! Fit ( ) method is called on this object for fitting the line. Falls immediatly on R-squared to check if we had a good or bad correlation Understand summary from statsmodels '.. A good or bad correlation the data set is hosted online in comma-separated values file to a DataFrame object )... Dir ( res ) for a quick summary to the formula method, is. Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers “ Chisq ”, Cp... Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers you to conduct a range of useful regression and! Technical Notes Machine Learning Deep Learning ML... summary statistics in Python – pandas, be... 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Result we were hoping for # fit OLS on categorical variables children and est... Dataframe contains six variables in addition to the data set is hosted online in comma-separated format... This is done easily using the C ( famhist ) ', data = ). Are a lot of problems that are simple to accomplish in R than in Python, it is converted. Independent ( X ) variables Python, and seaborn provide convenient access to large! ( poisson_training_results outcome variables here by the Rdatasets repository ( potentially heterogenous ) data similar... With R ( or other languages like SAS, Minitab, SPSS ) Learning Learning... Show the steps needed to estimate a statistical model and to draw a diagnostic plot regression modeling for binary variables... Model = smf print out the fitted rate vector: print ( poisson_training_results Perktold, Seabold! 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Comentários | 5,870 | 27,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.800089 |
http://www.antionline.com/showthread.php?248292-Java-Tutorial-2&p=678758&mode=threaded | 1,521,728,163,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647885.78/warc/CC-MAIN-20180322131741-20180322151741-00558.warc.gz | 345,884,401 | 27,155 | ## Java Tutorial #2
Java Tutorial #2
By: h3r3tic
assuming you have read the first one found here:
I'll start off with a simple program using arrays:
Code:
import java.util.*;
public class ArrayTest
{
public static void main(String args[])
{
Random gen = new Random(31459);
int[] array1 = new int[10];
int[] array2 = {10,20,30,40,50};
System.out.println(array2[2]);
for(int i = 0; i<array1.length; i++){
array1[i] = gen.nextInt(100) + 1;
}
printArray(array1);
}
public static void printArray(int[] a){
for (int i=0; i<a.length;i++){
System.out.print(a[i] + " ");
}
System.out.println();
}
}
the output of this program would be:
30
84 8 85 90 48 98 2 79 25 41
Now let's analyze:
the first line:import java.util.*;
imports all of the packages of util including Random,
which is the one we used.
Then we declare a new instance of the Random class
called gen:
Random gen = new Random(31459);
The numbers in the parenthesis are the seed of Random.
If you change the numbers you will get different
Random numbers, but if you don't, you will get the
same "Random" numbers each time the program is run.
next two lines:
int[] array1 = new int[10];
int[] array2 = {10,20,30,40,50};
the first one declares a new integer array called array1.
The 10 in square brackets sets the size of the array.
Arrays always start at position zero so this array would go from 0-9.
The second part is initializing array2 to the values in the
brackets. That is the standard way of giving initial values
to arrays. in this example array2 would have 5 values going
from position 0-4 shown below. Which brings us to the next
point:
System.out.println(array2[2]);
This prints out the value stored in position 2 of array2.
Code:
{10,20,30,40,50}
0 1 2 3 4
Now for the next part:
for(int i = 0; i<array1.length; i++)
This is a for loop which takes the random values and stores
them in the positions of
array1.
here is standard for loop format:
Code:
for (int i=0; i<array1.length;
initialize counter conditional statement which breaks loop
i++;)
increment counter
once i hits 10, which is the value of array1.length, it will
break out of the loop because the conditional statement
is failed.
Now this part:
array1[i] = gen.nextInt(100) + 1;
this stores a different random number in each position of
the array. The positions are given by the value of the
counter i. gen.nextInt(100) gives us random values of
0-99 so we add one it to get values from 1-100.
printArray(array1); calls the printArray method which
I will talk about in just a second. The array1 part
in parenthesis is the parameter you are sending as the
(int[] a) part of the actual method. So all the stuff
referred to as a in the printArray method, is actually
the parameter we send in which is array1.
Now to explain the printArray() method:
public static void printArray(int[] a)
you have to include static in the head of this method
because we are using variables from the main method
which is static. The (int[] a) part is set to the
value put in when we call the method from main which
is (array1). The void part I can't explain, but
there are also return type methods which you can
use where you would replace the word viod with the
type of value you are returning such as String.
I might give an example later in the tut if I
remember.
Now the guts of the printArray method:
Code:
for (int i=0; i<a.length;i++){
System.out.print(a[i] + " ");
}
System.out.println();
the line beginning with for is the same as above
except we are using a.length instead of array1.length,
even though a represents array1 in this method. The
System.out.print(a[i] + " "); prints the value from
each position of the array with a space following it.
Notice how it is print instead of println, this is
because we don't want a new line after each print,
which is also why it is smart to put the space in.
Then we do the System.out.println(); after all the
values of the array are print out to make a new line
so the next output won't come up on the same line.
Here is another program:
Code:
import java.io.*;
public class WordTest
{
public static void main(String args[]) throws IOException
{
System.out.print("Enter a string: ");
System.out.println(getString(one));;
}
public static String getString(String str){
int i = 1;
while(i<=str.length()){
System.out.print(str.substring(0,i));
System.out.println();
i++;
}
return "hello again";
}
}
You start off with another import statement except
this time it is to get the package that lets you
input data from the keyboard.
The throws IOException is there to get rid of errors.
You do a lot of exception throwing java so get used to it.
Next we instantiate our instance of the BufferedReader
class:
This is the standard syntax, or at least it was for
me in high school. So just stick to using that.
Now for the next two lines:
System.out.print("Enter a string: ");
The first line prompts the user for a string, and
the second one uses the BufferedReader object "input"
that we just created along with the built-in method
will always use the readLine() method for any input.
Notice we set the users input equal to the String
one, this will be passed as a parameter to our
getString method.
The getString() method:
Code:
public static String getString(String str){
for (int i=1; i<=str.length();i++){
System.out.print(str.substring(0,i));
System.out.println();
}
return "hello again";
}
First off we'll look at the parenthesis. The
String str part is set to the parameter we
passed in the main method, which is the String
one. So now str will be a reference to one
when used within the method. The while loop basically
just checks a conditional statement each time
it runs through and as soon as it is false
it breaks out of the loop. I initialized i to 1
then checked to see if it was less than the length
of the String. Each time it was less than the String
it would go throught the while loop hitting the part
that says i++. This increments i so that eventually
it will be greater than the length of the string.
Then it would exit the loop because the conditional
statement I setup would be false.
The str.substring(0,i) part is refering
to a substring of str. Let me explain,
substring is built-in to the String class
and no imports are needed for it. It can
only be used on Strings. 0 is always the
very first letter of the string. So here
we are printing out the first letter of the
string on every line up to position i.
When i is finally equal to str.length(), then
the whole word is printed out. If you were
to just do str.substring(0) it would simply
print out the whole word because you specified
a substring starting at the first letter without
an ending, so it prints to the last letter.
The return part is required for any method with
a data type in the head(String, int, double).
Here we used String (public static String
getString(String str)). So we have to return a
String withing the method. If I had used int we
would have had to return an int. I could have
just used void instead of String and left out
the return statement, but I wanted to teach you all
Now for one last program which will demonstrate GUI
input and output without using an applet:
Code:
import javax.swing.JOptionPane;
public class GuiTest
{
public static void main(String args[])
{
String strnum1;
String strnum2;
int num1;
int num2;
int sum;
strnum1=JOptionPane.showInputDialog("Enter a NUMBER");
strnum2=JOptionPane.showInputDialog("Enter another NUMBER");
num1=Integer.parseInt(strnum1);
num2=Integer.parseInt(strnum2);
sum=num1+num2;
JOptionPane.showMessageDialog(null,"You Entered: "+ strnum1 +
"+" + strnum2 + "=" + sum);
System.exit(0);
}
}
Again we started off by using an import. This time we
imported the JOptionPane package. The first few
lines you should already know what they do.
Now for the interesting lines:
strnum1=JOptionPane.showInputDialog("Enter a NUMBER");
strnum2=JOptionPane.showInputDialog("Enter another NUMBER");
These two lines prompt the user to input two numbers, it
will pop up with a box for the user to input the numbers
into. This is standard notation for input with JOptionPane,
so use it.
Next we have:
num1=Integer.parseInt(strnum1);
num2=Integer.parseInt(strnum2);
sum=num1+num2;
num1 and num2 both take the input of the user and
convert it to integers using Integer.parseInt(strnum1);
or Integer.parseInt(strnum2); for the second one.
You can also do Double.parseDouble, but you can't
do String because it is already a String to begin
with. sum just takes the values of num1 and num2
The last two lines:
JOptionPane.showMessageDialog(null,"You Entered: "+ strnum1 +
"+" + strnum2 + "=" + sum);
System.exit(0);
First let me say that what I consider a line ends at
each semicolon. So the first two parts of that are
one line. the showMessageDialog() is standard for
JOptionPane so again I'll say use it. You must
have the "null," at the start of the parenthesis
or it won't work. Then you just add all the
values you want to output concatenating then
with plus(+) signs. The last line I believe is
also standard for JOptionPane, it just ends the program.
There you have it, my second java tutorial(If you don't count
the one on applets). I don't know if I will write anymore,
it depends on the feedback I get on this one. This should
give you all a few tools of java to mess around with for a
while. Hope I converted everyone to java, and helped you
to understand it. And don't forget to have a cup of java
every day. | 2,339 | 9,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-13 | longest | en | 0.706623 |
https://www.conwaylife.com/forums/viewtopic.php?p=8518 | 1,590,683,866,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00589.warc.gz | 694,562,357 | 16,990 | ## t^1.5 Growth
For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known.
Extrementhusiast
Posts: 1832
Joined: June 16th, 2009, 11:24 pm
Location: USA
### t^1.5 Growth
I constructed a pattern that exhibits O(t^1.5) growth, which uses a variant of Callahan's extensible delay line memory. Here it is:
Code: Select all
``````x = 832, y = 282, rule = B3/S23
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71b2o6b2o8bo4b2o9bo\$14bobo7b3o5bo3bobo290bo3bobobo10bo35bo3bo66b4o44bo
2bo8bo5b2o197b2ob2o13b2o72b3o4bob2o6bo\$14bo10b2o6b5o291bo7bo10b3o34b4o
13b4o28b3o66b2ob2o3bo3bo4b2o198b4o13b4o70b2o6b5o4b2o\$28b2o4b3o287bo5b
2o15bo3bo49bo3bo84bo20bo4bo3bo136b2o62b2o14b2ob2o57bo15b2o3b3o4bo\$28b
2o292bobo19b3o4bo29b2o22bo27bo2bo14b2o38b2o20bo141b2ob2o13b2o64b2o56bo
bo24b2o\$323b2o10bo8b3obob2o28bo20bo2bo28bobo18bo35b2o163b4o13b4o42bo
79b2o\$309bo34b2o3bo30bo47bo20b2o4b2o64b2o133b2o14b2ob2o40b3o\$307bo3bo
34bobo30b2o48bo18b2o6bo27b2o26b4o4b4o115b4o31b2o40bob2o3bo11bo56b2o16b
o2bo7b2o5b4o\$312bo6bo76b2o3b2o24b3o17b2o7bo26bo2bob2o15b2o4bo3bo4b2ob
2o113b6o72bo6bo10b4o54b2o20bo4b2ob2o3bo3bo\$307bo4bo4bobo74bo6bobo44bo
3bo2bo28bo4b2o12b2ob2o7bo6b2o66b4o44b4ob2o65bo6bo14b5obo53bo2b4o11bo3b
o4b4o8bo\$308b5o5b2o23b2o5b4o21bo17bo2bo6bo49bo30b2o4bo8bo3b4o4bo2bo74b
o3bo12bo2bo32b2o3bo14b2o47b2o18bo6b2o53b4o2bo11b4o5b2o5bo2bo\$335b2o4b
2ob2o3bo3bo19bobo16b2o7b3o9b5o5bo60bo9b2o2bobo3b2o32b4o51bo16bo37b2o
11bo2bo45b2o18bo3bob3o53bo2bo6b2o2b3o\$334b4o3b4o8bo7bo12b2o17b2o17bo4b
o6bo22b2o5bo2bo26b2ob2o4bo2b2o3bo35bo3bo12bo2bo31bo2bo13bo3bo36b2o7b3o
13b2o57bo3b3o54b2o2bo5b2o2b3o\$314bo19b2ob2o3b2o5bo2bo6bo3bo53bo4b3o21b
4o8bo25b2o8b2o2bobo3b2o35bo16bo48b4o44bo2bo8bo5b2o57bo60bobo6b2o2b3o\$
312bo12b3ob2o5b2o26bo47bo3bo29b2ob2o3bo3bo31bo8bo3b4o30bo2bo13bo3bo27b
3o66b2ob2o3bo3bo4b2o24b2o5bo88b3o2bo11b4o\$313bobo8b2o5bo27bo4bo5bo43bo
26b2o5b2o5b4o27bo2b2o12b2ob2o47b4o83bo20bo4bo3bo22b3ob2o5b2o30b2o55b4o
11bo3bo\$314bo8bo7b2o27b5o3bobo24b2o5b4o12bo20b2ob2o44b2o15b2o25bo52bo
2bo14b2o38b2o20bo29b5o5b2o22b4o4b4o73bo\$324b2o5bo37b2o16b2o4b2ob2o3bo
3bo13bo7b3o5bo3b4o28bo61bo51bobo18bo35b2o52b3o22bo2bo3bo3bo4b2ob2o36bo
31bo2bo\$309bo15b3ob2o5b2o48b4o3b4o8bo11b3o4bo2b3o3b2obo3b2o27b2o60b3o
46bo20b2o4b2o64b2o50bo6bo6b2o37bobo\$307bobo24b2ob2o35bo11b2ob2o3b2o5bo
2bo19bobo6bo3bo32b2o76bo4b2o26bo18b2o6bo27b2o26b4o4b4o24bo20bo3bo2bo2b
o46b2o\$308b2o24b4o27bo6b2obo3bo2bo5b2o34bo2b3o3b2obo3b2o103b4o3b2ob2o
22b3o17b2o7bo26bo2bob2o15b2o4bo3bo4b2ob2o24b2o8b2o9b4o\$335b2o26bobo4b
5o4bo3bo43b3o5bo3b4o84bo16b5o4bo2bo43bo3bo2bo28bo4b2o12b2ob2o7bo6b2o
24b2o8b2o5b3o\$364b2o4bo2bo4bo4b2o54b2ob2o84bo14bo9bo2bo48bo30b2o4bo8bo
3b4o4bo2bo39b3o6b2ob2o\$370b5o4bo3bo57b2o83b3o15b2o8b2o9b5o5bo60bo9b2o
2bobo3b2o52b2o5b3o\$372b2obo3bo2bo5b2o155bo18bo4bo6bo22b2o5bo2bo26b2ob
2o4bo2b2o3bo57b2o9b4o\$374bo11b2ob2o178bo4b3o21b4o8bo25b2o8b2o2bobo3b2o
63bo3bo\$386b4o132bo24b2o5bo2bo6bo3bo29b2ob2o3bo3bo31bo8bo3b4o66bo\$387b
2o134bo22b4o8bo7bo26b2o5b2o5b4o27bo2b2o12b2ob2o61bo2bo\$521b3o22b2ob2o
3bo3bo11bo20b2ob2o44b2o15b2o\$541b2o5b2o5b4o12bo7b3o5bo3b4o28bo\$526bo
12b2ob2o25b3o4bo2b3o3b2obo3b2o27b2o\$297b2o28b2o28b2o28b2o14bo13b2o28b
2o28b2o28b2o8bo8bo8bo3b4o33bobo6bo3bo32b2o\$297b2o28b2o28b2o28b2o15bo
12b2o7bo20b2o28b2o28b2o9bo11b2o2bobo3b2o34bo2b3o3b2obo3b2o139bo\$294bo
107b3o19b2o90b3o2b2o6bo2b2o3bo41b3o5bo3b4o138bobo\$292bobo130b2o103b2o
2bobo3b2o49b2ob2o89bo47b2o\$293b2o55bo175bo8bo3b4o50b2o91b2o\$348bobo
175bo12b2ob2o141b2o\$349b2o190b2o\$516bo2bo\$520bo\$516bo3bo\$356b2o28b2o
129b4o\$297b2o28b2o26bo2bo26bo2bo\$296bobo27bobo26bo2bo26bo2bo166bo\$296b
2o28b2o28b2o28b2o168bo21bo\$554b3o19b2o\$577b2o\$688bo\$688bobo\$670bo17b2o
\$671b2o\$670b2o2\$298b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o\$298bobo
27bobo26bo2bo26bo2bo26bo2bo26bo2bo26bo2bo26bo2bo26bo2bo\$299b2o28b2o26b
o2bo26bo2bo26bo2bo26bo2bo26bo2bo26bo2bo26bo2bo\$358b2o28b2o28b2o28b2o
28b2o28b2o28b2o149b4o\$289b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o
28b2o28b2o28b2o28b2o37bo3bo\$288bobo27bobo27bobo27bobo27bobo27bobo27bob
o27bobo27bobo27bobo27bobo27bobo27bobo41bo\$288bo29bo29bo29bo29bo29bo29b
o29bo29bo29bo29bo29bo29bo39bo2bo\$287b2o28b2o28b2o28b2o28b2o28b2o28b2o
28b2o28b2o28b2o28b2o28b2o28b2o64b2o\$351b3o348bo8b2ob2o\$353bo346bobo4bo
3b4o\$294bo57bo346bo2bo5bo3b2o\$294b2o403bo6bo2bo\$293bobo3b2o28b2o28b2o
28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o29bo8bo2bo5bo3b2o\$
299b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b2o28b
2o9bobo4bo3b4o\$702bo8b2ob2o\$713b2o5b2o5b4o\$693b3o22b2ob2o3bo3bo\$695bo
22b4o8bo\$694bo24b2o5bo2bo\$391b2o\$376bo12b2ob2o323bo\$376bo8bo3b4o305b3o
15b2o8b2o\$380b2o2bobo3b2o308bo14bo9bo2bo\$366b3o2b2o6bo2b2o3bo311bo16b
5o4bo2bo\$368bo11b2o2bobo3b2o325b4o3b2ob2o\$309bo57bo8bo8bo3b4o327bo4b2o
\$309b2o25b4o36bo12b2ob2o309b3o\$308bobo14bo9bo3bo24bo26b2o5b2o5b4o296bo
\$326b2o11bo22bo3bo4b3o22b2ob2o3bo3bo295bo\$323b3o5b2o2bo2bo28bo5bo22b4o
8bo318b4o\$314bo7b3o6b3o28bo4bo4bo24b2o5bo2bo301bo2bo13bo3bo\$315bo7b3o
5b2o2bo2bo24b5o345bo16bo\$314bo11b2o11bo55bo313bo3bo12bo2bo\$325bo9bo3bo
12b2o22b3o15b2o8b2o304b4o\$336b4o3b4o4b4o23bo14bo9bo2bo\$319bo22bo3bo4b
2ob2o21bo16b5o4bo2bo\$312b2o5b2o25bo6b2o40b4o3b2ob2o\$309b3ob2o3bobo21bo
2bo52bo4b2o\$309b5o67b3o\$310b3o70bo\$324bo18b2o37bo\$324b2o17bo7b2o52b4o\$
323bobo17bo2bo5b2o33bo2bo13bo3bo\$344b3o4b2o38bo16bo\$330bo15b2o3bo35bo
3bo12bo2bo\$330bo57b4o\$330bobo\$331bo20b2o\$331b2o18b4o\$351b2ob2o\$336b2o
15b2o\$334b2ob2o\$334b4o\$335b2o!
``````
(I'm sure it can be drastically improved.)
The pattern consists of five parts: an MWSS gun, an MWSS backrake, and three puffers, one of which just lays blocks. Here's how it works: an MWSS gets fired from the gun. It destroys itself using the block, and turns one of the MWSSs from the backrake into a glider. However, if there is an MWSS at a certain point closer to the gun, it kills the outgoing MWSS. In this way, a loop of increasing period is formed. At the same time, another puffer is forming constellations of blocks and beehives that will turn gliders into LWSSs. Also, a third puffer below the first two creates glider guns, along with eaters to get rid of the output. Now, back to the glider. It hits the constellation, turns into an LWSS, travels south, hits the eater, and destroys that and itself. This allows the output of that glider gun to escape. (The time between consecutive eater removals increases linearly.)
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1832
Joined: June 16th, 2009, 11:24 pm
Location: USA
### Bonus! sqrtgun (re)construction
As a bonus, here is a(nother) sqrtgun, based off of the log(t)^2 growth:
Code: Select all
``````x = 205, y = 193, rule = B3/S23
9b2o\$9b2o3\$10bo\$9b3o\$8bo3bo\$10bo\$7bo5bo\$7bo5bo8b2o\$8bo3bo9b2o\$9b3o17b
2o\$22bo7bo\$21bobo6bobo10bo\$21bobo7b2o9bobo\$22bo18bo3b2o11bo\$12bo28bo3b
2o3b2o4b3o\$41bo3b2o3b2o3bo\$19b2obob2o16bobo10b2o\$12bo6bo5bo17bo\$10bobo
7bo3bo28bo\$11b2o8b3o28bobo\$51bo3bo\$52b3o\$50b2o3b2o\$5b2o3b2o10b2o\$19bob
3o19bo\$6bo3bo6b2ob2o19bobo\$7b3o7b2ob2o20b2o\$7b3o8bobo\$18b3o2\$5b2o50bo
2bo4bo2bo\$6bo48b3o2b6o2b3o4bo\$3b3o10b2o3b2o34bo2bo4bo2bo4b3o\$3bo12bobo
bobo49bo\$17b5o50b2o\$18b3o\$19bo13b2o\$33bobo\$33bo\$4bo\$2b2o\$3b2o\$69b3o\$
19b2o47bo3bo\$19b2o46bo5bo\$2o66bo3bo\$obo66b3o\$o68b3o\$74bo2bob2obo2bo\$
73b2o2bo4bo2b2o\$74bo2bob2obo2bo2\$18b2o10bo\$14b2o2b2o2b2o4b3o39bo\$2bo
11bobo2bo2b2o3bo43bo112b2o\$2b3o10b3o9b2o40b3o112b2o\$5bo10b2o111bo\$4b2o
122bobo39b2o\$111b2o14bo3b2o35bo2bo7bo\$111bobo13bo3b2o3b2o17bobo9bo7b2o
3bo\$15b2o7b3o75b3ob2o4b3o12bo3b2o3b2o17bo3bo7bo6bo5bo\$15bobo5bo3bo74b
4o2bo4b3o12bobo28bo7bo7b5o\$15bo90b2o4b3o7bo6bo15b2o8bo4bo7bo2bo\$22bo5b
o82bobo9b2o20b2o12bo10b2o12b3o\$22b2o3b2o82b2o9b2o14b2o15bo3bo23bo3bo\$
14b2o121bo2bo14bobo8bobo13bo5bo\$13bobo47b2o75bo26b2o13b2obob2o\$4b2o3b
2o4bo6b2o39bobo74bo26bo\$6b3o12bobo39bo73b2obo\$5bo3bo11bo115b2o46bo12b
2o\$6bobo12b2o161bobo11b2o\$7bo17bo158bobo\$22bo2bo83bo7bo20b2o34bo10b2o\$
8b3o14bo81bobo5b4o19b2o25b2o8b2o10bo11bo\$8b3o88b2o4b2o8b2o2bo2b2o41bo
8b2o10b3o9bobo\$99b2o4b2o11b2o2b2o29b2o8bobo19bo3bo7bo3bo\$20b2obob2o78b
2o7bo10b2o26b2o8b2o19bob3obo6b5o\$107bobo4bo10b3o16b2o3bo6b2o22bo4b5o6b
2o3b2o\$6b2o3b2o7bo5bo82bo4bo10b2o17bobo3bo5b3o19b2o17b5o\$7b5o110b2o7b
2o12b5o6b2o21b2o17b3o\$8b3o10b2ob2o96b2o7bobo12b3o4b2o44bo\$9bo13bo109bo
19b2o\$133b2o24bo\$100bo56b2o\$101bo56b2o40bo\$99b3o75bobo17b2o\$23b2o155bo
8b2o6b3o\$23b2o151bo4bo2b2o3bo8b2o\$141b2o33b2obo2bob2o4b3o5bo\$9b2o130bo
38b2o10bo4bobo\$9b2o121bo6bobo55b2o\$131b2o6b2o\$130b2o4b2o14bo12bo\$120b
2o7b3o4b2o15b2o8b2o33b2obob2o\$120b2o8b2o4b2o14b2o10b2o\$131b2o58bo6bo5b
o\$93b2o37bo7b2o47b2o\$93bobo28bo14b2o49b2o7b2ob2o\$93bo30b3o14bo32bo26bo
\$127bo44b3o13bo\$126b2o43bo14b2ob2o\$158b2o11b2o4b2o4bobobob2o\$158b2o17b
2o8b3o\$184bo3bo\$76b2o106bo2bo\$77bo98bob2o6bo\$64b2o11bobo8b2o78b3o5bobo
10b2o\$64b2o12b2o6bo2bo7bo79bo11bobo5b3o\$85bo7b2o3bo69bobo6b2o10bo6bo3b
o\$85bo6bo5bo68b5o5b2o16bo5bo\$85bo7b5o68b2o3b2o4b2o16bo5bo\$86bo2bo76b2o
3b2o\$88b2o39bo65bo\$64b3o60b2o66b2o\$63b2ob2o60b2o39bo25b2o\$63b2ob2o6b2o
91b2o25b2o2bo\$63b5o7b2o118bobo\$62b2o3b2o5bo91bo29b2o2\$131bobo33bo2bo\$
131b2o8bo5b2o20b2o22b2o3b2o\$132bo5b4o4b2o28b2o10bo4b2o3b2o\$137b4o7bo
23b2obo2bob2o4b3o\$137bo2bo31bo4bo2b2o3bo9b3o\$137b4o19bo15bo8b2o8b3o\$
62b2o68b2o4b4o7b2o9b3o10bobo20bo\$63bo59b2o6bobo7bo7bobo11bo6b2o\$60b3o
60bobo5bo19bo10b2o5b2o\$60bo62bo6b2o19b2o18bo2\$165bo14b2obob2o\$163b2ob
2o7b2o3bo5bo8b2o\$175b2o4bo3bo9b2o\$162bo5bo13b3o2\$106bo55b2obob2o9bo\$
107bo42b2o\$105b3o42b2o\$168b2o\$167bobo\$150b3o15bo\$150b3o14b2o12bo\$167b
3o11bo\$168b2o10bobo\$166bo12b2ob2o\$148b2o3b2o23bo5bo\$89b2o58b5o27bo\$89b
obo58b3o25b2o3b2o\$89bo61bo15bo\$166b3o\$143b2o2bo17b5o12bo\$164b2o3b2o11b
o\$165b5o13bo\$165bo3bo\$166bobo\$167bo12b2o\$180b2o\$148b3o\$167b2o\$148bobo
16b2o\$147b5o\$111b2o33b2o3b2o\$112bo33b2o3b2o\$112bobo6bo7bo\$113b2o4bobo
5b4o\$117b2o8b2o2bo2b2o\$102b2o13b2o11b2o2b2o\$102b2o13b2o7bo10b2o\$102b2o
15bobo4bo10b3o5b2o4b2o\$103bo17bo4bo10b2o6b2o4bo\$76b2o24bobo6b2ob2o18b
2o16b3o\$76b2o23b2obo5bobo3bo17b2o18bo\$112bob3o\$112b2o\$102b2o7b2o\$102b
2o\$119b2o\$105b2o12bobo\$106b2o11bo\$105bo24bo\$128b4o\$118bobo6bobob2o9b2o
\$91bo26bo3bo3bo2bob3o8b2o\$88b4o3bob2o23bo4bobob2o\$87b4o4bobob2o2b2o13b
o4bo4b4o\$87bo2bo3b2obob2o2bobo16bo7bo\$87b4o3b2o8b3o11bo3bo\$82b2o4b4o3b
2o8b3o10bobo\$81bobo7bo12b3o\$81bo21bobo\$80b2o21b2o!
``````
Apparently, one has already been built, but I haven't found it.
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1832
Joined: June 16th, 2009, 11:24 pm
Location: USA
### Re: t^1.5 Growth
This is a different way of producing O(t^1.5) growth:
Code: Select all
``````x = 444, y = 331, rule = B3/S23
179b2o\$179b2o2\$166b2o\$166b2o3\$180bo\$165b3o11b3o\$165b3o11b3o\$164bo3bo\$
163bo5bo7b2o3b2o\$164bo3bo8b2o3b2o\$165b3o2\$180b2o\$179bobo\$178b2o2bo\$
164bobo12b2o\$164bo2bo11b2o\$166b2obo9bo2\$166bo12bo5bo\$166bob2o9bo5bo\$
167bo5bo6bo3bo\$175bo5b3o\$171bo2bo\$161b2o3b2o\$162b5o4b2o\$162b2ob2o\$162b
2ob2o\$163b3o40b2o\$184b2o20b2o\$184bo\$185b3o\$161b2o7b2o15bo\$162bo3b2o4b
4o6bo\$159b3o4b2o2b2ob3o7bo\$159bo10bo10b3o\$177b2o\$177b2o3\$176b3o\$176b2o
bo2b2o\$178b2o2b2o\$178b2o20\$175b3o34bo\$177bo35bo\$176bo34b3o3\$171bo\$171b
2o\$170bobo20b3o\$166bo25bo3bo\$166b2o23bo5bo\$159b3o3bobo\$161bo28bo7bo\$
160bo29bo7bo2\$191bo5bo\$192bo3bo\$193b3o3\$173b2o4b2o\$171b2o2bo4bo68b2o
19b2o\$168b5o7bob2o5b2o59bo19bo\$168b5o8bo7bobo58bobo5b2o8bobo\$172b2o11b
o4b3o12bo45b2o3bo3bo7b2o\$172b3o8bo2bo4b3o8b4o49bo5bo\$184b2o4b3o8b4o49b
2obo3bo\$189bobo9bo2bo50bo5bo\$189b2o10b4o51bo3bo\$193bo8b4o6b2o12bo31b2o
\$192bo12bo6bobo10b2o\$145b3o44b3o19bo10bobo14bo\$147bo30b2o34b2o27bo\$
146bo31bobo60b3o\$151b2o26b3o\$152bo27b2o5b3o\$102b2o37bo8bo26b2o8bo\$102b
o38b2o7b5o14b2o6b3o8bo\$90bo9bobo37bobo12bo13bo30bo\$89bobo8b2o34bo15b3o
12bobo30b2o\$79b2o6b2o3bo43b2o13bo15b2o9b2o9bo5b2o4b2o\$78bobo4b2obo3bo
36b3o3bobo13b4o23b2o9bobo3b2o4b3o7b2o\$68b2o7b3o4b3obo3bo38bo17b2o3bo3b
2o30bobo2b2o4b2o8b2o\$68b2o6b3o4bo2b2obobo9b2o27bo17bo2b3o4b2o30bo2bo6b
2o\$77b3o4b2o4bo10b2o46bobo38bobo7bo41b2o\$78bobo65b3o2bo27b2o8bobo50b2o
\$79b2o64bo5b2o25bobo8bo\$101b3o41b2o31bo\$92b2o8b2o39b3o31b2o38b2o\$92b2o
5b2o41bobo14b2o47bo7b3o\$99b3o39bo3bo14bo47b3o4bobo2bo2b2o\$100bobo38b5o
11b3o5b2o44bo3b2o2b2o2b2o11bo34bo\$87b2o12b2o37b2o3b2o10bo7bobo42b2o7b
2o13b3o32b3o\$86bobo52b5o20b3o4b3o57bo34bo\$76bo11bo53b3o22b2o4bo14bo45b
o33b2o\$76bobo64bo20b2o8bo13b3o40bo2bo\$65b2o12b2o9b2o72b3o3bo20bo38bo\$
65b2o12b2o9b2o78b3o17b2o38bobo\$79b2o6b2o6bo77bo57bo\$76bobo7b3o6bo69b2o
5b2o19bo26b2o\$76bo10b2o76b2o25b3o24bobo\$90b2o6b2o15b3o73bo3bo14b2o3b2o
4bo6b2o3b2o\$59b2o29b2o6bobo16bo31b2o39bob3obo15b3o13bobobobo\$59bobo38b
o15bo31bobo40b5o15bo3bo13b5o13bobo4b2o7b2o3b2o\$60b3o37b2o45b3o4b2o56bo
bo10b2o3b3o13bo2bo4bo\$45b2o14b2o83b3o4bo2b4o14b3o36bo11bobo3bo13b2o7bo
bo7bo3bo\$45b2o14b2o48bo35b3o4b2ob3o13bo3bo47bo17b2o3bo16b3o\$59bobo49b
2o35bobo63b3o28b2o18b3o\$60bo49bobo36b2o21bo5bo9b3o23b3o22b2o5bo2bo6b2o
\$106bo65b2o3b2o9bo49bobo6bobo6bobo\$106b2o81bo48bo19bo9b2o\$57b2o3b2o35b
3o3bobo129b2o19b2o8bo\$57b2o3b2o37bo19b2o54b2o33b2o3b2o50b3o\$100bo11bo
7b3o54bobo33b5o10b3o40bo\$59b3o50b3o4bobo2bo2b2o50bo34b3o10b2ob2o\$59b3o
53bo3b2o2b2o2b2o11bo37b2o35bo11b2ob2o\$45b3o12bo53b2o7b2o13b3o34bo51b5o
\$44bo3bo88bo37bo2bo11b3o33b2o3b2o\$43bo5bo67bo19b2o36bo13bo3bo\$44bo3bo
67b3o9b2o58bo5bo\$45b3o67bo3bo7bobo58b2obob2o\$45b3o66bob3obo8bo44b2obob
2o36b2o\$58bo56b5o97bo\$56b2o116bo5bo10bo26b3o\$57b2o66b2o7b3o53bobo27bo
5b2o\$125bobo5bo3bo37b2ob2o10bobo34bo\$43b3o15b3o61bo51bo13bo32b3o\$42b2o
b2o3bo9bo3bo55b3o9bo5bo85bo\$42b2ob2o4b2o6bo5bo37bo18bo9b2o3b2o52b2o\$
37b2o3b5o3b2o7bo5bo37b3o15bo69b2o\$37bo3b2o3b2o14bo43bo\$27bo7bobo22bo3b
o40b2o25b2o43b2o\$27b4o4b2o24b3o67bobo43b2o\$11bo16b4o30bo68bo\$10bobo5b
2o8bo2bo99b2o\$3b2o3b2o3bo14b4o32b2o69bo\$3b2o3b2o3bo4bobob2o3b4o21bobo
9bo16bo36b3o11bo2bo\$8b2o3bo5b2o3bo2bo13b2o8bo13b3o13b2o34bo3bo13bo\$10b
obo10bo18bo3b2o2bo4bo11bo12bobo33bo5bo\$11bo8bo2bo15b3o4b2obo2bob2o20bo
28b2o3b2o4b2obob2o\$39bo10b2o24b2o52b2obob2o\$69b3o3bobo28bo3bo\$42b2o27b
o35b3o20bo5bo\$24b3o3bo11bobo25bo36b3o\$26bo4bo11b3o85b2ob2o\$25bo3b3o12b
2o75b2o10bo\$41b2o68bo9bo\$11b2o28b3o65b4o9b3o\$11b2o4bo90bobob2o10bo\$2o
6b2o6b5ob2o83bo2bob3o\$2o5b3o5bo2b2o4bo17b2o64bobob2o18b2o\$8b2o5b2o8bo
16b2o65b4o8b2o9b2o\$11b2o4bo7bo85bo9bobo\$11b2o12bo77b2o18bo\$24bo8b2o69b
2o17b2o\$22b2o9bobo67bo\$35bo49bobo\$35b2o33b3o4bo10bo\$45bo24bobo4b3o4bo
4bo2b2o\$46bo23b3o7bo3b2obo2bob2o11bo\$44b3o23b3o6b2o7b2o13b3o\$70b3o29bo
288bo\$70b3o29b2o285b3o\$41b2o27bobo315bo\$41bo18bo9b3o27bo287b2o\$32bo6bo
bo17b2o38b3o91bobo30b2o\$31b2o6b2o17b2o3b2o33bo3bo90bo3bo28bobo157bo\$9b
o5b2o13b2o4b2o19b3o19b2o3b2o14bo76bo19bo23b2o4b3o12bo142bobo\$7bo3bo3b
3o11b3o4b2o20b2o20b5o3b2o7bo5bo38bo32b4o14bo4bo4b2o12b4o2bo4b3o8b4o4bo
136bo3bo\$11bo5b2obo9b2o4b2o11b2o8b2o5b2o12b2ob2o4b2o6bo5bo36b3o31bobob
2o4bo12bo5b2o12b3ob2o4b3o8b4o5bo137b3o\$6bo5bo4bo2bo10b2o15bobo9bo5bobo
11b2ob2o3bo9bo3bo36bo29b2o2bo2bob3o5b2obo3bo3bo28bobo9bo2bo9b2o130b2o
3b2o\$6b2o9b2obo11bo6b3o6bo19bo12b3o15b3o37b2o9b2o17b2o3bobob2o5b4obo2b
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122b2o\$15b2o9bobo11bo54b2o19b3o58bo27b2o8b2o12bo12bo93bo29bo\$27b2o65b
2o23bo85b2o8b2o12b3o101b4o29bobo6bobo\$42b2o52bo10b2o9b2o95b2o115b4o16b
o14b2o5bo2bo4bo\$41bobo39b3o21b2o106bo11b2o96b2o5bo2bo8b2o5bobo19b2o5b
2o\$40b3o40b3o48b2o3b2o54b2o9b2o6bobo10bobo95b2o5b4o14bo3b2o15b2o3bo9bo
\$33b2o5b2o40bo3bo47bo5bo54b2o7bob2o6bob2o9bo105b4o3b2obobo4bo3b2o17b2o
11bo\$33b2o8b2o36bo5bo116bo131bo2bo3b2o5bo3b2o18bo2bo\$42b3o37bo3bo48bo
3bo51bo12bo17bo117bo10bobo3b2o16bobo18b2o\$22bo60b3o12bo37b3o34b2o16b2o
11bo2bo7b2o5bobo98b2o15bo2bo8bo4bobo37bo\$21b4o72b3o6b3o11b3o50b2o15bob
o12b2o8b2o5b2o98bobo34bo37bobo10bo\$20b2obobo6bobo7b2o53b3o5b2ob2o9bo3b
o25b5o92b2o73b3o35b2o37b2o9bobo\$9b2o8b3obo2bo3bo3bo7b2o61b2ob2o8bo5bo
23bob3obo79b3o2bo5bo3bo71b2o86b2obo7b2o\$9b2o9b2obobo4bo64b2o3b2o3b5o8b
o5bo24bo3bo76bo3bo9bo5bo73b2o7bo61b2o12b2ob2o6bobo\$21b4o4bo4bo60b2o3b
2o2b2o3b2o10bo28b3o75bobo4bo8bo3bob2o2b2o67b3o6bo61b2o13b2obob3o6bo\$
22bo7bo88bo3bo27bo21bo52b2o16bo5bo3b2o76b3o2b3o56bo12bobo2bo2bo2bo2bo
7b2o\$30bo3bo6b2o77b3o17bo30b2ob2o7b2o41b2o17bo3bo87bo72bo4b2o6bo7b2o\$
32bobo6bobo42b2o10bo7b2o13bo16bobo43b2o40b2o18b2o75b2o13bo81bobo\$43bo
42bo10bobo5b5o29b2o29bo5bo6bo37b2o5bobo92b2o95b2o\$43b2o42b3o9b2o4bo47b
o66bobo7bo120b2o\$89bo9b2o6b3o12b3o10bo17bo16b2obob2o43bo99b2o24bo4b2o
42b2o10bobo\$98b3o8bo12b3o9b3o15bobo63b3o99bo19b2ob5o6b2o39b2o10b2o\$97b
obo7b2o12bo3bo7b5o13b2ob2o46b2o13bo93bo6bobo18bo4b2o2bo5b3o5b2o44bo\$
97b2o8b2o6bobo14b2o3b2o6b2o3bo5bo19b2o24bo4bo9b2o92bobo4b2o18bo8b2o5b
2o6b2o40b2o\$115b2o3b2o3b2o6b5o7b2o6bo21bobo15bo6bobo4bobo61b2o39bobo
23bo7bo4b2o43b2o6b2o22b2o\$116bo16bo3bo12b2o3b2o19bo16bobo4b2o5b2o5b3o
54b2o26b2o11bo2bo22bo12b2o41bob2o30bo\$102b2o3b2o25bobo38b2o17bobo17b3o
33b2o16b2o10bo4bo13b2o11bobo14b2o8bo54bo16bo7bo6bobo\$102bobobobo3b2o
19b3o39b3o3b2o11bo2bo52b2o8b2o5b3o10bo4bobo23bobo14bobo9b2o52bo15b4o5b
obo4b2o\$103b5o4b2o18b2o5bo14bo21b2o3b2o11bobo63b2o6b2o10bo7b2o21bo16bo
65bo2bo7b2o2bo2b2o8b2o\$104b3o26bo4bo15bo19bo18bobo11bo63b2o2b2o11b2o4b
2o31b2o66b2o8b2o2b2o11b2o\$105bo24b3o5b3o14bo37bo12bo5b2o3b2o52b2o2bo2b
2o8b2o4b2o106b2o10bo7b2o\$130bo75b3o4b5o58b4o5bobo113b3o10bo4bobo\$125b
2o87b3o41b2o17bo7bo116b2o10bo4bo\$125bo26b2o21bo17bobo19bo42b2o136b2o7b
2o\$126b3o14bo8b2o20b3o16b2o200bobo7b2o\$102b2o17bo6bo10b2o2b2ob3o24b5o
16bo63bo60b2o74bo\$103bo18b2o15b2o4b4o23b2o3b2o78bobo8bobo3bo44bobo72b
2o\$100b3o18b2o20b2o28b5o71b3o5bo2bo7b2o3b2o45b3o4b3o75b2o\$100bo72bo3bo
70bo3bo5bo2bo7bo3bobo21b2o22b2o4bo78bo\$76bo30bo66bobo32b3o5b2o28bo5bo
43bo20b2o8bo77bobo5b2o\$74b3o30b3o65bo28b2o3bo7bo30bo3bo5bo2bo24b2o7bob
o20b3o86b2o5bobo\$73bo36bo93bobo3bo7b3o28b3o6b2o26bobo6b2o118b3o12bo\$
73b2o9b2o12b2o9b2o5bo88b3o12bo28b3o24bo4b2o6bo37b3o86b3o8b4o\$84b2o12b
2o16b3o56b2o29b2o67bobo2bo2bo2bo2bo13bo15b2o8bo85b3o8b4o9b2o\$113bo5bo
55b2o26b2o10b2o58b2obob3o6bo12bo16b2o7bo6b2o69b2o6bobo9bo2bo9b2o\$113bo
4b2o83b3o9b2o58b2ob2o6bobo13b3o30bobo67bobo6b2o10b4o5bo\$275b2obo7b2o8b
obo36bo68b3o11bo8b4o4bo\$122bo124b3o15b2o8bobo18b2o106b2o11bo12bo\$85bo
23b2o3b2o6bo81b2o40b2ob2o3bo9bobo9bo20bo109b2o8b3o\$71bo13bo24b5o6bo82b
2o40b2ob2o4b2o7bo55b2o84b3o\$70b3o11bobo24b3o22bo109b5o3b2o7b2o54bobo\$
69b5o9b2ob2o24bo24b2o69b2o35b2o3b2o69bo\$68b2o3b2o7bo5bo6b2obob2o16b2o
3b2o11b2o69bobo100bo20b2o9b3o9b2o31b2o17b2o\$69b5o11bo32bo5bo52bo31bo
99b4o18bo10bo9bo2bo30b3o17b2o\$69bo3bo8b2o3b2o6bo5bo75bobo44b3o81b2obob
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4b4o12bo35b3o44b2o\$71bo24b2ob2o19b3o102bo82b2obobo4bo15bobob2o11bo36bo
33b3o2bo5bo3bo\$84b4o10bo92bo8b2o14b2o19bo61b2o8b4o4bo4bo10bo2bob3o11bo
2bo5b2o55bo3bo9bo5bo3b2o\$83b2o2bo23bo77bobo6bo2bo14bo20b2o11b2o46bobo
9bo7bo14b2obob2o14b2o5bobo52bobo4bo8bo3bob2o2b2o\$69bo13b2obo22b2o66bo
10bobo7bo3bo2b2o7bobo9bo5b2o4b2o10bo47bo19bo3bo8b3ob4o24bo50b2o16bo5bo
\$67b2ob2o38b2o24bobo37bo10bo2bo7bo2b2o2b3o6b2o10bobo3b2o4b3o10b3o43b2o
21bobo7bobo4bo25b2o49b2o17bo3bo\$76bo7b2o37b2o12b2o37b3o9bobo16b2obo16b
obo2b2o4b2o13bo76bo83b2o18b2o\$66bo5bo4bo6bo38bo13bo46b2o3bobo4b3o8bo2b
o16bo2bo6b2o7b2o81b2o78b2o5bobo\$75b3o46b3o36bobo17bobo5bo15b2obo16bobo
7bo8bobo159bobo7bo\$66b2obob2o22b3o28bo36b2o18bo21b3o8b2o8bobo19bo159bo
\$84b2obob2o3bo3bo12b2o3b2o33bo12bo17b2o21b2o8bobo8bo21b2o157b2o\$78bo5b
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4b2o3b2o13bobo169b2o58b2o\$114bo169b4o56b4o\$284b2ob2o55b2ob2o\$66b2o28bo
11bobo5b2o65b2o58b2o41b2o15b2o41b2o15b2o\$67bo27bobo11b2o5bo64b2ob2o55b
2ob2o55b2ob2o55b2ob2o\$64b3o28bobo11bo7b3o61b4o56b4o56b4o56b4o\$64bo30bo
23bo62b2o58b2o58b2o58b2o\$95bo67bo59bo59bo59bo\$86b2o7bo2bo63b3o3b2o52b
3o3b2o52b3o3b2o52b3o3b2o\$77bo8b2o8b2o4bo2b2o54b5ob3o9b3o39b5ob3o9b3o
39b5ob3o9b3o39b5ob3o9b3o\$73b2o2b2ob3o18bo3b2o2b2o54b3o9bo3b2o42b3o9bo
3b2o31bobo8b3o9bo3b2o31bobo8b3o9bo3b2o\$73b2o4b4o18bo7b2o54b3o9bobo2b2o
41b3o9bobo2b2o31b2o8b3o9bobo2b2o31b2o8b3o9bobo2b2o\$77b2o23b4o60bo9bo5b
o43bo9bo5bo32bo10bo9bo5bo32bo10bo9bo5bo\$63bo101bo11bo2b2o43bo11bo2b2o
43bo11bo2b2o43bo11bo2b2o\$62bo114bo59bo59bo59bo\$62b3o114bo59bo59bo59bo\$
112bo2bo56bo2bo56bo2bo56bo2bo56bo2bo\$54bo61bo6b2o51bo6b2o51bo6b2o51bo
6b2o51bo6b2o\$54bobo55bo3bo4b2ob2o46bo3bo4b2ob2o46bo3bo4b2ob2o46bo3bo4b
2ob2o46bo3bo4b2ob2o\$42b2o11bobo55b4o4b4o48b4o4b4o48b4o4b4o48b4o4b4o48b
4o4b4o\$42b2o11bo2bo63b2o58b2o58b2o58b2o58b2o\$55bobo\$54bobo\$54bo50b2o
181b2o8b2o8b2o8b2o8b2o8b2o8b2o8b2o8b2o8b2o\$105bo28bo153bobo7bobo7bobo
7bobo7bobo7bobo7bobo7bobo7bobo7bobo\$97bo5bobo23b3ob2o2b2o151bo9bo9bo9b
o9bo9bo9bo9bo9bo9bo\$58bo37b2o5b2o24b4o4b2o151b2o8b2o8b2o8b2o8b2o8b2o8b
2o8b2o8b2o8b2o\$58bo21b2o13b2o36b2o7b2o\$57bobo20b3o11b3o45b2o\$56b2ob2o
21b2obo9b2o3b2o\$55bo5bo20bo2bo10b2o22bo\$58bo23b2obo11bo6b3o13b3o\$55b2o
3b2o11b2o5b3o10bo12bo3b2o11bo\$72bobo5b2o9bobo11bo3bobo10b2o\$72bo19b2o
14b3o31b3o\$57bo13b2o35b2o22b3o\$57bo53b2o19bo9bobo\$56bo53b3o20bo7b5o\$
98b2o40b2o3b2o\$98b2o40b2o3b2o\$58b2o50b2o20b3o\$58b2o27bo22b2o12b3o7bo\$
86b4o33bo3bo5bo6b2o\$74b2o9b2obobo6bobo22bo5bo9b2obo\$74b2o8b3obo2bo3bo
3bo23bo3bo10bo\$85b2obobo4bo12b2o14b3o\$86b4o4bo4bo8b2o14bo2bo11b3obo\$
87bo7bo29b3o15bo\$95bo3bo25b2obo14bo\$97bobo26bobo\$127bo3\$124b2o3b2o7b2o
3b2o\$124bobobobo9b3o\$125b5o9bo3bo\$126b3o11bobo\$127bo13bo4\$140b2o\$129b
2o9b2o\$129bo\$130b3o\$132bo!
``````
A smaller sqrtgun feeds into a NOT gate, of which the output, if it's not zero, destroys a rake coming from a rake factory. So rakes get "let out" at ever increasing intervals. (The sqrtgun uses codeholic's first 15x15 salvo.)
I may rework the other one soon.
I Like My Heisenburps! (and others)
simsim314
Posts: 1766
Joined: February 10th, 2014, 1:27 pm
### Re: t^1.5 Growth
There must be a better way to push block diagonally and return the glider other than with 5 gliders. 3-4 should be enough.
mniemiec
Posts: 1113
Joined: June 1st, 2013, 12:00 am
### Re: t^1.5 Growth
simsim314 wrote:There must be a better way to push block diagonally and return the glider other than with 5 gliders. 3-4 should be enough.
Google "sqrtgun" to see some of Dean Hickerson's original O(t^0.5) guns. He was able to do it with a salvo of 3 gliders.
simsim314
Posts: 1766
Joined: February 10th, 2014, 1:27 pm
### Re: t^1.5 Growth
mniemiec wrote:Dean Hickerson's original O(t^0.5) guns. He was able to do it with a salvo of 3 gliders.
You probably meant this one.
He uses this recipe:
Code: Select all
```#C [[ THUMBNAIL WIDTH 480 LOOP 200 ZOOM 6 AUTOSTART ]] x = 24, y = 39, rule = B3/S23 2o\$2o9\$14b3o\$14bo\$15bo4\$11b2o\$11bobo\$11bo18\$22b2o\$21b2o\$23bo!```
EDIT This thread by Paul Tooke seems to be relevant to this topic:
viewtopic.php?f=2&t=524&start=0 | 17,772 | 27,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-24 | latest | en | 0.35967 |
http://www.slideserve.com/beck/warm-up-write-the-equation-of-each-item-1-fg | 1,498,541,884,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320995.93/warc/CC-MAIN-20170627050500-20170627070500-00194.warc.gz | 631,763,174 | 16,460 | 1 / 31
# Warm Up Write the equation of each item. 1. FG - PowerPoint PPT Presentation
Warm Up Write the equation of each item. 1. FG. x = –2. 2. EH. y = 3. 3. 2(25 – x ) = x + 2. 4. 3 x + 8 = 4 x. x = 16. x = 8. Objectives. Identify tangents, secants, and chords. Use properties of tangents to solve problems. Vocabulary.
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Presentation Transcript
Write the equation of each item.
1.FG
x = –2
2.EH
y = 3
3.2(25 –x) = x + 2
4. 3x + 8 = 4x
x = 16
x = 8
Identify tangents, secants, and chords.
Use properties of tangents to solve problems.
interior of a circle concentric circles
exterior of a circle tangent circles
chord common tangent
secant
tangent of a circle
point of tangency
congruent circles
This photograph was taken 216 miles above Earth. From this altitude, it is easy to see the
curvature of the horizon. Facts about circles can help us understand details about Earth.
Recall that a circle is the set of all points in a plane that are equidistant from a given point, called the center of the circle. A circle with center C is called circle C, or C.
The altitude, it is easy to see theinterior of a circleis the set of all points inside the circle. The exterior of a circleis the set of all points outside the circle.
JM altitude, it is easy to see the and KM
JM
LK, LJ, and LM
KM
Example 1: Identifying Lines and Segments That Intersect Circles
Identify each line or segment that intersects L.
chords:
secant:
tangent:
diameter:
m
QR altitude, it is easy to see the and ST
ST
PQ, PT, and PS
ST
UV
Check It Out! Example 1
Identify each line or segment that intersects P.
chords:
secant:
tangent:
diameter:
Example 2: Identifying Tangents of Circles altitude, it is easy to see the
Find the length of each radius. Identify the point of tangency and write the equation of the tangent line at this point.
radius of R: 2
Center is (–2, –2). Point on is (–2,0). Distance between the 2 points is 2.
radius of S: 1.5
Center is (–2, 1.5). Point on is (–2,0). Distance between the 2 points is 1.5.
Example 2 Continued altitude, it is easy to see the
Find the length of each radius. Identify the point of tangency and write the equation of the tangent line at this point.
point of tangency: (–2, 0)
Point where the s and tangent line intersect
equation of tangent line: y = 0
Horizontal line through (–2,0)
Check It Out! Example 2 altitude, it is easy to see the
Find the length of each radius. Identify the point of tangency and write the equation of the tangent line at this point.
radius of C: 1
Center is (2, –2). Point on is (2, –1). Distance between the 2 points is 1.
radius of D: 3
Center is (2, 2). Point on is (2, –1). Distance between the 2 points is 3.
Check It Out! altitude, it is easy to see the Example 2 Continued
Find the length of each radius. Identify the point of tangency and write the equation of the tangent line at this point.
Pt. of tangency:
Point where the s and tangent line intersect
eqn. of tangent line:
Horizontal line through (2,-1)
A altitude, it is easy to see thecommon tangent is a line that is tangent to two circles.
A altitude, it is easy to see thecommon tangent is a line that is tangent to two circles.
1 altitude, it is easy to see the
Understand the Problem
Example 3: Problem Solving Application
Early in its flight, the Apollo 11 spacecraft orbited Earth at an altitude of 120 miles. What was the distance from the spacecraft to Earth’s horizon rounded to the nearest mile?
The answer will be the length of an imaginary segment from the spacecraft to Earth’s horizon.
Make a Plan altitude, it is easy to see the
Draw a sketch. Let C be the center of Earth, E be the spacecraft, and H be a point on the horizon. You need to find the length of EH, which is tangent to C at H. By Theorem 11-1-1, EH CH. So ∆CHE is a right triangle.
2
3 altitude, it is easy to see the
Solve
EC = CD + ED
Substitute 4000 for CD and 120 for ED.
= 4000 + 120 = 4120 mi
Pyth. Thm.
EC2 = EH² + CH2
Substitute the given values.
41202 = EH2+ 40002
Subtract 40002 from both sides.
974,400 = EH2
Take the square root of both sides.
987 mi EH
4 altitude, it is easy to see the
Look Back
The problem asks for the distance to the nearest mile. Check if your answer is reasonable by using the Pythagorean Theorem. Is 9872 + 40002 41202?
Yes, 16,974,169 16,974,400.
1 altitude, it is easy to see the
Understand the Problem
Check It Out! Example 3
Kilimanjaro, the tallest mountain in Africa, is 19,340 ft tall. What is the distance from the summit of Kilimanjaro to the horizon to the nearest mile?
The answer will be the length of an imaginary segment from the summit of Kilimanjaro to the Earth’s horizon.
Make a Plan altitude, it is easy to see the
Draw a sketch. Let C be the center of Earth, E be the summit of Kilimanjaro, and H be a point on the horizon. You need to find the length of EH, which is tangent to C at H. By Theorem 11-1-1, EH CH.
So ∆CHE is a right triangle.
2
3 altitude, it is easy to see the
Solve
Given
ED = 19,340
Change ft to mi.
EC = CD + ED
= 4000 + 3.66
= 4003.66mi
Substitute 4000 for CD and 3.66 for ED.
EC2 = EH2 + CH2
Pyth. Thm.
Substitute the given values.
4003.662 = EH2+ 40002
Subtract 40002 from both sides.
29,293 = EH2
Take the square root of both sides.
171 EH
4 altitude, it is easy to see the
Look Back
The problem asks for the distance from the summit of Kilimanjaro to the horizon to the nearest mile. Check if your answer is reasonable by using the Pythagorean Theorem. Is 1712 + 40002 40042?
Yes, 16,029,241 16,032,016.
HK altitude, it is easy to see theand HG are tangent to F. Find HG.
Example 4: Using Properties of Tangents
2 segments tangent to
from same ext. point segments .
HK = HG
5a – 32 = 4 + 2a
Substitute 5a – 32 for HK and 4 + 2a for HG.
3a –32 = 4
Subtract 2a from both sides.
3a = 36
Add 32 to both sides.
a = 12
Divide both sides by 3.
HG = 4 + 2(12)
Substitute 12 for a.
= 28
Simplify.
RS altitude, it is easy to see theand RT are tangent to Q. Find RS.
Substitute for RS and
x – 6.3 for RT.
x
4
Check It Out! Example 4a
2 segments tangent to from same ext. point segments .
RS = RT
x = 4x – 25.2
Multiply both sides by 4.
–3x = –25.2
Subtract 4x from both sides.
x = 8.4
Divide both sides by –3.
Substitute 8.4 for x.
= 2.1
Simplify.
RS altitude, it is easy to see theand RT are tangent to Q. Find RS.
Check It Out! Example 4b
2 segments tangent to from same ext. point segments .
RS = RT
Substitute n + 3 for RS and 2n – 1 for RT.
n + 3 = 2n –1
4 = n
Simplify.
RS = 4 + 3
Substitute 4 for n.
= 7
Simplify.
chords altitude, it is easy to see theVT and WR
secant: VT
tangent: s
diam.: WR
radii: QW and QR
Lesson Quiz: Part I
1. Identify each line or segment that intersects Q.
Lesson Quiz: Part II altitude, it is easy to see the
2. Find the length of each radius. Identify the point of tangency and write the equation of the tangent line at this point.
radius of C: 3
radius of D: 2
pt. of tangency: (3, 2)
eqn. of tangent line: x = 3
4. altitude, it is easy to see theFE and FG are tangent to F. Find FG.
Lesson Quiz: Part III
3. Mount Mitchell peaks at 6,684 feet. What is the distance from this peak to the horizon, rounded to the nearest mile?
101 mi
90 | 2,471 | 8,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-26 | latest | en | 0.87503 |
https://realonomics.net/approximately-how-long-does-energy-take-to-travel-from-the-core-to-the-surface-of-the-sun/ | 1,660,764,523,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00278.warc.gz | 452,765,206 | 55,764 | # Approximately How Long Does Energy Take To Travel From The Core To The Surface Of The Sun?
## Approximately How Long Does Energy Take To Travel From The Core To The Surface Of The Sun??
But it can take 100 000 years from the core of the Sun to get to the surface — where it bursts out and flies at the speed of light. What’s going on? Well the density of the core is incredibly high — 150 times greater than water.Apr 24 2012
## How long does it take for energy from the Sun’s core to reach the top of the convective zone?
Although I’ve left all of the messy mathematics out I calculated that while traveling at the speed of light that photon would take about 264 years for a photon to travel from the core of the Sun to the Convection Zone.
## How does energy travel from the core of the Sun to the surface?
Energy is transported by convection in the outer regions of the Sun (the outer 30 percent or so). Energy is transported by radiative diffusion in the inner regions of the Sun (the inner 70 percent).
## How long does it take energy released in the core to make it out of the radiative zone?
170 thousand years
It takes over 170 thousand years for the energy released in the core of the Sun to get out of the radiation zone!
## How long does it take heat from the Sun’s core to reach Earth?
about 8 minutes and 20 seconds
Explanation: They travel at speed of light and takes about 8 minutes and 20 seconds to reach earths surface.
## How is energy transferred from the core of the Sun to its outer layers?
How is energy transferred from the core of the Sun to its outer layers? Energy radiates outward from the core as visible light. … Energy is radiated out from the core first then carried to the outer layers by convection.
## How is energy created in the core of the Sun?
The sun generates energy in its core in a process called nuclear fusion. During nuclear fusion the sun’s extremely high pressure and hot temperature cause hydrogen atoms to come apart and their nuclei (the central cores of the atoms) to fuse or combine. Four hydrogen nuclei fuse to become one helium atom.
## How long does it take the energy to leave the center of the sun and arrive at the photosphere?
But it can take 100 000 years from the core of the Sun to get to the surface — where it bursts out and flies at the speed of light.
## How does enormous energy get released from the sun?
The correct answer is Nuclear fusion. Sun can generate enormous energy due to a process called nuclear fusion. In this process high temperature and pressure in the core of the sun cause the nuclei to separate from their electrons. These hydrogen nuclei fuse to form one Helium atom releasing a large amount of energy.
## How much energy comes from the sun?
At any moment the sun emits about 3.86 x 1026 watts of energy. So add 24 zeros to the end of that number and you’ll get an idea of how unimaginably large an amount of energy that is! Most of that energy goes off into space but about 1.74 x 1017 watts strikes the earth.
## Why does it take so long for energy to travel through the radiative zone?
Energy travels through the radiation zone in the form of electromagnetic radiation as photons. Matter in a radiation zone is so dense that photons can travel only a short distance before they are absorbed or scattered by another particle gradually shifting to longer wavelength as they do so.
## How long does it take for energy produced in the core to reach the photosphere 400000 miles )?
Light from the sun takes a mere eight minutes to travel to Earth across nearly 100 million miles of empty space. But it takes 120 000 years to travel just 400 000 miles from its origin in the sun’s core to its surface.
## How is energy transferred in the convection zone?
In the convection zone energy is transported by the movement of the plasma or the hydrogen and the helium ions. Plasma churns vigorously transporting heat energy to the surface of the Sun through convection.
## How long does it take to travel to the Sun?
It would be faster to fly to the sun: It would take 169 090 hours to fly there at 550 miles per hour. It would take 7 045 days to fly there at 550 miles per hour.
## How fast does heat travel?
Heat radiation travels along a straight line. But it travels only through transparent such as glass water etc. Hence heat radiation travels at a speed of 3×108m/s.
## How does heat travel from the Sun?
Heat moves in three ways: Radiation conduction and convection. Radiation happens when heat moves as energy waves called infrared waves directly from its source to something else. This is how the heat from the Sun gets to Earth. … All of these kinds of waves contain a lot of energy.
## How energy travels outward from the core and is emitted from the sun include what happens in each layer of the sun?
Explain how energy travels outward from the core and is emitted from the Sun. Include what happens in each layer of the Sun. Gamma rays are produced in the core. They are then transferred by radiation through the radiative zone and converted to other types of photons.
## How long is a sun day ie how long for the sun to rotate on its axis )?
The Sun rotates on its axis once in about 27 days. This rotation was first detected by observing the motion of sunspots. The Sun’s rotation axis is tilted by about 7.25 degrees from the axis of the Earth’s orbit so we see more of the Sun’s north pole in September of each year and more of its south pole in March.
## What is the layer of the sun through which energy is transferred away from the core by radiation?
Unit 4 Lesson 3: The Sun
A B
radiative zone layer of the sun which energy is transferred away from the core by radiation
core very dense center of the sun
nuclear fusion process by which tow or more low-mass atomic nuclei fuse to form another heavier nucleus.
radiation when energy is transferred as electromagnetic waves
## How long does it take for a photon to exit the Sun?
Very approximately this means that to travel the radius of the Sun a photon will have to take (696 000 kilometers/1 centimeter)^2 = 5 x 10^21 steps. This will take 5×10^21 x 3 x10^-11 = 1.5 x 10^11 seconds or since there are 3.1 x 10^7 seconds in a year you get about 4 000 years.
## How long does it take light to escape the solar system?
The outer edge is estimated to be around 100 000 AU. So it would take around 140 hours to reach the edge of the solar system a photon emitted by the Sun if we take the inner edge of the Oort cloud.
## How long does it take for a neutrino to escape?
[+] With a radius of approximately 432 000 miles (~700 000 km) neutrinos take less than three seconds to exit the Sun from the time they are produced.
## How much energy is released by the sun per second?
The sun releases energy at a mass–energy conversion rate of 4.26 million metric tons per second which produces the equivalent of 384.6 septillion watts (3.846×1026 W).
## How long does it take the sun to deliver to Earth the total amount of energy humankind uses in a year?
Photo: Abdul Rahman/Gulf News (story: Binsal) Image Credit: Abu Dhabi: As only two minutes of the sun provide the world with one year’s energy needs of entire humanity a day’s sun can do wonders according to a senior official.
## How long does it take light to travel from the sun to Earth?
The Sun is 93 million miles away so sunlight takes 8 and 1/3 minutes to get to us.
## How long does it take a photon to go through the convection zone?
Although it may have taken the photons a million years to reach the convection zone the energy they deliver rises through the entire convention zone in about three months. All the energy emitted at the surface of the sun is transported there by convection.
## How does the energy produced in the core reach the surface?
From the core first it moves outwards by radiation and then convection.
## How does energy move in the photosphere?
In the photosphere the gas pressure dominates the magnetic pressure and the kinetic energy of convection is transferred to the magnetic field in the form of magnetohydrodynamic waves or magnetic stresses (electric currents).
## How is energy transferred between the core and the convective zone of the sun?
Energy is generated in the core the innermost 25%. This energy diffuses outward by radiation (mostly gamma-rays and x-rays) through the radiative zone and by convective fluid flows (boiling motion) through the convection zone the outermost 30%.
## How does the energy travel through outer space to Earth?
Unlike conduction and convection radiation does not need matter to transfer heat. Energy is radiated from the sun through the vacuum of space at the speed of light. When this energy arrives at Earth some of it is transferred to the gases in our atmosphere.
## What happens to energy in the sun’s convection zone?
What happens to energy in the Sun’s convection zone? Energy is transported outward by the rising of hot plasma and sinking of cooler plasma. … We are seeing hot gas rising and cool gas falling due to the convection that occurs beneath the surface.
## Can you land on the Sun at night?
Yes NASA can land on the sun at night. But only when it’s dark in the Eastern Pacific time zone.
## Can you land on the Sun?
Sun gives us light and heat to warm our planet Earth. … You can’t stand on the surface of the Sun even if you could protect yourself. The Sun is a huge ball of heated gas with no solid surface. The Sun’s surface is always moving.
## How close can we get to the Sun without dying?
You can get surprisingly close. The sun is about 93 million miles away from Earth and if we think of that distance as a football field a person starting at one end zone could get about 95 yards before burning up. That said an astronaut so close to the sun is way way out of position. | 2,169 | 9,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-33 | latest | en | 0.942293 |
https://electronics.stackexchange.com/questions/278668/bjt-transistors-can-one-see-them-as-voltage-dividers-or-should-they-be-analyzed | 1,721,317,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00725.warc.gz | 199,896,149 | 44,454 | # BJT transistors: can one see them as voltage dividers or should they be analyzed differently?
I am currently reading The Art of Electronics 3rd edition and I am struggling to understand basic transistors.
On page 77 there is an example circuit with a PNP transistor driven by an NPN transistor. The second paragraph asks a question and in parenthesis it says "make sure you understand why", which I don't!
I am struggling to understand the voltages and currents present in this circuit. Can I treat the resistors as a simple voltage divider and perform ohms law and KVL/KCL as a normal circuit or do I have to analyze the circuit differently because of the transistor?
The βdividerβ formed by π
2π
3 may be confusing: π
3βs job is to keep π3 off when π2 is off; and when π2 pulls its collector low, most of its collector current comes from π3βs base (because only βΌ0.6β―mA of the 4.4β―mA collector current comes from π
3 - make sure you understand why). That is, π
3 does not have much effect on π3βs saturation. Another way to say it is that the divider would sit at about +11.6β―V (rather than +14.4β―V), were it not for π3βs base-emitter diode, which consequently gets most of π2βs collector current. In any case, the value of π
3 is not critical and could be made larger; the tradeoff is slower turn-off of π3, owing to capacitive effects.12
Figure 2.10. Switching the high side of a load returned to ground.
12) But donβt make it too small: π3 would not switch at all if π
3 were reduced to 100β―Ξ© (why?). We were surprised to see this basic error in an instrument, the rest of which displayed circuit design of the highest sophistication.
• silicon bjts need how much base-emitter voltage to turn ON? Commented Jan 5, 2017 at 16:26
• @bufo333 You can use impedance ratios for everything keeping in mind power limits and slew rate reduces with current. Normally data sheets give switch saturation for Vce(sat) which * I = Pd and times Rja ( gives temp rise. including sink if added) They standardize on saturation for Ic:Ib=10 ( regardless of hFE ) then report the Vce(sat for that current level and ratio. Try to remember this current ratio for BJT's and it can be relaxed to 50:1 for very low current or very high hFE devices >500 or in between ( my rule of thumb is 10% of hFE ) Commented Jan 5, 2017 at 17:23
• The footnote refers to R ratios on pull-up with 3k3/(100+3k3 being too high a ratio ~1 vs 3k3/(1k+3.3k) *15V drop from 15V with 0.75k =Req . But the R3 is important necessary for faster turn off. Thus Load R is derived from Ic:Ib ratio of 10:1 Commented Jan 5, 2017 at 17:30
• I really want to thank everyone for responding to this question, I was banging my head over and over this question, trying to puzzle out the currents and voltages and the cause and effects. I still feel confused but I am slowly getting it. I must have read each comment a dozen times over the last hour or so. I want to thank Tony Stewart, Olin Lathrop, petEEY, and Peter Smith. It really took me reading all of your comments for the answer to start to make sense for me. I guess if you say it enough times it eventually sinks in. Commented Jan 5, 2017 at 20:56
• Other questions about this circuit: PNP Circuit-The Art of Electronics, Why would the transistor not switch?
– CL.
Commented Jan 6, 2017 at 9:57
You seem to be asking multiple questions.
The first thing to understand is what the NPN in front of the PNP does for you. As the page correctly points out, the left circuit needs to hold the input at nearly 15 V for the transistor to be off. Understand why before proceeding.
The reason is that the B-E junction of these transistors looks like a diode to the driving circuit. This diode drops only 500-750 mV for normal currents. Therefore when current is drawn from the base of the PNP, that base is only about 700 mV below the emitter, so a bit above 14 V. As the book says, that is inconvenient when you want to control something with a 0 to 3.3 V digital logic signal.
The solution is the right circuit. The NPN has the same diode-looking characteristics, but flipped around in polarity. Now the base will be about 700 mV above the emitter, so 700 mV above ground, when the transistor is being turned on. That's a level that digital logic can do.
The two resistors R2 and R3 are to interface the easy to control NPN switch with the PNP switch you actually want to control.
R2 limits the current out the base of Q3 and into the collector of Q2. If it wasn't there, that current could get high enough to damage both transistors.
R3 isn't strictly necessary, but makes sure Q3 is really off when Q2 is off. Note that when Q2 is off, it's just not driving current thru the base of Q3. It doesn't actively hold the base of Q3 close to its emitter to make sure Q3 is really off and not randomly turning on from stray noise. That's what R3 does.
As the text says, there is wide latitude in picking the value of R3. However, it can't be so low that the current limited by R2 can't turn on Q3 anymore.
The book wants you to understand why most of the current through Q2 comes from Q3 and not from R3.
The voltage across R3 is the same as the voltage across the PNP's emitter and base. This emitter and base junction functions a bit like a diode because any voltage increase past ~0.6V increases current exponentially. ~0.6V across a 1K resistor is only ~0.6mA (the current mentioned in the book).
simulate this circuit – Schematic created using CircuitLab
Ic:Ib=10 minimum to 30:1 or 50:1 in extreme cases>500 since hFE drops rapidly when Vce < 1V.
There are several ways to quickly analyze this. From R Ratios or KVL-KCL as long as Vbe is forward biased and Vbe is included in KVL and treated as a voltage source and forward biased unlike when Rb pullup = 100 ohms in example of question.
(neglecting L1 and C1 and input frequency for this case but C has a non-zero ESR)
When Q2 is turned on, it will pull Q3s base to about 14.4V (a diode drop below its emitter).
Q2 therefore will have a collector current of 14.4V / 3.3k = 4.4mA.
The current through R3 will be 0.6V (as the resistor sees the same voltage as the base - emitter drop) / 1k = 0.6mA
As there is a total of 4.4mA in R2 is 4.4mA and the current in R3 is 0.6mA, the remaining current (3.8mA) must be flowing through the base of Q3
• You might want to add the calculation if R3 is reduced to 100R (see note 12 at bottom of page) Commented Jan 5, 2017 at 16:50
• @peter This finally clicked for me, let me see if I am understanding this correctly. The voltage source is 15v, the voltage through Q3 (Veb) has a 0.6v diode drop, which gives you 14.4v. That means that the current through R2 is 14.4v/3300 gives you ~4.4ma while the current through r2 is the difference in voltage 0.6 volts / 1000R3 = 0.6ma. Maybe I came to the right answer for the wrong reasons. I am confuses why the 0.6ma is not added to the 4.4ma, instead it is a part of the 4.4ma? Commented Jan 5, 2017 at 21:22 | 1,935 | 7,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.951616 |
https://www.codermerlin.com/wiki/index.php/W1175_Numbers_to_Words | 1,669,687,532,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710684.84/warc/CC-MAIN-20221128235805-20221129025805-00767.warc.gz | 767,067,391 | 13,619 | # W1175 Numbers to Words
Within these castle walls be forged Mavens of Computer Science ...
— Merlin, The Coder
## Introduction
Previously we built algorithms that converted a String to an Integer. Similarly, we can do the opposite and convert a number into any String that we want. For this exercise we will be taking an Integer and writing out in full the words that write out our number.
## Writing out Numbers
Instead of just converting 25 to "25", in this exercise we are going to convert 25 to "Twenty Five".
1 One
12 Twelve
123 One Hundred Twenty Three
1234 One Thousand Two Hundred Thirty Four
## Obtaining Individual Digits
To do this we need to deconstruct an Integer with multiple digits into individual digits. We did the opposite of in String to Integer, we took single digits and made it into a multi-digit number. Instead of multiplying digits by 10, we need to divide by 10. More specifically we need to get the remainder of each division, and that will be our digit.
1234 % 10 = 4
Now we have our "ones" digit. We can now try to get our "tens" digit, "hundreds", etc. To get these, remember back to Number Systems how we can get each place value using simple operations.
Once you have gotten a digit separated, you can then perform your conversion to make that digit into words to build a string.
## Conversion Functions
Now that you have a digit, you need to convert it from an Integer to the String of Words.
A good way to do this is to make a function for each place value that can take a number and return that number written out in words. So a "ones" function may take 5 and return "five". A "tens" function may take 8 and return "eighty".
Helpful Hint
Building functions will help keep your code modular and create sections of code that do one thing, and one thing well.
## Key Concepts
Key Concepts
• Using Remainders to obtain the current "Ones" place value of an Integer
• Using Functions to create code that does a specific action only
## Exercises
Exercises
• M1175-10 Complete Merlin Mission Manager Mission M1175-10. | 483 | 2,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-49 | latest | en | 0.86467 |
http://gmatclub.com/forum/understanding-prepositional-phrases-163345.html?fl=similar | 1,485,052,967,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281331.15/warc/CC-MAIN-20170116095121-00152-ip-10-171-10-70.ec2.internal.warc.gz | 114,184,133 | 52,273 | Understanding Prepositional phrases: : GMAT Sentence Correction (SC)
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# Understanding Prepositional phrases:
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Understand what prepositional phrases do in a sentence.
prepositional phrase will answer the question Which one?
The book on the bathroom floor is swollen from shower steam.
Which book? The one on the bathroom floor!
The sweet potatoes in the vegetable bin are green with mold.
Which sweet potatoes? The ones forgotten in the vegetable bin!
The note from Beverly confessed that she had eaten the leftover
pizza.
Which note? The one from Beverly!
As an adverb, a prepositional phrase will answer questions such as How? When? or
Where?
Freddy is stiff from yesterday's long football practice.
How did Freddy get stiff? From yesterday's long football practice!
Before class, Josh begged his friends for a pencil.
When did Josh do his begging? Before class!
Feeling brave, we tried the Dragon Breath Burritos at Tito's
Taco Palace.
Where did we eat the spicy food? At Tito's Taco Palace!
If you have any questions
New!
Understanding Prepositional phrases: [#permalink] 18 Nov 2013, 02:33
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2 Pronouns referring to a noun in a prepositional phrase 4 21 Mar 2012, 21:19
A prepositional phrase consists of a preposition, a noun or 4 30 Oct 2007, 13:50
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http://efofex.com/documentation/randnormal(meansdnumber).htm | 1,526,900,739,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122528-00022.warc.gz | 91,125,530 | 1,992 | RandNormal(mean, sd, number)
# RandNormal(mean, sd, number)
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Generates number normally distributed scores with a mean of mean and a sample deviation of sd. The RandNormal function can quickly produce normally distributed data sets and can produce data sets with almost exactly the given mean and standard deviation. For example, RandNormal(60,15,30) might return 71.14, 64.14, 53.42, 60.18, 85.18, 33.88, 77.8, 64.43, 54.79, 40.79, 42.7, 72.37, 69.64, 27.99, 61.92, 49.87, 64.42, 61.74, 71.67, 83.69, 70.52, 54.36, 63, 24.77, 57.12, 77.01, 57.06, 72.68, 66.49, 45.23 This set of thirty numbers has a mean of exactly 60, a standard deviation of 14.9995591 and has a distribution that looks like this: Note: If you type decimals using a comma (eg 3,2) you should enter this formula as RandNormal(60; 15; 30) - using the ; as a separator | 310 | 872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-22 | latest | en | 0.786274 |
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ACM-ICPC亚洲区预选赛(北京赛区)游戏对抗邀请赛正在报名中
Language:
Uyuw's Concert
Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 9329 Accepted: 3484
Description
Prince Remmarguts solved the CHESS puzzle successfully. As an award, Uyuw planned to hold a concert in a huge piazza named after its great designer Ihsnayish.
The piazza in UDF - United Delta of Freedom’s downtown was a square of [0, 10000] * [0, 10000]. Some basket chairs had been standing there for years, but in a terrible mess. Look at the following graph.
In this case we have three chairs, and the audiences face the direction as what arrows have pointed out. The chairs were old-aged and too heavy to be moved. Princess Remmarguts told the piazza's current owner Mr. UW, to build a large stage inside it. The stage must be as large as possible, but he should also make sure the audience in every position of every chair would be able to see the stage without turning aside (that means the stage is in the forward direction of their own).
To make it simple, the stage could be set highly enough to make sure even thousands of chairs were in front of you, as long as you were facing the stage, you would be able to see the singer / pianist – Uyuw.
Being a mad idolater, can you tell them the maximal size of the stage?
Input
In the first line, there's a single non-negative integer N (N <= 20000), denoting the number of basket chairs. Each of the following lines contains four floating numbers x1, y1, x2, y2, which means there’s a basket chair on the line segment of (x1, y1) – (x2, y2), and facing to its LEFT (That a point (x, y) is at the LEFT side of this segment means that (x – x1) * (y – y2) – (x – x2) * (y – y1) >= 0).
Output
Output a single floating number, rounded to 1 digit after the decimal point. This is the maximal area of the stage.
Sample Input
```3
10000 10000 0 5000
10000 5000 5000 10000
0 5000 5000 0
```
Sample Output
`54166666.7`
Hint
Sample input is the same as the graph above, while the correct solution for it is as below:
I suggest that you use Extended in pascal and long double in C / C++ to avoid precision error. But the standard program only uses double.
Source
POJ Monthly,Zeyuan Zhu
[Submit] [Go Back] [Status] [Discuss] | 675 | 2,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2016-44 | longest | en | 0.952428 |
http://mathhelpforum.com/pre-calculus/150825-inverse-functions-print.html | 1,529,936,333,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00051.warc.gz | 198,518,342 | 3,049 | # Inverse functions.
• Jul 12th 2010, 02:41 PM
MrGenuouse
I face another problem childish but still a problem
Find the inverse functions of
a) x ---> 2x+3
b)x---> 3x-4
c) x---> 2(x+5)
d) x----> 1 (x-1)
• Jul 12th 2010, 09:53 PM
earboth
Quote:
Originally Posted by MrGenuouse
I face another problem childish but still a problem
Find the inverse functions of
a) x ---> 2x+3
b)x---> 3x-4
c) x---> 2(x+5)
d) x----> 1 (x-1)
I'm going to show you how to get the inverse function with example a). The remaining examples have to be done similarly.
to #a):
From $\displaystyle f:x\to2x+3$ you'll get the equation of the function f:
$\displaystyle f(x)=y=2x+3$
To find the equation of the inverse function you have to swap the variables and solve this equation for y:
$\displaystyle f:y=2x+3~\implies~f^{-1}:x=2y+3$
$\displaystyle x=2y+3~\implies~x-3=2y~\implies~\boxed{y=\frac12 x - \frac32}$
(Remark: You'll get the graph of $\displaystyle f^{-1}$ by reflection of the graph of f at the line y = x)
... and now it's your turn!
• Jul 14th 2010, 04:41 AM
HallsofIvy
Another way of thinking about it: f(x)= 2x+ 3 tells you "whatever x is, first multiply by 2, then add 3". To find the inverse of f, do the opposite of each step, in the [b]opposite order. The opposite of "multiply by 2" is "divide by 2" and the opposite of "add 3" is "subtract 3". So the inverse of f is "subtract three, then divide by 2": $\displaystyle f^{-1}(x)= \frac{1}{2}(x- 3)= \frac{1}{2}x- \frac{3}{2}$ just as earboth got. | 529 | 1,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-26 | latest | en | 0.817769 |
https://studylib.net/doc/18162030/4-transient-analysis-of-markov-processes | 1,656,916,937,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00386.warc.gz | 598,443,513 | 13,016 | 4 Transient analysis of Markov processes
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```4
Transient analysis of Markov processes
In this chapter we develop methods of computing the transient distribution of a Markov
process.
Let us consider an irreducible Markov process with finite state space {0, 1, . . . , N } and
generator Q with elements qij , i, j = 0, 1, . . . , N . The random variable X(t) denotes the
state at time t, t ≥ 0. Then we want to compute the transition probabilities
pij (t) = P (X(t) = j|X(0) = i),
for each i and j. Once these probabilities are known, we can compute the distribution at
time t, given the initial distribution, according to
P (X(t) = j) =
N
X
P (X(0) = i)pij (t),
j = 0, 1, . . . , N.
i=1
4.1
Differential equations
It is easily verified that for t ≥ 0 the probabilities pij (t) satisfy the differential equations
N
X
d
pik (t)qkj
pij (t) =
dt
k=1
or in matrix notation
d
P (t) = P (t)Q,
dt
where P (t) is the matrix of transition probabilities pij (t). The above differential equations
are refered to as the Kolmogorov’s Forward Equations (Clearly, there are also Backward
Equations). The solution of this system of differential equations is given by
P (t) =
∞
X
Qn
n=0
n!
tn = eQt ,
t ≥ 0.
Direct use of the infinite sum of powers of Q to compute P (t) may be ineffcient, since Q
contains both positive and negative elements. An alternative is to reduce the infinite sum
to a finite one by using the spectral decomposition of Q. Let λi , 1 ≤ i ≤ N , be the N
eigenvalues of Q, assumed to be distinct, and let yi and xi be the orthonormal left and
right eigenvectors corresponding to λi . Further, let Λ be the diagonal matrix of eigenvalues
and X and Y the matrices of eigenvectors. Then we have
Q = Y ΛX,
and thus
P (t) = eQt
1
∞
X
Qn
tn
n!
n=0
∞
X
Y Λn X n
=
t
n!
n=0
=
= Y eΛt X,
where eΛt is just the diagonal matrix with elements eλi t . The difficulty with the above
solution is that it requires the computation of eigenvalues and eigenvectors. In the next
section we present a numerically stable solution based on uniformization.
4.2
Uniformization
To establish that the sojourn time in each state is exponential with the same mean, we
introduce fictitious transitions. Let ∆ satisfy
0 < ∆ ≤ min
i
1
.
−qii
In state i, 0 ≤ i ≤ N , we now introduce a transition from state i to itself with rate
qii + 1/∆. This implies that the total outgoing rate from state i is 1/∆, which does not
depend on i! Hence, transitions take place according to a Poisson process with rate 1/∆,
and the probability to make a transition from state i to j is given by
1 ≤ i, j ≤ N.
pij = ∆qij + δij ,
If we denote the matrix of transition probabilities by P and condition on the number of
transitions in (0, t), we immediately obtain
P (t) =
∞
X
e−t/∆
n=0
(t/∆)n n
P .
n!
(1)
Since this representation requires addition and multiplication of nonnegative numbers only,
it is suitable for numerical calculations, where we approximate the infinite sum by using
the first K terms. A good rule of thumb for choosing K is
K = max{20, t/∆ + 5 ·
q
t/∆}.
For a more elaborated algorithm we refer to [1]. It further makes sense to take the largest
possible value of ∆ (Why?), so
1
∆ = min
.
i −qii
2
4.3
Occupancy times
In this section we concentrate on the occupancy time of a given state, i.e., the expected
time in that state during the interval (0, T ).
Let mij (T ) denote the expected amount of time spent in state j during the interval
(0, T ), staring in state i. Then we have
Z
mij (T ) =
T
t=0
pij (t)dt,
or in matrix notation,
M (T ) =
Z
T
P (t)dt,
t=0
where M (T ) is the matrix with elements mij (T ). Substitution of (1) leads to
M (T ) =
∞ Z
X
T
e−t/∆
n=0 t=0
(t/∆)n
dtP n .
n!
By partial integration we get
Z
T
n
−t/∆ (t/∆)
e
t=0
n!
dt = ∆ 1 −
n
X
k
−t/∆ (T /∆)
e
k=0
k!
!
= ∆P (Y > n),
where Y is a Poisson random variable with mean T /∆. Hence, we finally obtain
M (T ) = ∆
∞
X
P (Y > n)P n .
n=0
The above representation provides a stable method for the computation of M (T ); for more
details see [1].
4.4
Exercises
Exercise 1.
Consider a machine shop consisting of N identical machines and M repair men (M ≤ N ).
The up times of the machines are exponential with mean 1/µ. When a machine fails, it is
repaired by a repair man. The repair times are exponential with mean 1/λ.
(i) Model the repair shop as a Markov process.
Suppose N = 4, M = 2, the mean up time is 3 days and the mean repair time is 2 hours.
At 8:00 a.m. all machines are operating.
(ii) What is the expected number of working machines at 5:00 p.m.?
(iii) What is the expected amount of time all machines are working from 8:00 a.m. till
5:00 p.m.?
3
References
[1] V.G. Kulkarni, Modeling, analysis, design, and control of stochastic systems,
Springer, New York, 1999.
4
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Question
# Aqueous Urea solution is 20% by mass of solution. Calculate percentage by mass of solvent.
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Solution
## Let the mass of the solution be 100gm So mass of urea is 20% of 100,so mass of urea is 20 gm So mass of water = 100 – 20 = 80gm So mass of urea by solvent(water) = 20/80 x 100% = 25%
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#### Resources tagged with Working systematically similar to Quadrilaterals Game:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
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Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
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##### Stage: 2 and 3
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### Isosceles Triangles
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### Ben's Game
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Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Twinkle Twinkle
##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### Where Can We Visit?
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### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
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### More on Mazes
##### Stage: 2 and 3
There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
### Counting on Letters
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### More Plant Spaces
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### Pair Sums
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### Extra Challenges from Madras
##### Stage: 3 Challenge Level:
A few extra challenges set by some young NRICH members.
### Triangles to Tetrahedra
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### Cuboids
##### Stage: 3 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Oranges and Lemons, Say the Bells of St Clement's
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##### Stage: 3 Challenge Level:
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### You Owe Me Five Farthings, Say the Bells of St Martin's
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### First Connect Three
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### Magic Potting Sheds
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### Creating Cubes
##### Stage: 2 and 3 Challenge Level:
Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour. | 2,376 | 10,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2015-22 | longest | en | 0.893377 |
https://www.ciandebia.it/Oct_23-25595.html | 1,642,975,301,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00371.warc.gz | 665,062,176 | 9,648 | 0%
Variable Roller Cfd
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UNITED s'rA'r 1:s \PATENT OFFICE s'rnruiss VARIABLE CHANGE-SPEED GEAR wrrn ROLLER BODIES Jean Kopp, Berne, Switzerland Application January 14,1946, Serial No. 641,039 In Switaerland February 1, 1945 l My present invention relates to stepless changespeed gearing comprising a driving and a driven Claims.
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Vertical Roller Mill In Niue | 2,772 | 12,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-05 | latest | en | 0.880487 |
https://mirror.codeforces.com/blog/entry/65079 | 1,713,919,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00217.warc.gz | 345,697,015 | 36,571 | Rating changes for last rounds are temporarily rolled back. They will be returned soon. ×
### cdkrot's blog
By cdkrot, history, 5 years ago,
The round has finished. I hope you liked it!
Credits to the round authors and developers:
1110A - Parity. Authored by _h_ and simonlindholm.
1110B - Tape. Authored by _h_ and simonlindholm, development: cdkrot
1110C - Meaningless Operations. Authored by GreenGrape
1110D - Jongmah. Authored by _h_ and simonlindholm, development: KAN and MikeMirzayanov
1110E - Magic Stones. Authored by _h_ and simonlindholm, developed by GreenGrape
1110F - Nearest Leaf. Authored by grphil, developed by vintage_Vlad_Makeev
1110G - Tree-Tac-Toe . Authored by cdkrot and KAN
1110H - Modest Substrings. Authored by _h_ and simonlindholm, development: budalnik
And now, the editorial:
• +155
| Write comment?
» 5 years ago, # | +16 In C, If we precalculate answers for all a=2^x−1 up to m=2^23−1. Shouldn't it be 2^25-1?
• » » 5 years ago, # ^ | +37 Corrected thanks!We indeed increased constraints slightly before the contest.
• » » » 4 years ago, # ^ | 0 Can you point me to where I am wrong in the third question? My approach was somewhat like this I took the intervals of the numbers ie A[i]-A[i-1] and stored them in one vector and sorted them in decreasing order Now I can make 'K' tape which means I can use K-1 use non attached tapes right... so I took the sum of all the intervals and deleted the first k-1 larger intervals from the sum that is sum-v[0]-v[1]-..v[k-2].. and then printed out (sum+k) as the final answer I got stuck in the test case number 9 ... Thanks in advance for your help.
» 5 years ago, # | +28 Probably one of the most well written editorial i have ever seen!
» 5 years ago, # | 0 Can someone please explain the solution of B in more detail?
• » » 5 years ago, # ^ | +2 Suppose, We have a stick with 100 cm length and 8 broken points,.........10.12.... .....30....35..... .......80.....86......93.........100...Since, there are 8 broken points and each point takes 1cm, at the very least we need 8cm tape to fix them. But with the constraint of number of tapes, we will have to cover a segment instead of a point. In the process, we will end up covering some unbroken points. So, total covered points will be the broken points + the points which fall under those segments that we picked to cover due to the constraint.Consider the following cases -Case 1: k = 8We have 8 tapes to use. So, there is no point wasting tape on unbroken points. Just cut 8 tapes of 1 cm, and fix the broken points. So, we are only using 8 cm of tape, no tape wasted. Cool!Case 2: k = 7We have 7 tapes to use. Now, we are 1 piece short. So, we will fix a segment with one tape (instead of just a point) and for rest of the broken points we can just put 1cm pieces.So, how do we choose that special segment? The goal is to waste as little as possible tape. Looking at the pipe above, we would like to cover 10.12, just wasting a 1cm tape. This will make us to cover point 11 which is not broken. But there is no better way.So, how much tape we are using now? 8 broken, 1cm points + 1 extra 1cm point = 9cmCase 3: k = 6We have 6 tapes to use. Using the same approach on Case 2, we have to pick another segment. 30....35 is our next best option, because we are justing wasting 4cm tape in the middle.Thus 8cm legit broken points + 1cm in the middle of 10.12 and 4cm in the middle of 30....35 = 13 cm So, we will be needing to cover up the short amount of tape, and we will choose the segments which come with little waste of tape. To find those segments, we need to list all b[i+1] — b[i] — 1, meaning the lengths of non-broken segments. Now, we will just cover the smallest (n-k) of the segments we need to.
• » » » 5 years ago, # ^ | 0 Thanks!
» 5 years ago, # | 0 Thanks for fast tutorials
» 5 years ago, # | 0 can someone explain tape (2 question) in better way
• » » 5 years ago, # ^ | +3 hey I will explain it with help of one example; let n=10, n=35, k=5; b[i]={1 2 5 6 9 11 13 15 23 27} broken segments; 1_2__5_6___9__11__13__15________23___27_ (where you have to repair I have put that integer no.)if you have value of k more than n (k>n), then you can easily cut the 10 segment of 1cm so it will very less tap require(only total 10cm tap required=10cm(1cm+1cm+1cm+1cm+1cm+1cm+1cm+1cm+1cm+1cm). But here the value of k is 5 so you have to cut 5 segment such that all broken part should cover. 1_2__5_6___9__11__13__15________23___27_ (where you have to repair I have put that integer no.) now if you cut 1cm of 10 segment tap it is less tap required but you can cut only 5 tap segment because k value is 5.so you will cut 5 segment such that segment cover max length with small tap. now a[i]=b[i+1]-b[i] So a[i]={1,3,1,3,2,2,2,8,4} so by sorting it a[i]={1,1,2,2,2,3,3,4,8} So you will repair first all that segment which are very near means 1 and 2, 5 and 6, and so on.. you can cut the tap at most five times(k value), so your target is with this five segment you have to repair the stick.
» 5 years ago, # | +6 Now consider the case when a=2x−1. This implies (a⊕b)+(a&b)=a, hence it's sufficient to find the largest proper divisor of a — it will be the desired answer. Why is that implied in this case?
• » » 5 years ago, # ^ | +1 When a=2^x-1,it's binary will have all bits as 1.So, a&b equals b and (a xor b) equals a-b.
• » » » 5 years ago, # ^ | 0 Thanks for quick response!Why is the largest proper divisor the answer? Does it relate to the Extended Euclidean algorithm?
• » » » » 5 years ago, # ^ | +6 Notice that when b is divisor of a,a-b will also be the divisor of a. So,maximum possible gcd(a-b,b)=largest divisor of a.
• » » » » » 5 years ago, # ^ | ← Rev. 3 → +1 I think you meant that, when b is a divisor of a, a - b will be divisible by b, didn't you?
• » » » » » » 5 years ago, # ^ | +3 Oh...Yes
• » » 5 years ago, # ^ | 0 gcd(a-b, b) eqaul to b when b is factor of a.
• » » 4 years ago, # ^ | 0 Can you point me to where I am wrong in the third question? My approach was somewhat like this I took the intervals of the numbers ie A[i]-A[i-1] and stored them in one vector and sorted them in decreasing order Now I can make 'K' tape which means I can use K-1 use non attached tapes right... so I took the sum of all the intervals and deleted the first k-1 larger intervals from the sum that is sum-v[0]-v[1]-..v[k-2].. and then printed out (sum+k) as the final answer I got stuck in the test case number 9 ... Thanks in advance for your help.
» 5 years ago, # | +21 I expected more than on G. Since it is just working with cases. The idea is obvious.
• » » 5 years ago, # ^ | +29 Well, yes, but there are several cute ideas along the way (that the degrees are small and and the idea about reducing the problem to uncolored vertices only, which is not mandatory to get AC, but helps greatly)
• » » » 5 years ago, # ^ | -9 can you explain part B
• » » » » 5 years ago, # ^ | +30 But there is an editorial written already...Ask some concrete question at least.
• » » » » » 5 years ago, # ^ | 0 What I want to know is, why have people stored the differences as b[i] — b[i — 1] — 1 and added n to the answer?
• » » » » » » 5 years ago, # ^ | 0 Looks fine to me, basically you need to spend tape to cover the broken segments themselves ( + n in the end) and then you only care about which gapes between segments are "collapsed" into single tape part and which are not.
• » » » » » » » 5 years ago, # ^ | 0 Got it, thank you!
• » » 4 years ago, # ^ | 0 Can you point me to where I am wrong in the third question? My approach was somewhat like this I took the intervals of the numbers ie A[i]-A[i-1] and stored them in one vector and sorted them in decreasing order Now I can make 'K' tape which means I can use K-1 use non attached tapes right... so I took the sum of all the intervals and deleted the first k-1 larger intervals from the sum that is sum-v[0]-v[1]-..v[k-2].. and then printed out (sum+k) as the final answer I got stuck in the test case number 9 ... Thanks in advance for your help.
» 5 years ago, # | +21 Anybody solved D with greedy?
» 5 years ago, # | 0 In D, Can you please explain the statement — So we can assume that there are at most 2 triples of type [x,x+1,x+2] for each x.
• » » 5 years ago, # ^ | ← Rev. 3 → +1 It means that given some x, it isn't necessary to try choosing the triplet [x, x+1, x+2] more than 2 times. Because for k > 2 (assume km = k mod 3), k[x, x+1, x+2] is equivalent to km[x, x+1, x+2] + [x, x, x] + [x+1, x+1, x+1] + [x+2, x+2, x+2].
• » » » 5 years ago, # ^ | ← Rev. 2 → 0 what do you mean by it doesn't make sense ? what will happen if we choose triplet (say) 3 times ?And how dp state is formulated ? can you describe a little bit.
• » » » » 5 years ago, # ^ | 0 As k[x, x+1, x+2] is equivalent to km[x, x+1, x+2] + (k-km)[x] + (k-km)[x+1] + (k-km)[x+2], you can use this observation to apply simple dp that tries to take a triplet [x, x+1, x+2] at most only 2 times, and gives you the same optimal answer.
• » » » » » 5 years ago, # ^ | 0 so , in the optimal answer there will be no triplet like [x,x,x] ? do we have to form triplets only using [x,x+1,x+2] . only it will give optimal answer ?
• » » » » » » 5 years ago, # ^ | ← Rev. 2 → +5 In the dp when you consider element x, you make 3 trials. In the jth trial (j ranges from 0 to 2), you try to take the triplet [x, x+1, x+2] j times (if minimum(count(x), count(x+1), count(x+2)) >= j), and you add to the jth trial's result. So basically, what is left from x in the jth trial is just consumed in the form of triplets [x, x, x].
• » » » » » » » 5 years ago, # ^ | 0 can you explain by writitng the recurrence relation for dp ?
• » » » » » » » » 5 years ago, # ^ | +16 One possible way:Let dp[i][j][k] be the maximum count of triplets that can be formed if you start at value i, j occurrences of value i and k occurrences of value i + 1 are consumed in the form of triplets [x, x + 1, x + 2] where x < i. And let co[i] be the initial count of value i occurrences.For 0 ≤ l ≤ 2, if min(co[i] - j - l, co[i + 1] - k - l, co[i + 2] - l) ≥ 0, (where dp[i][j][k] is initially 0). The answer is dp[1][0][0] (Loop order: i = m → i = 1).
• » » » » » » » » » 5 years ago, # ^ | 0 Why are you considering l upto 2 why not till min >=0 ?
• » » » » » » » » » 5 years ago, # ^ | 0 Read my first two replies to the top level comment.
• » » » » » » » » » 5 years ago, # ^ | 0 thanks alot
• » » » » » » » 5 years ago, # ^ | 0 Sir, why are you adding floor((count(x) — j)/3) to the jth trial?
• » » » » » » » » 5 years ago, # ^ | 0 Because you consumed j occurrences of x in form of triplets [x, x + 1, x + 2]. So you want to consume what is left from x in form of triplets [x, x, x] (and if 2 or 1 occurrences of x are left at the end, you will not use them). This is equivalent to taking .
• » » » » » » » » » 5 years ago, # ^ | +5 Got it now, thanks !!
» 5 years ago, # | +3 Fast editorial.Great!
» 5 years ago, # | ← Rev. 2 → +4 Hi everybody, In case anybody needs practice with B, AtCoder Beginner Contest 117 had a very similar problem here.Here is my editorial for Problem C. I believe it is easier to understand than the editorial provided here. Let me know if you have any doubts. :)Here is my editorial for E.
• » » 5 years ago, # ^ | +1 Thanks a lot!! nice explanation
» 5 years ago, # | +9 In C, if a=2^x . Then as per "Denote the highest bit of a as x (that is largest number x, such that 2^x≤a) and consider b=(2^x−1)⊕a. It's easy to see that if a≠2^x−1 then 0
• » » 5 years ago, # ^ | +9 It's probably a mistake. I think b should be .
• » » » 5 years ago, # ^ | 0 Thanks a lot !
• » » 5 years ago, # ^ | 0 I was wondering for a long time until I found that it's a mistake
» 5 years ago, # | +1 Has anyone solved "D" using greedy approach? If yes, can you please share!
» 5 years ago, # | 0 how to dp in AC automaton? how to calculate the answer of all the current suffixes? Could you explain in details. thanks cdkrot
» 5 years ago, # | 0 In C, we denote x as 2^x <= a, but then the explanation states the following: "consider the case when a=(2^x)-1". Could someone explain what they exactly mean here?
• » » 5 years ago, # ^ | ← Rev. 2 → 0 It should read "consider the case when a=2^(x+1)-1". I was also confused by that. Basically, a is a string of 1s in binary.
» 5 years ago, # | 0 can someone tall the c++ codes for 2nd and 3rd question, maybe with comments and steps will be more helpful
» 5 years ago, # | +2 I think in C editorial, meaning of x should be the smallest number such that 2x > a.
» 5 years ago, # | +4 Thanks for the editorial. I didn't realize the editorial was already out, cause I didn't see the "Recent actions" so much. Adding a link to the original announcement blog post ( http://mirror.codeforces.com/blog/entry/65059 ) would be helpful :)
» 5 years ago, # | -11 In fact, we can use a simple way to solve the problem C.Although this way is not very good. #include using namespace std; int wyj[25] = {3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,32767,65535,131071,262143,524287,1048575,2097151,4194303,8388607,16777215,33554431,67108863}; int qwq[25] = {1,1,5,1,21,1,85,73,341,89,1365,1,5461,4681,21845,1,87381,1,349525,299593,1398101,178481,5592405,1082401,22369621}; bool check[67108869]; int ans(int x) { if(x == 1) return 0; if(x == 2) return 3; if(check[x]) { for(int i = 0; i < 25; i++) if(wyj[i] == x) return qwq[i]; } else for(int i = 0; i < 25; i++) if(x <= wyj[i]) return wyj[i]; return -1; } int main() { for(int i = 0; i < 25 ; i++) check[wyj[i]] = 1; int q, tby; scanf("%d",&q); for(int i = 1; i <= q; i++) { scanf("%d",&tby); printf("%d\n",ans(tby)); } return 0; }
» 5 years ago, # | 0 Can anyone explain me the intuition behind the process in problem E ? Its hard to come up with such an idea
• » » 5 years ago, # ^ | +1 This comment and read the editorial to the problem: http://mirror.codeforces.com/blog/entry/65059?#comment-490669It's not obvious to me how to come up with such idea, it just seems like its a tricky known technique :S
» 5 years ago, # | 0 Please correct me if I'm wrong but it seems that in C the given editorial solution leads to wrong answer. For eg. if a=11 then given solution would give b=12. Instead it should have been b=4 (which we get by complimenting binary rep of a=11, considering significant bits).
• » » 5 years ago, # ^ | 0
» 5 years ago, # | 0 Can anyone please explain the solution to D in more detail? (Like the exact dp equation and its initial values) My mind got a bit mixed up...
• » » 5 years ago, # ^ | 0 I think this blog will help. https://mirror.codeforces.com/blog/entry/65092
» 5 years ago, # | ← Rev. 2 → 0 Could someone explain how to come up with such a dp state in problem D?I'm curious because I can't figure it out.
» 5 years ago, # | 0 Can someone elaborate more on E? I can't quite understand why this equations proves that you need to check array difference.
• » » 5 years ago, # ^ | 0 Hello!You can try some more complicate examples to find that it's obvious that the array difference with the first and the last number determines the result.By the way, there is a conclusion:if you can exchange any two neighbor elements in a array, you can obtain any permutation of this array.I help this helps:)
• » » » 5 years ago, # ^ | 0 Yes, thank you. In meantime I figured it out.
» 5 years ago, # | 0 please someone elaborate more on F
» 5 years ago, # | 0 I was trying to solve problem 1110F - Nearest Leaf offline, using centroid decomposition but is giving me TLE 49838550, any ideas? I know it might sound as an overkill but still it's complexity is something about O(NlgNlgN + QLgNlgN). thanks
» 5 years ago, # | 0 In Problem B, we will calculate the differences between every two adjacent broken points and store them in a sort array of size n-1. Now lets assume that we need to cover all the broken points with one piece of tape. For this the answer will be [difference of first and last element of the given array + 1] . Now we will optimize the answer as we have the chance to us at most k pieces of tape. We will subtract [x-1] form the answer for each of the last [k-1] elements of sorted array (where x is an element of the sorted array). Code Snippet int n,k; long long m; cin>>n>>m>>k; long long arr[n]; for(int i=0;i>arr[i]; } long long brr[n-1]; for(int i=0;in-k-1;i--){ ans-=brr[i]-1; } cout<
» 5 years ago, # | 0 In question 1110A-Parity , can someone explain what is the error in my code : long double b; long long int i,j,k,n=0; cin>>b>>k; long double a[k+5]; for(i=0;i>a[i]; for(i=k-1,j=0;i>=0 || j
• » » 5 years ago, # ^ | +13 For the sake of precision, don't use pow() with integers, ever.
» 5 years ago, # | ← Rev. 3 → 0 Problem D.Why does my solution in java fails in memory if it uses just MAX*5*3 of integers? https://mirror.codeforces.com/contest/1110/submission/51139125
• » » 4 years ago, # ^ | 0 Because recursive calls themself also consumes O(N) memory in total. Each call there are multiple ints used, such as the parameter passed in, local variable, etc.
» 4 years ago, # | ← Rev. 6 → 0 GreenGrape For 1110C — Meaningless Operations, Can you explain Why you take b=(a xor (2^x-1)) it looks bias,can you explain it more?
• » » 3 years ago, # ^ | ← Rev. 2 → 0 We try and make (a xor b) = 2^x-1 if possible{when a!= 2^x-1},since we notice that if (a xor b)=2^x-1, then (a & b) is 0, and gcd(x,0) = x Now, since (a xor b) = 2^x-1, b = (a xor 2^x-1)
» 3 years ago, # | 0 hey guys!!! can anyone pls tell why binary search doesnt work for the Problem B (Tape)My Code for the same
» 4 months ago, # | ← Rev. 2 → 0 IMO newbies,pupil should not be allowed to participate in global round, IMO it would be too much of pushing their limits and it may backfire their motivation It should be considere at par with Div1 | 5,737 | 18,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.892635 |
https://www.answerspile.com/ExpertAnswers/you-may-need-to-use-the-appropriate-technology-to-answer-this-question-consider-the-following-hypoth-pa792 | 1,713,185,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00035.warc.gz | 579,879,168 | 5,822 | # (Solved): You may need to use the appropriate technology to answer this question.Consider the following hypoth ...
You may need to use the appropriate technology to answer this question.
Consider the following hypothesis test.
H0: ???? = 100
Ha: ???? ≠ 100
A sample of 65 is used. Identify the p-value and state your conclusion for each of the following sample results. Use
???? = 0.05.
(a)
x = 103 and s = 11.4
Find the value of the test statistic. (Round your answer to three decimal places.)
p-value =
Do not reject H0. There is insufficient evidence to conclude that ???? ≠ 100.Reject H0. There is insufficient evidence to conclude that ???? ≠ 100. Do not reject H0. There is sufficient evidence to conclude that ???? ≠ 100.Reject H0. There is sufficient evidence to conclude that ???? ≠ 100.
(b)
x = 95.5 and s = 11.0
Find the value of the test statistic. (Round your answer to three decimal places.)
p-value =
Do not reject H0. There is insufficient evidence to conclude that ???? ≠ 100.Reject H0. There is insufficient evidence to conclude that ???? ≠ 100. Do not reject H0. There is sufficient evidence to conclude that ???? ≠ 100.Reject H0. There is sufficient evidence to conclude that ???? ≠ 100.
(c)
x = 102 and s = 10.4
Find the value of the test statistic. (Round your answer to three decimal places.)
p-value =
We have an Answer from Expert | 347 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.918478 |
https://includestdio.com/229.html | 1,603,965,946,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00203.warc.gz | 368,290,408 | 9,542 | # 问题内容:
Making a 4-sided pyramid is explained ‘lightly’ in a few texts, as follows:
``````pyramid.getPoints().addAll(0,0,0); //0 = top
pyramid.getPoints().addAll(0, height, -hypotenuse/2); //1 = closest
pyramid.getPoints().addAll(-hypotenuse/2, height, 0); //2 = leftest
pyramid.getPoints().addAll(hypotenuse/2, height, 0); //3 = furthest
pyramid.getPoints().addAll(0, height, hypotenuse/2); //4 = rightest
``````
It works, but I don’t understand it. See Pyramid with vertex numbering
. The 2nd face added has vertices 0, 1, 3, so (reference figure) … it slices the pyramid in half. It is not an external face as far as I can tell.
Same with the 4th face, only now the slice is orthogonal to the 2nd face. And then the last 2 faces, which should be the triangles making up the square base of the pyramid. The first one goes from vertex 4 to vertex 1 to vertex 2, so … that is the front triangle of the pyramid base (is what I think). So I expect vertices 2,3,4 to form the back triangle of the pyramid base, but in the last line of code we see vertices 4,3,1, which (according to my logic) make up the right triangle of the pyramid base, i.e. does not complement the front triangle of the pyramid base. Can someone please explain this on-the-face-of-it simple geometric puzzle? And is there a proper indepth resource out there I can study?
Much obliged – Michael
## 原文地址：
https://stackoverflow.com/questions/47755750/javefx-trianglemesh-pyramid-it-works-but-why
Tags:, | 415 | 1,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-45 | latest | en | 0.840046 |
http://spmaddmaths.onlinetuition.com.my/2016/10/26-quadratic-equations-spm-practice_89.html | 1,503,270,407,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00661.warc.gz | 407,745,739 | 22,526 | # 2.6 Quadratic Equations, SPM Practice (Paper 2)
2.6 Quadratic Equations, SPM Practice (Paper 2)
Question 3:
If α and β are the roots of the quadratic equation 3x2 + 2x – 5 = 0, form the quadratic equations that have the following roots.
Solutions:
3x2 + 2x – 5 = 0
a = 3, b = 2, c = –5
The roots are α and β.
$\begin{array}{l}\alpha +\beta =-\frac{b}{a}=-\frac{2}{3}\\ \alpha \beta =\frac{c}{a}=\frac{-5}{3}\end{array}$
(a)
Using the formula, x2 – (sum of roots)x + product of roots = 0
${x}^{2}-\left(\frac{4}{5}\right)x+\left(-\frac{12}{5}\right)=0$
$\begin{array}{l}=\alpha +\beta +\left(\frac{2}{\alpha }+\frac{2}{\beta }\right)=\alpha +\beta +\frac{2\alpha +2\beta }{\alpha \beta }\\ =\alpha +\beta +\frac{2\left(\alpha +\beta \right)}{\alpha \beta }\\ =\frac{4}{5}+\frac{2\left(\frac{4}{5}\right)}{-\frac{12}{5}}\\ =\frac{4}{5}-\frac{2}{3}=\frac{2}{15}\end{array}$
$\begin{array}{l}=-\frac{12}{5}+4+\frac{4}{-\frac{12}{5}}\\ =-\frac{12}{5}+4-\frac{5}{3}=-\frac{1}{15}\end{array}$
${x}^{2}-\left(\frac{2}{15}\right)x+\left(-\frac{1}{15}\right)=0$ | 451 | 1,057 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2017-34 | longest | en | 0.586404 |
https://us.metamath.org/mpeuni/climliminflimsup4.html | 1,719,150,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00670.warc.gz | 515,302,813 | 6,095 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > climliminflimsup4 Structured version Visualization version GIF version
Theorem climliminflimsup4 42392
Description: A sequence of real numbers converges if and only if its superior limit is real and equal to its inferior limit. (Contributed by Glauco Siliprandi, 2-Jan-2022.)
Hypotheses
Ref Expression
climliminflimsup4.1 (𝜑𝑀 ∈ ℤ)
climliminflimsup4.2 𝑍 = (ℤ𝑀)
climliminflimsup4.3 (𝜑𝐹:𝑍⟶ℝ)
Assertion
Ref Expression
climliminflimsup4 (𝜑 → (𝐹 ∈ dom ⇝ ↔ ((lim sup‘𝐹) ∈ ℝ ∧ (lim inf‘𝐹) = (lim sup‘𝐹))))
Proof of Theorem climliminflimsup4
StepHypRef Expression
1 climliminflimsup4.1 . . 3 (𝜑𝑀 ∈ ℤ)
2 climliminflimsup4.2 . . 3 𝑍 = (ℤ𝑀)
3 climliminflimsup4.3 . . 3 (𝜑𝐹:𝑍⟶ℝ)
41, 2, 3climliminflimsup2 42390 . 2 (𝜑 → (𝐹 ∈ dom ⇝ ↔ ((lim sup‘𝐹) ∈ ℝ ∧ (lim sup‘𝐹) ≤ (lim inf‘𝐹))))
53frexr 41958 . . . 4 (𝜑𝐹:𝑍⟶ℝ*)
61, 2, 5liminfgelimsupuz 42369 . . 3 (𝜑 → ((lim sup‘𝐹) ≤ (lim inf‘𝐹) ↔ (lim inf‘𝐹) = (lim sup‘𝐹)))
76anbi2d 631 . 2 (𝜑 → (((lim sup‘𝐹) ∈ ℝ ∧ (lim sup‘𝐹) ≤ (lim inf‘𝐹)) ↔ ((lim sup‘𝐹) ∈ ℝ ∧ (lim inf‘𝐹) = (lim sup‘𝐹))))
84, 7bitrd 282 1 (𝜑 → (𝐹 ∈ dom ⇝ ↔ ((lim sup‘𝐹) ∈ ℝ ∧ (lim inf‘𝐹) = (lim sup‘𝐹))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 209 ∧ wa 399 = wceq 1538 ∈ wcel 2114 class class class wbr 5042 dom cdm 5532 ⟶wf 6330 ‘cfv 6334 ℝcr 10525 ≤ cle 10665 ℤcz 11969 ℤ≥cuz 12231 lim supclsp 14818 ⇝ cli 14832 lim infclsi 42332 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2178 ax-ext 2794 ax-rep 5166 ax-sep 5179 ax-nul 5186 ax-pow 5243 ax-pr 5307 ax-un 7446 ax-cnex 10582 ax-resscn 10583 ax-1cn 10584 ax-icn 10585 ax-addcl 10586 ax-addrcl 10587 ax-mulcl 10588 ax-mulrcl 10589 ax-mulcom 10590 ax-addass 10591 ax-mulass 10592 ax-distr 10593 ax-i2m1 10594 ax-1ne0 10595 ax-1rid 10596 ax-rnegex 10597 ax-rrecex 10598 ax-cnre 10599 ax-pre-lttri 10600 ax-pre-lttrn 10601 ax-pre-ltadd 10602 ax-pre-mulgt0 10603 ax-pre-sup 10604 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2801 df-cleq 2815 df-clel 2894 df-nfc 2962 df-ne 3012 df-nel 3116 df-ral 3135 df-rex 3136 df-reu 3137 df-rmo 3138 df-rab 3139 df-v 3471 df-sbc 3748 df-csb 3856 df-dif 3911 df-un 3913 df-in 3915 df-ss 3925 df-pss 3927 df-nul 4266 df-if 4440 df-pw 4513 df-sn 4540 df-pr 4542 df-tp 4544 df-op 4546 df-uni 4814 df-int 4852 df-iun 4896 df-br 5043 df-opab 5105 df-mpt 5123 df-tr 5149 df-id 5437 df-eprel 5442 df-po 5451 df-so 5452 df-fr 5491 df-we 5493 df-xp 5538 df-rel 5539 df-cnv 5540 df-co 5541 df-dm 5542 df-rn 5543 df-res 5544 df-ima 5545 df-pred 6126 df-ord 6172 df-on 6173 df-lim 6174 df-suc 6175 df-iota 6293 df-fun 6336 df-fn 6337 df-f 6338 df-f1 6339 df-fo 6340 df-f1o 6341 df-fv 6342 df-isom 6343 df-riota 7098 df-ov 7143 df-oprab 7144 df-mpo 7145 df-om 7566 df-1st 7675 df-2nd 7676 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-1o 8089 df-oadd 8093 df-er 8276 df-pm 8396 df-en 8497 df-dom 8498 df-sdom 8499 df-fin 8500 df-sup 8894 df-inf 8895 df-pnf 10666 df-mnf 10667 df-xr 10668 df-ltxr 10669 df-le 10670 df-sub 10861 df-neg 10862 df-div 11287 df-nn 11626 df-2 11688 df-3 11689 df-n0 11886 df-z 11970 df-uz 12232 df-q 12337 df-rp 12378 df-xneg 12495 df-xadd 12496 df-ioo 12730 df-ico 12732 df-fz 12886 df-fzo 13029 df-fl 13157 df-ceil 13158 df-seq 13365 df-exp 13426 df-cj 14449 df-re 14450 df-im 14451 df-sqrt 14585 df-abs 14586 df-limsup 14819 df-clim 14836 df-rlim 14837 df-liminf 42333 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 2,157 | 3,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | latest | en | 0.136856 |
https://fhkma.org.tw/old-fashioned-cizuyb/ea4139-explain-the-various-classes-of-measuring-instruments-with-examples | 1,631,886,940,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055645.75/warc/CC-MAIN-20210917120628-20210917150628-00184.warc.gz | 296,785,254 | 9,812 | Interval Scales 4. Some can also be used to find solutions to highly complex equations. A micrometerr is used to measure the diameter of the given specimen only. In the last class, we had discussed Angular Measurements and Linear Instruments in a detailed way. types of Measuring Instruments and their Uses in a detailed way. Surface Measuring Instruments. 5) Do calculations: Some measuring instruments can also carry out a number of calculations like addition, subtraction, multiplication, division, etc. The various types of tools used in the Fitting Workshop are as follows. Sometimes these instrument show some readings even though there is no applied measurable parameter. Nowadays, the digital instruments are becoming very popular, which indicate the values directly in numerical form and even in decimals thus making them easy to read and more accurate. Nominal or Classificatory Scales 2. Below is the list of measuring instruments used in electrical and electronic work. The method used ought to be provable. -. The classification of measuring instruments according to their accuracy is to be effected by drawing up classes characterizing different levels of accuracy for instruments of the same category. In quantitative research, you have to consider the reliability and validity of your methods and measurements.. Validity tells you how accurately a method measures something. Instrumental – They are inherent in measuring instruments, because of their mechanical structure (e.g. This is the complete explanation of types of Measuring Instruments and their Uses in a detailed way. 1.2.1. There are many other functions performed by the instruments as indicated below: 1) Indicating the value of the physical quantity: The instruments are calibrated against the standard values of the physical quantities. In this blog we often discuss one form of metrology, but there are actually three major types of metrology used in today’s world. These Try squares come in various sizes from 75mm-300mm. There are basically three types of systematic errors, namely – a. Engineering Drawing Instruments and Their Usage [PDF], In this paper, I am going to familiarize you with the Engineering Drawing Instrument. There are also digital instruments for measurement of the blood pressure, heart beat rate, and others. The ability of the instrument to measure the accurate value is known as accuracy. They can give the readings in one or more decimal places. Gross errors are caused by mistake in using instruments or meters, calculating measurement and recording data results. There are also some disadvantages of the digital instruments. Instrumental Errors Similarly, the thermostat starts or stops the compressor of the refrigeration system depending on the temperature achieved in the evaporator. The movement of the pointer directly indicates the magnitude of the quantity, which can be whole numbers or also fractions. Copyright © 2020 Bright Hub PM. For measuring temperature we use different methods and devices. Types of Angular Measuring Instruments: There are various types of Angular Measurements which are based on various standards, and those are the following:…, Fitting Workshop Tools: Holding, Measuring, Marking, Cutting, Finishing, Striking & Power Tools [PDF], Fitting is the process of assembling various parts manufactured in the machine shop. 2. This is used for marking parallel lines from a finished edge and also for locating the center. The analogue instruments indicate the magnitude of the quantity in the form of the pointer movement. Name Purpose Ammeter (Ampermeter) Measures current Capacitance meter: Measures the capacitance of a component Current clamp: Measures current without physical connection Curve tracer: In case of high humidity and corrosive atmosphere the internal parts may get damaged and indicate the faulty values. Vernier Callipers is used to measure the dimensions of the given specimen like diameter(Outer Dia and Inner dia), length and depth etc. Working with absolute instruments for routine work is time consuming since every time a measurement is made, it takes a lot of time to compute the magnitude of quantity under measurement. The surface plate is made up of Cast Iron(C.I.) The instruments kept in unsafe locations like high temperature can be connected by wires and their output can be taken at some distant places which are safe for the human beings. Linear Measuring Instruments. Pressure is typically measured in units of force per unit of surface area.Many techniques have been developed for the measurement of pressure and vacuum.Instruments used to measure and display pressure in an integral unit are called pressure meters or pressure gauges or vacuum gauges. There are different types of measurement instruments that can be used by researchers for their studies; it depends on the nature of research that is to be carried out. People in ancient times used the different parts of their bodies to size things up. There are different types of measurement instruments that can be used by researchers for their studies; it depends on the nature of research that is to be carried out. Gauge – Reference to atmospheric pressure. Try to step back in time and imagine a world without measurement tools. Required fields are marked *, Engineering Drawing Instruments and Their Usage [PDF] In this paper, I am going to familiarize you with the Engineering Drawing Instrument. It is more precise than the marking table. The instruments indicate the value of these quantities, based on which we get some understanding and also take appropriate actions and decisions. Measuring instruments, measurement instruments, test equipment, measuring tools, measurement, hand held equipment like thermometers, tachometers... At the moment we provide about 500 different types and models of high quality measuring instruments to do tests and analysis of various physical, electrical, chemical and other parameters. ADVERTISEMENTS: This article throws light upon the four main types of scales used for measurement. The surface plate is used in conjunction with accessories such as a. Divider is used for marking circles, arcs, laying out perpendicular lines, bisecting lines, etc. These instruments are keep in safe place and maintaining the clean and dry places. The Surface plate is used for testing the flatness of a w/p.It is also used for Precision inspection, marking out layout and tooling setup. In this write up, we shall discuss various measurement instruments that can be used alongside with studies that are suitable for them. There are two main types of the measuring instruments: analogue and digital. 2) Measuring instruments used as the controllers: There are number of instruments that can be used as the controllers. Taper Measuring Instruments. Ratio Scales. You can find digital instruments in the cars, air planes, motor cycles, and also in places like air ports, railway stations, public places etc. For instance, when a certain value of the pressure is reached, the measuring instrument breaks the electrical circuit, which stops the running of compressor. Metrology, while often confused with the science of measuring weather (meteorology), is a very widely used field. Ordinal or Ranking Scales 3. Classification of Measuring Instruments The instrument used for measuring the physical and electrical quantities is known as the measuring instrument. The various types of tools used in the Fitting Workshop are as follows. Measuring Instruments Ranges. Here are some of the advantages of the digital instruments over the analogue instruments: Since there are very few moving parts in the electronic instruments, they are usually more accurate than the analogue instruments. Let’s take a look at the three different types of metrology. Type # 1. Nominal or Classificatory Scales: When numbers or other symbols are used simply to classify an object, person or […] Linear measuring instruments are divided into “4” types which are based on the utility. Here they are: Sometimes they tend to indicate erratic values due to faulty electronic circuit or damaged display. The term measurement means the comparison between the two quantities of the same unit. 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Convenient to use and also for locating the center markings on the temperature achieved in the last class, will... Transmitted throughout the system is classified into three types which are inverted to each other and connected! | 3,513 | 18,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-39 | latest | en | 0.949068 |
http://nrich.maths.org/public/leg.php?code=-334&cl=3&cldcmpid=2305 | 1,438,170,137,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986423.95/warc/CC-MAIN-20150728002306-00014-ip-10-236-191-2.ec2.internal.warc.gz | 179,715,894 | 9,755 | # Search by Topic
#### Resources tagged with Games similar to Buses:
Filter by: Content type:
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### There are 96 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Games
### 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.
### Behind the Rules of Go
##### Stage: 4 and 5
This article explains the use of the idea of connectedness in networks, in two different ways, to bring into focus the basics of the game of Go, namely capture and territory.
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Sufficient but Not Necessary: Two Eyes and Seki in Go
##### Stage: 4 and 5
The game of go has a simple mechanism. This discussion of the principle of two eyes in go has shown that the game does not depend on equally clear-cut concepts.
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Nim
##### Stage: 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### 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!
### Low Go
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players. Take turns to place a counter so that it occupies one of the lowest possible positions in the grid. The first player to complete a line of 4 wins.
### Learning Mathematics Through Games Series: 4. from Strategy Games
##### Stage: 1, 2 and 3
Basic strategy games are particularly suitable as starting points for investigations. Players instinctively try to discover a winning strategy, and usually the best way to do this is to analyse. . . .
### 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?
### Patience
##### Stage: 3 Challenge Level:
A simple game of patience which often comes out. Can you explain why?
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Turnablock
##### Stage: 4 Challenge Level:
A simple game for 2 players invented by John Conway. It is played on a 3x3 square board with 9 counters that are black on one side and white on the other.
### Winning Lines
##### Stage: 2, 3 and 4 Challenge Level:
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### One, Three, Five, Seven
##### Stage: 3 and 4 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### 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.
### Going First
##### Stage: 4 and 5
This article shows how abstract thinking and a little number theory throw light on the scoring in the game Go.
### Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Everthing you have always wanted to do with dominoes! Some of these games are good for practising your mental calculation skills, and some are good for your reasoning skills.
### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### 18 Hole Light Golf
##### Stage: 1, 2, 3 and 4 Challenge Level:
The computer starts with all the lights off, but then clicks 3, 4 or 5 times at random, leaving some lights on. Can you switch them off again?
### Advent Calendar 2010
##### Stage: 1, 2, 3 and 4 Challenge Level:
Advent Calendar 2010 - a mathematical game for every day during the run-up to Christmas.
### Sprouts
##### Stage: 2, 3, 4 and 5 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### Nim-like Games
##### Stage: 2, 3 and 4 Challenge Level:
A collection of games on the NIM theme
### Square it for Two
##### Stage: 1, 2, 3 and 4 Challenge Level:
Square It game for an adult and child. Can you come up with a way of always winning this game?
### The Unmultiply Game
##### Stage: 2, 3 and 4 Challenge Level:
Unmultiply is a game of quick estimation. You need to find two numbers that multiply together to something close to the given target - fast! 10 levels with a high scores table.
### Spiralling Decimals for Two
##### Stage: 2 and 3 Challenge Level:
Spiralling Decimals game for an adult and child. Can you get three decimals next to each other on the spiral before your partner?
### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### Some Games That May Be Nice or Nasty for Two
##### Stage: 2 and 3 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your oponent.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Tangram Pictures
##### Stage: 1, 2 and 3 Challenge Level:
Use the tangram pieces to make our pictures, or to design some of your own!
### Squayles
##### Stage: 3 Challenge Level:
A game for 2 players. Given an arrangement of matchsticks, players take it is turns to remove a matchstick, along with all of the matchsticks that touch it.
### Spiralling Decimals
##### Stage: 2 and 3 Challenge Level:
Take turns to place a decimal number on the spiral. Can you get three consecutive numbers?
### Snail Trails
##### Stage: 3 Challenge Level:
This is a game for two players. You will need some small-square grid paper, a die and two felt-tip pens or highlighters. Players take turns to roll the die, then move that number of squares in. . . .
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Ratio Pairs 3
##### Stage: 3 and 4 Challenge Level:
Match pairs of cards so that they have equivalent ratios.
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
### Domino Magic Rectangle
##### Stage: 2, 3 and 4 Challenge Level:
An ordinary set of dominoes can be laid out as a 7 by 4 magic rectangle in which all the spots in all the columns add to 24, while those in the rows add to 42. Try it! Now try the magic square...
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Making Maths: Snake Pits
##### Stage: 1, 2 and 3 Challenge Level:
A game to make and play based on the number line.
### Ratio Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
### Diamond Collector
##### Stage: 3 Challenge Level:
Collect as many diamonds as you can by drawing three straight lines.
### Twin Corresponding Sudoku III
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Quadruple Clue Sudoku
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Twin Corresponding Sudokus II
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Square It
##### Stage: 1, 2, 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.
### Connect Three
##### Stage: 3 Challenge Level:
Can you be the first to complete a row of three? | 2,299 | 9,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2015-32 | longest | en | 0.900744 |
https://community.rstudio.com/t/r-error-x-argument-is-missing-with-no-default/109299 | 1,652,912,190,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00094.warc.gz | 219,978,006 | 9,145 | # R (Error): x argument is missing, with no default
I am working with R.
I have some data ("train_data") below:
``````# create some data for this example
a1 = rnorm(1000,100,10)
b1 = rnorm(1000,100,5)
c1 = sample.int(1000, 1000, replace = TRUE)
train_data = data.frame(a1,b1,c1)
#view data
a1 b1 c1
1 110.36832 90.66670 662
2 96.28321 102.68244 810
3 101.95640 98.17639 956
4 121.58001 93.04896 697
5 95.08541 104.64527 712
``````
In this example, I am interested in performing an arbitrary task:
1. Choose 7 random numbers ( `"random_1"` (between 80 and 120), `"random_2"` (between "random_1" and 120) , `"random_3"` (between 85 and 120), `"random_4"` (between random_2 and 120), `"split_1"` (between 0 and 1), `"split_2"` (between 0 and 1), `"split_3"` (between 0 and 1 ))
2. Using these random numbers, perform a series of data manipulation procedures on "train_data" (these data manipulation procedures will be defined in the function below). For a specific set of 7 numbers, these data manipulation procedures will calculate a "total" mean.
3. Repeat steps 1) and 2) and see if you can find the set of these 7 numbers that produce the biggest value of the "total" mean.
Thus, I am trying to perform an optimization task.
Earlier, I was able to solve this task using a "random search":
``````# code for random search
results_table <- data.frame()
for (i in 1:10 ) {
#generate random numbers
random_1 = runif(1, 80, 120)
random_2 = runif(1, random_1, 120)
random_3 = runif(1, 85, 120)
random_4 = runif(1, random_3, 120)
#bin data according to random criteria
train_data <- train_data %>% mutate(cat = ifelse(a1 <= random_1 & b1 <= random_3, "a", ifelse(a1 <= random_2 & b1 <= random_4, "b", "c")))
train_data\$cat = as.factor(train_data\$cat)
#new splits
a_table = train_data %>%
filter(cat == "a") %>%
select(a1, b1, c1, cat)
b_table = train_data %>%
filter(cat == "b") %>%
select(a1, b1, c1, cat)
c_table = train_data %>%
filter(cat == "c") %>%
select(a1, b1, c1, cat)
split_1 = runif(1,0, 1)
split_2 = runif(1, 0, 1)
split_3 = runif(1, 0, 1)
#calculate random quantile ("quant") for each bin
table_a = data.frame(a_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_1)))
table_b = data.frame(b_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_2)))
table_c = data.frame(c_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_3)))
#create a new variable ("diff") that measures if the quantile is bigger tha the value of "c1"
table_a\$diff = ifelse(table_a\$quant > table_a\$c1,1,0)
table_b\$diff = ifelse(table_b\$quant > table_b\$c1,1,0)
table_c\$diff = ifelse(table_c\$quant > table_c\$c1,1,0)
#group all tables
final_table = rbind(table_a, table_b, table_c)
#create a table: for each bin, calculate the average of "diff"
final_table_2 = data.frame(final_table %>%
group_by(cat) %>%
summarize(
mean = mean(diff)
))
#add "total mean" to this table
final_table_2 = data.frame(final_table_2 %>% add_row(cat = "total", mean = mean(final_table\$diff)))
#format this table: add the random criteria to this table for reference
final_table_2\$random_1 = random_1
final_table_2\$random_2 = random_2
final_table_2\$random_3 = random_3
final_table_2\$random_4 = random_4
final_table_2\$split_1 = split_1
final_table_2\$split_2 = split_2
final_table_2\$split_3 = split_3
final_table_2\$iteration_number = i
results_table <- rbind(results_table, final_table_2)
final_results = dcast(setDT(results_table), iteration_number + random_1 + random_2 + random_3 + random_4 + split_1 + split_2 + split_3 ~ cat, value.var = 'mean')
#keep 5 largest resuts
}
``````
Now, we can view the results of the random search:
`````` #view results
final_results
iteration_number random_1 random_2 random_3 random_4 split_1 split_2 split_3 a b c total
1: 8 104.52182 104.8939 96.63609 99.14640 0.45389635 0.7970865 0.8264969 0.4560440 0.7954545 0.8265306 0.755
2: 10 119.04797 119.9907 93.13250 93.62925 0.27018809 0.5025505 0.6707737 0.2758621 0.5000000 0.6681465 0.632
3: 1 114.69535 117.7922 109.89274 116.39624 0.61857197 0.9609914 0.2661892 0.6180022 0.9615385 0.2702703 0.623
4: 6 85.64905 100.8127 94.02205 106.41212 0.00197946 0.7476889 0.1235777 0.2500000 0.7470588 0.1234568 0.442
5: 3 106.14908 119.7681 95.61753 100.73192 0.20678470 0.1787206 0.7166830 0.2111801 0.1802030 0.7146067 0.423
``````
According to the above table (for a very small random search of 10 iterations), the combination of " `random_1, random_2, random_3, random_4, split_1, split_2, split_3` " = ( 104.52182 104.8939 96.63609 99.14640 0.45389635 0.7970865 0.8264969) produces the highest "total" of 0.755 .
My Problem: The "random search" is not a very effective way at solving this problem. I am trying to use a different optimization algorithm to try and identify a set of `random_1, random_2, random_3, random_4, split_1, split_2, split_3` that produces the biggest value of `total` .
From the following link (A quick tour of GA), I decided to follow the example for optimizing this problem using an optimization algorithm called the "genetic algorithm":
``````#example of the genetic algorithm
library(GA)
#define function
Rastrigin <- function(x1, x2)
{
20 + x1^2 + x2^2 - 10*(cos(2*pi*x1) + cos(2*pi*x2))
}
x1 <- x2 <- seq(-5.12, 5.12, by = 0.1)
f <- outer(x1, x2, Rastrigin)
#run optimization algorithm
GA <- ga(type = "real-valued",
fitness = function(x) -Rastrigin(x[1], x[2]),
lower = c(-5.12, -5.12), upper = c(5.12, 5.12),
popSize = 50, maxiter = 1000, run = 100)
#view results of the genetic algorithm (the answer that optimizes the function in this example is (x1 = 5.4 e-05, x2 = 6.400 e-05)
summary(GA)
x1 x2
[1,] 5.41751e-05 6.400989e-05
``````
I now want to apply the "genetic algorithm" to my problem. This requires the user to define a "fitness function" that formalizes the requirements and directions for the "genetic algorithm". For my problem, I defined the "fitness function" as follows:
``````#define fitness function
fitness <- function(random_1, random_2, random_3, random_4, split_1, split_2, split_3) {
#bin data according to random criteria
train_data <- train_data %>% mutate(cat = ifelse(a1 <= random_1 & b1 <= random_3, "a", ifelse(a1 <= random_2 & b1 <= random_4, "b", "c")))
train_data\$cat = as.factor(train_data\$cat)
#new splits
a_table = train_data %>%
filter(cat == "a") %>%
select(a1, b1, c1, cat)
b_table = train_data %>%
filter(cat == "b") %>%
select(a1, b1, c1, cat)
c_table = train_data %>%
filter(cat == "c") %>%
select(a1, b1, c1, cat)
split_1 = runif(1,0, 1)
split_2 = runif(1, 0, 1)
split_3 = runif(1, 0, 1)
#calculate quantile ("quant") for each bin
table_a = data.frame(a_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_1)))
table_b = data.frame(b_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_2)))
table_c = data.frame(c_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_3)))
#create a new variable ("diff") that measures if the quantile is bigger tha the value of "c1"
table_a\$diff = ifelse(table_a\$quant > table_a\$c1,1,0)
table_b\$diff = ifelse(table_b\$quant > table_b\$c1,1,0)
table_c\$diff = ifelse(table_c\$quant > table_c\$c1,1,0)
#group all tables
final_table = rbind(table_a, table_b, table_c)
# calculate the total mean : this is what needs to be optimized
mean = mean(final_table\$diff)
}
``````
Just to test that this function works:
``````#call function for a specific set of the 7 numbers
a = fitness(85, 100, 90, 110, 0.5, 0.7, 0.3)
# view the corresponding "total mean"
a
[1] 0.845
``````
Now, I am trying to put everything together and instruct the "genetic algorithm" to optimize the "fitness function" I defined by considering different ranges of values for `"random_1, random_2, random_3, random_4, split_1, split_2, split_3"`
#genetic algorithm for my example:
``````GA <- ga(type = "real-valued",
fitness = fitness,
lower = c(80, 80, 80, 80, 0,0,0), upper = c(120, 120, 120, 120, 1,1,1),
popSize = 50, maxiter = 1000, run = 100)
``````
But this produces the following error:
``````Error: Problem with `mutate()` column `cat`.
i `cat = ifelse(...)`.
x argument "random_3" is missing, with no default
Run `rlang::last_error()` to see where the error occurred.
Error: Problem with `mutate()` column `cat`.
i `cat = ifelse(...)`.
x argument "random_3" is missing, with no default
Run `rlang::last_error()` to see where the error occurred.
``````
Does anyone know why this error is being produced? Can someone please show me what I am doing wrong?
Thanks
Your first fitness function has one argument: a vector.
Your second fitness function has seven arguments: all scalars.
And note that you don't use the last three split* arguments:
you redefine them in your function.
I am working with R. I am learning about how to optimize functions and estimate the maximum or minimum points of these functions.
For example, I created some random data ("train data):
``````#load libraries
library(dplyr)
# create some data for this example
a1 = rnorm(1000,100,10)
b1 = rnorm(1000,100,5)
c1 = sample.int(1000, 1000, replace = TRUE)
train_data = data.frame(a1,b1,c1)
``````
I also created the following function ("fitness") that takes seven inputs ( `"random_1"` (between 80 and 120), `"random_2"` (between "random_1" and 120) , `"random_3"` (between 85 and 120), `"random_4"` (between random_2 and 120), `"split_1"` (between 0 and 1), `"split_2"` (between 0 and 1), `"split_3"` (between 0 and 1 )), , performs a series of data manipulation procedures and returns a "total" mean:
``````fitness <- function(random_1, random_2, random_3, random_4, split_1, split_2, split_3) {
#bin data according to random criteria
train_data <- train_data %>% mutate(cat = ifelse(a1 <= random_1 & b1 <= random_3, "a", ifelse(a1 <= random_2 & b1 <= random_4, "b", "c")))
train_data\$cat = as.factor(train_data\$cat)
#new splits
a_table = train_data %>%
filter(cat == "a") %>%
select(a1, b1, c1, cat)
b_table = train_data %>%
filter(cat == "b") %>%
select(a1, b1, c1, cat)
c_table = train_data %>%
filter(cat == "c") %>%
select(a1, b1, c1, cat)
split_1 = runif(1,0, 1)
split_2 = runif(1, 0, 1)
split_3 = runif(1, 0, 1)
#calculate quantile ("quant") for each bin
table_a = data.frame(a_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_1)))
table_b = data.frame(b_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_2)))
table_c = data.frame(c_table%>% group_by(cat) %>%
mutate(quant = quantile(c1, prob = split_3)))
#create a new variable ("diff") that measures if the quantile is bigger tha the value of "c1"
table_a\$diff = ifelse(table_a\$quant > table_a\$c1,1,0)
table_b\$diff = ifelse(table_b\$quant > table_b\$c1,1,0)
table_c\$diff = ifelse(table_c\$quant > table_c\$c1,1,0)
#group all tables
final_table = rbind(table_a, table_b, table_c)
# calculate the total mean : this is what needs to be optimized
mean = mean(final_table\$diff)
}
``````
Just as a sanity check, we can verify that this function actually works:
``````#testing the function at some specific input:
a <- fitness(80,80,80,80,0.6,0.2,0.9)
a
[1] 0.899
``````
Now, using the following reference on optimization with R (https://cran.r-project.org/web/packages/optimization/optimization.pdf and https://cran.r-project.org/web/packages/optimization/vignettes/vignette_master.pdf), I am trying to perform some common optimization techniques on this function.
For example:
``````#load library
library(optimization)
``````
Nelder-Meade Optimization with an Initial Guess:
``````optim_nm(fitness, start = c(80,80,80,80,0,0,0))
``````
``````optim_nm(fun = fitness, k = 2)
``````
Optimization using Simulated Annealing:
``````ro_sa <- optim_sa(fun = fitness,
start = c(runif(7, min = -1, max = 1)),
lower = c(80,80,80,80,0,0,0),
upper = c(120,120,120,120,1,1,1),
trace = TRUE,
control = list(t0 = 100,
nlimit = 550,
t_min = 0.1,
dyn_rf = FALSE,
rf = 1,
r = 0.7
)
)
``````
But all of these procedures return a similar error:
``````Error: Problem with `mutate()` column `cat`.
i `cat = ifelse(...)`.
x argument "random_3" is missing, with no default
Run `rlang::last_error()` to see where the error occurred.
Error: Problem with `mutate()` column `cat`.
i `cat = ifelse(...)`.
x argument "random_3" is missing, with no default
Run `rlang::last_error()` to see where the error occurred.
``````
And this is preventing me from visualizing the results of these optimization algorithms :
``````#code for visualizations
plot(ro_sa)
plot(ro_sa, type = "contour")
``````
Can someone please show me what am I doing wrong? Is it possible to fix this?
Thanks
Han identified your key issue, I would guess he bowed out of the conversation because your response did not address his posting, but seemed to be a restatement of your initial post...
That said, here is some code that you can run and reflect on how it might be relevant to your scenario.
``````
fitness1 <- function(a,b,c,d,e,f,g){paste0(a,b,c,d,e,f,g)}
fitness1(80,80,80,80,0.6,0.2,0.9)
fitness1(c(80,80,80,80,0.6,0.2,0.9))
fitness2 <- function(v){paste0(v)}
fitness2(80,80,80,80,0.6,0.2,0.9)
fitness2(c(80,80,80,80,0.6,0.2,0.9))
``````
which of these fitness*() calls error and which don't, and why ?
This topic was automatically closed 21 days after the last reply. New replies are no longer allowed.
If you have a query related to it or one of the replies, start a new topic and refer back with a link. | 4,493 | 13,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.655027 |
http://vegasallnight.com/pdf/nonlinear-flow-phenomena-and-homotopy-analysis-fluid-flow-and-heat-transfer | 1,511,279,075,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806419.21/warc/CC-MAIN-20171121151133-20171121171133-00620.warc.gz | 317,149,164 | 9,002 | ## Nonlinear Flow Phenomena and Homotopy Analysis: Fluid Flow by Kuppalapalle Vajravelu,Robert A. Van Gorder
By Kuppalapalle Vajravelu,Robert A. Van Gorder
Since many of the difficulties coming up in technological know-how and engineering are nonlinear, they're inherently tough to resolve. conventional analytical approximations are legitimate just for weakly nonlinear difficulties and sometimes fail whilst used for issues of powerful nonlinearity. “Nonlinear stream Phenomena and Homotopy research: Fluid stream and warmth move” provides the present theoretical advancements of the analytical approach to homotopy research. This booklet not just addresses the theoretical framework for the tactic, but in addition provides a few examples of nonlinear difficulties which were solved via the homotopy research approach. the actual concentration lies on fluid circulate difficulties ruled by means of nonlinear differential equations. This publication is meant for researchers in utilized arithmetic, physics, mechanics and engineering.
Both Kuppalapalle Vajravelu and Robert A. Van Gorder paintings on the collage of relevant Florida, USA.
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Extra resources for Nonlinear Flow Phenomena and Homotopy Analysis: Fluid Flow and Heat Transfer
Example text | 588 | 3,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-47 | longest | en | 0.906438 |
https://www.attaccalite.com/lumen/thg_in_silicon.html | 1,723,150,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00322.warc.gz | 514,729,954 | 3,285 | ## Third Harmonic Generation in bulk Silicon
In this tutorial we will calculate third harmonic generation (THG) in bulk Silicon. Calculation of third harmonic generation proceeds in similar way to the SHG calculations. DFT input can be found here (abinit or quantum espresso). We import the wave-functions as explained in the SHG tutorial, then perform the setup using only 1000 plane waves.
We are interested in the $$\chi^{(3)}_{1111}$$ therefore we consider an external field in direction [1, 0, 0] (equivalent to [x,0,0]).
We remove all the symmetries not compatible with the external field plus the time-reversal symmetry by doing ypp -n:
fixsyms # [R] Reduce Symmetries
% Efield1
1.00 | 0.00 | 0.00 | # First external Electric Field
%
% Efield2
0.00 | 0.00 | 0.00 | # Additional external Electric Field
%
#RmAllSymm # Remove all symmetries
RmTimeRev # Remove Time Reversal
Then you go in the FixSymm folder, run again the setup (as explained in the first tutorial) and then generate the input file with the command lumen -u -V qp -F input.in:
nlinear # [R NL] Non-linear optics
% NLBands
1 | 7 | # [NL] Bands
%
NLstep= 0.0100 fs # [NL] Real Time step length
NLtime=-1.000000 fs # [NL] Simulation Time
NLintegrator= "CRANKNIC" # [NL] Integrator ("EULEREXP/RK4/RK2EXP/HEUN/INVINT/CRANKNIC")
NLCorrelation= "IPA" # [NL] Correlation ("IPA/HARTREE/TDDFT/LRC/JGM/HF/SEX")
NLLrcAlpha= 0.000000 # [NL] Long Range Correction
% NLEnRange
0.200000 | 2.000000 | eV # [NL] Energy range
%
NLEnSteps= 6 # [NL] Energy steps
NLDamping= 0.200000 eV # [NL] Damping
NLGvecs= 941 RL # [NL] Number of G vectors in NL dynamics for Hartree/TDDFT
NLInSteps= 1 # [NL] Intensity steps for Richardson extrap. (1-3)
% ExtF_Dir
1.000000 | 0.000000 | 0.000000 | # [NL ExtF] Versor
%
ExtF_Int=0.1000E+8 kWLm2 # [NL ExtF] Intensity
ExtF_kind= "SOFTSIN" # [NL ExtF] Kind(SIN|SOFTSIN|RES|ANTIRES|GAUSS|DELTA|PULSE)
% GfnQP_E
0.600000 | 1.000000 | 1.000000 | # [EXTQP G] E parameters (c/v) eV|adim|adim
%
We calculate the $$\chi^{(3)}_{1111}$$ only on 6 frequencies, but calculations can be extened to larger frequency range. We introduce a scissor operator of 0.6 eV in such a way to reproduce the gap of bulk silicon. Notice that we increased the field intensity in such a way to improve the ratio between non-linear response and numerical noise.
Now that you have created the file input.in you can run the non-linear optics calculation with the command:
lumen -F input.in
when the simulation will end, it will have produced the files SAVE/ndb.Nonlinear.
Results can be analized in the same way of the SHG, using the command ypp -u:
nonlinear # [R] NonLinear Optics Post-Processing
Xorder= 5 # Max order of the response functions
% TimeRange
40 | -1.00000 | fs # Time-window where processing is done
%
ETStpsRt= 200 # Total Energy steps
% EnRngeRt
0.00000 | 10.00000 | eV # Energy range
%
DampMode= "NONE" # Damping type ( NONE | LORENTZIAN | GAUSSIAN )
DampFactor= 0.10000 eV # Damping parameter
This time we increase the dephasing time to 40 fs in such a way to have a clean third-harminic response.
Hereafter we report the result with the 8x8x8 k-points sampling and the converged result with the 24x24x24 k-points sampling: | 1,065 | 3,549 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.645294 |
https://www.cnblogs.com/iwtwiioi/p/4160915.html | 1,722,905,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00073.warc.gz | 559,567,447 | 9,585 | # 【BZOJ】1089: [SCOI2003]严格n元树（递推+高精度/fft）
http://www.lydsy.com/JudgeOnline/problem.php?id=1089
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
const double PI=acos(-1.0);
struct big {
static const int N=10005;
static const ll M=10000000;
ll a[N];
void clr() { memset(a, 0, sizeof(ll)*(a[0]+1)); a[0]=1; }
void upd() { int len=a[0]; while(len>1 && a[len]==0) --len; a[0]=len; }
big() { a[0]=1; memset(a, 0, sizeof a); }
big(ll k) { a[0]=1; memset(a, 0, sizeof a); *this=k; }
big(const big &k) { a[0]=1; memset(a, 0, sizeof a); *this=k; }
big(char *k) { a[0]=1; memset(a, 0, sizeof a); *this=k; }
big & operator=(ll x) {
clr(); int len=0;
if(x==0) len=1;
while(x) a[++len]=x%M, x/=M;
a[0]=len;
return *this;
}
big & operator=(char *s) {
clr();
int n=strlen(s), len=1; ll k=1;
for3(i, n-1, 0) {
a[len]+=k*(s[i]-'0');
k*=10;
if(k>=M) { k=1; ++len; }
}
a[0]=len;
return *this;
}
big & operator=(const big &x) { clr(); memcpy(a, x.a, sizeof(ll)*(x.a[0]+1)); return *this; }
big & operator+(const big &x) {
static big t;
t.clr();
int len=max(x.a[0], a[0]), i=1; ll k=0;
for(; i<=len || k; ++i) {
t.a[i]=a[i]+x.a[i]+k;
k=t.a[i]/M;
if(t.a[i]>=M) t.a[i]%=M;
}
t.a[0]=i; t.upd();
return t;
}
big & operator-(const big &x) {
static big t;
t.clr();
for1(i, 1, a[0]) {
t.a[i]+=a[i]-x.a[i];
if(t.a[i]<0) --t.a[i+1], t.a[i]+=M;
}
t.a[0]=a[0]; t.upd();
return t;
}
big & operator*(const big &x) {
static big t;
t.clr();
fft(a+1, x.a+1, t.a+1, a[0], x.a[0], t.a[0]);
t.upd();
return t;
}
// big & operator*(const big &x) {
// static big t;
// t.clr();
// for1(i, 1, a[0]) for1(j, 1, x.a[0]) t.a[i+j-1]+=a[i]*x.a[j];
// int len=a[0]+x.a[0]-1, i=1; ll k=0;
// for(; i<=len || k; ++i) {
// t.a[i]+=k;
// k=t.a[i]/M;
// if(t.a[i]>=M) t.a[i]%=M;
// }
// t.a[0]=i; t.upd();
// return t;
// }
big & operator*(const ll &x) { static big t; t=x; t=(*this)*t; return t; }
big & operator+(const ll &x) { static big t; t=x; t=(*this)+t; return t; }
void P() {
printf("%lld", a[a[0]]);
for3(i, a[0]-1, 1) printf("%07lld", a[i]);
}
struct cp {
double r, i;
cp(double _r=0.0, double _i=0.0) : r(_r), i(_i) {}
cp operator + (const cp &x) { return cp(r+x.r, i+x.i); }
cp operator - (const cp &x) { return cp(r-x.r, i-x.i); }
cp operator * (const cp &x) { return cp(r*x.r-i*x.i, r*x.i+i*x.r); }
};
int rev[N];
void init(int &len) {
int k=1, t=0;
while(k<len) k<<=1, ++t;
len=k;
rep(i, len) {
k=t; int ret=0, x=i;
while(k--) ret<<=1, ret|=x&1, x>>=1;
rev[i]=ret;
}
}
void dft(cp *a, int n, int flag) {
static cp A[N];
rep(i, n) A[i]=a[rev[i]];
rep(i, n) a[i]=A[i];
int m, i, j, mid;
cp t, u;
for(m=2; m<=n; m<<=1) {
cp wn(cos(2.0*PI/m*flag), sin(2.0*PI/m*flag));
for(i=0; i<n; i+=m) {
cp w(1.0); mid=m>>1;
for(j=0; j<mid; ++j) {
u=a[i+j+mid]*w, t=a[i+j];
a[i+j]=t+u;
a[i+j+mid]=t-u;
w=w*wn;
}
}
}
if(flag==-1) rep(i, n) a[i].r/=n;
}
void fft(const ll *a, const ll *b, ll *c, const ll &la, const ll &lb, ll &lc) {
static cp x[N], y[N];
int len=la+lb-1;
init(len);
rep(i, len) x[i].r=a[i], x[i].i=0;
rep(i, len) y[i].r=b[i], y[i].i=0;
dft(x, len, 1); dft(y, len, 1);
rep(i, len) x[i]=x[i]*y[i];
dft(x, len, -1);
rep(i, len) c[i]=x[i].r+0.5;
rep(i, len) c[i+1]+=c[i]/M, c[i]%=M;
lc=len+1;
}
};
big f[35], k, r;
int main() {
int n=getint(), d=getint();
if(!d) { puts("1"); return 0; }
f[0]=1;
for1(i, 1, d) {
r=1; k=f[i-1];
int t=n;
while(t) { if(t&1) r=r*k; t>>=1; k=k*k; }
f[i]=r+1;
}
r=f[d]-f[d-1];
r.P();
return 0;
}
f[i]=f[i-1]^n+1
f[d]-f[d-1]
！！！为什么要剪掉呢？因为看我们的转移，并不是深度为i，而是深度最大为i，那么为什么要这样减呢？理由很简单吧。。。f[d]表示深度最大为d的数目，f[d-1]表示深度最大为d-1的数目，那么我们只要深度为d，那么剪掉之。。。
posted @ 2014-12-13 09:05 iwtwiioi 阅读(555) 评论(1编辑 收藏 举报 | 1,980 | 4,373 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.102095 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=10491&CurriculumID=25&Method=Worksheet&NQ=10&Num=7.26&Type=C | 1,553,480,690,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00330.warc.gz | 122,244,520 | 3,412 | Name: ___________________Date:___________________
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### Grade 8 - Mathematics7.26 Special Products: (a+b)(a2-ab+b2) = a3+b3
#### Formula : (a+b)(a2-ab+b2) = a3+b3
Example:
Find the product (2x+3y)(4x2-6xy+9y2) .
Solution:
Using above formula, Let a = 2x and b = 3y in the above equation
(a+b)(a2-ab+b2) = (2x+3y)(4x2-6xy+9y2)
= (2x)3+(3y)3
= 8x3+27y3
Directions: Solve the following problems. Also write at least ten examples of your own.
Q 1: Find the product (x+3y)(x2-3xy+9y2).3x3+27y3x3+27y3x3+9y3 Q 2: Find the product (x+2y)(x2-2xy+4y2).x3+4y32x3+8y3x3+8y3 Q 3: Find the product (3x+2y)(9x2-6xy+4y2).27x3+4y327x3+8y39x3+8y3 Q 4: Find the product (5x+2y)(25x2-10xy+4y2).25x3+8y3125x3+4y3125x3+8y3 Q 5: Find the product (4x+3y)(16x2-12xy+9y2).64x3+27y364x3+9y316x3+27y3 Q 6: Find the product (3x+y)(9x2-3xy+y2).9x3+y327x3+y327x3+3y3 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 472 | 1,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-13 | longest | en | 0.683892 |
http://stackoverflow.com/questions/1563188/comparisons-of-data-structures-algorithms-basic-computer-science-online-resou/1563216 | 1,441,113,245,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645176794.50/warc/CC-MAIN-20150827031256-00310-ip-10-171-96-226.ec2.internal.warc.gz | 227,285,918 | 19,265 | # comparisons of data structures, algorithms, basic computer science, online resources
I am searching for an online resource referring to data structures and algorithms. Basicaly what I am interested in, is a kind of comprehensive list of things such as: when a data structure or algorithm is used, pros and cons compared to each other, real life-problem usage, etc.
I have a couple of popular books on the subject, like "introduction to algorithms" and "algorithm design manual", and have found code implementations of various data structures and algorithms online. But what I can't find is a web page listing them all together.
So I am looking for a reference online page, where I can have an overview of all those structures (lists, maps, sets, trees, queues and their various implementations and ideally searching or sorting algos - merge sort, quick sort, their big-O performance, etc), perhaps all in a tabular format, with their features, comparisons and general uses listed together.
Not sure if such resource exists. I would appreciate any pointing to the right direction.
-
should be community wiki – SilentGhost Oct 13 '09 at 22:08
I am not sure I understand, you mean I can search for individual community wikis on each structure or algorithm? – user189441 Oct 13 '09 at 22:29
This site has good explanations on the topics. cprogramming.com/algorithms-and-data-structures.html Comparison of Sorting Algorithms cprogramming.com/tutorial/computersciencetheory/sortcomp.html – Chinmay Nerurkar Jan 27 '12 at 18:04
-
Yes, itl.nist.gov's list is impressive! – Bart Kiers Oct 14 '09 at 11:29
I am not sure of the kind of resource you are looking for exists but there are some very good university classes on www.youtube.com/edu which will surely help you to revise the concept of data structures and algorithms.
1. Data structure class by Prof. Jonathan at UC, Berkeley is really good on youtube is very good, link here.
-
It is best to search through University websites ... they have a lot of resources on that matter ... for example http://users.cs.fiu.edu/~weiss/dsaa%5Fc++/code/
-
Did you try on Wikipedia?
There are different good pages about data structures, like this one on queues.
Have fun!
-
MIT Online has a lot of good resources
-
Here is a website showing Big O analysis comparisons of algos http://bigocheatsheet.com/
- | 530 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2015-35 | latest | en | 0.902957 |
https://www.financereference.com/transaction-exposure/ | 1,695,426,611,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00505.warc.gz | 853,264,326 | 21,897 | # Transaction Exposure
## What is Transaction Exposure
Transaction exposure is the risk that a firm will incur losses due to changes in exchange rates. It can arise when a company makes purchases or sales in foreign currency, or when it has outstanding liabilities denominated in foreign currency. When transaction exposure arises, the company may choose to hedge its risk by entering into a forward contract or an options contract. By doing so, the company can lock in an exchange rate for the future transaction, and protect itself from losses if the value of the foreign currency declines.
## How to calculate Transaction Exposure
To calculate transaction exposure, you need to know the amount of the foreign currency transaction, the current exchange rate, and the expected future exchange rate. The formula for calculating transaction exposure is: Transaction Exposure = Amount of Foreign Currency Transaction x (Current Exchange Rate – Expected Future Exchange Rate).
For example, if your company plans to buy €100,000 of French-manufactured goods in 3 months, and the current exchange rate is €1 = \$1.10, but you expect the rate in 3 months to be €1 = \$1.15, your transaction exposure would be: €100,000 x (\$1.10 – \$1.15) = -€5,000. This means that if the actual exchange rate in 3 months is €1 = \$1.15, your company will lose €5,000 on the transaction. To hedge against this loss, your company could enter into a forward contract to sell €100,000 at an agreed-upon rate of €1 = \$1.15 in 3 months.
## Factors that affect Transaction Exposure
There are a number of factors that can affect transaction exposure, including the type of business, the currency in which transactions are denominated, the timing of transactions, and the hedging strategy used. The type of business is a major factor because some businesses are more likely to be affected by currency fluctuations than others. For instance, companies that export goods or services are typically more exposed than those that import goods or services. The currency in which transactions are denominated is also a significant factor.
Transactions denominated in a currency that is experiencing high volatility will be more exposed than those denominated in a currency that is relatively stable. Timing is also important because transactions that are scheduled to settle in the near future are more exposed than those that are not due to settle for some time.
Finally, the hedging strategy used can also affect transaction exposure. For example, forward contracts can be used to hedge exposure to a specific transaction, whereas currency options can be used to hedge against general exposure to currency fluctuations. By understanding these factors, companies can take steps to minimize their transaction exposure.
## Methods for reducing Transaction Exposure
There are a number of ways that companies can reduce their exposure to transaction risk. One is to use forward contracts to lock in the exchange rate at which future transactions will be conducted. Another is to use hedging instruments such as currency options to create a buffer against potential losses. By taking steps to reduce transaction exposure, companies can protect themselves from the volatility of the foreign exchange market.
## Examples of businesses that have been impacted by Transaction Exposure
Transaction exposure is the risk that a company will incur losses due to fluctuations in exchange rates. This risk can arise when a company engages in international trade or makes investments in foreign countries. For example, consider a U.S. company that exports widgets to Canada. If the value of the Canadian dollar decreases relative to the dollar, then the company will receive less revenue from widget sales. As a result, the company’s profits will be lower than expected. Conversely, if the Canadian dollar increases in value, then the company will see an increase in profits. Businesses that are engaged in international trade or have foreign investments are particularly vulnerable to transaction exposure.
## The benefits of mitigating your Transaction Exposure
If you’re doing business in a foreign currency, it’s important to take steps to mitigate your transaction exposure. There are a few different ways to do this. One is to use forward contracts, which allow you to lock in an exchange rate for a future transaction. Another is to use hedging instruments like options or futures contracts. These can help offset the risk of loss if the value of the currency changes. By taking steps to mitigate your transaction exposure, you can help protect your finances and ensure that you’re prepared for whatever the market may bring.
## The risks of not mitigating your Transaction Exposure
When companies conduct business internationally, they are exposed to the risk that changes in currency exchange rates will adversely affect the value of their assets or liabilities denominated in foreign currencies. This risk is known as “transaction exposure.” If a company does not take steps to mitigate its transaction exposure, it could potentially suffer significant losses. There are several ways to hedge against transaction exposure, including using forward contracts, currency options, and hedging with currency-denominated debt. By entering into hedging transactions, companies can reduce the risk that they will suffer losses due to adverse movements in exchange rates. | 1,019 | 5,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-40 | longest | en | 0.936183 |
https://www.coursehero.com/file/5769839/Review-of-Linear-Systems/ | 1,496,057,361,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612283.85/warc/CC-MAIN-20170529111205-20170529131205-00073.warc.gz | 1,073,635,231 | 24,864 | Review_of_Linear_Systems
# Review_of_Linear_Systems - Digital Image Processing 3rd ed...
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Unformatted text preview: Digital Image Processing, 3rd ed. www.ImageProcessingPlace.com © 1992–2008 R. C. Gonzalez & R. E. Woods Gonzalez & Woods Review of Linear Systems Objective To provide background material in support of topics in Digital Image Processing that are based on linear system theory. Review Linear Systems Digital Image Processing, 3rd ed. www.ImageProcessingPlace.com © 1992–2008 R. C. Gonzalez & R. E. Woods Gonzalez & Woods Review of Linear Systems Some Definitions With reference to the following figure, we define a system as a unit that converts an input function f ( x ) into an output (or response) function g ( x ), where x is an independent variable, such as time or, as in the case of images, spatial position. We assume for simplicity that x is a continuous variable, but the results that will be derived are equally applicable to discrete variables. Digital Image Processing, 3rd ed. www.ImageProcessingPlace.com © 1992–2008 R. C. Gonzalez & R. E. Woods Gonzalez & Woods Review of Linear Systems Some Definitions (Con’t) It is required that the system output be determined completely by the input, the system properties, and a set of initial conditions. From the figure in the previous page, we write where H is the system operator , defined as a mapping or assignment of a member of the set of possible outputs { g ( x )} to each member of the set of possible inputs { f ( x )}. In other words, the system operator completely characterizes the system response for a given set of inputs { f ( x )}. Digital Image Processing, 3rd ed. www.ImageProcessingPlace.com © 1992–2008 R. C. Gonzalez & R. E. Woods Gonzalez & Woods Review of Linear Systems Some Definitions (Con’t) An operator H is called a linear operator for a class of inputs { f ( x )} if for all f i ( x ) and f j ( x ) belonging to { f ( x )}, where the a 's are arbitrary constants and is the output for an arbitrary input f i ( x ) ∈ { f ( x )}. Digital Image Processing, 3rd ed. www.ImageProcessingPlace.com © 1992–2008 R. C. Gonzalez & R. E. Woods Gonzalez & Woods Review of Linear Systems Some Definitions (Con’t) The system described by a linear operator is called a linear system (with respect to the same class of inputs as the operator). The property that performing a linear process on the sum of inputs is the same that performing the operations individually and then summing the results is called the property of additivity ....
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## This note was uploaded on 01/27/2010 for the course ECE TECHCOMM taught by Professor Prof.soltanianzdeh during the Spring '10 term at Shahid Beheshti University.
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Ask a homework question - tutors are online | 775 | 3,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-22 | longest | en | 0.88632 |
http://www.huffingtonpost.com/simon-jackman/chart-dump-swings-probability_b_2077997.html?icid=hp_pollster_top_art | 1,503,465,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.92/warc/CC-MAIN-20170823035753-20170823055753-00351.warc.gz | 555,807,132 | 45,698 | THE BLOG
11/05/2012 02:32 pm ET Updated Jan 05, 2013
# Chart Dump!: Swings, Probabilities, Electoral College Counts, Raw Data
An election eve chart wrap... data is still coming in, and the models are still updating, but here's where things look this morning (Pacific time).
Anticipated swing from 2008, two-party terms. The left column of data shows Obama's two-party vote share by state in 2008; the right column shows the estimated Obama two-party vote share from my model, as of this morning:
I've posted versions of this graph at various stages of the campaign. Now, on Election Eve, we're back to a position we saw in September, with Romney simply not bringing enough states below over the 50-50 line to win the election. Indiana, North Carolina, and Florida are the only states to flip, at least according to the model and the polling data. Not shown on the graph is the uncertainty associated with the estimates; i.e., Florida is not much better than a 50-50 coin flip, but a state further up the tree, like Ohio, is estimated to stay with Obama with probability approaching 100 percent. See below.
No models, no house effects, just the battleground polls. Forget the modeling, forget the house effect corrections. Just take each battleground state poll appearing in the Pollster data base in the last seven days. Convert each poll to a two-party number for Obama: Obama/(Obama + Romney)*100. Plot in the "stacked" format shown below. Moreover, weigh each poll by its sample size so as to get a poll average. Obama 50.9 percent, +/- 0.4 percentage points. That is 4.5 sigma north of 50-50. Or more simply, "OTMOE" (Outside The Margin Of Error)...
A distribution over Obama Electoral College counts. The bottom line (literally): The probability of an Obama win in the Electoral College is now about 90 percent. Three-hundred-thirty-two and 303 are the two most likely Obama EV outcomes, closely followed by 347. Averaged over the distribution: 309.
Probabilities from poll average vs probabilities from forecast. The extra uncertainty that comes from converting a poll average to a forecast weakens the confidence of the calls; note the "shrinkage" back towards the 50-50 point, moving from left (poll average) to right (forecast). Note the implied Electoral College counts from the poll averaging model has the probability of an Obama win at close to 100 percent; this shrinks back to 90 percent in the forecast model.
Uncertainty, visualized. Finally, here is a look at nine swing states plus the national popular vote. Each panel shows a series of bell-curve (normal) densities in light grey, one for each poll; the polls shown are from the last seven days. Polls are converted to Obama two-party shares: Obama/(Obama + Romney) * 100. The less spread out the density, the bigger the sample size of the corresponding poll. Yesterday's massive YouGov national poll is clearly visible in the "USA" panel (and I've actually deflated its sample size, so as to not double count the state-by-state polls that contribute to the national figure).
The black density is what comes out of my forecast model, usually shifted a little from the poll average, and substantially more dispersed than the naive poll average, reflecting the extra uncertainty that creeps via house effects corrections, turning the poll average into a forecast and some temporal discounting of older polls.The probability that Obama wins is the area under the black density to the right of the 50-50 points.
The panels are ordered from bottom to top by the probability of an Obama win. North Carolina and Florida are the least likely Obama retains in the set I examine here; Ohio and Wisconsin the most likely. The uncertainty associated with the Florida and even the national popular vote forecast is particularly vivid when presented this way.
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How will Trump’s administration impact you? | 848 | 3,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-34 | latest | en | 0.924545 |
https://studylib.net/doc/8481214/kcc2-counting-forward-21-50 | 1,606,702,137,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00515.warc.gz | 487,763,153 | 11,967 | # KCC2-Counting-Forward-21-50
```CCSS Mathematics Assessment Task
Counting Forward 21-50
Mathematics Domain and Cluster:
Domain: Counting and Cardinality
Cluster: Know number names and the count sequence.
Common Core standard(s) being assessed (if the task is intended to assess only one part of the
standard, underline that part of the standard):
K.CC.2: Count forward beginning from a given number within the known sequence (instead of having
to begin at 1).
Student Materials:
None
Teacher Materials:
Counting Forward Checklist 21-50
Pencil
Teacher Note:
The emphasis of this standard is on the counting sequence to 100. Students should be able to count
forward from any number, 1-99. This assessment only works with numbers from 21-50.
Prompt:
For problem 1, say:
For problem 2, say:
For problem 3, say:
For problem 4, say:
For problem 5, say:
For problem 6, say:
What numbers come after 22?
What numbers come after 28?
What numbers come after 31?
What numbers come after 39?
What numbers come after 43?
What numbers come after 47?
1. 23, 24, 25
2. 29, 30, 31
3. 32, 33, 34
4. 40, 41, 42
5. 44, 45, 46
6. 48, 49, 50
Scoring Guide/Rubric (a score should be awarded for each criterion below)
Criteria (CCSS code)
0 points
1 Point
K.CC.2: Count forward
beginning from a given
number.
Student is unable to
accurately count forward
to 50 from any given
number or has significant
errors.
Student is able to count
forward to 50 with minor
errors.
2 Point
Student is able to
accurately count
forward to 50 from any
given number.
Name__________________________________________________Date_____________
Counting Forward Checklist 21-50
Kindergarten Mathematics Assessment
This checklist is for the children who are unable to write 21-50 because this standard focuses on
counting rather than number writing. The emphasis of this standard is on the counting sequence
to 100. Students should be able to count forward from any number, 1-99.
Count forward beginning from a given number within the
know sequence.
Date
Date
Say, “I’m going to say a number and I want you to tell me
the numbers that come after it. So if I said, “20” you
would say, “21, 22, 23, ”
Put a check in the box if student is able to answer
correctly.
1. “What numbers come after 22?” (23, 24, 25)
2. “What numbers come after 28?” (29, 30, 31)
3. “What numbers come after 31?” (32, 33, 34)
4. “What numbers come after 39?” (40, 41, 42)
5. “What numbers come after 43?” (44, 45, 46)
6. “What numbers come after 47?” (48, 49, 50)
In the boxes below, note any errors that the students make. Please remember to date.
Quarter 1:
Quarter 2:
Quarter 3:
Quarter 4:
Date Date
``` | 756 | 2,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-50 | latest | en | 0.885866 |
https://gmatclub.com/forum/floating-in-the-waters-of-the-equatorial-pacific-an-array-24696.html?kudos=1 | 1,488,257,150,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174124.9/warc/CC-MAIN-20170219104614-00044-ip-10-171-10-108.ec2.internal.warc.gz | 700,349,339 | 58,805 | Floating in the waters of the equatorial Pacific, an array : GMAT Sentence Correction (SC)
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# Floating in the waters of the equatorial Pacific, an array
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19 Dec 2005, 04:03
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Floating in the waters of the equatorial Pacific, an array of buoys collects and transmits data on long-term interactions between the ocean and the atmosphere, interactions that affect global climate.
A. atmosphere, interactions that affect
B. atmosphere, with interactions affecting
C. atmosphere that affects
D. atmosphere that is affecting
E. atmosphere as affects
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19 Dec 2005, 07:02
A
B: preposition phrase beginning with "with" is incorrect.
C/D: "that ..." incorrectly modifying atmosphere
E. as cannot be used as relative pronoun.
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19 Dec 2005, 07:11
I agree: A
If C would be ",that affect", would it be proper?
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19 Dec 2005, 07:46
It is important to note that interactions affect the global climate and not the atmosphere. So all other answers are wrong.
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19 Dec 2005, 16:28
A. atmosphere, interactions that affect ==> Correct!
B. atmosphere, with interactions affecting ==> Awkward construction
C. atmosphere that affects ==> changes meaning
D. atmosphere that is affecting ==> changes meaning
E. atmosphere as affects ==> changes meaning
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19 Dec 2005, 22:00
OA is A. very good job.
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20 Jun 2006, 01:33
I mistook the meaning
I thought atmostphere affects the climate
How can interactions affect, even if they affect, how do we know what the sentence actually means? certainly it cant expect us to have knowledge on climate .. etc
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20 Jun 2006, 01:44
We need A, as it's the only one that states explicitly that the interactions affect the climate.
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05 Feb 2008, 22:56
do you guys think that there could also be a case of plurals?
answer A...interactions that (affect) refer to the plural (long term interactions)
Re: SC # Atmosphere [#permalink] 05 Feb 2008, 22:56
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2 Floating in the waters of the equatorial Pacific, an array 11 08 Mar 2011, 07:03
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Floating in the waters of the equatorial Pacific, an array 1 07 Dec 2007, 16:54
24 Floating in the waters of the equatorial Pacific, an array 24 14 Jun 2007, 15:34
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https://stat.ethz.ch/pipermail/r-help/2021-May/471185.html | 1,702,029,359,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00467.warc.gz | 599,684,420 | 2,222 | # [R] if statement and for loop question
Rui Barradas ru|pb@rr@d@@ @end|ng |rom @@po@pt
Sun May 30 21:57:09 CEST 2021
```Hello,
You don't need a loop, the R way is a vectorized solution and it's also
clearer.
Create a logical index (note only one &) and assign b, c, d where it's TRUE.
i <- try\$a != "Positive" & try\$a != "VUS"
try <- within(try, {
b[i] <- ''
c[i] <- ''
d[i] <- ''
})
Hope this helps,
Às 17:28 de 30/05/21, Kai Yang via R-help escreveu:
> Hello List,I have a data frame which having the character columns:
>
> | a1 | b1 | c1 | d1 |
> | a2 | b2 | c2 | d2 |
> | a3 | b3 | c3 | d3 |
> | a4 | b4 | c4 | d4 |
> | a5 | b5 | c5 | d5 |
>
>
>
> I need to do: if a1 not = "Positive" and not = "VUS" then values of b1, c1 and d1 will be zero out. And do the same thing for the a2 to a5 series.
> I write the code below to do this. But it doesn't work. Would you please correct my code?
> Thank you,
> Kai
>
>
> for (i in 1:5)
> {
> if (isTRUE(try\$a[i] != "Positive" && try\$a[i] != "VUS"))
> {
> try\$b[i]== ''
> try\$c[i] == ''
> try\$d[i]== ''
> }
> }
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help | 463 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.658261 |
https://www.learnatnoon.com/s/category/cbse/class-12/rd-sharma-class-12/maths-rd-sharma-class-12/binary-operations/ | 1,709,221,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00849.warc.gz | 838,054,362 | 27,742 | Binary Operations
### For the binary operation ×10 set S = {1, 3, 7, 9}, find the inverse of 3.
Answer: 1 ×10 1 = remainder obtained by dividing 1 × 1 by 10 = 1 3 ×10 7 = remainder obtained by dividing 3 × 7 by 10 = 1 7 ×10 9 = remainder obtained by dividing 7 × 9 by 10 = 3 Composition table:...
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### Construct the composition table for ×5 on set Z5 = {0, 1, 2, 3, 4}
Answer: 1 ×5 1 = remainder obtained by dividing 1 × 1 by 5 = 1 3 ×5 4 = remainder obtained by dividing 3 × 4 by 5 = 2 4 ×5 4 = remainder obtained by dividing 4 × 4 by 5 = 1 Composition table: ×5 0 1...
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### Construct the composition table for ×6 on set S = {0, 1, 2, 3, 4, 5}.
Answer: 1 ×6 1 = remainder obtained by dividing 1 × 1 by 6 = 1 3 ×6 4 = remainder obtained by dividing 3 × 4 by 6 = 0 4 ×6 5 = remainder obtained by dividing 4 × 5 by 6 = 2 Composition table: ×6 0 1...
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### Construct the composition table for +5 on set S = {0, 1, 2, 3, 4}
Answer: 1 +5 1 = remainder obtained by dividing 1 + 1 by 5 = 2 3 +5 1 = remainder obtained by dividing 3 + 1 by 5 = 2 4 +5 1 = remainder obtained by dividing 4 + 1 by 5 = 3 Composition Table: +5 0 1...
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### Construct the composition table for ×4 on set S = {0, 1, 2, 3}.
Answer: Given, ×4 on set S = {0, 1, 2, 3} 1 ×4 1 = remainder obtained by dividing 1 × 1 by 4 = 1 0 ×4 1 = remainder obtained by dividing 0 × 1 by 4 = 0 2 ×4 3 = remainder obtained by dividing 2 × 3...
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### Let * be a binary operation on Z defined by a * b = a + b – 4 for all a, b ∈ Z. (i) Show that * is both commutative and associative. (ii) Find the identity element in Z
Answers: (i) Consider, a, b ∈ Z a * b = a + b – 4 = b + a – 4 = b * a a * b = b * a, ∀ a, b ∈ Z Then, * is commutative on Z. a * (b * c) = a * (b + c – 4) = a + b + c -4 – 4 = a + b + c – 8 (a * b)...
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### Let * be a binary operation on Q0 (set of non-zero rational numbers) defined by a * b = (3ab/5) for all a, b ∈ Q0. Show that * is commutative as well as associative. Also, find its identity element, if it exists.
Answer: Consider, a, b ∈ Q0 a * b = (3ab/5) = (3ba/5) = b * a a * b = b * a, for all a, b ∈ Q0 a * (b * c) = a * (3bc/5) = [a (3 bc/5)] /5 = 3 abc/25 (a * b) * c = (3 ab/5) * c = [(3 ab/5)...
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### Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then, (i) Show that * is both commutative and associative on Q – {-1} (ii) Find the identity element in Q – {-1}
Answers: (i) Consider, a, b ∈ Q – {-1} a * b = a + b + ab = b + a + ba = b * a a * b = b * a, ∀ a, b ∈ Q – {-1} a * (b * c) = a * (b + c + b c) = a + (b + c + b c) + a (b + c + b c) = a + b +...
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### Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then, Show that every element of Q – {-1} is invertible. Also, find inverse of an arbitrary element.
Answer: Consider, a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. a * b = e = b * a a * b = e and b * a = e a + b + ab = 0 and b + a + ba = 0 b (1 + a) = – a Q – {-1} b = -a/1 + a Q – {-1}...
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### Let A = R0 × R, where R0 denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R0 × R. (i) Show that ‘O’ is commutative and associative on A (ii) Find the identity element in A
Answers: (i) Consider, X = (a, b) Y = (c, d) ∈ A, ∀ a, c ∈ R0 b, d ∈ R X O Y = (ac, bc + d) Y O X = (ca, da + b) X O Y = Y O X, ∀ X, Y ∈ A O is not commutative on A. X = (a, b) Y = (c, d) a Z = (e,...
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### Let A = R0 × R, where R0 denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R0 × R. Find the invertible element in A.
Answer: Consider, F = (m, n) be the inverse in A ∀ m ∈ R0 and n ∈ R X O F = E F O X = E (am, bm + n) = (1, 0) and (ma, na + b) = (1, 0) Considering (am, bm + n) = (1, 0) am = 1 m = 1/a And bm + n =...
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### Let * be a binary operation on Z defined by a * b = a + b – 4 for all a, b ∈ Z. Find the invertible element in Z.
Answer: Consider, a ∈ Z and b ∈ Z be the inverse of a. a * b = e = b * a a * b = e and b * a = e a + b – 4 = 4 and b + a – 4 = 4 b = 8 – a ∈ Z Hence, 8 – a is the inverse of a ∈...
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### Find the identity element in the set of all rational numbers except – 1 with respect to * defined by a * b = a + b + ab
Answer: Consider, e be the identity element in I+ with respect to * such that a * e = a = e * a, ∀ a ∈ Q – {-1} a * e = a and e * a = a, ∀ a ∈ Q – {-1} a + e + ae = a and e + a + ea = a, ∀ a ∈ Q –...
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### Find the identity element in the set I+ of all positive integers defined by a * b = a + b for all a, b ∈ I+.
Answer: Consider, e be the identity element in I+ with respect to * a * e = a = e * a, ∀ a ∈ I+ a * e = a and e * a = a, ∀ a ∈ I+ a + e = a and e + a = a, ∀ a ∈ I+ e = 0, ∀ a ∈ I+ Hence, 0 is the...
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### For the binary operation set , find the inverse of 3 .
Solution: Here, $1 \times{ }_{10} 1=$ remainder obtained by dividing $1 \times 1$ by 10 $=1$ $3 \times{ }_{10} 7=$ remainder obtained by dividing $3 \times 7$ by 10 $=1$ $7 \times_{10} 9=$ remainder...
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### Construct the composition table for on set
Solution: Here, $1 \times_{5} 1=$ remainder obtained by dividing $1 \times 1$ by 5 $=1$ $3 \times_{5} 4=$ remainder obtained by dividing $3 \times 4$ by 5 $=2$ $4 \times_{5} 4=$ remainder obtained...
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### Construct the composition table for on set S = {0, 1, 2, 3, 4, 5}.
Solution: Here, $1 \times_{6} 1=$ remainder obtained by dividing $1 \times 1$ by 6 $=1$ $3 \times_{6} 4=$ remainder obtained by dividing $3 \times 4$ by 6 $=0$ $4 \times_{6} 5=$ remainder obtained...
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### Construct the composition table for on set S = {0, 1, 2, 3, 4}
Solution: $1+_{5} 1=$ remainder obtained by dividing $1+1$ by 5 $=2$ $3+{ }_{5} 1=$ remainder obtained by dividing $3+1$ by 5 $=2$ $4+_{5} 1=$ remainder obtained by dividing $4+1$ by 5 $=3$...
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### Construct the composition table for on set .
Solution: It is given that $x_{4}$ on set $S=\{0,1,2,3\}$ Here, $1 \times_{4} 1=$ remainder obtained by dividing $1 \times 1$ by 4 $=1$. $0 \times_{4} 1=$ remainder obtained by dividing $0 \times 1$...
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### Let , where denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: for all .(iii) Find the invertible element in A.
(iii) Assume $F=(m,n)$ be the inverse in $A\forall m\in {{R}_{0}}$and $n\in R$ $XOF=E$ and $FOX=E$ $(am,bm+n)=(1,0)$ and $(ma,na+b)=(1,0)$ Assuming $(am,bm+n)=(1,0)$ $am=1$ $m=1/a$ And $bm+n=0$...
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### Let , where denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: for all . (i) Show that ‘O’ is commutative and associative on A (ii) Find the identity element in A
(i) Assume $X=(a,b)$ and $Y=(c,d)$$\in A,\forall a,c\in {{R}_{0}}$ and $b,d\in R$ Now, $XOY=(ac,bc+d)$ Then $YOX=(ca,da+b)$ So, $XOY=YOX,\forall X,Y\in A$ So, O is not commutative on A. Then we have...
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### Let * be a binary operation on defined by for all . Then(iii) Show that every element of is invertible. Also, find inverse of an arbitrary element.
(iii) Assume $a\in Q-\left\{ -1 \right\}$ and $b\in Q-\left\{ -1 \right\}$ be the inverse of a. Then, $a*b=e=b*a$ $a*b=e$ and $b*a=e$ $a+b+ab=0$ and $b+a+ba=0$ $b(1+a)=–aQ–{-1}$ $b=-a/1+aQ–{-1}$...
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### . Let * be a binary operation on defined by for all . Then, (i) Show that * is both commutative and associative on (ii) Find the identity element in
(i) Firstly we all have to check commutativity of * Assume $a,b\in Q-\left\{ -1 \right\}$ So, $a*b=a+b+ab$ $=b+a+ba$ $=b*a$ Hence, $a*b=b*a$, $\forall a,b\in Q-\left\{ -1 \right\}$ Then we have to...
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### Let * be a binary operation on Q0 (set of non-zero rational numbers) defined by for all . Show that * is commutative as well as associative. Also, find its identity element, if it exists.
Firstly we all have to prove commutativity of * Assume $a,b\in {{Q}_{0}}$ $a*b=(3ab/5)$ $=(3ba/5)$ $=b*a$ Hence, $a*b=b*a$, for all $a,b\in {{Q}_{0}}$ Then we all have to prove associativity of *...
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### Let * be a binary operation on Z defined by for all .(iii) Find the invertible element in Z.
(iii) Assume $a\in Z$ and $b\in Z$ be the inverse of a. Now, $a*b=e=b*a$ $a*b=e$and $b*a=e$ $a+b–4=4$ and $b+a–4=4$ $b=8-a\in Z$ Therefore, $8–a$ is the inverse of $a\in Z$
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### Let * be a binary operation on Z defined by for all . (i) Show that * is both commutative and associative. (ii) Find the identity element in Z
(i) Firstly we have to prove the commutatively of * Assume $a,b\in Z$. Now, $a*b=a+b–4$ $=b+a–4$ $=b*a$ Hence, $a*b=b*a,\forall a,b\in Z$ So, * is commutative on Z. Then, we all have to prove...
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### Find the identity element in the set of all rational numbers except with respect to * defined by
Assume ‘e’ be the identity element in ${{I}^{+}}$ with respect to * such that $a*e=a=e*a,\forall a\in Q-\left\{ -1 \right\}$ $a*e=a$and $e*a=a,\forall a\in Q-\left\{ -1 \right\}$ $a+e+ae=a$and...
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### Find the identity element in the set of all positive integers defined by for all .
Assume ‘e’ be the identity element in ${{I}^{+}}$ with respect to * such that $a*e=a=e*a,\forall a\in {{I}^{+}}$ $a*e=a$and $e*a,\forall a\in {{I}^{+}}$ $a+e=a$ and $e+a=a$, $\forall a\in {{I}^{+}}$...
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### On the set Z of integers a binary operation * is defined by a 8 b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.
\[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b,\text{ }c\text{ }\in \text{ }Z \\ a\text{ }*\text{ }\left( b\text{ }*\text{ }c \right)\text{ }=\text{ }a\text{ }*\text{ }\left( bc\text{ }+\text{ }1...
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### Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?
\[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Z \\ a\text{ }*\text{ }b\text{ }=\text{ }3a\text{ }+\text{ }7b \\ b\text{ }*\text{ }a\text{ }=\text{ }3b\text{ }+\text{ }7a \\...
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### If the binary operation o is defined by a0b = a + b – ab on the set Q – {-1} of all rational numbers other than 1, show that o is commutative on Q – [1].
\[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q\text{ }\text{ }-\left\{ -1 \right\}. \\ Then\text{ }aob\text{ }=\text{ }a\text{ }+\text{ }b\text{ }-\text{ }ab \\ =\text{...
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### Check the commutativity and associativity of each of the following binary operations: (xv) ‘*’ on Q defined by a * b = gcd (a, b) for all a, b ∈ Q
(xv) to check: commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }N,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }gcd\text{ }\left( a,\text{ }b \right) ...
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### Check the commutativity and associativity of each of the following binary operations: (xiii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q (xiv) ‘*’ on Z defined by a * b = a + b – ab for all a, b ∈ Z
(xiii) to check :commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }\left( ab/4 \right) \\ =\text{ }\left(...
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### Check the commutativity and associativity of each of the following binary operations: (xi) ‘*’ on N defined by a * b = ab for all a, b ∈ N (xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z
(xi) to check : commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }N,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }{{a}^{b}} \\ b\text{ }*\text{ }a\text{...
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### Check the commutativity and associativity of each of the following binary operations: (vii) ‘*’ on Q defined by a * b = a + a b for all a, b ∈ Q (viii) ‘*’ on R defined by a * b = a + b -7 for all a, b ∈ R
(vii) to check : commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }+\text{ }ab \\ b\text{...
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### Check the commutativity and associativity of each of the following binary operations: (v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q (vi) ‘*’ on Q defined by a * b = ab2 for all a, b ∈ Q
(v) to check: commutativity of o \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then \\ a\text{ }o\text{ }b\text{ }=\text{ }\left( ab/2 \right) \\ =\text{ }\left(...
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### Check the commutativity and associativity of each of the following binary operations: (iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q (iv) ‘⊙’ on Q defined by a ⊙ b = a2 + b2 for all a, b ∈ Q
(iii) to check: commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Q,\text{ }then \\ a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }-\text{ }b \\ b\text{...
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### Check the commutativity and associativity of each of the following binary operations: (i) ‘*’ on Z defined by a * b = a + b + a b for all a, b ∈ Z (ii) ‘*’ on N defined by a * b = 2ab for all a, b ∈ N
(i) to check: commutativity of * \[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }Z \\ =>\text{ }a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }+\text{ }b\text{ }+\text{ }ab ...
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### Let A be any set containing more than one element. Let ‘*’ be a binary operation on A defined by a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?
\[\begin{array}{*{35}{l}} Let\text{ }a,\text{ }b\text{ }\in \text{ }A \\ =>\text{ }a\text{ }*\text{ }b\text{ }=\text{ }b \\ b\text{ }*\text{ }a\text{ }=\text{ }a \\ Therefore\text{ }a\text{...
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### Determine which of the following binary operation is associative and which is commutative: (i) * on N defined by a * b = 1 for all a, b ∈ N (ii) * on Q defined by a * b = (a + b)/2 for all a, b ∈ Q
(i) to prove: commutativity of * Let \[\begin{array}{*{35}{l}} a,\text{ }b\text{ }\in \text{ }N \\ a\text{ }*\text{ }b\text{ }=\text{ }1 \\ b\text{ }*\text{ }a\text{ }=\text{ }1 \\ =>a\text{...
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### Let ‘*’ be a binary operation on N defined by a * b = l.c.m. (a, b) for all a, b ∈ N (i) Find 2 * 4, 3 * 5, 1 * 6. (ii) Check the commutativity and associativity of ‘*’ on N.
(i) Since, \[\begin{array}{*{35}{l}} a\text{ }*\text{ }b\text{ }=\text{ }1.c.m.\text{ }\left( a,\text{ }b \right) \\ 2\text{ }*\text{ }4\text{ }=\text{ }l.c.m.\text{ }\left( 2,\text{ }4 \right) \\...
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### Let S = {a, b, c}. Find the total number of binary operations on S.
Number of binary operations on a set with n elements is ${{n}^{{{n}^{2}}}}$ Here, S = {a, b, c} Number of elements in S = 3 Number of binary operations on a set with 3 elements is...
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### Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.
LCM 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 5 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5 Since,, all the elements are not in the set {1, 2, 3, 4, 5}. If we consider a = 2 and b = 3, a * b = LCM...
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### Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.
\[\begin{array}{*{35}{l}} a~*~b~=\text{ }2a~+~b~-\text{ }3 \\ 3\text{ }*\text{ }4\text{ }=\text{ }2\text{ }\left( 3 \right)\text{ }+\text{ }4\text{ }-\text{ }3 \\ =\text{ }6\text{ }+\text{...
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### Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.(v) On Z+ define * by a * b = a (vi) On R, define * by a * b = a + 4b2
(v) Given on Z+ define * by a * b = a Let \[\begin{array}{*{35}{l}} a,\text{ }b\text{ }\in \text{ }{{Z}^{+}} \\ \Rightarrow \text{ }a\text{ }\in \text{ }{{Z}^{+}} \\ \Rightarrow \text{ }a\text{...
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### Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.(iii) On R, define * by a*b = ab2 (iv) On Z+ define * by a * b = |a − b|
(iii) Since, on R, define by a*b = ab2 Let \[\begin{array}{*{35}{l}} a,\text{ }b\text{ }\in \text{ }R \\ \Rightarrow \text{ }a,\text{ }{{b}^{2}}~\in \text{ }R \\ \Rightarrow \text{ }a{{b}^{2}}~\in...
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### Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this. (i) On Z+, defined * by a * b = a – b (ii) On Z+, define * by a*b = ab
(i)Since, On Z+, defined * by a * b = a – b If a = 1 and b = 2 in Z+, then \[\begin{array}{*{35}{l}} a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }-\text{ }b \\ =\text{ }1\text{ }-\text{ }2 \\...
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### Determine whether the following operation define a binary operation on the given set or not:(vii) ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q
(vii)Since, ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q If a = 2 and b = -1 in Q, \[\begin{array}{*{35}{l}} a\text{ }*\text{ }b\text{ }=\text{ }\left( a\text{ }-\text{ }1...
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### Determine whether the following operation define a binary operation on the given set or not: (v) ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b (vi) ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N
(v) Given ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b Consider the composition table, +6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 Here all...
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### Determine whether the following operation define a binary operation on the given set or not: (iii) ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N (iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a ×6 b = Remainder when a b is divided by 6.
(iii) Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N \[\begin{array}{*{35}{l}} If~a~=\text{ }1\text{ }and~b\text{ }=\text{ }1, \\ a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }+\text{...
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### Determine whether the following operation define a binary operation on the given set or not: (i) ‘*’ on N defined by a * b = ab for all a, b ∈ N. (ii) ‘O’ on Z defined by a O b = ab for all a, b ∈ Z.
(i) Given ‘*’ on N defined by a * b = ab for all a, b ∈ N. Let a, b ∈ N. Then, \[\begin{array}{*{35}{l}} {{a}^{b~}}\in ~N~~~~~~\left[ \because ~{{a}^{b}}\ne 0~and~a,\text{ }b~is~positive~integer...
read more | 7,091 | 18,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.919612 |
https://kelvinmao.github.io/367.-Valid-Perfect-Square/ | 1,643,280,617,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00568.warc.gz | 399,021,587 | 4,708 | # 367. Valid Perfect Square
Given a positive integer num, write a function which returns True if num is a perfect square else False.
Note: Do not use any built-in library function such as sqrt.
Example 1:
Input: 16
Returns: True
Example 2:
Input: 14
Returns: False
/*use Binary-Search，0ms*/
class Solution {
public:
bool isPerfectSquare(int num) {
if(num==0)
return false;
long long int low=0,high=num,mid=0;
while(low<=high){
mid=low+((high-low)>>1);
if(mid*mid<num)
low=mid+1;
else if(mid*mid>num)
high=mid-1;
else if(mid*mid==num)
return true;
}
return false;
}
};
### 学籍管理系统文档
#### 北邮教务系统评教脚本
Published on September 17, 2017
#### 72.Edit distance
Published on September 17, 2017 | 215 | 697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-05 | latest | en | 0.440257 |
https://certifiedcalculator.com/floor-truss-cost-calculator/ | 1,709,268,632,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00435.warc.gz | 160,880,519 | 44,381 | # Floor Truss Cost Calculator
When planning a construction project, it’s essential to have a clear understanding of the costs involved. Floor trusses are a crucial component of many building projects, and their cost can vary significantly depending on various factors. To simplify the process of estimating the cost of floor trusses, we have developed the Floor Truss Cost Calculator. This handy tool allows you to quickly determine the estimated cost of floor trusses based on key parameters.
Formula: The formula used by our calculator to estimate the cost of floor trusses is as follows: Estimated Cost = (Span Length × Truss Depth × Truss Spacing × Load Capacity) / 1000
How to Use:
1. Enter the Span Length (in feet) of the floor trusses you plan to use.
2. Input the Truss Depth (in inches).
3. Specify the Truss Spacing (in feet).
4. Enter the Load Capacity (in pounds per square foot) that the floor trusses need to support.
5. Click the “Calculate” button.
6. The estimated cost of the floor trusses will be displayed in the “Estimated Cost” field.
Example: Let’s say you are designing a building with floor trusses that have a span length of 30 feet, a truss depth of 14 inches, a truss spacing of 2 feet, and a load capacity of 40 lbs/ft². Using the Floor Truss Cost Calculator, the estimated cost would be calculated as follows: Estimated Cost = (30 × 14 × 2 × 40) / 1000 = \$33.60
FAQs:
1. What factors can affect the cost of floor trusses?
• Several factors can influence the cost, including span length, truss depth, truss spacing, and load capacity.
2. Is this calculator suitable for residential and commercial projects?
• Yes, it can be used for both residential and commercial construction projects.
3. Can I change the units of measurement (e.g., meters instead of feet)?
• Currently, the calculator only supports feet and inches for span length and truss depth. However, you can convert your measurements before inputting them.
4. Are there any specific load capacity requirements I should be aware of?
• Load capacity requirements can vary depending on local building codes and project specifications. Consult with a structural engineer for guidance.
5. What’s the significance of the “/ 1000” in the formula?
• This is a conversion factor used to ensure the result is in dollars rather than thousands of dollars.
Conclusion: Estimating the cost of floor trusses is an important step in any construction project. Our Floor Truss Cost Calculator simplifies this process, providing you with a quick and accurate estimate based on key parameters. By using this tool, you can make informed decisions about your construction budget and ensure the success of your project. | 588 | 2,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-10 | latest | en | 0.905083 |
https://usersquestions.com/library/lecture/read/2375-are-all-six-sides-of-a-hexagon-equal | 1,685,697,708,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00231.warc.gz | 645,293,719 | 6,932 | # Are all six sides of a hexagon equal?
### Are all six sides of a hexagon equal?
Regular Hexagons ' A regular hexagon has six sides that are all congruent, or equal in measurement. A regular hexagon is convex, meaning that the points of the hexagon all point outward. All of the angles of a regular hexagon are congruent and measure 120 degrees.
### How many equal sides does a hexagon have?
six equal sides A regular hexagon has six equal sides and six equal interior angles.
### Are all angles in a hexagon equal?
The sum of the interior angles of a hexagon must equal 720 degrees. Because the hexagon is regular, all of the interior angles will have the same measure. A hexagon has six sides and six interior angles. Therefore, each angle measures .
### What is a six sided shape with unequal sides called?
What is an Irregular Hexagon? An irregular hexagon is defined as a 6-sided polygon that is not regular—meaning that all of the sides and angles do not have the same measure.
### What is a 9 sided shape called?
Nonagon In geometry, a nonagon (/ˈnɒnəɡɒn/) or enneagon (/ˈɛniəɡɒn/) is a nine-sided polygon or 9-gon. The name nonagon is a prefix hybrid formation, from Latin (nonus, "ninth" + gonon), used equivalently, attested already in the 16th century in French nonogone and in English from the 17th century.
### What is the difference between a hexagon and a regular hexagon?
A hexagon is a six-sided polygon. A regular hexagon is one where all six sides and angles are equal. A hexagon is a six-sided polygon. A regular hexagon is one where all six sides and angles are equal.
### What is a 9 sided shape?
nonagon In geometry, a nonagon (/ˈnɒnəɡɒn/) or enneagon (/ˈɛniəɡɒn/) is a nine-sided polygon or 9-gon. The name nonagon is a prefix hybrid formation, from Latin (nonus, "ninth" + gonon), used equivalently, attested already in the 16th century in French nonogone and in English from the 17th century.
### What is special about a hexagon?
But what makes hexagons so special? ... A hexagon is the shape that best fills a plane with equal size units and leaves no wasted space. Hexagonal packing also minimizes the perimeter for a given area because of its 120-degree angles.
### What angle are the corners of a hexagon?
We know the three angles in a triangle add up to 180 degrees, and all three angles are 60 degrees in an equilateral triangle. A hexagon has six sides, and we can use the formula degrees = (# of sides – 2) * 180. Then degrees = (6 – 2) * 180 = 720 degrees. Each angle is 720/6 = 120 degrees.
### What is a 10 sided shape?
decagon In geometry, a decagon (from the Greek δέκα déka and γωνία gonía, "ten angles") is a ten-sided polygon or 10-gon. The total sum of the interior angles of a simple decagon is 1440°. A self-intersecting regular decagon is known as a decagram.
### Which is true of all sides of a hexagon?
All hexagons have six angles and six sides.If the sides are all equal and the angles are all equal, then it's called a regular hexagon. What has six sides and all sides are not equal? An irregular hexagon What has 6 sides and all of the sides are equal? A regular hexagon.
### What's the difference between irregular and regular hexagons?
In the case of a regular hexagon, all the sides are of equal length, and the internal angles are of the same value. The regular hexagon consists of six symmetrical lines and rotational symmetry of order of 6. Whereas in the case of the irregular hexagon, neither the sides are equal, nor the angles are the same.
### How to calculate the perimeter of a 6 sided hexagon?
Just calculate: perimeter = 6 * side, where siderefers to the length of any one side. As for the angles, a regular hexagon requires that all angles are equal and the sum up to 720º which means that each individual angle must be 120º. This proves to be of the utmost importance when we talk about the popularity of the hexagon shape in nature.
### How are the internal angles of a hexagon calculated?
Each internal angle of the hexagon has been calculated to be 120°. In the case of a regular hexagon, all the sides are of equal length, and the internal angles are of the same value. The regular hexagon consists of six symmetrical lines and rotational symmetry of order of 6. | 1,072 | 4,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-23 | latest | en | 0.923165 |
https://e-learnteach.com/a-block-initially-at-rest-on-a-horizontal-frictionless-surface/ | 1,685,654,754,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00179.warc.gz | 261,124,241 | 14,060 | # A block initially at rest on a horizontal frictionless surface
A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along the x-axis is applied to the block.A block initially at rest on a horizontal, frictionless surface is accelerated by a constant horizontal force of 5 newtons. if 15 joules of work. Question: A block initially at rest on a horizontal, frictionless surface is accelerated by a constant horizontal force of 5.0 newtons. If 15 joules of work is. A block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an axis is applied to the block.A block initially at rest on a horizontal, frictionless surface is accelerated by a constant horizontal force of 5.0 newtons. If 15 joules of work is done.
View this answer now! It’s completely free.
## a block initially at rest on a horizontal, frictionless surface is accelerated by a constant
A body initially resting on horizontal frictionless surface is accelerated by a constant force. It passes over a small region where it experiences a force. A block initially at rest on a horizontal, frictionless surface is accelerated by a constant horizontal force of 5.0 newtons. If 15 joules of work is done. 13A block initially at rest on a horizontal, frictionlesssurface is accelerated by a constant horizontalforce of 5.0 newtons. If 15 joules of work is doneon. 6.67J · A 20kg block on a horizontal surface is attached to a horizontal spring of spring constant k=4. · The maximum kinetic energy of the particle and the value. are initially at rest on a frictionless, horizontal surface. Frictionless surface. The magnitude of the acceleration of block B is. (Skill 20).
## a force f0 is applied continuously to a box initially at rest on a horizontal surface
A 100 N box is initially at rest on a rough horizontal surface. force on the box by the surface 1 second after the horizontal force is first applied.A 40 kg box initially at rest is pushed 5.0m along a rough horizontal floor with. up plane while acted upon by gravity, friction and an applied force.A constant horizontal force of mg/2 is applied to the box directed to the right. The coefficient of friction of the surface changes with the. mass m moving on a horizontal frictionless surface hits and sticks to an object of mass M > m, which is initially at rest on the surface?A 1.5 kg box is initially at rest on a horizontal surface when at t =0 a horizontal force F = (1.8t) iN (With t in seconds), is applied to the box.
## two objects, a and b, are held one meter
1. Two objects, A and B, are held one meter above the horizontal ground. The mass of Bis quadruple the mass of A. IF PE is the gravitational potential. 1.I can relate the relative gravitational potential energy of an object to the. Two objects, A and B, are held one meter above the horizontal ground.Compared to Object B, Object A has _____ the momentum. answer choices. Two times. Two objects, A and B, are held one meter above the horizontal ground.Two objects, A and B, are held one meter above the horizontal ground. The mass of B is twice as great as the mass of A. If PE is the gravitational potential.Two objects, A and B, are held one meter above the horizontal ground. The mass of B is twice as great as the mass of A. If PE is the gravitational potential.
## a constant horizontal force of 20 newtons east applied to a box
a constant speed of 20. meters per second. The. A constant horizontal force of 20.0 newtons east applied to a box causes it to move at a constant.The rope applies a horizontal force of 180 N to the box. A 10 kg box is pulled to the right by an applied force of 250 Newtons. Any.A) 80. J B) 120 J C) 240 J D) 480 J. 19. A student applies a 20.-newton force to move a crate at a constant speed of 4.0 meters per second across a rough.. force of 15 newtons is required to push the box at a constant speed of 1.5 meters per second. 3. magnitude of the horizontal force applied to the box.The diagram below shows a horizontal 12-newton force being applied to two blocks, A and B, initially at rest on a horizontal, frictionless surface. Block A has. | 981 | 4,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-23 | latest | en | 0.88452 |
https://physics.stackexchange.com/questions/423857/propagator-solution-to-klein-gordon-equation | 1,585,709,566,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00080.warc.gz | 645,670,290 | 31,584 | # Propagator solution to Klein-Gordon equation
We know that the Klein-Gordon operator is given by $(\partial^2+m^2)=(\partial_\mu\partial^{\mu}+m^2)$, which is used to describe the evolution of relativistic free particles.
How can we show that the relativistic propagator for the free particle $\langle x'|x\rangle$ is a solution to the Klein-Gordon equation? In other words, how do you apply the operators such that $(\partial^2+m^2)\langle x'|x\rangle=0$.
Additional information: the expression for the $\langle x'|x\rangle$ propagator can be written as $$\langle x'|x\rangle=\int\frac{d^3p}{(2\pi)^32\sqrt{\vec{p}^2+m^2}}e^{-\sqrt{\vec{p}^2+m^2}(t'-t)+ip\cdot(\vec{x}'-\vec{x})}=\int\frac{d^4}{(2\pi)^3}\delta(p^2-m^2)\theta(p^0)e^{-ip\cdot(x'-x)},$$ where the last expression is the Lorentz invariant form.
Edit: In a previous exercise I showed that the Feynman propagator in the form:
$$G_F(x'-x)=i\int\frac{d^4p}{(2\pi)^4}\frac{e^{-ip\cdot(x'-x)}}{p^2-^2+i\epsilon}$$
is a Green's function of the KG operator: $$(\partial^2+m^2)G_F(x'-x)=-i\delta(x'-x).$$
And now in this one I'm being asked to show that the propagator $\langle x'|x\rangle$ is a solution to the KG equation. However, the step function gives me a delta, so I'm not sure how to proceed. Another expression for the propagator in my notes is (after performing integration):
$$\langle x'|x\rangle=\frac{m}{(2\pi)^2}\frac{K_1(m\sqrt{-(x'-x)^2})}{\sqrt{-(x'-x)^2}}$$
Where $K_1$ is the modified Bessel function.
• The propagator is a Green’s function of the KG equation, i.e it satisfies $$(\partial^2 + m^2)\langle x | x’\rangle = \delta(x-x’)$$ – bapowell Aug 20 '18 at 22:05
• Thank you for your answer. I see where this comes, given the step function in the Lorentz invariant definition. However, for some reason the exercise I have says to show that $\langle x'|x\rangle$ is indeed a solution of the KG equation. In fact, I actually showed before that the Feynman propagator was a Green's function of KG, and now I must show that $\langle x'|x\rangle$ is the solution. – Charlie Aug 21 '18 at 0:00 | 660 | 2,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-16 | latest | en | 0.791266 |
https://phys.libretexts.org/Courses/HACC_Central_Pennsylvania's_Community_College/Introduction_to_Physical_Science/09%3A_Atoms_and_Periodic_Properties/9.03%3A_Lecture_3_-_The_Periodic_Table | 1,718,769,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00132.warc.gz | 404,755,624 | 28,151 | # 9.3: Lecture 3 - The Periodic Table
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
The lesson is based on section 2.5 in the OpenStax Chemistry 2e textbook. The lecture slides are provided in PowerPoint, Keynote, and pdf format.
9.3: Lecture 3 - The Periodic Table is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 742 | 1,965 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.300033 |
https://everything2.com/title/Finger+numbers+game | 1,544,926,816,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827175.38/warc/CC-MAIN-20181216003916-20181216025916-00497.warc.gz | 611,463,385 | 5,953 | This simple, yet diverting game requires nothing but you, your partner, and four hands with five digits each. The object of the game is to get the other player's hands out by adding the number of raised digits on one of your hands to the number raised digits on one of your opponent's until both his hands would be at or above five fingers up. Each player starts with one finger on each hand up:
```Player 1 Player 2
Left Right Left Right
1 1 1 1
Player 1 chooses a hand of his and a hand of his opponent's and adds
Turn 1: the number of raised fingers on his hand to the chosen opposing hand:
Player 1 Player 2
Left Right Left Right
1 1 2 1
^-----^
Turn 2: Player 2 now has a choice: he can either add two to one of his
2 1 2 1 opponent's raised digits, or he could add one. He adds one:
^---------------^
Turn 3: Player 1 now can add either 2 or 1 to either 2 or 1. He sticks with 1
2 1 2 2 again:
^-----------^
Turn 4: Player 2 can add 2 to 1 or 2 to 2. He adds 2 to one.
2 3 2 2
^-----^
Turn 5: Player 1 now has a chance to knock one of player 2's hands out by
2 3 X 2 bringing its raised finger count to five. He does so.
^-----^
Turn 6: Player 2 now has three choices: He can add his fingers to his
2 3 1 1 opponent's normally, or he can "split". A player can split when one of
*SPLIT* of his hands is out and the remaining hand has either two or four fingers
Turn 7: up. When you split, you halve the number of raised fingers on
2 3 3 1 your "in" hand and add the other half onto your "out" hand, bringing
^----------^ your "out" hand back into play.
Turn 8:
X 3 3 1
^----------^
Turn 9:
X 3 X 1
^-----^
Turn 10:
X 4 X 1
^-----------^
Turn 11:
X 4 X X At this point, the game ends and Player 1 is the victor.
^-----------^
```
Using this game as a decision-making tool in place of rock-paper-scissors can allow an experienced player to either get what he wants one hundred percent of the time or delay the decision indefinitely.
Log in or register to write something here or to contact authors. | 647 | 2,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-51 | latest | en | 0.95074 |
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