Calculate Square Root?
Can I use Simplified field notation to calculate the square root of another field? Not having any success.
No. You need to use the Custom Script option, with some code like this (assuming the name of the other field is Text1):
event.value = Math.sqrt(+this.getField("Text1").value);
Similar Messages
-
Calculate square root including sum and product
Hi,
I have to enter the square root of:
Text1*Text1+Text2*Text2-2*Text1*Text2*Text3
How do i do that?
I hope someone can help me.Thanks a million try67,
i found another way that worked as well:
event.value = Math.sqrt(+this.getField("Text1").value*this.getField("Text1").value+this.getField("Text2 ").value*this.getField("Text2").value-2*this.getField("Text1").value*this.getField("Text2" ).value*this.getField("Text3").value)
Mine is much more confuse:) -
I expect no replies to this thread - because there are no
answers, but I want to raise awareness of a faculty of other
languages that is missing in Flash that would really help 3D and
games to be built in Flash.
Below is an optimisation of the Quake 3 inverse square root
hack. What does it do? Well in games and 3D we use a lot of vector
math and that involves calculating normals. To calculate a normal
you divide a vector's parameters by it's length, the length you
obtain by pythagoras theorem. But of course division is slow - if
only there was a way we could get 1.0/Math.sqrt so we could just
multiply the vector and speed it up.
Which is what the code below does in Java / Processing. It
runs at the same speed as Math.sqrt, but for not having to divide,
that's still a massive speed increase.
But we can't do this in Flash because there isn't a way to
convert a Number/float into its integer-bits representation. Please
could everyone whinge at Adobe about this and give us access to a
very powerful tool. Even the guys working on Papervision are having
trouble with this issue.that's just an implementation of newton's method for finding
the zeros of a differentiable function. for a given x whose inverse
sq rt you want to find, the function is:
f(y) = 1/(y*y) - x;
1. you can find the positive zero of f using newton's method.
2. you only need to consider values of x between 1 and 10
because you can rewrite x = 10^^E * m, where 1<=m<10.
3. the inverseRt(x) = 10^^(-E/2) * inverseRt(m)
4. you don't have to divide E by 2. you can use bitwise shift
to the right by 1.
5. you don't have to multiply 10^^(-E/2) by inverseRt(m): you
can use a decimal shift of inverseRt(m);
6. your left to find the positive zero of f(y) = 1/(y*y) - m,
1<=m<10.
and at this point i realized what, i believe, is a much
faster way to find inverse roots: use a look-up table.
you only need a table of inverse roots for numbers m,
1<m<=10.
for a given x = 10^^E*m = 10^^(e/2) *10^^(E-e/2)*m, where e
is the largest even integer less than or equal to E (if E is
positive, e is the greatest even integer less than or equal to E,
if E is negative), you need to look-up, at most, two inverse roots,
perform one multiplication and one decimal shift:
inverseRt(x) = 10^^(-e) * inverseRt(10) *inverseRt(m), if
E-e/2 = 1 and
inverseRt(x) = 10^^(-e) * inverseRt(m), if E-e/2 = 0. -
How to find square root, log recursively???
I need to find the square root of a number entered recursively and log as well. Your help would be greatly appreciated. Thanks in advance!
import java.io.*;
/**Class provides recursive versions
* of simple arithmetic operations.
public class Ops2
private static BufferedReader in = null;
/**successor, return n + 1*/
public static int suc(int n)
return n + 1;
/**predecessor, return n - 1*/
public static int pre(int n)
if (n == 0)
return 0;
else
return n - 1;
/**add two numbers entered*/
public static int add(int n, int m)
if (m == 0)
return n;
else
return suc(add(n, pre(m)));
/**subtract two numbers entered*/
public static int sub(int n, int m)
if (n < m)
return 0;
else if (m == 0)
return n;
else
return pre(sub(n, pre(m)));
/**multiply two numbers entered*/
public static int mult(int n, int m)
if (m == 0)
return 0;
else
return add(mult(n, pre(m)), n);
/**divide two numbers entered*/
public static int div(int n, int m)
if (n < m)
return 0;
else
return suc(div(sub(n, m), m));
/**raise first number to second number*/
public static int exp(int n, int m)
if (m == 0)
return 1;
else
return mult(exp(n, pre(m)), n);
/**log of number entered*/
public static int log(int n)
if (n < 2)
return 0;
else
return suc(log(div(n, 2)));
/**square root of number entered*/
public static int sqrt(int n)
if (n == 0)
return 0;
else
return sqrt(div(n, ));
/**remainder of first number entered divided by second number*/
public static int mod(int n, int m)
if (n < m)
return 0;
else
return mod(div(n, pre(m)), m);
public static void prt(String s)
System.out.print(s);
public static void prtln(String s)
System.out.println(s);
public static void main(String [ ] args)
prtln("Welcome to the amazing calculator");
prtln("It can add, multiply and do powers for");
prtln("naturals (including 0). Note that all the");
prtln("HARDWARE does is add 1 or substract 1 to any number!!");
in = new BufferedReader(new InputStreamReader ( System.in ) );
int It;
while ( (It = getOp()) >= 0)
prt("" + It + "\n");
private static int getOp( )
int first, second;
String op;
try
System.out.println( "Enter operation:" );
do
op = in.readLine( );
} while( op.length( ) == 0 );
System.out.println( "Enter first number: " );
first = Integer.parseInt( in.readLine( ) );
System.out.println( "Enter second number: " );
second = Integer.parseInt( in.readLine( ) );
prtln("");
prt(first + " " + op + " " + second + " = ");
switch( op.charAt( 0 ) )
case '+':
return add(first, second);
case '-':
return sub(first, second);
case '*':
return mult(first, second);
case '/':
return div(first, second);
case '^':
return exp(first, second);
case 'v':
return log(first);
case 'q':
return sqrt(first);
case '%':
return mod(first, second);
case 's':
return suc(first);
case 'p':
return pre(first);
default:
System.err.println( "Need +, *, or ^" );
return -1;
catch( IOException e )
System.err.println( e );
return 0;
}Hi,
Is there any one to make a program for me in Turbo
C++ for Dos, which can calculate the square root of
any number without using the sqrt( ) or any ready
made functions.
The program should calculate the s.root of the number
by a formula or procedure defined by the user
(programmer).
Thanks.This is a Java forum!
If you want Java help:
1. Start your own thread.
2. Use code tags (above posting box) if you post code.
3. No one will write the program for you. We will help by answering your questions and giving advice on how to fix problems in code you wrote.
4. The formula you need to implement is given above by dizzy. -
Problems with square root approximations with loops program
i'm having some trouble with this program, this loop stuff is confusing me and i know i'm not doing this correctly at all. the expected values in the tester are not matching up with the output. i have tried many variations of the loop in this code even modifying the i parameter in the loop which i guess is considered bad form. nothing seems to work...
here is what i have for my solution class:
/** A class that takes the inputted number by the tester and squares it, and
* loops guesses when the nextGuess() method is called. The epsilon value is
* also inputted by the user, and when the most recent guess returns a value
* <= epsilon, then the hasMoreGuesses() method should return false.
public class RootApproximator
/** Takes the inputted values from the tester to construct a RootApproximator.
* @param val the value of the number to be squared and guessed.
* @param eps the gap in which the approximation is considered acceptable.
public RootApproximator(double val, double eps)
value = val;
square = Math.sqrt(val);
epsilon = eps;
/** Uses the algorithm where 1 is the first initial guess of the
* square root of the inputted value. The algorithm is defined by
* "If X is a guess for a square root of a number, then the average
* of X and value/X is a closer approximation.
* @return increasingly closer guesses as the method is continually used.
public double nextGuess()
final int TRIES = 10000;
double guess = 1;
for (double i = 1; i < TRIES; i++)
double temp = value / guess;
guess = (guess + temp) / 2.0;
return guess;
/** Determines if there are more guesses left if the difference
* of the square and current guess are not equal to or less than
* epsilon.
* @return the value of the condition.
public boolean hasMoreGuesses()
return (square - guess <= epsilon);
private double square;
private double value;
private double epsilon;
private double guess;
here is the tester:
public class RootApproximatorTester
public static void main(String[] args)
double a = 100;
double epsilon = 1;
RootApproximator approx = new RootApproximator(a, epsilon);
System.out.println(approx.nextGuess());
System.out.println("Expected: 1");
System.out.println(approx.nextGuess());
System.out.println("Expected: 50.5");
while (approx.hasMoreGuesses())
approx.nextGuess();
System.out.println(Math.abs(approx.nextGuess() - 10) < epsilon);
System.out.println("Expected: true");
and here is the output:
10.0
Expected: 1 // not sure why this should be 1, perhaps because it is the first guess.
10.0
Expected: 50.5 // (100 + 1) / 2, average of the inputted value and the first guess.
true
Expected: true
i'm new to java this is my first java course and this stuff is frustrating. i'm really clueless as to what to do next, if anyone could please give me some helpful advice i would really appreciate it. thank you all.i'm new to java this is my first java course and this
stuff is frustrating. i'm really clueless as to what
to do nextMaybe it's because you don't have a strategy for what the program is supposed to do? To me it looks like a numerical scheme for finding the squareroot of a number.
Say the number you want to squarerroot is called value and that you have an approximation called guess. How do you determine whether guess is good enought?
Well in hasMoreGuesses you check whether,
(abs(value-guess*guess) < epsilon)
The above decides if guess is within epsilon of being the squareroot of value.
When you calculate the next guess in nextGuess why do you loop so many times? Aren't you supposed to make just one new guess like,
guess = (guess + value/guess)/2.0
The above generates a new guess based on the fact that guess and value/guess must be on each side of value so that the average of them must be closer too value.
Now you can put the two together to a complete algoritm like,
while (hasMoreGuesses()) {
nextGuess();
}In each iteration of the loop a new "guess" of the squareroot is generated and this continues until the guess is a sufficiently close approximation of the squareroot. -
Sampling procedure-Square root n+1
Hi All,
I need to calculate the sample size based on the Containers & Square root n+1.
How to configure the system & how to maintain the master data to get the sample size.
User also insisted only the sample size has to be calculated, but he will enter only one composite result entry
Regards
SubbuHi Subbu,
you can use Physical Sample to control the sample Size.
You must to define a Sample Drawing Procedure (QPV2) as relevant for Container Number, then assign the formula 'Square Root n1´ ( TRUNC(SQRT(P2)1) ) in the partial sample. In this scenario, I suggest you to configure the "Container number' as a mandatory field during the goods receipt once the system will need this information to calculate the sample. For a single result recording, you can define a pooled sample.
I hope it can help you.
Best regards,
Robson -
Calculator square root event listener
I am trying to write a calculator using JButtons added onto Jpanels. Getting numbers to add , divide, multilpy etc. is simple enough, but i need to write an event handler for a button which calculates the square root of a given number. This is easy enough if you want the square root of 4,9,16,25 etc but if you enter a number which produces a decimal point value it doesn't work.
I am trying to do this using a variable "number" which is of type double, but have tried using it as a float, but it still doesn't work! here is my code for handling a click on the Square root JButton.
if(e.getActionCommand().equals("Sqrt")){
number = new Double(jtf.getText().trim()).doubleValue();
double rootNum = 1.000;
while(rootNum*rootNum <= number){
double rootTester = rootNum * rootNum;
if(rootTester == number){
jtf.setText(String.valueOf(rootNum));
rootNum = rootNum+1;How about
double answer = Math.sqrt(number); -
Besides using the math class, may I know if there is another way to find the square root of an integer?
thank you.Besides using the math class, may I know if there is
another way to find the square root of an integer? There are several numerical methods available. The most common is built on Newton-Raphson.
The idea is very simple. Say you want to calculate the squareroot of N. One can note that this squareroot will be somewhere between N and 1/N. So an approximate value will be in the middle like
N1 = (N + 1/N)/2.
Now you're closer and can get even closer by repeating this again with
N2 = (N1 + 1/N1)/2.
This is repeated again and again until Nx*Nx (where x are the numbers 1,2,3,4,5,6 etcetera) is sufficiently close to N. When it is Nx is the squareroot of N. -
The square root free Cholesky factorization
The Attached code calculates the square root free Cholesky factorization (LDL'), it is very useful to decompose matrices and in my specific case, to make observability analysis within electrical distribution networks. I'm publishing it because LV counts with other kind of decompositions like the LU decomposition, very useful but for my case, the values delivered by LDL' are more accurate and easier to calculate.
Best regards
Davis
Davis Montenegro
Attachments:
LDLT.zip 33 KBDavis,
How easy do you think it would be to port this over to the FPGA environment? Just a quick inspection shows the NaN check and I am not sure how this plays into the porting of the code.
This is exactly what I have been working on today and would have saved myself some time if I had known someone smarter than myself had already pieced this out. Thanks a lot!
Matt
Matt Richardson
Certified LabVIEW Developer
MSR Consulting, LLC -
Calculating the square root...
Hi all,
let me shortly sketch the situation.
I have a number of fields that I select from the database (coming from different queries)
that I need to do calculations on...
so far, I was able to do most of those calculations, but now I'm stuck.
One of my calculations contains a square root.
So I got something like:
VARIABLE1 + VARIABLE2 + SQRT(3*VARIABLE1)
with VARIABLE1 being a multiplication of several other fields.
my question: is there an xsl or xdofx function to calculate the square root? Or any other solutions for my problem?
Any help would be greatly appreciated :)
Wkr,
FilipYou can use like this.
<?xdoxslt:pow(10,2)?>
<?xdoxslt:pow(10,3)?>
<?xdoxslt:pow(10,4)?> -
ok, I used the search function and had a hard time understanding how to do things and I don't think any of the posts answered my question.
Right now, I am working on an assignment that makes me figure out the quadratic formula. I am having trouble with the square root part of the program.
For instance that part is: sqrt( b*b - 4*a*c)
I want my program to be able to tell if the square root is a perfect square then go ahead an calculate it, but if it is not, then we just want to reduce the square root function, or even if it is a perfect square, I need to make the program reduce fractions when possible.
For instance, here is an example that is a perfect square: sqrt(25) = 5 , the program would just spit out five automatically.
BUT
If it were: sqrt(24), then I would want it to reduce it to 2 sqrt(6) and that is what the program would pop out.
Also, I am having trouble on how to tell the program how to "reduce fractions" , because if I had the square root in the last example, I might be able to reduce fractions for the whole quadratic function.
example. ( 2 - 2 sqrt(6) ) / 2 would reduce to 1 - sqrt(6) and that is what the program would say.
basically it seems like we are avoiding actually dealing with decimals.
If you could just point me in the right direction or have a good link, it would be appreciated!!!It seems to me your real problem is taking a number and finding all prime factors of that number. Once you have a list of those, you know you can pull duplicates out of the square root. If all of them were duplicates, you had a perfect square.
Here's a simple program I scrawled out to factor and reassemble into the appropriate form:
import java.util.*;
class FactorTest {
public static void main(String[] args) {
int value = 96;
LinkedList<Integer> factors = new LinkedList<Integer>();
LinkedList<Integer> primeNumbers = new LinkedList<Integer>();
populatePrimeList(primeNumbers);
while (value > 1) {
boolean factorFound = false;
for (int prime : primeNumbers) {
if (value % prime == 0) {
value = value / prime;
factors.add(prime);
factorFound = true;
break;
if (!factorFound) {
expandPrimeList(primeNumbers);
int left = 1;
int right = 1;
int lastValue = 1;
for (int factor : factors) {
if (lastValue == factor) {
left *= factor;
lastValue = 1;
else {
right *= lastValue;
lastValue = factor;
right *= lastValue;
System.out.println(left + " sqrt " + right);
public static void populatePrimeList(LinkedList<Integer> primes) {
primes.add(2);
primes.add(3);
primes.add(5);
primes.add(7);
primes.add(11);
primes.add(13);
public static void expandPrimeList(LinkedList<Integer> primes) {
int testPrime = primes.get(primes.size() - 1);
boolean foundFactors;
do {
foundFactors = false;
testPrime += 2;
for (int prime : primes) {
if (testPrime % prime == 0) {
foundFactors = true;
break;
} while (foundFactors);
primes.add(testPrime);
}Edited by: jboeing on Jan 15, 2008 11:18 AM -
I am trying to use Diadem and still learning to do basic calculations.
Given a channel containing several values. How do I calculate that will give me another channel giving corresponding square root of values in original channel..?
Thanks. Himanshu.
Solved!
Go to Solution.In the Analysis panel you can bring up the calculator and enter the following equation Ch("[1]/results" )=Sqr(Ch("[1]/baseVals" )).
If you are doing this in script you can use the following.
Call ChnCalculate("Ch(""[1]/results"" )=Sqr(Ch(""[1]/baseVals"" ))" )
Hope this helps.
Message Edited by waynecj7 on 05-07-2009 08:51 PM -
Square root formula in Calculated Characteristics
Have a great days.
There are two MIC in my inspection plan. first (0010) is for entering number and another (0020) is calculated characteristics. I want to calculate the square root of 0010 charateristics in 0020.
What should be teh formula to put in 0020 char.
Please guide. thanks in advance.
Regards,
Dipesh Bhavsar
Edited by: dipeshbhavsar1982 on Jul 31, 2011 9:41 AMHI
0020 MIC should be created with control indicator for formula tab(calc.characteristic)
and in quality plan assign the first MIC as normal
and assign second MIC then system will ask a formula ...
there you can give the formula in terms of character
example (0010)2
Faisal -
Square root of an interval using Newton's method
Hello!
I am trying to create a method that calculates the square root of an interval, and I am having trouble with both the actual calculation part, as well as the base case for the recursion. I implemented a simple counter for the recursion, but was not seeing any kind of pattern for the values. (I am pretty sure the "better" values should converge to 0).
I was wondering if anybody wanted to take a swing at it and help me out. :)
Here is the code for my program, followed by the code for Newton's method for calculating square roots of doubles. I am supposed to use it as a reference.
I made the simple arithmetic methods with the help of http://en.wikipedia.org/wiki/Interval_arithmetic . They seem to work fine, so I am having issues with troubleshooting!
Thanks!
public class Interval {
double x1;
double x2;
public Interval(double newx1, double newx2){
x1 = newx1;
x2 = newx2;
public String toString(){
return "[" + this.x1 + ", " + this.x2 + "]";
//Add an interval to the current one.
public Interval add(Interval j){
double tempx1 = this.x1 + j.x1;
double tempx2 = this.x2 + j.x2;
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
//Subtract an interval from the current one.
public Interval sub(Interval j){
double tempx1 = this.x1 - j.x2;
double tempx2 = this.x2 - j.x1;
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
//Multiply an interval with the current one.
public Interval mul(Interval j){
double minx1 = Math.min(this.x1*j.x1, this.x2*j.x2);
double minx2 = Math.min(this.x1*j.x2, this.x2*j.x1);
double maxx1 = Math.max(this.x1*j.x1, this.x2*j.x2);
double maxx2 = Math.max(this.x1*j.x2, this.x2*j.x1);
double tempx1 = Math.min(minx1, minx2);
double tempx2 = Math.max(maxx1, maxx2);
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
//Divide the current interval by a new one.
public Interval div(Interval j){
double minx1 = Math.min(this.x1/j.x1, this.x2/j.x2);
double minx2 = Math.min(this.x1/j.x2, this.x2/j.x1);
double maxx1 = Math.max(this.x1/j.x1, this.x2/j.x2);
double maxx2 = Math.max(this.x1/j.x2, this.x2/j.x1);
double tempx1 = Math.min(minx1, minx2);
double tempx2 = Math.max(maxx1, maxx2);
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
static Interval step(Interval x, Interval y) {
// Compute a "better" guess than x for the square root of y:
// Code for doubles: Interval better = x - (x*x - y)/(2*x);
Interval two = new Interval(2.0, 2.0);
Interval better = x.sub( ( (x.mul(x)).sub(y) ).div(two.mul(x)) );
// For doubles:
if ( Math.abs(better.x2 - better.x1) < 0.001 ) { // base case
System.out.println(better.toString());
return better;
else {
return step(better, y); // try to get even better...
static Interval sqrt(Interval y) {
return step(y, y); //: start guessing at the square root
public static void main(String args[]){
Interval i = new Interval(4.0, 8.0);
Interval j = new Interval(4.0, 8.0);
Interval addij = i.add(j);
Interval subij = i.sub(j);
Interval mulij = i.mul(j);
Interval divij = i.div(j);
Interval sqrtj = i.sqrt(j);
System.out.println("Intervals:");
System.out.println(i.toString());
System.out.println(j.toString());
System.out.println("Add: " + addij.toString());
System.out.println("Sub: " + subij.toString());
System.out.println("Mul: " + mulij.toString());
System.out.println("Div: " + divij.toString());
System.out.println("Sqrt: " + sqrtj.toString());
}and newton's root finder for doubles:
public class SquareRoot {
static final double ALLOWED_ERROR = 0.001;
* Newton's method for finding square roots.
static double step(double x, double y) {
// Compute a "better" guess than x for the square root of y:
double better = x - (x*x - y)/(2*x);
// Are we close enough?
if ( Math.abs(x - better) < ALLOWED_ERROR ) { // => stop: base case
return better;
else {
return step(better, y); // try to get even better...
static double sqrt(double y) {
return step(y, y); //: start guessing at the square root
public static void main(String[] args) {
System.out.println(Math.sqrt(1234));
System.out.println(sqrt(1234));
// NOTE: you may need to adjust the error bound for these two to agree
}Nathron wrote:
Here is the code for my program, followed by the code for Newton's method for calculating square roots of doubles. I am supposed to use it as a reference.The only thing I can see that looks suspicious is the call to step(better, y) in your reference code.
Are you sure it shouldn't be step(y, better) or step(better, x))? Newton-Rhapson is supposed to be a progressive method, but as far as I can see the value of y can never change with the way you've got it. And if you've copied that to your new code, it might explain the problem.
Winston -
Hi Folks!
Can anybody tell me how to take square root of this value "bi"?
BigInteger bi = BigInteger.valueOf(2000000000);
Thanks in advance.I wrote this simple sqrt function for BigInteger.
* Returns the largest BigInteger, n, such that bigInt>=n*n.
* If round is true, the function returns n+1 if it is closer to actual square root.
* @param round if true, attempt to find a closer value by rounding up.
* @return <tt>round ? round(sqrt(bigInt)) : floor(sqrt(bigInt))</tt>
public static BigInteger sqrt(BigInteger bigInt, boolean round){
BigInteger op = bigInt;
BigInteger res = BigInteger.ZERO;
BigInteger tmp;
int shift = bigInt.bitLength()-1;
shift -= shift&1;
// set one to highest power of 4 <= bigInt
BigInteger one = BigInteger.ONE.shiftLeft(shift);
while(one.signum()>0){
tmp = res.add(one);
if(op.compareTo(tmp)>=0){
op = op.subtract(tmp);
res = res.add(one.shiftLeft(1));
res = res.shiftRight(1);
one = one.shiftRight(2);
if(round&&op.signum()!=0){
op = bigInt.subtract(res.pow(2));
if(op.compareTo(res)>0){
res = res.add(BigInteger.ONE);
return res;
}
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Photoshop Elements without CD drive?
In the past I purchased a copy of Photoshop Elements 9. I have the disks and the serial number. I just purchased a new macbook air and it has no CD drive. Is there a way I can download the software online and enter my serial code?