Square root function missing
Whether you use the Dashboard calculator widget or the Calculator that is built into the Apple OS as an application, there is no button for square root.
If you are using the Calculator application you can toggle the interface from Basic to Scientific to Programmer. Scientific includes buttons that square a number but no button for square root.
Why? Or am I missing a hidden keystroke or something?
Remember Square Root is the same as x^(1/2)
If you have an x^y button (showing x raised to a power) enter the second number as .5, and you'll have a square root. And if you are really starving, there is Emu:
http://emu48mac.sourceforge.net/
Similar Messages
-
Hi ,
I am wrting a business rule and am not able to get a function for square root.
Any ideas on this . Urgent pls
KhanKhan,
Use the function @POWER (expression, power) with 0.5 as the power parameter.
For more infor
http://download.oracle.com/docs/cd/E10530_01/doc/epm.931/html_esb_techref/techref.htm
Sandeep Reddy Enti
HCC
http://hyperionconsultancy.com/ -
ForLoop Square Root Calculator Program
* Programmer: R McBride
* Date: November 29, 2006
* Filename: SquareRootForLoop.java
* Purpose: This program uses for()loops to
* find the square root of a number
* up to four decimal places long.
import APCS.Keyboard;
class SquareRootForLoopTest1
public static void main (String[]args)
//define variables
double SquareRoot = 0; //holds initial value
double num1 = 0;
boolean done = false;
double tempReg1 = 0; //holds initial value of i
double tempReg2 = 0; //holds
double tempReg3 = 0;
//while loop start
// while (!done)
//program input
System.out.println("\t\t\tSQUARE ROOT CALCULATOR PROGRAM");
System.out.println(""); //skip line
System.out.println("Enter the number you would like to be square rooted or [0] to end the program.");
System.out.print("");
SquareRoot = Keyboard.readInt();
System.out.println(""); //skip line
System.out.println("You entered "+SquareRoot+".");
System.out.println(""); //skip line
//for loop function
for(int i = 0; i < 999999; i++)
tempReg1 = i * i;
tempReg2 = i;
// while (!done)
if(tempReg1 == SquareRoot)
System.out.println("The square root is "+tempReg2);
System.out.println("");
else
System.out.print("");
//while loop check to finish
// if(SquareRoot == 0)
// done = true;
// }//end while loop
}Here is my code for and of you who ever wants to use it. Our assignment was to make a square root calculator using ForLoop (a.k.a. no square root functions).
My question is, how would I make it increase by .0001 instead of 1 every loop? I need it to do that so I can find square roots of numbers of numbers that have decimals for squares.* Programmer: R McBride
* Date: November 29, 2006
* Filename: SquareRootForLoop.java
* Purpose: This program uses for()loops to
* find the square root of a number
* up to four decimal places long.
import APCS.Keyboard;
class SquareRootForLoopTest1
public static void main (String[]args)
//define variables
double SquareRoot = 0; //holds initial value
double num1 = 0;
boolean done = false;
double tempReg1 = 0; //holds initial value of i
double tempReg2 = 0; //holds
double tempReg3 = 0;
//while loop start
// while (!done)
//program input
System.out.println("\t\t\tSQUARE ROOT CALCULATOR PROGRAM");
System.out.println(""); //skip line
System.out.println("Enter the number you would like to be square rooted or [0] to end the program.");
System.out.print("");
SquareRoot = Keyboard.readInt();
System.out.println(""); //skip line
System.out.println("You entered "+SquareRoot+".");
System.out.println(""); //skip line
//for loop function
for(double i = 0; i < 999999; i+=.0001)
tempReg1 = i * i;
tempReg2 = i;
// while (!done)
if(tempReg1 == SquareRoot)
System.out.println("The square root is "+tempReg2);
System.out.println("");
else
System.out.print("");
//while loop check to finish
// if(SquareRoot == 0)
// done = true;
// }//end while loop
}Still doesn't work. =[ -
I was looking for a fast algorythm for integer square roots and I found this one http://medialab.freaknet.org/martin/src/sqrt/.
The algorithm comes from a book by Mr C. Woo on how to do maths on an abacus.
I post the javaized version here in case anyone finds it interesting.
I believe this is the fastest square root function in existance for integers (and in game programming much of the time you aren't interested in fractions)
/* Fast interger square root adapted from algorithm by Martin Guy @ UKC, June 1985.
* Origonally from a book on programming abaci by Mr C. Woo.
public static int fastSqrt2(int n)
int op, res, one;
op = n;
res = 0;
/* "one" starts at the highest power of four <= than the argument. */
one = 1 << 30; /* second-to-top bit set */
while (one > op) one >>= 2;
while (one != 0)
if (op >= res + one)
op = op - (res + one);
res = res + (one<<1);
res >>= 1;
one >>= 2;
return(res);
}public static double sqrt(double a){
if(a<0) throw new IllegalArgumentException("number<0");
double precision=0.001;
double x_nMinus1 = -1;
double x_n = 1;
while( Math.abs(x_n - x_nMinus1) > precision ) {
x_nMinus1 = x_n;
x_n = (x_nMinus1 * x_nMinus1 + a) / (2*x_nMinus1);
return x_n;
} -
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 -
Okay, two things...I've always kinda wondered what the algorithm for the square-root function is...where would I find that?
but the main thing is, I was making a class to store/deal with a complex/mixed number (a + b*i), and I was trying to make a square-root method for that. But I fiddled around with the variables in the equation, and I can't quite get any further.
This is what I got (algebraically: this isn't actual code):
( the variables a, b, c, d are all real numbers )
( the constant i is the imaginary unit, sqrt(-1) )
sqrt(a + b*i) == c + d*i
a + b*i == (c + di)^2
a + b*i == c*c - d*d + 2*c*d*i
a == c*c - d*d
b == 2*c*d
c == sqrt( a + d*d )
c == b / (2* d)
d == sqrt( c*c - a )
d == b / (2*c)
right now the only thing i can conclude from that, is that if you know (a or b) and (c or d) you can determine the other variables. but I can't figure out how to define c or d purely in terms of a and b, as the method would need to. so I'm stuck.Okay, two things...I've always kinda wondered what the
algorithm for the square-root function is...where
would I find that?
Math.sqrt()It's an extremely important skill to learn to read the API and become familiar with the tools you will use to program Java. Java has an extensive set of documentation that you can even download for your convenience. These "javadocs" are indexed and categorized so you can quickly look up any class or method. Take the time to consult this resource whenever you have a question - you'll find they typically contain very detailed descriptions and possibly some code examples.
http://java.sun.com/reference/api/index.html
http://java.sun.com/j2se/1.4.2/docs/api/ -
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. -
Does anyone have an clever ways to yield a square root result. I can use the ^ feature a business rule or calc script and it validates fine, but then I don't get a result
ex: I can say 100^.5 and should get a value of 10, but no go.....anyone having a similar type issue? Thanks for any assistance.Hyperacle,
You do not need a positive integer for the power argument in the @POWER(expession, power) function -- it can be any numeric expression. It sounds like you might have a block creation issue if you are seeing #missing. You can enter a value at the particular intersection to create the block. Your calc or business rule should calcluate correctly at this point. I just created a test business rule using the example @POWER(100, .5) and the formula returned 10 in the desired intersection.
Regards,
csdjr -
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;
} -
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. -
Cannot display square root symbol in cvi
I don't understand why this would be an issue, but if I'm writing in the source window (with the default font of NIEditor), I cannot display a square root symbol "√" - every time I type alt+251, I get "v". Ok, not a huge deal in the source window, but it is a big deal if THAT is what's being stored in a string variable I'm writing out to a file. Additionally, if I use the following to format a string:
"Fmt(setpointUOM, "%s<W%cT/P3", 251);", then setpointUOM = "WüT/P3". If I display this variable in a string control, it displays as "W√T/P3", which is correct, but if I use it in a Text Box or save it to a sql database, it displays "WüT/P3" which is unacceptable.
I've tried changing the fonts but nothing works or I get even stranger results. I've been dealing with/ignoring this since CVI7 and I'm currently at CVI10. Thanks for any help.
Solved!
Go to Solution.The difference in the display between W√T/P3 and WüT/P3 has to do with the character set that you select for the UI control. From what I can tell, only the OEM code pages map the √ symbol to character 251, so if you want to see √ for that character, you should pick the OEM character set, in that control.
Entering √ with the keyboard in a CVI window seems like a much more problematic task. When you type Alt+251 on a CVI window, the keyboard driver is converting the 251 to 118 (the letter v). I don't know why it does that, but I noticed that the code that it converts 251 to varies, depending on your input language (which you can change in Control Panel>>Region and Language>>Change keyboards or other input methods). When english is selected in the language bar, it converts it to 118. With other languages, it converts it to other codes. I tried entering the unicode value for √ directly, which is Alt+221A (to enter unicode characters using the keypad, you have to follow the steps described here). But it didn't work. It still converted it to 118. I suspect the keyboard driver is doing this because it tries to map 221A to some symbol that is valid in the code page that corresponds to the input language, isn't able to, and picks what it thinks is the closest match.
In lots of other applications (but not all) you can type Alt+251 or Alt+221A and it works just fine. This is because those applications accept unicode characters directly. Unfortunately, CVI isn't one of them. It uses code pages (a.k.a. character sets, or multibyte sequences) instead of unicode, and symbols don't have a universal meaning; they depend on the code page.
Even if all this were not an impediment, you'd still run into the problem that in CVI 2010 you cannot change the font of the CVI source window to use an OEM character set. So, you'd still see it displayed as a v or a ü. In CVI 2012 you can change the change it, and so if you were using CVI 2012 you could conceivably use the clipboard to paste the √, but there is a bug in CVI that is preventing OEM characters from being pasted correctly (I realize the description says that it affects tree tooltips, but it's also generally affecting clipboard operations of some less commonly used character sets).
So, to make a long story short, I think you should continue entering the code directly with the Fmt function, at least until this bug is fixed. And make sure that all UI controls that need to display this string are using the OEM character set.
Luis -
Hi, I'm suppose to write a programme that asks the user to enter a number. Find the square root of that number. Format theoutput to 4 decimal places. Use Math.sqrt(double a). i.e. to find the square root of 9.56, use: double sq = Math.sqrt(9.56);
This is what I have:
import java.io.*;
import java.math.*;
public class MethodEx2
static BufferedReader keyboard = new BufferedReader (new
InputStreamReader(System.in));
public static void main (String[] args) throws IOException
double num = 0;
System.out.print("Please enter a number");
System.out.flush();
num = Integer.parseInt(keyboard.readLine());
calsqrt(num);
public static double sqrt (double number);
double result = 0;
result = sqrt(number);
System.out.print(result);
}I keep getting an error saying that it cannot return a value from method whose result type is void at line 33, col 11.
How do I make the program to work?Oh okay, thanks. I also found another typo. I fixed it but it still doesn't work. Anyway, this is my new code:
import java.io.*;
import java.math.*;
public class MethodEx2
static BufferedReader keyboard = new BufferedReader (new
InputStreamReader(System.in));
public static void main (String[] args) throws IOException
double num = 0;
System.out.print("Please enter a number");
System.out.flush();
num = Integer.parseInt(keyboard.readLine());
sqrt(num);
public static double sqrt (double number)
double result = 0;
result = sqrt(number);
System.out.print(result);
}It says that MethodEx2.java is missing return statement at line 17, column 3. I'm not sure which variable I should return. I tried 'number' and 'result' but something like this happens:
"C:\Program Files\JBuilder9\jdk1.4\bin\javaw" -classpath "E:\Lili\School\GR12 Comp Sci\Period 1\Programs\MethodExs\Ex2\Ex2\classes;C:\Program Files\JBuilder9\jdk1.4\demo\jfc\Java2D\Java2Demo.jar;C:\Program Files\JBuilder9\jdk1.4\demo\plugin\jfc\Java2D\Java2Demo.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\charsets.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\dnsns.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\ldapsec.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\localedata.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\sunjce_provider.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\im\indicim.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jaws.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jce.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jsse.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\rt.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\sunrsasign.jar;C:\Program Files\JBuilder9\jdk1.4\lib\dt.jar;C:\Program Files\JBuilder9\jdk1.4\lib\htmlconverter.jar;C:\Program Files\JBuilder9\jdk1.4\lib\tools.jar" MethodEx2
Please enter a number9
java.lang.StackOverflowError
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19) -
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)?> -
Problem:
Using the Newton-Raphson method, find the square root of an input positive integer number (ensure that the input is positive). Compare the result with Math.sqrt (n).
Hints:
* Our function is f(x) = x^2 - n where n is the input number. Then, Newton-Raphson method sas that xk+1 is a better approximation for the square root of n than xk where
xk+1 = (xk + n / xk) ---------->all the k are below the x.
* You can take x0 as n.
* Stop when |xk+1 - xk| < e where e is a very small number (take as 1.0E-09).
Sample execution:
Type the integer whose square root you would like to compute: 2
Square root of 2 is 1.414213562373095 by Newton-Raphson method and
1.4142135623730951 by Math class.
I didnt understand what is formula of Newton-Raphson method exactly.Therefore, i couldnt apply the method into java.It is my lab assignment please help me until friday morning.Thanks.umutcan55 wrote:
I didnt understand what is formula of Newton-Raphson method exactly.You could [start with the Wikipedia article|http://en.wikipedia.org/wiki/Newton's_method] for example.
Therefore, i couldnt apply the method into java.So you don't actually have a Java question, right? So this is not the correct forum.
It is my lab assignment please help me until friday morning.That's irrelevant. -
Dear all
How can i get the square root of a number.for examle 4 from the number 16,as
16 = 4*4.
Best Regards
BuntyHii
To get the Square root: use the arithmetic Function SQRT( ).
Sample code: try...
DATA: V1 TYPE I VALUE 100.
DATA : V_ROOT TYPE I.
V_ROOT = SQRT( V1 ).
WRITE:/ 'square root = ', v_root.
<b>reward if Helpful.</b>
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