I'm a bit annoyed with a method I wrote to approximate sine function in Java. Here it is, it's based on Taylor's series.
static double PI = 3.14159265358979323846;
static double eps = 0.0000000000000000001;
static void sin(double x) {
x = x % (2 * PI);
double term = 1.0;
double res = 0.0;
for (int i = 1; term > eps; i++) {
term = term * (x / i);
if (i % 4 == 1) res += term;
if (i % 4 == 3) res -= term;
}
System.out.println(sum);
}
For little values, I got very good approximation of sine, but for large values (e.g pow(10,22)), results seems very very wrong.
Here are the results :
sin(pow(10,22)) // 0.8740280612007599
Math.sin(pow(10,22)) // -0.8522008497671888
Does someone have an idea ? Thank you !
Best regards,
Be reassured that the Java sin function will be off too.
You problem is that the Taylor expansion for sin has a small radius of convergence and convergence is slow even if you're within that radius.
There are floating point considerations too: a floating point double gives you about 15 significant figures of accuracy.
So for large arguments for sin, the accuracy will deteriorate significantly especially given that sin is a periodic function:
sin(x + 2 * pi * n) = sin(x) for any integer n.
Your answer is incorrect for big numbers because you accumulate a lot of rounding errors due to double presentation. When the number is big, then your for loop will iterate a lot before the term becomes smaller than epsilon. In each iteration, a rounding error is accumulated. The result is a very big error in the final value. Read some nice reference on "Numerical Analysis". Anyway, Tylor's series approximate sin near 0, by definition. So, it is normal not to be correct for very big numbers.
The difference actually has nothing to do with the radius of convergence of the Taylor Series and has to do with double precision not being accurate enough to hold the precision required for such big numbers. The radius of the Taylor series for the sine function is infinity.
10^22 is approximately 2^73. Since the mantissa for a double precision number is 52 bits, consecutive values that can be stored with double precision format will be 2^21 apart from each other. Since an evaluation of the sine function requires more resolution than that, you won't be able to reliably get an answer.
public class doublePrecision {
public static void main(String[] args) {
double total = 0;
total += 5.6;
total += 5.8;
System.out.println(total);
}
}
The above code prints:
11.399999999999
How would I get this to just print (or be able to use it as) 11.4?
As others have mentioned, you'll probably want to use the BigDecimal class, if you want to have an exact representation of 11.4.
Now, a little explanation into why this is happening:
The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction and a exponent.
More specifically, a double-precision floating point value such as the double type is a 64-bit value, where:
1 bit denotes the sign (positive or negative).
11 bits for the exponent.
52 bits for the significant digits (the fractional part as a binary).
These parts are combined to produce a double representation of a value.
(Source: Wikipedia: Double precision)
For a detailed description of how floating point values are handled in Java, see the Section 4.2.3: Floating-Point Types, Formats, and Values of the Java Language Specification.
The byte, char, int, long types are fixed-point numbers, which are exact representions of numbers. Unlike fixed point numbers, floating point numbers will some times (safe to assume "most of the time") not be able to return an exact representation of a number. This is the reason why you end up with 11.399999999999 as the result of 5.6 + 5.8.
When requiring a value that is exact, such as 1.5 or 150.1005, you'll want to use one of the fixed-point types, which will be able to represent the number exactly.
As has been mentioned several times already, Java has a BigDecimal class which will handle very large numbers and very small numbers.
From the Java API Reference for the BigDecimal class:
Immutable,
arbitrary-precision signed decimal
numbers. A BigDecimal consists of an
arbitrary precision integer unscaled
value and a 32-bit integer scale. If
zero or positive, the scale is the
number of digits to the right of the
decimal point. If negative, the
unscaled value of the number is
multiplied by ten to the power of the
negation of the scale. The value of
the number represented by the
BigDecimal is therefore (unscaledValue
× 10^-scale).
There has been many questions on Stack Overflow relating to the matter of floating point numbers and its precision. Here is a list of related questions that may be of interest:
Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273?
How to print really big numbers in C++
How is floating point stored? When does it matter?
Use Float or Decimal for Accounting Application Dollar Amount?
If you really want to get down to the nitty gritty details of floating point numbers, take a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic.
When you input a double number, for example, 33.33333333333333, the value you get is actually the closest representable double-precision value, which is exactly:
33.3333333333333285963817615993320941925048828125
Dividing that by 100 gives:
0.333333333333333285963817615993320941925048828125
which also isn't representable as a double-precision number, so again it is rounded to the nearest representable value, which is exactly:
0.3333333333333332593184650249895639717578887939453125
When you print this value out, it gets rounded yet again to 17 decimal digits, giving:
0.33333333333333326
If you just want to process values as fractions, you can create a Fraction class which holds a numerator and denominator field.
Write methods for add, subtract, multiply and divide as well as a toDouble method. This way you can avoid floats during calculations.
EDIT: Quick implementation,
public class Fraction {
private int numerator;
private int denominator;
public Fraction(int n, int d){
numerator = n;
denominator = d;
}
public double toDouble(){
return ((double)numerator)/((double)denominator);
}
public static Fraction add(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop + bTop, a.denominator * b.denominator);
}
else{
return new Fraction(a.numerator + b.numerator, a.denominator);
}
}
public static Fraction divide(Fraction a, Fraction b){
return new Fraction(a.numerator * b.denominator, a.denominator * b.numerator);
}
public static Fraction multiply(Fraction a, Fraction b){
return new Fraction(a.numerator * b.numerator, a.denominator * b.denominator);
}
public static Fraction subtract(Fraction a, Fraction b){
if(a.denominator != b.denominator){
double aTop = b.denominator * a.numerator;
double bTop = a.denominator * b.numerator;
return new Fraction(aTop-bTop, a.denominator*b.denominator);
}
else{
return new Fraction(a.numerator - b.numerator, a.denominator);
}
}
}
Observe that you'd have the same problem if you used limited-precision decimal arithmetic, and wanted to deal with 1/3: 0.333333333 * 3 is 0.999999999, not 1.00000000.
Unfortunately, 5.6, 5.8 and 11.4 just aren't round numbers in binary, because they involve fifths. So the float representation of them isn't exact, just as 0.3333 isn't exactly 1/3.
If all the numbers you use are non-recurring decimals, and you want exact results, use BigDecimal. Or as others have said, if your values are like money in the sense that they're all a multiple of 0.01, or 0.001, or something, then multiply everything by a fixed power of 10 and use int or long (addition and subtraction are trivial: watch out for multiplication).
However, if you are happy with binary for the calculation, but you just want to print things out in a slightly friendlier format, try java.util.Formatter or String.format. In the format string specify a precision less than the full precision of a double. To 10 significant figures, say, 11.399999999999 is 11.4, so the result will be almost as accurate and more human-readable in cases where the binary result is very close to a value requiring only a few decimal places.
The precision to specify depends a bit on how much maths you've done with your numbers - in general the more you do, the more error will accumulate, but some algorithms accumulate it much faster than others (they're called "unstable" as opposed to "stable" with respect to rounding errors). If all you're doing is adding a few values, then I'd guess that dropping just one decimal place of precision will sort things out. Experiment.
You may want to look into using java's java.math.BigDecimal class if you really need precision math. Here is a good article from Oracle/Sun on the case for BigDecimal. While you can never represent 1/3 as someone mentioned, you can have the power to decide exactly how precise you want the result to be. setScale() is your friend.. :)
Ok, because I have way too much time on my hands at the moment here is a code example that relates to your question:
import java.math.BigDecimal;
/**
* Created by a wonderful programmer known as:
* Vincent Stoessel
* xaymaca#gmail.com
* on Mar 17, 2010 at 11:05:16 PM
*/
public class BigUp {
public static void main(String[] args) {
BigDecimal first, second, result ;
first = new BigDecimal("33.33333333333333") ;
second = new BigDecimal("100") ;
result = first.divide(second);
System.out.println("result is " + result);
//will print : result is 0.3333333333333333
}
}
and to plug my new favorite language, Groovy, here is a neater example of the same thing:
import java.math.BigDecimal
def first = new BigDecimal("33.33333333333333")
def second = new BigDecimal("100")
println "result is " + first/second // will print: result is 0.33333333333333
Pretty sure you could've made that into a three line example. :)
If you want exact precision, use BigDecimal. Otherwise, you can use ints multiplied by 10 ^ whatever precision you want.
As others have noted, not all decimal values can be represented as binary since decimal is based on powers of 10 and binary is based on powers of two.
If precision matters, use BigDecimal, but if you just want friendly output:
System.out.printf("%.2f\n", total);
Will give you:
11.40
You're running up against the precision limitation of type double.
Java.Math has some arbitrary-precision arithmetic facilities.
You can't, because 7.3 doesn't have a finite representation in binary. The closest you can get is 2054767329987789/2**48 = 7.3+1/1407374883553280.
Take a look at http://docs.python.org/tutorial/floatingpoint.html for a further explanation. (It's on the Python website, but Java and C++ have the same "problem".)
The solution depends on what exactly your problem is:
If it's that you just don't like seeing all those noise digits, then fix your string formatting. Don't display more than 15 significant digits (or 7 for float).
If it's that the inexactness of your numbers is breaking things like "if" statements, then you should write if (abs(x - 7.3) < TOLERANCE) instead of if (x == 7.3).
If you're working with money, then what you probably really want is decimal fixed point. Store an integer number of cents or whatever the smallest unit of your currency is.
(VERY UNLIKELY) If you need more than 53 significant bits (15-16 significant digits) of precision, then use a high-precision floating-point type, like BigDecimal.
private void getRound() {
// this is very simple and interesting
double a = 5, b = 3, c;
c = a / b;
System.out.println(" round val is " + c);
// round val is : 1.6666666666666667
// if you want to only two precision point with double we
// can use formate option in String
// which takes 2 parameters one is formte specifier which
// shows dicimal places another double value
String s = String.format("%.2f", c);
double val = Double.parseDouble(s);
System.out.println(" val is :" + val);
// now out put will be : val is :1.67
}
Use java.math.BigDecimal
Doubles are binary fractions internally, so they sometimes cannot represent decimal fractions to the exact decimal.
/*
0.8 1.2
0.7 1.3
0.7000000000000002 2.3
0.7999999999999998 4.2
*/
double adjust = fToInt + 1.0 - orgV;
// The following two lines works for me.
String s = String.format("%.2f", adjust);
double val = Double.parseDouble(s);
System.out.println(val); // output: 0.8, 0.7, 0.7, 0.8
Doubles are approximations of the decimal numbers in your Java source. You're seeing the consequence of the mismatch between the double (which is a binary-coded value) and your source (which is decimal-coded).
Java's producing the closest binary approximation. You can use the java.text.DecimalFormat to display a better-looking decimal value.
Short answer: Always use BigDecimal and make sure you are using the constructor with String argument, not the double one.
Back to your example, the following code will print 11.4, as you wish.
public class doublePrecision {
public static void main(String[] args) {
BigDecimal total = new BigDecimal("0");
total = total.add(new BigDecimal("5.6"));
total = total.add(new BigDecimal("5.8"));
System.out.println(total);
}
}
Multiply everything by 100 and store it in a long as cents.
Computers store numbers in binary and can't actually represent numbers such as 33.333333333 or 100.0 exactly. This is one of the tricky things about using doubles. You will have to just round the answer before showing it to a user. Luckily in most applications, you don't need that many decimal places anyhow.
Floating point numbers differ from real numbers in that for any given floating point number there is a next higher floating point number. Same as integers. There's no integer between 1 and 2.
There's no way to represent 1/3 as a float. There's a float below it and there's a float above it, and there's a certain distance between them. And 1/3 is in that space.
Apfloat for Java claims to work with arbitrary precision floating point numbers, but I've never used it. Probably worth a look.
http://www.apfloat.org/apfloat_java/
A similar question was asked here before
Java floating point high precision library
Use a BigDecimal. It even lets you specify rounding rules (like ROUND_HALF_EVEN, which will minimize statistical error by rounding to the even neighbor if both are the same distance; i.e. both 1.5 and 2.5 round to 2).
Why not use the round() method from Math class?
// The number of 0s determines how many digits you want after the floating point
// (here one digit)
total = (double)Math.round(total * 10) / 10;
System.out.println(total); // prints 11.4
Check out BigDecimal, it handles problems dealing with floating point arithmetic like that.
The new call would look like this:
term[number].coefficient.add(co);
Use setScale() to set the number of decimal place precision to be used.
If you have no choice other than using double values, can use the below code.
public static double sumDouble(double value1, double value2) {
double sum = 0.0;
String value1Str = Double.toString(value1);
int decimalIndex = value1Str.indexOf(".");
int value1Precision = 0;
if (decimalIndex != -1) {
value1Precision = (value1Str.length() - 1) - decimalIndex;
}
String value2Str = Double.toString(value2);
decimalIndex = value2Str.indexOf(".");
int value2Precision = 0;
if (decimalIndex != -1) {
value2Precision = (value2Str.length() - 1) - decimalIndex;
}
int maxPrecision = value1Precision > value2Precision ? value1Precision : value2Precision;
sum = value1 + value2;
String s = String.format("%." + maxPrecision + "f", sum);
sum = Double.parseDouble(s);
return sum;
}
You can Do the Following!
System.out.println(String.format("%.12f", total));
if you change the decimal value here %.12f
So far I understand it as main goal to get correct double from wrong double.
Look for my solution how to get correct value from "approximate" wrong value - if it is real floating point it rounds last digit - counted from all digits - counting before dot and try to keep max possible digits after dot - hope that it is enough precision for most cases:
public static double roundError(double value) {
BigDecimal valueBigDecimal = new BigDecimal(Double.toString(value));
String valueString = valueBigDecimal.toPlainString();
if (!valueString.contains(".")) return value;
String[] valueArray = valueString.split("[.]");
int places = 16;
places -= valueArray[0].length();
if ("56789".contains("" + valueArray[0].charAt(valueArray[0].length() - 1))) places--;
//System.out.println("Rounding " + value + "(" + valueString + ") to " + places + " places");
return valueBigDecimal.setScale(places, RoundingMode.HALF_UP).doubleValue();
}
I know it is long code, sure not best, maybe someone can fix it to be more elegant. Anyway it is working, see examples:
roundError(5.6+5.8) = 11.399999999999999 = 11.4
roundError(0.4-0.3) = 0.10000000000000003 = 0.1
roundError(37235.137567000005) = 37235.137567
roundError(1/3) 0.3333333333333333 = 0.333333333333333
roundError(3723513756.7000005) = 3.7235137567E9 (3723513756.7)
roundError(3723513756123.7000005) = 3.7235137561237E12 (3723513756123.7)
roundError(372351375612.7000005) = 3.723513756127E11 (372351375612.7)
roundError(1.7976931348623157) = 1.797693134862316
Do not waste your efford using BigDecimal. In 99.99999% cases you don't need it. java double type is of cource approximate but in almost all cases, it is sufficiently precise. Mind that your have an error at 14th significant digit. This is really negligible!
To get nice output use:
System.out.printf("%.2f\n", total);
As I was trying to compute some very small simple precision and double precision floating numbers I encountered some issues.
Take a look at the following code sample:
public class FloatingLimits {
public static void doSimpleFloatingLimitDemo() {
float firstValue = 1.56F;
float newValue = 1.0F / ((float)Math.pow(2.0D, 150));
double doubleFirst = 2.56;
double doubleNew = 1.0F /Math.pow(2.0D, 150);
double doubleThird = 1.0F/Math.pow(2.0D, 589);
double doubleFourth = 1.0F/Math.pow(2.0, 1589);
System.out.println("float first value =" + firstValue);
System.out.println("float new value =" + newValue);
System.out.println("double first value =" + doubleFirst);
System.out.println("double new value =" + doubleNew);
System.out.println("double third value =" + doubleThird);
System.out.println("double fourth value =" + doubleFourth);
}
public static void main(String[] args) {
doSimpleFloatingLimitDemo();
}
}
It produces the following result:
There is therefore a representation issue or a display issue! Does this have anything to do with numbers precision? The very small numbers that I could not represent with a simple float precision type (32 bits), could be represented with double float precision numbers (64) bits, but the double float also is showing limits. So what would that limit be for very small numbers? Is there a workaround for this using float and double numbers or should I necessarily use BigDecimal to solve it. If I have to use BigDecimals, is there a certain limit to BigDecimal representation as well?
If you look at Double.MAX_VALUE
A constant holding the largest positive finite value of type double, (2-2-52).21023.
And if you run:
System.out.println(Math.pow(2.0, 1589));
You will see the expected result:
Infinity
If you need to represents decimal numbers with arbitrary precision you have to use BigDecimal.
Here is a link
http://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html
The internal binary representation of float and double can introduce errors. Another possibility is to work with int or long multiply the values by a factor (10, 100, 1000 ...) and treat them as non decimal values.
Hi I'm having an issue while creating a change calculator for a school assignment. It essentially is to calculate the lowest amount of change needed for a certain amount of money.
Ex. $5.36:
2 toonies (2$)
1 loonie (1$)
1 quarter
1 dime
0 nickels
1 penny
I've stated all of my variable to be doubles so I can calculate the values and the totals together. It seems to work fine on whole numbers (5.00, 6.00, 7.00) but messes up whenever I add a decimal place. Like when I say $5.25 it should say 2 toonies 1 loonie and 1 quarter. I think it could be an error in rounding or something wrong with my calculations. Any Help is appreciated. Here is the calculations of the code:
//Rounding to one number
DecimalFormat oneDigit = new DecimalFormat ("#,0");
//Ask user for input
String moneyinput = JOptionPane.showInputDialog ("Welcome to the Change Caluculator. "
+ "Please Enter your amount of money in Dollars ($): ");
//Take user input and create into string.
totmoney = Double.parseDouble (moneyinput);
//Calculate number of toonies
numtoonies = (totmoney/toonieval);
System.out.println ("There is a total of " + oneDigit.format (numtoonies) + " toonies.");
//Find new amount
totmoney = (totmoney%toonieval);
//Calculate the number of loonies
numloonies = (totmoney/loonieval);
//Find new amount
totmoney = (totmoney-numloonies);
System.out.println ("There is a total of " + oneDigit.format (numloonies) + " loonies.");
//Calculate number of quarters
numquarters = (totmoney/quarterval);
//State the about of Coins
System.out.println ("There is a total of " + oneDigit.format (numquarters) + " quarters.");
}
I don't quite understand why you are using the DecimalFormat at all. You should be able to do solve this with only mod % and division /.
This is how I would approach this problem:
Take the input from the user (as a double in the format 5.33)
Save that as int by moving the decimal place (int value = cost * 100)
Find the number of toonies by using division (numToonies = value / toonieval)
Find the remaining amount of money by (value = value % toonieval)
Repeat steps 3 and 4 for the other denominations
Note: you will have to modify the values of the toonies to reflect the fact that the price you used was multiplied by 100.
I don't think this is working properly for floating point numbers:
totmoney = (totmoney%toonieval);
Try
totmoney = totmoney - toonieval*numtoonies;
instead.
Also, be aware that floating point values are generally not suitable for handling monetary values (because of the possiblity of rounding errors). Use fixed point decimals instead (basically, store the value in cents and calculate everyting on a cents base).
Do not use floating point numbers when you need absolute accuracy. The IEEE floating point specification which java follows even states that there will be a loss of precision because the floating point spec can not properly represent all floating point operations it can only approximate them. The solution for this is that you need to store the value in 2 ints, one for the amount left of the decimal and one for the value to the right of the decimal
This question already has answers here:
How do I get whole and fractional parts from double in JSP/Java?
(18 answers)
Closed 9 years ago.
double d = 4.321562;
Is there an easy way to extract the 0.321562 on it's own from d? I tried looking in the math class but no luck. If this can be done without converting to string or casting to anything else, even better.
Well, you can use:
double x = d - Math.floor(d);
Note that due to the way that binary floating point works, that won't give you exactly 0.321562, as the original value isn't exactly 4.321562. If you're really interested in exact digits, you should use BigDecimal instead.
Another way to get the fraction without using Math is to cast to a long.
double x = d - (long) d;
When you print a double the toString will perform a small amount of rounding so you don't see any rounding error. However, when you remove the integer part, the rounding is no longer enough and the rounding error becomes obvious.
The way around this is to do the rounding yourself or use BigDecimal which allows you to control the rounding.
double d = 4.321562;
System.out.println("Double value from toString " + d);
System.out.println("Exact representation " + new BigDecimal(d));
double x = d - (long) d;
System.out.println("Fraction from toString " + x);
System.out.println("Exact value of fraction " + new BigDecimal(x));
System.out.printf("Rounded to 6 places %.6f%n", x);
double x2 = Math.round(x * 1e9) / 1e9;
System.out.println("After rounding to 9 places toString " + x2);
System.out.println("After rounding to 9 places, exact value " + new BigDecimal(x2));
prints
Double value from toString 4.321562
Exact representation 4.321562000000000125510268844664096832275390625
Fraction from toString 0.3215620000000001
Exact value of fraction 0.321562000000000125510268844664096832275390625
Rounded to 6 places 0.321562
After rounding to 9 places toString 0.321562
After rounding to 9 places, exact value 0.32156200000000001448796638214844278991222381591796875
NOTE: double has limited precision and you can see representation issue creep in if you don't use appropriate rounding. This can happen in any calculation you use with double esp numbers which are not an exact sum of powers of 2.
Use modulo:
double d = 3.123 % 1;
assertEquals(0.123, d,0.000001);