The question was basically to calculate e^x without inbuilt functions in Java.
The sequence to code was e^x = 1+x+x2/2!+x3/3!+x4/4!+ ...
Here was my attempt at the question:
public static void myexp(double x, double i){
double j = 1.0;
double sum = x;
while (j <= i){
j = j + 1;
sum = sum + (sum * (x / (j)));
System.out.println(sum + 1 + x);
}
}
public static void main(String[] args) {
double x = 1;
double i = 5.0;
myexp(x, i);
}
Now it wasn't working, and eventually I gave in and looked up what a model answer should look like (I know, I know). Here is what it is (in the style of my code):
public static void myexp(double x, double i){
double j = 1.0;
double sum = x;
double result = 1.0;
while (j <= i){
j = j + 1;
sum = sum * (x / (j));
result = result + sum;
System.out.println(result + x);
}
}
Now the difference is the inclusion of the 'results' variable, which delineates the summation of the sequence. However, I thought I had incorporated that when I wrote
"sum = sum+(sum*(x/(j)));".
But the machine recognises one style and not the other. What gives?
In each iteration you are supposed to add to the total sum the term
sum * x / j
where sum is the term added in the previous iteration.
This means you must store the term added in the previous iteration in a separate variable.
If you use the same variable for the total result (which is supposed to be the sum of all the terms of all iterations) and for the term of the current iteration (sum), you get an entirely different result.
In other words
sum = sum + (sum * (x / (j)));
is not equivalent to
sum = sum * (x / (j));
result = result + sum;
since the value of sum depends on the previous value of sum, and therefore you can't eliminate that variable.
I coded a very simple neural net with 1 neuron and 2 inputs and 1 bias in java, trying to classify dots on either the left or right side of a line. The problem is that the neural net recognizes the slope of the line but not the y-intercept (e.g. the function of the line may be y = m*x + c and the NN recognizes the m but not the c).
I tried to use a bias value=1 in order to enable the NN to calculate the weight for the bias which should be the y-intercept. But it doesnt. You will see that I am very new to Java programming. Unfortunately also new to NN. In that case I guess my problem is rather in the understanding of the underlying methodology of the bias in the NN.
Remark: In the output line at the very end of the code, I would expect in case of the function y = 3*x + 5 following figures: weight[0]=3 (which is the m) , weight[1]=1 (which is the factor for y) and weight[2]=5 (this is c). The weight[2] is always wrong.
package nn2;
public class anfang_eng {
public static void main(String[] args)
{
double[][] points = new double[5][10000];
double[] weights = new double[3];
double[][] normpoints = new double[5][10000];
// create 1000 dots with desired result for training afterwards
points = createPoints();
// the before randomly created x and y values of the 1000 dots
// shall be normalized between 0 and 1
normpoints = normalize(points);
// create two random initial weights
weights = createinitialWeights();
// training function, calculation of three different weights
calculateWeights(normpoints, weights);
testnewPoints(weights);
}
// thats the function of the line, that seperates all dots in
// two groups: all the dots at the left side of the line and all the dots
// at the right side.
static double function(double x, double y)
{
double result;
result = 3*x - y + 5;
return result;
}
static double[][] createPoints()
{
// 1. step: lets create for training reasons some dots and calculate
// the result for each dot (result is either "1" or "-1").
// point[0] is x, point[1] is y, point[2] is bias and point[3] is
// result (left or right side of the function above
int x;
int y;
int quantity= 1000;
double[][] point = new double[5][quantity];
for (int i=0; i<quantity; i++)
{
x = (int) (2000 * Math.random()-1000);
y = (int) (2000 * Math.random()-1000);
point[0][i] = x;
point[1][i] = y;
// point[2] is our bias
point[2][i] = 1;
// all dots which are at the right side of the function above get
// result "1". otherwise "-1"
if ( function(x,y) > 0)
point[3][i] = 1;
else
point[3][i] =-1;
// point[3] contains the result
}
// in the variable point, there are e.g. 1000 or 5000 dots with x, y,
// bias and the result (1=left side and -1=right side)
return point;
}
// normalize x and y values between 0 and 1
static double[][] normalize(double[][]points)
{
int quantity = points[0].length;
double minpoint_x=1000;
double minpoint_y=1000;
double maxpoint_x=-1000;
double maxpoint_y=-1000;
double[][] normpoints = new double[5][quantity];
minpoint_x= points[0][0];
minpoint_y = points[1][0];
maxpoint_x = points[0][0];
maxpoint_y = points[1][0];
for (int i=0; i<quantity;i++)
{
if (points[0][i]<minpoint_x)
minpoint_x=points[0][i];
if (points[1][i]<minpoint_y)
minpoint_y=points[1][i];
if (points[0][i]>maxpoint_x)
maxpoint_x=points[0][i];
if (points[1][i]>maxpoint_y)
maxpoint_y=points[1][i];
}
for (int u=0; u<quantity; u++)
{
normpoints [0][u]= (points[0][u]-minpoint_x)/(maxpoint_x-minpoint_x);
normpoints [1][u]= (points[1][u]-minpoint_y)/(maxpoint_y-minpoint_y);
normpoints [2][u] = 1; //bias is always 1
normpoints [3][u] = points[3][u];
}
return normpoints;
}
static double[] createinitialWeights()
{
// creation of initial weights between -1 and 1
double[] weight = new double[3];
weight[0] = 2*Math.random()-1;
weight[1] = 2*Math.random()-1;
weight[2] = 2*Math.random()-1;
return weight;
}
static void calculateWeights(double[][] normpoints, double[] weight)
// new weight = weight + error * input * learning constant
// c is learning constant
{
double c = 0.01;
double error = 0;
double sumguess = 0;
double guess = 0;
int quantity = normpoints[0].length;
for (int i=0; i < quantity; i++)
{
// normpoint[0][i] stands for the factor at x, normpoint[0][i] is
// for y and normpoint[2][i] is for bias
sumguess = normpoints[0][i] * weight[0] + normpoints[1][i]*weight[1] + normpoints[2][i]*weight[2];
if (sumguess > 0)
guess = 1;
else
guess = -1;
error = normpoints[3][i]- guess;
weight[0] = weight[0] + error * normpoints[0][i] * c;
weight[1] = weight[1] + error * normpoints[1][i] * c;
weight[2] = weight[2] + error * normpoints[2][i] * c;
System.out.println("i: " + i + " ;value_normpoint[0]:" + normpoints[0][i]+ " ;value_normpoint[1]" + normpoints[1][i]+ " ;value_normpoint[2]" + normpoints[2][i] + " result:" + normpoints[3][i]);
System.out.println("weight[0]: " + Math.round(weight[0]*100)/100.0 + " ;weight[1]: " +Math.round(weight[1]*100)/100.0 + " ;weight[2]: " + Math.round(weight[2]*100)/100.0 );
System.out.println("guess: "+ guess+ " result " + normpoints[3][i] + " error: " + error);
System.out.println();
}
System.out.println("final weights: x: " + weight[0] + " y: "+ weight[1] + " bias: " +weight[2]);
System.out.println("final weights normalized on y=1: x:" + weight[0]/weight[1] + " y: "+ weight[1]/weight[1] + " bias: " +weight[2]/weight[1]);
}
// lets test if the trained weights classify the test dot on the correct side of the line y=4*x+3
// again 500 random dots with "x", "y" and "results" are created and tested if the NN calculated correct weights
static void testnewPoints(double[] weights)
{
int x;
int y;
double[][] testpoint = new double[5][10000];
double[][] normalizedtestpoint = new double[5][10000];
int quantity = 500;
double sumcheck = 0;
double sumtest = 0;
int correct = 0;
int wrong = 0;
for (int i=0; i<quantity; i++)
{
// calculation of test points with x and y between -100 and 100
x = (int) (200 * Math.random()-100);
y = (int) (200 * Math.random()-100);
testpoint[0][i] = x;
testpoint[1][i] = y;
testpoint[2][i] = 1;
// lets classify the points: at the rights side of the line the result for each point is "1", on the left side "-1"
if (function(x,y) > 0)
testpoint[3][i] = 1;
else
testpoint[3][i] = -1;
// punkt[3] is the result
}
normalizedtestpoint= normalize(testpoint);
// are the test points with our calculated weights classified on the correct side of the line?
for (int i=0; i<quantity; i++)
{
sumcheck = normalizedtestpoint[0][i] * weights[0] + normalizedtestpoint[1][i] * weights[1] + normalizedtestpoint[2][i] * weights[2];
if (sumcheck > 0)
sumtest = 1;
else
sumtest = -1;
if (sumtest == normalizedtestpoint[3][i])
correct++;
else
wrong++;
}
System.out.println("correct: "+ correct + " wrong: " + wrong);
}
}
Please let me also know if you see some major issues in my coding style, quite an beginner style I guess.
Many thanks in advance!
Lonko
Note: Updated on 06/17/2015. Of course this is possible. See the solution below.
Even if anyone copies and pastes this code, you still have a lot of cleanup to do. Also note that you will have problems inside the critical strip from Re(s) = 0 to Re(s) = 1 :). But this is a good start.
import java.util.Scanner;
public class NewTest{
public static void main(String[] args) {
RiemannZetaMain func = new RiemannZetaMain();
double s = 0;
double start, stop, totalTime;
Scanner scan = new Scanner(System.in);
System.out.print("Enter the value of s inside the Riemann Zeta Function: ");
try {
s = scan.nextDouble();
}
catch (Exception e) {
System.out.println("You must enter a positive integer greater than 1.");
}
start = System.currentTimeMillis();
if (s <= 0)
System.out.println("Value for the Zeta Function = " + riemannFuncForm(s));
else if (s == 1)
System.out.println("The zeta funxtion is undefined for Re(s) = 1.");
else if(s >= 2)
System.out.println("Value for the Zeta Function = " + getStandardSum(s));
else
System.out.println("Value for the Zeta Function = " + getNewSum(s));
stop = System.currentTimeMillis();
totalTime = (double) (stop-start) / 1000.0;
System.out.println("Total time taken is " + totalTime + " seconds.");
}
// Standard form the the Zeta function.
public static double standardZeta(double s) {
int n = 1;
double currentSum = 0;
double relativeError = 1;
double error = 0.000001;
double remainder;
while (relativeError > error) {
currentSum = Math.pow(n, -s) + currentSum;
remainder = 1 / ((s-1)* Math.pow(n, (s-1)));
relativeError = remainder / currentSum;
n++;
}
System.out.println("The number of terms summed was " + n + ".");
return currentSum;
}
public static double getStandardSum(double s){
return standardZeta(s);
}
//New Form
// zeta(s) = 2^(-1+2 s)/((-2+2^s) Gamma(1+s)) integral_0^infinity t^s sech^2(t) dt for Re(s)>-1
public static double Integrate(double start, double end) {
double currentIntegralValue = 0;
double dx = 0.0001d; // The size of delta x in the approximation
double x = start; // A = starting point of integration, B = ending point of integration.
// Ending conditions for the while loop
// Condition #1: The value of b - x(i) is less than delta(x).
// This would throw an out of bounds exception.
// Condition #2: The value of b - x(i) is greater than 0 (Since you start at A and split the integral
// up into "infinitesimally small" chunks up until you reach delta(x)*n.
while (Math.abs(end - x) >= dx && (end - x) > 0) {
currentIntegralValue += function(x) * dx; // Use the (Riemann) rectangle sums at xi to compute width * height
x += dx; // Add these sums together
}
return currentIntegralValue;
}
private static double function(double s) {
double sech = 1 / Math.cosh(s); // Hyperbolic cosecant
double squared = Math.pow(sech, 2);
return ((Math.pow(s, 0.5)) * squared);
}
public static double getNewSum(double s){
double constant = Math.pow(2, (2*s)-1) / (((Math.pow(2, s)) -2)*(gamma(1+s)));
return constant*Integrate(0, 1000);
}
// Gamma Function - Lanczos approximation
public static double gamma(double s){
double[] p = {0.99999999999980993, 676.5203681218851, -1259.1392167224028,
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7};
int g = 7;
if(s < 0.5) return Math.PI / (Math.sin(Math.PI * s)*gamma(1-s));
s -= 1;
double a = p[0];
double t = s+g+0.5;
for(int i = 1; i < p.length; i++){
a += p[i]/(s+i);
}
return Math.sqrt(2*Math.PI)*Math.pow(t, s+0.5)*Math.exp(-t)*a;
}
//Binomial Co-efficient - NOT CURRENTLY USING
/*
public static double binomial(int n, int k)
{
if (k>n-k)
k=n-k;
long b=1;
for (int i=1, m=n; i<=k; i++, m--)
b=b*m/i;
return b;
} */
// Riemann's Functional Equation
// Tried this initially and utterly failed.
public static double riemannFuncForm(double s) {
double term = Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s);
double nextTerm = Math.pow(2, (1-s))*Math.pow(Math.PI, (1-s)-1)*(Math.sin((Math.PI*(1-s))/2))*gamma(1-(1-s));
double error = Math.abs(term - nextTerm);
if(s == 1.0)
return 0;
else
return Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s)*standardZeta(1-s);
}
}
Ok well we've figured out that for this particular function, since this form of it isn't actually a infinite series, we cannot approximate using recursion. However the infinite sum of the Riemann Zeta series (1\(n^s) where n = 1 to infinity) could be solved through this method.
Additionally this method could be used to find any infinite series' sum, product, or limit.
If you execute the code your currently have, you'll get infinite recursion as 1-(1-s) = s (e.g. 1-s = t, 1-t = s so you'll just switch back and forth between two values of s infinitely).
Below I talk about the sum of series. It appears you are calculating the product of the series instead. The concepts below should work for either.
Besides this, the Riemann Zeta Function is an infinite series. This means that it only has a limit, and will never reach a true sum (in finite time) and so you cannot get an exact answer through recursion.
However, if you introduce a "threshold" factor, you can get an approximation that is as good as you like. The sum will increase/decrease as each term is added. Once the sum stabilizes, you can quit out of recursion and return your approximate sum. "Stabilized" is defined using your threshold factor. Once the sum varies by an amount less than this threshold factor (which you have defined), your sum has stabilized.
A smaller threshold leads to a better approximation, but also longer computation time.
(Note: this method only works if your series converges, if it has a chance of not converging, you might also want to build in a maxSteps variable to cease execution if the series hasn't converged to your satisfaction after maxSteps steps of recursion.)
Here's an example implementation, note that you'll have to play with threshold and maxSteps to determine appropriate values:
/* Riemann's Functional Equation
* threshold - if two terms differ by less than this absolute amount, return
* currSteps/maxSteps - if currSteps becomes maxSteps, give up on convergence and return
* currVal - the current product, used to determine threshold case (start at 1)
*/
public static double riemannFuncForm(double s, double threshold, int currSteps, int maxSteps, double currVal) {
double nextVal = currVal*(Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s)); //currVal*term
if( s == 1.0)
return 0;
else if ( s == 0.0)
return -0.5;
else if (Math.abs(currVal-nextVal) < threshold) //When a term will change the current answer by less than threshold
return nextVal; //Could also do currVal here (shouldn't matter much as they differ by < threshold)
else if (currSteps == maxSteps)//When you've taken the max allowed steps
return nextVal; //You might want to print something here so you know you didn't converge
else //Otherwise just keep recursing
return riemannFuncForm(1-s, threshold, ++currSteps, maxSteps, nextVal);
}
}
This is not possible.
The functional form of the Riemann Zeta Function is --
zeta(s) = 2^s pi^(-1+s) Gamma(1-s) sin((pi s)/2) zeta(1-s)
This is different from the standard equation in which an infinite sum is measured from 1/k^s for all k = 1 to k = infinity. It is possible to write this as something similar to --
// Standard form the the Zeta function.
public static double standardZeta(double s) {
int n = 1;
double currentSum = 0;
double relativeError = 1;
double error = 0.000001;
double remainder;
while (relativeError > error) {
currentSum = Math.pow(n, -s) + currentSum;
remainder = 1 / ((s-1)* Math.pow(n, (s-1)));
relativeError = remainder / currentSum;
n++;
}
System.out.println("The number of terms summed was " + n + ".");
return currentSum;
}
The same logic doesn't apply to the functional equation (it isn't a direct sum, it is a mathematical relationship). This would require a rather clever way of designing a program to calculate negative values of Zeta(s)!
The literal interpretation of this Java code is ---
// Riemann's Functional Equation
public static double riemannFuncForm(double s) {
double currentVal = (Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s));
if( s == 1.0)
return 0;
else if ( s == 0.0)
return -0.5;
else
System.out.println("Value of next value is " + nextVal(1-s));
return currentVal;//*nextVal(1-s);
}
public static double nextVal(double s)
{
return (Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s));
}
public static double getRiemannSum(double s) {
return riemannFuncForm(s);
}
Testing on three or four values shows that this doesn't work. If you write something similar to --
// Riemann's Functional Equation
public static double riemannFuncForm(double s) {
double currentVal = Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s); //currVal*term
if( s == 1.0)
return 0;
else if ( s == 0.0)
return -0.5;
else //Otherwise just keep recursing
return currentVal * nextVal(1-s);
}
public static double nextVal(double s)
{
return (Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s));
}
I was misinterpretation how to do this through mathematics. I will have to use a different approximation of the zeta function for values less than 2.
I think I need to use a different form of the zeta function. When I run the entire program ---
import java.util.Scanner;
public class Test4{
public static void main(String[] args) {
RiemannZetaMain func = new RiemannZetaMain();
double s = 0;
double start, stop, totalTime;
Scanner scan = new Scanner(System.in);
System.out.print("Enter the value of s inside the Riemann Zeta Function: ");
try {
s = scan.nextDouble();
}
catch (Exception e) {
System.out.println("You must enter a positive integer greater than 1.");
}
start = System.currentTimeMillis();
if(s >= 2)
System.out.println("Value for the Zeta Function = " + getStandardSum(s));
else
System.out.println("Value for the Zeta Function = " + getRiemannSum(s));
stop = System.currentTimeMillis();
totalTime = (double) (stop-start) / 1000.0;
System.out.println("Total time taken is " + totalTime + " seconds.");
}
// Standard form the the Zeta function.
public static double standardZeta(double s) {
int n = 1;
double currentSum = 0;
double relativeError = 1;
double error = 0.000001;
double remainder;
while (relativeError > error) {
currentSum = Math.pow(n, -s) + currentSum;
remainder = 1 / ((s-1)* Math.pow(n, (s-1)));
relativeError = remainder / currentSum;
n++;
}
System.out.println("The number of terms summed was " + n + ".");
return currentSum;
}
public static double getStandardSum(double s){
return standardZeta(s);
}
// Riemann's Functional Equation
public static double riemannFuncForm(double s, double threshold, double currSteps, int maxSteps) {
double term = Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s);
//double nextTerm = Math.pow(2, (1-s))*Math.pow(Math.PI, (1-s)-1)*(Math.sin((Math.PI*(1-s))/2))*gamma(1-(1-s));
//double error = Math.abs(term - nextTerm);
if(s == 1.0)
return 0;
else if (s == 0.0)
return -0.5;
else if (term < threshold) {//The recursion will stop once the term is less than the threshold
System.out.println("The number of steps is " + currSteps);
return term;
}
else if (currSteps == maxSteps) {//The recursion will stop if you meet the max steps
System.out.println("The series did not converge.");
return term;
}
else //Otherwise just keep recursing
return term*riemannFuncForm(1-s, threshold, ++currSteps, maxSteps);
}
public static double getRiemannSum(double s) {
double threshold = 0.00001;
double currSteps = 1;
int maxSteps = 1000;
return riemannFuncForm(s, threshold, currSteps, maxSteps);
}
// Gamma Function - Lanczos approximation
public static double gamma(double s){
double[] p = {0.99999999999980993, 676.5203681218851, -1259.1392167224028,
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7};
int g = 7;
if(s < 0.5) return Math.PI / (Math.sin(Math.PI * s)*gamma(1-s));
s -= 1;
double a = p[0];
double t = s+g+0.5;
for(int i = 1; i < p.length; i++){
a += p[i]/(s+i);
}
return Math.sqrt(2*Math.PI)*Math.pow(t, s+0.5)*Math.exp(-t)*a;
}
//Binomial Co-efficient
public static double binomial(int n, int k)
{
if (k>n-k)
k=n-k;
long b=1;
for (int i=1, m=n; i<=k; i++, m--)
b=b*m/i;
return b;
}
}
I notice that plugging in zeta(-1) returns -
Enter the value of s inside the Riemann Zeta Function: -1
The number of steps is 1.0
Value for the Zeta Function = -0.0506605918211689
Total time taken is 0.0 seconds.
I knew that this value was -1/12. I checked some other values with wolfram alpha and observed that --
double term = Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s);
Returns the correct value. It is just that I am multiplying this value every time by zeta(1-s). In the case of Zeta(1/2), this will always multiply the result by 0.99999999.
Enter the value of s inside the Riemann Zeta Function: 0.5
The series did not converge.
Value for the Zeta Function = 0.999999999999889
Total time taken is 0.006 seconds.
I am going to see if I can replace the part for --
else if (term < threshold) {//The recursion will stop once the term is less than the threshold
System.out.println("The number of steps is " + currSteps);
return term;
}
This difference is the error between two terms in the summation. I may not be thinking about this correctly, it is 1:16am right now. Let me see if I can think better tomorrow ....
I'm looking for some method that takes or does not take parameters for calculate confidence interval.
I don't want the apache methods,
just a simple method or som type of code that does this.
My knowledge is restricted, it basically boils down to completing an online task against an expected set of answers (https://www.hackerrank.com/challenges/stat-warmup).
However, as far as I read up, there are mistakes in the given answer, and I'd like to correct these.
My source is pretty much wikipedia https://en.wikipedia.org/wiki/Confidence_interval#Basic_Steps
/**
*
* #return int[]{lower, upper}, i.e. int array with Lower and Upper Boundary of the 95% Confidence Interval for the given numbers
*/
private static double[] calculateLowerUpperConfidenceBoundary95Percent(int[] givenNumbers) {
// calculate the mean value (= average)
double sum = 0.0;
for (int num : givenNumbers) {
sum += num;
}
double mean = sum / givenNumbers.length;
// calculate standard deviation
double squaredDifferenceSum = 0.0;
for (int num : givenNumbers) {
squaredDifferenceSum += (num - mean) * (num - mean);
}
double variance = squaredDifferenceSum / givenNumbers.length;
double standardDeviation = Math.sqrt(variance);
// value for 95% confidence interval, source: https://en.wikipedia.org/wiki/Confidence_interval#Basic_Steps
double confidenceLevel = 1.96;
double temp = confidenceLevel * standardDeviation / Math.sqrt(givenNumbers.length);
return new double[]{mean - temp, mean + temp};
}
here is you go this is the code calculate Confidence Interval
/**
*
* #author alaaabuzaghleh
*/
public class TestCI {
public static void main(String[] args) {
int maximumNumber = 100000;
int num = 0;
double[] data = new double[maximumNumber];
// first pass: read in data, compute sample mean
double dataSum = 0.0;
while (num<maximumNumber) {
data[num] = num*10;
dataSum += data[num];
num++;
}
double ave = dataSum / num;
double variance1 = 0.0;
for (int i = 0; i < num; i++) {
variance1 += (data[i] - ave) * (data[i] - ave);
}
double variance = variance1 / (num - 1);
double standardDaviation= Math.sqrt(variance);
double lower = ave - 1.96 * standardDaviation;
double higher = ave + 1.96 * standardDaviation;
// print results
System.out.println("average = " + ave);
System.out.println("sample variance = " + variance);
System.out.println("sample standard daviation = " + standardDaviation);
System.out.println("approximate confidence interval");
System.out.println("[ " + lower + ", " + higher + " ]");
}
}
This question already has answers here:
Java: Inaccuracy using double [duplicate]
(4 answers)
Closed 9 years ago.
I am getting strange output when I add doubles together. Can someone tell me why I'm getting repeating decimals when I am adding 0.1 every time?
I have worked out the formula for adding these numbers together and I have done this myself on paper up to 3.3... The sum of all numbers (decreasing by 1 tenth) from 3.3 to 1 equals 51.6
3.3
3.2
3.1 +
3.0
...
1.0
_
51.6
There is an easier way to calculate this using two formulas:
The linear formula for the increasing number: Y = 0.1X + 1
And the sum of increasing numbers formula: [X * (Y + 1)]/2 = total
first solve for Y using any number (in this case 100)
11 = 0.1(100) + 1
Then solve for the total using X and Y
[100 * (11+1)]/2 = 600
The output of the following code should be 600 I believe. There is no question that it should not have a repeating decimal. What am I doing wrong here? There must be something I missed.
public static void main(String[] args) {
int days = 100;
double inc = 0.1;
double init = 1;
double total = 0;
for (int i = 1; i <= days; i++) {
if (i == 1) {
total = total + init;
} else {
init = init + inc;
total = total + init;
}
}
System.out.println("Total: " + total);
System.out.println("Daily: " + init);
}
Double does not have infinite precision (Neither does BigDecimal, but BigDecimal has sufficient precision for this implementation).
Try this,
public static void main(String[] args) {
int days = 100;
java.math.BigDecimal init = java.math.BigDecimal.ONE;
java.math.BigDecimal total = java.math.BigDecimal.ZERO;
java.math.BigDecimal oneTenth = new java.math.BigDecimal(
"0.1");
for (int i = 1; i <= days; i++) {
if (i != 1) {
init = init.add(oneTenth);
}
total = total.add(init);
}
System.out.println("Total: " + total);
System.out.println("Daily: " + init);
}
Which outputs
Total: 595.0
Daily: 10.9
Please read the link that Don Roby posted. In essence, double precision is not a good way to represent fractions. A number like 0.1 does not have an exact representation in binary float notation - because floating point numbers are written as "something times two to the power something else". And you cannot solve that exactly for 0.1. Thus, you are really getting a slightly smaller number - actually
0.0999999999999998
And that missing amount is enough to mess up the math…
See Jon Skeet's very excellent answer on this topic: https://stackoverflow.com/a/1089026/1967396
The formula should be
0.1 * (100 * (100 + 1) / 2)
except you start at 10 * 0.1 so the formula is more complicated.
In any case double precision is not exact esp for numbers like 0.1, so you should expect to get rounding error.
You can work around this by using numbers which can be represented accurately like 1 instead of 0.1 (or rounding the result)
public static void main(String... ignored) {
int days = 100;
double inc = 1;
double init = 10;
double total = 0;
for (int i = 1; i <= days; i++) {
if (i == 1) {
total = total + init;
} else {
init = init + inc;
total = total + init;
}
}
total /= 10;
init /= 10;
System.out.println("Total: " + total);
System.out.println("Daily: " + init);
}
or round the result.
public static void main(String... ignored) {
int days = 100;
double inc = 0.1;
double init = 1;
double total = 0;
for (int i = 1; i <= days; i++) {
if (i == 1) {
total = total + init;
} else {
init = init + inc;
total = total + init;
}
}
System.out.printf("Total: %.1f%n", total);
System.out.printf("Daily: %.1f%n", init);
}
both print
Total: 595.0
Daily: 10.9