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Double variable with 16+ digits gives error: Integer Number too Large

The following code is designed to factorize a number typed into the variable x.
public class testMod{
public static void main(String[]args){
double x = 11868681080091051216000;
StringBuilder output = new StringBuilder("1 * ");
for(double y = 2; y <= x; y++){
while (x % y == 0) {
System.out.print("Calculating... \n");
String printNumber = y + " * ";
x = x / y;
output.append(printNumber);
System.out.print(output.substring(0, output.length() - 2) + "\n");
}
}
}
}

The problem is that the compiler treats 11868681080091051216000 as an int, regardless of the attempt to assign it to a double. As such, it's out of range.
To specify a double literal, you can simply append D to the end – but do note that you'll lose precision this way:
double x = 11868681080091051216000D;
System.out.println(x); // prints 186868108009105E22
If you need the full precision, you can use a BigInteger instead, but you'll still need to specify that number in expressions that Java can handle, such as a product of its factors.

Approximate a derivative for a continuous function throughout certain step intervals

I am looking to write a method in Java which finds a derivative for a continuous function. These are some assumptions which have been made for the method -
The function is continuous from x = 0 to x = infinity.
The derivative exists at every interval.
A step size needs to be defined as a parameter.
The method will find the max/min for the continuous function over a given interval [a:b].
As an example, the function cos(x) can be shown to have maximum or minimums at 0, pi, 2pi, 3pi, ... npi.
I am looking to write a method that will find all of these maximums or minimums provided a function, lowerBound, upperBound, and step size are given.
To simplify my test code, I wrote a program for cos(x). The function I am using is very similar to cos(x) (at least graphically). Here is some Test code that I wrote -
public class Test {
public static void main(String[] args){
Function cos = new Function ()
{
public double f(double x) {
return Math.cos(x);
}
};
findDerivative(cos, 1, 100, 0.01);
}
// Needed as a reference for the interpolation function.
public static interface Function {
public double f(double x);
}
private static int sign(double x) {
if (x < 0.0)
return -1;
else if (x > 0.0)
return 1;
else
return 0;
}
// Finds the roots of the specified function passed in with a lower bound,
// upper bound, and step size.
public static void findRoots(Function f, double lowerBound,
double upperBound, double step) {
double x = lowerBound, next_x = x;
double y = f.f(x), next_y = y;
int s = sign(y), next_s = s;
for (x = lowerBound; x <= upperBound ; x += step) {
s = sign(y = f.f(x));
if (s == 0) {
System.out.println(x);
} else if (s != next_s) {
double dx = x - next_x;
double dy = y - next_y;
double cx = x - dx * (y / dy);
System.out.println(cx);
}
next_x = x; next_y = y; next_s = s;
}
}
public static void findDerivative(Function f, double lowerBound, double
upperBound, double step) {
double x = lowerBound, next_x = x;
double dy = (f.f(x+step) - f.f(x)) / step;
for (x = lowerBound; x <= upperBound; x += step) {
double dx = x - next_x;
dy = (f.f(x+step) - f.f(x)) / step;
if (dy < 0.01 && dy > -0.01) {
System.out.println("The x value is " + x + ". The value of the "
+ "derivative is "+ dy);
}
next_x = x;
}
}
}
The method for finding roots is used for finding zeroes (this definitely works). I only included it inside my test program because I thought that I could somehow use similar logic inside the method which finds derivatives.
The method for
public static void findDerivative(Function f, double lowerBound, double
upperBound, double step) {
double x = lowerBound, next_x = x;
double dy = (f.f(x+step) - f.f(x)) / step;
for (x = lowerBound; x <= upperBound; x += step) {
double dx = x - next_x;
dy = (f.f(x+step) - f.f(x)) / step;
if (dy < 0.01 && dy > -0.01) {
System.out.println("The x value is " + x + ". The value of the "
+ "derivative is "+ dy);
}
next_x = x;
}
}
could definitely be improved. How could I write this differently? Here is sample output.
The x value is 3.129999999999977. The value of the derivative is -0.006592578364594814
The x value is 3.1399999999999766. The value of the derivative is 0.0034073256197308943
The x value is 6.26999999999991. The value of the derivative is 0.008185181673381337
The x value is 6.27999999999991. The value of the derivative is -0.0018146842631128202
The x value is 9.409999999999844. The value of the derivative is -0.009777764220086915
The x value is 9.419999999999844. The value of the derivative is 2.2203830347677922E-4
The x value is 12.559999999999777. The value of the derivative is 0.0013706082193754021
The x value is 12.569999999999776. The value of the derivative is -0.00862924258597797
The x value is 15.69999999999971. The value of the derivative is -0.002963251265619693
The x value is 15.70999999999971. The value of the derivative is 0.007036644660118885
The x value is 18.840000000000146. The value of the derivative is 0.004555886794943564
The x value is 18.850000000000147. The value of the derivative is -0.005444028885981389
The x value is 21.980000000000636. The value of the derivative is -0.006148510767989279
The x value is 21.990000000000638. The value of the derivative is 0.0038513993028788107
The x value is 25.120000000001127. The value of the derivative is 0.0077411191450771355
The x value is 25.13000000000113. The value of the derivative is -0.0022587599505241585

The main thing that I can see to improve performance in the case that f is expensive to compute, you could save the previous value of f(x) instead of computing it twice for each iteration. Also dx is never used and would always be equal to step anyway. next_x also never used. Some variable can be declare inside the loop. Moving the variable declarations inside improves readability but not performance.
public static void findDerivative(Function f, double lowerBound, double upperBound, double step) {
double fxstep = f.f(x);
for (double x = lowerBound; x <= upperBound; x += step) {
double fx = fxstep;
fxstep = f.f(x+step);
double dy = (fxstep - fx) / step;
if (dy < 0.01 && dy > -0.01) {
System.out.println("The x value is " + x + ". The value of the "
+ "derivative is " + dy);
}
}
}

The java code you based on (from rosettacode) is not OK, do not depend on it.
it's expecting y (a double value) will become exactly zero.
You need a tolerance value for such kind of tests.
it's calculating derivative, and using Newton's Method to calculate next x value,
but not using it to update x, there is not any optimization there.
Here there is an example of Newton's Method in Java
Yes you can optimize your code using Newton's method,
Since it can solve f(x) = 0 when f'(x) given,
also can solve f'(x) = 0 when f''(x) given, same thing.

To clarify my comment, I modified the code in the link.
I used step = 2, and got correct results.
Check how fast it's, compared to other.
That's why optimization is used,
otherwise reducing the step size and using brute force would do the job.
class Test {
static double f(double x) {
return Math.sin(x);
}
static double fprime(double x) {
return Math.cos(x);
}
public static void main(String argv[]) {
double tolerance = .000000001; // Our approximation of zero
int max_count = 200; // Maximum number of Newton's method iterations
/*
* x is our current guess. If no command line guess is given, we take 0
* as our starting point.
*/
double x = 0.6;
double low = -4;
double high = 4;
double step = 2;
int inner_count = 0;
for (double initial = low; initial <= high; initial += step) {
x = initial;
for (int count = 1; (Math.abs(f(x)) > tolerance)
&& (count < max_count); count++) {
inner_count++;
x = x - f(x) / fprime(x);
}
if (Math.abs(f(x)) <= tolerance) {
System.out.println("Step: " + inner_count + ", x = " + x);
} else {
System.out.println("Failed to find a zero");
}
}
}
}

Java Float to "remove" comma

I need to convert a float to an int, as if the comma was removed.
Example:
23.2343f -> 232343
private static int removeComma(float value)
{
for (int i = 0; ; i++) {
if((value * (float)Math.pow(10, i)) % 1.0f == 0.0f)
return (int)(value * Math.pow(10, i));
}
}
The problem is with rounding up of the number. For example if I pass 23000.2359f it becomes 23000236, because it rounded up the input to 23000.236.

Java float doesn't have that much precision, which you can see with
float f = 23000.2359f;
System.out.println(f);
which outputs
23000.236
To get the output you want, you could use a double like
double d = 23000.2359;
String v = String.valueOf(d).replace(".", "");
int val = Integer.parseInt(v);
System.out.println(val);
Output is (the requested)
230002359

you must find a way to get the number of digit after decimal place 1st. Suppose it is n. then multiply the number with 10 times n
double d= 234.12413;
String text = Double.toString(Math.abs(d));
int integerPlaces = text.indexOf('.');
int decimalPlaces = text.length() - integerPlaces - 1;

Converting from JavaScript to Java but not giving same output

I have some lines of code that produces a number after calculation which is currently in JavaScript and here is the code:
if ( y2 != y1 )
{
// calculate rate
var f;
var y;
if ( y2 > y1 )
{
f = cpi[y2] / cpi[y1];
y = y2 - y1;
}
else
{
f = cpi[y1] / cpi[y2];
y = y1 - y2;
}
var r = Math.pow(f, 1/y);
r = (r-1)*100;
r = Math.round(r*100) / 100;
System.out.println( "number: " + r.toFixed(2) + "%." );
}
I converted the above JS code to Java and here is the code:
DecimalFormat decimalFormat = new DecimalFormat("0.##");
if (cpi[to] != cpi[from]) {
double f, y;
if (cpi[to] > cpi[from]) {
f = cpi[to] / cpi [from];
y = to - from;
}
else {
f = cpi[from] / cpi[to];
y = from - to;
}
q = Math.pow(f, 1/y);
q = (q-1)*100;
q = Math.round(q*100)/100;
Toast.makeText(getApplicationContext(), "number: " + String.valueOf(decimalFormat.format(q)), 2000).show();
}
The JavaScript code produces: number: 2.39
While the Java code produces number: 2
Why am I getting two different value? I will post what cpi[to], cpi[from], to and from values are if needed.

In this line
q = Math.round(q*100)/100;
both operands of the division operation are integral, therefore the result is also an integral type. Use 100.0 as the divisor to coerce the result to a double.

what is q type?
try to put 100.0 also

If you devide two integers in java the result is integer. Every operation with double and integer or with two doubles creates double.
1)In this line
1)q = Math.round(q*100)/100;
2)you are deviding two integers, so it has same output as:
2)q = (int) (Math.round(q*100)/100);
3)you can use casting to double for example:
3)q = Math.round(q*100)/(double)100;
4)or using the 100.0 which makes this number double:
4)q = Math.round(q*100)/100.0;
5)this should work too, because first the result of Math.round is converted to double and then devided by 100:
5)q = (double)Math.round(q*100)/100;
6)However this WILL NOT work, because first Math.round is devided by 100 and it creates integer, so the result of this operation is rounded down and AFTER then casted to double. So it will be double but still rounded down, because it was rounded before it becomes double.
6)q = (double)(Math.round(q*100)/100);

Most likely your issue is Math.round(double) returns a long value. So d would contain a long instead of a double at that point.

Getting accurate roots of quadratic equation

This is the code that I have thus far. The goal of the project is to have the user enter any integers for a, b, c for the ax^2+bx+c equation. For some reason I am not getting the correct roots for any numbers that are input into the program. Can anyone point out my wrong doings?
import java.util.*;
public class Quad_Form {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
double a = 0;
double b = 0;
double c = 0;
double discrim = 0;
double d = 0;
System.out.println("Enter value for 'a'");
String str_a = sc.nextLine();
a = Integer.parseInt(str_a);
System.out.println("Enter value for 'b'");
String str_b = sc.nextLine();
b = Integer.parseInt(str_b);
System.out.println("Enter value for 'c'");
String str_c = sc.nextLine();
c = Integer.parseInt(str_c);
double x1 = 0, x2 = 0;
discrim = (Math.pow(b, 2.0)) - (4 * a * c);
d = Math.sqrt(discrim);
if(discrim == 0){
x1 = (-b + d) / (2* a);
String root_1 = Double.toString(x1);
System.out.println("There is one root at: " + root_1);
}
else {
if (discrim > 0)
x1 = (-b + d) / (2 * a);
x2 = (-b - d) / (2 * a);
String root_1 = Double.toString(x1);
String root_2 = Double.toString(x2);
System.out.println("There are two real roots at:" + root_1 + "and" + root_2);
}
if (discrim < 0){
x1 = (-b + d) / (2 * a);
x2 = (-b - d) / (2 * a);
String root_1 = Double.toString(x1);
String root_2 = Double.toString(x2);
System.out.println("There are two imaginary roots at:" + root_1 + "and" + root_2);
}
}
}

#Smit is right about one of the issues, but there's a second one as well.
Math.sqrt(discrim) won't work when discrim is negative. You should be taking Math.sqrt(Math.abs(discrim)) instead.

a, b, c, d are double and you are parsing them as Integer. So this could be one of problem.
Use
Double.parseDouble();
Another problem is you can not make square root of negative numbers. This will result in NaN. For that use following, but you should handle that properly to get exact result.
Math.sqrt(Math.abs());
Moreover you should use following formula for getting roots
Taken from Wikipedia Quadratic equation
Class Double
Class Math