Java中的Riemann Zeta函数--函数形式的无限递归

Question

注:2015年6月17日更新。当然这是可能的。请参阅下面的解决方案。

即使有人复制和粘贴这些代码，您仍然需要进行大量的清理工作。还请注意，从Re(s) =0到Re(s) =1：，在关键条中会有问题。但这是个好的开始。

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);
}

}

Answer 1

我想我需要用另一种形式的zeta函数。当我运行整个程序时--

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;
}

}

我注意到插入zeta(-1)会返回-

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.

我知道这个值是-1/12。我用wolfram alpha检查了其他一些值，并观察到--

double term = Math.pow(2, s)*Math.pow(Math.PI, s-1)*(Math.sin((Math.PI*s)/2))*gamma(1-s);

返回正确的值。我只是每次都把这个值乘以zeta(1-s)。在Zeta(1/2)的情况下，这总是将结果乘以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.

我要看看能不能把这个角色替换成--

    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;
}

这种差异是求和中两个术语之间的误差。我可能没有正确地考虑这个问题，现在是凌晨1点16分。让我看看明天我能不能想得更好..。

Answer 2

好的，我们已经知道，对于这个特殊的函数，由于它的这种形式实际上不是一个无穷级数，所以我们不能用递归来近似。然而，Riemann级数(1\(n^s) where n = 1 to infinity)的无限和可以用这种方法求解。

此外，该方法还可用于寻找任何无穷级数的和、积或极限.。

如果您执行当前的代码，您将得到无限递归(例如，1-(1-s) = s，1-s = t，1-t = s，因此您只需在s的两个值之间无限地来回切换)。

下面我讲的是级数之和。现在看来，您正在计算该系列的产品。以下概念应适用于两者之一。

此外，Riemann函数是一个无穷级数函数。这意味着它只有一个极限，永远不会达到一个真正的和(在有限的时间)，所以你不能得到一个确切的答案通过递归。

然而，如果你引入一个“阈值”因子，你可以得到一个近似，这是你喜欢的。随着每一项的增加，这一总和将增加/减少。一旦和稳定，您可以退出递归，并返回您的近似和。“稳定”是使用您的阈值系数定义的。一旦和的变化量低于这个阈值因子(您已经定义)，您的和已经稳定。

阈值越小，近似越好，计算时间也越长。

(注意:只有当您的系列收敛时，如果它有可能不收敛，则可能需要在maxSteps变量中构建一个停止执行的方法，如果序列在递归的maxSteps步骤之后没有收敛到您的满意程度，则该方法才能工作。)

下面是一个示例实现，请注意，您必须使用threshold和maxSteps来确定适当的值：

/* 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);
    }
}

Answer 3

这是不可能的

Riemann Zeta函数的函数形式是--

zeta(s) = 2^s pi^(-1+s) Gamma(1-s) sin((pi s)/2) zeta(1-s)

这不同于标准方程，即在所有k=1到k=无穷大的情况下，从1/k^s计算出无限和。我们可以把它写成类似--

// 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;
}

同样的逻辑不适用于函数方程(它不是一个直接和，而是一个数学关系)。这需要一种相当聪明的方法来设计一个程序来计算Zeta(s)的负值！

这个Java代码的字面解释是

// 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);
}

对三到四个值的测试表明，这是行不通的。如果你写的东西和.

// 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));
}

我误解了如何通过数学来做到这一点。对于小于2的值，我必须使用zeta函数的不同近似。