Java中的Riemann-Zeta函数&函数形式的无限递归

注：于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

即使有人复制并粘贴了这段代码，您仍然有很多清理工作要做。还要注意的是，从Re（s）=0到Re（s）=1:）的关键条带内会出现问题。但这是一个良好的开端

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);
}

import java.util.Scanner；
公共类新测试{
公共静态void main（字符串[]args）{
RiemannZetaMain func=新的RiemannZetaMain（）；
双s=0；
双重启动、停止、总时间；
扫描仪扫描=新扫描仪（System.in）；
System.out.print（“在Riemann Zeta函数中输入s的值：”）；
试一试{
s=scan.nextDouble（）；
} 
捕获（例外e）{
System.out.println（“必须输入大于1的正整数”）；
}
start=System.currentTimeMillis（）；
如果（s=2）
System.out.println（“Zeta函数的值=“+getStandardSum”）；
其他的
System.out.println（“Zeta函数的值=“+getNewSum”）；
停止=System.currentTimeMillis（）；
总时间=（双倍）（停止-启动）/1000.0；
System.out.println（“所用的总时间为”+totalTime+“秒”）；
}
//标准形式的Zeta函数。
公共静态双标准Zeta（双s）{
int n=1；
双电流和=0；
双相对误差=1；
双误差=0.000001；
双倍余数；
while（相对错误>错误）{
currentSum=数学功率（n，-s）+currentSum；
余数=1/（（s-1）*数学功率（n，（s-1））；
相对误差=余数/当前和；
n++；
}
System.out.println（“合计的术语数为“+n+”）；
返回电流和；
}
公共静态双getStandardSum（双s）{
返回标准泽塔；
}
//新形式
//zeta（s）=2^（-1+2 s）/（-2+2^s）伽马（1+s））积分_0^无穷大t^s秒^2（t）dt对于Re（s）>-1
公共静态双积分（双起点、双终点）{
双电流积分值=0；
double dx=0.0001d；//近似值中的delta x的大小
double x=start；//A=积分起点，B=积分终点。
//while循环的结束条件
//条件#1:b-x（i）的值小于δ（x）。
//这将抛出一个越界异常。
//条件#2:b-x（i）的值大于0（因为从A开始并拆分积分）
//直到你到达δ（x）*n。
而（Math.abs（end-x）>=dx&&（end-x）>0）{
CurrnTrimalValue+=函数（x）*dx；//使用席（黎曼）矩形求和计算宽度*高度
x+=dx；//将这些总和相加
}
返回当前值；
}
专用静态双功能（双s）{
double sech=1/Math.cosh（s）；//双曲余割
双平方=数学功率（秒，2）；
返回值（（数学功率，0.5））*平方；
}
公共静态双getNewSum（双s）{
双常数=数学功率（2，（2*s）-1/（（数学功率（2，s））-2）*（伽马（1+s））；
返回常数*积分（0，1000）；
}
//Gamma函数-Lanczos近似
公共静态双伽马（双s）{
双[]p={0.99999999980993676.520368218851，-1259.139216722428，
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012、9.984369578019516E-6、1.5056327351493116e-7}；
int g=7；
如果（s<0.5）返回Math.PI/（Math.sin（Math.PI*s）*gamma（1-s））；
s-=1；
双a=p[0]；
双t=s+g+0.5；
对于（int i=1；in-k）
k=n-k；
长b=1；
对于（int i=1，m=n；i好的，我们已经发现，对于这个特殊的函数，因为它的这种形式实际上不是一个无穷级数，我们不能用递归来近似。然而，Riemann-Zeta级数的无穷和（
1\（n^s），其中n=1到无穷大
）可以通过这个方法来求解

此外，此方法可用于查找任何无穷级数的和、积或极限。

如果您执行当前拥有的代码，您将得到无限递归，如
1-（1-s）=s
（例如
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）

下面我讲的是级数的和。看起来你是在计算级数的乘积。下面的概念应该适用于这两种情况

除此之外，Riemann Zeta函数是一个。这意味着它只有一个极限，永远不会达到真和（在有限时间内），因此你无法通过递归得到确切的答案

但是，如果引入“阈值”因子，则可以得到一个近似值。随着每个项的添加，总和将增加/减少。一旦总和稳定，您可以退出递归并返回近似总和。“稳定”使用您的阈值因子定义。一旦总和的变化量小于此阈值因子（您已定义），则总和已稳定

阈值越小，逼近效果越好，但计算时间也越长

（注意：此方法仅在序列收敛时有效，如果它有可能不收敛，您可能还希望在
maxSteps
递归步骤后，如果序列没有收敛到令您满意的程度，则构建
maxSteps
变量以停止执行。）

这里是一个示例实现，请注意，您必须使用
threshold
和
/* 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);
    }
}

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;
}

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.

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

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.

    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;
}

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

// 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;
}

// 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);
}

// 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));
}