Java-寻找有关计算函数的最小值/最大值或步长间隔中的导数的建议

标签 java max min numerical-methods function-approximation

寻找有关已变成 Java 噩梦的数学问题的建议。我扫描了网络并找不到解决方案。我看过类似的程序,不幸的是找不到帮助。

问题总结:我希望在 Java 中实现一个方法,该方法将找到 Riemann-Siegel Z(t) 函数的最小值或最大值(我已经创建了计算 Z(t) 的代码)或它的衍生物。为了显示我正在尝试做的事情,从 0 < t < 100 开始的 Z(t) 图表看起来像这样。

enter image description here

直接查看 Wolfram Alphahere 中的函数会使我遇到的“Java 噩梦”看起来过于复杂。我所描述的问题并不是特别复杂,尽管这可能是由于我在数值分析方面缺乏经验。我要做的事情的大致轮廓是

  • 在 Java 中编写一个方法来计算该函数的导数为零的所有位置(在上图中,该函数在 0 < t < 100 之间大约有 30 个值)。
  • 在方法内部,定义一个步长区间,通过下限和上限来评估函数。
  • 以下三种方法之一 - 用一种方法计算最大值/最小值,用两种方法计算最大值/最小值,或计算导数为零的值。
  • 将此添加到我现有的程序中(我已经制作了一个测试程序以使问题更容易。测试程序查看 cos(x))

  • 我扫描了互联网并找到了 this 。我发现了很多其他不同的方法,但这些方法似乎都不起作用。提供的所有解决方案似乎只计算一个步长间隔内的一个最大值/最小值/导数。为了利用新方法,程序需要找到导数为零或函数具有最大值或最小值时的所有值。例如,cos(x) 在 0 < x < 50 之间有大约 16 个零(新方法会找到所有这些值)。

    为了使这更容易,我创建了一个可以针对 cos(x) 函数进行分析的测试程序。
    import java.math.*;
    
    public class Test {
        public static void main(String[] args){
            Function cos = new Function () 
            {
            public double f(double x) {
            return Math.cos(x);
            }
        };
    
    
            //findRoots(cos, 1, 1000, 0.001); 
            findDerivative(cos, 1, 100, 0.001);
        }
    
        // 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) {
    
        for (double x = lowerBound; x <= upperBound; x += step) {
            double fxstep = f.f(x);
            double fx = fxstep;
            fxstep = f.f(x+step);
            double dy = (fxstep - fx) / step;
            if (Math.abs(dy) < 0.001) {
                System.out.println("The x value is " + x + ". The value of the "
                        + "derivative is " + dy);
            }
        }
    

    测试程序的目的是检查 public static void findDerivative 的方法是否正确。它有点工作,尽管它返回两个值来近似导数。 cos(x) 的图表如下所示。

    enter image description here

    程序输出的值是
    The x value is 3.140999999999764. The value of the derivative is -9.265358602572604E-5
    The x value is 3.141999999999764. The value of the derivative is 9.073462475805982E-4
    The x value is 6.282000000000432. The value of the derivative is 6.853070969592423E-4
    The x value is 6.283000000000432. The value of the derivative is -3.1469280259432963E-4
    The x value is 9.424000000000216. The value of the derivative is -2.7796075396935294E-4
    The x value is 9.425000000000216. The value of the derivative is 7.220391380347024E-4
    The x value is 12.564999999998475. The value of the derivative is 8.706142144987439E-4
    The x value is 12.565999999998475. The value of the derivative is -1.2938563354047972E-4
    The x value is 15.706999999996734. The value of the derivative is -4.632679163618647E-4
    The x value is 15.707999999996733. The value of the derivative is 5.36731999623008E-4
    The x value is 18.849000000000053. The value of the derivative is 5.592153640154862E-5
    The x value is 18.850000000000055. The value of the derivative is -9.440782817726756E-4
    The x value is 21.990000000003892. The value of the derivative is -6.485750521090239E-4
    The x value is 21.991000000003893. The value of the derivative is 3.514248534397524E-4
    The x value is 25.132000000007732. The value of the derivative is 2.4122869812792658E-4
    The x value is 25.133000000007733. The value of the derivative is -7.587711848833223E-4
    The x value is 28.27300000001157. The value of the derivative is -8.338821652076334E-4
    The x value is 28.274000000011572. The value of the derivative is 1.6611769582119962E-4
    The x value is 31.41500000001541. The value of the derivative is 4.2653585174967645E-4
    The x value is 31.416000000015412. The value of the derivative is -5.734640621257725E-4
    The x value is 34.55700000001016. The value of the derivative is -1.9189476674341677E-5
    The x value is 34.55800000001016. The value of the derivative is 9.808103242914257E-4
    The x value is 37.69800000000284. The value of the derivative is 6.118430110335638E-4
    The x value is 37.69900000000284. The value of the derivative is -3.881568994001938E-4
    The x value is 40.83999999999552. The value of the derivative is -2.0449666182642545E-4
    The x value is 40.84099999999552. The value of the derivative is 7.955032111928162E-4
    The x value is 43.9809999999882. The value of the derivative is 7.971501513326373E-4
    The x value is 43.9819999999882. The value of the derivative is -2.028497212425151E-4
    The x value is 47.12299999998088. The value of the derivative is -3.8980383987308187E-4
    The x value is 47.123999999980875. The value of the derivative is 6.10196070671698E-4
    The x value is 50.26399999997356. The value of the derivative is 9.824572642092022E-4
    The x value is 50.264999999973554. The value of the derivative is -1.754253620145363E-5
    The x value is 53.405999999966234. The value of the derivative is -5.75111004597062E-4
    The x value is 53.40699999996623. The value of the derivative is 4.2488890927838696E-4
    The x value is 56.54799999995891. The value of the derivative is 1.6776464961676396E-4
    The x value is 56.54899999995891. The value of the derivative is -8.322352119671805E-4
    The x value is 59.68899999995159. The value of the derivative is -7.604181495590723E-4
    The x value is 59.68999999995159. The value of the derivative is 2.39581733230132E-4
    The x value is 62.83099999994427. The value of the derivative is 3.530718295507995E-4
    The x value is 62.831999999944266. The value of the derivative is -6.469280763310437E-4
    The x value is 65.97199999995095. The value of the derivative is -9.457252543310091E-4
    The x value is 65.97299999995096. The value of the derivative is 5.4274563066059045E-5
    The x value is 69.11399999996596. The value of the derivative is 5.383789610791112E-4
    The x value is 69.11499999996596. The value of the derivative is -4.616209549057615E-4
    The x value is 72.25599999998096. The value of the derivative is -1.3103257845425986E-4
    The x value is 72.25699999998096. The value of the derivative is 8.689672701400752E-4
    

    它接近了,尽管由于 findDerivative 方法中的 Math.abs(dy) < 0.001 需要计算两次导数。解决此问题的以下方法均未成功。
  • 建议通过牛顿法计算导数。我不知道任何应用牛顿的方法,因为我不知道 Z(t) 的导数。
  • 我在网上和其他网站上找到的所有程序都直接计算[a,b]区间内的“一个”最小值或最大值。在上图中和 Z(t) 函数的图中,我正在寻找所有最小值和最大值(或者,当函数为零时)。计算 [0, 100] 间隔之间的一个最小值或最大值没有帮助,我需要一种方法来计算所有这些值。
  • 原来我低估了这样做的难度。

  • 有人有建议吗?我能用 cos(x) 测试程序做什么?如果我得到这个工作,我可以自己去找出 Z(t) 程序。我花了很多时间思考这个问题,并且失眠了。我自己想不出办法来解决这个问题。

    这是我用来计算一般值的 Z(t) 函数的方法(不必理解下面的程序来解决这些困难)。
    /**************************************************************************
    **
    **    Riemann-Siegel Formula for roots of Zeta(s) on critical line.
    **
    **************************************************************************
    **    Axion004
    **    07/31/2015
    **
    **    This program finds the roots of Zeta(s) using the well known Riemann-
    **    Siegel formula. The Riemann–Siegel theta function is approximated 
    **    using Stirling's approximation. It also uses an interpolation method to
    **    locate zeroes. The coefficients for R(t) are handled by the Taylor
    **    Series approximation originally listed by Haselgrove in 1960. It is 
    **    necessary to use these coefficients in order to increase computational 
    **    speed.
    **************************************************************************/
    
    public class SiegelMain{
        public static void main(String[] args){
            SiegelMain();
        }
    
        // Main method
        public static void SiegelMain() {
            Function RiemennSiegelZ = new Function () 
            {
            public double f(double x) {
            return RiemennZ(x, 4);
            }
        };
            System.out.println("Zeroes inside the critical line for " +
                    "Zeta(1/2 + it). The t values are referenced below.");
            System.out.println();
            // Uncomment to find non-trivial zeroes for Zeta(1/2 + it)
        findRoots(RiemennSiegelZ, 1, 40000, 0.001);
            //findMax(RiemennSiegelZ, 1, 400, 0.001);
        }
    
        /**
         * Needed as a reference for the interpolation function.
        */
        public static interface Function {
        public double f(double x);
        }
    
        /**
         * The sign of a calculated double value.
         * @param x - the double value.
         * @return the sign in -1,  1, or 0 format.
        */
        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 a specified function through interpolation.
         * @param f - the function
             * @param lowerBound - the lower bound of integration.
             * @param upperBound - the upper bound of integration.
             * @param step - the step for dx in [a:b]
         * @return the roots of the specified function.
        */
        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;
        }
        }
    
        /**
         * Calculates the local maximum from a provided lower and upper bound.
         * @param f - the function
             * @param lowerBound - the lower bound of integration.
             * @param upperBound - the upper bound of integration.
             * @param step - the step for dx in [a:b]
         * @return the local maximum for the function.
        */
         public static void findMax(Function f, double lowerBound,
                      double upperBound, double step) {
        double x = lowerBound, next_x = x + step;
        double y = f.f(x), next_y = y + step;
    
        for (x = lowerBound; x <= upperBound ; x += step) {
                if (y > (next_y)) {
            System.out.println(y);
            }
            next_x = x; next_y = y;
        }
        }
    
        /**
         * Calculates the local minimum from a provided lower and upper bound.
         * @param f - the function
             * @param lowerBound - the lower bound of integration.
             * @param upperBound - the upper bound of integration.
             * @param step - the step for dx in [a:b]
         * @return the local minimum for the function.
        */
        public static double findMin(Function f, double lowerBound, double 
                upperBound, double step) {
        double minValue = f.f(lowerBound);
    
        for (double i=lowerBound; i <= upperBound; i+=step) {
            double currEval = f.f(i);
            if (currEval < minValue) {
                minValue = currEval;
            }
        }
    
            return minValue;
        }
    
        /**
         * Riemann-Siegel theta function using the approximation by the 
             * Stirling series.
         * @param t - the value of t inside the Z(t) function.
         * @return Stirling's approximation for theta(t).
        */
        public static double theta (double t) {
            return (t/2.0 * Math.log(t/(2.0*Math.PI)) - t/2.0 - Math.PI/8.0
                    + 1.0/(48.0*Math.pow(t, 1)) + 7.0/(5760*Math.pow(t, 3)));
        }
    
        /**
         * Computes Math.Floor of the absolute value term passed in as t.
         * @param t - the value of t inside the Z(t) function.
         * @return Math.floor of the absolute value of t.
        */
        public static double fAbs(double t) {
            return Math.floor(Math.abs(t));
    
        }
    
        /**
         * Riemann-Siegel Z(t) function implemented per the Riemenn Siegel 
             * formula. See http://mathworld.wolfram.com/Riemann-SiegelFormula.html 
             * for details
         * @param t - the value of t inside the Z(t) function.
             * @param r - referenced for calculating the remainder terms by the
             * Taylor series approximations.
         * @return the approximate value of Z(t) through the Riemann-Siegel
             * formula
        */
        public static double RiemennZ(double t, int r) {
    
            double twopi = Math.PI * 2.0; 
            double val = Math.sqrt(t/twopi);
            double n = fAbs(val);
            double sum = 0.0;
    
            for (int i = 1; i <= n; i++) {
              sum += (Math.cos(theta(t) - t * Math.log(i))) / Math.sqrt(i);
            }
            sum = 2.0 * sum;
    
            double remainder;
            double frac = val - n; 
            int k = 0;
            double R = 0.0;
    
            // Necessary to individually calculate each remainder term by using
            // Taylor Series co-efficients. These coefficients are defined below.
            while (k <= r) {
                R = R + C(k, 2.0*frac-1.0) * Math.pow(t / twopi, 
                        ((double) k) * -0.5);
                k++;
            }
    
            remainder = Math.pow(-1, (int)n-1) * Math.pow(t / twopi, -0.25) * R;
            return sum + remainder;
        }
    
        /**
         * C terms for the Riemann-Siegel formula. See 
             * https://web.viu.ca/pughg/thesis.d/masters.thesis.pdf for details.
             * Calculates the Taylor Series coefficients for C0, C1, C2, C3, 
             * and C4. 
         * @param n - the number of coefficient terms to use.
             * @param z - referenced per the Taylor series calculations.
         * @return the Taylor series approximation of the remainder terms.
        */
        public static double C (int n, double z) {
            if (n==0) 
                return(.38268343236508977173 * Math.pow(z, 0.0) 
                +.43724046807752044936 * Math.pow(z, 2.0) 
                +.13237657548034352332 * Math.pow(z, 4.0) 
                -.01360502604767418865 * Math.pow(z, 6.0) 
                -.01356762197010358089 * Math.pow(z, 8.0) 
                -.00162372532314446528 * Math.pow(z,10.0) 
                +.00029705353733379691 * Math.pow(z,12.0) 
                +.00007943300879521470 * Math.pow(z,14.0) 
                +.00000046556124614505 * Math.pow(z,16.0) 
                -.00000143272516309551 * Math.pow(z,18.0) 
                -.00000010354847112313 * Math.pow(z,20.0) 
                +.00000001235792708386 * Math.pow(z,22.0) 
                +.00000000178810838580 * Math.pow(z,24.0) 
                -.00000000003391414390 * Math.pow(z,26.0) 
                -.00000000001632663390 * Math.pow(z,28.0) 
                -.00000000000037851093 * Math.pow(z,30.0) 
                +.00000000000009327423 * Math.pow(z,32.0) 
                +.00000000000000522184 * Math.pow(z,34.0) 
                -.00000000000000033507 * Math.pow(z,36.0) 
                -.00000000000000003412 * Math.pow(z,38.0)
                +.00000000000000000058 * Math.pow(z,40.0) 
                +.00000000000000000015 * Math.pow(z,42.0)); 
            else if (n==1) 
                return(-.02682510262837534703 * Math.pow(z, 1.0) 
                +.01378477342635185305 * Math.pow(z, 3.0) 
                +.03849125048223508223 * Math.pow(z, 5.0) 
                +.00987106629906207647 * Math.pow(z, 7.0) 
                -.00331075976085840433 * Math.pow(z, 9.0) 
                -.00146478085779541508 * Math.pow(z,11.0) 
                -.00001320794062487696 * Math.pow(z,13.0) 
                +.00005922748701847141 * Math.pow(z,15.0) 
                +.00000598024258537345 * Math.pow(z,17.0) 
                -.00000096413224561698 * Math.pow(z,19.0) 
                -.00000018334733722714 * Math.pow(z,21.0) 
                +.00000000446708756272 * Math.pow(z,23.0) 
                +.00000000270963508218 * Math.pow(z,25.0) 
                +.00000000007785288654 * Math.pow(z,27.0)
                -.00000000002343762601 * Math.pow(z,29.0) 
                -.00000000000158301728 * Math.pow(z,31.0) 
                +.00000000000012119942 * Math.pow(z,33.0) 
                +.00000000000001458378 * Math.pow(z,35.0) 
                -.00000000000000028786 * Math.pow(z,37.0) 
                -.00000000000000008663 * Math.pow(z,39.0) 
                -.00000000000000000084 * Math.pow(z,41.0) 
                +.00000000000000000036 * Math.pow(z,43.0) 
                +.00000000000000000001 * Math.pow(z,45.0)); 
          else if (n==2) 
                return(+.00518854283029316849 * Math.pow(z, 0.0) 
                +.00030946583880634746 * Math.pow(z, 2.0) 
                -.01133594107822937338 * Math.pow(z, 4.0) 
                +.00223304574195814477 * Math.pow(z, 6.0) 
                +.00519663740886233021 * Math.pow(z, 8.0) 
                +.00034399144076208337 * Math.pow(z,10.0) 
                -.00059106484274705828 * Math.pow(z,12.0) 
                -.00010229972547935857 * Math.pow(z,14.0) 
                +.00002088839221699276 * Math.pow(z,16.0) 
                +.00000592766549309654 * Math.pow(z,18.0) 
                -.00000016423838362436 * Math.pow(z,20.0) 
                -.00000015161199700941 * Math.pow(z,22.0) 
                -.00000000590780369821 * Math.pow(z,24.0) 
                +.00000000209115148595 * Math.pow(z,26.0) 
                +.00000000017815649583 * Math.pow(z,28.0) 
                -.00000000001616407246 * Math.pow(z,30.0) 
                -.00000000000238069625 * Math.pow(z,32.0) 
                +.00000000000005398265 * Math.pow(z,34.0) 
                +.00000000000001975014 * Math.pow(z,36.0) 
                +.00000000000000023333 * Math.pow(z,38.0) 
                -.00000000000000011188 * Math.pow(z,40.0) 
                -.00000000000000000416 * Math.pow(z,42.0) 
                +.00000000000000000044 * Math.pow(z,44.0) 
                +.00000000000000000003 * Math.pow(z,46.0)); 
          else if (n==3) 
                return(-.00133971609071945690 * Math.pow(z, 1.0) 
                +.00374421513637939370 * Math.pow(z, 3.0) 
                -.00133031789193214681 * Math.pow(z, 5.0) 
                -.00226546607654717871 * Math.pow(z, 7.0) 
                +.00095484999985067304 * Math.pow(z, 9.0) 
                +.00060100384589636039 * Math.pow(z,11.0) 
                -.00010128858286776622 * Math.pow(z,13.0) 
                -.00006865733449299826 * Math.pow(z,15.0) 
                +.00000059853667915386 * Math.pow(z,17.0) 
                +.00000333165985123995 * Math.pow(z,19.0)
                +.00000021919289102435 * Math.pow(z,21.0) 
                -.00000007890884245681 * Math.pow(z,23.0) 
                -.00000000941468508130 * Math.pow(z,25.0) 
                +.00000000095701162109 * Math.pow(z,27.0) 
                +.00000000018763137453 * Math.pow(z,29.0) 
                -.00000000000443783768 * Math.pow(z,31.0) 
                -.00000000000224267385 * Math.pow(z,33.0) 
                -.00000000000003627687 * Math.pow(z,35.0) 
                +.00000000000001763981 * Math.pow(z,37.0) 
                +.00000000000000079608 * Math.pow(z,39.0) 
                -.00000000000000009420 * Math.pow(z,41.0) 
                -.00000000000000000713 * Math.pow(z,43.0) 
                +.00000000000000000033 * Math.pow(z,45.0) 
                +.00000000000000000004 * Math.pow(z,47.0)); 
          else 
                return(+.00046483389361763382 * Math.pow(z, 0.0) 
                -.00100566073653404708 * Math.pow(z, 2.0) 
                +.00024044856573725793 * Math.pow(z, 4.0) 
                +.00102830861497023219 * Math.pow(z, 6.0) 
                -.00076578610717556442 * Math.pow(z, 8.0) 
                -.00020365286803084818 * Math.pow(z,10.0) 
                +.00023212290491068728 * Math.pow(z,12.0) 
                +.00003260214424386520 * Math.pow(z,14.0) 
                -.00002557906251794953 * Math.pow(z,16.0) 
                -.00000410746443891574 * Math.pow(z,18.0) 
                +.00000117811136403713 * Math.pow(z,20.0) 
                +.00000024456561422485 * Math.pow(z,22.0) 
                -.00000002391582476734 * Math.pow(z,24.0) 
                -.00000000750521420704 * Math.pow(z,26.0) 
                +.00000000013312279416 * Math.pow(z,28.0) 
                +.00000000013440626754 * Math.pow(z,30.0) 
                +.00000000000351377004 * Math.pow(z,32.0) 
                -.00000000000151915445 * Math.pow(z,34.0) 
                -.00000000000008915418 * Math.pow(z,36.0) 
                +.00000000000001119589 * Math.pow(z,38.0) 
                +.00000000000000105160 * Math.pow(z,40.0) 
                -.00000000000000005179 * Math.pow(z,42.0) 
                -.00000000000000000807 * Math.pow(z,44.0) 
                +.00000000000000000011 * Math.pow(z,46.0) 
                +.00000000000000000004 * Math.pow(z,48.0));
        }     
    }
    

    最佳答案

    看起来您正在尝试进行数值优化。 Apache Commons Math 库有几个实现 optimizationroot-finding .即使您最终必须编写自己的实现,在您自己实现之前,先使用库中可用的算法对解决方案进行原型(prototype)设计以找到可行的解决方案可能会有所帮助。

    关于Java-寻找有关计算函数的最小值/最大值或步长间隔中的导数的建议,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32282183/

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