Java-寻找关于计算函数的最小值/最大值或步骤间隔中的导数的建议
寻找关于已成为Java噩梦的数学问题的建议。我扫描了网络，找不到解决方案。我查看了类似的程序，遗憾的是无法找到帮助。
问题摘要：我希望在Java中实现一个方法，它可以找到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）或者其衍生物的价值。为了显示我想要做的事情，Z（t）的图表从0 <0。 t＆lt; 100看起来像这样。
直接审核Wolfram Alpha或here中的功能会导致＆＃34; Java噩梦＆＃34;我看起来过于复杂。我所描述的问题并不是非常复杂，尽管可能是由于我在数值分析方面缺乏经验。我希望做的总体概要是
-
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
-
在方法内部，定义一个步骤间隔，通过下限和上限来评估函数。
-
以下三种方法之一 - 计算一种方法的最大值/最小值，计算两种方法的最大值/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））
-
要求建议通过牛顿法计算导数。我不知道应用牛顿的任何方法，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只能直接计算出一个＆＃34;一个＆＃34; [a，b]区间的最小值或最大值。在上图和Z（t）函数的图表中，我正在寻找所有的最小值和最大值（或者，当函数为零时）。计算间隔[0,100]之间的一个最小值或最大值并不会有帮助，我需要一个能够计算所有这些值的方法。
-
我原本低估了这样做的难度。
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，但这些方法似乎都没有用。所提供的所有解决方案似乎仅计算步进间隔内的一个最大值/最小值/导数。为了使用新方法，程序需要找到导数为零的所有值，或者当函数具有最大值或最小值时。作为示例，cos（x）在0 <0之间具有大约16个零。 x＆lt; 50（新方法会找到所有这些值）。
为了简化这一过程，我创建了一个可以根据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。虽然它为导数的近似值返回两个值，但它有点有用。 cos（x）的图表如下所示。
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，但由于Math.abs（dy）＆lt;它需要计算导数两次。在findDerivative方法中有0.001。以下几种方法都没有成功。
有人有建议吗？我可以用cos（x）测试程序做到这一点吗？如果我有这个工作，我可以自己去找出Z（t）程序。我花了很多时间思考这个并且失去了一些睡眠。我自己还没有想过解决这个问题的方法。
以下是我用来计算一般值的Z（t）函数的原因（没有必要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));
}
}
1 个答案:
答案 0 :(得分：2)
看起来你正在尝试进行数值优化。 Apache Commons Math库有optimization和root-finding的几种实现。即使您最终必须编写自己的实现，首先使用库中可用的算法对解决方案进行原型设计，以便在您自己实现之前找到一个有效的解决方案可能会有所帮助。
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