如果算的数小,可以放在int类型或者double类型的变量里.
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太大的,就要放到BigInteger类型的变量里.
先看看放到int类型的变量代码
---------------------------------------------------------------
/*
* 数量比较小的情况下可以用这个来计算
* 太大了就存放不了了
*/
public class PowerNumber {
public static void main(String[] args) {
PowerNumber power = new PowerNumber();
// 准备求第几项的值
int n = 10;
// 这里先打印第4项
System.out.println("第" + n + "项结果是:" + power.powerNumberN(n));
// 这里打印前4项的和
System.out.println("前" + n + "项的和是:" + power.powerNumberSum(n));
}
// 计算第n项的值
int powerNumberN(int n) {
// 用来存放第n个数
int sum = 1;
for (int i = 1; i = n; i++) {
sum = 2 * sum;
}
return sum;
}
// 计算前n项的和
int powerNumberSum(int n) {
// 存放前n项的和
int sumNum = 0;
int frontN = n;
for (int i = 1; i = n; i++) {
// 存放第n项的值
int sum = 1;
for (int j = 1; j = frontN; j++) {
// 第n项的值
sum = 2 * sum;
}
//求前一项
frontN--;
// 求出的值加到总和里
sumNum = sumNum + sum;
}
return sumNum;
}
}
------------------------------------------------------
前4项的结果:
第4项结果是:16
前4项的和是:30
--------------------
前20项的结果:
第20项结果是:1048576
前20项的和是:2097150
----------------------
前40项的结果......
第40项结果是:0
前40项的和是:-2
================================================
再看看放到BigInteger类型的变量里的程序
---------------------------------
import java.math.BigInteger;
public class Power {
public static void main(String[] args) {
Power power = new Power();
// 准备求第几项的值
int n = 4;
// 这里先打印第4项
System.out.println("第" + n + "项结果是:" + power.powerNum(n));
// 这里打印前4项的和
System.out.println("前" + n + "项的和是:" + power.sumPowerNum(n));
}
// 计算第n项的值
BigInteger powerNum(int n) {
BigInteger num = BigInteger.ONE;
for (int i = 1; i = n; i++) {
// 为了防止结果过大,将结果放在BigInteger中
// 每次对结果乘以2
num = num.multiply(new BigInteger(new Integer(2).toString()));
}
// 打印2^n结果
return num;
}
// 计算前n项的值
BigInteger sumPowerNum(int n) {
// 存放前n项的值
BigInteger sumNum = BigInteger.ZERO;
int frontN = n;
for (int i = 1; i = n; i++) {
// 存放第n项的值
BigInteger num = BigInteger.ONE;
for (int j = 1; j = frontN; j++) {
// 为了防止结果过大,将结果放在BigInteger中
// 每次对结果乘以2
num = num.multiply(new BigInteger(new Integer(2).toString()));
}
// 每次循环让最大值减掉1,以计算前面的值
frontN--;
// 计算出第n项的值,将其放入总和sumNum中
sumNum = sumNum.add(num);
}
return sumNum;
}
}
---------------------------------------------------
前4项的结果:
第4项结果是:16
前4项的和是:30
--------------------------------------------------
前40项的结果:
第40项结果是:1099511627776
前40项的和是:2199023255550
---------------------------------------------------------------------------------------------
前400项的结果:
第400项结果是:2582249878086908589655919172003011874329705792829223512830659356540647622016841194629645353280137831435903171972747493376
前400项的和是:5164499756173817179311838344006023748659411585658447025661318713081295244033682389259290706560275662871806343945494986750
-------------------------------------------------------------------------------------
前4000项的结果:
第4000项结果是:13182040934309431001038897942365913631840191610932727690928034502417569281128344551079752123172122033140940756480716823038446817694240581281731062452512184038544674444386888956328970642771993930036586552924249514488832183389415832375620009284922608946111038578754077913265440918583125586050431647284603636490823850007826811672468900210689104488089485347192152708820119765006125944858397761874669301278745233504796586994514054435217053803732703240283400815926169348364799472716094576894007243168662568886603065832486830606125017643356469732407252874567217733694824236675323341755681839221954693820456072020253884371226826844858636194212875139566587445390068014747975813971748114770439248826688667129237954128555841874460665729630492658600179338272579110020881228767361200603478973120168893997574353727653998969223092798255701666067972698906236921628764772837915526086464389161570534616956703744840502975279094087587298968423516531626090898389351449020056851221079048966718878943309232071978575639877208621237040940126912767610658141079378758043403611425454744180577150855204937163460902512732551260539639221457005977247266676344018155647509515396711351487546062479444592779055555421362722504575706910949376
前4000项的和是:26364081868618862002077795884731827263680383221865455381856069004835138562256689102159504246344244066281881512961433646076893635388481162563462124905024368077089348888773777912657941285543987860073173105848499028977664366778831664751240018569845217892222077157508155826530881837166251172100863294569207272981647700015653623344937800421378208976178970694384305417640239530012251889716795523749338602557490467009593173989028108870434107607465406480566801631852338696729598945432189153788014486337325137773206131664973661212250035286712939464814505749134435467389648473350646683511363678443909387640912144040507768742453653689717272388425750279133174890780136029495951627943496229540878497653377334258475908257111683748921331459260985317200358676545158220041762457534722401206957946240337787995148707455307997938446185596511403332135945397812473843257529545675831052172928778323141069233913407489681005950558188175174597936847033063252181796778702898040113702442158097933437757886618464143957151279754417242474081880253825535221316282158757516086807222850909488361154301710409874326921805025465102521079278442914011954494533352688036311295019030793422702975092124958889185558111110842725445009151413821898750
/**
* 最小二乘法计算类
*
* @author Administrator
*
*/
public class LeastSquareMethod {
private double[] x;
private double[] y;
private double[] weight;
private int m;
private double[] coefficient;
public LeastSquareMethod(double[] x, double[] y, int m) {
if (x == null || y == null || x.length 2 || x.length != y.length
|| m 2)
throw new IllegalArgumentException("无效的参数");
this.x = x;
this.y = y;
this.m = m;
weight = new double[x.length];
for (int i = 0; i x.length; i++) {
weight[i] = 1;
}
}
public LeastSquareMethod(double[] x, double[] y, double[] weight, int m) {
if (x == null || y == null || weight == null || x.length 2
|| x.length != y.length || x.length != weight.length || m 2)
throw new IllegalArgumentException("无效的参数");
this.x = x;
this.y = y;
this.m = m;
this.weight = weight;
}
public double[] getCoefficient() {
if (coefficient == null)
compute();
return coefficient;
}
public double fit(double v) {
if (coefficient == null)
compute();
if (coefficient == null)
return 0;
double sum = 0;
for (int i = 0; i coefficient.length; i++) {
sum += Math.pow(v, i) * coefficient[i];
}
return sum;
}
private void compute() {
if (x == null || y == null || x.length = 1 || x.length != y.length
|| x.length m || m 2)
return;
double[] s = new double[(m - 1) * 2 + 1];
for (int i = 0; i s.length; i++) {
for (int j = 0; j x.length; j++)
s[i] += Math.pow(x[j], i) * weight[j];
}
double[] f = new double[m];
for (int i = 0; i f.length; i++) {
for (int j = 0; j x.length; j++)
f[i] += Math.pow(x[j], i) * y[j] * weight[j];
}
double[][] a = new double[m][m];
for (int i = 0; i m; i++) {
for (int j = 0; j m; j++) {
a[i][j] = s[i + j];
}
}
coefficient = Algorithm.multiLinearEquationGroup(a, f);
}
/**
* @param args
*/
public static void main(String[] args) {
LeastSquareMethod l = new LeastSquareMethod(
new double[] { 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008 },
new double[] { 37.84, 44.55, 45.74, 63.8, 76.67, 105.59, 178.48, 355.27, 409.92 },
new double[] { 11, 12, 13, 14, 15, 16, 17, 18, 19 },
2);
double[] x = l.getCoefficient();
for (double xx : x) {
System.out.println(xx);
}
System.out.println(l.fit(2009));
}
}
用正交多项式与一般的多项式拟合结果一样，先前用正交多项式拟合的主要出发点是为了在求取多项式系数及进行假设测验时时简化计算，而这种简化在matlab条件下已经没有意义了，用polyfit可以进行任意多项式的拟合。大可不必再用正交多项式。
x2nbsp这其实就是一个多元线性回归问题;2; nbsp,1);
x2=rand(10;
gt，用regress函数就可以了： x1=rand(10.8651其中前四行的作用是产生测试数据，真正需要的只有最后一行代码，得到的a依次为a1、a2,1);
a1=1,1)])
a =
ones(10,1)*0.1;
a=regress(y,[x1 a2=2;a0=3;1;y=a1*x1+a2*x2+a0+randn(10;nbsp.1448
nbsp.1743
2
最小二乘发拟合的是直线吧，不是曲线，非线性曲线拟合你的有模型，知道曲线的方程式。或者就是差值了，三次样条或者B样条。
我是高级Java从业者，数学功底一般般，不给你翻译代码了。这类问题一般用matlab的多，用java处理计算问题的比较少。你可以参考国外大学的一些在线研究成果。如MIT的很多项目都直接放在网上。
他山之石，可以攻玉。这里给你找了一个模拟拟合的网站：