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java最小二乘法源代碼 最小二乘法例子

利用java實(shí)現(xiàn)2^1,2^2,2^3,2^4......,代碼！

如果算的數(shù)小,可以放在int類型或者double類型的變量里.

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太大的,就要放到BigInteger類型的變量里.

先看看放到int類型的變量代碼

---------------------------------------------------------------

/*

* 數(shù)量比較小的情況下可以用這個(gè)來計(jì)算

* 太大了就存放不了了

*/

public class PowerNumber {

public static void main(String[] args) {

PowerNumber power = new PowerNumber();

// 準(zhǔn)備求第幾項(xiàng)的值

int n = 10;

// 這里先打印第4項(xiàng)

System.out.println("第" + n + "項(xiàng)結(jié)果是:" + power.powerNumberN(n));

// 這里打印前4項(xiàng)的和

System.out.println("前" + n + "項(xiàng)的和是:" + power.powerNumberSum(n));

}

// 計(jì)算第n項(xiàng)的值

int powerNumberN(int n) {

// 用來存放第n個(gè)數(shù)

int sum = 1;

for (int i = 1; i = n; i++) {

sum = 2 * sum;

}

return sum;

}

// 計(jì)算前n項(xiàng)的和

int powerNumberSum(int n) {

// 存放前n項(xiàng)的和

int sumNum = 0;

int frontN = n;

for (int i = 1; i = n; i++) {

// 存放第n項(xiàng)的值

int sum = 1;

for (int j = 1; j = frontN; j++) {

// 第n項(xiàng)的值

sum = 2 * sum;

}

//求前一項(xiàng)

frontN--;

// 求出的值加到總和里

sumNum = sumNum + sum;

}

return sumNum;

}

}

------------------------------------------------------

前4項(xiàng)的結(jié)果:

第4項(xiàng)結(jié)果是:16

前4項(xiàng)的和是:30

--------------------

前20項(xiàng)的結(jié)果:

第20項(xiàng)結(jié)果是:1048576

前20項(xiàng)的和是:2097150

----------------------

前40項(xiàng)的結(jié)果......

第40項(xiàng)結(jié)果是:0

前40項(xiàng)的和是:-2

================================================

再看看放到BigInteger類型的變量里的程序

---------------------------------

import java.math.BigInteger;

public class Power {

public static void main(String[] args) {

Power power = new Power();

// 準(zhǔn)備求第幾項(xiàng)的值

int n = 4;

// 這里先打印第4項(xiàng)

System.out.println("第" + n + "項(xiàng)結(jié)果是:" + power.powerNum(n));

// 這里打印前4項(xiàng)的和

System.out.println("前" + n + "項(xiàng)的和是:" + power.sumPowerNum(n));

}

// 計(jì)算第n項(xiàng)的值

BigInteger powerNum(int n) {

BigInteger num = BigInteger.ONE;

for (int i = 1; i = n; i++) {

// 為了防止結(jié)果過大,將結(jié)果放在BigInteger中

// 每次對(duì)結(jié)果乘以2

num = num.multiply(new BigInteger(new Integer(2).toString()));

}

// 打印2^n結(jié)果

return num;

}

// 計(jì)算前n項(xiàng)的值

BigInteger sumPowerNum(int n) {

// 存放前n項(xiàng)的值

BigInteger sumNum = BigInteger.ZERO;

int frontN = n;

for (int i = 1; i = n; i++) {

// 存放第n項(xiàng)的值

BigInteger num = BigInteger.ONE;

for (int j = 1; j = frontN; j++) {

// 為了防止結(jié)果過大,將結(jié)果放在BigInteger中

// 每次對(duì)結(jié)果乘以2

num = num.multiply(new BigInteger(new Integer(2).toString()));

}

// 每次循環(huán)讓最大值減掉1,以計(jì)算前面的值

frontN--;

// 計(jì)算出第n項(xiàng)的值,將其放入總和sumNum中

sumNum = sumNum.add(num);

}

return sumNum;

}

}

---------------------------------------------------

前4項(xiàng)的結(jié)果:

第4項(xiàng)結(jié)果是:16

前4項(xiàng)的和是:30

--------------------------------------------------

前40項(xiàng)的結(jié)果:

第40項(xiàng)結(jié)果是:1099511627776

前40項(xiàng)的和是:2199023255550

---------------------------------------------------------------------------------------------

前400項(xiàng)的結(jié)果:

第400項(xiàng)結(jié)果是:2582249878086908589655919172003011874329705792829223512830659356540647622016841194629645353280137831435903171972747493376

前400項(xiàng)的和是:5164499756173817179311838344006023748659411585658447025661318713081295244033682389259290706560275662871806343945494986750

-------------------------------------------------------------------------------------

前4000項(xiàng)的結(jié)果:

第4000項(xiàng)結(jié)果是:13182040934309431001038897942365913631840191610932727690928034502417569281128344551079752123172122033140940756480716823038446817694240581281731062452512184038544674444386888956328970642771993930036586552924249514488832183389415832375620009284922608946111038578754077913265440918583125586050431647284603636490823850007826811672468900210689104488089485347192152708820119765006125944858397761874669301278745233504796586994514054435217053803732703240283400815926169348364799472716094576894007243168662568886603065832486830606125017643356469732407252874567217733694824236675323341755681839221954693820456072020253884371226826844858636194212875139566587445390068014747975813971748114770439248826688667129237954128555841874460665729630492658600179338272579110020881228767361200603478973120168893997574353727653998969223092798255701666067972698906236921628764772837915526086464389161570534616956703744840502975279094087587298968423516531626090898389351449020056851221079048966718878943309232071978575639877208621237040940126912767610658141079378758043403611425454744180577150855204937163460902512732551260539639221457005977247266676344018155647509515396711351487546062479444592779055555421362722504575706910949376

前4000項(xiàng)的和是:26364081868618862002077795884731827263680383221865455381856069004835138562256689102159504246344244066281881512961433646076893635388481162563462124905024368077089348888773777912657941285543987860073173105848499028977664366778831664751240018569845217892222077157508155826530881837166251172100863294569207272981647700015653623344937800421378208976178970694384305417640239530012251889716795523749338602557490467009593173989028108870434107607465406480566801631852338696729598945432189153788014486337325137773206131664973661212250035286712939464814505749134435467389648473350646683511363678443909387640912144040507768742453653689717272388425750279133174890780136029495951627943496229540878497653377334258475908257111683748921331459260985317200358676545158220041762457534722401206957946240337787995148707455307997938446185596511403332135945397812473843257529545675831052172928778323141069233913407489681005950558188175174597936847033063252181796778702898040113702442158097933437757886618464143957151279754417242474081880253825535221316282158757516086807222850909488361154301710409874326921805025465102521079278442914011954494533352688036311295019030793422702975092124958889185558111110842725445009151413821898750

已知一組數(shù)據(jù)，用JAVA JFRAME利用最小二乘法求出該組數(shù)據(jù)的多項(xiàng)式擬合公式

這個(gè)問題，我好像回答過！

/**

* 最小二乘法計(jì)算類

*

* @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("無效的參數(shù)");

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("無效的參數(shù)");

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));

}

}

誰能給一個(gè)java編寫的利用最小二乘法進(jìn)行曲線擬合的算法？

最小二乘發(fā)擬合的是直線吧，不是曲線，非線性曲線擬合你的有模型，知道曲線的方程式?；蛘呔褪遣钪盗耍螛訔l或者B樣條。

最小二乘法，y=a0+a1*x1+a2*x2，要求使用java編寫

因?yàn)橹蓝嘟M數(shù)據(jù)，就是按照矩陣的方式來求.建立矩陣乘法之后，通過在兩邊同時(shí)乘以原矩陣的逆矩陣，然后就能得到三元方程組的解了.如果不會(huì)的話，可以上網(wǎng)查一下java實(shí)現(xiàn)矩陣乘法以及java實(shí)現(xiàn)逆矩陣..然后你的這個(gè)功能就完成了.

java 九九乘法表

肯定的啊.第二個(gè)程序循環(huán)

for (int j=1;j==i;j++){

System.out.print(i+"*"+j+"="+(i*j)+"\t");

}

i=1時(shí),j=1,好吧,出來了1*1=1

j=2時(shí),i==j不成立了,所以j不++了.所以j永遠(yuǎn)是2了.永遠(yuǎn)不等于,所以不會(huì)打印了.

i=2,3,4,5,6,7,8,9時(shí)

j開始等于1,結(jié)果j永遠(yuǎn)不會(huì)等于i,所以j永遠(yuǎn)是1了,后面的也就不會(huì)執(zhí)行,不會(huì)打印了

誰能幫忙用java解這個(gè)最小二乘法建立的求參數(shù)的方程，已知x1,x2,y。求參數(shù)a,b1,b2三個(gè)參數(shù)的值

我是高級(jí)Java從業(yè)者，數(shù)學(xué)功底一般般，不給你翻譯代碼了。這類問題一般用matlab的多，用java處理計(jì)算問題的比較少。你可以參考國外大學(xué)的一些在線研究成果。如MIT的很多項(xiàng)目都直接放在網(wǎng)上。

他山之石，可以攻玉。這里給你找了一個(gè)模擬擬合的網(wǎng)站：


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