我开发了一种实现Newton-Raphson方法的算法来查找五次函数的根。我必须反映的结果是303.6。但是，我的实施无法衡量。
参数
g = 9.81; Ds = 0.198; uj = 805.9; W = 0.0557;
公式
0.024 *（（g Ds / uj ^ 2）^（1/3））（Y ^（5/3））+ 0.2 *（Y ^（2/3）） - （（2.85 / W）^（2/3））= 0
Y的衍生物：
（0.04 * d ^（1/3）⋅g（1/3）⋅y^（2/3））/ u（2/3）+ 2/15 * y ^（1/3）
解决Y
import java.lang.*;
public class InvokeNewton {
public static void main(String argv[]) {
double del = 1e-5,
double xx = 0 ;
double dx =0,
double x= Math.PI/2;
int k = 0;
while (Math.abs(xx-x) > del && k<10 && f(x)!=0) {
dx = f(x)/d(x);
xx=x;
x =x - dx;
k++;
System.out.println("Iteration number: " + k);
System.out.println("Root obtained: " + x);
System.out.println("Estimated error: " + Math.abs(xx-x));
}
}
// Method to provide function f(x)
public static double f(double x) {
return 0.024*(Math.pow(g * Ds / Math.pow(uj, 2.0),(1.0/3.0)) * (Math.pow(Y,5.0/3.0))+ 0.2*(Math.pow(Y,2.0/3.0)) - (Math.pow((2.85/W)(2.0/3.0))));
}
// Method to provide the derivative f'(x).
public static double d(double x) {
return (0.04*Math.pow(Ds,1.0/3.0)*Math.pow(Y,2.0/3.0)) / Math.pow*uj,2.0/3.0) + 2 / 15*Math.pow(Y,1.0/3.0);
}
}
Iteration number: 1
Root obtained: 3.65373153496716
Estimated error: 2.0829352081722634
Iteration number: 2
Root obtained: 5.2246000232674215
Estimated error: 1.5708684883002615
Iteration number: 3
Root obtained: 6.618389759316356
Estimated error: 1.3937897360489346
Iteration number: 4
Root obtained: 7.906164279270034
Estimated error: 1.287774519953678
Iteration number: 5
Root obtained: 9.119558352547333
Estimated error: 1.213394073277299
Iteration number: 6
Root obtained: 10.27633029334909
Estimated error: 1.1567719408017574
Iteration number: 7
Root obtained: 11.387769167896339
Estimated error: 1.1114388745472485
Iteration number: 8
Root obtained: 12.461641418739712
Estimated error: 1.0738722508433725
Iteration number: 9
Root obtained: 13.503592201954325
Estimated error: 1.041950783214613
Iteration number: 10
Root obtained: 14.517895007865569
Estimated error: 1.0143028059112442
已经检查了方程式和导数并进行了双重检查，但是我仍然没有得到所需的输出
答案 0 :(得分：2)
Y ^（2/3）的导数是2/3 * Y ^（ - 1/3）。
一般来说，首先用X = Y ^（1/3）代替X = Y ^（1/3）会使算法更加稳定，因此你得到一个普通的多项式五次，为X求解，然后用结果设置Y = X ^ 3。
A=0.024*(g*Ds/uj^2)^(1/3)), B=0.2, C= ((2.85/W)^(2/3))
0=F(X)=A*X^5+B*X^2-C
F'(X)=4A*X^4+2B*X
对于X正数，这是一个很好的凸函数，从负值到正值。 X=0是最小值，因此初始值不是很好。如果X=(C/A)^(1/5)，则F(X)中的第一个和最后一个字段取消，第二个字段为正数，这表示起始值很高。
public class InvokeNewton {
// Define the given constants
static double g = 9.81, Ds = 0.198, uj = 805.9, W = 0.0557;
// Combine the constants into to coefficients of the polynomial
// Executed during class creation
static double A=0.024*Math.pow(g*Ds/(uj*uj), 1.0/3), B=0.2, C= Math.pow(2.85/W,2.0/3);
/* The original problem asks for the solution of
* A*y^(5/3)+B*y^(2/3)-C, introduce x=y^(1/3) <=> y=x^3
* The equation to solve now is f(x)=Ax^5+Bx^2-C=0,
* a polynomial with derivative f'(x)=5Ax^4+2Bx
*/
public static void main(String argv[]) {
// set the precision with some buffer to the full 1e-16
double del = 1e-12;
double xx = 0 ;
double dx =0;
/* -------------------
* initial point obtained by solving Ax^5-C=0
* Now f(0)<0, f(x)>0 and f is convex monotone increasing
* for x>0, so Newtons method produces a decreasing
* sequence of points quadratically converging to
* the root of the equation.
*/
double x= Math.pow(C/A,1.0/5);
int k = 0;
while (Math.abs(xx-x) > del && k<20) {
// Standard Newton method
dx = f(x)/d(x);
xx=x;
x =x - dx;
k++;
System.out.println("Iteration number: " + k);
System.out.print("Root obtained: " + x);
// the number of interest is y=x^3
System.out.println(" solution Y="+Math.pow(x,3));
System.out.println("Estimated error: " + Math.abs(xx-x));
}
}
// Method to provide function f(x)
public static double f(double x) {
return A*Math.pow(x,5)+B*Math.pow(x,2)-C;
}
// Method to provide the derivative f'(x).
public static double d(double x) {
return 5*A*Math.pow(x,4)+2*B*x;
}
}
结果
Iteration number: 1
Root obtained: 7.127382504549578 solution Y=362.06804746760736
Estimated error: 1.192264780096913
Iteration number: 2
Root obtained: 6.7530415629870015 solution Y=307.962806003808
Estimated error: 0.37434094156257647
Iteration number: 3
Root obtained: 6.722154364534209 solution Y=303.75640454448063
Estimated error: 0.030887198452792752
Iteration number: 4
Root obtained: 6.72196107618877 solution Y=303.73020272815836
Estimated error: 1.932883454385248E-4
Iteration number: 5
Root obtained: 6.721961068677178 solution Y=303.7302017099299
Estimated error: 7.511592237108289E-9
Iteration number: 6
Root obtained: 6.721961068677178 solution Y=303.7302017099299
Estimated error: 0.0
答案 1 :(得分：0)
代码已相应修改。该代码提供了注释，以简化对特定案例的修改。
public class RootFinder {
// Method to be called to calculate the root
public void InvokeNewton(){
double del = 1e-10; // Delta or precision;
double xx = 0 ; // Storage for previous root
double dx = 0; // Storage to hold a derivative of a pre-defined function;
double x= Math.sqrt(beta/alpha); // Initial guess;
while (Math.abs(xx-x) >= del && f(x)!= 0.0) { // Math.abs(xx-x) - Estimated error;
dx = f(x)/d(x); // Derivative
xx=x; // New xx value set to previous root for compersion;
x = x - dx; // Root obtained;
}
}// Method to provide function f(x)
// !!! Provide your function bellow !!!
public double f(double x) {
return (alpha * (Math.pow(x,5.0/3.0)) + 0.2*(Math.pow(x,2.0/3.0)) - beta);
}
// Method to provide the derivative f'(x).
// !!!Provide the prime derivative of your function bellow!!!
public double d(double x) {
return (0.04*Math.pow(jf.getSourceDiameter(),1.0/3.0)*Math.pow(x,2.0/3.0)) / Math.pow(jf.getJetVelocity(),2.0/3.0) + 2.0 / 15.0*Math.pow(x,1.0/3.0);
}
}