牛顿插值java代码,牛顿插值法c语言编程

牛顿插值m文件编辑改错

给你一段代码，已验证通过！

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你可以按照其中的算法进行相应的修改。

x=[0.40,0.55,0.65,0.80,0.90,1.05];%插值节点

y=[0.41075,0.57815,0.69675,0.88811,1.02652,1.25382];%插值

n=length(x);%节点个数

z=input('请输入需要计算的函数值点z=');

newton=[x',y'];%给出差商表的前两列

for j=2:1:n

for i=n:-1:j

y(i)=(y(i)-y(i-1))/(x(i)-x(i-j+1));

end

end

u=y(n);

for i=n-1:-1:1

u=y(i)+u*(z-x(i));

end

fprintf('f(%8.5f)=%8.5f\n',z,u)

牛顿插值计算的c++代码

double newton(double *x, double *y, int n, double num, int cur, int pointNum, double answer)

{

//计算均差

for(int i = pointNum -1; icur; i--)

{

y[i] = ( y[i] - y[i-1] ) / ( x[i] - x[i-1] );

}

//已经计算完cur自加

cur++;

//temp进行临时计算

double temp = y[cur];

for(int i= 0; icur; i++)

{

temp *= ( num - x[i] );

}

//将临时计算的结果加到answer

answer += temp;

//如果得到想要的结果就返回答案 否则继续计算

if(cur==n)

return answer;

else

return newton(x,y,n,num,cur,pointNum,answer); /// 少了 return

}

100分求高手帮我写下牛顿插值和样条插值的VB代码，急用啊！！！

自编的，都弄上来了，缺样条插值。这里仅是函数，什么控件的编程你自己弄，那实在太简单了。

Dim aa As Double, bb As Double '分别接收findway有根区间两端值的变量

Dim x(1) As Double '分别接收ercigenway的根

'1.0 ercigenway 求二次方程实根 -已测试

Private Sub ercigenway(a As Single, b As Single, c As Single) 'a、b、c对应为二次方程的系数

Dim d As Double

d = b ^ 2 - 4 * a * c

If d 0 Then

MsgBox "Δ小于0，没有实根", , "消息"

x(0) = 0: x(1) = 0

ElseIf d = 0 Then

x(0) = -b / (2 * a): x(1) = x(0)

Else

x(0) = (-b - Sgn(b) * Sqr(d)) / (2 * a): x(1) = c / (a * x(0))

End If

End Sub

'2.1 findway 等步长扫描有根区间 -已测试

Private Sub findway(ByVal a As Single, ByVal b As Single, h As Double) 'a、b分别为待扫描区间端点，h为步长

Dim a1 As Double

a1 = a

Do

If f(a1) * f(a1 + h) = 0 Then

aa = a1: bb = a1 + h

Exit Sub

End If

a1 = a1 + h

Loop While a1 b

If a1 b Then

MsgBox "没有找到有根区间，请换更小的步长试一下"

Exit Sub

End If

End Sub

'2.2 erfenfun 二分法求根 -已测试

Private Function erfenfun(ByVal a As Single, ByVal b As Single, eps As Double) 'a、b为有根区间端点，eps为误差

Dim x0 As Double, x1 As Double, x2 As Double, f0 As Double, f1 As Double, f2 As Double

x1 = a: x2 = b

Do

x0 = (x1 + x2) / 2

f0 = f(x0)

If f0 = 0 Then

Exit Do

Else

f1 = f(x1): f2 = f(x2)

If f0 * f1 0 Then

x2 = x0

Else

x1 = x0

End If

End If

Loop While Abs(x1 - x2) eps

x0 = (x1 + x2) / 2

erfenfun = x0

End Function

'2.4 newtonfxfun Newton切线法 -已测试

Private Function newtonfxfun(ByVal x0 As Double, eps As Double) As Double 'x0为附近根，eps为误差

Dim x1 As Double, f0 As Double, f1 As Double

x1 = x0

Do

x0 = x1

f0 = f(x0): f1 = fd(x0) 'fd表示f的导函数

If Abs(f1) eps Then

x1 = x0: Exit Do

End If

x1 = x0 - f0 / f1

Loop Until Abs(x1 - x0) eps

newtonfxfun = x1

End Function

'2.3 stediedaifun Seffensen加速迭代法 (方程形式为x-f(x)=0) -已测试

Private Function stediedaifun(ByVal x0 As Double, eps1 As Double, eps2 As Double) As Double 'x0为解析解附近的根，eps1为输出结果误差，eps2为迭代能否继续判断标准

Dim y As Double, z As Double, x1 As Double

x1 = x0

Do

x0 = x1

y = f(x0): z = f(y)

If Abs(z - 2 * y + x0) eps2 Then

MsgBox "为满足eps2条件，不能继续迭代"

Exit Function

End If

x1 = x0 - (y - x0) ^ 2 / (z - 2 * y + x0)

Loop Until Abs(x1 - x0) eps1

stediedaifun = x1

End Function

'2.5 newtonfxnfun n次代数方程Newton切线法 -已测试

Private Function newtonfxnfun(a() As Single, eps As Double, x0 As Double) As Double 'a()分别存储按降幂排列的方程的n个系数，eps为误差，x0为附近根

Dim k As Integer, n As Integer, f0 As Double, f1 As Double, x1 As Double

n = UBound(a)

x1 = x0

Do

x0 = x1

f0 = a(0): f1 = f0

For k = 1 To n - 1

f0 = a(k) + f0 * x0

f1 = f0 + f1 * x0

Next k

f0 = a(n) + f0 * x0

x1 = x0 - f0 / f1

Loop Until Abs(x1 - x0) eps

newtonfxnfun = x1

End Function

'2.6 linecutfun 弦截法 -已测试

Private Function linecutfun(ByVal x0 As Double, ByVal x1 As Double, eps As Double, n As Long) As Double 'n为迭代次数限制，x0、x1为有根区间端点，eps为误差

Dim f0 As Double, f1 As Double, f2 As Double

Dim x2 As Double, i As Long

f0 = f(x0): f1 = f(x1)

For i = 1 To n

x2 = x1 - (x1 - x0) * f1 / (f1 - f0)

f2 = f(x2)

If Abs(f2) eps Then

Exit For

End If

x0 = x1: x1 = x2: f0 = f1: f1 = f2

Next i

If i = n + 1 Then

MsgBox "要求的计算次数太低，没有达到精度要求"

End If

linecutfun = x2

End Function

'4.1 lagrangeczfun 拉格朗日插值法 -已测试

Private Function lagrangeczfun(a() As Double, ByVal u As Double) As Double 'a(1,n)存储n+1个节点，u为插值点

Dim i As Integer, j As Integer, n As Integer

Dim l As Double, v As Double

v = 0

n = UBound(a, 2)

For j = 0 To n

l = 1#

For i = 0 To n

If i = j Then GoTo hulue

l = l * (u - a(0, i)) / (a(0, j) - a(0, i))

hulue:

Next i

v = v + l * a(1, j)

Next j

lagrangeczfun = v

End Function

'4.2 newtonczfun newton插值法 -已测试

Private Function newtonczfun(a() As Double, u As Double) As Double 'a(1,n)存储n+1个节点，u为插值点

Dim n As Integer, i As Integer, j As Integer, k As Integer

Dim z() As Double, f() As Double, v As Double

n = UBound(a, 2)

ReDim z(n), f(n)

For i = 0 To n

z(i) = a(1, i)

Next i

For i = 1 To n

k = k + 1

For j = i To n

f(j) = (z(j) - z(j - 1)) / (a(0, j) - a(0, j - k))

Next j

For j = i To n

z(j) = f(j)

Next j

Next i

f(0) = a(1, 0)

v = 0

For i = n To 0 Step -1

v = v * (u - a(0, i)) + f(i)

Next i

newtonczfun = v

End Function

'4.3 hermiteczfun Hermite插值法 -已测试

Private Function hermiteczfun(a() As Double, fd() As Double, u As Double) As Double 'a(1,n)存储n+1个节点，fd(n)存储n+1个节点处导数值，u为插值点

Dim l() As Double, ld() As Double, g() As Double, h() As Double, aim As Double

Dim n As Integer, i As Integer, j As Integer

n = UBound(a)

ReDim l(n), ld(n), g(n), h(n)

aim = 0

For i = 0 To n

l(i) = 1: ld(i) = 0

For j = 0 To n

If j = i Then GoTo hulue

l(i) = l(i) * (u - a(0, j)) / (a(0, i) - a(0, j))

ld(i) = ld(i) + 1 / (a(0, i) - a(0, j))

hulue:

Next j

g(i) = (1 + 2 * (a(0, i) - u) * ld(i)) * l(i) * l(i)

h(i) = (u - a(0, i)) * l(i) * l(i)

aim = aim + g(i) * a(1, i) + h(i) * fd(i)

Next i

hermiteczfun = aim

End Function

'5.2.1 tixingjffun 变步长梯形积分法 -已测试

Private Function tixingjffun(a As Single, b As Single, eps As Double, m As Long) As Double 'a、b分别为积分上下限，eps为误差，m为最大计算次数

Dim h As Double, t1 As Double, t2 As Double, t As Double, hh As Double

Dim n As Long: n = 1

h = b - a: t1 = h * (f(a) + f(b)) / 2

Do

t = 0

For i = 1 To n

t = t + f(a + (i - 0.5) * h)

Next i

hh = h * t

t2 = (t1 + hh) / 2

If Abs(t2 - t1) eps Then Exit Do

t1 = t2: h = h / 2: n = 2 * n

Loop Until n 2 * m

If n 2 * m Then

MsgBox "计算次数预定太小，不能达到误差要求"

End If

tixingjffun = t2

End Function

'5.2.2 simpsonjffun 变步长Simpson积分法 -已测试

Private Function simpsonjffun(a As Single, b As Single, eps As Double, m As Long) As Double 'a、b分别为积分上下限，eps为误差，m为最大计算次数

Dim n As Long, i As Long

Dim h As Double, t1 As Double, t2 As Double, hh As Double, s1 As Double, s2 As Double

n = 1: h = b - a: t1 = h * (f(a) + f(b)) / 2

hh = h * (f((a + b) / 2)): s1 = (t1 + 2 * hh) / 3

Do

n = 2 * n: h = h / 2: t2 = (t1 + hh) / 2

t = 0

For i = 1 To n

t = t + f(a + (i - 0.5) * h)

Next i

hh = t * h

s2 = (t1 + 2 * hh) / 3

If Abs(s2 - s1) eps Then Exit Do

t1 = t2: s1 = s2

Loop Until n m

If n m Then MsgBox "计算次数预定太小，不能达到误差要求"

simpsonjffun = s2

End Function

'5.3 Rombergjffun Romberg积分法

Private Function rombergjffun(a As Single, b As Single, eps As Double) As Double

Dim k As Integer, n As Integer, h As Double

k = 0: n = 1: h = b - a

End Function

'5.5.1 ds1fun 求一阶导数 -已测试

Private Function ds1fun(x0 As Single, eps As Double) As Double 'x0为求导点，eps为误差

Dim h As Double, t1 As Double, t2 As Double

h = 1: t1 = (f(x0 + h) - f(x0 - h)) / (2 * h)

h = h / 2: t2 = (f(x0 + h) - f(x0 - h)) / (2 * h)

Do While Abs(t2 - t1) eps

t1 = t2

h = h / 2

t2 = (f(x0 + h) - f(x0 - h)) / (2 * h)

Loop

ds1fun = t2

End Function

'5.5.2 ds2fun 求二阶导数 -已测试

Private Function ds2fun(x0 As Single, eps As Double) As Double 'x0为求导点，eps为误差

Dim h As Double, t1 As Double, t2 As Double

h = 1: t1 = (f(x0 + h) + f(x0 - h) - 2 * f(x0)) / (h * h)

h = h / 2: t2 = (f(x0 + h) + f(x0 - h) - 2 * f(x0)) / (h * h)

Do While Abs(t2 - t1) eps

t1 = t2

h = h / 2

t2 = (f(x0 + h) + f(x0 - h) - 2 * f(x0)) / (h * h)

Loop

ds2fun = t2

End Function

求用c语言编写牛顿插值法

牛顿插值法：

#includestdio.h

#includealloc.h

float Language(float *x,float *y,float xx,int n)

{

int i,j;

float *a,yy=0.0;

a=(float *)malloc(n*sizeof(float));

for(i=0;i=n-1;i++)

{

a[i]=y[i];

for(j=0;j=n-1;j++)

if(j!=i)a[i]*=(xx-x[j])/(x[i]-x[j]);

yy+=a[i];

}

free(a);

return yy;

}

void main()

{

float x[4]={0.56160,0.5628,0.56401,0.56521};

float y[4]={0.82741,0.82659,0.82577,0.82495};

float xx=0.5635,yy;

float Language(float *,float *,float,int);

yy=Language(x,y,xx,4);

printf("x=%f,y=%f\n",xx,yy);

getchar();

}

2.牛顿插值法#includestdio.h

#includemath.h

#define N 4

void Difference(float *x,float *y,int n)

{

float *f;

int k,i;

f=(float *)malloc(n*sizeof(float));

for(k=1;k=n;k++)

{

f[0]=y[k];

for(i=0;ik;i++)

f[i+1]=(f[i]-y[i])/(x[k]-x[i]);

y[k]=f[k];

}

return;

}

main()

{

int i;

float varx=0.895,b;

float x[N+1]={0.4,0.55,0.65,0.8,0.9};

float y[N+1]={0.41075,0.57815,0.69675,0.88811,1.02652};

Difference(x,(float *)y,N);

b=y[N];

for(i=N-1;i=0;i--)b=b*(varx-x[i])+y[i];

printf("Nn(%f)=%f",varx,b);

getchar();

}

留下个邮箱，我发给你：牛顿插值法的程序设计与应用

用Java实现牛顿插值 求大神帮下

class Lagrange

{

public static double esitimate(double x,int n,double k[][])

{

double y=0;

int i;

int j;

double w;

double w1;

for(i=0;in;i++)

{

w=1;

w1=1;

for(j=0;jn;j++)

{

if(j==i)

continue;

else

{

w=w*(x-k[j][0]);

w1=w1*(k[i][0]-k[j][0]);

}

}

//System.out.printf("y=%f,w=%f,w1=%f\n",k[i][1],w,w1);

y=y+k[i][1]*w/w1;

}

return y;

}

}

class Newton

{

double ad[][];//均差表

int n;

Newton(int n,double k[][])

{

int i,j;

this.n=n;

ad=new double[n][n];

for(i=0;in;i++)

ad[i][0]=k[i][1];

for(i=1;in;i++)

ad[i][1]=(ad[i][0]-ad[i-1][0])/(k[i][0]-k[i-1][0]);

for(i=2;in;i++)

for(j=i;jn;j++)

{

ad[j][i]=(ad[j][i-1]-ad[j-1][i-1])/(k[j][0]-k[j-i][0]);

//System.out.printf("%f %f %f %f\n", ad[j][i-1],ad[j-1][i-1],k[j][0],k[j-i][0]);

}

show();

}

void update(double x,double y,double k[][])

{

int i,j;

n++;

k[n-1][0]=x;

k[n-1][1]=y;

double temp[][]=new double[n][n];

for(i=0;in-1;i++)

for(j=0;j=i;j++)

temp[i][j]=ad[i][j];

temp[n-1][0]=y;

for(i=1;in;i++)

temp[n-1][i]=(temp[n-1][i-1]-temp[n-2][i-1])/(x-k[n-1-i][0]);

ad=temp;

show();

}

void show()

{

int i,j;

System.out.println("均差计算可列均差表如下：");

for(i=0;in;i++)

{

for(j=0;j=i;j++)

System.out.printf("%f   ", ad[i][j]);

System.out.printf("\n");

}

System.out.println("*******************************************");

}

double esitimate(double x,double k[][])

{

int i,j;

double y=ad[0][0];

double w;

for(i=1;in;i++)

{

w=ad[i][i];

for(j=0;ji;j++)

w=w*(x-k[j][0]);

y=y+w;

}

return y;  

}

}

public class interpolate 

{

public static void main(String args[])

{

int n=20;

double k[][]=new double[n][2];

int i;

n=3;

k[0][0]=0.32;

k[0][1]=0.314567;

k[1][0]=0.34;

k[1][1]=0.333487;

k[2][0]=0.36;

k[2][1]=0.352274;

System.out.println("拉格朗日插值的节点：");

for(i=0;in;i++)

System.out.printf("%f    %f\n", k[i][0],k[i][1]);

System.out.println("估算0.3367处的函数值：");

n=2;

System.out.printf("采用%d次插值得：%f\n",n-1,Lagrange.esitimate(0.3367,n,k));

n=3;

System.out.printf("采用%d次插值得：%f\n",n-1,Lagrange.esitimate(0.3367,n,k));

System.out.println("*******************************************");

k[0][0]=0.4;

k[0][1]=0.41075;

k[1][0]=0.55;

k[1][1]=0.57815;

k[2][0]=0.65;

k[2][1]=0.69675;

k[3][0]=0.80;

k[3][1]=0.88811;

k[4][0]=0.90;

k[4][1]=1.02652;

k[5][0]=1.05;

k[5][1]=1.25382;

Newton nt=new Newton(3,k);

nt.update(0.80, 0.88811, k);

nt.update(0.90, 1.02652, k);

nt.update(1.05, 1.25382, k);

nt=new Newton(5,k);

System.out.println("估算0.596处的函数值：");

System.out.printf("%f",nt.esitimate(0.596, k));

}

}