C语言二次插值法求f(x)=3x^3-4x+2=0

Python012

C语言二次插值法求f(x)=3x^3-4x+2=0,第1张

#include<stdio.h>

double fun(double x)

{//多项式

double a[]={2,-4,0,3}

double sum

int n=4

n--

sum=a[n]

while(--n >= 0)

sum=sum*x+a[n]

return sum

}

typedef double (*Function)(double)

double cal(Function f,double a,double b,double e)

{

double a1,a2,a3,t_a2,t

a1=aa3=ba2=(a+b)/2

while(1)

{

t=((f(a3)-f(a2))/(a3-a2)-(f(a2)-f(a1))/(a2-a1))/(a3-a1)

t_a2=((a1+a2)*t-(f(a2)-f(a1))/(a2-a1))/(2*t)

if(t_a2-a2 <e &&t_a2-a2 >-e)break

if(t_a2 >a2)a1=a2

else a3=a2

a2=t_a2

}

return f(t_a2)

}

void main()

{

printf("%lf\n",cal(fun,0,2,0.01))

}希望能帮到你 望采纳 谢谢

void find(float start,float end,float &best)

{

float m=start

float n=0

float the_best=0

while(m<end)

{

n=(float)(m+(float)(20/m))

if(n<the_best)

the_best=n

m+=0.01

}

best=the_best

}

//start end分别代表初始点左右坐标，best是最终的最优解