编写程序计算cosx近似值

#includemain(){doublex,y;scanf("%lf",&x);if(x

#include"stdio.h"voidmain(){inti=2,s=1;doublee=2;while(1.0/s>1e-7){s=s*i;e=e+1.0/s;i++;}printf("%lf"

^的意思就是让你一直加下去,知道有一项足够小未知另外,你的公式少了一个1,如果要算出正确的e值,应该把下面程序中result的初始值改成1objectTest{defmain(args:Array[S

x=[3,4,5,6,7,8,9,10];y=[4,5,6,7,8,9,10,11];z=21.89-9.87*x+8.09*y-7.98*x.*y-1.87*x.^2-7.98*y.^2z=1.0e

e=1+1/1!+1/2!+1/3!+.C代码：#includevoidmain(){doublee=1;doublejc=1;//求阶乘,并存入jc中inti=1;while(1/jc>=1e-6)

doublef(doublex){doubles=.0;if(x

#include"stdio.h"#include"math.h"main(){intt=-1,i=2,j;doubleh,a=1,b,x,s=1,m;printf("intputX:");scanf

#includeintmain(){intn,i,t=1;floate=1;scanf("%d",&n);for(i=1;i

是泰勒级数吧.sinx：#include#include//fabs()intmain(){doublex,s,a;//为了提高精度,我把它们定义成双精度的.inti;scanf("%lf",&x);

手打的,最好自己再上机敲下,#include<stdio.h>#include<math.h>int fun(int n){  &nbs

#includemain(){intn,i;doublet,sum;/*1*/printf("请输入n的值\n");scanf("%d",&n);sum=2;i=1;t=2;/*2*/while(i

DimsignAsInteger,aAsLong,piAsDoublesign=-1Fori=1To100000a=2*i-1sign=-1*signpi=pi+(1/a)*signNextpi=pi

#include <stdio.h>int main(){  float x, tax = 0; 

阶乘函数,不能直接修改m的值.不过此处无影响.主要是没加1.#include#include#includeintfact(intm){intt=1;inti;//if(m==0)//return1;

6题：//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){doubles,tmp;intx;for(s=0,tmp=x

intmain(){doublesum=1.;doublet=3.;while(fabs(1/t)>=10E-9){t=-1/t;sum+=t;if(t{t=1/t-2;}else{t=1/t+2;}

y=cos(2x);z=cos(x);plot(x,y,x,z,)再问：是cosx除以cos2x再答：ezplot('cos(x)/cos(2x)')ezplot(‘f(x)’,[a,b])表示在a

//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){doublex,dec,tmp;inti,k;printf("Ty

;MOVAX,AANDAX,B;AX=aANDbMOVBX,AXORBX,B;BX=aXORbADDAX,BXADDAX,BX;AX=2*(aXORb)+aANDbADDAX,A;AX=a+2*(aX

'cosx=1-x^2/2!+x^4/4!.+x^(2n)/(2n)!我写的代码：Private Sub Command1_Click()