vb求e^x的近似值

#include <stdio.h> //头文件置顶#include <math.h>double fact(int n){\

#include#include//这个函数需要返回double类型,不然结果很快就溢出了doublef(intn){\x05inti;\x05doublem=1.0;\x05for(i=1;i=e)

#include#includeintmain(void){intfun(inti);inti;floatsum=0.0,x;inttemp=1;scanf("%f",&x);sum=x;for(i=

g再问：shenmeyisi什么意思？再答：youaresobed.itisfgasdwswimmingdoingin.再问：汗英语也说得这么糊涂你还是用拼音说吧还有这程序是我自己写的但我实在查不出哪

PrivateSubForm_Load()Fori=1To50n=1Forj=2Toin=n*jNexts=s+1/nNextMsgBoxsEndSu

Private Sub Command1_Click()   '如果是题目要求用循环镶嵌个人觉得应该这样写   

#include<stdio.h>#include<math.h>int main(void){ double x, e,cosx;&n

指出几个问题吧..(1)doublefuncos(doublee,doublex);函数申明应该放在main函数外面;(2)doublefuncos(doublee,doublex)函数里面的sum变

假设级数表达式为f(i),随i值的变化而减小,则可在循环时利用级数f(i)和f(i-1)的差值来与10^-6比较,当两次计算的结果满足dpp=Abs(dc)b=b*-1n=n+1mv=rad^(2*(

DimaAsDoubleDimsAsDoubleDimeAsSingleDimtAsSinglea=1s=1e=1DoWhile1/s>=10^(-0.4)s=s*at=1/se=e+ta=a+1Lo

1#include2#include3#include4doubleabs(doublea){5returna>0?a:-a;6}7intfun(inta){8if(a==1)return1;9ret

公式应该是：e=1+1/1!+1/2!+1/3!+..+1/n!代码：#include "iostream"using namespace std;i

把A换成double或者float都行

Privatefunctione()Dime1,n,jcn=0e1=1jc=1Don=n+1jc=n*jce1=e1+1/jcLoopUntil1/jc

PrivateSubCommand1_Click()x=InputBox("请输入x：")s=1fact=1Doi=i+1fact=fact*it=x^i/facts=s+tLoopUntilt

DimaAsLong,sAsLong,DAsString,eAsLong,bAsLongD=1:a=1:s=1DoUntile>6D=((2*a)/(2*a+1))*((2*a)/(2*a-1))s=

functionjiecheng(iasinteger)asdoubleforj=1toijiecheng*=inextendfunctiondime=0dimi=0doe+=1/jiecheng(i

#include#includevoidmain(){doubleterm=1.0,e=1.0,eps=1e-05;inti=0;printf("inputeps1e-05:\n");scanf("%

e是个无理数相当于2.718…….它的x次方,x又是小数,是个递增函数.当x趋近于1时有最大值就是e,当x趋近于0时有最小值就是1

e=1+1+1/2!+1/3!+1/4!+……,取前八项,即可使误差小于10^(-4)