用vb求π的近似值

给定精确度ξ,用二分法求函数f(x)零点近似值的步骤如下:1确定区间[a,b],验证f(a)0,给定精确度ξ.2求区间(a,b)的中点c.3计算f(c).(1)若f(c)=0,则c就是函数的零点;(2

PrivateSubCommand1_Click()x=Text1.Texts=1a=1Dok=k+1a=a*-1*x^2/((2*k-1)*(2*k))s=s+aLoopUntilAbs(a)Tex

公式:π/4=1-1/3+1/5-1/7+...求值n=0sn=0an=1dountilABS(an)

1、循环前面加一条：pi=02、i=-1改为i=-1*i

应该是Pi/4=1-1/3+1/5-1/7+…+（-1）^(n-1)/(2*n-1)吧,PrivateSubForm_Load()DimPiAsSingle,iAsLongForm1.AutoRedr

PrivateSubForm_click()p1=1p2=1p3=p1i=1DoWhileAbs(p3)>0.0000001'p3要取绝对值p2=(-1)*p2p3=p2/(i+1)p1=p1+p3'

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>void main (void){int i,j,k=-1;float p=0.0;for (i=

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

x^n-(x-Δx)^n≈d(x^n)=nx^(n-1)dx≈nx^(n-1)Δx所以(x-Δx)^n≈x^n-nx^(n-1)Δx1000^0.1=(1024-24)^0.1≈1024^0.1-0.

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

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

PrivateSubCommand1_Click()a=1Don=n+1m=2*n-1s=s+a*1/ma=-aLoopUntilAbs(1/m)pi=4*sPrint"pi="&piEndSu

PrivateSubCommand1_Click()DimNAsDouble,PiAsDoublePi=2ForN=1To1000Pi=Pi*(2*N)^2/((2*N-1)*(2*N+1))IfN=

PrivateSubCommand1_Click()Dimn,pi,ipi=2n=Val(InputBox("请输入一个数"))Fori=1Tonpi=pi*((2*i)^2/((2*i-1)*(2*

(arctan)'=1/(1+x^2),arctan(1+0.01)=arctan(1)+1/2*0.01=pai/4+0.005-sin31=-sin(pi/6+pi/180)=-sinpi/6-p