int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+
int s=1; double n=1,t=1,pi=0; while((fabs(t)) >1e-7) {pi=pi+
T=[-1 1] q=cos(pi/8*(cos(pi*t/T)).^3-3pi/8*cos(pi*t/T))+pi/4
>> t=(0:pi/100:pi)';
高中数学f(t)=1+Cos(2000*Pi*t)+Sin(4000*Pi*t)
求pi的近似值#include #include void main() { int i=1; double sum=0
#include #include { int p=1;float eps,n=1.0,t=1,pi=0; eps=1*
#include main() {int s,n; float pi=0,t; t=float(s)/n; for(s=
以下是matlab中的语句,syms t y=sin(pi*t)*exp(-s.*t) z=int(y,t,0,1) 求
证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1)
关于求π的近似值,为什么我输出的值为0?double pi=0,a,b; double i=1; int s=1;
求pi问题c语言#include#includevoid main(){ int a, b ; double pi=0;
x_t=sin(2*pi*f1*t)+sin(2*pi*f2*t)+sin(2*pi*f3*t);n_t=wgn(1,M