题目
解决代码及点评
/************************************************************************/
/*
19. 在一个程序中计算出给定误差小于0.1,0.01,0.001,0.0001,0.00001 时,下式的值:
*/
/************************************************************************/
#include <stdio.h>
#include <stdlib.h>
void Go19(float num)
{
float n=1;
float J1=2*n*(2*n+2)/(2*n+1)/(2*n+1);
float J2=2*n*(2*n+2)/(2*n+1)/(2*n+1);
do
{
n=n+1;
J1=J2;
J2*=2*n*(2*n+2)/(2*n