Рекурсия на всех путях выполнения, функция вызовет переполнение стека

#include "stdafx.h"
#include <math.h>
#include <iostream>
const int n=200;
using namespace std;
double T(double V)
{double T1;
T1=T(V);
return 0;}
double R(double V)
{double R1;
R1=R(V);
return 0;}
double fV(double V)
{double m=100000;
return (1/m)*(T(V)-R(V));}
int main()
{
    double a=0,h,b=250,S[n],t[n],V[n]={0,0.694,1.389,2.083,2.778,3.472,4.267,4.861,5.556,6.25,6.944,7.639,8.333,9.028,9.722,10.417,11.111,11.806,12.5,13.194,13.611},
        T[n]={140500,139500,139000,138300,137700,137100,136200,135000,134000,132200,130000,127500,124700,120600,116700,111800,106000,100000,92500,85000,80000},
        R[n]={0,1000,3800,8000,15000,23700,33000,43500,55000,67200,80000,92200,100800,104700,105000,103800,100000,97000,92500,89000,87500};
    int i;
    setlocale(LC_ALL,"Russian");
    h=(b-a)/n;
    V[1]=0;S[1]=0;t[1]=0;V[0]=12.921;
    for(i=1;i<=n;i++)
    {t[i+1]=t[i]+h/2;
    S[i+1]=S[i]+(V[i]+(h/2)*fV(V[i]))*(h/2);
    V[i+1]=V[i]+(h/2)*(fV(V[i])+fV(V[i]+(h/2)*fV(V[i])));
    }
    for(i=1;i<=n;i++)
    {cout<<"t["<<i<<"]="<<t[i]<<" ";}
    cout<<endl;
    for(i=1;i<=n;i++)
    {cout<<"V["<<i<<"]="<<V[i]<<" ";}
    cout<<endl;
    for(i=1;i<=n;i++)
    {cout<<"S["<<i<<"]="<<S[i]<<" ";}
    cout<<endl;
    return 0;
}