where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
Sample Output
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n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333
Analysis
The range of the number is very small, only 0 to 9, store all the factorial of n, and then calculate then one by one.
Code
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#include<iostream>
usingnamespace std;
intmain(){ printf("n e\n- -----------\n"); int f[15]; f[0] = 1; double ans = 1; cout << 0 << " " << 1 << endl; for(int i = 1; i <= 9; i ++) { f[i] = f[i - 1] * i; ans += 1.0 / f[i]; if(i == 1) printf("%d %.0f\n", i, ans); elseif(i == 2) printf("%d %.1f\n", i, ans); elseprintf("%d %.9f\n", i, ans); } return0; }