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I hope these example will help you: static int Direct (int n) { if (n<=0) return 0; else return n + Direct (n-1); } static int InDirect (int n) { if (n<=0) return 0; else return n + Buddy (n-1); } int Buddy (int n) { return InDirect (n); }
// Print prime number series. class prime { void main () { int i = 1; int C = 0; int n = 1; int X = 0; while (X != 1000) { if ( n % i n ) { C = 0; n++ ; i = 0; } } i++ ; } } }
for (int i=0; i<3; ++i) printf ("1 1 0 "); printf("1\n");
Here is a good answer for recursion Fibonacci series. #include <stdio.h> #include <conio.h> long Fibonacci(long n); int main() { long r, n,i; printf("Enter the value of n: "); scanf("%ld",&n); for(i=0;i<=n;i++) { printf(" Fibonacci(%ld)= %ld\n", i,Fibonacci(i)); } getch(); return 0; } long Fibonacci(long n) { if(n==0 n==1) return n; else { return (Fibonacci(n-1)+Fibonacci(n-2)); } } for n=5; Output: Fibonacci(0)=0 Fibonacci(1)=1 Fibonacci(2)=1 Fibonacci(3)=2 Fibonacci(4)=3 Fibonacci(5)=5
for (int n = 0; n <=100; n++) // you may start n = 1, same result, 1 less iteration { if (n % 6) // or, n % 6 != 0 printf(n); }
Any number to the power 0 is 1. Because 1 = x^n/x^n = x^(n-n) = x^0
Let n be any number and n/n = 1 and n1/n1 = n1-1 which is n0 that must equal 1
The primary answer is 0/1. However, you can multiply the numerator and denominator of the primary answer by any non-zero integer to obtain an equivalent fraction. If the selected integer is n, the equivalent fraction will be (0*n)/(1/n) = 0/n. Thus any number of the form 0/n is a fraction equivalent to 0.
The simple answer is that it is defined to be 1. But there is reason behind the decision.As you know, the factorial of a number (n) is equal to:n! = n * (n-1) * (n-2) ... * 1Another way of writing this is:n! = n * (n-1)!Suppose n=1:1! = 1 * 0!or1 = 1 * 0!or1 = 0!So by defining 0! as 1, formula involving factorials will work for all integers, including 0.
x0 = x(n -n), which is equal to xn/xn by the law of powers. This obvoiusly = 1
Any number to the power '0' equals '1'. Proof ; Let a^(n) = b Then dividing a^(n) / a^(n) = b/b a^(n-n) = b/b a^(0) = 1
Geometric series may be defined in terms of the common ratio, r, and either the zeroth term, a(0), or the first term, a(1).Accordingly,a(n) = a(0) * r^n ora(n) = a(1) * r^(n-1)
This is equal to 1. On the Wikipedia page for imaginary numbers, they have a table, but here is a summary for in: n value of i^n -- ------ -4, 1 -3, i -2, -1 -1, -i 0, 1 1, i 2, -1 3, -i 4, 1 Notice there is a repeating pattern.
A recursive formula for the factorial is n! = n(n - 1)!. Rearranging gives (n - 1)! = n!/n, Substituting 'n - 1' as 0 -- i.e. n = 1 -- then 0! = 1!/1, which is 1/1 = 1.
Zero factorial is one because n! = n-1! X n. For example: 4! = (4-1) X 4. If zero factorial was zero, that would mean 1! =(1-1) X 1 = 0 X 1=0. Then if 1!=0, then even 999! would equal zero. Therefore, zero factorial equals 1.
Well, 0! (the mathematical sign for "factora") is actually 0 itself. ... So that leads us to the question "Is 1 equal to 0?" Although many people have tried to make proof that this is true, it's not. The only way that I can disprove this is that 0+0=0, and 1+1=2. If 0 really did equal 1, then it would also equal 2 (1+1). With that being said, all numbers would then equal 0. So this pattern would include numbers like pi, 5.7892, 0.75, and every other decimal. HOWEVER, this is NOT possible because, for example, 0.75 (which supposedly equals 0) can be converted to the fraction 3/4. Any fraction just simply means *numerator* divided by *denominator*. So, if 0=3&4, the fraction is really 0/0 (which means 0 divided by 0). It is absolutley IMPOSSIBLE to divide by zero. So, no. 0! does not equal 1.
Properties of Division: n/n =1, If n ≠ 0. Any number other than zero divided by itself is one.