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Can someone explain here the above question. If possible can you prove it via Matlab code? ;)

Thorsten
on 4 Oct 2016

"In mathematics, an empty product, or nullary product, is the result of multiplying no

factors. It is by convention equal to the multiplicative identity 1."

0! is such an empty product. You cannot prove it using Matlab, it's a mathematical convention.

gubertoli
on 4 Oct 2016

You might go to definition of factorial:

n! = n*(n-1)! for n>0

Calculating for n = 1, to prove 0! = 1 1! = 1*(1-1)! //by def. 1! = 1 1 = 0! 0! = 1

Considering definitions, you can implement:

function fac=fact(n)

fac = 1

if (n==1)

fac = 1

elseif n >= 1

fac = n*fact(n-1)

else

fac = 1

end

ayyappa rudrasimha yedida
on 7 Oct 2016

Edited: ayyappa rudrasimha yedida
on 7 Oct 2016

Simple program:

n = 0;

f = 1;

for i = 1:n

f = f*i;

end

f

you can get 0!=1 for n=0; Theoratical proof: he rigorous answer to this question is that the factorial operation is extended to non-integer arguments by what is called the Gamma function, defined as

Γ(x)=∫z^(x−1)*e^−z dz.limits 0 to infinity.

A bit of integration by parts can get you to the relation that for n∈N (natural numbers, where the factorial operation is naturally defined)

n!=Γ(n+1).

Therefore

0!=Γ(1)=∫e^−z dz=1.

Walter Roberson
on 8 Oct 2016

Abhishek Jain
on 18 Oct 2016

A Factorial of a number n can defined as the total number of ways of selecting n different objects. For Example, total number of ways of selecting 5 different objects is 5!=120.

Since, there is only one way of selecting none of the objects.

Hence, 0!=1.

Hope that helps.

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