Why is the factorial of 0 equal to 1?
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Can someone explain here the above question. If possible can you prove it via Matlab code? ;)
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回答 (6 件)
Thorsten
2016 年 10 月 4 日
"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.
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gubertoli
2016 年 10 月 4 日
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
2016 年 10 月 7 日
編集済み: ayyappa rudrasimha yedida
2016 年 10 月 7 日
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.
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Walter Roberson
2016 年 10 月 8 日
Factorial is the number of distinguishable ways you can arrange a particular number of objects. The empty set (zero objects) is distinguishable from sets that are not empty, so the number of ways to arrange the empty set has to be at least 1. But there are not multiple distinguishable ways to arrange the emptiness, so the count is not more than 1. Therefore the factorial must be 1.
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Abhishek Jain
2016 年 10 月 18 日
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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