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zero factorial

why value of 0! =1

Note by G J
3 years, 10 months ago

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Have you taken calculus?

There is a way to extend the factorial function to complex numbers. It is called the Gamma function. For positive real numbers, it satisfies \(\displaystyle \Gamma(x)=\int\limits_{[0,+\infty)}t^{x-1}e^{-t}\mathrm{d}t\). By "extending the factorial function," I mean that, if \(x\) is a positive integer, then it also satisfies the property \(\Gamma(x+1)=x!\).

\(\displaystyle 0!=\Gamma(1)=\int\limits_{[0,+\infty)}t^{0}e^{-t}\mathrm{d}t=\int\limits_{[0,+\infty)}e^{-t}\mathrm{d}t=1\). That's why.

Jacob Erickson - 3 years, 10 months ago

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Think of it like this:

\[n!= \frac {(n+1)!}{n+1}\]

Putting \(n=0\) :

\[0!= \frac {1!}{1} = 1\]

Snehdeep Arora - 3 years, 10 months ago

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"Factorial" is a mathematical function. If f(x) means factorial function, f(x) is defined to give the value 0 when x=0 Just like x^0=1. .

Biswaroop Roy - 3 years, 10 months ago

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This is among my favorite "explanations" for the definition of factorial. http://www.zero-factorial.com/whatis.html

Matthew Feng - 3 years, 10 months ago

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