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# zero factorial

why value of 0! =1

Note by G J
3 years, 8 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. · 3 years, 8 months ago

Think of it like this:

$n!= \frac {(n+1)!}{n+1}$

Putting $$n=0$$ :

$0!= \frac {1!}{1} = 1$ · 3 years, 8 months ago

"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. . · 3 years, 8 months ago