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factorial_function [2015/11/14 03:08] nikolaj |
factorial_function [2015/12/14 18:26] nikolaj |
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| ===== Factorial function ===== | ===== Factorial function ===== | ||
| - | ==== Set ==== | + | ==== Function ==== |
| | @#FFBB00: definiendum | @#FFBB00: $!: \mathbb N\to \mathbb N$ | | | @#FFBB00: definiendum | @#FFBB00: $!: \mathbb N\to \mathbb N$ | | ||
| | @#FFBB00: definiendum | @#FFBB00: $n\mapsto n!:=\prod_{k=1}^n\ k $ | | | @#FFBB00: definiendum | @#FFBB00: $n\mapsto n!:=\prod_{k=1}^n\ k $ | | ||
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| === Discussion === | === Discussion === | ||
| - | Thinking of $n!=\frac{{\mathrm d}^n}{{\mathrm d}x}x^n$ and Fermat theory, I though there must be an expression for $n!$ which is more algebraic and indeed I found | + | Thinking of $n!=\left.\frac{{\mathrm d}^n}{{\mathrm d}x^n}\right|_{x=0}x^n$ and Fermat theory, I though there must be an expression for $n!$ which is more algebraic and indeed I found |
| <code> | <code> | ||
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| The binomial coefficients use the factorial of course, so there's not real computational benefit. | The binomial coefficients use the factorial of course, so there's not real computational benefit. | ||
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| + | The theorem underlying here is that, for all $n$ | ||
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| + | $\sum_{k=0}^n\dfrac{(-1)^k (-k)^n}{k!\,(n - k)!}=1$ | ||
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