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factorial_function [2015/11/14 03:07] 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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| 0! = + 0^0 | 0! = + 0^0 | ||
| 1! = - 0^1 + 1*1^1, | 1! = - 0^1 + 1*1^1, | ||
| - | 2! = + 0^2 - 2*1^2 + 2^2, | + | 2! = + 0^2 - 2*1^2 + 2^2, |
| - | 3! = - 0^3 + 3*1^3 - 3*2^3 + 3^3, | + | 3! = - 0^3 + 3*1^3 - 3*2^3 + 3^3, |
| - | 4! = + 0^4 - 4*1^4 + 6*2^4 - 4*3^4 + 4^4, | + | 4! = + 0^4 - 4*1^4 + 6*2^4 - 4*3^4 + 4^4, |
| - | 5! = - 0^5 + 5*1^5 - 10*2^5 + 10*3^5 - 5*4^5 + 5^5, | + | 5! = - 0^5 + 5*1^5 - 10*2^5 + 10*3^5 - 5*4^5 + 5^5, |
| - | 6! = + 0^6 - 6*1^6 + 15*2^6 - 20*3^6 + 15*4^6 - 6*5^6 + 6^6, | + | 6! = + 0^6 - 6*1^6 + 15*2^6 - 20*3^6 + 15*4^6 - 6*5^6 + 6^6, |
| - | 7! = - 0^7 + 7*1^7 - 21*2^7 + 35*3^7 - 35*4^7 + 21*5^7 - 7*6^7 + 7^7, | + | 7! = - 0^7 + 7*1^7 - 21*2^7 + 35*3^7 - 35*4^7 + 21*5^7 - 7*6^7 + 7^7, |
| 8! = + 0^8 - 8*1^8 + 28*2^8 - 56*3^8 + 70*4^8 - 56*5^8 + 28*6^8 - 8*7^8 + 8^8 | 8! = + 0^8 - 8*1^8 + 28*2^8 - 56*3^8 + 70*4^8 - 56*5^8 + 28*6^8 - 8*7^8 + 8^8 | ||
| </code> | </code> | ||
| 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 $n$ | ||
| + | |||
| + | $\sum_{k=0}^n\dfrac{(-1)^k (-k)^n}{k!\,(n - k)!}=1$ | ||
| ----- | ----- | ||
