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Is there a notation for addition form of factorial? $$5! = 5\times4\times3\times2\times1$$ That's pretty obvious. But I'm wondering what I'd need to use to describe. $$5+4+3+2+1$$ like the factorial $5!$ way. EDIT: I know about the formula. I want to know if there's a short notation.
3. The theorem that (n k) = n! k!(n−k)! (n k) = n! k! (n − k)! already assumes 0! 0! is defined to be 1 1. Otherwise this would be restricted to 0 <k <n 0 <k <n. A reason that we do define 0! 0! to be 1 1 is so that we can cover those edge cases with the same formula, instead of having to treat them separately.
The derivative of a function of a discrete variable doesn't really make sense in the typical calculus setting. However, there is a continuous variant of the factorial function called the Gamma function, for which you can take derivatives and evaluate the derivative at integer values. In particular, since n! = Γ(n + 1), there is a nice formula ...
The Factorial of a Rational number is defined by the Gamma function. A link is in the comments. Since,
Γ(x) is related to the factorial in that it is equal to (x − 1)!. The function is defined as. Γ(z) = 1 z ∞ ∏ n = 1(1 + 1 n)z 1 + z n. Simply use this to compute factorials for any number. A handy way of calculating for real fractions with even denominators is: Γ(1 2 + n) = (2n)! 4nn!√π. Where n is an integer.
And if you take a look at this graph, you will see that. (1 / 2)! = √π 2. This extends beyond negative numbers as well. Indeed, you could even take complex numbers into the scheme: i! = lim n → ∞ n!(n + 1)i (i + 1)…(i + n) Other forms of the extended factorial (Gamma function) may be found on Wikipedia:
To explain it more precisely, n! grows very fast when compared to a power n. Because the greater number is multiplied with the product each time: (n + 1)! = 1 ⋅ 2⋯n ⋅ (n + 1). But in case of exponential function, an + 1 = a ⋅ a⋯a, the term a remains constant. Share.
Since the factorial function is defined recursively, $(n+1)!=n! \cdot (n+1)$, your question boils down to whether or not the recurrence relation has a closed form solution, which it doesn't have. You want to be able to skip around calculating $1!$ through $9!$. However $10!$ is defined by $9!$, so there isn't a way of skipping the intermediate ...
29. You can just divide the "answer" by consecutive positive integers, and when the result is 1, the last number you divided by is the number that the "answer" is factorial of. For example: 120 / 2 = 60, 60 / 3 = 20, 20 / 4 = 5, 5 / 5 = 1, so the number that 120 is the factorial of is 5. This would probably be the simplest and quickest way to ...
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