Why isn't there a simple algebraic expression for the nth factorial like there is for the nth triangular number?
i.e. 1+2+3...+n = n(n+1)/2
but 1*2*3...*n = ???
n!=integral from 0 to inf x^ne^-x dx
It's too fast for polynomials or exponentials to keep up with it.
>>7657290
*simple* as in using only the basic algebraic operations, nth roots, logarithms, etc. and no transcendental numbers.
>>7657303
>integrals are not simple enough
Good luck with triple integrals kid
>>7657299
It just seems strange that such a simple recursive algorithm would be so difficult to calculate in a nonrecursive way.
I hoped it could be defined as the product derivative of another simple function, but if it increases faster than polynomials and exponentials it doesn't seem possible. Is there at least a way to define it non-recursively without transcendentals?
>>7657320
Gamma function
You can use Stirling's approximation which is a series approximation or any tool used for the gamma function, e.g. Lanczos' approximation which is useful because you can set error arbitrarily.
>>7657283
>1*2*3...*n = ???
n!
>>7657303
>logarithms, etc. and no transcendental numbers
>logarithms
>no transcendental numbers
son
>>7657346
I mean that it would be in terms of integers and the superfunctions of the successor function and their inverses.
Is there a better word for that set of functions?
Interesting question. Keep in mind that the reason why the triangle number is reducible is because 1 + n = 2 + (n - 1) .... whereas 1 * n and 2 * (n-1) have no linearly reducible relation (this is an algebraic consequence). So I believe you would need some type of algorithm or limit.
>>7657299
This isn't particularly convincing considering n! < n^n in the same way that T_n < n*n
>>7657299
>>7657361
So there would only be a reduced expression if n the values for n were members of a geometric sequence.
>>7657372
https://www.youtube.com/watch?v=hU7EHKFNMQg
>>7657283
Everyone here is full of shit. The factorial function is not analytical, therefore there is no real algebraic expression for it. Other examples of non-analytical functions are the complex conjugate function and the modulo function.
>>7657283
at least stirling's approximation... but it's an aproximation
\lim_{n \to +\infty} {n\,!\over \sqrt{2 \pi n} \; \left({n}/{\rm e}\right)^{n} } = 1
https://en.wikipedia.org/wiki/Stirling%27s_approximation
>>7657350
It's like hyperalgebraic functions on the natural numbers. That isn't standard, but there is no standard word for it and there really should be.
>>7657283
I have a more advanced question. Why cant you integrate the gamma function?
>>7657798
Here's a question for anyone who has some math background. Can every real-valued function be written as a composition of some finite collection of hyperalgebraic functions? Hyperalgebraic functions consist of addition, multiplication, exponentiation, tetration, pentation, etc. (the hyperoperation hierarchy starting with addition) and their left and right inverse functions? (Superroots/superlogarithms)
>>7657816
>addition, multiplication, exponentiation, tetration etc
>their inverse functions
they're all continous, their composition will be continous too
hence you can't do that for any real-valued function
>>7657812
Who says you can't? Isn't it clearly continuous?
>>7657853
have you never seen a plot of the gamma function
>>7657316
This.
>>7657572
>he doesnt know about the gamma function
Interesting question.
For some reason, what came to my mind was Fermat theory
http://ncatlab.org/nlab/show/Fermat+theory
and it lead me do derive a formula of the form
[math] n! = \sum_{k=0}^n \, R_{nk} \, (-k)^n [/math]
where the matrix coefficients [math]R_{nk}[/math] are related to the Pascal's triangle numbers.
0! = + 0^0
1! = - 0^1 + 1*1^1,
2! = + 0^2 - 2*1^2 + 2^2,
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,
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,
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
etc.
>>7657283
1*2*3*4*...*n = exp(ln(1*2*3*...*n)) = exp(ln(1) + ln(2) + ln(3) + ... + ln(n)) = exp(ln(1+0) + ln(1+1) + ln(1+2) + ... + ln(1+n-1))
Taylor expansion of ln(1+x) is x - 1/2x^2 + 1/3x^3 - ... 1/m x^m
the sum becomes
1 - 1/2 * 1^2 + 1/3 * 1^3 ... +
2 - 1/2 * 2^2 + 1/3 * 2^3 ... +
3 - 1/2 * 3^2 + 1/3 * 3^3 ...
...
each column is a generalized harmonic number H_n^(-x)
so it becomes
H_n^(-1) - 1/2H_n^(-2)+ 1/3H_n^(-3) - ...
Thus you can calculate the factorial with
n! = exp(H_n^(-1) - 1/2H_n^(-2)+ 1/3H_n^(-3) - ... )
in latex
[eqn] n! = exp({\sum_{i = 1}^{\infty} i^{-1}H_n^{(-i)}} )[/eqn]
>>7658609
This would be far better the other way around i.e calculating harmonic numbers from factorials.
Of course, the factorial grows faster than any polynomial and n! is already very easy to compute, easier than hyper-operations.
>>7658609
the exponential function is an infinite sum of terms involving 1/k!.
Also, there are exact versions of Sterlings formula with look like that too.
>>7658469
He obviously meant on the positive reals, This thread being factorials and all
>>7657866
You clearly don't know shit about the gamma function desu senpai
Read his post again. Then go read a calculus book.
anybody know how to solve
[math] \int_{-\infty}^\infty \dfrac {1} {(2x)^{2n} + 1} \dfrac {1} {(2(t-x))^{2n} + 1} \, {\mathrm d}x [/math]
with t real and n large but finite?
