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Following is a question spun off from a comment I received:

is a factorial an elementary function and an algebraic function?

From elementary functions by Wikipedia

By starting with the field of rational functions, two special types of transcendental extensions (the logarithm and the exponential) can be added to the field building a tower containing elementary functions.

So isn't a factorial a multiplication of finite polynomials, and therefore a polynomial, a rational function, an algebraic function, and an elementary function?

Added: Now I realized a factorial cannot be a polynomial, for that it doesn't make sense to talk about its degree while it does for a polynomial


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From my perspective your question may be good, but its not phrased properly, I can't understand your main doubt, may be the question can be reformulated more neatly , but is this problem only for me ?, if so sorry !! @Tim –  Iyengar Jan 12 '12 at 15:45
How a factorial is a multiplication of finite polynomials ? –  Sasha Jan 12 '12 at 15:48
@Sasha: $n!=n \times (n-1) \times \cdots \times 1$. I am not claiming it is true. I am not sure, but it seems like a multiplication of finite polynomials. –  Tim Jan 12 '12 at 15:50
The gamma function (factorial), of course. I quoted and linked to Dave's answer; you might wish to peruse the references he gave. –  J. M. Jan 12 '12 at 16:09
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1 Answer

up vote 6 down vote accepted

To start basically factorial is really a function, and generally the Gamma function extends the factorial to the non-integer values.

To your best reference I have one beautiful article with me, let me suggest it , its here. The article is by Manjul Bhargava, it has the precise information what you are looking for.

Please read it and give your feed-back.

Edit : After thinking much on Mr.Srivatsan's comment , I came to a conclusion that $n!$ is not a polynomial, in fact $n!$ grows faster than $a^n$ for any $a$. Once you go out $a$ steps, adding $1$ to $n$ multiplies $a^n$ by $a$, while it multiplies $n!$ by $a$ by (at least) $a+1$.

And to add some interesting points,

  1. The falling factorial $(x)_n$ is a binomial polynomial which is defined as $$(x)_n=x(x-1)....(x-(n-1))$$ for $n\ge 0$, and it can be related to the raising factorial $(x)^n$ ( defined as $(x)^n=x(x+1)...(x+n-1)$ ) as $$(x)_n=(-1)^n(-x)^{(n)}$$
  2. The usual factorial ( that OP is talking about ) can be written as $$n!=(n)_n$$ which is not a polynomial anymore.

( Credits : Thanks Mr.Srivatsan for letting me know the difference ) .

Thank you,

Yours truly,


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Your first sentence is incorrect: the factorial function is not a polynomial. [In fact, it grows faster than any polynomial.] I think you just assumed what the OP claimed to be true. Added note: I did not downvote. –  Srivatsan Jan 12 '12 at 16:37
@Srivatsan : I am not an expert to argue further, but a polynomial in sense I mean a function of a single variable and its a function by definition as its a product that gives an $n$-th degree polynomial of single variable . –  Iyengar Jan 12 '12 at 16:44
Yes, it is of course a function of $n$ - no doubt. But it is not a polynomial: because the number of terms in the product grows unbounded. –  Srivatsan Jan 12 '12 at 16:49
Ok thank you sir, I have fixed it, and I came to know about the difference , Thanks a lot !! @Srivatsan –  Iyengar Jan 12 '12 at 16:52
@Tim: A polynomial has a fixed degree, independent of its argument ($n$ in this case). The product $n(n-1)(n-2) \cdots 1$ contains $n$ terms, which is not allowed. –  Srivatsan Jan 12 '12 at 21:37
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