Производная факториала как найти

Первое: факториал как операция определена только для целых чисел. Поэтому х! для произвольных значений х просто не имеет смысла.

Второе: производная может браться только от непрерывной, или хотя бы кусочно-непрерывной функции. Если функция определена лишь на дискретной совокупности точек (ну вот на целых числах, к примеру), то брать от неё производную нельзя.

Третье: существует гамма-функция Эйлера, которая определяется как интеграл от t^(z-1)*exp(-t)dt в пределах от 0 до бесконечности. Вообще говоря, z тут комплексное число, но для простоты можно рассматривать только значения с нулевой мнимой частью (чисто действительные), то есть считать, что z=x, и тогда

Г(x) = интеграл от t^(x-1)exp(-x)dt в пределах от 0 до бесконечности.

Фишка тут в том, что а) эта функция определена для любых положительных значений х, и даже для некоторых отрицательных (интеграл расходится лишь для целых отрицательных значений); б) она непрерывна; и в) для целых чисел x=n Г(n) = n!.

То есть гамма-функцию Эйлера можно рассматривать как обобщение факториала на случай произвольных значений переменной. И для неё таки да, можно вычислить производную: Г'(x) = интеграл от (ln t)*t^(x-1)*exp(-t)dt в пределах от 0 до бесконечности.

Start from

$$x!=x(x-1)!$$
$$x!’=x(x-1)!’+(x-1)!$$

So we are looking for a function that satisfies

$$f(x)=xf(x-1)+(x-1)!$$

Replacing we have

$$f(x)=x((x-1)f(x-2)+(x-2)!)+(x-1)!$$

Again

$$f(x)=x((x-1)((x-2)f(x-3)+(x-3)!)+(x-2)!)+(x-1)!$$

Notice that we have at this stage

$$f(x)=x(x-1)(x-2)f(x-3)+x(x-1)(x-3)!+x(x-2)!+(x-1)!$$

So we can extend

$$f(x)=x(x-1)(x-2)…(x-(k-1))f(x-k)+$$
$$x(x-1)…(x-(k-2))(x-k)!+…+x(x-1)(x-3)!+$$
$$x(x-2)!+(x-1)!$$

or

$$f(x)=x(x-1)(x-2)…(x-(k-1))f(x-k)+sum_{m=x}^{x-(k-1)}frac{x!}{m}$$

Taking $k=x$ and $x$ integer we have

$$f(x)=x!f(0)+sum_{m=x}^{1}frac{x!}{m}$$

or

$$f(x)=x!(f(0)+sum_{m=1}^{x}frac{1}{m})$$

Notice that this must be completely valid no matter what extension of factorial we take. And we could essentially stop here.

Still, since we can, it all now comes to defining $f(0)$ which is $0!’$, first derivative of factorial at $0$.

What is the value of $0!’$ ?

Well $f(0)$ is a constant so there is no harm of replacing it with $f(0)=-gamma+c$. (We use $gamma$ so we could argue about the asymptotic evaluation as it is obviously needed to reach $ln(x)$)

$$f(x)=x!(-gamma+c+sum_{m=1}^{x}frac{1}{m})$$

Asymptotically then

$$f(x) sim x!(ln(x)+c)$$

Now

$$ln(x!)’=frac{1}{x!}x!’ sim frac{1}{x!}x!(ln(x)+c)=ln(x)+c$$

Yet in the simplest

$$ln(x!)’ approx frac{ln(x!)-ln((x-1)!)}{1}=ln(x)$$

meaning there is no problem to take $c=0$, even though it can be any other value. (We are just trying to give some interpretation for having $c=0$. In the notes there are more of it.)

That makes:

$$x!’=x!(-gamma+sum_{m=1}^{x}frac{1}{m})$$

or

$$n!’=n!(H_n-gamma)$$

shortly written as

$$ln(n!)’= H_n-gamma$$

(Of course, this is just one of the possibilities that happens to match the standard Gamma function factorial extension as well. We explain further other implications of taking $c=0$ and how the solution might not correspond to the standard Gamma function at all.)

Notes

To conclude this all, if we require $x!=x(x-1)!$, then any other possible extension of factorial function has a form $x!=g(x)Gamma(x+1)$ where $g(x+1)=g(x)$, meaning the additional multiplier is any periodic function with period $1$ .

This is reveling the format of all possible values for $c$ no matter what extension we have.

$$f(0)=-gamma g(0)+ g'(0)$$

So if for a periodic function at integers $g(n)=1$ and $g'(n)=0$, that is our choice. The simplest possible, since we do want to have naturally $1!=1$, for example, leaving:

$$f(0)=-gamma + g'(0)$$

which makes $c=g'(0)$.

There is nothing we could say about the derivative at integers $g'(n)$ without some additional requirement. We have chosen it to be $0$.

However, an additional argument is that asymptotically it is not possible to have any other constant value for $c$ as it is not difficult to find that $ln(n!) = nln n – n +O(ln(n))$ yet an integral of $ln(n)+c$ would add one more linear term beyond $-n$.

This argumentation requires that an extension of factorial, as there is no other way of defining first derivative, conforms with its asymptotic properties even locally. Other versions of extended factorial might not follow this requirement.

This requirement is in line with so called logarithmically convex function that fulfills for any $x,y$

$$ln f(x) geq ln f(y) + frac{f'(y)}{f(y)}(x – y)$$

meaning

$$ln((n+1)!) geq ln(n!) + frac{n!’}{n!}((n+1) – 1)$$
$$ln(n+1) geq ln(n)+c$$

and this asymptotically requires $c=0$.
So we could say that $c$ is equal to $0$, if our choice of an extension for factorial is at least (asymptotically, i.e. for sufficiently large integer) logarithmically convex at integers, even though logarithmically convex is not usefully defined just for integers, since it is a global property.

However, $0!’=-gamma$ does not necessarily define a classical Gamma function neither it is a prerequisite to have a solution. For integer factorial, any value of $0!’$ would do. We are just trying to connect dots a little bit more in depth.

Now that we are there, it is not difficult to establish for any extension of factorial an illustrative connection:

$$ln(x!)’=H_{[x]}-ln({x}!)+0!’$$
where $[x]$ is integer and ${x}$ fractional part of $x=[x]+{x},0leq{x}<1$.

(This is all far more interesting than it may seem at first. Choosing a periodic

$$g(x)=frac{1}{Gamma({x}+1)}$$

we get this as a possible factorial extension

$$x!=frac{Gamma(x+1)}{Gamma({x}+1)}$$

and that is a “linear” version of $x!$ for $x geq 0$. This is probably the most direct extension of integer factorial one could think of. It is a completely acceptable extension.)

Решение

Gamma(1 + n)*polygamma(0, 1 + n)

/         2                                             
polygamma (0, 1 + n) + polygamma(1, 1 + n)/*Gamma(1 + n)

/         3                                                                                         
polygamma (0, 1 + n) + 3*polygamma(0, 1 + n)*polygamma(1, 1 + n) + polygamma(2, 1 + n)/*Gamma(1 + n)

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