Newton's Interpolation formula (series) on the factorial

Question

I was wondering what would happen if we took Newton's Interpolation formula (series) and applied it the factorial.

It is given as

$$f(x)=\sum_{k=0}^\infty\binom{x-a}k\Delta^k[f](a)$$

where

$$\Delta^n[f](a)=\sum_{k=0}^n\binom nk(-1)^{n-k}f(a+k)$$

and I want $f(x)=x!$

Putting this all in with $a=0$, I get

$$x!=\sum_{k=0}^\infty\sum_{n=0}^k\binom xk\binom nk(-1)^{n-k}k!$$

which, notably, is capable of taking non-integer values for $x$.

So of course, I am concerned with its radius of convergence and especially if it comes out to equal the gamma function.

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As a side question, if this fails, then are there any interpolations of the factorial that come out to the factorial?

Answer 1

Notice that the inner sum can be reduced to

$$ \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k} k! = n! \sum_{j=0}^{n} \frac{(-1)^j}{j!} =: d_n, $$

which is the $n$-th derangement. It is easy to check that $d_n \sim n!/e$ as $n \to \infty.$ On the other hand, if $x \notin \{0, 1, 2, \cdots \}$, then we have

$$\left| \binom{x}{n} \right| = \left| \frac{\Gamma(n-x)}{n!\Gamma(-x)} \right| \sim \frac{1}{|\Gamma(-x)| n^{\Re(x)+1}}$$

as $n \to \infty$. So it follows that the series

$$ \sum_{n=0}^{\infty} \binom{x}{n} \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k} k! = \sum_{n=0}^{\infty} \binom{x}{n} d_n \tag{1} $$

diverges unless $x$ is a non-negative integer.

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We can possibly assign a value to the formal series $\text{(1)}$ under a suitable renormalization. (Of course, the meaningfulness of this assignment is thus purely one's taste of matter.) To this end, consider the following representation of $d_n$:

$$ d_n = \int_{0}^{\infty} (t-1)^n e^{-t} \, dt. $$

Although totally invalid, we may attempt the following ''computation'':

$$ \sum_{n=0}^{\infty} d_n \ ``\text{=''} \int_{0}^{\infty} \sum_{n=0}^{\infty} \binom{x}{n} (t-1)^n e^{-t} \, dt \ ``\text{=''} \int_{0}^{\infty} t^x e^{-t} \, dt = x!. $$

One may try to give a meaning to this nonsense by considering a suitable renormalization. So we consider the following function

$$ d_n(\epsilon) := e^{-\epsilon n} \int_{0}^{1} (t-1)^n e^{-t} \, dt + \int_{1}^{\infty} (t-1)^n e^{-\epsilon n t} e^{-t} \, dt. $$

For each fixed $n$, we have $\lim_{\epsilon \to 0^+} d_n(\epsilon) = d_n (0)$. Now we consider the following renormalized sum:

$$ f(x, \epsilon) := \sum_{n=0}^{\infty} \binom{x}{n} d_n(\epsilon). $$

Then on the region $\mathcal{D} = \{\epsilon \in \Bbb{C} : \Re(\epsilon) > W_0(\frac{1}{e}) \}$, we have

$$ \sup_{t \in [1, \infty)} |(t-1) e^{-\epsilon t}| \leq \frac{e^{-1-\Re(\epsilon)}}{\Re(\epsilon)} < 1 $$

and thus $d_n(\epsilon)$ decays exponentially. Thus the function $\epsilon \mapsto f(x, \epsilon)$ is holomorphic on $\mathcal{D}$. Moreover, using Fubini's theorem and the binomial series we can write

$$ f(x, \epsilon) = \int_{0}^{1} (1 + (t-1) e^{-\epsilon})^x e^{-t} \, dt + \int_{1}^{\infty} (1 + (t-1) e^{-\epsilon t})^x e^{-t} \, dt $$

It is not hard to prove that the right-hand side defines a holomorphic function near an open neighborhood of $(0, \infty)$. So we can use this expression to extend the function $\epsilon \mapsto f(x, \epsilon)$ and we may define

$$ \sum_{n=0}^{\infty} \binom{x}{n} d_n := \lim_{\epsilon \to 0^+} f(x, \epsilon) = \int_{0}^{\infty} t^x e^{-t} \, dt = x!. $$

Answer 2

Let me rewrite the Newton expansion as: $$ \begin{gathered} f(x) = \sum\limits_{0\, \leqslant \,j} {\;\left( \begin{gathered} x \hfill \\ j \hfill \\ \end{gathered} \right)\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,j} \right)} {\left( { - 1} \right)^{j - k} \left( \begin{gathered} j \\ k \\ \end{gathered} \right)\;k!} } = \hfill \\ = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,k\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\;\left( \begin{gathered} x \\ j \\ \end{gathered} \right)\left( { - 1} \right)^{j - k} \left( \begin{gathered} j \\ k \\ \end{gathered} \right)\;k!} \hfill \\ \end{gathered} $$ which, when developed further gives: $$ \begin{gathered} f(x) = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,k\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\;\left( \begin{gathered} x \\ j \\ \end{gathered} \right)\left( { - 1} \right)^{j - k} \left( \begin{gathered} j \\ k \\ \end{gathered} \right)\;k!} = \hfill \\ = \sum\limits_{\begin{array}{*{20}c} {k\, \leqslant \,j} \\ {0\, \leqslant \,k} \\ \end{array} } {\;\left( \begin{gathered} x \\ k \\ \end{gathered} \right)\left( { - 1} \right)^{j - k} \left( \begin{gathered} x - k \\ j - k \\ \end{gathered} \right)\;k!} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\;\left( \begin{gathered} x \\ k \\ \end{gathered} \right)\left( {1 - 1} \right)^{\,x - k} \;k!} = \sum\limits_{0\, \leqslant \,k} {\;\left( \begin{gathered} x \\ k \\ \end{gathered} \right)0^{\,x - k} \;k!} \hfill \\ \end{gathered} $$ or:
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$$ \begin{gathered} f(x) = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,k\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\;\left( \begin{gathered} x \\ j \\ \end{gathered} \right)\left( { - 1} \right)^{j - k} \left( \begin{gathered} j \\ k \\ \end{gathered} \right)\;k!} = \hfill \\ = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,k\, \leqslant \,j} \\ \end{array} } {\;\left( {\frac{{\left( { - 1} \right)^k }} {{\left( {j - k} \right)!}}} \right)\left( { - 1} \right)^j x^{\,\underline {\,j\,} } } = \hfill \\ = \sum\limits_{0\, \leqslant \,j} {\;\left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,j} {\frac{{\left( { - 1} \right)^k }} {{k!}}} } \right)x^{\,\underline {\,j\,} } } = \hfill \\ = \sum\limits_{0\, \leqslant \,l} {\left( {\sum\limits_{0\, \leqslant \,j\, \leqslant \,u} {\;\left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,j} {\frac{{\left( { - 1} \right)^k }} {{k!}}} } \right)\left( { - 1} \right)^{j - l} \left[ \begin{gathered} j \\ l \\ \end{gathered} \right]} } \right)x^{\,l} } \hfill \\ \end{gathered} $$ where the upper bound $u$ in the summation in $j$ is $u=x$ if $x$ is a non-negative integer, otherwise $u= \infty$.

Either the first derivation - which practically gives $f(x) = \left[ {0 \leqslant \text{integer}\,x} \right]x!$ - and the second indicate that the Newton interpolation is valid only for integral $x$.

In the second expression we have that the coefficients , given by the sum in $k$, are $1, \; 0, \; 1/2, \; 1/3, \; 3/8, \cdots \to \;1/e$. The Stirling Numbers of the first kind are increasing with the upper index. Their alternate sum, if extended to infinity, is undefined.

If you plot a partial sum from the above expansion against $\Gamma(x+1)$ you will notice that the approximant polynomial oscillates among the interpolation points, which is known as Runge's phenomenon.
The Gamma function has the requisites for this to occur since it has poles distributed all over the negative $x$ axis.
And in fact, the Newton interpolation works much better for 1/Gamma, i.e. $$ \begin{gathered} \frac{1} {{\Gamma (x + 1)}} = \sum\limits_{\begin{array}{*{20}c} {0\, \leqslant \,j} \\ {0\, \leqslant \,k\,\left( { \leqslant \,j} \right)} \\ \end{array} } {\;\left( \begin{gathered} x \\ j \\ \end{gathered} \right)\left( { - 1} \right)^{j - k} \left( \begin{gathered} j \\ k \\ \end{gathered} \right)\;\frac{1} {{k!}}} = \hfill \\ = \sum\limits_{0\, \leqslant \,j} {\left( {\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,j} \right)} {\;\;\frac{{\left( { - 1} \right)^k }} {{k!\left( {\left( {j - k} \right)!} \right)^2 }}} } \right)x^{\,\underline {\,j\,} } } \hfill \\ \end{gathered} $$