# Are all discrete-analytic funtions as defined here also natural?

Let's define a discrete-analytic function as a function that is equal to its Newton expansion:

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)=\sum_{m=0}^{\infty} \binom {x}m \sum_{k=0}^m\binom mk(-1)^{m-k}f(k)$$

Let's define a natural function such a function whose shift $f(x+n)$ is also descrete-analytic for any natural n.

The question is whether there exist discrete-analytic functions that are not natural.

• I wonder why the first horizontal line is not visible and the formulas collide with the text. Please feel free to correct it you can. – Anixx Nov 5 '13 at 4:08
• Crossposted to math.se : math.stackexchange.com/questions/553795/… – Andy Putman Nov 6 '13 at 3:20
• I claim (to my initial amazement) that the true function $g(x)$ is $1$ at $0$ but $0$ for all positive real $x$! The convergence can be slow. You may confirm that the $k$th partial sum ( of the Newton expansion ) for $g(x)$ is $(-1)^k\binom{x-k}{k}$ i.e. $\prod_{j=1}^k(1-\frac{x}{j})$. Not that that leaves me much more confident. – Aaron Meyerowitz Nov 6 '13 at 3:29
• @Andy Putman i posted it there after a day since i posted it here because of littele number of answers here. Is it you who downvoted? Just because i crossposted it? – Anixx Nov 6 '13 at 3:45
• You are right, now I see it. As I said above, the value at $x$ is the limit on $N$ of $(-1)^N\binom{x-N}{N}=\prod_{j=1}^N(1-\frac{x}{j})$ so $x=0$ is easy, for any $x \gt 0$ the terms are positive and decrease fast enough to have limiting product $0$, and for any $x\lt 0$ the terms are positive and decrease to $1$ slowly enough that the product diverges to $\infty$ like $\ln{N}$. – Aaron Meyerowitz Nov 7 '13 at 4:37

Updated I think that for non-polynomials we need to restrict to non-negative $x$. With this restriction, are there discrete analytic functions which are not natural functions? I think not, at least under rather lax conditions.

Consider an arbitrary expansion $f(x)=\sum_0^{\infty}a_k \binom{x}{k}$ with the $a_k$ real. It is defined for all non-negative integer $x=n$ since only the first $n+1$ terms are non-zero. However for negative integral $x$ we have $\binom{x}{k}=(-1)^k\binom{|x|+k-1}{k}$ so divergence is quite possible for negative $x$ values. I think that as long as $\lim_{k \to \infty}\frac{a_k}{k!}=0$ (or at least if $a_k$ has at worst polynomial growth) then we also have $f(x)$ convergent for all real $x \ge 0$ and, by your definition, $f(x)$ is discrete-analytic (on that range). Then $f$ is determined by it's values at the non-negative integers.

Here is a slightly more formal version of my previous comments (which only deal with the first sentence of your post, which isn't even the main question.) I like it well although it is not original. It has that nice quality of seeming mystifying until the moment when it seems trivially obvious.

Consider a member $\mathbf{y}=(y_0,y_1,y_2,\cdots)$ of the space of sequences. We also write $\mathbf{y}(x)=y_x$ for non-negative integer $x$. Aside from the identity operator $I$ we have the difference operator with $\Delta\mathbf{y}=(y_1-y_0,y_2-y_1,\cdots)$ and the shift operator with $S\mathbf{y}=(y_1,y_2,y_3,\cdots)$. So $y_n=\left(S^n\mathbf{y}\right)(0).$

Since $S=I+\Delta,$ and the operators commute, we have $S^n=(I+\Delta)^n=\sum_{k=0}^n\binom{n}{k}\Delta^k$ and, as you said, $$y_n=\sum_{k=0}^{\infty}\binom{n}{k}\left(\Delta^k\mathbf{y}\right)(0)=\sum_{k=0}^{\infty}a_k\binom{n}{k}$$ for the coefficients $a_k=\left(\Delta^k\mathbf{y}\right)(0).$ Also, $$\mathbf{y}=\sum_{k=0}^{\infty}a_k\mathbf{\binom{x}{k}}$$ Where, for example, $\mathbf{\binom{x}{3}}=(0,0,0,1,4,10,20,\cdots)$ is a sequence.

Note too that the $\mathbf{\binom{x}{k}}$ are a natural basis for the space of sequences with $\Delta\mathbf{\binom{x}{k}}=\mathbf{\binom{x}{k-1}}$ and $S\mathbf{\binom{x}{k}}=\mathbf{\binom{x}{k}}+\mathbf{\binom{x}{k-1}}$ for $k \ge 1,$ while $\mathbf{\binom{x}{0}}$ is sent to the $0$ sequence by $\Delta$ and itself by $S$. Accordingly the shift of $\mathbf{y}$ is $$S\mathbf{y}=\sum_{k=0}^{\infty}(a_k+a_{k+1})\mathbf{\binom{x}{k}}.$$

Going back to the expansion for $\mathbf{y}$, we then have a corresponding discrete-analytic function $$f(x)=\sum_{k=0}^{\infty}a_k\binom{x}{k}$$ defined for all non-negative real $x.$ Here convergence is not an issue (under mild conditions) and $\mathbf{y}$ is the restriction of $f$ to the integers. Also, the discrete-analytic function corresponding to $S\mathbf{y}$ is $$\sum_{k=0}^{\infty}(a_k+a_{k+1})\binom{x}{k}. \tag{*}$$

For $f(x)$ to satisfy the definition of a natural function we need the shift $f(x+n)$ to be discrete-analytic for all positive $n.$ It is sufficient to establish this just for $n=1$ provided that we do this for all discrete-analytic functions.

But $$f(x+1)=\sum_{k=0}^{\infty}a_k\binom{x+1}{k}=a_0\binom{x+1}{0}+\sum_{k=1}^{\infty}a_k\left( \binom{x}{k-1}+\binom{x}{k}\right).$$ Thus $$f(x+1)=\sum_{k=0}^{\infty}(a_k+a_{k+1})\binom{x}{k}. \tag{**}$$

To show that $f(x+1)$ is indeed discrete-analytic, examine the sequence arising from the restriction of $g(x)=f(x+1)$ to the non-negative integers. This restriction is $S\mathbf{y}$ so, comparing $( * )$ and $( ** )$, we are done.

• I think your statement that given the limit, f(x) converges for all real x is wrong. Consider a function which is 1 in x=0 and 0 otherwise. Its Newton expansion diverges for non-natural x, even though all the deltas are limited. – Anixx Nov 5 '13 at 7:21
• That is a counterexample, but maybe not as strong as you say. A function which is $1$ at $x=0$ and $0$ at every positive integer will lead to the expansion $\sum(-1)^k\binom{x}{k}$. This is convergent for all $x \gt -1$ but indeed $\lim_{n \to \infty}|\binom{x}{n}|=\infty$ for $x \lt -1.$ What I ask is a small part of your question (I don't understand the motive of mixing discrete and continuous "differences/derivative") but I wonder if a version of what I said is true restricted to positive reals. – Aaron Meyerowitz Nov 6 '13 at 0:40
• is not the Newton expansion of this function an example of discrete-analytic, but non-natural function? Consider Newton expansion f(x) of the function which is 1 in x=0 and otherwise 0. It is by definition discrete-analytic. Yet if we shift it by 1 to f(x+1), this function is no longer discrete-analytic because its Newton expansion is constant zero which is not equal to the function f(x+1). – Anixx Nov 6 '13 at 1:48
• I expanded the question to account for this counterexample. – Anixx Nov 6 '13 at 2:11
• Since you answer sheds light to an interesting aspect of the matter not directly connected with the main question, I am going to split the question to better reflect your answer. – Anixx Nov 7 '13 at 2:38