0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Anixx Nov 5 '13 at 4:08
  • $\begingroup$ Crossposted to math.se : math.stackexchange.com/questions/553795/… $\endgroup$ – Andy Putman Nov 6 '13 at 3:20
  • $\begingroup$ 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. $\endgroup$ – Aaron Meyerowitz Nov 6 '13 at 3:29
  • $\begingroup$ @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? $\endgroup$ – Anixx Nov 6 '13 at 3:45
  • 1
    $\begingroup$ 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 $\ln{N}$. $\endgroup$ – Aaron Meyerowitz Nov 7 '13 at 4:37
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Anixx Nov 5 '13 at 7:21
  • $\begingroup$ 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. $\endgroup$ – Aaron Meyerowitz Nov 6 '13 at 0:40
  • $\begingroup$ 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). $\endgroup$ – Anixx Nov 6 '13 at 1:48
  • $\begingroup$ I expanded the question to account for this counterexample. $\endgroup$ – Anixx Nov 6 '13 at 2:11
  • $\begingroup$ 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. $\endgroup$ – Anixx Nov 7 '13 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.