I read about the system of $n$ equations in the link below. I wonder how it behaves for growing $n$. Does it converge ?


Here it is explicit :

Consider the polynomial $f_n(x) = x^n + a_1 x^{(n-1)} + a_2 x^{(n-2)} + ...$

Now solve $f_n(x+1) = f_n(x)$ without solving for the $x^{(n-1)}$ term and with $f(0)=0$. See example below for clarification.

The idea seems to find a " periodic polynomial ".

Does this system of equations converge to a truncated Taylor series of a known periodic function such as sine ? Or even does it converge to a periodic function ? Or maybe it converges to a periodic function + a (fixed) polynomial of fixed degree ( indep of $n$ ) ? Or even a closed form ?

Im not sure how to solve this kind of problems. Matrices do not seem to help or do they ?

$$ **example** $$

Since there appears to be some confusion I give the example for $n=3$.

$f_3(x) = x^3 + a_1 x^{2} + a_2 x + a_3$

We solve $f_3(x+1) = (x+1)^3 + a_1 (x+1)^{2} + a_2 (x+1) + a_3 = f_3(x) = x^3 + a_1 x^{2} + a_2 x^{1} + a_3$

Thus we expand $f_3(x+1)$ :

$x^3 + (3+a_1) x^2 + (3+2 a_1+a_2) x + 1 + a_1 + a_2 + a_3$

Hence we get the set of equations :

$3+a_1 = a_1$

$3 + 2 a_1 + a_2 = a_2$

$1 + a_1 + a_2 + a_3 = a_3$

Now solving this set starting with $3+a_1 = a_1$ is not possible. We cannot solve these equations for all variables. So we solve the remaining equations : (for instance this way below)

$1 + a_1 + a_2 + a_3 = a_3$ => $1 + a_1 + a_2 = 0$ => $a_2 = -1 - a_1$ and $a_1 = - a_2 - 1$

Now we plug this into the other remaining equation ($3 + 2 a_1 + a_2 = a_2$) to get :

(first reduce) $3 + 2 a_1 + a_2 = a_2$ => $3 + 2 a_1 = 0$

plug in

$3 - 2 a_2 - 2 = 0$ => $1 - 2 a_2 = 0$ => $a_2 = 1/2$ and $a_1 = - 3/2$

So $f_3(x) = x^3 -3/2 x^{2} + 1/2 x $ ( the restriction $f(0) = 0$ forces the constant term $a_3$ to be $0$. )

I guess that clarifies alot.

I also wonder if there are (other) known systems of $n$ equations that converge to a Taylor series that is real analytic and real periodic as $n$ grows to infinity.

  • $\begingroup$ Possibly related ? : math.stackexchange.com/questions/839951/… $\endgroup$ – mick Jun 20 '14 at 11:23
  • $\begingroup$ Or maybe math.stackexchange.com/questions/839966/… $\endgroup$ – mick Jun 20 '14 at 11:24
  • $\begingroup$ What does "without solving for the $x^{(n-1)}$ term" mean? $\endgroup$ – Antonio Vargas Jun 20 '14 at 14:53
  • $\begingroup$ @AntonioVargas It means we do not solve the associated equation. You see for instance if n = 3 we already see we cannot solve for the x^2 term. So to relax the system of equations we ignore that term and solve the other equations. $\endgroup$ – mick Jun 20 '14 at 18:59
  • $\begingroup$ I edited the OP for clarity. $\endgroup$ – mick Jun 22 '14 at 21:17


So your question asks:

Given the functional: $G[n,x] = x^n + a_0 x^{n-1} ... a_{n-2}x + a_{n-1}$ Such that

$$G[n,x+1] = G[n,x]$$

What is a closed form or accurate description of the expressions $G[n,x]$ as n approaches infinity whereas $G[n,0] = 0$

Now to begin with it is evident that no such $G$ can exist if we are dealing with finite polynomials (since finite sized polynomials) do not display periodicity. But that is not what the question is asking.

Instead you appear to be interested in the system of equations (involving coefficients) that arises from the aforementioned functional equation and the behavior of this system, after discluding certain inconsistent equations.

So we begin naturally at the start: How do we expand this into a system of equations in the general form for the coefficients.

$$(x+1)^n + a_0(x+1)^{n-1} + a_1(x+1)^{n-2} ... a_{n-1} = (x+1)^n + \sum_{i=1}^{n}[a_{n-i}(x+1)^{i-1}]$$

Can be expanded via the Binomial theorem which states for positive integers k

$$(x+r)^k = \sum_{j=0}^{k}\lbrace\frac{k!}{j!(k-j)!}x^{k-j}r^j\rbrace $$

Thus we at this point know that our left hand side for the equation $f(x+1) = f(x)$ can be expressed as

$$(x+1)^n + \sum_{i=1}^{n}[a_{n-i}(x+1)^{i-1}] = \sum_{j=0}^{n}\lbrace\frac{n!}{j!(n-j)!}x^{n-j}\rbrace + \sum_{i=1}^{n}[a_{n-i}\sum_{j=0}^{i-1}\lbrace\frac{(i-1)!}{j!(i-1-j)!}x^{i-1-j}\rbrace] $$

This is of course being equated with:

$$x^n + a_0x^{n-1}+ a_1x^{n-2} ... a_{n-1}$$

On the right hand side. Now our objective is to go to the left hand side expression and determine what the coefficients will be in front of the individual powers of x ranging from the constant $x^0$ terms to the $x^{n-2}$ since the OP does not want to consider $x^{n-1}$ terms (I will attempt but not guarantee a proof that these terms will always result in inconsistent equations) and it is obvious that the $x^n$ terms drop from both sides.

We start with the constant terms which occurs only if $n -j=0$ and if $i - 1 - j = 0 $ (since these expressions determine the powers of x). In that case:

$$\sum_{j=0}^{n}\lbrace\frac{n!}{j!(n-j)!}x^{n-j}\rbrace $$

contributes a single $1$ and $$\sum_{i=1}^{n}[a_{n-i}\sum_{j=0}^{i-1}\lbrace\frac{(i-1)!}{j!(i-1-j)!}x^{i-1-j}\rbrace]$$

reduces to

$$ \sum_{i=1}^{n}[a_{n-1}]$$

Thus our first equation is:

$$ 1 + \sum_{i=1}^{n}[a_{n-i}] = a_{n-1} = 0$$

Using similar logic (try and derive it yourself) we can state that the '$k^{th}$' equation in our system is generated by removing nested sums and setting the first appearance of index j to $j = n-k$ and setting the next appearances of j each to $j = i - 1 - k$ based on this substitution we end up seeing that the '$k^{th}$' equation is

$$ \frac{n!}{(n-k)!k!} + \sum_{i=1}^{n}[a_{n-i}\frac{(i-1)!}{(i-1-k)!k!}] = a_{n-1-k}$$

For all terms that are defined. (If an expression such as (i - 1 - k)! is not defined for some i then that term is not included). We can revise this to:

$$ \frac{n!}{(n-k)!k!} + \sum_{i=k+1}^{n}[a_{n-i}\frac{(i-1)!}{(i-1-k)!k!}] = a_{n-1-k}$$

Notice here if $k = n-1$ we get:

$$ \frac{n!}{(1)!(n-1)!} + \sum_{i=n}^{n}[a_{n-i}\frac{(i-1)!}{(i-n)!(n-1)!}] = a_{0}$$


$$ \frac{n!}{(1)!(n-1)!} + [a_{0}\frac{(n-1)!}{(n-n)!(n-1)!}] = a_{0}$$


$$n + a_0 = a_0$$

So now we have a proof that the equation involving the coefficient $a_0$ of your $x^{n-1}$ term always results in an inconsistent system.

Now to understand the longterm behavior of the equations appears to be out of my ability level BUT at least I can repose your question (to someone more technically qualified than me) as:

GIVEN A SYSTEM Of LINEAR EQUATIONS INVOLVING VARIABLES $a_{i,j}$ such that the $k^{th}$ equation of the $n^{th}$ system of equations is given by:

$$ \frac{n!}{(n-k)!k!} + \sum_{i=1}^{n}[a_{n-i,n}\frac{(i-1)!}{(i-1-k)!k!}] = a_{n-1-k,n}$$

And $n \in Z, n>0, 0 \le k < n-1, k \in Z$

What is a general formula for the expression $a_{i,j}$ and furthermore what is the long term behavior of terms $a_{i,n}$ as $n$ approaches infinity and $a_{n-1} = 0$


Well given that

$$ f(x+1) = f(x), $$

you get

$$ f(x) = f(x+k), $$

where $k$ is an integer. Let us consider

$$ g(x) = f(x) - f(0). $$

Then it is clear that

$$ g(k) = 0, $$

where $k$ is an integer. Whence

$$ g(x) = \prod_k \Big( x - k\Big), $$

so you cannot have a finity polynomial that satisfy $f(x+1) = f(x)$

  • $\begingroup$ I never claimed that such a polynomial exists. If we increase the degree of a polynomial we might be getting a Taylor series. So the question should be considered a bit more from that viewpoint. $\endgroup$ – mick Jun 22 '14 at 21:43

If I understand correctly, your equations say $$f(x+1) - f(x) = c x^{n-1} \quad (\ast)$$ where $f(x)$ is a degree $n$ polynomial with leading term $x^n$ and constant term $0$. In particular, $c$ must be $n$.

Induction then shows that, for any integer $k$, $$f(k) =n \left( 1+2^{n-1} + 3^{n-1} + \cdots + (k-1)^{n-1} \right).$$ Plugging in Faulhaber's formula: $$f(x) = - n x^{n-1} + \sum_{j=0}^{n-1} (-1)^j \binom{n}{j} B_j x^{n-j}$$ where $B_j$ are the Bernouli numbers.

I don't see any sense in which this is approaching a periodic function.

For example, you take $f(x) = x^3 - (3/2) x^2 + (1/2) x$. We compute $$f(x+1) - f(x) = 3 x^2.$$ And, sure enough, for $x$ an integer, we have $$f(x) = 3 (1+2^2+3^2+ \cdots + (x-1)^2).$$ We can use this to find the formula for $f$.

  • $\begingroup$ I hesitate If you understand the question correctly. My guess is not. Sorry if I confused. Have you tried the example given in the OP to see if it matches what you said ? I deliberately gave an example to avoid confusion, I recommend that people take a look at it. $\endgroup$ – mick Jun 28 '14 at 18:51

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