On certain functional equations defining polynomials

1. 1.
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     Either$f(x)$a function defined in the closed interval ($a,b$),$a<b$, and verifying the linear functional equation

     Report issue for preceding element

for everything$x$and for everything$h>0$such as$a ≤ x, x + nh ≤ b$, THE$a_{i}$being constants.

To the functional equation (1) we attach the characteristic equation

We can assume, without restricting the generality, that the polynomial$F(z)$not be rational with respect to a positive integer power of$z$.

We can assume$a_{0}\neq 0,a_{n}\neq 0$These conditions can be expressed by saying that equation (1) is of degree n. It is easy to see, by appropriately adding relations of the form (1), that if a function$f(x)$verifies a functional equation of degree$n$it also satisfies an equation of degree$n+m$, whatever the positive integer$m$.

We will also say that equation (1) is of order$k$when 1 is a root of order$k$of multiplicity of the characteristic equation.

If$f(x)$verifies an equation of degree$n$order number$k$it also verifies an equation of degree$n+m$and of an order at least equal to$k$, whatever the positive integer$m$2.
- For equation (1) to be of order$k$it is necessary and sufficient that

The function$x^{m}$is a solution of equation (1) if$m=0,1,\ldots k-1$, but$n$it's not one for$m\geq k$since

Therefore,
the functional equation (1) is satisfied by any polynomial of degree k-1, but not by a polynomial whose effective degree exceeds$k-1$.

3 - Let's consider the polynomial$G_{p}(z)$which is obtained by the formula

Or$\alpha$is a primitive root of order$p$of unity.
Suppose that the polynomial$F(z)$verifies identity

We then have$F(z)=a_{0}(1-z)^{k}\prod_{i=1}^{n-k}\left(1-\rho_{1}z\right)$, Or$\rho_{i}$are roots, different from 1, of unity. If$p_{1}$is root of order$p$From the unit, one can immediately see that$\mathrm{G}_{\rho}(z)$cannot be identical to$a_{0}^{p-1}.\mathrm{F}(z)$.

It follows that a number can always be determined$p$so that we have
(3)
$\mathrm{G}.(z)\equiv\equiv a_{0}^{p-1}.\mathrm{F}(z)$
except in the case where$\mathrm{F}(z)=a_{0}(1-z)^{n}$
The general form of polynomials $\mathrm{F}(z)$satisfying (2), or the more general relation$\mathrm{G}_{p}(z)=a_{0}^{p-1}.\mathrm{F}(z)$, can be obtained easily, but we don't need it later.
4. - Let's return to equation (1) and assume that the order$k$be less than n. The function$f(x)$also check the functional equation whose characteristic equation is$\mathrm{G}_{p}(z)=0$the number$p$being determined
such that we have relation (3). The function also satisfies the functional equation whose characteristic equation is$\mathrm{G}_{p}(z)+\lambda\mathrm{F}(z)=0$,$\lambda$being a constant.

We can determine the constant$\lambda\left[G_{\rho}^{(k)}(1)+\lambda F^{(k)}(1)=0\right]$in such a way that this last functional equation is of order at least equal to$k+1$We therefore deduce that

If the function$f(x)$verifies equation (11), of degree$n$and order$k$ : it also satisfies an equation of degree$n$of at least equal order.$\grave{a}k+1$.

The property relating to degree follows from the remark made at the end of No. 1.5.
- It follows that if the function$f(x)$verifies equation (1); it also verifies a functional equation of order$n$and degree$n$.

An equation (1) of degree and order$n$is of the form

and we know that, if$f(x)$is assumed to be continuous, the general solution of this equation is an arbitrary polynomial of degree$n-1(1)$.

We therefore deduce the following final property:
The general continuous solution of equation (1) of order k is an arbitrary polynomial of degree$k-1$.

1. 6.
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     Now consider a function$f(x,y)$of two variables$x$And$y$, defined in the rectangle$a\leq x\leq b,c\leq y\leq d$and verifying the linear functional equation

     Report issue for preceding element

for everything$x,y$and for everything$h>0,h^{\prime}>0$such as$a\leq x,x+nh\leq b_{\text{\penalty 10000\thinspace; }}$.$c\leq y,y+mh^{\prime}\leq d$, THE$a_{i,j,}$being constants.
⁽¹⁾ We can subject the function to more general conditions than: continuity. See in my Thesis (Mathematica t. VIII sp. p. 57.) the generalization of a theorem of MW Sierrinski.

To this equation we attach the characteristic equation

Let us first define the degree and order of the functional equation (4). We say that equation (4) is of degree$(n,m)$if we have

We will say that equation (4) is of order ($k,k^{\prime}$) if the equation$\mathrm{li}\mathrm{F}(z,t)=0$admits the root$z=1$of order$k$of multiplicity, identically in$t$and the root$t=1$of order$k^{\prime}$of multiplicity, identically in$z$. The number$tk$is therefore such that we have

identically in$t$And

for at least one value of$t$In other words$k$is the number of times the polynomials$F_{f}(z)$they all have in common the factor ($1-z$).

If$f(x,y)$verifies equation (4) of degree ($n,m$) and order ($k,k^{\prime}$) it also satisfies an equation of degree$\left(n+n_{1},m+m_{1}\right)$and of at least equal order$\left(k,k^{\prime}\right)\left({}^{2}\right)$, regardless of whether the integers are positive or zero$n_{1}$And$m_{1}$7.
- If the function$f(x,y)$checks equation (4) of order ($k,k^{\prime}$) it also satisfies the functional equation whose characteristic equation is$\mathrm{G}_{p}(z,t)=0$Or

"$\alpha$being still a primitive root of order$p$of the unit.
The function will also satisfy the functional equation whose characteristic equation is$\mathrm{G}_{p}(z,t)+\mathrm{H}(t)\mathrm{F}(z,t)=0,\mathrm{H}(t)$being a polynomial in$t$.

We can choose the number$p$so that

(2) The order ($k_{1},k_{1}^{\prime}$) is at least equal to the order ($k,k^{\prime}$) if$k_{1}\geq k,k_{1}\geq k^{\prime}$and then
we can determine the polynomial$\mathrm{H}(t)$such that the new functional equation obtained is of order at least equal to ($k+1,k^{\prime}$a. The degree of this equation is of the form$\left(n^{\prime},m^{\prime}\right)$Or$n^{\prime}\leq n$.

We therefore deduce the following property:
If the function$f(x,y)$verifies equation (4) of degree ($n,\dot{m}$) and: of order ($k,k^{\prime}$) it also satisfies an equation of degree ($n,m^{\prime}$) and: of order at least equal to ($k+1,k^{\prime}$8.
- We deduce that$f(x,y)$verifies an equation of degree ($n,m^{\prime}$) and of order at least equal to ($n,k^{\prime}$).

We arrive at the property:
If the function$f(x,y)$verifies equation (4) of degree ($n,m$) it: also satisfies an equation of degree ($n,m^{\prime}$) and order ($n,m^{\prime}$).

We show in exactly the same way that$f(x,y)$check. an equation of degree ($n$',$m$) and order ($n$',$m$).

We don't need to specify the numbers:$n^{\prime}$And$m^{\prime}$9.
- An equation (4) of degree$(n,m)$and order$(n,m)$is of: the form

and we know that the general continuous solution of this equation is a pseudo-polynomial of order ($n-1,m-1$), that is to say a function of the form

THE$A_{i}(y)$being continuous functions of$y\in t$THE$B_{i}(x)$continuous functions of$x$alone.

It follows that the general continuum solution of the functional equation (4) is a pseudo-polynomial.
⁽³⁾ See: A. Marchaud, "On the derivatives and differences of functions of real variables." See also my thesis (Mathematica, vol. VIII), p. 65.
10. - So that$x^{r}y^{s}$to be a solution of equation (4), it is necessary that

If ($k,k^{\prime}$) is the order of the equation; we can immediately see that:$x^{k}y^{m}$And$x^{n}y^{k}$cannot be solutions.

If we are looking for the condition for which$x^{r}\mathrm{\penalty 10000\ A}(y)$one solution is found to be the function$\Lambda(y)$must check the equations

If$r\geq k\mathrm{\penalty 10000\ A}(y)$is necessarily a polynomial of degree at most equal to$m-1$(or it is identically zero) and if$r\geq n\mathrm{\penalty 10000\ A}(y)$is a polynomial of degree at most equal to$k^{\prime}-1$. When$r=0,1,2,\ldots k-1$the function$\mathrm{A}(y)$can be arbitrary. Solutions of the form$y^{s}\mathrm{\penalty 10000\ B}(x)$enjoy similar properties. Therefore,

The general continuous solution of the functional equation(4) of order ($k,k^{\prime}$) is the sum of a pseudo-polynomial of order ($k-1,k^{\prime}-1$) and a certain polynomial of degree at most equal to ($n-1,m-1$) ( ⁴ ).

It easily follows from the above that the necessary and sufficient condition for equation (4) to admit only polynomials as continuous solutions is that this equation be of order (0,0).
11. - Let us determine in particular the equations of degree$(n,m)$whose continuous general solution is an arbitrary polynomial of degree$(n-1,m-1)$.

It is immediately apparent that these equations have the characteristic equation

Or$F(z)$is a polynomial of degree$n$in$z$And$G(t)$a polynomial of degree$m$in$t$These two polynomials must satisfy the inequalities$\mathrm{F}^{(i)}(1)\neq 0,\quad i=0,1,2,\ldots,n-1,\mathrm{G}^{(j)}(1)\neq 0,j=0,1,2,\ldots,m-1(-1)^{m}m!\mathrm{F}^{(n)}(1)+(-1)^{n}n!\mathrm{G}^{(m)}(1)\neq 0$
( ⁴ ) We say that a polynomial is of degree ($n,m$) if it is of degree$n$in$x$and degree$m$in$y$A polynomial has a degree at most equal to ($n,m$) if it is of degree$\leq n$in$x$and degree$\leq m$in$y$.

Inequalities$\mathrm{F}(1)\neq 0,\mathrm{G}(1)\neq 0$express precisely that the equation is of order (0,0).

The simplest of these equations is the one with the characteristic equation

It therefore follows that the general continuous solution of the functional egation

is any polynomial of degree ($n-1,m-1$In particular ,
the general solution of the equation

$a,b,c,d$being constants.
We also see that the general continuous solution of the equation