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Ricardo Andrade
Ricardo Andrade
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Polynomials and divided dfferencesdifferences

Hi,

I would greatly appreciate any hint for proving the following.

Question  : Let $f:[0, 1] \to {\bf R}$. It canCan it be proved that if    $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$, for all $m=1,2, 3,\dots$ then , then $f$ necessarily is necessarily a polynomial of degree less than or equal to $N$ $N$?

(hereHere $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of  $f$ onon the knots nodes $0, 1/(N+m), \dots, 1$).)

Remark.: In order to obtainmake it easier to obtain a proof, it can be supposed that  $f$ satisfiessatisfies some smoothness properties, as for example to be infinitely differentiable or analytic on $[0, 1]$.

Thank you,

G

Polynomials and divided dfferences

Hi,

I would greatly appreciate any hint for proving the following.

Question  : Let $f:[0, 1] \to {\bf R}$. It can be proved that if  $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$, for all $m=1,2, 3,\dots$ then $f$ necessarily is a polynomial of degree less or equal to $N$ ?

(here $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of  $f$ on the knots $0, 1/(N+m), \dots, 1$).

Remark. In order to obtain easier a proof, it can be supposed that  $f$ satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on $[0, 1]$.

Thank you,

G

Polynomials and divided differences

I would greatly appreciate any hint for proving the following.

Question: Let $f:[0, 1] \to {\bf R}$. Can it be proved that if  $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$ for all $m=1,2, 3,\dots$, then $f$ is necessarily a polynomial of degree less than or equal to $N$?

(Here $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of $f$ on the nodes $0, 1/(N+m), \dots, 1$.)

Remark: In order to make it easier to obtain a proof, it can be supposed that $f$ satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on $[0, 1]$.

Thank you,

G

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Gerry Myerson
Gerry Myerson
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Hi,

I would greatly appreciate any hint for proving the following.

Question : Let f:[0, 1] ---- R$f:[0, 1] \to {\bf R}$. It can be proved that if [0, 1/(N+m), ... , (N+m)/(N+m) ; f ]=0$[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$, for all m=1,2, 3, ...., ....$m=1,2, 3,\dots$ then f $f$ necessarily is a polynomial of degree less or equal to N $N$ ?

(here [0, 1/(N+m), ..., (N+m)/(N+m) ; f]$[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of f $f$ on the knots 0, 1/(N+m), ...., 1$0, 1/(N+m), \dots, 1$).

Remark. In order to obtain easier a proof, it can be supposed that f $f$ satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on [0, 1]$[0, 1]$.

Thank you,

G

Hi,

I would greatly appreciate any hint for proving the following.

Question : Let f:[0, 1] ---- R. It can be proved that if [0, 1/(N+m), ... , (N+m)/(N+m) ; f ]=0, for all m=1,2, 3, ...., .... then f necessarily is a polynomial of degree less or equal to N ?

(here [0, 1/(N+m), ..., (N+m)/(N+m) ; f] denotes the divided difference of f on the knots 0, 1/(N+m), ...., 1).

Remark. In order to obtain easier a proof, it can be supposed that f satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on [0, 1].

Thank you,

G

Hi,

I would greatly appreciate any hint for proving the following.

Question : Let $f:[0, 1] \to {\bf R}$. It can be proved that if $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$, for all $m=1,2, 3,\dots$ then $f$ necessarily is a polynomial of degree less or equal to $N$ ?

(here $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of $f$ on the knots $0, 1/(N+m), \dots, 1$).

Remark. In order to obtain easier a proof, it can be supposed that $f$ satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on $[0, 1]$.

Thank you,

G

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George
George
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Polynomials and divided dfferences

Hi,

I would greatly appreciate any hint for proving the following.

Question : Let f:[0, 1] ---- R. It can be proved that if [0, 1/(N+m), ... , (N+m)/(N+m) ; f ]=0, for all m=1,2, 3, ...., .... then f necessarily is a polynomial of degree less or equal to N ?

(here [0, 1/(N+m), ..., (N+m)/(N+m) ; f] denotes the divided difference of f on the knots 0, 1/(N+m), ...., 1).

Remark. In order to obtain easier a proof, it can be supposed that f satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on [0, 1].

Thank you,

G