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Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO questionMO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

deleted 2 characters in body
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Andrey Rekalo
Andrey Rekalo
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Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the LavrentievLeont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Lavrentiev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

deleted 2 characters in body
Source Link
Andrey Rekalo
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Lavrent'evLeont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Lavrentiev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Lavrent'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Lavrentiev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

Problem.

Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients

$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?

Background and motivation.

A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).

The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).

Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Lavrentiev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

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Andrey Rekalo
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122
added 1293 characters in body
Source Link
Andrey Rekalo
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122
Source Link
Andrey Rekalo
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122