Let $f(t)=\sum_{k=1}^n c_k e^{ia_k t}$ and $g(t)=\sum_{k=1}^n d_k e^{ib_k t}$ be exponential polynomials with complex coefficients and having $n$ terms; and suppose they are equal in modulus: \begin{equation} |f(t)|=|g(t)| \qquad \text{for } t\in\mathbb{R} \end{equation} What does this imply about their exponent sets $A:=\{a_k\}$ and $B:=\{b_k\}$? In particular, does it follow that $A-A=B-B$ and how would you prove this?

Here's a start: $|f(t)|^2=|g(t)|^2$ is equivalent to \begin{equation} \sum_{k,j} c_k c_j^* e^{i(a_k - a_j)t} =\sum_{k,j} d_k d_j^* e^{i(b_k - b_j)t} \end{equation}

which can be written \begin{equation} \sum_l C_l \cos(A_l t+ \phi_l) = \sum_l D_l \cos(B_l t+ \psi_l) \end{equation} where the $A_l$ and $B_l$ are of the form $|a_j-a_k|$ and $|b_j-b_k|$ respectively, and appear only once in their respective sums. The desired result $A-A=B-B$ follows only if none of the coefficients $C_l$ and $D_l$ vanish.

Now, they can't all vanish of course, so the above argument at least implies the equality of some subsets of $A-A$ and $B-B$.


Hint. You only need to use the following fact:

If $a_1,\ldots,a_n$ are distinct real numbers, then the functions $\exp(i a_1 t),\ldots,\exp(i a_n t)$ are linearly independent over $\mathbb R$.

This can be shown inductively. For $n=1$ it is obvious. Assume that it holds for $n=k$, and $a_1,\ldots,a_k,a_{k+1}$ are distinct, and $$ c_1\exp(i a_1 t)+\cdots+c_k\exp(i a_k t) +c_{k+1}\exp(i a_{k+1} t)=0, $$ for all $t$, where $c_1,\ldots,c_{k+1}\in \mathbb R$. Then $$ c_1\exp(i (a_1-a_{k+1}) t)+\cdots+c_k\exp(i (a_k-a_{k+1}) t) +c_{k+1}=0, $$ and differentiating we obtain $$ i (a_1-a_{k+1})c_1\exp(i (a_1-a_{k+1}) t)+\cdots+i (a_k-a_{k+1})c_k\exp(i (a_k-a_{k+1}) t)=0, $$ which by the inductive hypothesis implies that $$ i (a_1-a_{k+1})c_1=\cdots=i (a_k-a_{k+1})c_{k+1}=0, $$ and since $a_j\ne a_\ell$, then $$ c_1=\cdots=c_k=0, $$ and hence $c_{k+1}=0$, as well.

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  • $\begingroup$ I don't see how this is useful... For instance, |e^(it)+e^(3it)| = |1+e^(2it)| even though the consitutent functions are linearly independent. Also, please say whether you concur with the conjecture that A-A = B-B. Thanks the response. $\endgroup$ – MathManM Apr 2 '14 at 21:17
  • $\begingroup$ It can be used with the equation $$ \sum_{k,j} c_k \bar c_j \mathrm{e}^{i(a_k - a_j)t} =\sum_{k,j} d_k \bar d_j \mathrm{e}^{i(b_k - b_j)t}. $$ $\endgroup$ – Yiorgos S. Smyrlis Apr 2 '14 at 21:30
  • $\begingroup$ Okay, group like terms and rewrite this equation as \begin{equation} \sum_{l}C_l e^{i A_l t}= \sum_{l}D_l e^{i B_l t}= \end{equation} where \begin{equation}C_l = \sum_{k,j:a_k-a_j=A_l} c_k c_j^*\end{equation} and \begin{equation} D_l = \sum_{k,j:b_k-b_j=B_l} d_k d_j^*\end{equation} Linear independence implies that sets $\{a_k-a_j: C_l \neq 0\}=\{b_k-b_j: D_l \neq 0 \}$, but this is less than the desired result $A-A=B-B$ in the notation above. Please elaborate. $\endgroup$ – MathManM Apr 2 '14 at 21:41

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