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Can anyone tell me the following statement is true or not? Thank you.

There are two polynomials: \begin{align} p(r,\theta) &=\sin(n_0\theta) + \sum_{j=1}^{\ell}a_j r^{n_j}\sin(n_j\theta), \quad r>0\\ \widetilde{p}(r,\theta) &=\sin(n_0\theta) + \sum_{j=1}^{\ell}\widetilde{a}_j r^{\widetilde{n}_j}\sin(\widetilde{n}_j\theta), \quad r>0 \end{align} where $1\le n_0 < n_1 <\cdots<n_{\ell}$, $1\le n_0 < \widetilde{n}_1 <\cdots<\widetilde{n}_{\ell}$.$\quad$ $n_j,\ \widetilde{n}_j\in\mathbb{N}$,

$a_{\ell}, \widetilde{a}_{\ell}>0$, $a_j\geq0$, $\widetilde{a}_j\geq0$, $j=1,\cdots,\ell-1$.

If these two polynomials are different, then can there exists point $(r_0,\theta_0)$ such that $$ p(r_0,\theta_0)\cdot \widetilde{p}(r_0,\theta_0)<0. $$

The following is the original problem: $$ \sum_{k=1}^{\infty} {\sigma_k} \frac{\sum_{j=1}^{\ell}a_js^{n_j}} {\sum_{j=1}^{\ell}a_js^{n_j}+\lambda_k} =\sum_{k=1}^{\infty} \widetilde{\sigma}_k \frac{\sum_{j=1}^{\ell}a_js^{\widetilde{n}_j}} {\sum_{j=1}^{\ell}a_js^{\widetilde{n}_j}+\widetilde{\lambda}_k},\quad \forall |s|<\epsilon. $$ where $$a_j>0, n_j,\ \widetilde{n}_j\in\mathbb{N}, 1\leq n_1<n_2<\cdots<n_{\ell}, 1\leq \widetilde{n}_1<\cdots<\widetilde{n}_{\ell}. $$

$$ \sigma_k,\ \widetilde{\sigma}_k>0, \mbox{and } \sigma_k\rightarrow\sigma_0(>0), \widetilde{\sigma}_k\rightarrow\widetilde{\sigma}_0(>0), \mbox{as } k\rightarrow\infty. $$ $$ 0<\lambda_1<\cdots<\lambda_k<\cdots\rightarrow\infty, 0<\widetilde{\lambda}_1<\cdots<\widetilde{\lambda}_k<\cdots\rightarrow\infty, \mbox{as } k\rightarrow\infty. $$ Can we derive $\{n_j\}_{j=1}^{\ell}=\{\widetilde{n}_j\}_{j=1}^{\ell}$, $\lambda_k=\widetilde{\lambda}_k$, $\sigma_k=\widetilde{\sigma}_k$, $k\in\mathbb{N}$.

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    $\begingroup$ Where does this statement come from? $\endgroup$ – Igor Rivin Nov 11 '13 at 14:11
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    $\begingroup$ Maybe this old result of Fejer will help you get started: if $f(\theta)$ is a trigonometric polynomial with real coefficients such that $f(\theta)\geq 0$, $\forall \theta$, then there exists a polynomial $p(z)$ such that $f(\theta)=|p\bigl(\; e^{\sqrt{-1}\; \theta}\;\bigr)|^2$, $\forall \theta$. $\endgroup$ – Liviu Nicolaescu Nov 11 '13 at 14:54
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    $\begingroup$ I added the original problem. $\endgroup$ – CooLee Nov 12 '13 at 10:44

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