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Let $e(x)=e^{2\pi i x}$ and consider the following functions defined for $x\in [0,1]$: $$ f_1(x)=\frac{e(10x)-e(x)}{e(x)-1}, \quad f_2(x)=\frac{e(110x)-e(11x)}{e(11x)-1}, $$ and $$ f_3(x)=\frac{e(1010x)-e(101x)}{e(101x)-1}\cdot\frac{e(100x)-1}{e(10x)-1} $$ I would like to show bounds of the form $|f_1(x)+f_2(x)| \ll |f_1(x)|$ and $|f_1(x)+f_2(x)+f_3(x)|\ll |f_1(x)+f_2(x)|$.

For the first bound, the zeros of $f_1(x)$ occur at $x=n/9$ for $n\in\{0,\ldots 9\}$ and now note that $f_2(x)$ is also zero at those points. It is then possible to show that $\lim_{x\to n/9} |f_2(x)/f_1(x)|$ exists, so that $|f_2(x)/f_1(x)|$ is continuous and therefore attaches its maximum on $[0,1]$. Hence $$ \frac{|f_1(x)+f_2(x)|}{|f_1(x)|}\le 1+ \frac{|f_2(x)|}{|f_1(x)|} \ll 1. $$ I don't know how to prove the second bound. My intuition tells me that if $f_1(x)+f_2(x)$ is zero, then $f_1(x)+f_2(x)+f_3(x)$ is zero as well but I don't know how to prove this without explicitely computing the zeros. I'm also interested in the implied constant. By plotting the functions on sage, $$ \frac{|f_1(x)+f_2(x)|}{|f_1(x)|}\le 40, $$ and $$ \frac{|f_1(x)+f_2(x)+f_3(x)|}{|f_1(x)+f_2(x)|}\le 60 $$ but I don't know how to prove this without a numeric computation. This functions appear on the context of the Fourier transform of some set of integers

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Consider $\displaystyle g_1(t) = \frac{t^{10} - t}{t -1}$ and $\displaystyle g_2(t) = \frac{t^{110} - t^{11}}{t^{11} - 1}$, so that $g_j(e^{2\pi i x}) = f_j(x)$ for $j = 1,2$. We have $$ g_1(t) = t\frac{t^9 - 1}{t - 1} = t\Phi_3(t)\Phi_9(t), \\ g_2(t) = t^{11}\frac{t^{99} - 1}{t^{11} - 1} = t^{11}\Phi_3(t)\Phi_9(t)\Phi_{33}(t)\Phi_{99}(t), \\ g_2(t) = g_1(t)t^{10}\Phi_{33}(t)\Phi_{99}(t), $$ where $\Phi_n$ are cyclotomic polynomials. Hence to bound the ratio $f_2(x)/f_1(x)$ from above one has to bound $\Phi_{33}(t)\Phi_{99}(t)$ from above for $|t| = 1$. All of the coefficients of $\Phi_{33}(t)\Phi_{99}(t)$ are $0$ and $\pm 1$ and there are $49$ non-zero coefficients, which gives the bound $f_2/f_1\le 49$.

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  • $\begingroup$ To show the claim for $f_1 + f_2 + f_3$ you have to show that $$g_3 = \frac{t^{1010} - t^{101}}{t^{101 - 1}}\cdot \frac{t^{100} - 1}{t^{10} - 1}$$ is divisible by $g_1 + g_2 = t\Phi_3\Phi_9(1 + t^{10}\Phi_{33}\Phi_{99})$ $\endgroup$ Commented Mar 29, 2023 at 18:27
  • $\begingroup$ Oh, not divisible. The correct claim is: if $t$ is a root of $t\Phi_3\Phi_9(1 + t^{10}\Phi_{33}\Phi_{99})$ with $|t | = 1$ then it is also a root of $g_3$. $\endgroup$ Commented Mar 29, 2023 at 18:43
  • $\begingroup$ Using cyclotomic polynomials is very clever! However the bounds then start getting more complicated because not neccesarily all coefficients are $\pm 1$ or $0$. Also I don't know how to deal with the case when $1+t^{10} \Phi_{33}\Phi_{99}=0$, one root is for instance $t=i$, and then the reason $g_3=0$ is because of the term $t^{100}-1$ (contrary to what I thought it would have something to do with $t^{909}-1$) $\endgroup$
    – Itachi
    Commented Mar 30, 2023 at 10:16

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