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I have observed that the expression $$\frac{\sin(\pi f_n t)}{f_n \sin(\pi t)}.\Pi(t)\approx\text{sinc}(\pi f_n t).\Pi(t)$$ as $f_n$ becomes large and $\Pi(t)$ is the rect($t$) function.

Without the $\Pi(t)$, the left hand term is periodic so the effect of $t$ in the denominator is limited when compared to a sinc() function. However I struggle to see how $f_n$ in concert with sin($\pi t$) drives the observed behaviour.

Would appreciate any insights on this.

Dave L. Renfro
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  • This is basically just replacing, in the denominator, $f_n \sin(\pi t)$ by $f_n \pi t$ on a bounded interval, right? In terms of relative error that has nothing to do with $f_n$ at all (since the $\sin(\pi f_n t)$ is there on both sides). $f_n$ being large reduces the absolute error by just straight up making both sides smaller, but in relative terms it does the same to both of them. – Ian Feb 11 '23 at 17:00
  • If $\lim_{t \to \pm\infty} h(t)=0$, $\sup_t |g(t)|<\infty, \lim_{t\to 0} g(t)=1$ then $\lim_{n\to \infty}\sup_{t\in \Bbb{R}}|g(t)h(nt)- h(nt)| = 0$. Here $h(t)=sinc(t)$ and $g(t)=1_{|t|<1}/sinc(t)$. – reuns Feb 11 '23 at 19:23

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