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rect(x) is replaced by rect(x/2).
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u136536
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A simplest case is a rectangular function defined as $${\text{rect}}\left( x \right) = \left\{ \begin{aligned} 1, & \quad - 1/2 < x < 1/2\hfill \\ 1/2, & \quad \left| x \right| = 1/2 \hfill \\ 0, & \quad {\text{otherwise}}{\text{.}} \hfill \\ \end{aligned} \right.$$ We can see that if $f(x) = \text{rect}(x)$$f(x) = \text{rect}(x/2)$, then: $$\sum_{n=-\infty}^\infty \text{rect}(x+n)=1.$$$$\sum_{n=-\infty}^\infty \text{rect}(x/2+n)=1$$ Note thatand the requirement: $$f(x)=C-\sum_{n\in\mathbb Z\setminus\{0\}}f(n+x), \qquad x\in[0,1)$$ is satisfied. The function $f(x) = \text{rect}(x/2)$ is analytic when $x\in[0,1)$.

The rectangular function is the Fourier transform of the sinc function.

A simplest case is a rectangular function defined as $${\text{rect}}\left( x \right) = \left\{ \begin{aligned} 1, & \quad - 1/2 < x < 1/2\hfill \\ 1/2, & \quad \left| x \right| = 1/2 \hfill \\ 0, & \quad {\text{otherwise}}{\text{.}} \hfill \\ \end{aligned} \right.$$ We can see that if $f(x) = \text{rect}(x)$, then: $$\sum_{n=-\infty}^\infty \text{rect}(x+n)=1.$$ Note that the rectangular function is the Fourier transform of the sinc function.

A simplest case is a rectangular function defined as $${\text{rect}}\left( x \right) = \left\{ \begin{aligned} 1, & \quad - 1/2 < x < 1/2\hfill \\ 1/2, & \quad \left| x \right| = 1/2 \hfill \\ 0, & \quad {\text{otherwise}}{\text{.}} \hfill \\ \end{aligned} \right.$$ We can see that if $f(x) = \text{rect}(x/2)$, then: $$\sum_{n=-\infty}^\infty \text{rect}(x/2+n)=1$$ and the requirement: $$f(x)=C-\sum_{n\in\mathbb Z\setminus\{0\}}f(n+x), \qquad x\in[0,1)$$ is satisfied. The function $f(x) = \text{rect}(x/2)$ is analytic when $x\in[0,1)$.

The rectangular function is the Fourier transform of the sinc function.

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u136536
u136536
  • 35
  • 2

A simplest case is a rectangular function defined as $${\text{rect}}\left( x \right) = \left\{ \begin{aligned} 1, & \quad - 1/2 < x < 1/2\hfill \\ 1/2, & \quad \left| x \right| = 1/2 \hfill \\ 0, & \quad {\text{otherwise}}{\text{.}} \hfill \\ \end{aligned} \right.$$ We can see that if $f(x) = \text{rect}(x)$, then: $$\sum_{n=-\infty}^\infty \text{rect}(x+n)=1.$$ Note that the rectangular function is the Fourier transform of the sinc function.