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And how to express $\{ x \} = x - \lfloor x \rfloor$ as function of $sin(x)$ and $sign(x)$?

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The infinitely many jump discontinuities in the floor function preclude the possibility of a Taylor series for it. –  J. M. Dec 20 '10 at 9:02
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As for the second one: $\frac12-\frac1{\pi}\arctan(\cot(\pi x))$ –  J. M. Dec 20 '10 at 9:09
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This might also interest you. –  J. M. Dec 20 '10 at 10:07

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up vote 9 down vote accepted

To express ⌊x⌋ in terms of Fourier series for $x \in \mathbb{R}/\mathbb{Z}$, let f(x) = x - ⌊x⌋, then the Fourier series for f(x) is quite straightforward, and so we start by showing that f(x) is indeed periodic:

For any integer $n \in \mathbb{Z}$.

f(x) = x - ⌊x⌋ = x + n - n - ⌊x⌋

= x + n - ⌊x + n⌋ = f(x+n) = f(x + n*1)

Hence f(x) is periodic with fundamental period $T_0 = 1$, so $\omega_0 = \frac{2\pi}{T_0} = 2\pi$, and f(x)=x at $0 \leq x < 1$.

A plot of f(x)

The periodic function f(x) can be expressed by the Trigonometric Fourier series,

$f(x) = a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(m\omega_0 x) + b_{m}\sin(m\omega_0 x))$

$ = a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(2\pi mx) + b_{m}\sin(2\pi m x))$

where the coefficients are calculated as follows

$a_0 = \frac{1}{T_0}\int_{<T_0>}f(x)dx = \int_{0}^{1}x dx = \frac{1}{2}$

$a_m = \frac{2}{T_0}\int_{<T_0>}f(x)\cos(m 2\pi x)dx = 0$

$b_m = \frac{2}{T_0}\int_{<T_0>}f(x)\sin(m 2\pi x)dx = -\frac{1}{\pi m}$

Hint. Use integration by parts...

Plot of $f(x) = a_0 + \sum_{m=1}^{10} b_m\sin(2\pi m x)$, (summation from m=1 to 10).

alt text

Now that we obtained the Fourier series representation of f(x) = x - ⌊x⌋

Then ⌊x⌋ $ = x - f(x) = x - 0.5 + \frac{1}{\pi}\sum_{m=1}^{\infty} \frac{\sin(2\pi m x)}{m}$.

A plot of ⌊x⌋, where summation from m=1 to m=10

alt text

Note this is only valid for $x \in \mathbb{R}/\mathbb{Z}$. Because although ⌊x⌋ is continuous on $\mathbb{R}/\mathbb{Z}$. It has a jump of one unit at each integer.

⌊x⌋ consists of an infinite number of steps, and also can be represented as a summation of shifted unit step functions u(x).

⌊x⌋ $ = \sum_{m=1}^{\infty}(u(x-m)-u(-x-m+1))$

Note that u(x) = Sign(x)/2 + 0.5

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+1. Nice and pretty. –  Yonatan N Dec 20 '10 at 16:25
    
@Yonatan N: Thank you :) –  Weaam Dec 20 '10 at 16:40
    
@Weann: You should use \sin and \cos (instead of just sin and cos) inside math formulas so that they are typeset correctly. –  Arturo Magidin Dec 20 '10 at 18:41
    
@Arturo Magidin: Fixed. Thank you. –  Weaam Dec 20 '10 at 18:59

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