How to express the whole part $\lfloor x \rfloor$ as analytical function or Taylor/Fourier series?

Question

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

Answer 1

To express ⌊x⌋ in terms of the Fourier series for $x \in \mathbb{R}/\mathbb{Z}$ let f(x) := x - ⌊x⌋, which is easier to compute the Fourier series for. Hence, start by showing that f(x) is periodic. Given any integer $n \in \mathbb{Z}$.

$$\begin{align}f(x) &= x - ⌊x⌋ \\&= x + n - n - ⌊x⌋\\ &= x + n - ⌊x + n⌋ \\&= f(x+1\times n)\end{align}$$

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

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

$$\begin{align}f(x) &= a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(m\omega_0 x) + b_{m}\sin(m\omega_0 x))\\ &= a_0 + \sum_{m=1}^{\infty} (a_{m}\cos(2\pi mx) + b_{m}\sin(2\pi m x))\end{align}$$

where the coefficients are calculated as follows

$$\begin{align}a_0 &= \frac{1}{T_0}\int_{<T_0>}f(x)dx = \int_{0}^{1}x dx = \frac{1}{2} \\a_m &= \frac{2}{T_0}\int_{<T_0>}f(x)\cos(m 2\pi x)dx = 0 \\b_m &= \frac{2}{T_0}\int_{<T_0>}f(x)\sin(m 2\pi x)dx = -\frac{1}{\pi m}\end{align}$$

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).

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⌋ (using finite number of terms as above)

Notice also, it is valid only 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⌋ can also be represented as a summation of shifted unit step functions u(x) = Sign(x)/2 + 0.5, in particular

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