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About fractional parts and floor functions...

DreamWeaver

Well-known member
Sep 16, 2013
337
Hi all! :D

I wasn't really sure where to post this, but Analysis seemed a fair bet.

While searching on-line recently, I came across the following expressions for the Fractional Part \(\displaystyle \{x\}\) and Floor Function \(\displaystyle \lfloor x \rfloor\) respectively:


\(\displaystyle \{x\}=\frac{i\log\left(-e^{-2\pi i x}\right)}{2\pi}+\frac{1}{2}\)


\(\displaystyle \lfloor x \rfloor=x-\frac{i\log\left(-e^{-2\pi i x}\right)}{2\pi}-\frac{1}{2}\)



I can't remember where I found them, but just made a note of them... I seem to recall that a condition of both of the above was that the principal branch of the complex logarithm must be taken.

And so, finally, the question: can any of you shed intuitive light on the above? I've been trying to divine some sense out of those expressions, but sadly for me, I'm not Euler... :eek::eek::eek:


All the best, and thanks in advance! (Sun)

Gethin
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
The imaginary argument of the exponential function has a period of $2 \pi$, just like the sine and the cosine. That means that:
$$e^{2\pi i x} = e^{2\pi i (\lfloor x \rfloor + \{x\})} = e^{2\pi i \{x\}}$$

The log function of this expression is:
$$\log(e^{2\pi i x}) = i (2\pi x \text{ mod }{2\pi}) = i (2\pi \{x\} \text{ mod }{2\pi})$$

The principal branch is:
$$\text{Log}(e^{2\pi i x}) = 2\pi i \{x\}$$
 

DreamWeaver

Well-known member
Sep 16, 2013
337
The imaginary argument of the exponential function has a period of $2 \pi$, just like the sine and the cosine. That means that:
$$e^{2\pi i x} = e^{2\pi i (\lfloor x \rfloor + \{x\})} = e^{2\pi i \{x\}}$$

The log function of this expression is:
$$\log(e^{2\pi i x}) = i (2\pi x \text{ mod }{2\pi}) = i (2\pi \{x\} \text{ mod }{2\pi})$$

The principal branch is:
$$\text{Log}(e^{2\pi i x}) = 2\pi i \{x\}$$


Doh!!! But of course.... :eek:

Thank you!!! (Hug)