MHB What is the domain of f(x)=ln(sin(pi/x))?

[Yankel]

Hello guys,

I need some assistance in calculating the domain of this function:

f(x)=ln(sin(pi/x))

I started with sin(pi/x)>0 due to the ln function.

From here 0<(pi/x)<pi. That lead me to some calculations giving x>1, but obviously I have periods of 2*pi to include.

The answer is: x>1, 1/(2k+1)<x<1/2k for k=1,2,3,... and 1/2k<x<1/(2k+1) for k=-1,-2,-3,...

I don't understand why or how to get to this solution.

thanks !

 

[MarkFL]

I would write:

$\displaystyle 2k\pi<\frac{\pi}{x}<(2k+1)\pi$

where $\displaystyle k\in\mathbb{Z}$

$\displaystyle 2k<\frac{1}{x}<(2k+1)$

When $\displaystyle k=0$ we have $\displaystyle 1<x$

When $\displaystyle 0<k$ we have $\displaystyle \frac{1}{2k+1}<x<\frac{1}{2k}$

When $\displaystyle k<0$ we have $\displaystyle \frac{1}{2k}<x<\frac{1}{2k+1}$