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$f(x)=\int_1^{x} \frac{1}{\sqrt[m]{P(t)}}\;dt$

$P(x)$ is polynomial with degree $n$.

$m$ is an positive integer and $m>1$

What is the algoritm to determine $f^{-1}(x)$ is periodic function via using $P(x)$ and $m$ without evaluting the integral ? Is there also a way to find period without evaluting the integral?


Example 1: $P(x)=x^2$ , $m=2$

$f_1(x)=\int_1^{x} \frac{1}{t}\;dt=\ln x$

$f_1^{-1}(x)=e^x$

$f_1^{-1}(x+2k\pi i)=e^{x+2k\pi i}=e^x=f_1^{-1}(x)$ $k$ is an integer $f_1^{-1}(x+2k\pi i)=f_1^{-1}(x)$ $f_1^{-1}(x)$ is a periodic function


Example 2: $P(x)=1-x^2$ , $m=2$ $f_2(x)=\int_1^{x} \frac{1}{\sqrt{1-t^2}}\;dt=\arcsin x-\frac{\pi}{2}$

$f_2^{-1}(x)=\sin {(x+\frac{\pi}{2})}$

$f_2^{-1}(x+2k\pi )=\sin{(x+\frac{\pi}{2}+2k\pi )}=\sin{(x +\frac{\pi}{2})}=f_2^{-1}(x)$ $k$ is an integer $f_2^{-1}(x+2k\pi )=f_2^{-1}(x)$ $f_2^{-1}(x)$ is a periodic function


Example 3: $P(x)=(1+x^2)^2=1+2x^2+x^4$ , $m=2$ $f_3(x)=\int_1^{x} \frac{1}{1+t^2}\;dt=\arctan x -\frac{\pi}{4}$

$f_3^{-1}(x)=\tan (x+\frac{\pi}{4})$

$f_3^{-1}(x+k\pi )=\tan{(x+\frac{\pi}{4}+k\pi )}=\tan{(x+\frac{\pi}{4} )}=f_3^{-1}(x)$ $k$ is an integer $f_3^{-1}(x+k\pi )=f_3^{-1}(x)$ $f_3^{-1}(x)$ is a periodic function


Example 4: $P(x)=x^4$ , $m=2$

$f_4(x)=\int_1^{x} \frac{1}{t^2}\;dt=\frac{x-1}{x}$

$f_4^{-1}(x)=\frac{1}{1-x}$

$f_4^{-1}(x)$ is not periodic function


Thanks a lot for answers

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    I don't understand your question, sorry. $x$ is real? Local inverse is ok, but how do you define the periodicity in this case? $f$ is real-valued or can be complex? What does happen when range of the local inverse is a bounded set?2013-01-26

0 Answers 0