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I'm trying to find the inverse function of $$U_{k-1}(\cos(\frac{\pi}{x}))=\sum_{n=0}^{\left\lfloor\frac{k-1}2\right\rfloor}\frac{(-1)^n \Gamma(k-n)}{n!\Gamma(k-2n)} \left(2\cos\left(\frac\pi x\right)\right)^{k-2n-1}$$ When I type it into Wolfram Alpha, it can't figure out the inverse.enter image description here I have no idea how to solve this so I would really like to know the solution.

HarryXiro
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    Why do you expect this function to admit an inverse? Or do you have a certain domain in mind, on which this function is injective? – frog Jun 16 '23 at 08:48
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    I do think there may be an inverse for this function because this equation is equal to another equation from another post I've made. If you read through my other post you might have a better understanding. Here's that post: https://math.stackexchange.com/questions/4712552/p-s-left-lfloor-t-right-rfloor-frac-sin-frac-pi-left-lfloor-t-right – HarryXiro Jun 16 '23 at 10:41
  • @frog I found this document and it mentions the inverse equation of the Chebyshev polynomial second kind https://hal.science/hal-01705040/document. Do you know if it mentions the inverse equation of the Chebyshev polynomial second kind in this article because that can help to solve my variation of it. – HarryXiro Jun 26 '23 at 14:27
  • @ТymaGaidash I'm not too sure how to use the equation in the link to solve $\frac{\sin(a x)}{\sin(x)}=b$, is it ok if you explain how to solve it using the method in the link? – HarryXiro Dec 10 '23 at 11:22

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