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I've wanted to know the solution to this for years since I couldn't figure out how to get the answer, not even from wolfram alpha and this has also made me question whether if it was even possible to make t the subject of this equation. In other words, I've been trying to change it from:

$$p = s(\left \lfloor t\right \rfloor+\frac{\sin(\frac{\pi\left \lfloor t\right \rfloor}{t})}{\sin(\frac{\pi}{t})})$$

To:

$t =\text{equation in terms of }p\text{ and }s$

Тyma Gaidash
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HarryXiro
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    Have you drawn the graphical repesentation of your function ? Is it bijective ? – Jean Marie Jun 04 '23 at 17:50
  • @HarryXiro You just need to invert $\frac{\sin(nx)}{\sin(x)}$, like in this graph. There could be an answer, but please include more detail like what you have tried so far. – Тyma Gaidash Jun 05 '23 at 19:05
  • I have made a graphical representation of the equation. I know when t is an integer that t = p, but when t is not an integer then t does not equal p. I don't know how to solve equations with the floor function in it, so I quickly got stuck trying to solve it algebraically. Here's the graph of the equation https://www.desmos.com/calculator/bpgnmuy944 – HarryXiro Jun 08 '23 at 05:34

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As shown by this graph:

$$p = s\left(\left \lfloor t\right \rfloor+\frac{\sin(\frac{\pi\left \lfloor t\right \rfloor}{t})}{\sin(\frac{\pi}{t})}\right)\iff\frac ps-k=\frac{\sin\left(\frac{\pi k}{x}\right)}{\sin\left(\frac\pi x\right)},0<x-k<1$$

Rewriting with a Chebyshev U series:

$$\frac ps-k=U_{k-1}\left(\cos\left(\frac\pi x\right)\right)=\sum_{n=0}^{\left\lfloor\frac{k-1}2\right\rfloor}\frac{(-1)^n \Gamma(k-n)}{n!\Gamma(k-2n)} \left(2\cos\left(\frac\pi x\right)\right)^{k-2n-1} $$

Now it is just about inverting a $\cos\left(\frac\pi x\right)$ polynomial of degree $\left\lfloor\frac{k-1}2\right\rfloor$. There are elementary inverses for the $k=1,\dots 5,7,9$ cases. However, cases, like when $k=7$ and $k=9$ do not show major simplifications in radical form.

Тyma Gaidash
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  • Yes, I would like a series solution to that since it will probably get us closer to solving the problem. Also, I tried plugging the equation in this website https://www.mathenomicon.net/mathematics-parabola/factoring/make-x-the-subject-calculator.html but it doesn't have the floor function as an option. Do you know if there are other websites that do this that have the floor function? – HarryXiro Jun 08 '23 at 14:55
  • @HarryXiro You do not need the floor function. If you know $k<t<k+1,k\in\Bbb Z$, then $\frac ps-k=\frac{\sin\left(\frac{\pi k}t\right)}{\sin\left(\frac\pi t\right)}$. Anyways your site does not solve the equation – Тyma Gaidash Jun 08 '23 at 15:35
  • When you answered the question, what does n represent? Also when you answered the question, it didn't appear to solve for t. Is that because it's not possible to solve for t? – HarryXiro Jun 09 '23 at 03:55
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    @HarryXiro $n$ is a sum index. This graph shows the inverse for $2<x<4$. Do you understand the process of how this inverse in the graph was found? – Тyma Gaidash Jun 09 '23 at 12:00
  • I don't understand the process of how the inverse was found but I know now that you asked Wolfram Alpha to invert the series and get solutions for t when $2 < p < 3$ and $3 < p < 4$, thank you for that. I don't understand how to get an equation of the inverse of the Chebyshev U series that works for any value of p. Is there an equation that is the inverse of the Chebyshev U series that works for any value of p, or do you have to use Wolfram Alpha to invert the series and spit out an equation that only works when t is bound between 2 values? – HarryXiro Jun 10 '23 at 04:56
  • Sorry I meant to say "when p is bound between 2 values" instead of "when t is bound between 2 values" – HarryXiro Jun 11 '23 at 02:03
  • This Wikipedia page looks like it shows the equation for the inverse of the Chebyshev U series. https://en.wikipedia.org/wiki/Chebyshev_polynomials#Explicit_expressions Now it's just editing the equation so that it works for our specific case, which I don't know how to do unfortunately. – HarryXiro Jun 11 '23 at 14:06
  • @HarryXiro That article is for the reciprocal, not inverse function, of a polynomial. It cites this arxiv source – Тyma Gaidash Jun 11 '23 at 14:24
  • Oh sorry, my bad. Anyways, do you know if there is an equation for the inverse function for $U_{k-1}(cos(\frac{\pi}{x}))$ or do you have to ask Wolfram Alpha to invert it for a specific value of k each time? – HarryXiro Jun 11 '23 at 14:46
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    I think if I make a new post asking to solve for the inverse function of $U_{k-1}(cos(\frac{\pi}{x}))$, then someone else might know how to find the inverse function because solving this is a bit too advanced for me. If someone else finds a solution I couldn't have got that solution without your help, big thanks. – HarryXiro Jun 13 '23 at 07:18
  • I found a math document that says it solved the inverse formulas of the Chebyshev polynomials of the first and second kinds: https://hal.science/hal-01705040/document

    Do you know if these formulas are actually the inverse?

     – HarryXiro Jul 08 '23 at 04:01
  • @HarryXiro I saw no inverse function $f^{-1}(x)$ in the article, but I did see $\sum_k a_k=U_k(x)$ and the article wrote $\sum_{k=1}^n U_k(x) c_k=n (2x)^{(2n)}$ meaning the “inverse” probably referred to switching the $U_k(x)$ function from being outside a to being inside a sum. – Тyma Gaidash Jul 08 '23 at 12:15
  • Correction, the degree is $\begin{cases} k-1&k\text{ even}\\frac{k-1}2&k\text{ odd}\end{cases}$ – Тyma Gaidash Dec 17 '23 at 14:25