How would one find the inverse of the function $y=\frac{\sin(x)}{x}$? Here are my steps: $y=\frac{\sin(x)}{x}$, $x=\frac{\sin(y)}{y}$, $xy=\sin(y)$, $\arcsin(xy)=y$, After that step, I can’t find a way to isolate $y$.
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1It is probably not elementary, you can just solve numerically. – Jair Taylor Apr 15 '19 at 23:30
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3This function is not invertible. You'll have to restrict your focus to a certain set of $x$ values to make it invertible. – Cameron Williams Apr 15 '19 at 23:30
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What would those values be? – ItIsLastThursday Apr 15 '19 at 23:44
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1@ItIsLastThursday By analyzing the derivative of $\frac{\sin x}{x}$ it's easy to see that it is monotonicaly decreasing on an interval $[0,a]$ where $a$ is the smallest positive solution of equation $a = \tan a$. This interval is one of the possible domains you may restrict the function to make it injective, and thus invertible. But any set on which the function will be injective works. – Adam Latosiński Apr 15 '19 at 23:52
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Can't you use the first few terms of the Maclaurin series expansion and solve algebraically for $x$ ? – user1952500 Apr 16 '19 at 02:40
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1Possible duplicate of https://math.stackexchange.com/questions/2175174/inverse-sinc-approximation ? – Florian Apr 16 '19 at 08:10
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See inverse function of sinc – Тyma Gaidash Sep 18 '22 at 20:53
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If you are concerned by the inverse function, you could use the usual Taylor series of $\sin(x)$ and use series reversion to get $$x=t+\frac{1}{40}t^3+\frac{107 }{67200}t^5+\frac{3197 }{24192000}t^7+\frac{49513 }{3973939200}t^9+O\left(t^{11}\right)$$ where $t=\sqrt{6(1-y)}$.
To see how good or bad it is, give $x$ a value from which you obtain $y$ and recompute $x$ from the expansion. Below are given some results using the above truncated series $$\left( \begin{array}{ccc} x_{given} & y & x_{calc} \\ 0.0 & 1.00000 & 0.00000 \\ 0.1 & 0.99833 & 0.10000 \\ 0.2 & 0.99335 & 0.20000 \\ 0.3 & 0.98507 & 0.30000 \\ 0.4 & 0.97355 & 0.40000 \\ 0.5 & 0.95885 & 0.50000 \\ 0.6 & 0.94107 & 0.60000 \\ 0.7 & 0.92031 & 0.70000 \\ 0.8 & 0.89670 & 0.80000 \\ 0.9 & 0.87036 & 0.90000 \\ 1.0 & 0.84147 & 1.00000 \\ 1.1 & 0.81019 & 1.09997 \\ 1.2 & 0.77670 & 1.19995 \\ 1.3 & 0.74120 & 1.29989 \\ 1.4 & 0.70389 & 1.39980 \\ 1.5 & 0.66500 & 1.49964 \\ 1.6 & 0.62473 & 1.59937 \\ 1.7 & 0.58333 & 1.69896 \\ 1.8 & 0.54103 & 1.79834 \\ 1.9 & 0.49805 & 1.89741 \\ 2.0 & 0.45465 & 1.99608 \\ 2.1 & 0.41105 & 2.09421 \\ 2.2 & 0.36750 & 2.19165 \\ 2.3 & 0.32422 & 2.28819 \\ 2.4 & 0.28144 & 2.38362 \\ 2.5 & 0.23939 & 2.47768 \\ 2.6 & 0.19827 & 2.57009 \\ 2.7 & 0.15829 & 2.66053 \\ 2.8 & 0.11964 & 2.74866 \\ 2.9 & 0.08250 & 2.83412 \\ 3.0 & 0.04704 & 2.91653 \end{array} \right)$$
For sure, we couls make it better using more terms.
Another possibility could be to tansform the above series as a Padé approximant to get $$x=t\,\frac {1-\frac{2927561 }{27485040}t^2+\frac{193184137 }{138524601600}t^4 } {1-\frac{3614687 }{27485040}t^2+\frac{428067253 }{138524601600}t^4 }$$
Claude Leibovici
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What is series reversion/how did you get that first series from the taylor series of $sin(x)$? Wouldn’t I need to use the taylor series of $\frac{sin(x)}{x}$? – ItIsLastThursday Apr 16 '19 at 10:26
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I tried this for sinc(90/2) and it didn't work -- I got 0.9898 for the Padé approximation and 3.026 for the Taylor series method. Why didn't I get 45? – Levi Lesches Apr 16 '19 at 16:03
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@LeviLesches. We work with radians not degrees. So $\text{sinc}\left(\frac{\pi }{4}\right)=\frac{2 \sqrt{2}}{\pi }=0.900316$. Using the Taylor series, we get $0.785398$; using the Padé approximation, we get the same number and $\frac \pi 4=0.785398$ too ! – Claude Leibovici Apr 17 '19 at 01:57
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Oh wow thank you so much I'm usually more careful about this stuff, but now that I converted to radians the formulas work! Thank you so much! – Levi Lesches Apr 17 '19 at 03:14
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Ok, I know this is very late, but we still have more questions, and I was unable to find a way to PM you. Could you please explain "Series Reversion", explain the substitution involving "t", the big "O" (O(n)) notation, how the taylor series is related to the series you gave above, and how you get the approximation from all of the information provided. If you find a way to PM either me or @LeviLesches, please do so. Thank you! – ItIsLastThursday Aug 06 '19 at 01:26
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@ItIsLastThursday. For series reversion, have a look at http://mathworld.wolfram.com/SeriesReversion.html. This a quite simple. When you make it, you find terms $\sqrt{6(1-y)}$ which I replaced by $t$ for more simple expressions for $x$. Transforming series to Padé approximants is quite simple since it is almost the same problem since you know all derivatives required from the series. Is this clearer ? – Claude Leibovici Aug 06 '19 at 02:18
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Yes, thank you, and in this case, does $y=sin(x)$ or does $y=sinc(x)$? – ItIsLastThursday Aug 06 '19 at 12:02
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Elementary functions:
$$f(x)=\frac{\sin(x)}{x}$$ $$f(x)=-\frac{1}{2x}i(e^{ix}-e^{-ix})$$
We see, this function is an algebraic function in dependence of both $x$ and $e^x$. Liouville proved that such kind of functions (over a complex domain without isolated points) don't have (partial) inverses that are elementary functions: How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
Lambert W, Generalized Lambert W:
The defining equation for the inverse $\frac{\sin(x)}{x}$ can be rearranged to a polynomial equation of both $x$ and $e^x$ which is quadratic for $e^x$. This equation is therefore not in a form to apply Lambert W or Generalized Lambert W.
"Leal-functions":
The partial inverses of the function mentioned in the question can be represented in terms of the function $\text{Lcsc}$ presented in [Vazquez-Leal et al. 2020].
$$\frac{\sin(x)}{x}=y$$ $\sin(x)=\frac{1}{\csc(x)}$: $$\frac{1}{x\csc(x)}=y$$ $$x\csc(x)=\frac{1}{y}$$ $$x=\text{Lcsc}\left(\frac{1}{y}\right)$$
We can take the "Leal functions" as closed-form functions because some of their algebraic properties and their applicability for some other kinds of equations are presented in the cited article.
IV_
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