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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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    $\begingroup$ It is probably not elementary, you can just solve numerically. $\endgroup$ – Jair Taylor Apr 15 at 23:30
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    $\begingroup$ This function is not invertible. You'll have to restrict your focus to a certain set of $x$ values to make it invertible. $\endgroup$ – Cameron Williams Apr 15 at 23:30
  • $\begingroup$ What would those values be? $\endgroup$ – ItIsLastThursday Apr 15 at 23:44
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    $\begingroup$ @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. $\endgroup$ – Adam Latosiński Apr 15 at 23:52
  • $\begingroup$ Can't you use the first few terms of the Maclaurin series expansion and solve algebraically for $x$ ? $\endgroup$ – user1952500 Apr 16 at 2:40
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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 $$x=t\,\frac {1-\frac{2927561 }{27485040}t^2+\frac{193184137 }{138524601600}t^4 } {1-\frac{3614687 }{27485040}t^2+\frac{428067253 }{138524601600}t^4 }$$

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  • $\begingroup$ 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 $\frac{sin(x)}{x}$? $\endgroup$ – ItIsLastThursday Apr 16 at 10:26
  • $\begingroup$ 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? $\endgroup$ – Levi Lesches Apr 16 at 16:03
  • $\begingroup$ @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 ! $\endgroup$ – Claude Leibovici Apr 17 at 1:57
  • $\begingroup$ 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! $\endgroup$ – Levi Lesches Apr 17 at 3:14

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