1

I understand that to find the inverse you switch the $x$ and $y$ then solve for $y$.
However, in this case you end up with $x=3y + \sin(\pi y)$. I know that the inverse exists and is a function since the $f$ is monotonic for all $x$.
Is there any way for me to isolate the $y$ and solve for the inverse?

Bumblebee
  • 18,220
  • 5
  • 47
  • 87
Prandals
  • 135
  • 2
    The function $f$ is not one-to-one on $\mathbb{R}$ and hence not invertible on $\mathbb{R}$. – K. Miller Nov 08 '16 at 18:43
  • 2
    The function is not monotonic. It decreases on very small intervals. I think the function you're looking for is $f(x) = πx + \sin(πx)$ – Kaynex Nov 08 '16 at 18:44
  • If you had said f(x) = 3x + sin (x) you would have had a more interesting question. As it is, we can say that f(x) is not injective (1-1) and so an inverse does not exist, and weasel our way out of answering the question. But, even if you changed the function so that it was 1-1, the answer is unsatisfying. The inverse cannot be expressed as elementary functions. – Doug M Nov 08 '16 at 18:44
  • is it really not one to one? when graphed, it appears so. – Prandals Nov 08 '16 at 18:45
  • it is deceasing around odd integers, and increasing nearly everywhere else. – Doug M Nov 08 '16 at 18:45
  • 2
    $f(0.9)=3.009...$ while $f(1)=3$ and $f(1.2)=3.01...$. – Matthew Conroy Nov 08 '16 at 18:46
  • 1
    It decreases whenever $\cos(πx) < -\frac3π$, which is not often, but it does happen. Note that cos can go down to -1. – Kaynex Nov 08 '16 at 18:47
  • Ok I see, thanks for the help guys – Prandals Nov 08 '16 at 18:48
  • Anyway, assuming $f(x) = πx + \sin(πx)$, the inverse exists, but can't be written out as an elementary function. We can give you a series solution for it, but I'm not sure you'd find that useful. – Kaynex Nov 08 '16 at 18:49
  • This is Kepler’s equation which has known series solutions – Тyma Gaidash Jun 12 '23 at 17:27

1 Answers1

0

$$3x+\sin(\pi x)=y$$ $x\to\frac{t}{\pi}$: $$\frac{3}{\pi}t+\sin(t)=y$$ $$t+\frac{1}{3}\pi\sin(t)=\frac{1}{3}\pi y$$

This isKepler's equation.

see How to solve Kepler's equation $M=E-\varepsilon \sin E$ for $E$?

IV_
  • 6,964