1

I found the equation $$ \frac{(y-x)\sqrt{2}}{2}=\sin\left(\frac{(y+x)\sqrt{2}}{2}\right) $$ for a sine wave rotated $45^\circ$. Based on the shape, I know the relation is a one to one function. It stands to reason, then, that you could isolate y and describe it in terms of $x$. The best I could do with my knowledge of algebra was the infinite iteration: $$ y=x+\sqrt{2}\sin(x\sqrt{2}+\sin(x\sqrt{2}+\sin(x\sqrt{2}+...))). $$

Is there a way to derive a simpler expression?

Aaron Hendrickson
  • 6,042
Nate
  • 11
  • 1
    If it does indeed describe a one to one function, why should that imply an elementary solution for $y$? Consider $y=xe^x$ which is one to one for $x>0$ but whose inverse is described by the nonelementary Lambert W function. – Golden_Ratio Mar 02 '22 at 19:19
  • Have a look at the solutions https://www.wolframalpha.com/input?i=ContourPlot%5B%28-x+%2B+y%29%2FSqrt%5B2%5D+-+Sin%5B%28x+%2B+y%29%2FSqrt%5B2%5D%5D+%3D%3D+0%2C+%7Bx%2C+-15%2C++++15%7D%2C+%7By%2C+-15%2C+15%7D%5D – Claude Leibovici Mar 03 '22 at 08:09

1 Answers1

2

$$\frac{y-x}{\sqrt{2}}=\sin \left(\frac{y+x}{\sqrt{2}}\right)\tag 1$$

Let $$x=\rho \,\sqrt 2\,\cos^2(\theta)\qquad \text{and} \qquad y=\rho \,\sqrt 2\,\sin^2(\theta)\tag 2$$ Equation $(1)$ becomes $$\rho \cos (2 \theta)+\sin (\rho)=0 \tag 3$$ the contour plot of which being interesting (have a look here).

Equation $(3)$ is explicit in $\theta$ but implicit in $\rho$. So, for a given value of $\rho$, we have the various possible values of $\theta$ then $(x,y)$.

But, equation $(3)$ also leads to the parametric equation $$x=\frac{\rho -\sin (\rho )}{\sqrt{2}}\qquad\qquad\qquad y=\frac{\rho +\sin (\rho )}{\sqrt{2}}$$

So, for example, the arc length $$L=\int_0^t \sqrt{1+\cos ^2(\rho )}\,d\rho=\sqrt{2}\, E\left(t \left|\frac{1}{2}\right.\right)$$

Similarly, the surface area under the curve $$A=\int_0^t \frac{1}{2} (\rho +\sin (\rho )) (1-\cos (\rho ))\,d\rho=\frac{1}{8} \left(2 t^2-4 t \sin (t)-8 \cos (t)+\cos (2 t)+7\right)$$

Claude Leibovici
  • 260,315