Why is $y = \sin \left(\frac{y^s}{t} - k \right)$ outstandingly approximated by $y = - \sin k$, even for large $k$?

Question

Equations of the form $$y = \sin \left(\frac{y^s}{t} - k \right)$$ are surprisingly well approximated as $$y = - \sin k$$ for a large range of $s$ $t$ and $k$ values:

Why? I understand why this is so for $k \approx 0$. But why is it true for moderate or large $k$? It's true that since $|y| \leq 1$, then $\frac{y^s}{t}$ is bounded. But the nature of $\sin$ is that the change in the $\sin$ of a quantity is a function of the absolute delta, not the relative (percent) change. And an absolute change of e.g. $1/3$ (as shown in the plot) is enough to change the $\sin$ by close to $1/3$.

Here's a very rough, imprecise, initial crack: Let $$u := \frac{y^s}{t} - k.$$ When $y \approx 0$, then $\sin u \approx - \sin k$, as expected. What about larger $y$? Well, as $|y| \to 1$, then $\frac{dy}{dk} \to 0$, so the discrepancies aren't "important".

In other words: When $|\frac{y^s}{t}|$ is small, the approximation is clearly good. And when $|\frac{y^s}{t}|$ is comparatively large (e.g. $1/3$), then $|y|$ needs to be near unity, and so the curve goes flat, such that absolute changes to the input don't change the output very much. I believe this intuition is correct -- though I don't know how to make it precise.

This observation arose investigating The surface $x + y + z = \sin xyz$: Can it be visualized? , where it seems that $$ h = \frac{1}{\sqrt 3} \sin \left( \frac{\sqrt 3}{18}(2h^3 - 3hr^2 - \sqrt 2 r^3 \cos 3 \theta) \right) $$ is well approximated by $$ h \approx - \frac{1}{\sqrt 3} \sin \left( \frac{\sqrt 6}{18} r^3 \cos 3 \theta) \right).$$ Proving, bounding, or otherwise explaining this is well within scope of this question.

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By Claude Lebovici's request, here is a zoomed in plot (blue is $y = -\sin x$, red is $y = \sin (y^2/3 - x)$).

Answer 1

Short answer:

Because $|y|\le 1$,

For large $s,t$, $|y^s/t|$ becomes very tiny (fraction raised to a big power divided by a big number).

$y=\sin\!\left(\frac{y^s}{t}-x\right)$ reduces to $y \approx \sin(-x)= -\sin x.$

Thus the curve must closely follow $-\sin x$ for large $s,t$.

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Long Answer:

Quantitative justification via mean squared error (MSE).

Let the curve $$y = \sin\!\left(\frac{y^{s}}{t} - x\right)$$ be described implicitly by a function $y = f(x)$.

Now this implicit equation is multivalued for some conditions of s,t , but for brevity we can conclude that it isn't multivalued for the large values of s,t under study $(s>1,t>1)$.

In particular, it isn't multivalued if, $$ t^{2} \;>\; s\left(1 - \frac{1}{s}\right)^{\,s-1}, \qquad s>1, $$

Within the domain $0 \le x \le 2\pi$, the function $y = f(x)$ can be represented by three monotone branches.
Solving the defining equation for $x$ in terms of $y$ gives the following parametrizations:

\begin{aligned} x &= \frac{y^{s}}{t} - \arcsin(y), && 0 \le x \le \frac{\pi}{2} - \frac{1}{t}, \\[4pt] x &= \frac{y^{s}}{t} + \pi + \arcsin(y), && \frac{\pi}{2} - \frac{1}{t} \le x \le \frac{3\pi}{2} + \frac{1}{t}, \\[4pt] x &= \frac{y^{s}}{t} - \arcsin(y) + 2\pi, && \frac{3\pi}{2} + \frac{1}{t} \le x \le 2\pi . \end{aligned}

These three pieces together cover the entire graph of $y = f(x)$ over one full period.

Now, the mean squared error (MSE) of the approximation $y \approx -\sin x$ over one full period is $$ \mathrm{MSE} = \frac{1}{2\pi} \int_{0}^{2\pi} \bigl( f(x) + \sin x \bigr)^{2} \, dx . $$ Expanding the square in the integrand, we can write

$$ \mathrm{MSE} = \frac{1}{2\pi} \int_{0}^{2\pi} f^2(x) \, dx + \frac{1}{2\pi} \int_{0}^{2\pi} 2 f(x) \sin x \, dx + \frac{1}{2\pi} \int_{0}^{2\pi} \sin^2 x \, dx. $$

For convenience, we define

$$ I_1 := \frac{1}{2\pi} \int_{0}^{2\pi} f^2(x) \, dx, \qquad I_2 := \frac{1}{2\pi} \int_{0}^{2\pi} 2 f(x) \sin x \, dx, \qquad I_3 := \frac{1}{2\pi} \int_{0}^{2\pi} \sin^2 x \, dx. $$

Then $\mathrm{MSE} = I_1 + I_2 + I_3.$

We now split $I_1$ into three contributions, corresponding to the three monotone branches of $y=f(x)$:

$$ I_1=I_{1a}+I_{1b}+I_{1c}, $$ where $$ I_{1a} := \frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}-\frac{1}{t}} f^2(x) \, dx = \frac{1}{2\pi} \int_{0}^{-1} y^2 \, \frac{dx}{dy} \, dy, $$ $$ I_{1b} := \frac{1}{2\pi} \int_{\frac{\pi}{2}-\frac{1}{t}}^{\frac{3\pi}{2}+\frac{1}{t}} f^2(x) \, dx = \frac{1}{2\pi} \int_{-1}^{1} y^2 \, \frac{dx}{dy} \, dy, $$ $$ I_{1c} := \frac{1}{2\pi} \int_{\frac{3\pi}{2}+\frac{1}{t}}^{2\pi} f^2(x) \, dx = \frac{1}{2\pi} \int_{1}^{0} y^2 \, \frac{dx}{dy} \, dy. $$

For each branch, $dx/dy$ is given by differentiating the corresponding parametrization:

$$ \frac{dx}{dy} = \begin{cases} \frac{d}{dy}\Big(y^s/t - \arcsin(y)\Big) = \frac{s y^{s-1}}{t} - \frac{1}{\sqrt{1-y^2}}, & \text{for branch } I_{1a},\\[2mm] \frac{d}{dy}\Big(y^s/t + \pi + \arcsin(y)\Big) = \frac{s y^{s-1}}{t} + \frac{1}{\sqrt{1-y^2}}, & \text{for branch } I_{1b},\\[1mm] \frac{d}{dy}\Big(y^s/t - \arcsin(y) + 2\pi\Big) = \frac{s y^{s-1}}{t} - \frac{1}{\sqrt{1-y^2}}, & \text{for branch } I_{1c}. \end{cases} $$

Now, $$ I_{1a} = \frac{1}{2\pi} \int_{0}^{-1} \frac{s y^{s+1}}{t} - \frac{y^2}{\sqrt{1-y^2}} \, dy, \qquad I_{1b} = \frac{1}{2\pi} \int_{-1}^{1} \frac{s y^{s+1}}{t} + \frac{y^2}{\sqrt{1-y^2}} \, dy, \qquad I_{1c} = \frac{1}{2\pi} \int_{1}^{0} \frac{s y^{s+1}}{t} - \frac{y^2}{\sqrt{1-y^2}} \, dy. $$

Evaluating these integrals, we see that the contributions from the $\frac{s y^{s+1}}{t}$ terms cancel across the three branches due to symmetry,
and the remaining terms sum exactly to give $$ I_1 = \frac{1}{2}. $$

Also, $$I_3 := \frac{1}{2\pi} \int_{0}^{2\pi} \sin^2 x \, dx= \frac{1}{2}$$

We now split $I_2$ into three contributions, corresponding to the three monotone branches of $y=f(x)$:

$$ I_2 = I_{2a} + I_{2b} + I_{2c}, $$ where $$ I_{2a} := \frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}-\frac{1}{t}} 2f(x)\sin(x) \, dx = \frac{1}{\pi} \int_{0}^{-1} y\sin(x) \, \frac{dx}{dy} \, dy, $$ $$ I_{2b} := \frac{1}{2\pi} \int_{\frac{\pi}{2}-\frac{1}{t}}^{\frac{3\pi}{2}+\frac{1}{t}} 2f(x)\sin(x) \, dx = \frac{1}{\pi} \int_{-1}^{1} y\sin(x) \, \frac{dx}{dy} \, dy, $$ $$ I_{2c} := \frac{1}{2\pi} \int_{\frac{3\pi}{2}+\frac{1}{t}}^{2\pi} 2f(x)\sin(x) \, dx = \frac{1}{\pi} \int_{1}^{0} y\sin(x) \, \frac{dx}{dy} \, dy. $$

Now, $$ I_{2a} = \frac{1}{\pi} \int_{0}^{-1} \left(\frac{s y^{s}}{t} - \frac{y}{\sqrt{1-y^2}}\right)\sin\left(\frac{y^s}{t} - \arcsin(y)\right) \, dy$$ $$I_{2b} = \frac{1}{\pi} \int_{-1}^{1} \left(\frac{s y^{s}}{t} + \frac{y}{\sqrt{1-y^2}}\right)\left(-\sin\left(\frac{y^s}{t} + \arcsin(y)\right)\right) \, dy$$ $$I_{2c} = \frac{1}{\pi} \int_{1}^{0} \left(\frac{s y^{s}}{t} - \frac{y}{\sqrt{1-y^2}}\right)\sin\left(\frac{y^s}{t} - \arcsin(y)\right) \, dy. $$

Similar to before, when $I_{2a}$ and $I_{2c}$ are combined and added with $I_{2b}$, Some integrals cancel out, to finally yield, $$I_2 := I_{2a}+I_{2b}+I_{2c}= \frac{-2}{\pi} \int_{-1}^{1} \frac{sy^s}{t}\sqrt{1-y^2} \sin\left(\frac{y^s}{t}\right) \, dy+\frac{-2}{\pi} \int_{-1}^{1} \frac{y^2}{\sqrt{1-y^2}} \cos\left(\frac{y^s}{t}\right) \, dy$$

And finally, $$MSE=I=I_1+I_2+I_3$$ $$\implies MSE=\frac{1}{2}+\frac{-2}{\pi} \int_{-1}^{1} \frac{sy^s}{t}\sqrt{1-y^2} \sin\left(\frac{y^s}{t}\right) \, dy+\frac{-2}{\pi} \int_{-1}^{1} \frac{y^2}{\sqrt{1-y^2}} \cos\left(\frac{y^s}{t}\right) \, dy+\frac{1}{2}$$ $$\implies MSE=1+\frac{-2}{\pi} \int_{-1}^{1} \frac{sy^s}{t}\sqrt{1-y^2} \sin\left(\frac{y^s}{t}\right) \, dy+\frac{-2}{\pi} \int_{-1}^{1} \frac{y^2}{\sqrt{1-y^2}} \cos\left(\frac{y^s}{t}\right) \, dy$$

Numerically, it can be observed that the MSE is typically close to $0$.

Also, For large $s,t$, note that $|y|\le1$ implies $\frac{y^s}{t}\to 0$, hence, $$ MSE \approx 1-0-\frac{2}{\pi}\int_{-1}^{1} \frac{y^2}{\sqrt{1-y^2}} \, dy=0$$

Answer 2

You really should just look at $$y=\sin\left(\frac{y^s}t-k\right)=\sin\frac{y^s}t\cos k-\cos\frac{y^s}t\sin k$$

You have already correctly realised that $y=\sin$ something means that $|y|\leqslant 1$, so that $$\forall s\in\mathbb R^+\forall|t|>1\qquad\left|\dfrac{y^s}t\right|\ll\dfrac\pi2$$ and then the behaviour that you have noticed is quite expected, due to small angle approximations.

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In fact, you know a lot more. Your equation is really $$y=-\sin(k-\delta)\qquad\bigwedge\qquad\delta=\frac{y^s}t$$ and so the graph is really all about horizontal shifts of the sine function. It cannot do anything much more than that, and for fixed $s$ and $t$, you should be able to see that $\delta=0$ whenever $y=0$ and that happens $\forall n\in\mathbb Z\qquad k=n\pi$

This then implies the much stronger behaviour, namely, that you really have a Fourier series in terms of $k$ because the equation is strictly periodic with period $2\pi$ in $k$ just as $\sin k$ is. Of course, the particular Fourier series will depend upon $s$ and $t$, but you get the gist.

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The picture will only change if $s<0$ and/or $|t|<1$, in which case things will be much more complicated.

Answer 3

The function is obviously $2\pi$-periodic. If you already proved that it is decreasing on $I=(\frac{-\pi}2+\frac 1t,\frac{\pi}2-\frac 1t)$ and increasing on $J=(\frac{\pi}2-\frac 1t,\frac{3\pi}2-\frac 1t)$, it is easy to see that the inverse functions are $g:y\mapsto \frac{y^s}t-\arcsin(y)$ on $I$ and $h:y\mapsto \pi-\frac{y^s}t+\arcsin(y)$ on $J$.

The graphs of $g$ and $h$ converge to those of the inverse function of $-\sin$ over $(-1,1)$ as $t$ tends to $+\infty$, with a gap less than $\frac 1t$. Therefore the graph of $f$ gets also closer to the graph of $-\sin$.

Answer 4

When $s$ is much larger than $t$ (for example, $s=50,$ $1 < t < 4$) there's a peculiar effect. Plotting $y$ versus $k,$ each crest and trough of the sine wave is replaced by a horizontal spike. Essentially, the minimum and maximum points are displaced a horizontal distance $1/t$ and the end of the spike projects a little farther. Obviously the horizontal displacement is less when $t$ is also large. And of course for large $s$ the effect is not easily visible until $y$ is close to $1.$

You then have a situation where there are points of $y = \sin\!\left(\frac{y^{s}}{t} - k\right)$ and $y = -\sin k$ very close together for most values of $k$ and not terribly far away for any value of $k,$ but there are also parts of $y = \sin\!\left(\frac{y^{s}}{t} - k\right)$ that are much further from $-\sin k.$

In particular, while $y = -\sin k$ determines $y$ as a function of $k,$ the equation $y = \sin\!\left(\frac{y^{s}}{t} - k\right)$ does not always specify $y$ as a function of $k,$ even if we specify that $s$ and $t$ are large.

Answer 5

For the approximation of $h$.

Consider the function $$f(h)= \frac{1}{\sqrt 3} \sin \left( \frac{\sqrt 3}{18}(2h^3 - 3hr^2 - \sqrt 2 r^3 \cos (3 \theta)) \right)-h$$ and expand it as a series around $h=0$.

Using $$x=\frac{r^3 \cos (3 \theta )}{3 \sqrt{6}}$$ $$f(h)=-\frac 1 {\sqrt 3}\sin(x)-\left(1+\frac{1}{6} r^2 \cos(x)\right) h+\frac{ r^4 \sin (x)}{24 \sqrt{3}}h^2+O(h^3)$$

So, a "better" approximation could be $$\large h \sim -\frac{2 \sqrt{3} \sin \left(\frac{r^3 \cos (3 \theta )}{3 \sqrt{6}}\right)}{6+r^2 \cos \left(\frac{r^3 \cos (3 \theta )}{3 \sqrt{6}}\right)}$$