Is there a closed form for $\sin(x +\cos(x + \sin(x + \cos(x + \cdots))))$?

Question

Is there a closed form expression to represent the function obtained by taking nested repeating sums of alternating sin and cos functions?

Something looking like this:

$$\sin(x +\cos(x + \sin(x + \cos(x + \cdots))))$$

I guess it could be defined this way:

$$L(x)=\sin(x+\cos(x+L(x)))$$

I have plotted several iterations of the function here: https://www.desmos.com/calculator/h2aj27gveg

Although you can't algebraically solve for it using that definition, maybe there's some other clever way to get to a closed form expression I'm overlooking.

If there's not a closed form solution, is there a name for the numerical approximation of this function? It looks like it converges based on the graphs.

Answer 1

If the iterations $$y_{n+1}=\sin(x+\cos(x+y_{n}))$$

converge, they converge to some $y$ that satisfies

$$y=\sin(x+\cos(x+y)).$$

This is the implicit equation of the curve.

There is no analytical expression of $y(x)$ nor $x(y)$ as the equation is highly transcendental.

For rigor, analysis of pointwise convergence using the fixed-point theorem should be carried out. It seems that given the contracting nature of the sine and cosine, convergence is granted everywhere.

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You can get faster convergence by using Newton's iterations instead of simple fixed-point. You obtain the value of the derivative of $y$ by implicit derivation.

Answer 2

There is almost likely no closed form. However, to find the Fourier series for $y=L(x)$ in $y=\sin(x+\cos(x+y))$, substitute $y=\frac\pi2-w-x$, set $x=a$, and subtract $\sin(a)$ so the curve passes through $(\pm\pi,\pm\pi)$:

$$f(w)=\sin(\sin(w)+a)+w-\sin(a)=\frac\pi2-x-\sin(a)$$

Setting up the Fourier series for $g(x)=f^{-1}(x)$, substituting $g(t)\to t$, and integrating by parts gives:

$$g(x)=\sum_{n\in\Bbb Z}a_n e^{i nx},a_n=\frac1{2\pi}\int_{-\pi}^\pi g(t)e^{-i nt}dt=\frac in\cos(\pi n)-\frac in\int_{-\pi}^\pi e^{-inf(t)}dt$$

One extracts $a_0=\frac1{2\pi}\int_{-\pi}^\pi t df(t)=\lim\limits_{n\to0}a_n $ becoming $\sin(a)-\frac1{2\pi}\int_{-\pi}^\pi\sin(\sin(t)+a)dt=\sin(a)(1-J_0(1))$, evaluates $\sum\limits_{n\in\Bbb Z/0}\frac ine^{inx}\cos(\pi n)=x$, and takes the real part. Thereafter, a Jacobi Anger expansion is applied with the Bessel function of the first kind $J_n(x)$:

$$g(x)=\sin(a)(1-J_0(1))+x-\operatorname{Re}\sum_{n\in\Bbb Z/0}\sum_{k\in\Bbb Z}\frac{iJ_k(-n)}{2\pi n}e^{in(\sin(a)+x)} \int_{-\pi}^\pi e^{i(k(\sin(t)+a)-n t)}dt$$

The inner integral gives $J_n(x)$:

$$\boxed{f(x)=\sin(\sin(x)+a)+x-\sin(a)\implies f^{-1}(x)=\sin(a)(1-J_0(1))+x+\sum_{n\in\Bbb Z/0}\sum_{m\in\Bbb Z}\frac{(-1)^k}nJ_k(n)J_n(k)\sin(nx+n\sin(a)+ak)}$$

Since $L(x)=\frac\pi2-x-g\left(\frac\pi2-x-\sin(a)\right)$, the Fourier series is therefore:

$$\bbox[2px, border: 3px blue groove]{y=\sin(x+\cos(x+y))\implies y=L(x)=\sin(x)J_0(1)-\sum_{n\in\Bbb Z/0}\sum_{m\in\Bbb Z}\frac{(-1)^k}nJ_k(n)J_n(k)\sin\left(x(k-n)+\frac{\pi n}2\right)}$$

shown here by plotting $y=L(x)$ and the truncated series: