# Is there any use for $\sin(\sin x)$?

The convention that $$\sin^2 x = (\sin x)^2$$, while in general $$f^2(x) = f(f(x))$$, is often called illogical, but it does not lead to conflicts because nobody uses $$\sin(\sin x)$$.

But is this really true? Or is there a real-world application in which $$\sin(\sin x)$$ occurs? Or maybe something a bit more general, like $$\sin(C \sin x)$$ for some constant $$C \neq 0$$?

• Bessel functions are expressed as integrals of this form, for example $\int_0^\pi \cos(\cos x)dx=\pi J_0(1);$ I would actually never use the notation $f^2(x)$ for $f(f(x))$, it is too confusing. Feb 1, 2021 at 18:27
• It is well known that if $x > 0$ is very small, then $\sin(x) \approx x$. Therefore, if you have any formula involving $\sin(x)$ and you can assume $x$ is very small, then you can replace $\sin(x)$ with $\sin(\sin(x))$ in your formula. :) (Sorry, I couldn't help myself -- but in all seriousness, welcome to MathOverflow.) Feb 1, 2021 at 18:38
• I'm not sure whether this really counts, but if you consider $\sinh(\sinh(x)) = -i\sin(\sin(i x))$ instead, you get the exponential generating function for set partitions with an odd number of parts, each of which has odd cardinality. Feb 1, 2021 at 20:20
• @CarloBeenakker If one works in discrete dynamics, it's completely standard for $f^n$ to denote the $n$th iterate. Sometimes people use $f^{\circ n}$ to avoid ambiguity, but in a paper that all about iteration, it really is very convenient to use exponentiaion. Further, if one views the set of functions $X\to X$ as a semi-group, then the $n$th power of an element is most naturally the $n$th iterate; indeed, "squaring" in any other way doesn't really make sense unless the set $X$ has a multiplication map. :) Feb 1, 2021 at 21:37
• As a side comment, people still use the notation $\log^2 x$ even though $\log \log x$ arises frequently. So even if $\sin \sin x$ somehow becomes common, I expect that the notation $\sin^2 x$ will continue to be used. Feb 1, 2021 at 23:23

The intensity of light diffracted at a slit as a function of the angle actually involves a term $$\sin\left(\frac{\alpha\beta}{2}\sin(\theta)\right)$$, see

https://en.wikipedia.org/wiki/Fraunhofer_diffraction

(I'm no physicist at all, but this has been stuck in my head since high school just because it is such an unusual term to encounter naturally)

• Could you point me to the precise location of this formula? Feb 2, 2021 at 9:10
• It's in the subsection "Single-slit diffraction of Electric Field using Huygens' Principle" of the article I linked. Feb 2, 2021 at 12:01
• Direct link to "Single-slit diffraction using Huygens' Principle" referenced by @‍AchimKrause. Oct 1, 2021 at 21:49

Since $$\sinh(x) = i\sin(i x)$$ is the odd part of the exponential function, we can interpret it (for example within the framework of combinatorial species) as the (exponential) generating function for sets of odd size.

Thus, $$\sinh(\sinh(x)) = -i\sin(\sin(ix))$$ is the (exponential) generating function for set partitions with an odd number of parts, each of which has odd cardinality.

We can slightly refine this by interpreting $$\sin(\omega\sin(x))$$ as the generating function of a weighted species, giving each set partition the weight $$(-1)^{n-1} w^b$$, where $$b$$ is the number of blocks and $$n$$ is the size of the ground set.

Similarly, $$\cosh(\sinh(x))$$ is the generating function for set partitions with an even number of parts, all of which are of odd cardinality, and $$\cosh(\cosh(x)-1)$$ is the generating function for set partitions with an even number of (nonempty) parts, all of which are of even cardinality. Note however, that the coefficients of $$\cos(\cos(x)-1)$$ and $$\cosh(\cosh(x)-1)$$ are very different, whereas the coefficients of $$\sin(\sin(x))$$ and $$\sinh(\sinh(x))$$ only differ in sign.

As an aside, $$\sin(\sin(\cdot))$$ satisfies a nice differential equation: $$(f^2-1)^2 (f''' + f') - 3 (f^2-1) f f' f'' + (2f^2+1) f'^3 = 0$$ while $$\sinh(\sinh(\cdot))$$ satisfies: $$(f^2+1)^2 (f''' - f') - 3 (f^2+1) f f' f'' + (2f^2-1) f'^3 = 0$$

• For $\sin(\sin(\cdot))$ there is also the 2d-order equation $$(f'' - f '' f^2 + f f'^2)^2 - (1 - f^2 - f'^2) (1 - f^2)^2=0$$ which may or may not be nice. Do you know any of the other solutions for any of these equations?
– user44143
Feb 2, 2021 at 15:05
• How did you find this beautiful equation? (I missed it, because my program looks for smaller powers first) No, I don't know of any solutions. It is very interesting that in this case, throwing away the zero coefficients makes the equation harder to find. Feb 2, 2021 at 16:10
• My process was to: 1) write down the formulas for $f, f’, f’’$, 2) define $\cos(x)^2$ in terms of $f$ and $f’$, 3) express $f’’$ in terms of $f$ and this expression for $\cos (x)^2$, 4) simplify
– user44143
Feb 2, 2021 at 17:30

People in complex dynamics consider the behavior of all sorts of functions under iteration. For example, here is the Julia set of $$\sin(z)$$. In that context, it makes perfect sense to talk about $$\sin(\sin(\cdots \sin(z)))$$.

• Technically, shouldn't you write $\sin(\sin(\cdots \sin(z)\cdots))$? Feb 2, 2021 at 12:29
• @YaakovBaruch I have become wild and uncivilized in my year of isolation :). Feb 2, 2021 at 20:53

A frequency-modulated (FM) signal (like those used in FM radio), with a sinusoidal input, can be represented as

$$x_\mathrm{c}(t) = A\cos(\omega_\mathrm{c}t+\beta\sin \omega_\mathrm{m}t), \tag{1}$$

where $$\omega_\mathrm{c}=2\pi f_\mathrm{c}$$ is the carrier angular frequency (e.g. $$f_\mathrm{c}$$ ranges from 87.5 MHz to 108 MHz for the FM radio), $$\omega_\mathrm{m}$$ is the modulating angular frequency (e.g. that of an audio tone) and $$\beta$$ is the modulation index (a normalized measure of the intensity of the modulating signal).

If you decompose (1) with the usual trigonometric identity for the sum of angles you get the term $$\sin(\beta\sin \omega_\mathrm{m}t)$$. This decomposition can be expanded in a Fourier series with coefficients generated by Bessel functions, thus obtaining the spectrum associated to the FM signal.

• It might be fun to interpret this combinatorially. Essentially, set partitions with an even number of blocks, each of which has odd cardinality, and there are singleton blocks of two kinds. Feb 2, 2021 at 9:31
• @MartinRubey, of what is that a combinatorial expansion? The function $x_c$? The coefficients in a Fourier series? Something else? Aug 25, 2021 at 22:45
• Up to sign, it should be the expansion of $x_c$. The cosine is (up to sign) the exponential generating function for sets with an even number of ellements. The sine is (up to sign) the exponential generating function for sets with an odd number of elements. Thus, the composition is the generating function for set partitions with an even number of blocks, all of which have odd cardinality. Aug 26, 2021 at 6:54
• For example, $[t^4] \cos(\sin(t)) = 5/4!$. The set partitions are $\left[\{\{1, 2, 3\}, \{4\}\}, \{\{1, 2, 4\}, \{3\}\}, \{\{1, 3, 4\}, \{2\}\}, \{\{1\}, \{2, 3, 4\}\}, \{\{1\}, \{2\}, \{3\}, \{4\}\}\right]$. Aug 26, 2021 at 6:56
• Adding $t$ amounts to colouring singletons with two colours. For example, $[t^4] \cos(t+\sin(t)) = 24/4!$. The first four of the set partitions above come in two kinds, the final one having all singleton blocks comes in 16 kinds, and indeed $24 = 2\cdot 4 + 16\cdot 1$. Aug 26, 2021 at 7:02

I think Carlo Beenakker's comment deserves to be upgraded to an answer:

Bessel functions have integral representations involving such iterates of trigonometric functions. For example, $$\int_0^1\sin(x\sin(\pi t))dt=\mathbf{H}_0(x)$$ is the zeroth Struve function, one in the crowded family of various Bessel and Bessel-like functions.

• I agree that this should be promoted to an answer, but Carlo only left that comment 1 hour ago and he hasn't been online since an hour ago to see the comment back by the OP saying it is an answer. I think it is important that we leave original commenters a reasonable amount of time to convert their comments to answers, and 1 hour when the commenter is offline seems short. Feb 1, 2021 at 19:58
• @AlecRhea As soon as I deleted it, he responded with "please feel free to undelete your answer", which I did. And it is a cw, so... Feb 2, 2021 at 23:21
• Now that he’s requested it I have no qualms with this, and I never meant to imply you were trying to get rep or anything, apologies if it came across that way. I’m just someone who regularly leaves for hours between interacting with a post, so I’d like the general precedent to be for us to allow some time for conversion to answers. Feb 2, 2021 at 23:24
• @AlecRhea No need to apologize, you were absolutely right, and I am glad to have closed it and reopened only after his OK Feb 2, 2021 at 23:30

The characteristic function for the Poisson distribution is given by

$$e^{\lambda\left(e^{i\theta}-1\right)}=e^{\lambda\left(\cos\theta-1\right)}\left(\cos(\lambda\sin\theta)+i\sin(\lambda\sin\theta)\right)$$

Just another cw, in way of an illustration: level plot for modulus and argument of $$\sin(\sin(z))$$ in the complex plane (Thanks to Michael E2 from Mathematica.SE for helping out with the plot)

Valerii Salov, in Notation for Iteration of Functions, Iteral, has written a 23 page (!) paper on how to denote the iteration of a function, specifically for the issue raised by the OP in Inevitable Dottie Number. Iterals of cosine and sine.

The notation proposed by Salov uses the first Cyrillic letter И of the Russian word Итерация for iteration, as a variation on the product symbol $$\prod$$, so that И$$_v^n f$$ is the n-fold iteration of the function $$f(x)$$ for initial value $$x=v$$.

Some examples of the use of this compact notation: This can be typeset in MathJax as a LARGE version of Unicode character 1048:

$$\large{ {\LARGE\unicode{1048}}_0^n(x+1)=n,\ {\LARGE\unicode{1048}}_{x=1}^n ax=a^n,\ {\LARGE\unicode{1048}}_{1}^\infty \frac{1}{x+1}=\lim_{n\to\infty} {\LARGE\unicode{1048}}_{1}^n \frac{1}{x+1}=\frac{2}{1+\sqrt{5}} }$$

• we should start a petition to add И to the LaTeX code base as an operator primitive like $\prod$. Feb 2, 2021 at 12:44
• Use \mathop to get the correct behaviour. The following definition of \iter seems about right for a 10pt document, but you can adjust the sizes to your liking: \input cyracc.def \font\fourtncyr=wncyr10 scaled \magstep2 \font\twelvcyr=wncyr10 scaled \magstep1 \font\tencyr=wncyr10 \font\sevencyr=wncyr7 \def\iter{\mathop{\mathchoice{\doiter\fourtncyr}{\doiter\twelvcyr}{\doiter\tencyr}{\doiter\sevencyr}}} \def\doiter#1{\vcenter{\hbox{#1\cyracc I}}}. Remove the \vcenter if you prefer the symbol to sit on the text baseline, as in the example image. Feb 2, 2021 at 14:35
• I see that in Salov’s paper, sub/superscripts are written at the side even in display style. Put \nolimits at the end of the definition of \iter if you want that. Feb 2, 2021 at 14:46
• $\text{И}_{1}^n x=1$. $\mathop{И}_{1}^n x=1$. the first is \text{И}_{1}^n, the second \mathop{И}_{1}^n Feb 2, 2021 at 20:47
• It seems very rare that there's much need to iterate formulae rather than 'functions'; the notations $f^{(2)}()$ or (if there's any concern about multiple derivatives) $f^{\circ 2}()$ are common enough that I'm hard-pressed to see the need for a notation like this. Aug 25, 2021 at 21:44