Relationships between half-iterates of trig functions

Question

Let the function $\operatorname{hsin}(x)$ be defined as a continuous function with the property $$\operatorname{hsin}(\operatorname{hsin}(x))=\sin x$$ and let other half-trig functions be defined analogously, so that $$\operatorname{hcos}(\operatorname{hcos}(x))=\cos x$$ $$\operatorname{htan}(\operatorname{htan}(x))=\tan x$$ $$\operatorname{hsec}(\operatorname{hsec}(x))=\sec x$$ $$\operatorname{hcsc}(\operatorname{hcsc}(x))=\csc x$$ $$\operatorname{hcot}(\operatorname{hcot}(x))=\cot x$$ There are a vast multitude of identities involving trig functions, like $$\sin^2x+\cos^2x=1$$ and $$\tan x=\frac{\sin x}{\cos x}$$ to name just a few. Can anybody find any interesting identities that can be determined about the half-trig functions, without just taking normal trig identities and substituting in the definitions?

For example, can we find a way to express $\operatorname{htan}(x)$ in terms of $\operatorname{hsin}(x)$ and $\operatorname{hcos}(x)$? Or, in other words, can we find a function $f$ such that $$\operatorname{htan}(x)=f(\operatorname{hsin}(x),\operatorname{hcos}(x))$$

Answer 1

For invertible functions without constant term one can use a "q&d"-method to compute/approximate the half-iteratives: the adaption of the iterative Newton-method to find square-roots.

In Pari/GP this is especially convenient, because you can work with formal powerseries as variables and their inversion.

The following method approximates the powerseries for the halfiterative of the $\sin()$-function

\ps 32  \\ define default order for powerseries

\\ initialization
hsin = x + O(x^32)     \\ assume hsin being initially a simple powerseries

\\ iteration
hsin = (sin(serreverse(hsin)) + hsin )/2)  \\ iterate this Newton-adaption 
                                           \\ until convergence

\\ show the formal powerseries for the half-iterate
hsin
   x - 1/12*x^3 - 1/160*x^5 - 53/40320*x^7 - 23/71680*x^9 - ... + O(x^32)

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I implemented this in the following way and show a small example-value $x_0=0.2$
hsin()

hsin = x + O(x^32)
for(k=1,20,hsin =(sin(serreverse(hsin )) + hsin )/2) \\20 iterations suffice
                                      \\ for "\ps 32"  and 12 visible digits

x0 = 0.2
x05=subst(Pol(hsin ),x, x0)   \\ inserting the x0 in the powerseries hsin
x1 =subst(Pol(hsin ),x,x05)
[x0,x05,x1;x0,x0,sin(x0)]
\\ output
\\  x0              x05           x1
[0.200000000000 0.199331316342 0.198669330795]  by hsin(x0)
[0.200000000000 -------------- 0.198669330795]  by sin(x0)

Similarly with the $\tan()$-function:
htan()

htan=x+ O(x^32)
for(k=1,20,htan=(tan(serreverse(htan))+htan)/2)

x0=0.2
x05=subst(Pol(htan),x,x0)
x1=subst(Pol(htan),x,x05)
[x0,x05,x1;x0,x0,tan(x0)]
\\ output  
\\  x0              x05           x1
[0.200000000000 0.201341376827 0.202710035509] by htan(x0)
[0.200000000000 -------------- 0.202710035509] by tan(x0)

hcos(): Unfortunately, the $\cos()$-function has a constant term and is thus not simply invertible. The assumption, that hsin(x)/htan(x) = hcos(x) however, can be checked (and fails!), so this is not a way to compute the half-iterate for the $\cos()$:

x0=0.2
x05=subst(Pol(hsin/htan),x,x0)
x1 =subst(Pol(hsin/htan),x,x05)
[x0,x05,x1;x0,x0,cos(x0)]
\\ output  
\\  x0              x05           x1
[0.200000000000 0.990016654715 0.764529818658]  by hsin(x0)/htan(x0)
[0.200000000000 -------------- 0.980066577841]  by cos(x0)

Didn't check the other possible functions.

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Important remarks: Note that the powerseries of the half-iterates have terms whose absolute value diverges strongly, so strictly spoken we could not evaluate the powerseries for the half-iterates. But the divergence of the coefficients begins only after a handful of leading terms, so using a powerseries "precision" \ps 32 should not be extended much or might even better be slightly reduced to get more accurate approximations. (We have so called "asymptotic series" which when appropriately truncated give more-or-less rough estimates to some dozen decimal digits)