2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ?

[Gottfried]

I was studying the function
\( f_a(x) = 2 \cdot \sinh ( a \cdot \sinh^{-1}(x/2)) \)
finding that for odd a this gives (finite) polynomials in x with integer coeffcients - thus for integer x and odd a this is integer and for rational x this is also rational. (For even a replace sinh by cosh). So this is somehow interesting.



Formally this looks like the Schroeder-function of some unknown function \( f_a(x) \) (with some parameter a), where also the iteration-height h is introduced:
\( f_a^{[h]}(x) = 2 \cdot \sinh( a^h \cdot \sinh^{-1}(x/2)) \)

Is this function \( f_a(x) \) something common in our usual, daily toolbox?

[tommy1729]

Hi Gottfried

Once again we see this 2sinh function !!

Im thinking about this function too, together with my friend mick.

I assume you know very well that when given \( f(x) \) we can compute that it is the superfunction of \( g(x) \) where \( g(x)=f(f^{-1}(x)+1) \). Although that formula does not simplify the expression if that is even possible.

There are many named (odd) polynomials in math and I am reminded that the similar looking (problem) \( sin (n arcsin(x)) \) and related ones are connected to ChebyshevPolynomials (of the First or Second Kind for example) and other named polynomials that are usually defined by differential equations.

So I guess we could say it is an iteration of the g-formula above for some named polynomials.

But I doubt that answers your questions.

Notice that - just like the Riemann hypothesis - the zero's of \( 2sinh((2n+1)arcsinh(z/2)) = 0 \) All lie on a critical line !!

Hint: \( cos(pi*m/(4n+2)) i \) !!

Im not sure how to answer your question more.
I guess we need to look at some named polynomials and their recursions.

Further just like the addition formula for sine was usefull , the same is probably true for this \( 2sinh \) !!

Btw did you read my fermat superfunction post ?

Elementary functions are still underrated !

regards

tommy1729

[tommy1729]

My g formula is used to find f(x+k)=g^[k](x).
What Gottfried asked is rather f(k x) = g^[k](x).

This can be fixed ofcourse.

[mike3]

Letting \( P_n(x) = \sinh((2n+1) \mathrm{arsinh}(x)) \) (so your \( f_{2n+1}(x) = 2 P_n(x/2) \)), I noticed

\( P_0(x) = x \)
\( P_1(x) = 4x^3 + 3x \)
\( P_2(x) = 16x^5 + 20x^3 + 5x \)
\( P_3(x) = 64x^7 + 112x^5 + 56x^3 + 7x \)
\( P_4(x) = 256x^9 + 576x^7 + 432x^5 + 120x^3 + 9x \)
...

Now look at the Chebyshev polynomials \( T_n(x) \) for odd \( n \)...

\( T_1(x) = x \)
\( T_3(x) = 4x^3 - 3x \)
\( T_5(x) = 16x^5 - 20x^3 + 5x \)
\( T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x \)
\( T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x \)
...

So \( P_n(x) = \frac{T_{2n+1}(ix)}{i} = -i T_{2n+1}(ix) \). Then,

\( f_{2n+1}(x) = 2 \sinh((2n+1) \mathrm{arsinh}(x/2)) = -2i T_{2n+1}\left(\frac{ix}{2}\right) \).

This, I suppose, is as "close as we can get to something in the 'usual' toolbox", at least if your "usual" toolbox has enough to include well-known named sequences of polynomials like the Chebyshev polynomials.

Note also that

\( T_n(x) = \cos(n \mathrm{arccos}(x)) = \cosh(n \mathrm{arcosh}(x)) \)

and, for \( x > 0 \),

\( \begin{align}
\mathrm{arcosh}(ix) &= \log(ix + \sqrt{(ix)^2 - 1})\\
&= \log(ix + \sqrt{-x^2 - 1})\\
&= \log(ix + \sqrt{-(x^2 + 1)})\\
&= \log(ix + i \sqrt{x^2 + 1}) \\
&= \log(i(x + \sqrt{x^2 + 1})\\
&= \log(i) + \log(x + \sqrt{x^2 + 1})\ (\mathrm{note\ that\ }-\pi < \arg(i) + \arg(x + \sqrt{x^2 + 1}) < \pi\ \mathrm{so\ the\ log\ identity\ is\ OK})\\
&= \log(i) + \mathrm{arsinh}(x)
\end{align}
\)

. Now

\( T_{2n+1}(ix) = \sinh((2n+1) (\log(i) + \mathrm{arsinh}(x))) = \cosh((2n+1) \log(i)) \sinh((2n+1) \mathrm{arsinh}(x)) + \sinh((2n+1) \log(i)) \cosh((2n+1) \mathrm{arsinh}(x)) \)

Taking \( \log(i) = \frac{i \pi}{2} \) and using the correspondence between hyperbolic and trigonometric functions gives \( \sinh\left((2n+1)\frac{i \pi}{2}\right) = \sinh\left(i\left(n + \frac{1}{2}\right) \pi\right) = i \sin\left(\left(n + \frac{1}{2}\right)\pi\right) = i (-1)^n \). Also, for cosh we get \( \cosh\left((2n+1)\frac{i \pi}{2}\right) = \cos\left(\left(n + \frac{1}{2}\right)\pi\right) = 0 \). Thus the result above simplifies to \( T_{2n+1}(x) = i \sinh((2n+1) \mathrm{arsinh}(x)) \) (the \( (-1)^n \) drops out due to the evenness of cosh) and so we have a formal proof of the relation to the Chebyshev polynomials we just gave.

[Gottfried]

Hi Mike -

Letting \( P_n(x) = \sinh((2n+1) \mathrm{arsinh}(x)) \) (so your \( f_{2n+1}(x) = 2 P_n(x/2) \)), I noticed

\( P_0(x) = x \)
\( P_1(x) = 4x^3 + 3x \)
\( P_2(x) = 16x^5 + 20x^3 + 5x \)
\( P_3(x) = 64x^7 + 112x^5 + 56x^3 + 7x \)
\( P_4(x) = 256x^9 + 576x^7 + 432x^5 + 120x^3 + 9x \)
...

Now look at the Chebyshev polynomials \( T_n(x) \) for odd \( n \)...

- yes I've also put a question in MSE but retracted it because just after posting it I had found the entry in wikipedia... which is a really good one btw. That solved also the problem of the connection between the sinh/asinh cosh/acosh and sin/asin and cos/acos versions (cosh(x) = cos(ix) and the resulting change of signs in the polynomials).

Well, this did not give some "usual,more common" simple function of which the asinh is the Schröder-function and only a family of polynomials instead (perhaps there might be some expression by elementary functions, anyway). But what this continuous iterable function \( 2\sinh(a^h \cdot \operatorname{asinh}(x/2)) \) gives at least is then a fractional interpolation for the index of the Chebychev-polynomials, which in turn are specifically useful for polynomial interpolation... What will this give to us...?

At the moment I've put it aside and I'll take a breath to look at it later again: what it has originally been for (for me in my notepad) in the bigger picture.

Gottfried

[Gottfried]

ahh, stupid....

Of course the basic Chebychev-polynomials are "the-functions-of which-acosh(x)-is-the-Schröderfunction".

Let for instance \( f_2(x) = T_2(x) = -1 + 2x^2 \) then of course the iterates \( f_2^{\circ 2}(x) = T_2(T_2(x)) = 1 -8x^2+8x^4 \) and the iterates to any height h are \( f_2^{\circ h}(x)= T_2^{\circ h}(x) \) which can then be evaluated/interpolated using the cosh/acosh-pair as \( f_2^{\circ h}(x) = \cosh(2^h \operatorname{acosh}(x)) \) .

So my question is then -very simple- answered for "base=2" that the "more common" function "in our toolbox" is the quadratic polynomial \( f_2(x)=-1 + 2x^2 \) (where the fractional iterates become power series instead...).

And if I take any higher-indexed Chebychev-polynomial as the base-function , the iterate is simply computable by the cosh/acosh-mechanism
\( f_b^{\circ h}(x) = \cosh(b^h \operatorname{acosh}(x)) \)
(And for my initial question using sinh/asinh it is simply the same except with other polynomials)


Should have seen this before... -

Gottfried

---

[update]: And -oh wonder- this is just related to a question in MSE where I was involved this days without knowing that this two questions are in the same area; I just added the information about the cosh/arccosh-composition there... :-) http://math.stackexchange.com/questions/...965#490965