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Gottfried · (This post was last modified: 09/10/2013, 12:31 PM by 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 · (This post was last modified: 09/10/2013, 09:56 PM by 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 · 09/10/2013, 10:06 PM

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 · (This post was last modified: 09/11/2013, 10:51 AM by 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 · (This post was last modified: 09/12/2013, 07:25 AM by Gottfried.)

Hi Mike -

(09/11/2013, 10:49 AM)mike3 Wrote: 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 · (This post was last modified: 09/12/2013, 07:28 AM by 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

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