# Tetration Forum

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Well, before I can begin our trip into abstract algebra, I must first explain what a "circular" operator is, but to do that, I must first explain what a "hyperbolic" operator is.

hyperbolic exponentiation, is what we normally refer to as exponentiation. I've coined it "hyperbolic" because of the following:

$\exp(x) = e^x = \cosh(x) + \sinh(x)$

and then therefore, hyperbolic multiplication is normal multiplication; it being that hyperbolic exponentiation is the super function of hyperbolic multiplication. And also therefore hyperbolic multiplication is the superfunction of hyperbolic addition. Very, simple.

Circular exponentiation, is a different form of exponentiation, defined by the following:
$\text{cxp}(x) = \delta^{\circ x} = \cos(x) + \sin(x)$ where $\delta$ is a constant to be revealed shortly.

Interestingly, for imaginary arguments, it behaves the same as hyperbolic exponentiation, only reversed.
$\text{cxp}(xi) = \delta^{\circ xi} = \cosh(x) + i\sinh(x)$

Now that was all fairly simple to understand, but now we come to the part where we must define circular multiplication. It's still fairly simple, but may seem a bit awkward.

$x \odot (x^{\circ n}) = x^{\circ n+1}$

but seeing as $x^{\circ n}$ is still unknown to us, we better stick to using $x = \delta$

$\delta \odot (\delta^{\circ n}) = \delta^{\circ n+1}$

therefore
$\delta \odot (\delta^{\circ 0}) = \delta^{\circ 1}$

and now, to make things simple, since $\delta^{\circ 0} = \cos(0) + \sin(0) = 1$ we can say that:

$\delta \odot 1 = \delta^{\circ 1}$ and if we let $\delta = \text{cxp}(1) = cos(1) + sin(1)$, 1 becomes the identity of circular multiplication and circular exponentiation.

Also, we know that circular multiplication, and circular exponentiation behave like normal multiplication and exponentiation.

$a^{\circ x} \odot a^{\circ y} = a^{\circ x+y}$
$(a^{\circ x})^{\circ y} = a^{\circ x \cdot y}$

and so therefore we get the beautiful, circular logarithm laws, (cln is taken to mean natural circular logarithm, which is just the circular logarithm base $\delta$) The principal branch always returns between $[0, 2\pi)$:

$\text{cln}(x \odot y) = \text{cln}(x) + \text{cln}(y)$
$\text{cln}(x^{\circ y}) = y \cdot \text{cln}(x)$

where: $\delta^{\circ \text{cln}(x)} = x$

And now, with these laws, defining circular multiplication is as simple as:

$x \odot y = \delta^{\circ \text{cln}(x)} \odot \delta^{\circ \text{cln}(y)} = \delta^{\circ \text{cln}(x) + \text{cln}(y)} = \cos(\text{cln}(x) + \text{cln}(y)) + \sin(\text{cln}(x) + \text{cln}(y))$

circular division, the inverse of circular multiplication is:
$x\oslash y = \delta^{\circ \text{cln}(x) - \text{cln}(y)}$

And circular exponentiation is as simple as:
$x^{\circ y} = \delta^{\circ y \cdot \text{cln}(x)} = \cos(y \cdot \text{cln}(x)) + \sin(y \cdot \text{cln}(x))$

And now, with all three of these defined, we come to the issue of defining circular addition. Since circular multiplication is it's superfunction we can write the equation as such:

$x \oplus y = y \odot (x \oslash y + 1)$
and
$x \ominus y = y \odot (x \oslash y - 1)$
or generally:
$x \oplus (y \odot n) = y \odot (x \oslash y + n)$

This gives the very strange equation that "converts" addition into circular addition:

$a \odot (x + y) = (a \odot x) \oplus (a \odot y)$,

which also implies:
$a \odot (x \oplus y) \neq (a \odot x) \oplus (a \odot y)$

Circular addition has identity $\upsilon$, therefore:

$a \oplus \upsilon = a$
$a \odot \upsilon = \upsilon$
$a \oslash \upsilon = \infty$

These equations become necessary to observe when defining the circular derivative. Which we shall do now.

$\frac{c}{cx} f(x) = \lim_{h\to \upsilon} (f(x+h) \ominus f(x))\oslash h$

Therefore:
$\frac{c}{cx} k \odot x = \lim_{h\to \upsilon} (k \odot (x + h) \ominus k \odot x) \oslash h = \lim_{h\to \upsilon} (k \odot x \oplus k\odot h \ominus k \odot x) \oslash h = \lim_{h\to \upsilon} (k\odot h) \oslash h = k$

and:
$\frac{c}{cx} (f(x) \oplus g(x)) = \frac{c}{cx} f(x) \oplus \frac{c}{cx} g(x)$

and also:
$\frac{c}{cx} x^{\circ n} = n \odot x^{\circ n-1}$

We'll probably find:
$\frac{c}{cx} f(g(x)) = \frac{c}{cg} f(g(x)) \odot \frac{c}{cx} g(x)$, but I don't want to just state it and I cannot prove it.

Therefore, with this, we can now create an infinite termed "circular" polynomial that will be it's own circular derivative.

if
$n \dagger = 1 \odot 2 \odot 3 ... \odot n-1 \odot n$

and
$\oplus \sum_{n=0}^{R} f(n) = f(0) \oplus f(1) \oplus f(2) ... \oplus f(n-1) \oplus f( R )$

If:
$\frac{c}{cx} J(x) = J(x)$

$J(x) = \oplus \sum_{n=0}^{\infty} x^{\circ n} \oslash n\dagger$

And this is where I'm stuck and I need help. If J(x) turns out to equal $\delta^{\circ x} = \text{cxp}(x)$ I believe we may have a sort of symmetry between circular tetration and hyperbolic, or normal tetration. We should be able to, by regular iteration (of the normal equation for cxp), create the superfunction of cxp, or "circular" tetration. It should be simpler to do than exp because it has a real fixpoint.

I have the equations sort of worked out, but they rely on confirmation that $J(x) = \text{cxp}(x)$

The only way I can properly confirm it is if I had a way of calculating $\text{cln}(x)$. I tried using Lagrange's inversion theorem to solve for its power series, but this was to no avail. It's necessary that the values work for imaginary numbers as well since the range without imaginary numbers is $[-\sqrt{2}, \sqrt{2}]$.

If anybody is curious, the test I'd be doing is seeing if:
$J(1) = \oplus \sum_{n=0}^{\infty} 1 \oslash n\dagger = \oplus \sum_{n=0}^{\infty} (n\dagger)^{\circ -1} = \cos(1) + \sin(1)$,

if it does, then we're in business and there's a whole lot more I can post.
This one can be solved via trigonometric identities. We have

$\cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b)$

If we can find a value $b$ for which $\cos(b) = \sin(b)$, then we can use division to make the right-hand side equal your "$\mathrm{cxp}(a)$". Solving the equation yields $1 = \frac{\sin(b)}{\cos(b)} = \tan(b)$, $b = \arctan(1) = \frac{\pi}{4}$. Then, $\cos(b) = \sin(b) = \frac{1}{\sqrt{2}}$, and

$\mathrm{cxp}(x) = \sqrt{2} \cos\left(x - \frac{\pi}{4}\right)$.

Then, the inverse, the "circular logarithm", is given by

$\mathrm{cln}(x) = \arccos\left(\frac{x}{\sqrt{2}}\right) + \frac{\pi}{4}$.

Of course, since $\mathrm{cxp}(x)$ is not injective (i.e. not "one-to-one"), then this is actually a multivalued "function" (relation). But if we choose the principal branch of $\arccos$, then the above will range in $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$, and the domain is $\left[-\sqrt{2}, \sqrt{2}\right]$ if we interpret as a real-valued function of a real number. The circular logarithm will not return in $[0, 2\pi)$ if taken as a single-valued branch, because $\mathrm{cxp}$ is not injective over that interval.
(06/23/2011, 12:58 AM)mike3 Wrote: [ -> ]This one can be solved via trigonometric identities. We have

$\cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b)$

If we can find a value $b$ for which $\cos(b) = \sin(b)$, then we can use division to make the right-hand side equal your "$\mathrm{cxp}(a)$". Solving the equation yields $1 = \frac{\sin(b)}{\cos(b)} = \tan(b)$, $b = \arctan(1) = \frac{\pi}{4}$. Then, $\cos(b) = \sin(b) = \frac{1}{\sqrt{2}}$, and

$\mathrm{cxp}(x) = \sqrt{2} \cos\left(x - \frac{\pi}{4}\right)$.

Then, the inverse, the "circular logarithm", is given by

$\mathrm{cln}(x) = \arccos\left(\frac{x}{\sqrt{2}}\right) + \frac{\pi}{4}$.

Of course, since $\mathrm{cxp}(x)$ is not injective (i.e. not "one-to-one"), then this is actually a multivalued "function" (relation). But if we choose the principal branch of $\arccos$, then the above will range in $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$, and the domain is $\left[-\sqrt{2}, \sqrt{2}\right]$ if we interpret as a real-valued function of a real number. The circular logarithm will not return in $[0, 2\pi)$ if taken as a single-valued branch, because $\mathrm{cxp}$ is not injective over that interval.

Wow that's incredible how you did that. That's awesome that cxp can be a closed form expression of only cos, that means for imaginary arguments it should be purely positive. I thought it would be way harder to solve for it... I guess sometimes the answers just so simple and right infront of you that you can't think of it.

Thanks a lot for your help. I'll try to see if this pans out to anything