On to C^\infty--and attempts at C^\infty hyper-operations

[sheldonison]

As Sheldon has thoroughly convinced me of non-holomorphy of my tetration. I thought I'd provide the proof I have that it is \( C^{\infty} \) on the line \( (-2,\infty) \). I sat on this proof and didn't develop it much because I was too fixated on the holomorphy part. But, I thought it'd be nice to have a proof of \( C^{\infty} \).

Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that,

\( \tau^{(k)}(t)|(-2,\infty)\to\mathbb{R}\,\,\text{for}\,\,k<k;\;\;\; \) edit the : didn't work replaced with |
\( \sum_{m=1}^\infty||\tau^{(k)}_{m+1}(t)-\tau^{(k)}_{m}||_{a \le t \le b}<\infty\\ \)

Where,
\( \tau^{(k)}_0(t)=0 \)
\( \tau^{(k)}_m(t) = \frac{d^k}{dt^k} \log(1+\frac{\tau_{m-1}(t+1)}{\phi(t+1)}) \)

And \( ||...||_{a\le~t\le~b} \) is the sup-norm across some interval \( [a,b]\subset (-2,\infty) \). As a forewarning, this is going to be very messy...

Now to begin we can bound,

\( ||\phi^{(j)}(t)||_{a\le~t\le b}\le~M\,\,\text{for}\,\,j\le k\\ \)

And that next,

\( \phi^{(k)}(t+1)+\tau_m^{(k)}(t+1)=\frac{d^k}{dt^k}e^{\phi(t)+\tau_{m+1}(t)} \)
\( =\sum_{j=0}^k\binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)}) \)
\( =\sum_{j=0}^{k-1}\binom{k}{j}(\frac{d^{k-j}}{dt^{k-j}}\phi(t+1)e^{-t})(\frac{d^j}{dt^j}e^{\tau_{m+1}(t)})+\phi(t+1)e^{-t}(\frac{d^k}{dt^k}e^{\tau_{m+1}(t)}) \)

Now,

\( \frac{d^k}{dt^k} e^{\tau_{m+1}(t)}=e^{\tau_{m+1}(t)}(\tau_{m+1}^{(k)}(t)+\sum_{j=0}^{k-1} a_j \tau_{m+1}^{(j)}(t)) \)


So, we ask you to put on your thinking cap, and excuse me if I write,

\( \tau_{m}^{(k)}(t+1)=A_m+C_m\tau_{m+1}^{(k)}(t) \)

And by the induction hypothesis,

\( \sum_{j=1}^\infty ||A_{j+1}-A_j||_{a\le t \le b}<\infty \)
\( \sum_{j=1}^\infty ||C_{j+1}-C_j||_{a\le t \le b}<\infty \)

Which is because these terms are made up of finite sums and products of \( \tau_m^{(j)} \) and these are said to be summable.  Now the proof is a walk in the park.

\( \tau_{m}^{(k)}(t)=\frac{\tau_{m-1}^{(k)}(t+1)-A_{m-1}}{C_{m-1}}...=\sum_{j=0}^{m-1}(\prod_{k=0}^{m-1-j}C_{m-1-k}^{-1})A_j \)

Where, we've continued the iteration and set \( \tau_0 = 0 \) and \( \tau_1 = 0 \) for \( k>1 \), and \( \tau_1 = 1 \) for \( k=1 \) (but we're tossing this away because we know it's differentiable). Therefore,

\( \sum_{m=1}^\infty||\tau_{m+1}^{(k)}(t)-\tau_m^{(k)}(t)||_{a\le~t\le b}<\infty \)

Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code.

EDIT: I'll do it properly as I correct my paper and lower my expectations of the result.


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As to the second part of this post--now that we have \( C^\infty \) out of the way, we ask if we can continue this iteration and get pentation.  Now, \( \text{slog} \) will certainly be \( C^{\infty} \) and \( \frac{d}{dt}e \uparrow \uparrow t > 0 \) so it's a well defined bijection of \( \mathbb{R} \to (-2,\infty) \). So, first up to bat is to get another phi function...

Hey James, I reposted your post, given the limitation that the forums Tex support has bugs with spaces.  I don't know if it was always this bad  I might write a oneline perl script to remove spaces or substiture a ~ and replace new lines with a new tex /tex pair and remove \\ at the end of a line.  Of course if I do all that, then I could also have perl trivially replace \(...\) combos with a tex /tex pair too  and then I'd automatically convert Latex to math.eretandre Tex .... Anyhow, I'm reading your equations now; are you satisified with proving \( C^\infty \) for the Tetration case?   I've seen Walker's proof of \( C^\infty \), but I've always been curious about how to do it for other functions.