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 Weak theorem about an almost identical function: PROOF Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/14/2008, 01:20 AM (This post was last modified: 11/14/2008, 01:22 AM by Kouznetsov.) Hello. This topic is related to http://math.eretrandre.org/tetrationforu...hp?tid=208 In that branch, I explained why the lemma (when we prove it) about "alomost identical finction" $J(z)=z+f(z)$ leads to the uniqueness of the holomorphic tetration, assuming, that tetration has no singularities at $\Re(z)>-2$. "almost udentical" means, that function $h$ is 1-periodic and "small enough" so that $J^{\prime}(x)>0 \forall x\in \mathbb{R}$ Here, I present the weak version of that theorem. This weak version can be used to prove the uniqueness of tetration, assuming that this tetration is holomorphic outside the real axis. It can be applied to the case of base $b\ge\exp(1/\mathrm{e})$, considered in http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf ; in this case, all the singulatities of tetration are at the real axis. My today's proof does not match exactly all the criteria suggested in http://math.eretrandre.org/tetrationforu...hp?tid=208 (that range allows tetration to have singulatities at $\Re(z)<-2$ ). To avoid confusions, I open the new branch for this. I call it "weak theorem", because it cannot be applied as is to the tetation at base $b<\exp(1/\mathrm{e})$, reserving the name "strong theorem" for the more general case that will work for all $b>1$. I begin with so long introduction in order to make this piece readable without background about previous posts. Weak theorem about almost identical functions. Let $M=\{ n\in \mathbb{N} : n<-1 \}$ $C=\mathbb{C} \backslash \{ x \in \mathbb{R}$ $h$ is entire 1-periodic function $\exists x \in \mathbb{R} h(x)\ne 0$ $h(0)=0$ $h\big(z^*\big)=h(z)^*$ $J(z) = z+h(z) ~ \forall z\in \mathbb{C}$ $J^{\prime}(x)>0~\forall x\in \mathbb{R}$ Then $\exists z \in C ~:~ J(z) \in M$ Vulgarization: if function $h$ is somehow "small", then, function $J$ looks as "almost identical". The almost identical function $J$ must have negative integer values in some points outside the real axis. Proof. Let $E=\{z \in \mathbb{C} : J(z)\in M \}$ Hypothesis 0: For each integer $m<1$, there exists only one $z\in \mathbb{C} : J(z)=m$. Below, I make some deduction based on Hypothesis 0 and show that this hypothesis is not consistent with Theorem 0. From Hypothesis (0) it follows that $v_m(z)=\frac{z-m}{J(z)-m}$ is entire function of $z$. $v_m(z)=\frac{z-m}{z-h(z)-m}= \frac{z-m}{z-m-h(z-m)}$ Define $v(z)=\frac{z}{z+h(z)}$ Then $V_m(z)=v(z-m)$ Then $v(x)=V_m(z+m)$ Theredore $v$ is also entire function. Function $v$ grows up to infinity along some contours $\zeta(t)$ at $t\rightarrow \infty$. Let contour $\zeta$ provides the fastest growth, id est, realizes the maximal $\frac {\mathrm{d}\left| v\big(\zeta(t)\big)\right|} {\left|\mathrm{d}\zeta(t)\right|}= \frac{ \frac{\mathrm{d}} {\mathrm{d} t} \left| v \big(\zeta(t) \big) \right| } { \left|\zeta^{\prime}(t) \right| }$ While $v(z)=\frac{z}{z+h(z)}$, the combined function $h(\zeta(t))$ should approach $\zeta(t)$: $\lim_{t\rightarrow \infty}\Big( h\big(\zeta(t)\big)-\zeta(t)\Big)=0$ The limit $\lim_{t \rightarrow \infty} \zeta(t)$ is not allowed to exist because $\zeta(\infty)$ would be singularity of function $v$. Hence, $\zeta(t)$ grows to infinity. Hence, $h(z)$ should have asymptotically linear growth at infinity. Among entire functions, the only polynomials are allowed to have polynomial growth at infinity; all other entire funcitons grow faster. Hence, $v$ is linear function. Then, $h(z)=\frac{z}{v(z)-z}$ is rational function. This property contradicts the conditions of the Theorem, because $h$ is non-trivial periodic function. In such a qay, Hypothesis 0 contradicts conditions of the Theorem. Hence, there exist more than one point $z$ such that $J(z)\in M$. Due to the monotonous growth of function $J$ along the real axis, such a point is outside of the real axis. (end of proof). The theorem above provides the uniqueness of tetration at least for base $b\be \exp(1/\mathrm{e})$; in this case, all the singulatities of tetration are along the real axis. It colleagues agree with the proof of the theorem abovr, we can copypast here the condinuation, id est, the proof of uniqueness of tetration for this case. P.S. Although Henryk suggests word "superexponentiation" instead of "tetration", I keep here "tetration" because I hope to make also unique holomorphic pentation in the similar way; in the case of superexponentiation, the pentation would have to be called supersuperexponentiation which is too long. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/14/2008, 11:15 AM (This post was last modified: 11/14/2008, 01:07 PM by bo198214.) Hi Dmitrii, though I didnt completely follow your proof it triggered coming up with a much more simplified proof. As a prerequisite we only need Picard's big theorem: Picard's big theorem applied to entire functions. Each entire non-polynomial function takes on every complex number with at most one exception infinitely often. First it can be shown that $J$ (as an entire function) does not even omit one value from $\mathbb{C}$. (I give the proof in a next post, though this assertion is not really necessary for the following conclusions.) We know that $J(k)=k$ for each $k\in\mathbb{Z}$, but by the above theorem for each $k\in\mathbb{Z}$ there have to be infinitely other $z\in \mathbb{C}\setminus \mathbb{Z}$ with $J(z)=k$. Hence $J(\mathbb{C}\setminus\mathbb{Z})=\mathbb{C}$. That means if we have a function $g(z)=f(J(z))$, where $f$ is a superexponential with singularities only at $\{z\in\mathbb{Z}:z\le -2\}$, then $g$ has singularities outside $\{z\in\mathbb{Z}:z\le -2\}$. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/14/2008, 01:05 PM bo198214 Wrote:First it can be shown that $J$ (as an entire function) does not even omit one value from $\mathbb{C}$. (I give the proof in a next post, though this assertion is not really necessary for the following conclusions.) Ok, here the proof. We first show that: Each 1-periodic entire function $p$ has a fixed point. Proof. Assume that $p$ has no fixed point i.e. that there is no $z$ such that $p(z)=z$. In this case the entire function $q(z):=p(z)-z$ omits 0. But then there is also no $z$ such that $p(z+1)=z+1$, i.e. $p(z)-z=1$ hence $q$ omits 1. But then $q$ is an entire function that omits 2 values and must be a constant (little Picard): $p(z)=z+c$. But this function is not 1-periodic for any $c$. Contradiction.$\boxdot$ Now lets look whether there is a $z$ for each $w$ such that $w=J(z)=p(z)+z$. We see that $w-p(z)$ is also 1-periodic entire and hence has a fixed point.$\boxdot$ Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/14/2008, 01:11 PM bo198214 Wrote:Hi Dmitrii, though I didnt completely follow your proof it triggered coming up with a much more simplified proof. As a prerequisite we only need Picard's big theorem: Picard's big theorem applied to entire functions. Each entire non-polynomial function takes on every complex number with at most one exception infinitely often. First it can be shown that $J$ (as an entire function) does not even omit one value from $\mathbb{C}$. (I give the proof in a next post, though this assertion is not really necessary for the following conclusions.) We know that $J(k)=k$ for each $k\in\mathbb{Z}$, but by the above theorem for each $k\in\mathbb{Z}$ there have to be infinitely other $z\in \mathbb{C}\setminus \mathbb{Z}$ with $J(z)=k$. Hence $J(\mathbb{C}\setminus\mathbb{Z})=\mathbb{C}$. That means if we have a function $g(z)=f(J(z))$, where $f$ is a superexponential with singularities only at $\{z\in\mathbb{Z}:z\le -2\}$, then $g$ has singularities outside $\mathbb{C}\setminus\{z\in\mathbb{Z}:z\le -2\}$.Henryk, I like your proof. Indeed, it is shorter. Now, please, prove that some of singularities are in the right hand side of the complex plane. Small hint: From the asymptotic behavior, at 1

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