open problems survey bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/17/2008, 10:03 AM (This post was last modified: 10/24/2010, 07:46 AM by bo198214.) This thread is for mathematicians looking for challenges. Rules for posting in this thread: a) One open problem/conjecture per post (except very closely related problems) b) Put a meaningful title together with a succeeding problem id into the post subject. c) Every problem shall be stated clearly and basicly, if necessary by giving an introduction before and linking to relevant resources. d) Avoid problems like: "find the best algorithm for ...", "investigate more on ...", "is there a connection between ...", etc. Make effort to culminate the problem in a yes/no question or conjecture. e) If you found a proof or have a comment then dont post it here but open a new thread containing the string "TPID n" where n is replaced by the problem number. This way comments and proofs for a problem can by found by searching the forum for that string, while the problems thread stays clean and readable. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/17/2008, 10:23 AM (This post was last modified: 06/29/2008, 10:57 AM by bo198214.) Conjecture Let $B_n$ be the Bell/Carleman matrix of $f(x)=b^x$, $1 0$ for $x \in S$ where S is an open interval on the real line that contains 1 but not 0. For a complex analytic extension, being bounded may be sufficient. Conjecture (Part 2): This is contingent on part 1, and if it is uniquely determined by these conditions, then $EF(0) = \gamma$, the Euler-Macheroni constant. Discussion: Starting with the definition of the exponential factorial $EF(x)=x^{EF(x-1)}$, and differentiating we get $EF'(x)=x^{EF(x-1)}(EF(x-1)/x+\ln(x)EF'(x-1))$ and evaluating at one, we get $EF'(1)=EF(0)$. So by finding EF(0) we are really finding the first coefficient of the power series expansion of EF (about x=1). A numerical approximation of the power series of EF using only these first principles will give a value of $EF(0) \approx 0.57$. Aside from the numerical approximations, if we assume that EF is invertible at one, then that means that EF'(1) is nonzero, which means that EF(0) is nonzero. Here is the first few real solutions. There seems to always be exactly 2 real solutions for every approximation, but the number of complex solutions increases with the approximation number. Code:M = Number of coefficients N = Number of solutions M N Solutions 2 2 {{0.575571, 0.151142}, {10.4244, 19.8489}} 3 2 {{0.575571, 0.229718, 0.0785761}, {10.4244, 39.0172, 19.1683}} 4 6 {{0.570807, 0.232292, 0.114153, 0.0234736}, {4.22694, 18.1158, -0.652778, -11.3147}, ...} These coefficients correspond to the functions: $EF(x)_2 = 1 + 0.575571(x-1) + 0.151142(x-1)^2$ $EF(x)_3 = 1 + 0.575571(x-1) + 0.229718(x-1)^2 + 0.0785761(x-1)^3$ $EF(x)_4 = 1 + 0.570807(x-1) + 0.232292(x-1)^2 + 0.114153(x-1)^3 + 0.0234736(x-1)^4$ Below I have attached of some graphs made with these approximations. Notice that values of the function for $x<-1$ are required to be complex. Attached Files   expfac2.pdf (Size: 13.25 KB / Downloads: 1,338)   expfac4.pdf (Size: 45.15 KB / Downloads: 1,234) bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 06/29/2008, 12:36 PM (This post was last modified: 06/29/2008, 01:32 PM by bo198214.) Conjecture Let $C_N$ be the Carleman matrix of $(x+s)^p-s$ (truncated to N rows and columns), $s>0$, $p>0$ real. Then the set of eigenvalues of $C_N$ converges to the set $\{p^k:k\ge 0\}$ for $N\to\infty$ in the sense that there exist an enumeration $v_{N,k}$ of the Eigenvalues of $C_N$ such that $\lim_{N\to\infty} v_{N,k} = p^k$ for each $k$. Discussion This is about the function $f(x)=x^p$ shifted by $s$. The fixed point 0 is a singularity for $f$ (for non-natural $p$), so $f$ has to be developed at the different point $s$. In the particular case $s=1$ we have the fixed point at 0 and the first derivative is $p$. So the Carleman matrix is triangular and we can solve it exactly, getting $f^{\circ t}(x)=x^{p^t}$. The conjecture is again about the independence of the matrix function method with respect to the development point. $f$ can even be developed at the fixed point 0 in the particular case $p\in\mathbb{N}$. However in this case $f'(0)=0$ except $p=1$ and regular iteration can not be applied, which makes sense as $x^{p^t}$ can for most t not be developed at 0. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 10/08/2008, 04:22 PM We know that the recurrence for $b>0$ (1) $f(0)=1$ (2) $f(z+1)=b{f(z)}$ has $f(z)=b^z$ as the only entire solution that is bounded on the strip $S=\{z: 0<\Re(z)\le 1\}$. The image of $S$ under $b^z$ is an annulus for $b>0$ and so bounded. We know that for complex $b\not\ge 0$ the function $f(z)=b^z=\exp((\log(z)+2\pi i k)z)$ is not bounded on $S$ (the image is kinda infinite spiral) for any $k$. The question remains whether Conjecture There is no entire solution $f$ that satisfies (1) and (2) and is bounded on $S$ for complex $b\not\ge 0$. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 05/01/2009, 09:20 AM (This post was last modified: 05/01/2009, 09:24 AM by bo198214.) For a discussion of the topic see http://math.eretrandre.org/tetrationforu...11#pid2411 Conjecture Let $b=\sqrt{2}$. Every real function $f$ on $(-2,\infty)$ that satisfies: $f(0)=1$ $f(x+1)=b^{f(x)}$ $f(-f(x))=-x$ is not continuous at any point. andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 10/07/2009, 12:03 AM (This post was last modified: 10/24/2010, 07:48 AM by bo198214.) Conjecture $\lim_{n\to\infty} f(n) = e^{1/e}$ where $f(n) = x$ such that ${}^{n}x = n$ Discussion To evaluate f at real numbers, an extension of tetration is required, but to evaluate f at positive integers, only real-valued exponentiation is needed. Thus the sequence given by the solutions of the equations$x = 1$ $x^x = 2$ $x^{x^x} = 3$ $x^{x^{x^x}} = 4$ and so on... is the sequence under discussion. The conjecture is that the limit of this sequence is $e^{1/e}$, also known as eta ($\eta$). Numerical evidence indicates that this is true, as the solution for x in ${}^{1000}x = 1000$ is approximately 1.44. The conjecture is proven to be true. Search the forum for "TPID 6". andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 10/23/2009, 05:27 AM (This post was last modified: 10/25/2009, 07:49 AM by andydude.) Conjecture The following holds for regular tetration: $0 \,< \,({}^{y}x)({}^{-y}x) \,< \,1$ for all $1 < x < e^{1/e}$ and 0 < |y| < 1. The following holds for intuitive tetration: $0 \,< \,({}^{y}x)({}^{-y}x) \,< \,1$ for all $x > e^{1/e}$ and 0 < |y| < 1. Discussion This would be interesting in its own right, partly because it is symmetric, but also because it is useful in demonstrating the difference between reciprocal heights and super-roots. Another notable aspect of this set of bounds is that it is very extension-dependent. This does not hold for linear tetration. Appended are several graphs of the function $f(y) = ({}^{y}x)({}^{-y}x)$ for x = 1.001, 1.1, eta, e, 10. The first three (1.001, 1.1, eta) were calculated with regular tetration, and the last two (e, 10) were calculated with intuitive tetration. More discussion of this is here Attached Files   xty-times-xtny-base-1p001.pdf (Size: 6.39 KB / Downloads: 1,116)   xty-times-xtny-base-1p1.pdf (Size: 6.28 KB / Downloads: 1,111)   xty-times-xtny-base-eta.pdf (Size: 6.26 KB / Downloads: 1,090)   xty-times-xtny-base-e.pdf (Size: 6.2 KB / Downloads: 1,096)   xty-times-xtny-base-10.pdf (Size: 6.31 KB / Downloads: 1,091) bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 04/25/2010, 10:53 AM (This post was last modified: 04/25/2010, 01:48 PM by bo198214.) Is there an elementary real function $F$, such that $F(1+F^{-1}(x))$ is a real polynomial of degree at least 2 without real fixed points. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 07/20/2010, 05:50 AM (This post was last modified: 10/24/2010, 07:55 AM by bo198214.) Let the sequence $(a_n)_{n\in\mathbb{N}}$ be defined recursively in the following way for $b>0$: $a_1 = 1$ and $a_n = \frac{1}{1-b^n}\sum_{m=1}^{n-1} a_m \left(n\\m\right) (1-b)^{n-m} b^m$ for $n\ge 2$ Is $\lim_{n\to\infty} a_n = \frac{b-1}{\ln(b)}$? The conjecture is proven to be true. See http://arxiv.org/abs/1008.1409 and also the discussions: Logarithm reciprocal True or false logarithm There is also a result-less (as of this writing) thread in sci.math.research and sci.math called "Logarithm reciprocal". « Next Oldest | Next Newest »

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