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.

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 \( f(x)=b^x \), \( 1<b<e^{1/e} \), truncated to \( n \) rows and columns.
There is an enumeration \( (v_{n,k})_{1\le k\le n} \) of the eigenvalues of \( B_n \) such that \( \lim_{n\to\infty} v_{n,k} = \ln(a)^k \) where \( a \) is the lower real fixed point of \( b^x \).

Explanation
This is a key question for deciding whether the diagonalization/matrix power method is independent of its development point. A more general question would be under which circumstances the eigenvalues of the Carleman matrix converge to the powers of an attracting fixed point. It seems they dont do for \( f(x)=x^2+x-1/16 \) which has an attracting fixed point at \( -1/4 \) and a repelling fixed point at \( 1/4 \). Also the behaviour of the eigenvalues for \( e^{-e}<b<1 \) seems unclear. However there is only one real repelling fixed point in this case.

Notes
\( b=a^{1/a} \)
\( \ln(a) \) is the derivative of \( b^x \) at \( a \): \( \frac{\partial b^x}{\partial x}|_{x=a}=\ln(b)b^x|_{x=a}=\ln(b)a=\ln(b^a)=\ln(a) \)

05/26/2008, 03:24 PM (This post was last modified: 05/26/2008, 06:47 PM by bo198214.)

Conjecture (Part 1):
The exponential factorial (EF) is uniquely determined by the assumptions:

\( EF(1) = 1 \)

\( EF(x) = x^{EF(x-1)} \)

EF(x) is invertible for \( x \in S \)

EF(x) is real analytic for \( x \in S \)

\( \frac{d^n}{dx^n}EF(x) > 0 \) 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

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.

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 \( b\not\ge 0 \).

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:
\( f(0)=1 \)
\( f(x+1)=b^{f(x)} \)
\( f(-f(x))=-x \)

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".

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

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 \( \lim_{n\to\infty} a_n = \frac{b-1}{\ln(b)} \)?