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juan
  • Member for 14 years, 5 months
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Asymptotic analysis of $x_{n+1} = \frac{x_n}{n^2} + \frac{n^2}{x_n} + 2$
@RiverLi I do not assume that $x_n=n f(1/n)$, I assume $x_{2n}=nf(1/n)$ and $x:_{2n+1}=n g(1/n)$. My power series exist. If they have positive radius of convergence, and I think this is not very difficult to prove, then my assumptions about $x_{2n}$ and $x_{2n+1}$ are proved and all is justified.
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Injectivity of an integral transform
@Jun This remind me of Salem criterium for the Riemann hypothesis in: Salem, R., Sur une proposition equivalente à l'hypothèse de Riemann, C. R. Acad. Sci. Paris vol 236 (1953) pp. 1127--1128.
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On RH in the Clay Institute list
@GHfromMO RH could be unproved by a complicated argument, for example by a very careful study of the behaviour of $\pi(x)$ something as the elementary proof of the prime number theorem but more complicated. In this case the prize will be well deserved I think.
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Is $Q_n(x)=\sigma_{n+1}(x)/\sigma_n(x)$ logarithmically convex on $\mathbf{R}$?
Nice start, I wrote similar things but I do not recall to use Fourier coefficients here. Jan van de Lune is the real proponent of this problem, he asked me to propose it in MathOverflow. I have sent mails to him about your answer, but his health is not good and I have not received any answer until now. I am sure he will like it, his PhD-thesis is full of trapezoidal sums for $z^x$.
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Is $Q_n(x)=\sigma_{n+1}(x)/\sigma_n(x)$ logarithmically convex on $\mathbf{R}$?
@fedja Yes I will be interested. I do not know this, I think Jan never mention this to me. I will ask him.
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Find an element with given periodicity
@KungYao Your use of the word boundary is not clear to me. Are you taking the elements in $[0,1]^2$ mod 1? It is your first condition, for example, equivalent to $f(0, x_2)=e^{-\pi i x_2} f(1,x_2)$? This is what I will have called a boundary condition. Or can we use $x_1=1/3$? In this case what is the meaning of $x_1+1$?
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The unproved formulas of Ramanujan
@GerryMyerson I was confused. It is true that there is an unproved assertion of Ramanujan at the point I referred but it is unrelated to the $2^x$, $3^x$ question.
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The unproved formulas of Ramanujan
@Wojowu In the paper "Highly composite numbers" point 36 in page 114 of Ramanujan's Collected Papers, it is said that the quotient of two consecutive superior highly composite numbers is a prime. This is not proved, but follows it the problem have a positive answer. I think this is not known.
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Banach algebra of smooth functions
@AymanMoussa Your function is not continuous at $\mathbf{T}$.
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Banach algebra of smooth functions
@YemonChoi Thanks, I see. The examples $f$ will be weird functions.
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Banach algebra of smooth functions
@AymanMoussa I think that it is not closed by differentiation. Although it is not easy to find $f\in A$ with $f\not\in A$.
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Banach algebra of smooth functions
@AymanMoussa Consider the case $d=1$, and let $A$ be the set of function of $\mathcal{C}^\infty$ such that $$\Vert f\Vert=\sum_{n=0}^\infty \frac{\Vert f^{(n)}\Vert_\infty}{n!}<+\infty.$$ It is not this an example of what you want?
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An example that the sum of two Borel sets which is not a Borel set in n-dimensional Euclidean space
Yes, there are Borel sets with projection non Borel. Lebesgue, in one of his paper pretended to proof that the projection of a Borel set is a Borel set. Lusin detected the error. This started the Theory of analytic sets. But the examples, I think, are always difficult.
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An example that the sum of two Borel sets which is not a Borel set in n-dimensional Euclidean space
A closed set in $R^2$ is a countable union of compacts. So its projection is a countable union of compact sets. Therefore it is a Borel set.
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Derivatives of Riemann $\xi$ and traces of zeros
added link to my paper that is today (June 11,2020) in arXiv
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revised
Derivatives of Riemann $\xi$ and traces of zeros
I removed an expression about Coffey computations.
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