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I am a professor of Mathematics at the University of Pisa, Italy.
2h

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Solving a non linear equation
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2h

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Solving a non linear equation
It's unimodality. Automatic corrector again. I leave it. 
3h

answered  Solving a non linear equation 
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Periodicity with irrational numbers
Note that by density reasons, $T$ is necessarily $x/(x+1)$, and α is necessarily a multiple of $T$. So everything reduces to prove: for an irrational number $0 < T < 1$ and any (large enough) $k\in\mathbb{N}^+$ the inequalities $Tk< i < T(k+1)$ and $(1T)k<j<(1T)(k+1)$ are satisfied either by exactly one $i\in\mathbb{N}^+$ and no $j\in\mathbb{N}^+$ or by exactly one $j\in\mathbb{N}^+$ and no $i\in\mathbb{N}^+$. Which is easy to verify. Your theorem is nice anyway. 
Sep 6 
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Circumscribing simplex to convex body?
Precisely: it is not obvious (to me) that a simplex growing inside $L$ can in fact keep growing till all its vertices have reached the boundary of $L$. But the positive answer to Q implies a positive answer to Q' by duality, right? 
Sep 6 
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Circumscribing simplex to convex body?
Note the dual question: Q' Does every compact convex body $L$ in $\mathbb{R}^d$ admit an inscribed regular simplex, each vertex of which lies in the boundary of $L$ ? While Q admits the simple construction by moving planes described above, it is not obvious to me how the dual construction should work for the answer of Q'. 
Sep 5 
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Chebyshev centres of a bounded closed convex set in a strictly convex Banach space
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Sep 5 
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Rediscovery of lost mathematics
Part of Galois' work itself  the proof of impossibility of solving the quintic by radicals, can also be considered a piece of rediscovered lost mathematics, since Paolo Ruffini's essentially complete proof had been ignored (possibly because times were not ready to such a revolutionary idea as an impossibility result). 
Sep 5 
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Extending pointwise bound to uniform bound
Note that the continuous dependence from $x$ is neither sufficient nor necessary for the positive answer to hold. If $\{f(\cdot, x)\}_x$ is an equicontinuous family of functions on $[0,1]$, then either $C(t)=\infty$ for all $t$ or $C$ is a continuous real valued function of $t$ (thus bounded). 
Sep 5 
answered  Extending pointwise bound to uniform bound 
Sep 5 
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Chebyshev centres of a bounded closed convex set in a strictly convex Banach space
As a general rule, titles more to the point are preferred (it make it easier to reach potentially interested people). Also, you can use more tags, and more precise (banachspaces, convexity,..). I suggest to change the title accordingly. 
Sep 3 
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Is the sequence of Apéry numbers a Stieltjes moment sequence?
Moreover, from this point of view, the boundary conditions in my answer should be sufficient to determine the solution $w$ just looking at the point $c$, arguing by exclusion, since as I showed a base of solutions is $u_1^2$ $u_1u_2$ and $u_2$ &cetera. 
Sep 3 
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Is the sequence of Apéry numbers a Stieltjes moment sequence?
Yes I think so, even with no numerics. Indeed I think I could turn the argument in my answer into a regularity result for the measure that solves the moment problem. Therefore, assuming Suvrit's existence proof, it follows that the measure whose moments are the Apéry numbers, does admit a density satisfying the third order EDO that I wrote. Thus, no wonder that the numerics suggest that there is such a solution. 
Sep 1 
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Is the sequence of Apéry numbers a Stieltjes moment sequence?
Well, logarithmic is not bad  I was scared of stronger singularities that could make $u^2$ nonintegrable... 
Sep 1 
revised 
Is the sequence of Apéry numbers a Stieltjes moment sequence?
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Sep 1 
revised 
Is the sequence of Apéry numbers a Stieltjes moment sequence?
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Sep 1 
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Is the sequence of Apéry numbers a Stieltjes moment sequence?
But yes, $(c_0,c)$ would make everything nicer. (btw the 2order DE can be written in a form of SturmLiouville operator, $(gu')'+hu=0$, dividing $2Pu''+P'u'+Qu=0$ by $P^{1/2}$ ). 
Sep 1 
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Is the sequence of Apéry numbers a Stieltjes moment sequence?
I thought each term $A(x), B(x), C(x)$ should vanish at $x=0$ and $x=c$, and be continuous at $x=c_0$, in order that the boundary terms cancel, because they are coefficients with different powers of $n$ in front. Also, by linearity $2u_1u_2=((u_1+u_2)^2u_1^2u_2^2$ is also a solution of the 3rd order DE, so we have $3$ linearly independent solutions ODE. 
Aug 31 
revised 
Is the sequence of Apéry numbers a Stieltjes moment sequence?
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Aug 31 
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Is the sequence of Apéry numbers a Stieltjes moment sequence?
The next step should be, checking if the boundary conditions in 2 are compatible with the solutions $u_1$ and $u_2$. And, of course, if they are in $L^2$. Note that (iii) would be satisfied if $u(c)=0$ because $c$ is then a root of $pw$, $qw$, $rw$ with multiplicity resp. $3,2,2$. Unfortunately this w.e. I'm in a place with no Maple! 