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conjugate analysis
ironball   5
N 41 minutes ago by tobiSALT
Given $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$, prove that $f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n$ if and only if $f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n$, where $f^*(x)$ denotes the conjugate function.
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ironball
Nov 11, 2024
tobiSALT
41 minutes ago
Nice limit
Snoop76   1
N 3 hours ago by Snoop76
Source: Own
If $a_n={\sqrt[n]{n!!}}$ $,$$ $ find :$\lim_{n \to \infty}   \sqrt{n}|a_{n+1}+a_{n-1}-2a_n|$.
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Snoop76
Nov 11, 2024
Snoop76
3 hours ago
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