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 Mar 18 revised Prove that $\vert E \vert=0$ if $\frac{x+y}{2}\notin E$ for $x,y \in E$ added 283 characters in body Mar 18 revised Prove that $\vert E \vert=0$ if $\frac{x+y}{2}\notin E$ for $x,y \in E$ added 283 characters in body Mar 18 answered Prove that $\vert E \vert=0$ if $\frac{x+y}{2}\notin E$ for $x,y \in E$ Mar 15 revised An upper bound for $\sum_{n=1}^{\infty}e^{-n^2}$ added 4 characters in body Mar 15 answered An upper bound for $\sum_{n=1}^{\infty}e^{-n^2}$ Jul 14 awarded Yearling Sep 30 awarded Explainer Sep 28 comment Relationship between 2 L-p spaces Hint: (1) use that $f$ is bounded, (2) prove that for every bounded $f$, there exists $C$ such that $|f(x)|^{p_2} \leq C |f(x)|^{p_1}$ for every $x$. Sep 28 comment Relationship between 2 L-p spaces For this statement to be true, we should have $p_1 \leq p_2$. Sep 28 comment Relationship between 2 L-p spaces This is false if $f$ is bounded and $\mu(E) = \infty$. Consider $f(x) = 1$. We have, $f\in L_{p_1}({\mathbb R})$ for $p_1=\infty$, but $f\notin L_{p_2}({\mathbb R})$ for $p_2 = 1 Q -> R as P -> (Q -> R).” Mar 16 comment Difference between$A\to B\to C$and$A\to(B\to C)$See e.g. springer.com/cda/content/document/cda_downloaddocument/… : “Although the implication operator is assumed to be right associative, so that$p \to q \to r$unambiguously means$p \to (q \to r)$, we will write the formula with parentheses to avoid confusion with$(p\to q)\to r$.” Mar 16 answered Difference between$A\to B\to C$and$A\to(B\to C)$Mar 16 revised On the importance of independence in the central limit theorem added 187 characters in body Mar 16 answered On the importance of independence in the central limit theorem Feb 18 revised Estimating certain order statistics of a set. added the normalization factor k/n to the formula for E[Y] Feb 18 answered Estimating certain order statistics of a set. Feb 1 reviewed Approve Prove that$\sum_{k=1}^{n}\binom{n}{k}=\sum_{k=0}^{n-1}2^{k}$Feb 1 comment Random walk on$\mathbb{Z}$with more than two possible steps Sounds good. Regarding question 2, even if$A=\{-1,1\}$and$p_{-1} > p_1$, we have$\lambda_1=1$and$\lambda_2 > 1$. Consider another example:$A=\{-3,-1,2\}$and$p_{-3} = 1/4$,$p_{-1} = 1/4$, and$p_2 = 1/2$. Then values of$\lambda$are given by$\lambda^{-2}/2 + \lambda/4 + \lambda^3/4 = 1\$. The roots of this equation have absolute values approximately 1, 0.6388969200, and 1.769292354 (some are equal to 1, some are greater than 1, and some are less than 1).