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 Yearling
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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 $p_2 = 1<p_1$.
Jul
14
awarded  Yearling
Mar
16
comment Difference between $A\to B\to C$ and $A\to(B\to C)$
Or cs.nott.ac.uk/~txa/g52ifr/html/Prop.html : “Implication is right associative, i.e. we read P -> 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).