# Igäria Mnagarka

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 May28 accepted On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$? May27 awarded Popular Question May27 asked Why to see that $\overline{B}(x;r)$ is closed if it was just defined? May26 revised On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$? edited title May26 comment On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$? So basically I must look for some $\epsilon$ that $\forall\epsilon\gt 0,\exists N\in\Bbb N :\forall n\ge N:|a_n-1|\lt\epsilon$ does not hold? The question seems to be utterly trivial, but I was thinking that some algorithmic process should be applied in order to automatically obtain those $\epsilon$'s. May26 revised On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$? edited body May26 asked Are there functions that converge to $P$ when $f:\mathbb{N}\to\mathbb{R}$ and to $Q$ when $f:\mathbb{R}\to\mathbb{R}$ with $Q\neq P$? May26 asked On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$? May19 accepted $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. May14 awarded Notable Question May12 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers $h$ is just an aribtrary name for the vector. When I used it, I was trying to refer to $v$, to avoid confusion with the $v_n$'s. May12 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers I guess I got it. $h$ is the vector $(c_1,c_2,c_3)$. Then I multiply each of it's coordinates for one vector of the basis in order to obtain $v$ which is: $c_1v_1+c_2v_2+c_3v_3$? May12 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers Any orthonormal basis is a basis for $\mathbb{R}$? I'm still a little confused because I've read a little about subspaces and there are subspaces that are different of the primary vector space. May12 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers I don't understand where are you taking the $c_n$'s from. I'm thinking like this: $v=(v_1,v_2,v_3)$, $v_1=(v_{1,1},v_{1,2},v_{1,3})$, $v_2=(v_{2,1},v_{2,2},v_{2,3})$, $v_3=(v_{3,1},v_{3,2},v_{3,3})$. May12 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers I'm still stuck. I've written $\langle v,v_1 \rangle+\langle v,v_2 \rangle + \langle v,v_3 \rangle=0$ but I still can't see it. May11 comment What are good questions that could be used to demonstrate the nature of mathematics study? @PandaBear Yes. Timothy Gowers tried to answer it and the answer was his fabulous 1000 page Princeton Companion to Mathematics. May11 asked What are good questions that could be used to demonstrate the nature of mathematics study? May10 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers Yes, but there could be also unit vectors that are not perpendicular to each other, isn't it? May10 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @T.Bongers The orthonormality of the basis means that the sum of all vectors in the basis equals $1$, right? I've thought only this by summing $(0,1)$ and $(1,0)$ as an example. May10 comment $\mathcal{B}=\{v_1,v_2,v_3\}$ is an orthonormal basis, find all the vectors $v\in \mathbb{R}^3$ such that $\langle v, v_1+v_2-v_3\rangle=0$. @FlybyNight Yes. He's right. Some time later I realized that it's useless to find all the unit vectors.