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 Oct 10 revised $S^{-1}R$ is local then it is of the form $R_P$ for some prime ideal $P$ added 12 characters in body Oct 10 accepted $S^{-1}R$ is local then it is of the form $R_P$ for some prime ideal $P$ Oct 8 awarded Popular Question Oct 6 asked $S^{-1}R$ is local then it is of the form $R_P$ for some prime ideal $P$ Jul 25 awarded Popular Question May 31 awarded Peer Pressure May 5 comment $\frac{1}{n}\int\limits_0^1 f(x) dx=\int\limits_0^{\theta(n)} f(x) dx+\int\limits_{1-\theta(n)}^1 f(x) dx$ Thank you so much @Elaqqad....:) May 5 accepted $\frac{1}{n}\int\limits_0^1 f(x) dx=\int\limits_0^{\theta(n)} f(x) dx+\int\limits_{1-\theta(n)}^1 f(x) dx$ May 5 comment min $\int_0^1 (f''(x))^2 \, dx$ So minimum is 12 right? May 5 accepted min $\int_0^1 (f''(x))^2 \, dx$ May 4 comment Convergence of $\sum\limits_{n=1}^{\infty}\left (1-n\sin \frac{1}{n}\right)^\alpha$ for parameter $\alpha$ Thanks a lot.....:) May 4 accepted Convergence of $\sum\limits_{n=1}^{\infty}\left (1-n\sin \frac{1}{n}\right)^\alpha$ for parameter $\alpha$ May 3 awarded Yearling May 3 comment Convergence of $\sum\limits_{n=1}^{\infty}\left (1-n\sin \frac{1}{n}\right)^\alpha$ for parameter $\alpha$ Then the answer should be $\alpha \geq 1$, right? May 3 accepted normal matrix A $= \left( \begin{array}{cc} 2 & i \\ i & 2 \end{array} \right)$ May 3 comment normal matrix A $= \left( \begin{array}{cc} 2 & i \\ i & 2 \end{array} \right)$ thanks...fool me, I was taking product without considering transpose. May 3 awarded Self-Learner May 3 comment min $\int_0^1 (f''(x))^2 \, dx$ can somebody provide me the other link? May 3 comment normal matrix A $= \left( \begin{array}{cc} 2 & i \\ i & 2 \end{array} \right)$ but in the question it is asked that first column and second row has all positive entries, so if we change the eigenvectors accordingly, it is not giving a diagonal matrix. There is something wrong in my way, please check it once. May 3 reviewed Approve $\frac{1}{n}\int\limits_0^1 f(x) dx=\int\limits_0^{\theta(n)} f(x) dx+\int\limits_{1-\theta(n)}^1 f(x) dx$