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 Apr 8 awarded Popular Question Aug 26 awarded Popular Question Nov 6 awarded Notable Question Feb 10 awarded Popular Question Nov 20 comment Fourier transform of $f(x) = e^{-x^2}$ Yes, but in my book and a few other sources is other (the same as Mathematica) result. Nov 20 awarded Editor Nov 20 revised Fourier transform of $f(x) = e^{-x^2}$ added 9 characters in body; edited tags Nov 20 asked Fourier transform of $f(x) = e^{-x^2}$ Nov 1 accepted Extended transition function of a DFA - a proof Nov 1 accepted Taylor series for different points… how do they look? Oct 31 comment Taylor series for different points… how do they look? OK. I did something like this on wolframalpha wolframalpha.com/input/?i=series%5Bsinx%2C%7Bx%2C2%2C4%7D%5D why are there these 3 curves? Oct 31 comment Taylor series for different points… how do they look? I know it, but I don't know when it should be done around 0, and when around any another point. And why. Oct 31 asked Taylor series for different points… how do they look? Oct 24 comment Extended transition function of a DFA - a proof Ohh.. the last line should be $$L = \delta^{+}(q,PQa) = \delta(\delta^{+}(q,PQ),a) = \delta(\delta^{+}(\delta^{+}(q,P),Q),a) = P$$ the check is for $|Q| = 0$: $$|Q| = 0 \rightarrow Q = \epsilon$$ $$L = \delta^{+}(q,P\epsilon) = \delta(\delta^{+}(q,P),\epsilon) = \delta^{+}(q,P)$$ $$P = \delta^{+}(\delta^{+}(q,P),\epsilon) = \delta^{+}(q,P)$$ $$L = P$$ Oct 24 awarded Commentator Oct 24 comment Extended transition function of a DFA - a proof Ohh... I tried to do it like in my notes from the lecture (there was only a proof for P only). OK... maybe it should be like this: the assumption: $$\delta^{+}(q,PQ) = \delta^{+}(\delta^{+}(q,P),Q)$$, the argument: $$\delta^{+}(q,PQa) = \delta(\delta^{+}(\delta^{+}(q,P),Q),a)$$ and the proof: $$L = \delta^{+}(q,PQa) = \delta(\delta^{+}(q,PQ),a) = \delta(\delta^{+}(q,P),Q),a) = P$$ Is is OK now? Oct 23 comment Extended transition function of a DFA - a proof The assumption is $q_{0}(PQ) = \delta^{+}(\delta^{+}(q_{0},P),Q)$ and the argument is $q_{0}(PQa) = \delta^{+}(\delta^{+}(q_{0},P),Q)a$ $$L=q_{0}(PQa)=\delta(\delta^{+}(q_{0},PQ),a)=\delta(\delta^{+}(\delta^{+}(q_{0}‌​,P),Q),a)=\delta^{+}(\delta^{+}(q_{0},P),Q)a=P$$ Is it done right? I used $q_{0}(Pa) = \delta(\delta^{+}(q_{0},P),a)$ Oct 21 asked Extended transition function of a DFA - a proof Jun 23 comment Integral with a substitution @user6312, @Américo Tavares I know it :), but it didn't make any difference then. I didn't know what I should do next. Jun 23 awarded Supporter