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 Aug26 awarded Popular Question Nov6 awarded Notable Question Feb10 awarded Popular Question Nov20 comment Orthogonal functions - a coupling of the function So if $g(x) = x - 1, x \in \mathbb{C}$, then $\overline{g(x)} = x + 1$ or $\overline{g(x)} = \overline{x} - 1$ ? Nov20 comment Orthogonal functions - a coupling of the function Nov20 asked Orthogonal functions - a coupling of the function Nov20 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. Nov20 awarded Editor Nov20 revised Fourier transform of $f(x) = e^{-x^2}$ added 9 characters in body; edited tags Nov20 asked Fourier transform of $f(x) = e^{-x^2}$ Nov1 accepted Extended transition function of a DFA - a proof Nov1 accepted Taylor series for different points… how do they look? Oct31 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? Oct31 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. Oct31 asked Taylor series for different points… how do they look? Oct24 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 $|Q| = 0$: $$|Q| = 0 \rightarrow Q = \epsilon$$ $$L = \delta^{+}(q,P\epsilon) = \delta(\delta^{+}(q,P),\epsilon) = \delta^{+}(q,P)$$ $$P = \delta^{+}(\delta^{+}(q,P),\epsilon) = \delta^{+}(q,P)$$ $$L = P$$ Oct24 awarded Commentator Oct24 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? Oct23 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 $q_{0}(Pa) = \delta(\delta^{+}(q_{0},P),a)$ Oct21 asked Extended transition function of a DFA - a proof