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Aug
26
awarded  Popular Question
Nov
6
awarded  Notable Question
Feb
10
awarded  Popular Question
Nov
20
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 $\overline{g(x)} = \overline{x} - 1$ ?
Nov
20
comment Orthogonal functions - a coupling of the function
At the top of faculty.wmi.amu.edu.pl/~domanski/dani2/angielski/wyk5a.pdf
Nov
20
asked Orthogonal functions - a coupling of the function
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