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 Tumbleweed
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  • 0 posts edited
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  • 45 votes cast
May
4
comment Finding the matrix representation of a linear transformation $ T: P_{3} \to \text{M}_{2 \times 2} $.
By linearity it suffices to check what $T$ does to the basis elements, how about calculating $T(1), T(x),T(x^2),T(x^3)$ and writing it as a linear combination of those? ($T(a_0+a_1x+\dots)=a_0T(1)+a_1T(x)+\dots$)
Jan
18
awarded  Tumbleweed
Jan
14
accepted How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
Jan
14
comment How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
Ahh, that explains quite a lot, now (at least in 1D, but that's enough to convince me for now) it reduces to $e^{-\beta U}\mathcal{L}^* g$! Do you want to post this as an answer so I can mark this question as answered?
Jan
13
revised How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
added case in 1D
Jan
13
revised How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
suggested approach
Jan
13
comment How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
Yes, we are. That's just what I tried, but I'm stuck with said term (added it in the question) and I don't see a way to rescue this ansatz, so I wanted to ask if there'd be a more sophisticated approach.
Jan
13
revised How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
added my current progress
Jan
13
asked How to prove $\mathcal{L}e^{-U}g = e^{-U}\mathcal{L}^*g$ for Fokker-Planck operators
Jan
11
asked Finding a homeomorphism between quadratic polynomials
Dec
13
awarded  Citizen Patrol
Sep
24
awarded  Autobiographer
Sep
21
comment What can I think of the function $F$ that's being used for most(?) explicit first order ODEs?
@"Are there any such sentences in the book? Did the professor say anything at all while writing those formulas?" There aren't any explanations in the book and I couldn't attend the lecture (as well as multivariate calculus which might explain the above problem) because I had to do alternative service. Maybe he said something in the lectures, but my fellow students didn't have a clue either.
Sep
21
comment What can I think of the function $F$ that's being used for most(?) explicit first order ODEs?
Thank you very much! Finally I have an idea of what $F$ looks like! The last equation makes perfectly sense now and the second last the problem reduced to the question why there's an $F$ after $\frac{\Delta F}{\Delta y}$ and not after $\frac{\Delta F}{\Delta t}$, but I guess this has something to do with $y$ being a vector. However, this is not part of the original question anymore.
Sep
21
accepted What can I think of the function $F$ that's being used for most(?) explicit first order ODEs?
Sep
20
comment What can I think of the function $F$ that's being used for most(?) explicit first order ODEs?
That still leaves me without a clue what the last term (and others) means though, isn't there any intuitive explanation what the single arguments are for?
Sep
20
asked What can I think of the function $F$ that's being used for most(?) explicit first order ODEs?
Jul
2
awarded  Curious
Apr
26
accepted Is every invertible matrix over an algebraically closed field diagonalisable?
Apr
23
answered Is every invertible matrix over an algebraically closed field diagonalisable?