@MikeMiller @TedShifrin Here's a similar type of mindfuck example: $1+z^2$ has no logarithm in $\Bbb C \setminus [-i, i]$, but it has a square root. Obvious if one knows theory, not so easy otherwise.
I mean it took me 10 (maybe 15) years of life being self conscious about other people's skills before I realized that is counterproductive.
Those 10-15 years were not exclusively spent in my undergrad or PhD, no, it's just a general life phenomenon that people are too quick to compare achievement to others
@MikeMiller: Well, I remember that once I asked my topology professor that what is the intuitive meaning of "compact space" he answered that "it has no intuition and is abstract". This happened to me a lot. And the result I am self-study from beginning after 8 years.
We are like attack ships on fire off the shoulder of Orion, or C-beams glittering in the dark near the Tannhäuser Gate. Soon to be lost in time, like tears in rain
Milnor's book preface: Pontrjagin's many contributions to mathematics are the more remarkable in that he is totally blind, having lost his eyesight in an accident at the age of fourteen.)
"The editors must emphasize, however, that neither the lecturers nor the note-takers have any responsibility for any inaccuracies which may remain: they are an act of God."
Consider two matrices acting on points in $(0,1)^2$ in the real plane, $$h_s=\begin{pmatrix} e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix}.$$
and $$g_s=\begin{pmatrix} 1-e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix}.$$
How do you compose these transformations?
I tried to do $h_s...
$$ \int_{a}^{\infty} p\left(y, s \mid v^{\prime}, t^{\prime}\right) L_{y} p(v, t \mid y, s) d y=\int_{a}^{\infty} p(v, t \mid y, s) L_{y}^{*} p\left(y, s \mid v^{\prime}, t^{\prime}\right) d y $$
Here, p is a conditional density
Here, L is the differential operator from the fokker-planck equation