1,842 reputation
2622
bio website
location
age
visits member for 2 years, 3 months
seen 7 hours ago

Aug
25
awarded  Nice Answer
Aug
25
answered Factoring $(x+1)(x+2)(x+3)(x+4)+1$
Jun
11
answered prove that $f(x)=\sum _{n=0}^{\infty}\frac{\cos(nx)}{2^n}$ is continuous
Jun
2
awarded  Yearling
Apr
9
revised Integral of 1/(8+2x^2)
added 1 characters in body
Apr
9
answered Integral of 1/(8+2x^2)
Nov
11
comment Calculate $i ^ {i+1}$ and also $i^{i^{i^{\dots}}}$
Similar question: math.stackexchange.com/questions/191572/…
Sep
29
answered Number of path from $A$ to $B$ in a grid that can be traveled either rightward or upward
Sep
26
comment Rating changes and probability calculations for chess world championship
Not an answer to your question, but you might find it interesting: chessbase.com/Home/TabId/211/PostId/4011007/…
Sep
25
answered A function continuous in both arguments
Sep
17
comment Physical interpretation of $q$-deformation
Thanks, but I have a few questions: So I see there is a difference since in a quantum system we have $[x_i, x_j] = 0$, while we have $ x_ix_j=qx_jx_i$ for $q$-deformation. But why are there no exponentials here (like in your comparison)? Shouldn't there be a physical interpretation of $x_ix_j=qx_jx_i$? If it depends on the situation, I would like to see some references to that. Also, could you please give me references about your answer to my first question?
Sep
14
comment Help with limit $\displaystyle\lim_{t\to 0^{+ }}\sup_{x\in[0, \infty)} |e^{-t^2-2tx}f(x+t)-f(x)|=0$..
Good! You learn more that way :)
Sep
14
answered Help with limit $\displaystyle\lim_{t\to 0^{+ }}\sup_{x\in[0, \infty)} |e^{-t^2-2tx}f(x+t)-f(x)|=0$..
Sep
14
answered Vector Field Problem
Sep
14
comment Vector Field Problem
Do you know the definition of $\hat{r}$?
Sep
13
awarded  Promoter
Sep
10
revised Physical interpretation of $q$-deformation
added 12 characters in body
Sep
10
asked Physical interpretation of $q$-deformation
Sep
2
awarded  Scholar
Sep
2
comment Show that an operator is bounded (from Reed and Simon)
Very nice! Thanks for your patience.