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Mar
29
awarded  Popular Question
Feb
23
comment The definition of open set in metric space and general topological zpace
Could u be more concrete for general metric for topological space
Feb
23
asked The definition of open set in metric space and general topological zpace
Jan
26
accepted The orthogonal operator onto $ran(T)$
Jan
26
accepted Why for dirac function $\int_{-\infty}^{\infty}\delta(x)\ dx=1$
Jan
18
awarded  Yearling
Jan
7
comment Why for dirac function $\int_{-\infty}^{\infty}\delta(x)\ dx=1$
@AdamHughes we always use it, but up till now I find that it is bad, since how can we say the function defined like that has integration is 1 but not 2?
Jan
7
asked Why for dirac function $\int_{-\infty}^{\infty}\delta(x)\ dx=1$
Dec
13
asked The orthogonal operator onto $ran(T)$
Dec
12
asked Integrate $\int_0^\pi e^{-ik\cos\theta}\sin^2{\theta} \, \mathrm{d}\theta$
Dec
11
asked The dimension of the operator if the domain has dimension 2
Dec
11
asked Lebesgue integrable discontinuity points
Dec
11
comment $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
sorry for modification again
Dec
11
revised $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
added 11 characters in body
Dec
11
comment $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
can u write more explicitly, I don't understand
Dec
11
comment $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
it is $(cos(\theta))^n$
Dec
11
comment $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
Edited. Please relook it
Dec
11
revised $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
added 29 characters in body
Dec
11
asked $\int_0^\pi g(\theta)\sin(\theta)\cos^n(\theta)\equiv 0$ imply $g(\theta)=0$?
Dec
10
awarded  Caucus