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1d
accepted The orthogonal operator onto $ran(T)$
1d
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
Nov
18
asked Sufficient condition for the block matrix $\big(\begin{smallmatrix} B & A^T \\ A & 0 \end{smallmatrix} \big)$ to be invertible
Nov
14
awarded  Nice Question
Oct
22
asked Evaluate the angle between two curves at their intersection: $y=x^2+1, x^2+y^2=1$