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Jul
6
comment If a sequence of self-adjoint linear operators is convergent, show that its limit is self-adjoint.
Convergent in what sense?
Jul
6
comment Efficient Test For Commuting Matrices
... and $n^3$ can be approved: Using Strassen multiplication, we have $O(n^{2.8\cdots})$
Jul
6
comment How to solve this combinations with repetitions problem using generating functions?
Added somehing @idandi
Jul
6
revised How to solve this combinations with repetitions problem using generating functions?
added 108 characters in body
Jul
6
answered How to solve this combinations with repetitions problem using generating functions?
Jul
6
comment How to solve this combinations with repetitions problem using generating functions?
Not the explanations! The lines starting with '%...' in the source!
Jul
3
answered How to manually determine big number congruences
Jul
2
reviewed Edit Relation between dot product and magnitude of vector product
Jul
2
revised Relation between dot product and magnitude of vector product
edit formatting and written text
Jul
2
answered Proving that $V = U_1 \oplus U_2 \oplus \ldots \oplus U_k$.
Jul
2
answered $f: [0,1] \to \mathbb R$ is continuous and $\int_0^x f(t) dt = \int_x^1 f(t)dt$ for all $x \in [0,1]$, then $f(x) = 0$ for all $x \in [0,1]$.
Jul
2
answered Find the missing dollar
Jun
30
reviewed Approve Determine the roots of equation if possible
Jun
26
comment Hilbert space and uncountable cardinal
You should ask this as a seperate question, but the answer is yes. Consider $$ \mathbf K^{(\kappa)} := \left\{ x \colon \kappa \to \mathbf K \biggm| x^{-1}[\mathbf K \setminus \{0\}] \text{ is finite} \right\} $$ this space has $e^i$, $i \in \kappa$ (defined as above) as basis.
Jun
26
answered Hilbert space and uncountable cardinal
Jun
26
answered Method for finding intersection between two basis
Jun
25
awarded  Good Answer
Jun
24
answered Why does the matrix product of jacobian of coordinate transformation and jacobin of reverse coordinate transformation equals the identity matrix
Jun
22
comment Graph vertex set with a certain property
@Arthur Doesn't $V_1$ allready have the stated property? Let $v \in V(G)$ be connected to all vertices in $V_1$. Then $v$ is connected to all vertices in $V_0$ (as $V_0 \subseteq V_1$), hence $v \in V_1$.
Jun
22
revised About the exact form of a gaussian kernel
fixed LaTeX