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1d
comment Solving integral $\int \frac{3x-1}{\left(x^2+16\right)^3}$
You have already computed the partial fraction decomposition. This fraction cannot by decomposed any more. Go on to the next step ... integrate ... see above!
1d
comment if f is continuous almost everywhere , must there exist a function g such that g=f almost everywhere and g is continuous?
@user156441 Feel free to ask this as a new question ...
Apr
14
comment Understanding Asymptotic Notation of a constant
In what sense is Knuth's definition stronger? If $c := \limsup |f(n)| > 0$, we have that finally $|f(n)| > \frac c2$. That is, finally $1 \le \frac 2c|f(n)|$ or $1 = O(f)$.
Apr
14
comment Matrix norm optimization problem : $\min_{\textit{ }x} \| A x B \|_4$, $x$ in the “unit” circle
$x=0$ (in the correct format) will do it.
Apr
13
comment Every set with more than point admits a permutation with no fixed point and the Axiom of Choice
Sageev, G.: Notices Amer. Math. Soc. 20 (1973), A22.
Apr
13
comment Showing that an operator is the projection
1-codimensional subspace. As $L \ne 0$ (from $w \ne 0$), we have $L(w) = w$, We will show that $V = \ker(L) + span(w)$. Let $v \in V$, then $v = v - L(v) + L(v)$, with $(v - L(v)) \in \ker(L)$, $L(v) \in span(w)$. Hence $\ker(L)$ is 1-codimensional
Apr
13
comment Showing that an operator is the projection
Sry. Had an error in writing, I intended to write $w^\bot \subseteq \ker(L)$, as you state ... now use, that both are hyperolanes ...
Apr
13
comment Diffeomorphism between a sphere and ellipsoid in $\mathbb R^3$.
$f^{-1}(x) = (ax_1, bx_2, cx_3)$?!
Apr
13
comment Diffeomorphism between a sphere and ellipsoid in $\mathbb R^3$.
If $S = \{x \in \mathbf R^3 \mid \|x-x_0\| = r\}$ is any sphere, note that $g \colon S^2 \to S$, $x \mapsto r(x+x_0)$ is a diffeomorphism between $S$ and $S^2$. Compose if with $f$.
Apr
13
comment Continuity dot product
Somewhere you'll have to use that $\mathbf R$'s topology is given by the absolute value or the order ...
Apr
10
comment Cardinality of $\{ (x, y) \in \mathbb{R}^2 \mid \left| x \right| + \left| y \right| = 1 \}$ and $\{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1 \}$
I like the total overkill way :)
Apr
10
comment Limit of a composition of functions
@Johann Added something.
Apr
10
comment Limit of a composition of functions
It holds, yes ... ${}{}{}$.
Apr
9
comment How do I show that the sequence below is a Cauchy sequence?
Replace $=0$ and the end with $\to 0$ for $n \to \infty$ ... yes.
Apr
9
comment Can find the determinant of a matrix A of size $n$ in terms of the traces of $A^m$
see here: math.stackexchange.com/questions/668374/…
Apr
9
comment Volume Integral Discriminant
It is.${}{}{}{}$
Apr
9
comment Is there a name for the integral used to define square integrable functions?
... the square of the $L^2$-norm, the norm itself is given by $$ \|f\|_{L^2} = \left(\int_{\mathbf R^n} |f(x)|^2\, dx\right)^{1/2} $$
Apr
9
comment Choosing numbers without consecutive numbers.
What exactly is $k$?
Apr
9
comment Linear forms which vanish on commutators
How do you define the trace for elements of $L(E)$?
Apr
9
comment Duality (conjugate) function for $f \in L^\infty$
Note that $\delta_x$ is not absolutely continuous with respect to Lebesgue measure, but you can take $\lambda(A) = \frac 1{\mathrm{Leb}(A_i)}\cdot \mathrm{Leb}(A \cap A_i)$, for $i$ such that $|c_i|$ is maximal, which has density $\frac 1{\mathrm{Leb}(A_i)}\cdot \chi_{A_i}$