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Jul
26
comment length of continuously differentiable curves
An informal interpretation: think of $\lvert{\gamma'(t)}\rvert$ as your speed at time $t$ as you move along the curve. The distance you travel is your speed multiplied the time traveled -- with changing speed this is formulated as exactly that integral.
Jul
26
accepted The Riesz transform kernel satisfies Hörmander's condition
Jul
26
asked The Riesz transform kernel satisfies Hörmander's condition
Jul
12
awarded  Yearling
Jun
27
comment Property of complex borel measures with absolutely convergent Fourier series (Wiener algebra)
Thanks. If you don't mind, do you know how an expression $f \ast \mu$ is interpreted? Where $f$ is continuous and $\mu$ is a complex Borel measure. And by extension we can have $f \in L^p$.
Jun
27
accepted Property of complex borel measures with absolutely convergent Fourier series (Wiener algebra)
Jun
27
asked Property of complex borel measures with absolutely convergent Fourier series (Wiener algebra)
May
18
revised Non measurable subset of a positive measure set
added 106 characters in body
May
18
answered Non measurable subset of a positive measure set
Apr
20
comment Characterization of measurable sets $E$ with $|E|_e<\infty$
Late to the party but: $\lvert E{\rvert}_e < \infty$ is needed for the other direction to show that when writing an open set $G \supset E$, $\lvert G-E\rvert < \epsilon$ as a countable union of nonoverlapping intervals, the tail of the union has small measure. This direction does not appear to need this, since the finite union of intervals is so nice and the other $N_1$ and $N_2$ are already small.
Mar
7
comment How could I prove whether the infinite series $n!/n^n$ converges or diverges?
@JulianRachman "By this result, we can see that the series does not diverge, however it is still inconclusive that the series converges." This is not right at all. The theorem you used gives sufficient conditions for convergence, not necessary. And "converges xor diverges" is a tautology, so the statement itself does not make sense. (edit: it's fixed now, though only the ratio test is necessary, undid -1).
Feb
22
answered this is not a vector space,is it?
Dec
21
awarded  Caucus
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Jul
12
awarded  Yearling
Apr
9
comment Prove that if b is coprime to 6 then $b^2 \equiv 1 $ (mod 24)
Since $b^2$ is a perfect square, we have that $b^2 = 1,4,9,16$ modulo $24$. All of these, besides $1$, require that $2 | b$ or $3 | b$.
Mar
9
comment Is this fraction even possible to put into partial fractions?
@Wolff Think inverse trigonometric functions.
Feb
6
comment Real Analysis Question concerning existence of curve and derivative?
Suppose $f(0) \ne 0$. Use continuity of $f$ to obtain a contradiction.
Jan
31
answered How to find the inverse function?