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Feb
22
comment Evaluating $f(x) f(x/2) f(x/4) f(x/8) \cdots$
@EgorMaximenko: this was an old comment. After a few minutes, you can no longer edit comments. The actual answer isn't from me. So unfortunately, I cannot fulfill your request.
Nov
19
awarded  Yearling
May
9
awarded  Caucus
Apr
27
comment “adding” numbers $\in\mathbb{U\left(2\right)}\times \mathbb{R}^+_0$
@WimC but that is directly the quaternions and reduces the space I'm working in. It's what you get when $p=q=0$. I explicitly want to allow arbitary real $p$ and $q$
Apr
27
revised “adding” numbers $\in\mathbb{U\left(2\right)}\times \mathbb{R}^+_0$
deleted 8 characters in body
Apr
27
asked “adding” numbers $\in\mathbb{U\left(2\right)}\times \mathbb{R}^+_0$
Mar
26
revised Two-point Taylor expansion with one assymptotic point?
added 7 characters in body
Mar
25
revised Two-point Taylor expansion with one assymptotic point?
edited title
Mar
25
revised Two-point Taylor expansion with one assymptotic point?
Taylor is a name and thus should be capitalized.
Mar
25
asked Two-point Taylor expansion with one assymptotic point?
Mar
23
comment $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$?
and I didn't downvote
Mar
23
comment $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$?
Heh. The answer is fine but that final step is so minimal, it just feels weird to keep it like that.
Mar
22
comment $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$?
$=\sqrt{2}\sqrt{a}$
Mar
9
comment Parallel functions.
I do not have a formal proof for this but that's what my intuition would say about the topic. I guess you could get some other piecewise cases, like a periodicly repeating half-circle which would result in a mirrored and shifted version of itself in the cusping limit. Such cases have to be considered more carefully, but even then, they won't never form a cusp. (double negative)
Mar
9
comment Parallel functions.
the only way that a cusp is reached at the very same time uniformly everywhere is to have a curve of constant curvature, e.g. a circle or line. All others will not form equal cusps. (The circle forms a point eventually. The line will only produce more lines or, if you prefer, it'll form cusps at infinity.
Mar
9
comment Parallel functions.
The limit distance should simply be the minimum radius of curvature of the curve, right? - at that point, you'd get a cusp. After that, you get cross-overs.
Jan
13
revised Recursive Integration over Piecewise Polynomials: Closed form?
info about image reliability
Dec
29
comment Show $\lim\limits_{n\to\infty} n \ln\left(1-\frac{1}n\right) = -1$
@Limitless well, the space of possible events is limitless. (not the same as infinite, I know...)
Dec
15
comment Evaluating $f(x) f(x/2) f(x/4) f(x/8) \cdots$
@hardmath yeah, just like the Fouriertransform of it, which my question was all about. I wonder if there are algorithmic ways to quickly approximate the limit more directly than doing a finite sum or product, stopping at an arbitrary term. I suspect the Fabius function to be potentially valuable for interpolation and/or as wavelet kernel. Googling something along those lines yielded nothing though.
Dec
15
comment Evaluating $f(x) f(x/2) f(x/4) f(x/8) \cdots$
This approach can not hold beyond the first zero of $F\left(x\right)$ though, can it? Because the series has to become $-\infty$ at that that point to satisfy it. And beyond that point, it would have to become complex to go negative, which it definitely does... (see my graph of it here )