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#average

1 post1 participant0 posts today
Public *
Marie 🇸🇪

The saying, that you’re the average of the five people you spend most time with, suggests that people we surround ourselves with can significantly influence our thoughts, behaviours & overall mindset. The idea is that our closest relationships can positively shape our attitudes, habits & even our aspirations.
On the other hand, if your circle is negative or unambitious, it can affect your outlook & goals. Does this resonate with your experiences?

YOU'RE THE AVERAGE OF THE FIVE PEOP YOU SPEND THE MOST TIME WITH. Text on beige background.
#average#people#influence
Public
Bharath Krishnan

In a magazine article [1] on problems and progress in quantum field theory, Wood writes of Feynman path integrals, “No known mathematical procedure can meaningfully average an infinite number of objects covering an infinite expanse of space in general. The path integral is more of a physics philosophy than an exact mathematical recipe.”

This article [2] provides a method for averaging an arbitrary collection of objects; however, the average can be any number in the extension of the range of these objects. (Note, an arbitrary collection of these objects is a function.)

Question: Suppose anything meaningful has applications in quantum field theory. Is there a way to meaningfully choose a unique, finite average of a function whose graph matches the description in Wood's quote?

For more info, see this post [3].

[1]: quantamagazine.org/mathematici

[2]: arxiv.org/pdf/2004.09103

[3]: math.stackexchange.com/q/50520

Quanta Magazine · Mathematicians Prove 2D Version of Quantum Gravity Really Works | Quanta MagazineIn three towering papers, a team of mathematicians has worked out the details of Liouville quantum field theory, a two-dimensional model of quantum gravity.
#PathIntegral#quantum#FeynmanPathIntegral
Public *
Bharath Krishnan

I finally know what I want.

Let \(n\in\mathbb{N}\) and suppose function \(f:A\subseteq\mathbb{R}^{n}\to\mathbb{R}\), where \(A\) and \(f\) are Borel. Let \(\text{dim}_{\text{H}}(\cdot)\) be the Hausdorff dimension, where \(\mathcal{H}^{\text{dim}_{\text{H}}(\cdot)}(\cdot)\) is the Hausdorff measure in its dimension on the Borel \(\sigma\)-algebra.

§1. Motivation

Suppose, we define everywhere surjective \(f\):

Let \((A,\mathrm{T})\) be a standard topology. A function \(f:A\subseteq\mathbb{R}^{n}\to\mathbb{R}\) is everywhere surjective from \(A\) to \(\mathbb{R}\), if \(f[V]=\mathbb{R}\) for every \(V\in\mathrm{T}\).

If f is everywhere surjective, whose graph has zero Hausdorff measure in its dimension (e.g., [1]), we want a unique, satisfying [2] average of \(f\), taking finite values only. However, the expected value of \(f\):

\[\mathbb{E}[f]=\frac{1}{{\mathcal{H}}^{\text{dim}_{\text{H}}(A)}(A)}\int_{A}f\, d{\mathcal{H}}^{\text{dim}_{\text{H}}(A)}\]

is undefined since the integral of \(f\) is undefined: i.e., the graph of \(f\) has Hausdorff dimension \(n+1\) with zero \((n+1)\)-dimensional Hausdorff measure. Thus, w.r.t a reference point \(C\in\mathbb{R}^{n+1}\), choose any sequence of bounded functions converging to \(f\) [2, §2.1] with the same satisfying [2, §4] and finite expected value [2, §2.2].

[1]: mathoverflow.net/questions/476

[2]: researchgate.net/publication/3

#HausdorffMeasure #HausdorffDimension
#EverywhereSurjectiveFunction
#ExpectedValue
#Average
#research

MathOverflowIs there an explicit, everywhere surjective $f:\mathbb{R}\to\mathbb{R}$ whose graph has zero Hausdorff measure in its dimension?Suppose $f:\mathbb{R}\to\mathbb{R}$ is Borel. Let $\text{dim}_{\text{H}}(\cdot)$ be the Hausdorff dimension, where $\mathcal{H}^{\text{dim}_{\text{H}}(\cdot)}(\cdot)$ is the Hausdorff measure in its
Public
gentlegardener

Important interview with #privileged #treasurysec #bessent on #BloombergTV
#federalworkforce #dogees #tusk #ukraine #zelensky

#Protip #media interviews: Never call anyone an “ #averageAmerican”. Nor their kids #average. You have no experience of life outside your bubble of #privilege. Working for the gummint was steady work with steady benefits. Your idea is to make everyone grovel.

And learn to stop smirking on camera. Save it for #SiliconValley @bloomberg

Public
ReallyCanadianFly

Shoot for Good Enough doesn't fit as well on motivational influencers' t-shirts though. I once read "Perfectionism is the ultimate form of self-abuse", which was probably in Chicken Soup for the Soul or some such dreck, but it does strike a certain chord of truth too.

#mediocrity #average #perfectionism #goodenough #happiness

getpocket.com/explore/item/it-

PocketIt’s Okay to Be Good and Not GreatWhat if striving to be great is what’s holding you back?
Public
Ingo Schramm

>> So, there you have it. The interiors of our homes, coffee shops and restaurants all look the same. The buildings where we live and work all look the same. The cars we drive, their colours and their logos all look the same. The way we look and the way we dress all looks the same. Our movies, books and video games all look the same. And the brands we buy, their adverts, identities and taglines all look the same.

#average #culture #bored

alexmurrell.co.uk/articles/the

Alex MurrellThe age of average — Alex MurrellIn the early 1990s, two Russian artists named Vitaly Komar and Alexander Melamid hired a market research firm to survey the public on what they wanted in a work of art. Across 11 countries they then set about painting a piece that reflected the results. Each piece was intended to be a unique a colla
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