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# Ben Reiniger@bmreiniger@mathstodon.xyz

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"Study shows for first time that a free, online course can change students' mindsets towards their mathematical abilities, leading to increased academic achievement" sciencedaily.com/releases/2018

can reduce waste in cab fleets:
nature.com/articles/d41586-018
(OK, that itself is not so surprising, but the 30% drop in cabs is pretty sweet)

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sometimes i sit at my desk scrolling through mastodon with a very serious expression to give the impression that im working

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We built the Setteract.

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Pros of not being a doctor:
Being able to instantly identify as such any spams referring to me as "Dr Lastname"

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Just wondering, is there some cultural influence on the preference in the type of optimization objective (min vs max)? So far my lecturers tended to go for min, while many online lectures I've watched went for max...

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Correlations

Mercator projection of the contour plot of the sum of the absolute values of the dot products of a point on the sphere with the vertices of a rotating octahedron.

Source code and explanation: community.wolfram.com/groups/-

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for a,b in a group G: if ab != ba then aba != e , equivalently the proof that aba = e implies ab = ba. @ColinTheMathmo Thought you would appreciate

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New proof of triangle inequality (in inner product spaces): rjlipton.wordpress.com/2018/05

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of course there are lots of different $\mathcal{N}(0, 1)$ random variables so clearly the next step is to designate one of them as the "canonical standard normal random variable"

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New entry!
Nice Neighbors: A Brief Adventure in Mathematical Gamification
Article by Chris Staecker
In collections: Easily explained, Puzzles
Last year I came across a strange graph theory problem from digital topology. I turned it into a video game to help wrap my mind around it. It was fun to play, so I made it into a web game that other people...
URL: cstaecker.fairfield.edu/~cstae

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A from Matthew Doty, via John Baez:

Let a∘b = ⌊a/b⌋, for natural numbers a and b.
Show that (a∘b)∘c = a∘(b×c) for all a,b,c.

There's a proof using category theory! forum.azimuthproject.org/discu

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(Inviting someone to a baby shower) would you like to be at the unboxing?

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@amiloradovsky looks like the answer is yes: $\require{AMScd} \begin{CD} K(X) @>{ch}>> H(X;\mathbb Q);\\ @VVV @VVV \\ K(Y) @>{ch}>> H(Y;\mathbb Q); \end{CD}$

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Last night my friend asked to use a USB port to charge his cigarette, but I was using it to charge my book. The future is stupid.

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Thanks to @sam_hartburn, we now have a emoji