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User:GregariousMadness

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Welcome to my user page! I'm GregariousMadness, nice to meet you! ๑(◕‿◕)๑

About me

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I'm a graduate student at Duke University focusing on deep learning and mathematical foundations of artificial intelligence. My research interests lie at the intersection of topology, differential geometry, and neural network architecture design. I also love competition math, having done it since middle school, and am currently trying to make the article Olympiad mathematics.

I also like going through the list of requested Mathematics articles at Wikipedia:Requested_articles/Mathematics and making new articles when I can. I've created Taniyama's problems, Thurston's 24 questions, Darboux cyclide, Yang-Baxter operator, Ultradistribution, Brokard's theorem (projective geometry), and Kobayashi's theorem thanks to that list.

Wikipedia experience

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I'm still a new editor, but I try my best! I'm a gregarious person to a fault, and that may leak into my edits at times. Please be patient with me, and I'll be patient with you!

Articles I created

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Articles I'm helping clean up

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My favorite math proofs

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Proof of the infinitude of the primes using topology

Consider the set for and . Each set is a two-way infinite arithmetic progression. Now call a set open if either O is empty, or if to every there exists some with . Clearly, the union of open sets is open again. If are open, and with and , then . So, any finite intersection of open sets is again open. Thus, this family of open sets induces a topology on . Any number has a prime divisor p, and so is contained in Thus .

Suppose for the sake of contradiction that is finite; then would be a finite union of closed sets, and hence closed. Thus, would be an open set, a contradiction since any nonempty open set is infinite.

Proof of the fundamental theorem of algebra using Liouville's theorem

Suppose for the sake of contradiction that there is a nonconstant polynomial with no complex root. Note that as . Take a sufficiently large ball ; for some constant there exists a sufficiently large such that for all .

Because has no roots, the function is entire and holomorphic inside , and thus it is also continuous on its closure . By the extreme value theorem, a continuous function on a closed and bounded set obtains its extreme values, implying that for some constant and .

Thus, the function is bounded in , and by Liouville's theorem, is constant, which contradicts our assumption that is nonconstant.