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Two students who discovered a seemingly impossible proof to the Pythagorean theorem in 2022 have wowed the math community again with nine completely new solutions to the problem.

While still in high school, Ne'Kiya Jackson and Calcea Johnson from Louisiana used trigonometry to prove the 2,000-year-old Pythagorean theorem, which states that the sum of the squares of a right triangle's two shorter sides are equal to the square of the triangle's longest side (the hypotenuse). Mathematicians had long thought that using trigonometry to prove the theorem was unworkable, given that the fundamental formulas for trigonometry are based on the assumption that the theorem is true.

Jackson and Johnson came up with their "impossible" proof in answer to a bonus question in a school math contest. They presented their work at an American Mathematical Society meeting in 2023, but the proof hadn't been thoroughly scrutinized at that point. Now, a new paper published Monday (Oct. 28) in the journal American Mathematical Monthlyshows their solution held up to peer review. Not only that, but the two students also outlined nine more proofs to the Pythagorean theorem using trigonometry.

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Calculator: https://www.omnicalculator.com/everyday-life/dilution-ratio

If I type in the dilution ratio and final volume it calculates the concentrate amount and water amount but I don't know how it does that and want to find out how it does that

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submitted 10 months ago* (last edited 10 months ago) by dysprosium@lemmy.dbzer0.com to c/math@lemmy.ml
 
 

Then I am stuck. I think the provided answer contains an error. But even if they are right, why does this last step equal f(x,y) + g(y) ????

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(For the sake of intuition, 1/√0=0)

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Algebra question (lemmy.world)
submitted 1 year ago* (last edited 1 year ago) by drawerair@lemmy.world to c/math@lemmy.ml
 
 

I'm thinking re the latest vid of @mindyourdecisions

No need to view his vid. Here's the problem –

Brian has some boxes of paper clips. Some boxes hold 10 clips and some boxes hold 100. He has some paper clips left over. He has 3 more boxes with 100 paper clips than he has boxes with 10 paper clips. He has 2 fewer paper clips left over than he has numbers of boxes with 100 paper clips. What number of paper clips could he have?

  • let x1 be the number of boxes with 10 clips
  • x2 be the number of boxes with 100 clips
  • n be the number of leftover clips

I thought of 100x2 = 10x1 + 300

Is that equation right? Something tells me I shouldn't equate 100x2 to 10x1 plus 300. Something tells me I shouldn't make an equation re number of clips as it isn't explicit in the problem. I'm confused.

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Hi,

I found online a nice (and seemed easy) math problem.

Rocket A travel from Mars to Earth in 200 days
Rocket B travel from Earth to Mars in 150 days, but take off 30 days later

When they cross each other, which one is the closet to the earth ?

So they give a "flat" answer, without giving any explanation on how they reach this conclusion.

What would be your simplest Mathematical solution for this ?

Thanks.

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submitted 1 year ago* (last edited 1 year ago) by dataprolet@lemmy.dbzer0.com to c/math@lemmy.ml
 
 

So this is bugging me for a while and I'm just do dumb to get how I solve this, but here's the situation:
Given I take a local backup of my system daily and have a retention policy that keeps a backup of the past 7 days each, a backup of the past 4 weeks each and a backup of the past 6 month each. That's either 17 backups or less if you consider some backups being counted as a daily and weekly or as a weekly and monthly. But that's not that important.
The interesting part is, that I also take a remote backup of my local backup daily, which has the same retention policy, so it's cascading. Here there is obviously a huge overlap of backups, but I can't wrap my head around, how I calculate this.
Is anybody willing and/or interested to solve this for and with me?

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This year's Abel Prize has just been awarded to Michael Talagrand. I didn't knew about his work, but it seems really interesting and he made an effort to make it really accessible both to read and access.

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submitted 1 year ago* (last edited 1 year ago) by wargreymon2023@sopuli.xyz to c/math@lemmy.ml
 
 

Isn't it just "composite"?

Every arrow in category can be composed, the set(or class or whatnot..) of that is composite.

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submitted 1 year ago* (last edited 1 year ago) by wargreymon2023@sopuli.xyz to c/math@lemmy.ml
 
 

We have "triangle" "rectangle" "pentagon"...etc "tetrahedral" "cube" "octahedron" ..etc

Instead of having to say "group" all the time, like "dihedral group" "cyclic group", if we make it into one word it will sound more like an elementary mathematical object.

What would be a nice suffix for group?

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Hello.

I am currently inventing a language, and have created a base 4 number system for it. Unfortunately, I am horrible with numbers, even in decimal. So it was a hard slog. But I finally got there.

It would be great if I could know of any practical applications quaternary has (if any), so I can incorporate it into the language and make it more naturalistic. Thanks.

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I feel like this has to be a math/logic thing that has a name already and I wanna know what it's called so I can look it up when I'm no longer extremely drunk.

In this phone game the objective is to get all the people on all the same color floors with as few stops at any floor as possible. When the last few moves look like this, you just have to go through in the right order and only stop at each stop once (except the first/last floor).

But sometimes there's different little sub-sets of pairs inside the bigger set of pairs that are self-contained, and for each one of those there's another floor that has to be started and stopped on to complete that loop. That makes the minimum number of moves to solve: the sum of the number of pairs in both sub-sets together plus the number of subsets. (And only counting the number of pairs in both subsets because if one of the pairs is already matched it won't count for the moves).

So like these two are all one big continuous loop: A-E, B-A, C-B, D-C, E-D and A-B, B-E, C-A, D-C, E-D

And this one has one already matched leaving a single complete loop in need of matching: A-B, B-E, C-A, D-D, E-C

These ones, however, have two loops. one loop that's three floors long (four moves) and one that's two floors long (three moves): A-B, B-C, C-A, D-E, E-D and A-D, B-E, C-A, D-C, E-B

And these ones have one already matched pair, and two sub-sets of two that still need to be matched: A-B, B-A, C-C, D-E, E-D and A-D, B-B, C-E, D-A, E-C

What is this called?

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@math The West Virginia University Provost's Office is recommending closing the MS and Ph.D. programs in Math. It is the *only* Ph.D. program in Math in the entire state, and about 10% of all WVU Ph.D.'s are in Math.

Please consider signing this petition to save the program: https://chng.it/yPZDTTsfBk

#ProtectWVUMath

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submitted 2 years ago* (last edited 2 years ago) by JohnDClay@sh.itjust.works to c/math@lemmy.ml
 
 

Wiki: https://en.wikipedia.org/wiki/Mandelbrot_set

Here are a bunch of other visualizations: I don't know how artistic or data-driven some of these are, but they look very interesting. I think the nebula-looking one measures how often a point is visited?

Black and Green mandelbrot set

The Bulbic Mandelbrot Set

Bulbic Mandelbrot Set

https://www.deviantart.com/metafractals/art/The-Bulbic-Mandelbrot-Set-811453986

A Nebulabrot

Nebula looking mandelbrot set

https://mathematica.stackexchange.com/questions/89458/how-to-make-a-nebulabrot

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So I'm gearing up to take a calculus 1 exam, and this question is on the sample test. My initial thought was that since we are looking for F(9), and F(x) is an antiderivative of f(x), I can just use the integral of the equation of f(x) at 9, which is f(x) = -2x/3 + 5, which, when integrated, becomes -x^2/3 + 5x + 2 (C = 2 because F(0) = 2). Thing is, though, that won't give me any of the answers listed. And even after taking the integral of all of the equations of f(x), I still have no idea how to produce any of the answers in the multiple choice.

I'm super stumped on this one. Any help would be welcome!

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BOINC is a free tool you can download to participate in several different math research projects. It runs on Windows, MacOS, Linux, and even Android. Each project gives you fun stats and graphs about your participation, many of them will even credit you individually for your discoveries (such as finding a new prime) on their website or in their published papers.

Here's a few of the projects available (emoji legend at bottom of post):

🏆💚❤️✖️✒️🔓 Amicable Numbers Independent research project that uses Internet-connected computers to find new amicable pairs. Currently searching the 10^20 range.

🎓🔓✖️ NFS@Home - Lattice sieving step in Number Field Sieve factorization of large integers. Many public key algorithms, including the RSA algorithm, rely on the fact that the publicly available modulus cannot be factored. If it is factored, the private key can be easily calculated.

🏆🎓💚❤️✖️🔓 Numberfields@home - Research in number theory. Number theorists can mine the data for interesting patterns to help them formulate conjectures about number fields.

🔓 ODLK1 - Building a database of canonical forms of diagonal Latin squares of the 10th order

🔓💚❤️ SRBase - Attempting to solve Sierpinski / Riesel Bases up to 1030.

🔓✖️PrimeGrid - Find new prime numbers!

Gerasim@home - research in discrete mathematics and logic control. Testing and comparison of heuristic methods for getting separations of parallel algorithms working in the CAD system for designing logic control systems

🔓✖️ Loda@home - LODA is an assembly language, a computational model, and a distributed tool for mining programs. You can use it to generate and search programs that compute integer sequences from the On-Line Encyclopedia of Integer Sequences® (OEIS®). The goal of the project is to reverse engineer formulas and efficient algorithms for a wide range of non-trivial integer sequences.

🔓🎓Rakesearch - The enormous size of the diagonal Latin squares space makes it unfeasible to enumerate all its objects straightforwardly in reasonable time. So, in order to discover the structure of this space, sophisticated search methods are needed. In RakeSearch project, we implement an application that picks up separate pairs of mutually orthogonal DLSs, which allows to reconstruct full graphs of their orthogonality.

🔓✒️ Ramanujan machine - Discover new mathematical conjectures

Legend:

🔓 - Publishes data openly and regularly. Note many projects publish papers detailing the results of their work, this icon means that they regularly publish the source materials as well/the results of the computation in an open fashion.

🏆 - Credits individual crunchers for discoveries, such as finding a new black hole or prime number

🎓 - Sponsored by major university or research institute.

💚 - Supports NVIDIA GPU/graphics card (all projects should be assumed to support CPU unless otherwise stated)

❤️ - Support AMD GPU (all projects should be assumed to support CPU unless otherwise stated)

✖️ - Supports OS X (all projects should be assumed to support Windows & Linux unless otherwise stated)

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ChatGPT will gobble up every symbolic manipulation task I give to it. At worst, sometimes I have to check its output and point out anything weird, then it'll correct it.

I'm writing pages over pages of scary differential equations and the damn thing is saving me lots of time on it. And everything checks out! I wonder about GPT 4, since it is supposed to give correct answers without help as often as the average calculus student...

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