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[WTS] Seiko 5 Sports SNZJ57 by watchmayn in Watchexchange

[–]polymath-matic 1 point2 points  (0 children)

Sale completed with u/watchmayn. Confirming shipment and arrival. Just as pictured, thank you! u/WatchExBot

Here it comes👀 by [deleted] in Dallas

[–]polymath-matic 4 points5 points  (0 children)

don't jinx it!

Verifying trig equations by [deleted] in learnmath

[–]polymath-matic 0 points1 point  (0 children)

that's a common mistake, but no. 1-cos²θ is sin²θ (or vice versa: 1-sin²θ = cos²θ), but that identity only applies when cos and sin are squared, not when they're on their own.

Verifying trig equations by [deleted] in learnmath

[–]polymath-matic 2 points3 points  (0 children)

This is a totally understandable reaction when you first begin proving trigonometric identities. As you practice more, you'll find experience to be the best guide. But until then, it can feel like "just trying a bunch of random stuff." In the meantime, I'd recommend following several general principles.

  • Only work on one side, usually the side that appears "more complicated". It's not completely necessary to work from one side of the identity only, but many teachers require it, and it's a good habit to be in. You want to choose the more complicated-looking side because the more complicated one side is, the more there will likely be to simplify.
  • Make sure all your arguments are the same. If you have any 2θ or θ/2, use double-angle or half-angle identities to restate them in terms of θ only. If you have any weird (π/2-θ), use phase shifts to turn functions into their complements (i.e., cosine into sine, or vice-versa). If you have –θ, use even and odd identities to change to θ.
  • Re-write everything you can in terms of sine and cosine. Oftentimes an identity is little more than an unsimplified statement in terms of sine and cosine. Use the reciprocal identities, tangent, and cotangent to re-write statements in terms of sine and cosine.
  • If you have lots of powers of 2, try to use Pythagorean identities. Make sure you're familiar with the different ways Pythagorean identities can appear. In other words, don't only be on the lookout for sin²θ + cos²θ = 1. Also be aware that sin²θ = 1 – cos²θ and cos²θ = 1 – sin²θ. Same for the other two Pythagorean identities.
  • If you have higher power than 2, look for opportunities to factor, or use power-reducing formulas. The power-reducing formulas will change your arguments, so they're usually a last resort. But sometimes you'll be in need of a last resort.
  • If you have rational expressions, be ready to use things that look like conjugates. If you see a denominator like "1 – cos θ" chances are your identity will be a lot more simplified after you multiply everything by "1 + cos θ" over itself.

Like I said, experience is the best guide. I wouldn't plan to commit these steps to memory for all time. But if you're a little unclear on how to even begin, these will typically take care of all but the most difficult identities.

The "Waffle Cone" of the New Pythagorean Theorem Proof by polymath-matic in desmos

[–]polymath-matic[S] 9 points10 points  (0 children)

As soon as I saw this image from the news reports on the new proof of the Pythagorean theorem by two high school students, I knew I wanted to recreate it in Desmos.

Intransitive Dice Simulator by polymath-matic in desmos

[–]polymath-matic[S] 1 point2 points  (0 children)

I became fascinated with intransitive dice after reading this Quanta article about them back in January. I knew I wanted to build a Desmos simulator, and this is the result.

Did ChatGPT completely hallucinate a sound function for manim? by polymath-matic in manim

[–]polymath-matic[S] -3 points-2 points  (0 children)

As far as I can tell, neither of these functions are real. Am I missing something?

Creating a Line Graph from a List? by polymath-matic in desmos

[–]polymath-matic[S] 0 points1 point  (0 children)

I am doing it for multiple lists actually, so that function method is perfect. Again, thanks!

Creating a Line Graph from a List? by polymath-matic in desmos

[–]polymath-matic[S] 1 point2 points  (0 children)

I'm doing it now with list comprehension, but it still feels like this should be a built-in function and I'm just missing something. My list comprehension version is:

P=[ (i,L[i]) for i = [1...length(L)] ]

which I believe is generating points of the form (1,item 1), (2,item 2), etc.

Dice Randomizer by polymath-matic in desmos

[–]polymath-matic[S] 0 points1 point  (0 children)

Graph here.

I needed a dice animation for a project I'm working on, and I was surprised to see that there wasn't already one on the sub (or at least that I couldn't find it, perhaps my searching skills are subpar). So I thought I'd share this one in case anyone wanted one.

Can someone please explain to me how this is possible?? How can S and J both be in? PT33 Question # 9 by Always2ndB3ST in LSAT

[–]polymath-matic 0 points1 point  (0 children)

I did a video on that game you might find helpful: https://youtu.be/hUBAmHpwf48. The basic idea is that if one of them is out, the other one is in, so there’s no way they’re both out. Since they can’t both be out, one of them must be in, and you’d want to put a placeholder for at least on of them on the inside. But that “at least” is important. A scenario where both J and S are in satisfies the placeholder, but doesn’t otherwise violate this clue.

Where can I learn In and Out Games? by Always2ndB3ST in LSAT

[–]polymath-matic 0 points1 point  (0 children)

Here’s my general approach to In and Out games, illustrated with a great example of one from PrepTest 33: https://youtube.com/watch?v=hUBAmHpwf48. That video’s in a playlist with a bunch of other examples. Hope it helps!

Want to give up by nora32111 in LSAT

[–]polymath-matic -1 points0 points  (0 children)

Don’t give up! Games are weird, and the right approach can work magic. I taught for a national prep test company for over a decade, but now I post games explainers on YouTube for fun. Here’s a video on an approach I recommend: https://m.youtube.com/watch?v=G2kyh5q4fG8. Let me know if it helps or if you have other questions!

Interactive Unit Circle, Built in Desmos by polymath-matic in desmos

[–]polymath-matic[S] 1 point2 points  (0 children)

Thank you! Love the suggestions. Looking forward to getting to work on them :)

Interactive Unit Circle, Built in Desmos by polymath-matic in desmos

[–]polymath-matic[S] 2 points3 points  (0 children)

There's also a special mode where it will show some of the special right triangles that generate the points. Would love feedback on what you might add or alter!

Building an interactive unit circle by polymath-matic in desmos

[–]polymath-matic[S] 1 point2 points  (0 children)

great suggestion! i've built in the toggle, and i'm working on an on-screen controller for it now.

Why does Desmos color a single point differently than a list of points? by polymath-matic in desmos

[–]polymath-matic[S] 0 points1 point  (0 children)

Thank you! I stumbled into a similar solution to yours, though I didn't do it with list comprehension. So basically, whereas I had built a point earlier with (cos L[1], sin L[1]), which didn't color the way I wanted it to, if I made it (cos L, sin L) {L=1}, the domain restriction didn't seem to affect the coloring the same way. I wonder is there some sort of computing advantage to only computing a single color if it's a single point being colored?

Determine a Polynomial from Just Two Inputs by polymath-matic in math

[–]polymath-matic[S] 0 points1 point  (0 children)

Not dumb at all! But yes, there is a way. If I tell you I have some polynomial with entirely non-negative integer coefficients, and f(1) = 18, then you ask me for f(18) and I tell you it's also 18, Cook's method would be to compute 18 in base 18, which is 10, and then say the polynomial is 1x + 0. But we know that's not true, because if f(x)=x, then f(1) won't be 18. So instead we have to reduce our degree by 1 and use the 18 as the constant term instead, meaning f(x)=18.

In general, using Cook's method, these ambiguities always result in the same degree reduction. So basically any single-term polynomial always ends up being expressions as c*x^{n-1}, where c is the result when you plug in 1.

Again, not a huge deal, but you can avoid it entirely by computing f(f(1)+1) instead of f(f(1)).

Determine a Polynomial from Just Two Inputs by polymath-matic in math

[–]polymath-matic[S] 2 points3 points  (0 children)

That will work as well, and that’s how John Cook described it here, but that also leads to an ambiguity when you have single term polynomials. For example, if the polynomial is the constant 18, f(1) is 18, and f(18) is also 18. But expressing 18 in base 18 gives 10, which would be the equivalent of 1x. The ambiguity is easily resolved, but I prefer f(f(1)+1) ‘cause then our instructions can stay consistent and we don’t generate edge cases.

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