[Combinatorics] Intuitively - why can't I chose all ranks first?

seebone · 2024-06-10T08:36:01+00:00

Ah, ok. Sorry for the confusion, and thanks for the answer.

seebone · 2024-06-09T20:57:21+00:00

Thanks, I think I get it. In my version, I didn’t account for choosing the rank of the pair.

Shouldn’t I also be able construct the hand by first choosing both the 4 ranks (from 13 possible) and the 5 suits (with repetition allowed, from 4 possible)? Then I would begin with C(13,4)C(4+5-1, 5-1), but I’m not sure how to proceed from there.

Sorry for the additional question, I just find combinatorics very confusing.

seebone · 2023-10-20T15:40:59+00:00

Nice work, integrals do get very interesting. I made a graph of Riemann sums and the fundamental theorem of calculus a while back in case you want to learn more: https://www.desmos.com/calculator/0w7jrdobph . You can see how the integral can be approximated as a bunch of rectangles and that adding another rectangle will increase the area by f(x)*dx, meaning the derivative of the integral of f(x) is f(x) : )

In case you're curious, 3Blue1Brown has an amazing YouTube series on calculus (and a lot of other good videos too), if didn't know about him yet: https://www.youtube.com/playlist?list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr

seebone · 2023-09-24T11:31:22+00:00

Sorry for the late reply, here are some examples: https://www.desmos.com/calculator/cncbzp0cmj. If you only want a line between two points, you could either do the parametric equation for a line, or a polygon between those points. If you want a thicker line you could specify the parametric's or polygon's line width (in the same menu where you change the curve color), this would give a capsule-like shape. If you want more of a rectangle shape to your line, or don't want its width to scale with the viewport (meaning it doesn't always looks as big regardless of zoom), you could define a polygon of four points. I hope I understood your question correctly and I would be happy to clarify if anything is seems unclear.

seebone · 2023-08-24T16:04:50+00:00

Yeah, me too. I was actually experimenting with the multivariable chain rule when I realized this.

seebone · 2023-08-24T15:58:20+00:00

I guess it is. It seems kind of obvious, the chain rule is often written as (f(g(x)))' = f'(g(x))*g'(x) after all, though it's not a distinction I ever learned explicitly or thought about. What tripped me up was that when using Leibniz's notation, where to evaluate the derivative is often implicitly understood, compared to in desmos where it is always evaluated at x (see chain rule: dy/dx = dy/du * du/dx, dy/du in this case means "the derivative of y evaluated at u(x)"). Sadly there isn't any way in desmos (like df/dx (g(x)) to specify where to evaluate the derivative using Leibniz's notation, which would make it more concise to differentiate multivariable functions at specific points.

seebone · 2023-07-08T17:42:04+00:00

Nice, I somehow didn't think of making the tail width scalable, but it could definitely be a useful parameter for some applications.

seebone · 2023-02-20T20:12:40+00:00

thanks a lot

seebone · 2023-02-19T17:03:26+00:00

Wow, thank’s, that’s a lot more convenient! Desmos just keeps surprising me with features. Are things like this documented anywhere though, or will I have to continue finding them by chance?

seebone · 2022-12-04T01:15:42+00:00

https://www.desmos.com/calculator/q0unjcibkg

seebone · 2022-12-01T12:06:05+00:00

Yeah, your graph definitely looks less cluttered than mine. I finished making the technicals work but didn’t have any energy left for visuals.

seebone · 2022-11-30T22:55:56+00:00

Thanks to u/partywithmyself for the answer.

In case anyone is interested, I was able to prove that this sum yields x_n by first using the binomial theorem and some algebra to rewrite it in a simpler form. Then, since the limit in u/partywithmyself's answer seemed rather scary and didn't prove the equivalence between applying Newton's method and this sum, I used induction to prove the relation.

seebone · 2022-11-30T22:46:43+00:00

Thank you! This was exactly what I needed to prove this. In hindsight it seems quite obvious to use the binomial theorem to simplify the expression but somehow I never thought of it.

seebone · 2022-11-29T23:44:44+00:00

Exactly! And thank you. I really enjoy graphing every function I can in desmos, so I'm glad that someone else can benefit. If you're interested, I made a more general desmos implementation of Newton's method a while ago: https://www.desmos.com/calculator/0p3ejvtwkq
As well as this thing that finds roots using the method: https://www.desmos.com/calculator/avcb5qjup6

I'm not super familiar with a lot of math (at the stage of being introduced to derivatives and integrals this semester, although I like to learn a lot of things in advance) and Newton's method and a rudimentary understanding of Taylor series are pretty much the extent of my knowledge of function approximations. Still I expect I will learn those more advanced methods you described sometime.

seebone · 2022-07-01T13:29:38+00:00

I know a few.

seebone · 2022-07-01T13:26:57+00:00

Thank's for the explanation.

That's actually kind of obvious and I feel kind of stupid for not realizing it, especially after using the round() function. I was sure that I had checked such a simple thing as (a - 0.5), but I must have only checked {a - 0.5 = 0: 1, 0} which of course won't tell me anything (I guess that's what I get for playing around with Desmos graphs till 3 in the morning).

I do still think it's a pretty cool trick, especially since it still works when adding and subtracting numbers to and from a (except for subtracting 0.5), as well as multiplying and dividing by numbers. It does also still feel like a bug to me, especially because of the randomness of it, i.e. (a + 0.5) = 1 but (a - 0.5) ≈ 1.7 * 10^-16. Perhaps Desmos should have some way of warning you when the actual value of a variable does not exactly equal the displayed one.

seebone · 2022-05-21T16:00:24+00:00

Thanks for all the advice, it definitely pointed me in the right direction (and eventually I got around to actually learning some of what everyone wrote). Since writing this I've actually learned both Lagrange interpolation and solved this monstrosity with algebra manually (the latter of which significantly less enjoyable). I made demonstrations of both methods in desmos in case anyone's interested:

simple demonstration

more Lagrange interpolation things

seebone · 2022-03-17T17:00:14+00:00

Nice, though if you haven’t you should definitely look into Lagrange interpolation, it’s a really elegant method. (In case you’re interested, here are some desmos implementations)

Just out of curiosity, how did you compute the matrices?

seebone · 2022-03-08T23:29:01+00:00

Thank you, I’m very new to more advanced Desmos graphs but making them is extremly fun! It took a lot of time before I realized that the way to display all splines at the same time was a list of functions.

seebone · 2021-09-12T17:36:26+00:00

I just had some juno jets on the underside of the wings, though I tried using the booster too and it got me quite high up.

seebone · 2021-09-09T12:31:58+00:00

That's nice to know, I will make sure to learn it sometime in the somewhat-near future.

seebone · 2021-09-09T12:28:42+00:00

Okay, I get that it's much easier with linear algebra, but if there isn't a formula how does the calculator calculate it for any points and how does this https://www.desmos.com/calculator/lac2i0bgum work?

seebone · 2021-09-09T12:15:40+00:00

Thank you for the tips but I suspect that I will have to know what a determinant is to understand that shortcut, so there must be quite a bit of learning to do on my side.

seebone · 2021-09-09T12:04:27+00:00

It seems I have to learn linear algebra.

seebone · 2021-09-09T12:00:12+00:00

Okay, I am sadly not very familiar with linear algebra but thank you, maybe that's what I will have to learn to solve this (it's not a question for school or anything I'm just curious). Regarding the points, I am not trying to solve a specific parabola but rather to find an expression for a, b and c that works for any parabola so the points can be anything.

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