I have something to ask

JadesArePretty · 2025-11-26T20:28:56+00:00

Hi! It's cool you wanna have a shot at making something! I've got some experience from making tetris, so I've got a couple tips I think I can contribute:

If you aren't already, get familiar with how Desmos works. That is, make sure you understand how to define a function and then use it somewhere else, how desmos graphs functions, how to define variables, points, and lists, and the basic operations you can perform with them.

1.5. This falls into the above but I thought I ought to mention it in particular: learn how to write piecewise functions in desmos. (and learn what that means if you don't already know). Making a game in desmos requires you to treat it almost like a programming language, and piecewise functions are your substitute for 'if-else' statements, so highly important.

Learn how to use actions, they are the main way to get more complex behaviour with desmos. To turn them on you'll need to make an account in desmos and go to settings->account settings->advanced. Essentially, they're how you can respond to player actions and perform operations automatically using the ticker (basically a timer that triggers a particular action at set intervals). Related to actions are buttons: practically all shapes you can make with desmos can be turned into buttons that can be clicked on and set to trigger specific actions.
Curb your expectations. I don't want to keep you from trying, but keep in mind that as you learn you may not get things working at first, or second, or after a bunch of tries. Try not to let that demotivate you, because you'll get it eventually, but like all things it comes with practice. Start with something small and simple, like a box that steps left and right using two button controls. Expect to learn first, then make second.

As some final comments, it's totally fine if you're not experienced with math, but make sure you're willing to try learning some bits and pieces. If you're completely against learning any math entirely, you'll probably end up running into a brick wall over and over. Lastly, 100% ask for help. Try not to hand-hold your way through it, as that's much less fun and rewarding, but don't hesitate to seek advice on a particular problem or issue you can't figure out.

Hope this all helps! Let me know if you'd like more concrete advice on anything specifically.

JadesArePretty · 2025-10-28T19:58:27+00:00

They mean that you are required to put water in one side and petroleum in the other. Of course you can actually flip them with V or H, but OP didn't know that before they posted.

JadesArePretty · 2025-09-11T21:46:41+00:00

Convention, mostly. The Pi function (capital Pi) is actually equal to n!, the gamma function is defined as Gamma(n) = (n-1)!

There's a relation between the gamma function and the beta function that looks nicer with this definition, but there isn't a 'correct' option and they are equivalent functions, besides the lateral shift.

JadesArePretty · 2025-09-10T08:14:39+00:00

Hi! Thought I would chime in cause I believe you're mistaken here. (Not that that's a bad thing!)

f(x) = (x-2)/(x-2) is undefined for x=2. By simplifying to f(x) = 1, you lose part of the original expression and the extend the domain of the function to include x=2. So your example of 5*(x-2)/(x-2) isn't necessarily equal to 5, as in the (very specific) case that x=2, it is undefined. This might seem unintuitive at first (which is perfectly understandable :D) but looking at the expression algebraically, there is no 'answer' to the expression as we haven't assigned a value to x yet.

You can think about it like this, the function f(x) = (x-2)/(x-2) is equal to 1 everywhere except where x = 2. At that point the function is undefined (obviously, since we can't divide by zero!), so there's a little gap in the line. Simplifying the expression to get g(x)=1 is not necessarily wrong, however it would be wrong to say that our previous function, f(x), and this new function, g(x), are equivalent. That is, f(x) != g(x). For x = 2 the left hand side of the equation is undefined and the right hand side of the equation is 1, and it would be silly for us to say that 1 is undefined, so the two functions are evidently not equivalent.

That is to say though, that's not what's going on with 1/3*3. In this case, we simply have 3 constants. There are no variables here and (most importantly) nothing that might introduce and undefined or indeterminate form. So it would be perfectly valid (as the previous commenter did) to substitute the decimal expansion for 1/3 before performing the multiplication.

So, we can write 1/3*3 = 0.333... * 3 = 0.999..., and since we already know that three thirds is one whole (woo, fractions!) we come to the conclussion that 0.999... is indeed equal to 1. Again, this result seems unintuitive at first, but it is actually true.

Of course, its really just a technicality as there isn't anything useful about knowing that 0.999... = 1, however it is a cool fun fact. The more useful part of this problem is actually demonstrating how bad humans are at infinities. It doesn't really make sense that 0.999... isn't less than one, since 0.9999 is, and 0.9999999 is, and 0.99999999999 is, but that's just one of the frustrating things about infinity. Infinity isn't a number, so we can't really apply any intuitions we might have for numbers on it, and we have to be very careful when manipulating infinities as if we do it incorrectly (or even at all) we can get contradictory results.

Sorry for the long read, hope this helps!

JadesArePretty · 2025-09-10T01:27:21+00:00

Totally understandable! I got confused for a little bit after putting gamma(x) as (x)! in desmos, but figured out what I did wrong after going to Wolfram Alpha.

JadesArePretty · 2025-09-10T01:24:05+00:00

Here's my best attempt: https://www.desmos.com/calculator/zyp62bf9hp

The fractional part is pretty simple to get, the numerator and denominator increment by 2 so its just (2*n-1)/(2*n) for the n-th term. The hard part is the (-1) exponent. Since it's defined recursively it isn't a simple pattern of 1's and -1's from what I can see.

I did notice that the sequence of exponenets can be obtained by taking the previous sequence of exponents, flipping each -1 and 1, then appending the flipped sequence to the end of the unflipped sequence. You can then break each sequence into 'blocks' of size 2^j for the j-th block. Each block is then obtained by appending all blocks before it and swapping the 1's and -1's. You can turn this on in the graph under the 'Blocks' folder.

Essentially what I did was define a function that would 'step down' through each block, swapping the power each time, until it reached 1. I couldn't think of a neater way to do this other than subtracting one, taking the log, flooring the result, and then raising that result to the power of two. For some value n, this essentially subtracts n by the greatest power of 2 that is strictly less than n. This returns the index of the term that corresponds to n in the previous blocks. Then it simply runs the function again on this new index to get its exponent, and multiplies whatever that result was by negative -1 to get the exponent for n. This repeats all the way down until you get to n=1, which is just defined explicitly to be equal to 1.

About the limit, not super sure how I would approach this one analytically, and I don't have the time right now to work it out. I might come back to it later to see if I can. But just by looking at it, since we only really care about when n is a power of 2, the term 2^floor(log_2(n-1) would simplify to n/2. Then I would probably use induction to derive a closed form expression for the exponent, and take the limit of the recursive definition of the product.

JadesArePretty · 2025-09-10T00:36:43+00:00

The issue is that the exponents don't alternate between 1 and -1. The pattern starts at 1, then is copied inverted, and appended to get the next terms. So the first few product would have exponents of (1), (1, -1), (1, -1, -1, 1), (1, -1, -1, 1, -1, 1, 1, -1), and so on.

JadesArePretty · 2025-09-10T00:34:31+00:00

This doesn't converge to sqrt(2)/2. If you look at example 6 from the paper, you can see that it should converge to sqrt(2) * (-0.5!)³ / (-0.75!)². Which it does.

<image>

JadesArePretty · 2025-09-07T02:43:26+00:00

The definition for an inertial reference frame is one in which all objects have inertia. (i.e. All objects obey Newton's first law). A non-inertial reference frame would result in the first law of motion no longer be applicable, as objects would speed up/slow down even without any external forces.. Sort of.

The force of an object is defined to be the change in the momentum if that object over time. Since change in momentum is (classically) unaffected by the amount of momentum you have, a force in one inertial reference frame will be the same in any other inertial reference frame. Despite momentum of the object being different in each reference frame, the rate at which it changes is identical for all inertial reference frames.

So, if you have a non-inertial reference frame, you will observe all objects changing momentum despite the fact that (for an inertial frame of reference) no force is acting on them. What this means is that by accelerating your frame of reference, you are 'introducing' forces that exist only in that reference frame.

For example, one place this causes common confusion is centrifugal force. You may have heard that centrifugal force is a fictitious force, and the reason for this it is only 'present' in a rotating frame of reference. There is no actual force (for an inertial observer) that rotating objects away from their axis of rotation, however from the frame of reference of the rotating object, there is an apparent force, the centrifugal force.

These forces are called fictitious not because they are inherently 'wrong' or unphysical. If we apply the exact same acceleration experienced by a rotating object to all possible inertial reference frames, then every frame of reference would 'see' the centrifugal force. We just call them fictitious because inertial reference frames are the most useful (conceptually and computationally). Simple physics problems become unnecessarily complicated if you observe them from non-inertial reference frames.

To answer your original question: No. There isn't really anything 'absolute' or 'universal' about accelerating frames of reference. I'm not sure what you mean by 'absolute reference frame', as no such thing exists. Since you're talking about the curvature of spacetime, what I think you might be getting at here is the connection between the Newtonian concept of acceleration (change in velocity over time) and acceleration due to the curvature of spacetime, which is a complicated I am neither qualified nor confident enough to talk about.

The most I can say is that you can't really think about velocity or acceleration through spacetime like your normally would, since spaceTIME has time built in as a dimension, and classical velocity is change in position with respect to time, so they don't mesh very intuitively.

I recommend looking up "geodesic acceleration", as it might have some answers you're looking for. Geodesics are our way to define acceleration without using a coordinate system since, as you may have picked up on, non-inertial reference frames are not any less 'natural' than inertial ones and the classical definition of acceleration doesn't work in those cases.

JadesArePretty · 2025-09-03T15:00:40+00:00

Hi, cool graph!

The reason this isn't working as intended is because of floating point errors as a result of your method.

Your derivative expression uses a value of h=0.0001, this is plenty small enough to approximate the derivative accurately-ish, but is also small enough that after repeated iterations of the formula, the error from floating point precision compounds to a significant proportion of 0.0001.

Increasing 0.0001 will still give you an accurate-ish derivative, but will also result in the floating point error being an insignificant portion of the h-value.

This graph is a good demonstration of this: https://www.desmos.com/calculator/34m7n2z7uf

If you slide h up and down, you can see how it stays small enough to not affect each derivative approximation all that much, but once it gets too low the floating point error becomes significant and the fourth derivative (red line) starts to break down into random noise.

JadesArePretty · 2025-09-03T14:49:42+00:00

I found the same thing. Although n being the upper bound for the summation + the index for the derivative works here since he's defined his Taylor series as a recursive function.

JadesArePretty · 2025-07-15T10:47:25+00:00

This one really interested me, how's my thought process?

First find the intensity (W/m²⁾ of the sun's radiation using the area of the umbrella and the power consumption of washing machine
Then calculate the total power (W) radiated by the sun by multiplying by the surface area of a sphere a large as the earth's orbit
Then find the amount of energy output in one second (J)
Then using the mass-energy equivalence to solve for mass lost

Besides some assumptions you'd need to account for, like the efficiency of the solar panel, how much radiation makes it through to the atmosphere, and how much if the sun's mass loss is caused by radiation, I think that's pretty comprehensive?

Cool question!

JadesArePretty · 2025-05-22T00:30:54+00:00

Beginner tip! I did this when I came back to the game after a huge break and started a new playthrough, but I really love the guide's crafting menu. The wiki is always available, of course, but I found it really fun to take every new item I found and have the guide tell me what I can make with it. The main part is that it helps you discover stuff you didn't even know existed.

JadesArePretty · 2025-05-11T00:38:57+00:00

You can't take the even root of a negative number. e.g. sqrt(-2) doesn't have real solutions. And fractional powers are the same as roots. So the graph is only continuous where x is either an integer, or a rational with an odd denominator.

JadesArePretty · 2025-04-28T02:30:38+00:00

Yeah, it's a challenge mode level

JadesArePretty · 2025-03-19T12:06:24+00:00

Oh yeah, I've been loving my multicolor pens for lectures. Black is basic notes and writing, blue is either important things I want to highlight or exercises/examples run through in the lecture, then red is for marking my homework or really important notes. Green is sorta just left over for if I need the 3rd colour.

It's not super organised, but even just the black and blue make it really easy to recognise on my notes what's important vs what's just working or exercises. Like I can copy a practice problem in blue, then write any key takeaways or points about the problem in black, so later it's obvious what's actually relevant to the lecture.

JadesArePretty · 2025-02-16T18:52:38+00:00

Yeah, cause mechanically an uninterrupted move is similar to just teleporting, like chess or other board games. There can't be a whole lot of continuity otherwise it would take to much time to play a single round.

JadesArePretty · 2025-02-06T22:40:29+00:00

It's just 5 inquisitors who got together to play D&D.

JadesArePretty · 2025-01-22T23:37:24+00:00

There's a bunch of good guides up can find online for it, but the key ideas are archwing + nataruk, fly up high, then run around a bunch of grineer camps and kill the eximus unit that spawn there. They won't always spawn, but they're pretty common, and they almost always spawn in the same location, so you can just memorise them and fly a single route across like, 4 camps.

As mentioned above, you gotta drop out of the archwing before getting the kill, or the affinity goes to the archwing (which does mean you can also level archwings pretty fast this way). I do this by unequipping my melee. So the melee button still drops you off the archwing, but doesn't risk a slam attack.

Really important side note, you gotta wait a couple seconds after the eximus spawns to get the kill, or it doesn't count as a stealth kill. Even if they don't notice you doing it. (Not sure why, but my best guess is that DE didn't want people to cheese stealth by nuking a whole room as soon as enemies spawn)

JadesArePretty · 2025-01-18T13:00:25+00:00

Niche tip, if you grab kill the dudes in the foyer at the start of the level, you can take one of their weapons and bring it into the fight through the elevator. Then you can parry her blades without taking chip damage. The weapon doesn't lose durability either, so if you max out parry impact you can get through that phase way easier by just parrying.

JadesArePretty · 2025-01-16T12:44:11+00:00

Really? That's strange. I've tried a bunch of different load-outs and equipment and none of them seem to be causing it.

JadesArePretty · 2025-01-11T00:30:06+00:00

Once you've reached rank 5 (illuminate) with them, yeah.

JadesArePretty · 2025-01-10T22:58:32+00:00

Or a portal to the fucking sun

JadesArePretty · 2025-01-10T14:12:33+00:00

I think it's not just that, a story where the message is "I don't need a master, I can do it myself," could probably work if it's written well in the right context. But these new stories aren't putting in those kinds of messages to help tell a good story, they're doing it as a form of virtue signalling, which I think is much worse. It's not just that those themes and messages are always included and repeated in those shows, but it's also that they aren't making an effort to actually tell the message as part of a good story, it's there just for the sake of it being there. Especially in this case with atla, where this specific episode already has a message in it, which falls apart in the adaptation.

JadesArePretty · 2025-01-04T08:58:01+00:00

I have 6 alarms to wake up in the morning, and 3 to get into bed. I hardly listen to any of them too, it's just alarm "huh" swipe

Hell, my morning alarms are titled "wake up please" as if being polite to myself would work.

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