Anonymous at Fri, 31 May 2024 09:55:02 UTC No. 16202253

I just listened to one of her podcasts wheres she mentions she's now a customer support of some magnetic cell separation company because she felt it was high tech like in the movies. She also mentions she was tired of bench work and gave her burnouts so she chose this. She was so interested in this that she gave job interview from airport hotel at Ghana! And now she wants to inspire girls to be a scientist!
I don't know if she's actually legit retarded lmao.

Anonymous at Fri, 31 May 2024 13:17:30 UTC No. 16202473

why doesn't this work?

Anonymous at Fri, 31 May 2024 13:29:30 UTC No. 16202483

>>16202473
There are two solutions to the square root of 1, -1 and 1, but when made a function we take the positive branch. If you wanted to avoid the sort of problem your line of reasoning demonstrates, you should disallow squaring negative numbers.

Anonymous at Fri, 31 May 2024 13:43:52 UTC No. 16202491

>>16202473
[math]x^{a/b} \neq (x^a)^{1/b}[/math] when x is negative. So you cannot write that middle term. It's similar to the fact [math]\sqrt{a} \times \sqrt{b} = \sqrt{ab}[/math] only works when a and b are both positive.

Anonymous at Fri, 31 May 2024 14:29:44 UTC No. 16202540

if -r<a<b<r and g=max{ l a l , l b l }
then how do I show that [-g,g] \subseteq (-r,r)?

Anonymous at Fri, 31 May 2024 14:36:47 UTC No. 16202547

Just to doublecheck 108 miles multiplied by 144 miles are 15552 square miles, right?

Anonymous at Fri, 31 May 2024 14:45:06 UTC No. 16202551

did gpt-4o got it right?

Anonymous at Fri, 31 May 2024 14:45:26 UTC No. 16202552

>>16202136
Planning on doing a summer course for Linear Algebra in order to do Quadratic Equations next semester.
Thing is i struggle a lot with integration and have very shoddy math fundamentals.
Any tips as to how to reinforce my math in the next couple of months in order to not get ass raped next semester?

Anonymous at Fri, 31 May 2024 15:28:31 UTC No. 16202611

>>16202136
Hello Anons, I hope you are all doing alright.

I just read about this https://en.wikipedia.org/wiki/Multiple_integral#Particular_cases. And I think somewhat grasped it but can't be sure. So I made up something up to reflect my idea into a form of writing and got the validation from the ""AI"" but I want an actual intelligence to validate my understanding of this concept. I am a newbie, so I am sorry if this is like the dumbest thing you will read today. Thanks.

Anonymous at Fri, 31 May 2024 16:45:54 UTC No. 16202699

>>16202136
i miss the image of the hot chick

Anonymous at Fri, 31 May 2024 16:49:57 UTC No. 16202705

>>16202540
either 0 < a < b, a < 0 < b, or a < b < 0
You only have three cases to check

Anonymous at Fri, 31 May 2024 20:36:27 UTC No. 16202960

>>16202540
[math]|a| < |b| \iff a < b \land -a < b[/math]

Anonymous at Fri, 31 May 2024 20:37:51 UTC No. 16202963

>>16202960
*[math]|a| < b \iff a < b \land -a < b[/math]

Anonymous at Fri, 31 May 2024 20:44:03 UTC No. 16202969

>>16202699
Twum is hot, you only don't like her because you're racist

Anonymous at Fri, 31 May 2024 20:47:38 UTC No. 16202974

>>16202699
me too :(
>>16202969
>being racist
the way you say it like that's a bad thing

Anonymous at Fri, 31 May 2024 20:49:42 UTC No. 16202977

>>16202136
What are examples of (same-size) square matrices A,B with det(A)=det(B)=det(A+B)=1 ?

Anonymous at Fri, 31 May 2024 21:46:17 UTC No. 16203049

>>16202969
jesus, i only glanced at the image, and i swear, i thought it was a messed up pic of a dude

Anonymous at Fri, 31 May 2024 21:48:50 UTC No. 16203051

>>16202969
Twum is so hot that none of the fire alarms in her house can function correctly

Anonymous at Sat, 1 Jun 2024 07:20:03 UTC No. 16203759

I had a theory about pedophilia
what if it's actually the human instinct to be a father crossing with a large malformed libido due to excessive sexual frustration?
Assuming the subject is human, unlike demonic politicians

Anonymous at Sat, 1 Jun 2024 09:05:16 UTC No. 16203847

>>16202977
[math]
A = \begin{pmatrix} 1 & \sqrt{3} \\ 0 & 1\end{pmatrix}, B = \begin{pmatrix} 1 & 0 \\ \sqrt{3} & 1\end{pmatrix}
[/math]

Anonymous at Sat, 1 Jun 2024 13:02:40 UTC No. 16204094

let's say I have the equation of a line Ax + By + C = 0
I'm aware that if I have the coordinates of a point I can calculate the distance between the point and the line by plugging the x and y values into the equation and then normalise the result by dividing it by the length of the normal vector of the line.
my question is what does the result you get when you plug those numbers in mean?
is there a visual representation of the scalar value (the one before the normalisation) on a plane?
I'm pretty sure it's supposed to be the length of a vector that you project onto the shortest line that connects the point to the original line or something like that.

I haven't done math in years, I swear I've been trying to solve this problem for hours

Anonymous at Sat, 1 Jun 2024 13:31:54 UTC No. 16204144

Is there enough information to solve this? This was posted on /v/.

Anonymous at Sat, 1 Jun 2024 13:56:05 UTC No. 16204173

>>16204144
Is there a radius to that circle or any part of the sides?

Anonymous at Sat, 1 Jun 2024 13:59:47 UTC No. 16204185

>>16204173
Nope. OP probably posted it as bait.

Anonymous at Sat, 1 Jun 2024 14:01:13 UTC No. 16204189

>>16204144
From that setup, no. [math]\alpha[/math] depends on how far the centre of the circle is from the point at angle [math]\theta[/math]

Anonymous at Sat, 1 Jun 2024 14:55:17 UTC No. 16204303

>>16202136
Picrel is from the wikipedia entry for empirical measures and I just noticed that the RHS of the equation looks like a stochastic kernel (provided measurability and so on...) Am I right? I don't think it's very relevant, I just nooticed.

Anonymous at Sat, 1 Jun 2024 16:11:31 UTC No. 16204509

>>16204303
Yeah, you're right, though it's not really dependent on a parameter [math]x[/math] and is `just' a random measure.
In fact, this is also what's happening in some kernel methods in machine learning / nonparametric statistics.
For example in https://en.wikipedia.org/wiki/Kernel_density_estimation.
Here what's called a kernel is really the function on which the stochastic kernel [math](x,A)\mapsto P_n(A+x)[/math] acts by integration (with a scale parameter for good measure).

Anonymous at Sat, 1 Jun 2024 16:24:38 UTC No. 16204582

>>16204303
Actually, now that I think about it you probably meant that [math](\omega,A)\mapsto \frac1n \sum\limits_{i=1}^n\delta_{X_i(\omega)}(A)[/math] is a stochastic kernel, which is also true.
There is a 1-1 correspondence between taking a random element of some suitable set of measures and such stochastic kernels.

Anonymous at Sat, 1 Jun 2024 16:45:46 UTC No. 16204717

How can I prove that the polynomial in pic rel is orthogonal on the interval [-1,1] wrt to the weight function ϕ(t)=(1−t^2) ^ (λ - 1/2)?

Anonymous at Sat, 1 Jun 2024 16:52:18 UTC No. 16204744

>>16204094
>Have point (1,1)
>Have line x + y + 0 = 0
>Want to calculate the distance between point and line
>Plug in (1,1), get 2 = 0
??

Anonymous at Sat, 1 Jun 2024 16:54:46 UTC No. 16204751

>>16204717
show what you did first

Anonymous at Sat, 1 Jun 2024 16:55:33 UTC No. 16204753

>>16204751
I did nothing, I have no idea how to solve it

Anonymous at Sat, 1 Jun 2024 17:03:30 UTC No. 16204782

>>16204753
Write shit out and evaluate it yo.

Anonymous at Sat, 1 Jun 2024 17:10:48 UTC No. 16204805

>>16204753
>I did nothing
as I expected

Anonymous at Sat, 1 Jun 2024 17:17:48 UTC No. 16204825

>>16204805
Can (You) give me a hint at least?

Anonymous at Sat, 1 Jun 2024 17:28:02 UTC No. 16204856

>>16204825
Follow >>16204782. Do you even know what orthogonal means? How do you not even write out the integral?

Anonymous at Sat, 1 Jun 2024 18:19:09 UTC No. 16204917

>>16204582
Yeah, that's what I meant, thanks. I should've been more clear desu

Anonymous at Sat, 1 Jun 2024 18:51:32 UTC No. 16204946

why is the Gauss-Legendre quadrature the best fit for this integral? I know that the interval of the G-L quad is [-1, 1] and that its weight is 1 but how else can I figure that out? is it because sin(x^2) is continuous over [-1, 1]?

Anonymous at Sat, 1 Jun 2024 19:05:50 UTC No. 16204963

>>16202136
Does anyone know whether a closed-form solution for the inverse logarithmic integral exists or has it been proven that it doesn't exist?:
https://en.wikipedia.org/wiki/Logarithmic_integral_function

I am playing around trying to find a closed-form solution, but if it doesn't exist I'm wasting my time.

Anonymous at Sat, 1 Jun 2024 19:19:56 UTC No. 16204989

How do I prove that the Klein subgroup [math] K = (\mathrm{Id}, (12)(34), (13)(24), (14)(23)) [/math] is a normal subgroup of [math]S_4[/math] using the fact that cycles are invariant under conjugation?

Anonymous at Sat, 1 Jun 2024 19:42:42 UTC No. 16205035

>>16204989
Am I mistaken or is the Klein group everything that satisfies [math]g^2 = 1[/math] and isn't a transposition?
If it is you can argue that, since [math]h^2 = 1 \implies (ghg^{-1})^2 = gh^2g^{-1} = 1[/math] and [math]ghg^{-1}[/math] cannot be a transposition (since those are cycles, and hence [math]h[/math] would also be a transposition), the subgroup has to be closed under conjugation.

Anonymous at Sat, 1 Jun 2024 20:39:21 UTC No. 16205135

>>16204963
No, it's out there, the closed-form solution, but we just don't know. Why even live? I will make an attempt to find it.

Anonymous at Sat, 1 Jun 2024 22:09:20 UTC No. 16205225

>>16204946
are you fucking stupid?

Anonymous at Sat, 1 Jun 2024 22:32:34 UTC No. 16205261

>>16204946
Gauss has the best accuracy for the number of nodes used. The precalculated weights for Gauss is based on the interval [-1,1], so there's no need to rescale.

Anonymous at Sat, 1 Jun 2024 23:40:04 UTC No. 16205394

what material should i look at to get good at maths

Anonymous at Sun, 2 Jun 2024 00:34:30 UTC No. 16205485

>>16205394
Have you tried looking at math?
Truth is you need to solve problems
Studying just builds up an intuition which could be false and you won't remember most of that trash either
You gotta put yourself into a math context and build structure in yer head. Brain data transference isn't a thing yet so you're just gonna have to figure out yourself, plus it beats accidentally downloading 1 trillion yottabytes of furry porn into your brain.
Anyone that tells you that you can "learn math" is bullshitting you

Anonymous at Sun, 2 Jun 2024 04:01:42 UTC No. 16205736

Is there a sign error in picrel? The stress-energy tensor is defined as:
[math]T_\mu{}^\nu = \frac{\partial \mathcal{L}}{\partial \left ( \partial_\nu \eta_\rho \right )} \partial_\mu \eta_\rho - \mathcal{L}\delta_\mu{}^\nu[/math]
Lowering the nu index gives
[math]T_{\mu \nu} = g_{\lambda \nu} \frac{\partial \mathcal{L}}{\partial \left ( \partial_\lambda \eta_\rho \right )} \partial_\mu \eta_\rho - g_{\mu \nu} \mathcal{L}[/math]
Now, the Lagrangian density that's given is that of the complex Klein-Gordon field:
[math]\mathcal{L} = c^2 \partial_\lambda \phi \partial^\lambda \bar{\phi}-\mu_0^2 c^2 \phi \bar{\phi}[/math]
My concern is with the last part of the equation that's multiplied by the metric:
[math]\cdots +c^2 \left ( \partial_\lambda \phi \partial^\lambda \bar{\phi}+\mu_0^2 c^2 \phi \bar{\phi} \right )g_{\mu \nu} [/math]
which should correspond to the
[math]- g_{\mu \nu} \mathcal{L}[/math]
part of the tensor. But should this part not be
[math]\cdots +c^2 \left ( -\partial_\lambda \phi \partial^\lambda \bar{\phi}+\mu_0^2 c^2 \phi \bar{\phi} \right ) g_{\mu \nu}[/math]
given that there's a minus sign in the Lagrangian density? Where does that minus sign go?

Anonymous at Sun, 2 Jun 2024 04:03:15 UTC No. 16205739

>>16205736
Forgot to mention that this is the kinda shitty notation that they use to denote [math]\partial_\nu[/math] etc.

Anonymous at Sun, 2 Jun 2024 06:59:20 UTC No. 16205846

>>16205225
No, I'm just new to this. Douchebag.
>>16205261
Ah alright, thanks

Anonymous at Sun, 2 Jun 2024 07:00:10 UTC No. 16205847

>>16202960

🗑️ Anonymous at Sun, 2 Jun 2024 13:40:13 UTC No. 16206122

>>16205847
Yes, yes, it's slightly wrong, I've already posted the correct equivalence.
[math]|a| < |b| \iff (a < b \land -a < b) \lor (-a < b \land -a < -b)[/math]

Anonymous at Mon, 3 Jun 2024 04:40:19 UTC No. 16207257

>>16203051
Twum is so hot that if you threw her into the deep end of a swimming pool, she'd drown.

🗑️ Anonymous at Mon, 3 Jun 2024 06:36:03 UTC No. 16207365

>>16202253
>from airport hotel at Ghana
Was she in Ghana to study black magic and how they control lightning?
https://www.youtube.com/watch?v=C9SiRNibD14

Anonymous at Mon, 3 Jun 2024 10:29:46 UTC No. 16207570

For modern portfolio theory, why is it that when you want to calculate the variance of a portfolio it comes to this equation?

[eqn]\sigma^{2}_p = \sum_{i=i}^n{w_i}^2 \sigma^2_i + \sum_{i=1}^n\sum_{j=1}^n{w_i}{w_j}\text{Cov}_{ij}[/eqn]

But the expected return of the same portfolio is simply a weighted sum?

Anonymous at Mon, 3 Jun 2024 10:36:06 UTC No. 16207577

>>16207570
Basic probability and behavior of averages/variances under sum of random variables.

🗑️ bibi !!ZEC9/+ffJN3 at Mon, 3 Jun 2024 17:38:16 UTC No. 16208053

Hi can someone help me understand something?
I'm trying to teach myself math and am reading Logic: the Laws of Truth and am stumped at one part about implications.

It says if someone says something that is translated into PL as A, then that implies A ∨ B. and that
A
∴
A ∨ B
is a valid argument.
But this doesn't make sense to me because if the argument A is false then there can still be a conclusion that's true if B is true.
So what am I missing?

Anonymous at Mon, 3 Jun 2024 18:15:10 UTC No. 16208127

>>16208053
You're mixing up implication and equivalence.
If A, then [math]A \lor B[/math], but [math]A \lor B[/math] does not imply A.

Anonymous at Mon, 3 Jun 2024 19:45:49 UTC No. 16208270

[math]\lambda[/math] is a complex number of modulus 1 s.t. [math]|\lambda^k - 1|[/math] is small (say, [math]< \epsilon[/math]). intuitively, there should be some [math]k[/math]th root of unity [math]\omega = e^{2 \pi i j / k}[/math] s.t. [math]|\lambda - \omega|[/math] is small (say, [math]<[/math] some function of [math]\epsilon, k [/math]). how to actually prove this?

Anonymous at Mon, 3 Jun 2024 20:24:15 UTC No. 16208325

>>16208270
Draw a picture. Draw out the circle |lambda^k - 1| and draw the unit circle.

Anonymous at Mon, 3 Jun 2024 20:40:20 UTC No. 16208341

>>16202136
I've seen quite a lot of speculation about ammonia as solvent in astrobiology, but nothing beyond that. Has anyone formally outlined or speculated on biochemistry of life using ammonia as a solvent, given the challenges posed by ammonia being such a potent nucleophile?

Anonymous at Tue, 4 Jun 2024 00:50:38 UTC No. 16208784

>>16208270
If [math]| \lambda^k - 1 | < \epsilon[/math] then immediately [math]r^k - 1 < r - 1 < \epsilon[/math] and [math]2 j \pi - \epsilon < k \theta < 2 j \pi + \epsilon[/math] for some [math]j[/math]
>inb4 those are terrible estimates
Terrible estimates are the best estimates.

Anonymous at Tue, 4 Jun 2024 00:52:42 UTC No. 16208788

>>16208784
>If [math] \lambda^k - 1 | < \epsilon[/math] then immediately [math]r^k - 1 < r - 1 < \epsilon[/math]
My bad, correction:
[math]|r - 1| \leq |r^k - 1| < \epsilon[/math]

Anonymous at Tue, 4 Jun 2024 01:02:23 UTC No. 16208817

>>16207577
41% probability you kys

Anonymous at Tue, 4 Jun 2024 08:25:11 UTC No. 16209403

ok, /sci/ I'm coming to you because I'm desperate. I already searched around for this. But the information I found is too complicated for me. I'm too retarded to grasp trigonometry. I have the FAINTEST understanding of cos, sin, and tan. All I know is that side is divided by the other to get a result, and then magic stuff happens, and somehow angles are found.

The question is this: How do I take Point C, and scale it to point B which lies on the circle. I have most of the information in the image. I have the circle's center point. I have the radius. I have the position of C and the position of A. B is missing. So how do I get C to scale to B?

Please explain it to me like I'm a retard.

Anonymous at Tue, 4 Jun 2024 08:34:05 UTC No. 16209412

>>16209403
>So how do I get C to scale to B?
What does that even mean? What kind of answer are you expecting?

🗑️ Anonymous at Tue, 4 Jun 2024 08:40:07 UTC No. 16209418

>>16209412
Like a scalar type of equation I suppose? Because I have A and C already. That makes a line, no? So I guess AC scalared up to match AB. Keeping in mind that I don't have B's position. B is just where the magnitude of AB intersects with the circle.

Anonymous at Tue, 4 Jun 2024 08:41:34 UTC No. 16209420

>>16209412
Like a scalar type of equation I suppose? Because I have A and C already. That makes a line, no? So I guess AC scalared up to match AB. Keeping in mind that I don't have B's position. B is just where the magnitude of AC intersects with the circle.

Anonymous at Tue, 4 Jun 2024 08:44:23 UTC No. 16209421

>>16209420
> Because I have A and C already.
I still don't get it. Knowing C is irrelevant, it doesn't affect anything. All you need to know to work out the position of B is A.

Anonymous at Tue, 4 Jun 2024 08:45:50 UTC No. 16209423

>>16209421
oh. I guess that could work. If I could find B, I think I could work out the rest. How do I find B then?

Anonymous at Tue, 4 Jun 2024 08:55:13 UTC No. 16209430

>>16209403
The line segment [DB] has length R because it is a radius, and DAB is a right-angled triangle. So by pythagoras [AB]^2+[AD]^2=R^2 so [AB] =sqrt(R^2-[AD]^2).

Anonymous at Tue, 4 Jun 2024 09:16:02 UTC No. 16209454

My lecturer introduced a new way to do Fourier series, but I don't completely get it.
So for [math]f(x) = x[/math] in [math][-\pi,\pi][/math], [math]\hat{f}(n) = \dfrac{1}{2\pi}\int\limits_{-\pi}^\pi x e^{\frac{-2\pi inx}{2\pi}} dx = \dfrac{i (-1)^n}{n}, n \in |Z\backslash \{0\}[/math].
If I plug that into a Fourier Series: [math]f(x) = \sum\limits_{n \in |Z\backslash \{0\}} \hat{f}(n) e^{inx}= \sum\limits_{n \in |Z\backslash \{0\}} \dfrac{i (-1)^n}{n} (cos(nx)+isin(nx))[/math].

My problem ist, the real part of the Fourier series converges towards x pretty well, but the imaginary part cannot be removed as far as I see. I haven't seen this way to do a FS before, and i can't find anything online about complex fourier series

Anonymous at Tue, 4 Jun 2024 09:24:16 UTC No. 16209461

>>16209430
I think I see. Well plugging it into a calculator appears to be getting correct numbers. But I'm a little slow on how it works. I get you rearranged the equation to solve for [AB]. It's just not making intuitive sense how that arrangement works.

In any case, I don't have to understand it entirely for my purposes. So thanks. I'm going to try and apply it now, and see what happens.

Anonymous at Tue, 4 Jun 2024 09:39:40 UTC No. 16209479

what in God's name are splines

Anonymous at Tue, 4 Jun 2024 10:15:42 UTC No. 16209506

>>16209454
The imaginary parts for [math]n[/math] and [math]-n[/math] cancel each other out.

Anonymous at Tue, 4 Jun 2024 11:24:39 UTC No. 16209580

>>16209479
Piecewise polynomial C^n functions.

Anonymous at Tue, 4 Jun 2024 11:29:45 UTC No. 16209587

>>16209479
>>16209580
Correction: a spline is a function [math]f[/math] such that, for some [math]n \geq 0[/math], [math]D^n f[/math] is piecewise linear continuous.

Anonymous at Tue, 4 Jun 2024 11:40:57 UTC No. 16209605

>>16202552
Quadratic equations are middle school-level algebra. Linear algebra is first/second year of college algebra. Despite the fact that linear equations are by themselves an easy concept, linear algebra requires an ability to reason abstractly about mathematical objects in a way beyond anything in K-12 math.
So well, I suggest you to pick up a high school/college algebra book and go through as many problems as possible before you go onto Linear Algebra.

Anonymous at Tue, 4 Jun 2024 12:06:22 UTC No. 16209637

>>16209430
>>16209461
It worked: >>>/3/985442
Setting up the equation was easy. But then I spend the next 2 hours failing to think what to do with the solution. It was incredibly obvious what to do, my brain just wasn't working. Anyway, thanks again.

Anonymous at Tue, 4 Jun 2024 13:37:02 UTC No. 16209752

did Pythagoras actually contribute anything to maths or other science?

Anonymous at Tue, 4 Jun 2024 13:42:56 UTC No. 16209756

Is it normal to feel sleepy when reading about limits?

Anonymous at Tue, 4 Jun 2024 14:14:46 UTC No. 16209801

>>16209580
>>16209587
Ok this actually kind of makes sense (to me), thanks anon

Anonymous at Tue, 4 Jun 2024 15:19:26 UTC No. 16209884

>>16208784
>>16208788
thanks but I'm super confused.. I suppose your [math]r[/math] is my [math]\lambda[/math], and [math]\theta[/math] is such that [math]\lambda = e^{i \theta}[/math]? why is it the case that [math]|\lambda - 1| \le |\lambda^k - 1|[/math] (I mean consider [math]k = 2, \lambda = -1[/math])? how do you deduce that [math]\lambda[/math] is close to some [math]e^{2 \pi i j / k}[/math]? I must be missing something stupid. what is it?

Anonymous at Tue, 4 Jun 2024 16:19:25 UTC No. 16209953

>>16209884
[math]\lambda = re^{i \theta}[/math], the usual polar decomp.

Anonymous at Tue, 4 Jun 2024 19:40:05 UTC No. 16210268

How would you prove that P(A U B) >= P(A) or that P(A ∩ B) <= P(A)?

Anonymous at Tue, 4 Jun 2024 20:18:16 UTC No. 16210350

>>16210268
you can prove the latter with conditional probabilities
and the former by applying the latter in the formula of the probability of an union of sets

Anonymous at Wed, 5 Jun 2024 01:28:14 UTC No. 16210854

What are the implications of me enjoying symbolic logic problems?

Anonymous at Wed, 5 Jun 2024 09:34:16 UTC No. 16211272

what if i weigh 0 grams

Anonymous at Wed, 5 Jun 2024 10:51:38 UTC No. 16211305

>>16211272
if by weight you mean mass then you're gonna become a bunch of radiation and "explode" in all directions at light speed

Anonymous at Wed, 5 Jun 2024 14:30:51 UTC No. 16211602

>>16203847
Is this true for all transpose matrices? New to linear algebra.

If det(A) = 1

det(A + A') = 1?

Anonymous at Wed, 5 Jun 2024 15:02:06 UTC No. 16211779

>>16211602
Where the prime means transpose? No
Just off hand, I'm thinking about the 1-by-1 matrix [1]
det [1] = 1
det ([1] +[1]) = det [2] = 2

Anonymous at Wed, 5 Jun 2024 15:33:47 UTC No. 16211898

>>16211779
The prime symbol is used in matlab to denote transpose. Ah right, 1x1 matrices. But it might be true for all 2x2 matrices?

Anonymous at Wed, 5 Jun 2024 15:37:44 UTC No. 16211911

>>16211898
Are you joking? It's not true for any order of matrices, just look at the identity matrices.

Anonymous at Wed, 5 Jun 2024 15:42:11 UTC No. 16211914

Does a mass spectrometer "weigh" atoms?

For a mass to have weight, it traditionally must exert a force on a scale. Weight involves force. Mass can be "weightless" in space, where gravitational forces are negligible.

But does gravity need to be the force?

If the magnetic/electric field in a mass spectrometer exerts a force on the atoms to measure them, does that count as "weight"?

Does "weight" imply gravity? Does it imply force? Or is it simply the act of measuring mass? Can you "weigh" something simply by counting atoms?

Anonymous at Wed, 5 Jun 2024 15:50:36 UTC No. 16211932

>>16211911
Idfk, just spit balling. Except the identity matrices though, those don't count.

Anonymous at Wed, 5 Jun 2024 17:30:49 UTC No. 16212065

How do you calculate the exponential constant of the radial component a Slater-type Orbital? Only thing I've found says it's "related" to effective nuclear charge, but that doesn't say anything about actually finding it.

Anonymous at Wed, 5 Jun 2024 18:58:39 UTC No. 16212187

>>16211914
>But does gravity need to be the force?
no. acceleration can also be the force
in fact, gravity can't even be called a force in a classical sense

Anonymous at Wed, 5 Jun 2024 19:35:59 UTC No. 16212226

>>16212187
>gravity can't even be called a force in a classical sense
Uh what?

Anonymous at Wed, 5 Jun 2024 19:45:56 UTC No. 16212237

Will she replace Twum?

Anonymous at Wed, 5 Jun 2024 19:51:13 UTC No. 16212244

>>16212226
if we're to believe general relativity, and because graviton is not confirmed yet, gravity is a consequence of motion in curved space. there's no "pulling" force here

Anonymous at Wed, 5 Jun 2024 20:08:56 UTC No. 16212281

>>16212244
Ah ok, I thought you meant "in classical physics"

Anonymous at Wed, 5 Jun 2024 21:13:34 UTC No. 16212435

>>16211914
Strictly speaking "weight" is the amount of force produced by an object due to gravity, so the act of weighing something is a measure of the force. Mass spectrometry does not weigh atoms since gravity is not involved, instead it uses magnetic fields to measure the mass ratio of different ions.

Anonymous at Wed, 5 Jun 2024 21:57:39 UTC No. 16212525

>>16212237
Considering that possibility of replacement, we must consider her
on the rigorous standards of this board especially in terms of her
meme-ability, representation (in all counts), known thesis and article/book
entries and any other shit we might come up.

With that, I wish her the best of luck.

Anonymous at Thu, 6 Jun 2024 01:38:46 UTC No. 16212829

If there is no introduction section on the rubric, does that mean you should skip the introduction in your essay or is it assumed that you should have one?

Anonymous at Thu, 6 Jun 2024 01:50:01 UTC No. 16212839

>>16212244
>if we're to believe general relativity
Anon, not all of us are retards.

Anonymous at Thu, 6 Jun 2024 01:55:25 UTC No. 16212842

>>16210854
autism
>>16209756
no
>>16209752
yes, hugely
>>16202552
If you still struggle with trigonometry and algebra, download "Precalculus" by Stitz and Zeiger (available for free download online). It's a long book, maybe more than you need for a short refresher, but you can just scroll through it quickly to refresh yourself and take more time at any section that involves stuff you forgot. It is a long thorough book and a very good book.

Anonymous at Thu, 6 Jun 2024 07:21:16 UTC No. 16213219

I was reading an interview with Peter Brooks about the book "Darwinian Survival Guide" (https://thereader.mitpress.mit.edu/the-collapse-is-coming-will-humanity-adapt/). This caught my eye:
>P.W: (...) But yeah, a widely adhered-to view of evolution is a reactive one— the pool is drying up, and evolution looks at that and says, "oh my goodness, the pool is drying up! We should probably get those fish to evolve lungs". Whereas what evolution actually does is say, "oh look, the pool is drying up! Good thing that fish over in the corner that everybody picked on has a perforated swim bladder; it might be able to, like, breathe air long enough to make it over to the next pool. Too bad about all those other poor bastards who are going to die". And to hone that down to a specific example that you guys cite in the book, you’re saying “high fitness equals low fitness” — that you need variation to cope with future change.
The last part is especially interesting, because after I thought about it some more, it made perfect sense (being too well adapted to some specific conditions leaves you vulnerable to change). However, is this a valid stance (or the consensus) in the field?

Anonymous at Thu, 6 Jun 2024 08:22:15 UTC No. 16213293

Sorry for the mass ping, but my stupid question is:
how do you anons type out all of these things?
>>16202960
>>16202963
>>16203847
>>16204582

Anonymous at Thu, 6 Jun 2024 08:30:11 UTC No. 16213300

>>16213219
Yes it is the consensus, and by a large margin. In fact I don't know of anyone that fundamentally disagrees. It's pretty much the cause for most of the major extinction events throughout history. A sudden and massive change to the environment meaning much of life was no longer well adapted. For example it wasn't the asteroid impact that directly killed off the non-avian dinosaurs, it was the resultant climate change, and the rapid loss of abundant food and water their huge bodies required to live. Going large was a great adaptation for them, until over-night it wasn't.

Anonymous at Thu, 6 Jun 2024 08:38:33 UTC No. 16213311

I've calculated that in a 3-phase symetrical AC system in a Y configuration, if line currents are 11.5, 0 (disconnected) and 23 amps, respectively, and the generator generates a phase and a line voltage of 230 and 400 volts, respectively, that the neutral line current will be 19.92 amps and the total power of the system 7935 watts. Someome please check those results. And how would the total power change if the neutral line were grounded or not and what would be its values in both cases? Thanks!

Btw, the pic is Maserati Biturbo.

Anonymous at Thu, 6 Jun 2024 08:44:19 UTC No. 16213320

>>16213300
Oh, okay - thank you. Looking back I also used to think about the evolution this way, but over time I gradually forgot about it, mainly because I lost touch with the topic. Then I think the phrase "survival of the fittest" really is overused and mischaracterized in our culture today, as the article indicates.

Anonymous at Thu, 6 Jun 2024 09:07:24 UTC No. 16213350

>>16212842
>yes, hugely
elaborate further

Anonymous at Thu, 6 Jun 2024 10:26:44 UTC No. 16213465

Is a Bachelors of (Biology/Chemistry/Physics) enough to get a decent job or do you need a Masters?

Anonymous at Thu, 6 Jun 2024 12:51:07 UTC No. 16214069

>>16213293
[math]\mathbb{R}ead\, the\, \mathbb{OP} [/math]
>>16213465
you can get a job teaching high school or an entry level tech job if you can demonstrate coding proficiency, but it's not ideal
if you want to employed after bachelors major in engineering

Anonymous at Thu, 6 Jun 2024 14:38:33 UTC No. 16216284

In 2024, does it still make sense for ovens to have timers? I honestly never trust the oven timer, I set it to a random amount of time and then use my phone, because I can hear the phone timer from anywhere.

Anonymous at Thu, 6 Jun 2024 17:43:09 UTC No. 16217243

why is this unsolvable?
Is there a way to tell easily that some problems do not have analytical solutions? here i am trying to solve a voltage divider problem:
You have Vin and and Vout. And the voltage divider formula is Vout=Vin * R2 / (R1 + R2).
Now imagine you have a load connected to it that will divert some current, so Vout will equal to Vin * R2 (R1 + R2 || Rl). So the error can be calculated. But I am trying to solve for R1 and R2 given Rload and the error threshold, such as find me R1 and R2 to drop voltage from 5v to 3v if the load resistance is 100k and the error should not exceed 5%. I tried solving for different parts but I am just going in circles.

Anonymous at Thu, 6 Jun 2024 17:45:10 UTC No. 16217252

>>16217243
Here is what I mean to illustrate the problem

Anonymous at Thu, 6 Jun 2024 17:55:33 UTC No. 16217274

>>16216284
the timer turns the oven off so if your unable to return to the stove in time, for any reason, it wont burn down your house

Anonymous at Thu, 6 Jun 2024 17:56:47 UTC No. 16217279

>>16217243
you haven't assigned eqn to the equation to solve.

Anonymous at Thu, 6 Jun 2024 18:11:31 UTC No. 16217341

>>16217252
This. The word problem variety from >>16217243 is completely unnecessary.
if you are writing a program, for the love of god only use node or mesh analysis.

Anonymous at Thu, 6 Jun 2024 18:12:58 UTC No. 16217347

>>16217279
duh, thanks anon. now i am getting this madness. why is it a matrix?

Anonymous at Thu, 6 Jun 2024 18:14:20 UTC No. 16217353

Kind of a weird question, but I guess it belongs here. what is it called when you look at an object so long that you forget it exists? For instance: Say i have a program on my desktop that I see all the time, but I've become so used to it being in the background that when I actually go to use it, I don't even remember where it is. I feel like there must be a phenomenon, but I don't know what it is.

Anonymous at Thu, 6 Jun 2024 18:15:18 UTC No. 16217358

>>16217341
Not sure what you mean. I did use a standard analysis for the direct calculation input -> output. But I don't know how to do the opposite: calculate the inputs based on the output, generally speaking. I don't even know if that is possible.

Anonymous at Thu, 6 Jun 2024 18:23:30 UTC No. 16217401

>>16217347
Because there are two possible solutions (it's a quadratic equation). Note the +/- on the sqrt term.

Anonymous at Thu, 6 Jun 2024 18:36:57 UTC No. 16217480

>3 = -(-3)
Why is it not 0?

Anonymous at Thu, 6 Jun 2024 18:37:43 UTC No. 16217485

Any advice for a beginner comp sci major?(I've no experience programming but I'm good at math)

Anonymous at Thu, 6 Jun 2024 18:47:32 UTC No. 16217539

>>16217401
Ah I see. Why are there so many terms?

Anonymous at Thu, 6 Jun 2024 18:47:38 UTC No. 16217541

>>16217485
Don't become religious about things like programming languages and frameworks. It's fine to have preferences and things you like more than others but don't be an ass who goes around lecturing anyone who makes a different technical choice. This goes 10x if you're in consulting and it's the client's tech stack. Save those arguments for /g/.

Anonymous at Thu, 6 Jun 2024 18:49:43 UTC No. 16217549

>>16217485
Switch to a more useful major while you still have the chance. AI will destroy all comp sci jobs.

Anonymous at Thu, 6 Jun 2024 18:53:36 UTC No. 16217564

>>16217539
Erm... because math. If you think the equation for r3 should not be that complex then maybe your original equation is wrong, but I don't know anything about circuits to answer that.

Anonymous at Thu, 6 Jun 2024 19:01:46 UTC No. 16217603

How can I approximate this integral with at least 6 exact decimals using a Gauss-Laguerre quadrature?
I split it into two integrals, one from [-inf,0] and another from [0, inf],the second one already being in a Gauss-Laguerre form, but the first one wasn't. I got stuck at the first one, I applied the x = -t substitution to try and flip the interval but that left me with exp(x) which isn't Gauss-Laguerre. How else could I have approached this?

Anonymous at Thu, 6 Jun 2024 19:05:02 UTC No. 16217613

>>16217603
exp(x) = exp(2x) * exp(-x)

Anonymous at Thu, 6 Jun 2024 19:06:16 UTC No. 16217620

>>16217613
Oh my God... how did I not see that

Anonymous at Thu, 6 Jun 2024 19:46:56 UTC No. 16217772

>>16214069
I thought you could be like a lab tech with a degree.

Anonymous at Thu, 6 Jun 2024 19:51:31 UTC No. 16217795

>>16217549
I tried mechanical engineering engineering but it was too much for me. The people I was surrounded by were complete assholes and made me lose most of my motivation. I did put in the work to get straight As in all three of my general physics classes though. Comp sci seems better in terms of workload and I plan on supplementing it with either data science minor or stats ninor.

Anonymous at Thu, 6 Jun 2024 19:52:43 UTC No. 16217799

>>16213293
[math]\text{[math]text[/math]}[/math]

Anonymous at Thu, 6 Jun 2024 20:21:32 UTC No. 16217904

>>16217541
I don't plan on going to /g/ except for pc building advice. I would much rather come here or my profs for programming/theoretical advice

Anonymous at Thu, 6 Jun 2024 20:30:33 UTC No. 16217928

>>16217904
There's some competent and nice people on /g/, just like here. Just stick to the generals (just like here).

Anonymous at Thu, 6 Jun 2024 20:44:28 UTC No. 16217978

Plausibility check: I know the effects of VX in The Rock are completely inaccurate, nerve agent is a shitty way to go but it doesn't corrode/burn exposed flesh like that (and also doesn't come in giant gel caps). What I'm curious about is if you COULD develop a chemical weapon with similar effects? My first thought is some kind of toxin that rapidly attacks cell membranes and basically turns organic tissue into a cytoplasmic goop, but I'm not clear if anything like that exists. Thoughts?

Anonymous at Thu, 6 Jun 2024 20:52:20 UTC No. 16217994

>>16217978
The nerve gas they warned us about caused muscle spasms strong enough to break your back. Not sure if true.

Anonymous at Thu, 6 Jun 2024 22:29:33 UTC No. 16218202

Let [math]f: \mathbb{R} \to \mathbb{R}[/math] C1
and [math]\forall _{x \in \mathbb{R}} |f(x) - cos(x) | < 1[/math].
Show that every solution to the differential equation [math] x' = f(x) [/math] is bounded.

I'm honestly not even sure how to approach this. I thought about using stability since that is what we were learning recently but doesn't work.
I know that this implies the x' is within (-1,1), has an infinite amount of critical points, etc, but I honestly don't know how to show it is bounded.

🗑️ Anonymous at Thu, 6 Jun 2024 22:42:58 UTC No. 16218235

>>16218202
[math]f(x) = cos(x) + 0.5[/math]
[math]\int _0^{2 \pi} f(x) = 1[/math]
Set [math]x(t) = \int_0^t f(y) dy[/math] and it's false, no?

Anonymous at Thu, 6 Jun 2024 22:46:11 UTC No. 16218242

>>16218235
For your function you have
x'(t) = f(t)
not
x'(t) = f(x(t))

Anonymous at Thu, 6 Jun 2024 22:51:42 UTC No. 16218261

>>16218242
Yeah, my bad. Been a while since I've done ODEs.
If [math]f(x(0)) \neq 0[/math], the interval [math][x(0) - 2 \pi, x(0) + 2 \pi[/math] has to have at least two points surrounding [math]x(0)[/math] where [math]f(x) = 0[/math] (by Bolzano's theorem). That's that.

Anonymous at Thu, 6 Jun 2024 23:36:40 UTC No. 16218367

>>16217978
Sounds a little like lewisite or phosgene oxime to me, though I haven't watched it

Anonymous at Thu, 6 Jun 2024 23:41:42 UTC No. 16218380

>>16217480
imagine " - " as multiplying by (-1)

Anonymous at Fri, 7 Jun 2024 00:03:38 UTC No. 16218436

>>16218202
>cosX - 1 < f(x) < cosX + 1
0 < f(0) < 2 -> f is positive
-2 < f(pi) < 0 -> f is negative
0 < f(2pi) < 2 -> f is positive

There are at least one point on each interval (0,pi), (pi,2pi) etc. where f(x) = 0, so there are asymptotes for 1/f(x) at those points. This means at those points, [math]\displaystyle t(x_i) = \int^{x_i} \frac{1}{f(x)} \ dx = \pm \infty [/math]. So, if you tilt your graph sideways, this shows that x(t) are bounded at those points since the curve never crosses those asymptotes.

There has to be a better way to do this though. Something more slick. But drawing it out, this made sense to me.

Anonymous at Fri, 7 Jun 2024 08:59:01 UTC No. 16219125

I came across this picture, it's a photo of a 1977 book which I suspect is telling a bullshit story.
Question: can the age of wood be estimated or inferred from x-ray scans? From what I could find x-rays are used to measure the density of wood, but maybe they are related.
Thanks.

Anonymous at Fri, 7 Jun 2024 09:37:16 UTC No. 16219163

>>16219125
> From what I could find x-rays are used to measure the density of wood
Which is the same thing as showing tree-rings, they are formed from differing densities of wood. Find the tree-ring pattern of the wood and you can match that to other known samples.

Anonymous at Fri, 7 Jun 2024 11:06:42 UTC No. 16219313

>>16218436
You don't need to argue that the solution doesn't cross the critical points, it just doesn't.
For whatever [math]C^1 ( \mathbb{R} )[/math] function [math]f[/math], [math]x(t)[/math] a solution of [math]f(x(t)) = x'(t)[/math], if there's some [math]t_0[/math] such that [math]f(x(t_0)) = 0[/math] then we can produce a new solution to the differential equation [math]y(t) = x(t_0)[/math] and by uniqueness of first order ODE solutions we have [math]x(t) = y(t)[/math]

🗑️ Anonymous at Fri, 7 Jun 2024 15:18:29 UTC No. 16219850

>>16219313
Ahh, you're right. I was thinking of a way to use Picard-Lindelof to prove cyclic solutions (since it's C^1 and autonomous).
Thanks.

Anonymous at Fri, 7 Jun 2024 23:12:33 UTC No. 16220855

In terms of the grades of similarity between two objects, it is : isomorphism, isometry, homology, from the strongest to the weakest?

Anonymous at Sat, 8 Jun 2024 04:02:28 UTC No. 16221210

Is vacuum energy really unreachable?

🗑️ B4RK0N (300 IQ) &#039;kneel&#039; at Sat, 8 Jun 2024 04:05:14 UTC No. 16221213

>>16221210
You asleep are you?

Anonymous at Sat, 8 Jun 2024 04:42:01 UTC No. 16221284

>>16221210
Can you extract work from a watermill where all the water is flat and stagnant?

Anonymous at Sat, 8 Jun 2024 12:45:13 UTC No. 16221792

>>16220855
Not really. Isomorphism is just contextual equivalence. In the context of metric spaces or Riemannian manifolds, isometries are isomorphisms.
Homotopy/homology equivalence (I don't think I've ever seen homology equivalence be used as an expression, but it's easy to extend the concept) would be weaker, yes, but it's also important to pay attention to whether or not you're just changing the context (strong homotopy equivalences are isomorphisms in the category of topological spaces where you modulo out homotopical continuous maps).

Anonymous at Sat, 8 Jun 2024 12:50:27 UTC No. 16221798

>>16220855
>>16221792
Btw you could argue that canonical isomorphisms make objects similar-er than random isomorphisms (because then your objects aren't just interchangeable, they are interchangeable in a specific, immutable way, and this may help you with the axiom of choice at some point), but that's kind of a weird opinion to have.

🗑️ Anonymous at Sat, 8 Jun 2024 13:03:16 UTC No. 16221813

Is the wikipedia article wrong? How is this not a Hamiltnian cycle that I drew in red

🗑️ Anonymous at Sat, 8 Jun 2024 13:05:51 UTC No. 16221818

Wikipedia claims that this graph doesn't have a Hamiltonian cycle. But how is this not a Hamiltonian cycle that I drew in red?

🗑️ Anonymous at Sat, 8 Jun 2024 13:07:34 UTC No. 16221821

>>16221818
Anon, it's literally written right there. Hamiltonian path, not cycle.

Anonymous at Sat, 8 Jun 2024 13:09:02 UTC No. 16221827

>>16221818
Edge I've marked in purple doesn't exist.

Anonymous at Sat, 8 Jun 2024 17:27:07 UTC No. 16222337

I'm looking for examples of "naturally occurring" countable groups that are isomorphic to a free product of two non-trivial groups, at least one of which is infinite.

Example: Free groups of rank [math]\ge 2[/math].
Non-example: Any Abelian group.

Anonymous at Sat, 8 Jun 2024 17:31:47 UTC No. 16222342

[eqn]\sum_{k=0}^{\infty}\frac{x^{3}-1}{(x^{2}+k)(x^{2}+k+1)}[/eqn] How do I find the sum of this series?

Anonymous at Sat, 8 Jun 2024 17:45:29 UTC No. 16222371

>>16222342
-1/12

Anonymous at Sat, 8 Jun 2024 17:54:51 UTC No. 16222388

>>16222342
Use partial fraction decomposition.

Anonymous at Sat, 8 Jun 2024 18:42:44 UTC No. 16222497

>>16222388
I get that a(x^2+k+1)+b(x^2+k)=x^3-1, but how do you get the 1 from 1 * x^3 to equal anything from the other side?

Anonymous at Sat, 8 Jun 2024 18:59:00 UTC No. 16222531

>>16222497
You compare the coefficients of the stuff with k and the stuff without k
[eqn]
a + b = 0 \\
a(x^2 + 1) + b x^2 = x^3 - 1
[/eqn]
So
[eqn]
a = x^3 - 1 \\
b = -(x^3 - 1)
[/eqn]

It becomes a simple telescoping series.
[eqn]\sum_{k=0}^\infty \frac{x^{3}-1}{(x^{2}+k)(x^{2}+k+1)} = \sum_{k=0}^\infty \left( \frac{x^3 - 1}{x^2 + k} - \frac{x^3 - 1}{x^2 + k + 1} \right) = \frac{x^3 - 1}{x^2} [/eqn]

Anonymous at Sat, 8 Jun 2024 21:15:40 UTC No. 16222850

>>16202136
Fundamental theorem of arithmetic states that every integer greater than 1 can be represented uniquely as a product of prime numbers. Prime numbers are integers too. For example, 5 can only be represented as 5*1. But 1 is not considered to be a prime number. Should it be in this case?

Anonymous at Sat, 8 Jun 2024 21:16:40 UTC No. 16222853

>>16222850
>Should it be in this case?
Are you really asking this?
No, it shouldn't, and you've given the exact reasoning in your lead-in to this question.

Anonymous at Sat, 8 Jun 2024 21:23:10 UTC No. 16222863

>>16222853
I can hardly understand. An integer of 5 can be represented only as a product of 5 (a prime number) and 1 (not a prime number).

Anonymous at Sat, 8 Jun 2024 21:25:51 UTC No. 16222869

>>16222863
Consider what would happen to the fundamental theorem of arithmetic if we were to consider 1 to be prime.
Hint: [math]5 = 5 \times 1 = 5 \times 1 \times 1[/math], etc.

Anonymous at Sat, 8 Jun 2024 21:33:18 UTC No. 16222891

>>16222869
So why the theorem's application shouldn't be narrowed only to composite numbers?

Anonymous at Sat, 8 Jun 2024 21:41:06 UTC No. 16222908

>>16222891
[math]4 = 2 \times 2 = 2 \times 2 \times 1[/math] etc. etc. etc.
if 1 is prime, then literally no integer has a unique prime factorisation

Anonymous at Sat, 8 Jun 2024 21:41:28 UTC No. 16222909

>>16222850
yes, it was coined and worded retardedly and it also forced mathematicians to stop considering 1 as a prime despite it perfectly following their prime definition
it's a systemd of maths

Anonymous at Sat, 8 Jun 2024 21:49:49 UTC No. 16222930

>>16222908
What would be then a unique factorisation for any prime number which is an integer too? Composite numbers can do it without "1" unlike prime ones.

Anonymous at Sat, 8 Jun 2024 21:51:34 UTC No. 16222933

>>16222930
[math]5 = 5[/math] is the prime factorisation in and of itself.

Anonymous at Sat, 8 Jun 2024 22:04:56 UTC No. 16222969

>>16222933
But it still would not fit the definition and the criteria of "product".

Anonymous at Sat, 8 Jun 2024 22:07:45 UTC No. 16222975

>>16222969
Sure it does. See also the empty product, which has zero terms being multiplied together.
Still, you might find it better to rephrase it in the more formal sense you find in number theory.

Consider the infinite product [math]2^a \times 3^b \times 5^c \times 7^d \times ...[/math]
the theorem then states that for any integer greater than 1, there is precisely one unique set of values [math]a,b,c,d,...[/math] that returns that integer

Anonymous at Sat, 8 Jun 2024 22:13:46 UTC No. 16222985

>>16222969
> Every positive integer n > 1 can be represented in exactly one way as a product of prime powers
[eqn]n = \prod_{i=1}^{k}{p_i}^{n_i}[/eqn]The exponents can be zero.

Anonymous at Sat, 8 Jun 2024 22:16:31 UTC No. 16222991

>>16222985
>mfw there are only k prime numbers

Anonymous at Sat, 8 Jun 2024 22:22:04 UTC No. 16222999

>>16222933
factorisation means decomposing an entity into a product of several entities
5 = 5 is nothing more than an equation

Anonymous at Sat, 8 Jun 2024 22:27:12 UTC No. 16223003

>>16222975
Thank you, the empty product argument probably solves the question. The rephrasing

Anonymous at Sat, 8 Jun 2024 22:29:41 UTC No. 16223008

>>16222999
Yes. Such as decomposing 5 into the product of itself with the empty product. Perhaps even equating them.

Anonymous at Sat, 8 Jun 2024 22:31:22 UTC No. 16223012

>>16222999
As wiki states, "As another example, the fundamental theorem of arithmetic says that every positive integer greater than 1 can be written uniquely as a product of primes. However, if we do not allow products with only 0 or 1 factors, then the theorem (and its proof) become longer.". I'll still have to find out why the proof and the theorem would become longer though.

Anonymous at Sat, 8 Jun 2024 22:40:35 UTC No. 16223028

>>16223008
i feel that this kind of factorisation is beyond the scope of FTA

Anonymous at Sun, 9 Jun 2024 01:36:06 UTC No. 16223409

>>16222991
Kek.
>>16223008
IIRC you formally define [math]\displaystyle \prod _{k = a}^b c_k = c_b \prod _{k = a}^{b - 1} c_k = c_a \prod _{k = a + 1}^b c_k[/math] and [math]\displaystyle \prod_{k = a}^a = c_a[/math] or something similar to that.
So you can just do stuff like [math]\displaystyle 5 = \prod_{k = 1}^1 5 = 5 * \prod_{k = 1}^0[/math], yes.
The indexes may seem weird but that's because mathematicians closed the intervals on both ends decades ago.

Anonymous at Sun, 9 Jun 2024 06:30:09 UTC No. 16223926

Are there any forums where people knowledgeable in science and/or math discuss things?
I only know of stackexchange.

Anonymous at Sun, 9 Jun 2024 14:18:10 UTC No. 16224458

Total mathlet here who believes maths is the best tool we have to discovering what is conceptually possible
Are there any mathematical concepts where something does not have a beginning but has an end?

Anonymous at Sun, 9 Jun 2024 14:32:29 UTC No. 16224476

>>16224458
depending upon precisely what you mean, the p-adic numbers (or more generally the n-adic numbers) technically meet that definition

Anonymous at Sun, 9 Jun 2024 14:35:18 UTC No. 16224481

>>16224458
Actually makes me think of descending chains before anything else.

Anonymous at Sun, 9 Jun 2024 14:42:57 UTC No. 16224490

>>16224458
AFAIK the simplest example would just be the negative integers, which have a bijective function between them and the natural numbers[math] f(z) = -z [/math], hence they have the same cardinality as the natural numbers [math]\aleph_0 [/math]. In a certain sense, you can take any infinite sequence and say "look, I flipped it around" to get something with no beginning but an end but there isn't much value in that since it is much more straightforward to deal with it in the traditional way.

Anonymous at Sun, 9 Jun 2024 14:54:07 UTC No. 16224507

>>16224476
>>16224481
>>16224490
Its a relief that there are alot of answers and i even somewhat understood the negative numbers one as a cherry on top. Im not sure what i mean specifically, i can try explain
As far as im aware, for something to exist, it has to either have a beginning or have always existed. If something has a beginning, it proves that a beginning is possible and something that is possible will eventually happen again after it ends. I also thought that if something has always existed, it should continue to exist forever. So existence would essentially be indestructible, either literally or by virtue of always coming back
If something has no beginning but an end, it could be the only way to actually destroy something permanently so im quite happy to hear that such things are atleast possible in a vague sense. This universe is dreadful

Anonymous at Sun, 9 Jun 2024 14:58:51 UTC No. 16224523

is dark matter a real topic in physics or it is a meme? can i major in dark matter?

Anonymous at Sun, 9 Jun 2024 15:06:47 UTC No. 16224534

>>16224507
Sounds like you should be reading more about causation in philosophy rather than science. For the most part science has nothing to do with causation of existence in general because the only causative relations one can deal with scientifically are those which can be described under models of physics, which pretty much take it axiomatically that forces cause changes in motion.
Be careful about drawing conclusions about reality from mathematics. Most of the time we are just playing logic games when talking about mathematics and as such the purview of mathematics includes all concepts which can possibly be expressed in a system of formal logic (which I'm highly confident includes every possible concept), while the complex interdependencies of material reality mean that just because a thing can be thought doesn't mean that which the thought signifies is real in any material way.

Anonymous at Sun, 9 Jun 2024 15:28:30 UTC No. 16224587

>>16224534
I agree but its alot more comforting to know something is atleast conceptually possible in a logical system than not. Its the difference between total hopelessness and a sliver of hope
I believe that philosophy can be detached from actual reality in a way that is similar to mathematics so i think its okay to ask math wizards some questions from time to time. But you are right, i should focus on reading philosophy more

Anonymous at Sun, 9 Jun 2024 15:39:29 UTC No. 16224617

On a PVT thermodynamic surface of an ideal gas, isothermal, isobaric and isochoric processes are intersections of the PVT surface with planes orthogonal to the three pvt axis.
But what about adiabatic process? Is it also an intersection of some (possibly sloped) plane with the surface? Or is its shape more complicated than that?

Anonymous at Sun, 9 Jun 2024 17:07:14 UTC No. 16224844

>>16224507
Anon, that's literally unironically word-for-word Parmenides.

Anonymous at Sun, 9 Jun 2024 17:21:01 UTC No. 16224878

>>16202136
1. Why do some anons here think entropy is the worst thing ever to exist (as a law of physics)?

2. Isn't biological life creating order in chaos and thus working against entropy?

Anonymous at Sun, 9 Jun 2024 17:46:32 UTC No. 16224941

>>16224523
What do you mean by a real topic?

Anonymous at Sun, 9 Jun 2024 17:50:47 UTC No. 16224949

>>16224878
1. No idea, can't say I've ever seen such posts.
2. Yes. Life uses the energy of the sun to decrease entropy.

🗑️ Anonymous at Sun, 9 Jun 2024 17:55:06 UTC No. 16224960

Erabe!
https://vndb.org/g464
https://vndb.org/g1486

Anonymous at Sun, 9 Jun 2024 18:17:41 UTC No. 16225041

>>16224878
>2. Isn't biological life creating order in chaos and thus working against entropy?
"Creating order" as you described it requires entropy. It is entropy that creates life. And life itself acts like a fast forward button making entropy rise even faster than it would without it

Anonymous at Sun, 9 Jun 2024 19:44:01 UTC No. 16225205

What am I doing wrong? Is it related to the warning or it is a bug in sympy, some null object is referenced or something?

Anonymous at Mon, 10 Jun 2024 00:25:37 UTC No. 16225859

I have a particle with position [math]p[/math] and rotation [math]q[/math] (the latter described as a quaternion). For initial position [math]p_0[/math], initial velocity [math]v_0[/math], and constant acceleration [math]a[/math] it's fairly clear that [math]p(t) = p_0 + v_0t + \frac{1}{2}at^2[/math].
What I don't understand is how to do this for rotation quaternion [math]q[/math] given initial angular velocity [math]\omega_0[/math] and constant angular acceleration [math]\alpha[/math]. I've seen that when angular velocity is a constant [math]\omega[/math] then [math]q(t) = q(0) + \frac{1}{2}\omega q(t)[/math], but I don't understand what's meant by scalar multiplication of a quaternion, or addition of quaternions. Additionally I have no idea what this would be when angular acceleration [math]\alpha \neq 0[/math].
Is the derivative [math]\frac{d}{dt}q(t) \equiv \frac{1}{2}\omega q(t)[/math]? How do I determine the second derivative [math]\frac{d^2}{dt^2}q(t)[/math] for constant angular acceleration [math]\alpha \neq 0[/math]?

Anonymous at Mon, 10 Jun 2024 00:28:15 UTC No. 16225864

>>16225859
God damn it, [math]q(t) = q(0) + \frac{1}{2}\omega q(t)[/math] is missing a term. It should read [math]q(t) = q(0) + \frac{1}{2}t\omega q(t)[/math]. Note the multiplication by [math]t[/math].

Anonymous at Mon, 10 Jun 2024 00:49:45 UTC No. 16225896

is the "if you sit down for a certain amount of time, you will lose 20 years of your life" pseudo-science? I know if all you do is sit down, you'e probably not going to live the longest. Still, I'd like to know if i was overreacting when I heard about it when I was 12. Hell, it's the reason why i don't have a chair in my room.

Anonymous at Mon, 10 Jun 2024 01:50:08 UTC No. 16225980

>>16224617
Look up adiabats. They connect isothermal level curves, and ate PV^\gamma = const, where gamma is 1+2/f, where f is the degrees of freedom

Anonymous at Mon, 10 Jun 2024 01:52:19 UTC No. 16225987

>>16225205
>asks if the problem is with yourself, or the package
it's probably you

Anonymous at Mon, 10 Jun 2024 02:24:19 UTC No. 16226035

>>16225859
w is the angular speed, so if there's an angular acceleration a(t), then do the integral w = S a(t) dt. Obviously, a(t) is constant, then w = at.

wt is how many angles passed in this amount of time, i.e. the length of the rotation, theta.

A unit quaternion q is made up of the unit vector u, about which the object rotates, and theta = wt, the amount it rotates about the axis u.
q is the equal to Euler's formula for quaternions, so q=e^(theta/2 u). If any of these variables are a function of t, then do the derivative with respect to t for those variables.

Anonymous at Mon, 10 Jun 2024 02:47:44 UTC No. 16226063

>>16226035
What is w * q? w is a vector and q is a quaternion. I've read that you're supposed to treat angular velocity w as a quaternion with angle = 0. But then the quaternion multiplication w * q doesn't rotate anything. A 0-angle rotation * q is just q, unchanged.

Anonymous at Mon, 10 Jun 2024 03:07:30 UTC No. 16226089

Is the "magnetic flux quantum" literally a single unit of flux (discrete), or just the lower limit for flux passing through a superconductor (continuous)?

Anonymous at Mon, 10 Jun 2024 03:11:11 UTC No. 16226097

>>16223926
researchgate, scientific conferences, various specialty blogs

Anonymous at Mon, 10 Jun 2024 03:11:27 UTC No. 16226099

>>16226063
Guy imagine a rotating object. It rotates about an axis. Imagine what the w vector looks like in this scenario. Now, imagine what q looks like in this scenario. They point in the same direction - why would you need to incorporate the axis into both w and q?

w is a number, you already have the quaternion to store information about the axis of rotation -_-

Anonymous at Mon, 10 Jun 2024 04:19:48 UTC No. 16226158

>>16226089
The latter. Using a superconductor means the amount of flux has to be quantized, which means there is a minimum discrete amount.

Anonymous at Mon, 10 Jun 2024 04:25:26 UTC No. 16226168

>>16226099
Because the quaternion has no information about which way the object is rotating, only how the object is currently oriented. Imagine you have a top spinning on a table. The y-axis is perpendicular to the table, so the top is spinning about the y-axis. Now lie the top down on its side and spin it. Angular velocity w is still about the y-axis but its orientation quaternion q at any time t is not the same as before because it's lying on its side.

Anonymous at Mon, 10 Jun 2024 04:41:40 UTC No. 16226182

>>16226168
The OP is using q to describe the rotation, anon, and he's using p as position.

Anonymous at Mon, 10 Jun 2024 05:33:53 UTC No. 16226215

>>16226182
I'm the one who posted the original question. Unless I'm doing something wrong, p describes the object's current position and q describes how it is currently rotated. It has no information about which way the object is rotating, only its current orientation (just as p has no information about which way the object is moving, only where it is right now). I want to take an initial way the object is facing and determine which way it will be facing after [math]t[/math] seconds, but I don't understand what is meant by [math]q = q_0 + \frac{1}{2}t\omega q_0[/math], where [math]q_0[/math] is the initial orientation of the object and [math]\omega[/math] is an angular velocity (rate and direction of rotation).

See section 4 here: https://www.cs.cmu.edu/~baraff/sigcourse/notesd1.pdf
>the multiplication ω(t)q(t)is a shorthand for multiplication between the quaternions [0, ω(t)] and q(t)
Wouldn't a multiplication between "quaternion" [0, ω] and quaternion q just be q?

Anonymous at Mon, 10 Jun 2024 05:52:00 UTC No. 16226249

>>16226158
So is it possible for 1.1x the flux quanta, 1.11x the flux quanta, and 1.111...x the flux quanta to exist? Or is the next smallest amount 2x?

Anonymous at Mon, 10 Jun 2024 06:57:33 UTC No. 16226325

>>16226249
It's quantized, so integer steps: 2x, 3x, ... etc

However there is something called the Fractional Quantum Hall Effect where in 2D systems certain rational values can be obtained.

Anonymous at Mon, 10 Jun 2024 10:30:08 UTC No. 16226554

>>16226325
>It's quantized, so integer steps: 2x, 3x, ... etc
Interesting, is there experimental evidence of this? That sounds amazing if so, but it'd have to be tested

Anonymous at Mon, 10 Jun 2024 12:04:18 UTC No. 16226647

>>16225980
in other words, what I was really asking about was if you cut the PVT surface with an arbitrary flat plane, what would you get as a resulting intersection?
Since you get isotherms, isobars and isochores as a result of cutting the PVT surface with flat planes perpendicular to the axis, I was wondering, that maybe adiabats might be analogously results of cutting it with an arbitrary flat plane?

Anonymous at Mon, 10 Jun 2024 12:44:50 UTC No. 16226693

>>16226554
yes there is tons of evidence for it, you encounter flux quanta constantly if you do any kind of experiment with type II superconductors
look up
>Abrikosov lattice
>Bitter decoration
>superconducting quantum interference device
>>16226647
don't think so, isotherms can be obtained by intersecting with a family of planes [math]T = c [/math] parameterized by [math]c [/math] for example, but the adiabatic constraint [math]PV^\gamma =c [/math] defines a family of curved surfaces not planes

Anonymous at Mon, 10 Jun 2024 12:46:04 UTC No. 16226694

>>16226647
>>16226693
if you intersect with an arbitrary plane you can only ever get a linear constraint of the type
[math] \alpha P + \beta V + \gamma T = c[/math]

Anonymous at Mon, 10 Jun 2024 12:58:43 UTC No. 16226712

>>16226693
>defines a family of curved surfaces not planes
>>16226694
>if you intersect with an arbitrary plane you can only ever get a linear constraint of the type
yeah, I just checked by substituting PVT equation into arbitrary plane equation, and except for some boundary conditions where the plane represents an isochoric and isobaric process, the intersection of the arbitrary plane and PVT surface is always some shifted hyperbola, so I suppose it can't be an equation for an adiabat
thanks

Anonymous at Mon, 10 Jun 2024 17:25:52 UTC No. 16227071

Is this notation correct? Or am I missing something?
Because I know that in SU(n) n is the number of dimensions, but I don't know what the second number in SU(1,3) is

Anonymous at Mon, 10 Jun 2024 18:05:27 UTC No. 16227141

>>16227071
For SO(1,3), it means 1 time and 3 space

Anonymous at Mon, 10 Jun 2024 18:55:25 UTC No. 16227215

>>16227141
follow up question: why is qcd represented with a su(3) lie group? is it because there are three colors?

Anonymous at Mon, 10 Jun 2024 18:57:03 UTC No. 16227216

Maybe I'm not asking this question correctly, but what is meant by the fact that an electron can have an infinite number of excited states?

In hydrogen (Balmer Series) there are only 6 lines on the visible spectrum meaning there are only a maximum of 6 energy levels that correspond to wavelengths on the visible light scale right?

If a hydrogen electron is able to to go to some energy level, say the 79th level, shouldn't that imply that there are 79 different emissions on the electromagnetic spectrum when it goes back to is ground state? How would an electron even get to such a high energy state without ionizing the Hydrogen? Each energy level corresponds to some quanta of energy absorbed by the electron right? How can there be an infinite number of energy levels if at a certain point the energy would just ionize the hydrogen?

Anonymous at Mon, 10 Jun 2024 18:58:57 UTC No. 16227217

>>16226215
They literally tell you how to multiply quaternions in the pdf. They also tell you to look up the Appendix for the answer to your questions. Have you looked up the appendix?

Anonymous at Mon, 10 Jun 2024 19:01:14 UTC No. 16227220

>>16227215
yes

Anonymous at Mon, 10 Jun 2024 19:02:56 UTC No. 16227223

>>16227217
They sorta do it by the dq/dt = q(t+dt) - q(t) / dt way.

Anonymous at Mon, 10 Jun 2024 19:13:40 UTC No. 16227243

>>16227216
Because the difference in energy between the states is proportional to: [math]E_{diff} \propto 1/n^2[/math]. So the higher energy levels become closer and closer together and become essentially indistinguishable.

Anonymous at Mon, 10 Jun 2024 19:20:34 UTC No. 16227251

Can someone explain to me in reasonably easy terms how the FUCK am I supposed to integrate a differential form like [math]xzdx - yzdy + \frac{xyz}{\sqrt{x^2 + y^2 + z^2}} dz [/math] over the surface of the (faces of) tetrahedron with edges (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,1,1)?

Am I supposed to construct a parameterization for each of the faces? This seems impossible

Anonymous at Mon, 10 Jun 2024 19:21:28 UTC No. 16227252

>>16227216
>>16227243
This, and also because when trying to teach fundamental QM principles to gen chem students most textbook writers are going to try to avoid talking about the treatment of free electrons, which is necessary if you're going to talk about ionization.

Anonymous at Mon, 10 Jun 2024 20:21:39 UTC No. 16227355

>>16227243
>>16227252
Thanks alot. Makes sense.

So n approaches infinity as the energy absorbed by the electron increases up the ionization energy. Spectroscopy seems rad

🗑️ Anonymous at Mon, 10 Jun 2024 20:51:17 UTC No. 16227411

>>16227251
>seems impossible
dude, you only have 6 faces. Literally do each face. I'll do one face for you:

The xy plane is just z is constant at z=0. For your integral then, you have 0*xdx - 0*ydy + 0/rho*0 = 0. Wasn't that hard

Anonymous at Mon, 10 Jun 2024 20:53:57 UTC No. 16227413

>>16227355
If you want a good general overview of spectroscopic methods I highly recommend Skoog et. al.'s Principles of Instrumental Analysis, very well written and an authoritative reference. Covers a lot more as well.
Aside from the usual suspects, there is an older version available through archive.org if you want to browse.

Anonymous at Mon, 10 Jun 2024 21:46:07 UTC No. 16227485

Can the appearance of Jesus be modeled as a Poisson distribution? Since it is a rare event that could have happened over a certain period of time and supposedly more than once.

Anonymous at Mon, 10 Jun 2024 22:00:46 UTC No. 16227506

>>16227251
What you have right now is F * dL. You're question is probably not integrate over the faces (2D), but integrate over the edges (1D).

Anonymous at Mon, 10 Jun 2024 22:11:00 UTC No. 16227524

>>16227217
The appendix is cut off.

>they tell you how to multiply quaternions
Here's what they say:
>[s1, v1][s2, v2] = [s1s2 − v1 · v2, s1v2 + s2v1 + v1 × v2].
>A rotation of θ radians about a unit axis u is represented by the unit quaternion [cos(θ/2),sin(θ/2)u].
>ω(t)q(t)is a shorthand for multiplication between the quaternions [0, ω(t)] and q(t)

Now suppose q is unrotated. Then its orientation quaternion is [cos(0), sin(0)u] = [1, 0u], which is the unit quaternion (1, 0, 0, 0). Am I wrong?

Then:
[math][0, \omega][1, 0u] = [0(1) - \omega \cdot 0u, 0(0u) + 1\omega + \omega \times 0u][/math]
[math]= [-\omega \cdot 0u, \omega + \omega \times 0u][/math]
[math]= [0, \omega][/math]
Which is a quaternion that rotates by 0 radians. So there is no resultant rotation, which I don't understand.

Anonymous at Mon, 10 Jun 2024 22:27:20 UTC No. 16227557

>>16227251
Use the divergence theorem.

Anonymous at Mon, 10 Jun 2024 22:40:28 UTC No. 16227580

>>16227251
ignore >>16227506. also, that is a triangular bipyramid, not a tetrahedron.

you have all the differentials. determine the limits of integration from the edges given (if these aren't edges parallel to your axes, these will be expressions with respect to at least one other variable). for example, the bottom integral will have:

z = 0; 0 <= y <= -x+1; 0 <= x <= 1

integrating over this yields 0 for your integral.

rinse and repeat. this problem should be straightforward.

Anonymous at Tue, 11 Jun 2024 02:23:17 UTC No. 16227817

>>16225896
To be fair, unless you are too poor for a standing desk, there's no reason to sit except for leisure (and then, laying is better).

Anonymous at Tue, 11 Jun 2024 09:59:05 UTC No. 16228177

Are there any math concepts you tried to understand repeatedly but found no explanation of it to be convincing, gave up and just accepted it as truth?

Anonymous at Tue, 11 Jun 2024 10:02:18 UTC No. 16228184

>>16228177
This existence of spin in fundamental particles.

Anonymous at Tue, 11 Jun 2024 10:06:04 UTC No. 16228192

>>16228184
Isn't it still somewhat disputed?

Anonymous at Tue, 11 Jun 2024 10:10:51 UTC No. 16228197

>>16228177
Many still can't accept Monty Hall paradox.

Anonymous at Tue, 11 Jun 2024 10:18:25 UTC No. 16228202

>>16228192
No, at least not as far as I know. Maybe convincing isn't the right word. The math that explains it, and I've come across several different methods, is always too complicated for me to understand.

Anonymous at Tue, 11 Jun 2024 10:26:46 UTC No. 16228207

How do I make meth?

Anonymous at Tue, 11 Jun 2024 10:38:21 UTC No. 16228214

Where can I go if I want to re-learn maths as an adult? Nothing in particular, just want to sharpen up some arithmetic and get some basics back but ideally with some structure.

Even stuff like multiplying/dividing on paper and the best methods I need a refresher on.

Anonymous at Tue, 11 Jun 2024 11:56:55 UTC No. 16228254

>>16228214
khan academy

Anonymous at Tue, 11 Jun 2024 14:10:29 UTC No. 16228398

>>16228202
QM and QFT in general have this thing where either the presentation is absolutely crystalline or it's this nightmare where you have to reason around experiment outcomes but what's the experiment is actually really opaque.

Anonymous at Tue, 11 Jun 2024 14:19:54 UTC No. 16228406

Is this real?

Anonymous at Tue, 11 Jun 2024 14:31:22 UTC No. 16228418

>>16228406
>Black-Scholes, QM, entropy
It's meme stuff for a movie background.

Anonymous at Tue, 11 Jun 2024 16:50:47 UTC No. 16228566

How can photons mediate the electromagnetic force if they don't have an electric charge?

Anonymous at Tue, 11 Jun 2024 17:50:21 UTC No. 16228701

>>16227524
>appendix is cut off
have you looked it up?

Anonymous at Tue, 11 Jun 2024 17:56:56 UTC No. 16228720

>>16227580
Ok, then tell me how would you solve the integral F * dl over a flat surface like the unit square on the xy plane?

Anonymous at Tue, 11 Jun 2024 18:23:29 UTC No. 16228780

Looking through a proof on the Peano remainder in Taylor series, they prove it by induction and in the step they assume that f.D->R is differentiable n times and for those n it is true now they need to show if f is differentiable n+1 times then the peano remainder stuff also hold, now they take the derivative of f and since it is differentiable n+1 times then f' is differentiable n times, meaning you can use the induction assumption, but I don't get how this works since the derivative function is different from the function f itself and the induction assumption says that if F is differentiable n times(not f') then the peano remainder stuff works. Is it just beacuse f' taylor series is kinda similar to f's taylor series?

Anonymous at Tue, 11 Jun 2024 18:27:08 UTC No. 16228787

>>16228566
If a single electron emitted radiation that had electric charge, then the electron would lose it's charge.

Anonymous at Tue, 11 Jun 2024 19:17:53 UTC No. 16228898

>>16227506
Yeah, this turned out to be right.
I partially misunderstood the question and thought it was about a different shape. It turned out to be more of a folded triangle thing, not a tetrahedron.
Then I just had to integrate it along 4 edges and the result was 0.

Anonymous at Tue, 11 Jun 2024 21:03:32 UTC No. 16229098

>>16228787
An electric field doesn't contain just one electron. They could be split +ve and -ve, and charge would still be conserved.

Anonymous at Tue, 11 Jun 2024 21:09:54 UTC No. 16229112

>>16229098
So if an electron emits radiation, you're saying it should emit virtual electrons?

Anonymous at Tue, 11 Jun 2024 21:15:37 UTC No. 16229126

Does anyone know why for spherical coordinates, why does
[math] r\sin\theta \tfrac{\partial \phi}{\partial x} = \tfrac{1}{r\sin\theta} ``\tfrac{\partial x}{\partial \phi}" [/math]
work, where [math] \tfrac{\partial x}{\partial \phi} [/math] is if you write out x as a function of r, phi, and theta)? Like, it doesn't have to be phi or x.

Like, is the an intuitive explanation for this?

Anonymous at Tue, 11 Jun 2024 21:24:30 UTC No. 16229140

>>16229112
That's the original question. Why not? How can virtual photons mediate the force?

Anonymous at Tue, 11 Jun 2024 21:34:01 UTC No. 16229162

>>16219313
I don't understand at all why uniqueness or that [math]f(x(t_0)) = 0[/math] matters here. Can you elaborate further?
I assume that [math] y(t) = x(t_0 + t)[/math] in which case they differ by a constant and don't necessarily cross.

Anonymous at Tue, 11 Jun 2024 21:41:56 UTC No. 16229176

>>16229140
An electron moving to the right emits radiation, where should it's two emitted particles be thrown out? In which directions? Choose any that makes sense to you. What's going to happen to the original electron now? What SHOULD happen when an electron emits radiation?

Anonymous at Tue, 11 Jun 2024 21:43:41 UTC No. 16229179

>>16229176
it's "supposed" two emitted...

Anonymous at Tue, 11 Jun 2024 21:58:10 UTC No. 16229206

Can I use FX rates to convert currencies at different points in time?

For example
I have an item purchased at 10 USD in 2023 and then another at 15 USD in 2024, let's say I want to find the total spend across the two years in terms of JPY, using the current rate of 1 USD = 157 JPY.

Can I apply the conversion to both numbers OR do I first convert the 10 USD at 2023 into 2024 USD then apply the conversion, since 2023 USD =/= 2024 USD

Anonymous at Tue, 11 Jun 2024 22:06:31 UTC No. 16229227

>>16228566
There are two ways to induce a change in the electric field: Either with an electric charge, or a change in the magnetic field.
The easiest way to induce a change in the magnetic field is with a change in the electric field.
This self-propagating feedback loop manifests as the photon. It is able to mediate electromagnetism while being electrically neutral, because being electrically charged is only a requirement for an initial change in the EM field. The photon does not do that. It IS the change.

Anonymous at Tue, 11 Jun 2024 22:16:44 UTC No. 16229243

>>16203051

Anonymous at Wed, 12 Jun 2024 02:53:59 UTC No. 16229657

Could an absolute bro please help me access this paper?
https://www.cell.com/cell/abstract/S0092-8674(24)00108-9

Anonymous at Wed, 12 Jun 2024 08:55:09 UTC No. 16230003

>>16229227
Thanks, that makes some sense.

Anonymous at Wed, 12 Jun 2024 10:31:45 UTC No. 16230087

>>16228197
And how could they?
Your odds don't actually improve to 2/3: you're still picking one from three and then one from two.

Anonymous at Wed, 12 Jun 2024 14:33:27 UTC No. 16230363

>>16230087
Because the second of the two encompasses two different scenarios.
Monty always eliminates a wrong answer that you didn't pick when he reduces it to two. If either of the two you didn't pick was correct (which is a 2/3 chance), then that's going to be the one you'd get to by switching post-elimination. And as you've noted, the odds don't change.

Anonymous at Wed, 12 Jun 2024 14:45:29 UTC No. 16230381

I'm playing a russian roulette game, where you can pull the trigger 2 or 3 times in a row.
What's the math of the highest win rate strategy looks like?

Barkon. at Wed, 12 Jun 2024 14:47:22 UTC No. 16230382

>>16230381
I'm playing 'reply to a faggot', and I just won.

Anonymous at Wed, 12 Jun 2024 14:48:32 UTC No. 16230385

>>16230382
congratz, I just won too

Anonymous at Wed, 12 Jun 2024 19:33:17 UTC No. 16230847

What are the energetic costs for plants to produce fruits? I have a calamondin at home and I recently noticed that she started to drop leaves (I kinda messed up, because I forgot to fertilize her...), but not fruits, which I thought are more costly in terms of energy for plants to produce. Maybe it's because of how this particular species behaves, but I find it strange it'd get rid of leaves first instead of fruits. Is there a reason for that?

Anonymous at Wed, 12 Jun 2024 20:35:46 UTC No. 16230946

What should I review for an intro level linear algebra class? I had to take a break from college for family reasons.

Anonymous at Wed, 12 Jun 2024 20:43:48 UTC No. 16230954

>>16230946
Linear algebra is one of those courses that doesn't have much in the way of prerequisite knowledge, especially if you're doing it at an introductory level.
At most you should only really need to brush up on matrices and solving systems of equations

Anonymous at Wed, 12 Jun 2024 20:49:15 UTC No. 16230963

>>16230954
Oh then this will be piss easy then for me. I wonder why people have trouble with it though

Anonymous at Wed, 12 Jun 2024 23:03:37 UTC No. 16231166

I want to represent increase or decrease in a rank as a percentage so I can sort by trending-ness whether they rise from rank 1 million to 0.75 million (ez) or rank 2 to 1 (very hard).

ideas?

Anonymous at Thu, 13 Jun 2024 00:05:57 UTC No. 16231353

>>16231166
Logarithms?

Anonymous at Thu, 13 Jun 2024 00:24:06 UTC No. 16231383

>>16231353
seems a lot more reasonable than what I had. Thank you

Anonymous at Thu, 13 Jun 2024 00:55:30 UTC No. 16231434

How can I get a comprehensive understanding of mathematics? from what i understand stuff like algebra came from Euclid? Should I start there? if i wanted to start somewhere where would that be? All I was taught was do step 1-5 etc to get the answer in skool

🗑️ Anonymous at Thu, 13 Jun 2024 10:04:16 UTC No. 16232200

would a small machine powered suction device not be perfect to remove acne? I always wondered why this doesnt exist

Anonymous at Thu, 13 Jun 2024 12:32:08 UTC No. 16232425

>>16228214
>>16228254
OK, I'm on KhanAcademy
One thing I don't seem to see is methods of long multiplication or division by hand. I have forgot how to do it.
I know for multiplying there's the box method and the lattice method. I can't remember how to do it for division. Any advice on how to divide by hand, or if there are better ways to multiply? I just need a refresher on all this stuff

Anonymous at Thu, 13 Jun 2024 13:07:44 UTC No. 16232466

>>16231166
>>16231353
>>16231383
Although Google hitting 70% just for going rank 2 to rank 1 on 1yr seems a little bit extreme

Anonymous at Thu, 13 Jun 2024 13:10:59 UTC No. 16232471

>>16232466
playing around with base now

Anonymous at Thu, 13 Jun 2024 14:25:06 UTC No. 16232569

>>16232466
>>16232471
Yeah, you should finnagle with the base a bit. Remember that [math]\displaystyle \log _a b = \dfrac{\log b}{\log a}[/math]

Anonymous at Thu, 13 Jun 2024 15:10:58 UTC No. 16232640

>>16232569
Finally came up with something that seems to make sense at all ranks with reasonably inbuilt assumptions of traffic levels at both ends

Anonymous at Thu, 13 Jun 2024 15:34:36 UTC No. 16232679

I'm trying to show that, for a finite group [math]G[/math] with subgroup [math]H[/math], [math]|G|=|H| \cdot |G/H|[/math], where [math]|A|[/math] is the number of elements of the set [math]A[/math].
I know that I can write [math]G=\bigcup\limits_{[g] \in G/H} [g][/math] as a disjoint union, and conclude.
Is it possible to find, explicitly, a bijection between [math]G[/math] and [math]H \times G/H[/math]?
That is, some pair [math](h,[x])[/math] associated to each [math]g \in G[/math].
Of course we can send [math]g \mapsto [g] \in G/H[/math], and we have a bijection [math]f_g:[g] \to H[/math] by [math]f_g(x)=g^{-1}x[/math], but I'm having trouble finding [math]h[/math] in terms of the [math]f_g[/math] and [math]g[/math].
I don't really know algebra so don't know if this is a normal thing to want to do.

Anonymous at Thu, 13 Jun 2024 16:12:00 UTC No. 16232752

I blew the biggest, sloppiest, messiest load of my life yesterday. For as long as I can remember, I would off-and-on have a discomfort in my groin, and I always just thought it was an old injury. But it's gone. Really gone. Is it possible I just had a decade-long case of blue-balls?

Anonymous at Thu, 13 Jun 2024 16:30:31 UTC No. 16232801

Is there a good resource for arguements for nuclear energy? If not I want to start collecting information around it

Anonymous at Thu, 13 Jun 2024 17:48:46 UTC No. 16232973

What is the word that describes this pattern?

Anonymous at Thu, 13 Jun 2024 20:03:46 UTC No. 16233186

>>16202136
Are those books about "learning how to learn" like a mind for numbers are good for getting good or should I just aim to solve 50 problems a day to become a math guy?

Anonymous at Thu, 13 Jun 2024 20:12:25 UTC No. 16233199

>>16233186
Doing problems isn't enough, but it is almost there.
Build a toolbox.
When you learn something new, make a place for it within your toolbox.
Everytime you approach a problem, look at all of the tools in your toolbox and pick the best option.
Periodically test that you remember everything in your toolbox.
Toolbox in this case is just a mnemonic device, method of loci style. To assist my toolbox, I have attached the number of tools in each cordoned area. This lets me know that I am remembering all of the tools without cheating.

Math is a fractal form where simple ideas in the are isomorphic to higher level ones. For example, all of the basic algebra operations are learned again in matrix format. One of these number niggers should be able to help you out, I am electrical fag.
Oh yeah, separate your tools between disciplines.

Anonymous at Thu, 13 Jun 2024 23:07:01 UTC No. 16233439

Does anyone please know the heat transfer effect in Watt per meter for the
>BEACON MORRIS Hydronic Baseboard Heater Element: Hydronic Baseboard Heater Elements, 4 ft Lg, Silver

Anonymous at Thu, 13 Jun 2024 23:09:55 UTC No. 16233449

>>16233439
Because this little 400mm bugger clocks in at 400W/m.

(And in case someone wonders, I'm going to build an air cooled distiller.)

Anonymous at Thu, 13 Jun 2024 23:10:36 UTC No. 16233453

>>16233439
>>16233449
Not a /sci/ question.

Anonymous at Thu, 13 Jun 2024 23:17:07 UTC No. 16233463

>>16233453
>Not a /sci/ question.
Then what is it?

Anonymous at Thu, 13 Jun 2024 23:26:57 UTC No. 16233481

>>16232973
"bullshit"

Anonymous at Thu, 13 Jun 2024 23:30:29 UTC No. 16233484

if i superimpose a right handed triangle onto a 2D grid, how do i know whether a specific (x,y) point is inside the triangle?

Anonymous at Thu, 13 Jun 2024 23:45:24 UTC No. 16233513

If mass curves spacetime then why doesn't Earth end up a year in the past every time it completes an orbit of the Sun?

Anonymous at Fri, 14 Jun 2024 02:20:55 UTC No. 16233726

What does this notation mean?
[math]\sum_{i < j}[/math]
Looking at this specifically:
[math]\sum_{i < j}P(A_i\cap A_j)[/math]

Anonymous at Fri, 14 Jun 2024 02:33:22 UTC No. 16233738

>>16233726
Would it be correct to interpret
[math]\sum_{i=1}^n\sum_{i < j}a_{ij}[/math]
as
[math](a_{12}+a_{13}+a_{14}+...)+(a_{23}+a_{24}+a_{25}+...)+...[/math]
?

Anonymous at Fri, 14 Jun 2024 09:20:42 UTC No. 16234229

>>16233726
>>16233738
Yes, in your first example the set from which i and j can be taken must have been defined somewhere before
Note also that
[math]\sum_{i<j} a_{ij} = \frac{1}{2}\sum_{i \neq j} a_{ij}[/math]
and some people prefer to use the latter notation where you double-count and then divide by two

Anonymous at Fri, 14 Jun 2024 09:39:36 UTC No. 16234243

>>16233738

Anonymous at Fri, 14 Jun 2024 09:59:55 UTC No. 16234259

What programs are better than matlab for physical world simulation using user-defined models? I am looking into radio behaviors and matlab seems to be too slow. Should I just get a new PC?

Anonymous at Fri, 14 Jun 2024 10:14:45 UTC No. 16234275

Gotcha, thanks anon. For the tip about [math]\frac{1}{2}\sum_{i \neq j} a_{ij}[/math] as well, looks interesting. If I understand correctly, this would only be equal if the elements are diagonally symmetric, right?
Also, somewhat irrelevant, but this is the equation that I failed properly copy:
[math]P\biggl(\bigcup_{i=1}^n A_i\biggr) =\sum_{i=1}^n P(A_i)-\sum_{i < j}P(A_i\cap A_j)[/math]
[math]+\sum_{i < j < k}P(A_i\cap A_j\cap A_k)-\ \cdots\ +(-1)^{n-1}\, P\biggl(\bigcap_{i=1}^n A_i\biggr)[/math]

Anonymous at Fri, 14 Jun 2024 10:16:58 UTC No. 16234276

>>16234275
Meant for >>16234229
Also
>failed to properly copy*

Anonymous at Fri, 14 Jun 2024 10:38:55 UTC No. 16234301

>>16203051
Lost

Anonymous at Fri, 14 Jun 2024 10:46:42 UTC No. 16234306

>>16202136
Why is picking two people simultaneously from a group different from picking one person then picking another person
I know the probability for the second is
[eqn]\frac{1}{n}*\frac{1}{n-1}[/eqn]
But I don't remember the equation for the first one, it had something to do with combinations, high school was a long time ago

Anonymous at Fri, 14 Jun 2024 10:47:49 UTC No. 16234307

>>16234306
Why the fuck is my tex code not working
[eqn] \frac{1}{n}*\frac{1}{n-1} [/eqn]

Anonymous at Fri, 14 Jun 2024 11:00:02 UTC No. 16234318

>>16234307
Suppose that you enter a raffle and 2 winners are drawn simultaneously. Which ticket probability is equivalent to either sequential draw case? (n^-1, (n - 1)^-1)

Anonymous at Fri, 14 Jun 2024 12:01:51 UTC No. 16234353

>>16202136
at what point you can say you understood something?
i'm learning about Fourier series and transform. i understand what they are supposed to solve and how to use them but i would not be able to prove them or come up with it by myself. now what? do i spend un undefinite amount of time just to understand it more or just move on and learn more stuff?

Anonymous at Fri, 14 Jun 2024 12:27:19 UTC No. 16234383

How do you solve this, I tried trig. substitution and now my house is on fire

Anonymous at Fri, 14 Jun 2024 13:00:24 UTC No. 16234424

>>16234275
>this would only be equal if the elements are diagonally symmetric
yes you're right

Anonymous at Fri, 14 Jun 2024 13:02:17 UTC No. 16234429

>>16232425
Try corbettmaths. It has videos and practice questions for most high school topics.

Anonymous at Fri, 14 Jun 2024 13:09:22 UTC No. 16234440

>>16234424
I see. Thanks for helping me out.

🗑️ Anonymous at Fri, 14 Jun 2024 13:17:50 UTC No. 16234452

>>16234306
It is the same, unless you specifically care about the order in which the people are picked [math]

Anonymous at Fri, 14 Jun 2024 13:22:15 UTC No. 16234458

>>16234306
It is the same, unless you specifically care about the order in which the people are picked

🗑️ Anonymous at Fri, 14 Jun 2024 13:54:55 UTC No. 16234511

>>16234383
Have you tried a simple u-sub followed by collecting the powers into a neat package?

Anonymous at Fri, 14 Jun 2024 14:05:14 UTC No. 16234531

>>16234511
I think I've practically tried everything under the sun, I've gotten close to the function geogebra gives when integrating, but I'm stuck on how to do tan(arcsinx/2), I know that tan(arcsinx) is a fraction with x and stuff, but for it to be halved, I'm not sure

Anonymous at Fri, 14 Jun 2024 14:06:56 UTC No. 16234533

>>16234429
thank you, will look through

Anonymous at Fri, 14 Jun 2024 14:14:32 UTC No. 16234547

>>16234383
Starting with the u-sub x=sin(u) seems like a good start. It does get messy after that. It seems to require another u-sub and a partial fraction decomposition.

Anonymous at Fri, 14 Jun 2024 14:17:02 UTC No. 16234555

>>16234547
That's exactly what I did

Anonymous at Fri, 14 Jun 2024 14:18:03 UTC No. 16234556

>>16234547
Sorry, my results are here >>16234531

Anonymous at Fri, 14 Jun 2024 14:50:09 UTC No. 16234586

>>16233484
You could use an inequality.

Let's say the hypotenuse of the right triangle is
of the form [math] ax+by<c [/math] and also
[math] x,y>0 [/math] like in your drawing.

Then with each point you put them in both
of these inequalities to see if they make
comparative sense. If either or both inequalities
don't make sense, the point is outside the triangle.

Anonymous at Fri, 14 Jun 2024 15:35:17 UTC No. 16234644

>>16234531
[math]\tan \arcsin {m} = \frac { m } { \sqrt { 1 - m^2 } }[/math] (most easily proven by sketching it out)
therefore by substitution, [math]\tan \arcsin { \frac { t } { 2 } } = \frac { t } { 2 \sqrt { 1 - \frac { t ^ 2 } { 4 } } } [/math]

Anonymous at Fri, 14 Jun 2024 15:37:29 UTC No. 16234648

>>16233484
there are many answers on google...

Anonymous at Fri, 14 Jun 2024 21:16:31 UTC No. 16235198

Hey sci! Im a neurosciecne phd. Whats wrong with my brilliant idea: https://medium.com/@lunde.anders/training-neurons-in-a-dish-can-biological-neural-networks-be-trained-in-silicon-4927da492ca6

Anonymous at Fri, 14 Jun 2024 21:51:33 UTC No. 16235246

How would you calculate the bits of entropy of this password?

>stagnant-squid-possible-legal

Entropy = log2(S^L)
Where S is symbol pool size and L is the length. Log base 2 of 26 alphabets + 1 hyphen, raised to the power of 32, because 32 characters long.

Chatgpt says the bits of entropy are 152. But that's assuming an attacker doesn't know our password scheme and is bruteforcing.

What if the attacker knows we are picking 4 random words separated by a hyphen?

Then the symbol pool is the size of random words you are pulling from. Let's say 7,000 words. The length then is 4, because 4 words. The hyphens are negligible.
So the actual entropy, assuming our attacker knows the scheme is, 51 bits.

Not quite weak, but not strong either.

My goal here is to use 4 random words, because it's easy to type and to remember, and then harden it by randomly inserting a special character and a digit.

The problem now is I am too dumb to calculate the bits of entropy for the new scheme and therefore cannot measure how much stronger the password is.

>stag?nant-squid-pos7sible-legal

What is the entropy now that I have randomly placed one special char and one digit? Is this measurable?

I asked chatgpt and it said to account for the increased complexity of the possible insertions the entropy is 67.

And then I asked what if the attacker doesn't know the exact length of each word, and it basically said fuck you calculate it yourself.