/sqt/ /qtddtot/ - stupid questions thread

[Anonymous No.14661473]

Previous thread: >>14629605

>what is /sqt/ for?
Questions regarding math and science. Also homework.
>where do I go for advice?
>>>/sci/scg or >>>/adv/
>where do I go for other questions and requests?
>>>/wsr/ >>>/g/sqt >>>/diy/sqt etc.
>how do I post math symbols (Latex)?
rentry.org/sci-latex-v1
>a plain google search didn't return anything, is there anything else I should try before asking the question here?
scholar.google.com
>where can I look up if the question has already been asked here?
>>/sci
https://eientei.xyz/
>how do I optimize an image losslessly?
trimage.org
pnggauntlet.com

>where can I get:
>books?
libgen.rs
z-lib.org
stitz-zeager.com
openstax.org
>articles?
sci-hub.st
>book recs?
sites.google.com/site/scienceandmathguide
4chan-science.fandom.com/wiki//sci/_Wiki
math.ucr.edu/home/baez/physics/Administrivia/booklist.html
>charts?
imgur.com/a/pHfMGwE
imgur.com/a/ZZDVNk1
>tables, properties and material selection?
www.engineeringtoolbox.com
www.matweb.com

Tips for asking questions here:
>attach an image (animal images are ideal. Grab them from >>>/an/)
>avoid replying to yourself
>ask anonymously
>recheck the Latex before posting
>ignore shitpost replies
>avoid getting into arguments
>do not tell us where is it you came from
>do not mention how [other place] didn't answer your question so you're reposting it here
>if you need to ask for clarification fifteen times in a row, try to make the sequence easy to read through
>I'm not reading your handwriting
>I'm not flipping that sideways picture
>I'm not google translating your spanish
>don't ask to ask
>don't ask for a hint if you want a solution
>xyproblem.info

[Anonymous No.14661780]

hey STEMlords, /ic/ peasant here. I want to make a set of calipers out of a couple popsicle sticks such that whatever one side measures, the other side measures 1.168x that, i.e. a golden ratio caliper. So where should I put my fulcrum? Let's assume a standard length of 112.7mm for the popsicle sticks, but I assume the fulcrum's position would just be a function of the variable length. And yes I know I could just buy a pair of golden means calipers but I think it would be fun to make a set and it's a hobby of mine to spend as little money as possible ;)

[Anonymous No.14661840]

>>14661780
>>>/diy/2427052
would be much more helpful for you i guess

[Anonymous No.14661846]

>>14661840
I guess, I mean I don't need help building the things, just need to know where to stick the pin, which I figured was /sci/'s domain. I'll ask there as well.

[Anonymous No.14661936]

How do they know that space itself is expanding? What if gravity (or another, perhaps even multiple of the other forces) just becomes repulsive after a certain distance?

[Anonymous No.14661960]

>>14661936
>How do they know that space itself is expanding?
Because of red shift.
>What if gravity (or another, perhaps even multiple of the other forces) just becomes repulsive after a certain distance?
There are many theories for explaining this phenomenon and repulsive gravity is one of them but dark energy is by far the most popular one.

[Anonymous No.14662005]

I don't understand the validity of the following claim about [math]M[/math] being an upper bound.
Let [math]\mathbb{R}^k[/math] be the euclidean space with the square metric [math]\rho[/math]. Let [math](x_n)[/math] be a Cauchy sequence in [math](\mathbb{R}^k , \rho)[/math]. Then the set [math]\{x_n \}[/math] is a bounded subset of [math](\mathbb{R}^k , \rho)[/math]. For if we choose [math]N[/math] so that [eqn]\rho(x_n,x_m) \leq 1 [/eqn] for all [math]n,m \geq N [/math], then the number [eqn] M = \max \{ \rho(x_1,\mathbf{0}),\ldots,\rho(x_{N-1},\mathbf{0}),\rho(x_N,\mathbf{0})+1\}[/eqn] is an upper bound for [math] \rho(x_n,\mathbf{0})[/math].
It's probably something as simple as using triangle inequality but I don't see it yet.

[Anonymous No.14662032]

>>14661473
Integer arithmetic keeps filtering me. How the fuck do I absorb all these theorems after theorems? I can't even finish the chapter let alone attempt the exercise.

[Anonymous No.14662160]

Everyone on this board likes to talk about how brainlets are a blight on humanity, but what about the opposite side of the spectrum? I'm wondering what is the biggest possible brain size that a creature can have? I think blue whales and elephants rank amongst the biggest. I've also read somewhere, but I forgot where, that there is a limit to brain size. Namely that after a certain size, the body just can't cool it, this is of course ignoring the issue of weight and calorie required. What I'm wondering is just how big can brains physically get?

[Anonymous No.14662267]

why are there no water distillers that use thermoelectric heat pumps

[Anonymous No.14662647]

>>14662005
Doing it for [math]k=1[/math], in which case [math]\rho(x_n, x_m)=|x_n-x_m|[/math].
Let [math]N[/math] be as in your post and fix [math]n\geq N[/math]. For all [math]m\geq N[/math], we have
[math]|x_n|-|x_m|\leq |x_n-x_m|\leq 1[/math].
In particular, for [math]m=N[/math]
[math]|x_n|\leq 1 + |x_N|[/math].
This takes care of all [math]n\geq N[/math].
But what happens if [math]n< N[/math]? Well there's only a finite amount of terms so we can just take the maximum of those too.

[Anonymous No.14662723]

>>14662647
thanks you

[Anonymous No.14662868]

>>14660424
I'll start from the top. Hopefully this clears your confusion a bit.
1) Why can we reduce the problem to proving that every finite abelian group has a cyclic tower?
Assume this last statement to be true and let the finite, non-necessarily abelian, group [math]G[/math] have the following abelian tower
[math]1=G_0\triangleleft G_1\triangleleft \cdots \triangleleft G_n = G[/math].

Since [math]G_1/G_0[/math] is finite abelian, there is a cyclic tower
[math]0=H^1_0\triangleleft H^1_1\triangleleft \cdots \triangleleft H^1_{m_1} = G_1/G_0 = G_1 [/math].

The same goes for [math]G_2/G_1[/math], we have a cyclic tower
[math]0=H^2_0\triangleleft H^2_1\triangleleft \cdots \triangleleft H^2_{m_2} = G_2/G_1 [/math].
Now look at the canonical projection [math]\pi: G_2\rightarrow G_2/G_1[/math]. Taking the inverse image of the above cyclic tower yields a cyclic tower for [math]G_2[/math] ending at... [math]G_1[/math],
[math]G_1=\pi^{-1}(H^2_0)\triangleleft \pi^{-1}(H^2_1)\triangleleft \cdots \triangleleft \pi^{-1}(H^2_{m_2}) = \pi^{-1}(G_2/G_1 ) = G_2[/math].
To prove the quotients are cyclic: write the natural projection [math]\pi^{-1}(H^2_i)\rightarrow H^2_i\rightarrow H^2_i/H^2_{i-1}[/math], look at the kernel and use the isomorphism theorem.

Now we can continue this process and end up with a cyclic tower for [math]G[/math]
[math]0=H^1_0\triangleleft H^1_1\triangleleft \cdots \triangleleft H^1_{m_1} = G_1 = H^2_0\triangleleft \cdots \triangleleft H^2_{m_2}=G_2\triangleleft\cdots\triangleleft G_{n-1} = H^n_0\triangleleft \cdots \triangleleft H^n_{m_n} = G_n = G[/math].

[Anonymous No.14662871]

>>14662868
2) Every finite abelian group admits a cyclic tower.
Let [math]G[/math] be a finite abelian group of cardinal [math]m[/math]. The statement is true when [math]m=1[/math].

Assume now that the statement is true for abelian groups of order less than [math]m-1[/math], take [math]x\in G[/math] with [math]x\neq e[/math] and let [math]X = <x>[/math].
As [math]x\neq e[/math] we have [math][G:X]<m[/math] and since [math]G'=G/X[/math] is finite abelian, we can apply our statement by induction:
There exists a cyclic tower for [math]G'[/math] denoted
[math]0=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n = G'[/math].
Once again, consider the projection [math]\pi:G\rightarrow G'[/math] and the inverse image of the above tower:
[math]X=\pi^{-1}(H_0)\triangleleft \pi^{-1}(H_1)\triangleleft \cdots \triangleleft \pi^{-1}(H_n) = \pi^{-1}(G') = G[/math].
The same argument as in part 1 shows that this is a cyclic tower ending at [math]X[/math].

Now [math]X[/math] is... cyclic! So simply complete this tower by [math]1[/math] and we get a cyclic tower for [math]G[/math].

[Anonymous No.14662874]

>>14661473
One that left the cavern and saw the light of the world, how can he testify to the light when cave dwellers only ever knew the shadows?

[Anonymous No.14663271]

I'm trying to do the Contacts problem in Hackrerank. It looks like I need to use a Trie. I can't find anything that walks me through this on youtube or google from the last year.

I tried to import trie from pytrie but it seemed impossible in Hackerrank.

So I guess I have to make the class and define the add and find functions in it?

Any ideas on how to do this?

[Anonymous No.14663293]

Mathchads, I have a burning question that's been killing me since this morning.

I think it has to do with my poor understanding of summation division.

I'm not going to explain the variables but they're all constants when you plug them in. The value that is being divided is the left number and the right number on the right side of the last two matrixes that I poorly drew a '/' for.

After canceling out V(kx,ky) from the numerator and denominator, I get exactly: Deltax (kx + 0.5) which in the 2nd matrix is 50.

I tried doing it for the next column (green) and now it's 59.6 rather than 150.

Does anyone know the math explanation behind this?

[Anonymous No.14663302]

>>14662871
thank you so much! I somehow managed to forget about the actual induction part and was lost in the weeds before, your answer makes it very clear

[Anonymous No.14663303]

>>14663271
>>>/g/

[Anonymous No.14663333]

>>14663293
Okay I think I get it now, it's because everything accumulates to the next step as that's how summation works and if I were to take the difference between the top and the bottom and do the division like before, I'll get 150. Shit, I need to retake calculus 2.

[Anonymous No.14663338]

>>14663293
>After canceling out V(kx,ky) from the numerator and denominator
You absolutely cannot do that: [math]V(k_x, k_y)[/math] depends on [math]x[/math] and [math]y[/math].
But [math]\Delta_x(k_x+0.5)[/math] too, as such you cannot factor and simplify.
Counter example: does [math](5*2 + 6*3)/(2 + 3) = (5 + 6)[/math]?

[Anonymous No.14663378]

>>14663338
Ohhhhh okay that helps reduce my itch a lot more. It's because all of them depends on a particular variable and the only way I could cancel V(kx,ky) is if I also cancel out the deltax portion as well so that there is no dependency on the x,y?

[Anonymous No.14663470]

>>14663378
Kind of yes. I'm having a hard time explaining this somehow, maybe some other anon will do a better job.
If you want to simplify, you have to make sure you can split the sum. For instance if you have two functions [math]f[/math] and [math]g[/math] independent of each other regarding their inputs:
[math]\sum_{k=0}^{N}\sum_{l=0}^M f(k)g(l) / \sum_{l=0}^M g(l) = (\sum_{k=0}^{N}f(k))( \sum_{l=0}^Mg(l)) /\sum_{l=0}^M g(l)) = \sum_{k=0}^{N}f(k) [/math].
In your example you could see [math]f(k_x) = \Delta_x(k_x + 0.5)[/math] and [math]g(k_x, k_y) = V(k_x, k_y)[/math]. But then these functions are not independent.
Now this is assuming both functions ACTUALLY depend on these inputs. It could very well be that there is another way of writing [math]f\cdot g[/math] as a product of independent functions or that your function [math]V[/math] doesn't even use [math]k_y[/math].

[Anonymous No.14663551]

I was looking into this aryl hydrocarbon signaling pathway as a way of detecting phenyl groups, This image and all the other papers I looked into referred to the part of DNA which the aryl hydrocarbon receptor (AhR) binds to once it grabs an aryl hydrocarbon as a "dioxin responsive element" (DRE in pic rel). Does that mean that this only binds to dioxins and not, say, raw phenyl groups attached to AhR?

Also, should this be its own thread?

[Anonymous No.14663558]

>>14662874
fuck off plato, you never experimented anything in your whole goddamn career, philosophers circlejerk about how smart you all are and literally nothing you said was true because you thought you were too smart to have to prove any of it with an experiment.

[Anonymous No.14663651]

>>14662032
It seems to me that that entire page is just restating the statement of Bezout, or at least the consequences of Bezout when a, b are coprime. So really none of it is saying anything, there's nothing to "absorb". Learn Bezout's identity and make sure you know how to apply it - which is what that page is demonstrating to you.

[Anonymous No.14663684]

>>14663551
>Does that mean that this only binds to dioxins and not, say, raw phenyl groups attached to AhR
The DRE does not bind to dioxins directly. It's called a Dioxin Response Element because of its role in mediating the cells response to dioxin toxicity. It binds to the ARNT/AhR dimer; which forms after the AhR is activated by binding to the dioxin. You can see it in your diagram - the little black ovals are the dioxins; they bind to the AhR which then enters the nucleus and forms a dimer with ARNT - the red and green bits. That's what actually binds to the DNA.
Several ligands other than dioxins can bind to and activate AhR, so it's possible they could trigger the same response, but I don't know if that has ever been demostrated.
>Also, should this be its own thread?
No point - /sci/ is shit at biology.

[Anonymous No.14663752]

>>14663684
Ok, thank you so much for being like the only other biologist I've ever seen on /sci. and being helpful <3

[Anonymous No.14664222]

i don't understand "kinetics" when i'm reading papers about chemistry. i took two semesters of organic chem, and a biochem course. i know a bit about kinetics with regard to enzymatic reactions, and how with most reactions, the rate correlates with the concentration of one reactant.
but i don't really understand how to interpret charts of reaction kinetics, or equations derived from them.
like, pic related is from this paper:
https://doi.org/10.1155/2014/701092
on the kinetics of thermal degradation of PVC

i don't know how to extract meaningful information from the data here. looking at these charts, and looking at the equations, it doesn't mean anything. i can read the conclusions and see what the guy says, but i don't understand the applicability.

with this one in particular, if i have an equation detailing the kinetics of the degradation at a given temperature, say 120C, how would i determine what percent conversion of the PVC happens at a given point in time, or how would i predict how long it would take for full conversion? is that something i can infer from this data?

i'm not particularly invested in just PVC degradation, this is just an example of a paper that deals with kinetics. i don't really understand the concept on the whole.

[Anonymous No.14664232]

>>14661780
the calipers form two similar isosceles triangles, so if the sides form a golden ratio then so will the bases.
>>14661840
did you even read the post?

[Anonymous No.14664234]

What property am I forgetting here?

[Anonymous No.14664242]

>>14664234
the property that plus and minus cancels out? whats confusing you?

[Anonymous No.14664304]

>>14664234
the property that [math]2x-x=x[/math] and [math]-x+x=0[/math]

[Anonymous No.14664306]

>>14664234
what problem is this btw?

[Anonymous No.14664335]

>>14664306
partial segment of the derivation of a shortcut for the variance of a pareto distribution
E(x^2) - E(x)^2 = E(x)^2 - (α/(α-2))

[Anonymous No.14664415]

>>14664335
mmm I don't like pbb too much but post it anyway

[Anonymous No.14664418]

>>14664415
There's not very much more to it.
I'm just a retard who doesn't see obvious thing sometimes

[Anonymous No.14664480]

>>14664418
>I'm just a retard who doesn't see obvious thing sometimes
it happens to everyone anon, the important part is noticing it and make corrections, sooner the better.

[Anonymous No.14664501]

Why does it seem almost forbidden to have threads about the dangers of Electromagnetic radiation and microwave technologies on this board?
They seemed to get pruned rapidly.

[Anonymous No.14664517]

>>14664501
Never paid attention to this. Maybe because it quickly turns into a propaganda / /x/ / fanatics argument, like what sometimes happens with the 'real numbers are not real' threads.

[Anonymous No.14664565]

>>14661473
What is the trace of the hat matrix

[Anonymous No.14664728]

What is the difference between variance and distribution?

[Anonymous No.14665160]

Scientifically speaking, what is the best place online for trolling?

[Anonymous No.14665328]

>>14664517
They barely even establish themselves as a thread. even flat earth threads seem to last longer.
I don't know if it's lack of interest or active suppression?

[Anonymous No.14665447]

>>14664728
Distribution is the function that maps outcomes to the probabilities of said outcomes happening. Variance is the expected value of the squared deviation of a random variable.

[Anonymous No.14665455]

>>14665328
Nobody's suppressing you schizo

[Anonymous No.14665460]

>>14664501
non ionizing radiation is non ionizing. feel free to live in a faraday cage though so your brain waves are free of interference. foil hats may also be effective.

[Anonymous No.14665585]

This is an Engineering Drawing Approximate the dimesions of the drawing

[Anonymous No.14665675]

If you can solve one of these 6 major math problems, you'll win a $1 million prize???

[Anonymous No.14665723]

>>14665455
>>14665460
meds

[Anonymous No.14665930]

since earth is protected from space aids by its magnetic field... could a super duper strong (electro)magnet be used to shield astronauts, space stations and space craft from space aids?

[Anonymous No.14665956]

>>14661473
How does a simple theoretical physics brainlet like myself learn ergodic theory? Cutting through all of the technical proof bullshit, what are the important theorems that are actually necessary to understand and describe reality?

[Anonymous No.14665995]

>>14661473
How am I supposed to solve this when the vapor pressure of pure ethylene glycol is not given? Is the question complete?

[Anonymous No.14666018]

Not really a stupid question, but I dont' want to make a thread for this.

I'm a science fiction writer and trying to figure out if something would work. I want a guy in 2020 to make a time capsule to store the access method to his crypto wallet, but getting back to it takes significantly longer than he ever dreamt. It's a hundred years later, 2120, when his great-grandson finally gets it.

But I don't think a smartphone has a method of retaining data on it after 100 years unpowered? Or does it? From briefly googling how NAND flash works, it seems like the whole thing would just corrode and short out after a few decades.

Do I need a different storage device entirely?

[Anonymous No.14666161]

>>14666018
>But I don't think a smartphone has a method of retaining data on it after 100 years unpowered? Or does it?
It should be possible for NAND flash to retain data for 100 years provided that it rarely gets hot and the number of erase cycles is low (each erase cycle increases leakage, temperature increases leakage). The JEDEC standard says 20 years at 55°C, but 55°C is quite extreme (e.g. interior of a car parked in the sun all day in summer). It will last much longer at more sensible temperatures. The electronics could also last that long in a dry environment.

[Anonymous No.14666222]

>>14666161
Thanks. Hopefully buried underground in a vacuum sealed bag is plausible enough

[Anonymous No.14666498]

>>14665995
I think you're supposed to use Antoine equation for that, have you looked the book's appendix for EG's Antoine coefficients?

[Anonymous No.14666623]

Is there any lifehack for putting on weight when you don't like cooking nor eating, apart from gomad? I'm worried about fucking up my kidneys

[Anonymous No.14666687]

>>14665956
>Cutting through all of the technical proof bullshit, what are the important theorems that are actually necessary to understand and describe reality?
Just google ergodic theory for physicists and eenie meenie some pdf.
>inb4 doesn't anyone here have any specific recs specifically for physicists
I doubt it.
If you're fine with books for mathematicians https://link.springer.com/book/10.1007/978-1-4471-7287-1 is fine.

[Anonymous No.14666821]

>>14665930
"in theory" yes it would work. in reality it's not possible to build a field both strong enough and large enough - at least not with current technologies.

[Anonymous No.14666847]

>>14666623
Sounds like a guaranteed way to fuck up your entire body. You gain weight in that method by essentially ingesting bucket loads of milk fats. I seem to remember an article about someone who did that for a few years and ended up dying from a heart attack.

Lots of pasta and ice cream, while not exactly healthy, seems to be a hell of a later safer and nutritionally balanced.

[Anonymous No.14666861]

>>14666821
well that's a shame. is there any math/formula behind the amount of kiloniggagiga volts needed to power such a magnet? eg. I want the field to protect an astronauth who is 2m tall with suit on while the magnet isn't "too big".

[Anonymous No.14666866]

>>14665930
Good question, how do i prevent getting space aids from other astronauts

[Anonymous No.14666950]

What temperature does oil (say butane or whatever oil is oil man) boil at in a vacuum? I wonder if its simply computable?

[Anonymous No.14666993]

>>14666847
Would it be safer to just drink whey protein instead?

[Anonymous No.14667114]

Can anyone explain vectors?

[Anonymous No.14667124]

>>14666018
As far as I can tell, a standard crypto key is one or two kilobytes; the most straightforward way to store it long-term is to print base64 or even hex on a piece of plastic, laminated paper or similar.
Regular-ass paper has been proven to survive centuries if not millenia in proper storage, NAN memory hasn't been around long enough to prove itself, and we already know it degrades.
100 years in a non-postapocalyptic future you could just image scan the result, potentially allowing megabytes of data to be stored without risking human copying error.

[Anonymous No.14667126]

>>14665930
Rotating magnets is literally how saucercraft fly.

[Anonymous No.14667135]

>>14665460
Why does something have to be ionising in order to cause damage? That feels like an irrelevant point. Blue light from screens causes damage by disrupting our hormonal patterns, so there is no reason that only ionising radiation can cause harm.
https://radiationdangers.com/2021/03/26/why-is-arthur-firstenberg-not-telling-you-that-5g-is-a-weapon-by-claire-edwards/

[Anonymous No.14667153]

>>14667114
All these diagrams are just a buncha faery bs meant to confuse you.
A vector is just the sum of the x and ys of some directions and movement.
>3 up, 5 across, then 7 across and 6 down
The vector would be (12,-3) I think because x before y, and it works in as many dimensions as needed.

[Anonymous No.14667177]

>>14667153
What are they used for?

[Anonymous No.14667183]

>>14666950
I think at zero pressure butane cannot boil at any temperature. But I'd have to find a phase diagram somewhere

[Anonymous No.14667205]

Why am I so dumb ?
square root of -2i should be [math]\sqrt{-2i} = \sqrt{2}* e^{- \frac{\pi}{4} }[/math] according to Wolfram alpha. but the whole transform to phi is kinda difficult for me. What is my issue? Could someone please quickly calculate this, so I have a reference how this works?

[math]\sqrt{-2i} = \sqrt{2}* e^{\frac{2\pi}{2} + \frac{\frac{3\pi}{2}}{2} }[/math]

[Anonymous No.14667213]

No cool math questions so I'll ask one myself.

I quite well understand how to construct a space-filling curve that maps [math][0,1][/math] to [math][0,1]^n[/math] but I keep reading that such a construction is easily extendable to the real line but I don't see how to do this, I even struggle to find a space filling map for the two dimensional case i.e [math]f:\mathbb{R}\to\mathbb{R}^2 [/math]

[Anonymous No.14667232]

>>14667205
what you wrote down is correct recall that [math] -i = e^{-\pi/2 + 2\pi n} [/math] for all integers [math] n[/math]. This follows directly from [math]e^{ix}=\cos(x) + i\sin(x) [/math]

[Anonymous No.14667236]

>>14667213
The trick is to combine it with a [math]\mathbb Z\to \mathbb Z^n[/math] surjection, and noticing that R and R^n are split into fractional and integer parts.
Let f be the space-filling curve and g the integer surjection above, then [math]x\mapsto f(x - \lfloor x \rfloor) + g(\lfloor x \rfloor)[/math] surjects.
If f bijects [math][0,1)\to[0,1)^n[/math] and g bijects Z -> Z^n, then this bijects R->R^n

[Anonymous No.14667245]

>>14667236
Space filling curves cannot be bijections

[Anonymous No.14667247]

Good morning sirs,
I have this data set: https://www.kaggle.com/datasets/urstrulyvikas/lending-club-loan-data-analysis
What's the difference between these two graphs? i already printed out the lists individually so I get what the bottom graph would be. I thought hue = credit.policy did the same thing. can someone break this down for a retard?

[Anonymous No.14667249]

>>14667247
I meant to get layered instead of stacked but you get the gist.

[Anonymous No.14667264]

>>14667213
A nicer alternative is to take a nice invertible [math]g:(0,1)\to\mathbb R[/math], extend it arbitrarily to [0,1],
and take [math]g \circ f \circ g^{-1}[/math] (componentwise in the outer application)

>>14667245
I assume "space-filling map" doesn't need to be a curve

[Anonymous No.14667277]

>>14667264
my mistake of poor phrasing, I do sneed it to be a curve sorry.

[Anonymous No.14667362]

>>>/wsr/1234339
Scientifically speaking, is latino a race?
I always assumed it was.

[Anonymous No.14667410]

I found the equation on (a), but I don't know how he got the answer in (b) assuming E^2 >> ka.
The solution also uses the equal symbol with a dot, which I assume is defined?
Basically I factored out the E^2 from inside the second term, and them tried to do a Taylor expansion in the resulting sqrt.

[Anonymous No.14667433]

Theres something really off about Lex Fridman and I cant put my finger on it, I usually notice it when he asks a really silly question and wonder if he's putting on an act. The best way I can describe it is he feels like an actual robot trying to be human. Everything around him from his questions and beliefs to his guests to his comment section of his videos feels really off. I only bring this up because hes the only one in this space that has such a strange aura. Sorry for bringing up e-celebs but it really urks me

[Anonymous No.14667463]

>>14667213
>space-filling curve that maps [0,1] to [0,1]n
redpill me on this

[Anonymous No.14667480]

>>14666687
Thanks, this looks like a good source.

[Anonymous No.14667525]

>>14667126
wtf aliems are real! where's the magnet tho?

[Anonymous No.14667526]

>>14667410
[math]x\dot{=}\sqrt{\frac{2E}{k}}\left(\sqrt{\frac{ak}{4E^2}}+\cos\Bigl(\sqrt{\frac km}t+\theta_0/2\Bigr)\right)[/math]
The first term is negligible compared to the second.

[Anonymous No.14667538]

>>14667177
The simplest example is position. Like Anon said, any point in space (2D, 3D, you name it) can be written down in vector form. The vector is often written down as an arrow but what's really only important is the tip. Using vectors, we can now talk more easily about angles between vectors, and magnitudes. Another example of a vector is velocity: a particle doesn't just occupy a certain position in space, but it's also going in some direction with a certain speed. If you're going 3m/s in the x-direction and 4m/s in the y-direction, the velocity vector can be written down as (3, 4), and the total magnitude, by the Pythagorean Theorem, is 5 (m/s).

But up until here I've just been talking about a fancy way of representing data. What vectors are really studied with, is linear transformations. That is, you take a vector, and you construct a new vector by taking combinations of the values of the old vector. For instance, I can transform the velocity vector (3, 4) into a new vector with x-coordinate a*3 + b*4 and y-coordinate c*3 + d*4 which is then written (3a + 4b, 3c +4d), where a,b,c,d are values depending on the problem. They together are called a matrix. Why would you want to do this? I'll give two examples: 1. Rotation. If we have a point on a circle we can write its position as a vector. Suppose that we want to rotate it by some angle around the circle; what's its new position? It can be shown quite easily that such a transformation is linear, and thus can be performed using a matrix. 2. Change of basis. Imagine we're looking at points on a triangular lattice. Standard Cartesian coordinates are neat, but in this specific scenario it would be much more useful to say how many steps I need to take to the right, and how many along a diagonal. A matrix can translate between these two different representations of position.

There is a lot, lot more to be said about this. Linear algebra is one of the most elementary (and fun) fields of mathematics.

[Anonymous No.14667698]

>>14667236
You forgot about continuity.
Which I don't blame you for because the construction is ass.
>>14667277
Choose a space filling curve [math]f: [0, 1] \to [0, 1]^n[/math] such that [math]f(0) = f(1) = 0[/math] and a surjection [math]g: \mathbb{Z} \to \mathbb{Z}^n[/math].
Define [math]h: \mathbb{R} \to \mathbb{R}^n[/math] by [math]f(2(x - floor(x))) + g(floor(x))[/math] whenever [math]x \leq floor(x) + 0.5[/math] and by [math]g(floor(x)) + 2 (x - floor(x) - 0.5) (g(floor(x + 1) - g(floor(x))[/math] otherwise (literally a line going from[math]g(n)[/math] to [math]g(n + 1)[/math]).

[Anonymous No.14667731]

I've been working as a EE since I got out of college in 2015. I never took the FE because I'm a fucking idiot. Anyone in a similar situation to me? I've been working through a practice test and all the questions seem very easily answered based on the equation handbook you are allowed to use. Is it really that easy?

[Anonymous No.14667797]

Mathematically, it is unlikely I will fail a class. My average would have to crater and my final has to be a laughable 60%+.
How do I keep motivation to study these last two units. I find myself skimming material and day dreaming.

[Anonymous No.14667819]

I dont understand k in the big O stuff, its extremely aggravating.

Show 2^x +17 is O(3^x)

2^x + 17 <= 2^x + 2^x = 2*2^x < 2*3^x.

ok so C = 2 because its 2*3^x

but k supposedly equals 5, meaning the inequality is true when x > 5.

when is 17 <= 2^x? when x is 5. so why does k = 5? k should be 4. when x is 5, 2^x +17 is 49 and 2*2^x is 64. why does x have to be greater than 5.

im sick of being off by one on k, literally every ducking time, it makes no sense to me.

[Anonymous No.14667854]

Hello, please help me with my question about curve fitting
>>>/wsr/1234676

[Anonymous No.14667866]

>>14667854
I need the mathematical function and not the pictures.

[Anonymous No.14667880]

>>14667819
I don't understand what [math]k[/math] is in your question.

Though, I would try to directly show that for a polynomial [math]P[/math] it holds that [math]P(x) = O(\alpha^x)[/math] for any [math]\alpha>1[/math].
If [math]P[/math] is of degree [math]d[/math] then [math]P(x) = O(x^d)[/math] so you just have to show that [math]x^d=O(\alpha^x)[/math].
Write both sides using [math]exp[/math] and conclude.
Remember that exact moment after which your function is less than [math]C\cdot3^x[/math] for some constant [math]C[/math] doesn't matter, it just needs to exist.

[Anonymous No.14667894]

>>14667880

i thought i understood it but i am constantly getting k wrong, usually my k is one less than the answer.

like in this question they say it has to be that x>5 so k=5, but it should be x>4 so k = 4. because when x is 5, 2^x + 17 is smaller than 2*3^x.

i dont get why they say x>5 and k =5

[Anonymous No.14667918]

>>14667894
>because when x is 5, 2^x + 17 is smaller than 2*3^x.
yeah, its supposed to be.

[Anonymous No.14667929]

>>14667894
>because when x is 5, 2^x + 17 is smaller than 2*3^x.
That is literally what you want.
>i dont get why they say x>5 and k =5
you need this to for the inequality [math] 17 \leq 2^x [/math] to proceed with the proof

[Anonymous No.14667930]

>>14667894
You seem to be confused by the quantifiers. When the definition says "there are constants [math]C[/math] and [math]k[/math] such that..." it means that there exists constants verifying the following property.
Now there might be more than one [math]C[/math] or one [math]k[/math] that satisfy this property. And indeed, if the property is true whenever [math]k\geq 5[/math], obviously it's also going to be true when [math]k>5[/math].
When you work with these [math]O[/math] and [math]o[/math] and [math]\sim[/math] what you're interested in is the behavior of the function at infinity (or around a point). So whether we need to look for [math]x[/math] after [math]k=4[/math], [math]k=5[/math] or [math]k=10^{100}[/math], it doesn't matter.

[Anonymous No.14667947]

>>14667698
i will check this out tomorrow
>>14667463
It is fairly simple to construct a space filling curve [math]g:[0,1]\to [0,1]^n[/math] if you know about the existence of a space filling curve (like peano's curve) [math]f:[0,1]\to [0,1]^2 [/math] using induction. Space filling curves are unironically interesting they are also nowhere differentiable and continuous.

[Anonymous No.14667978]

Do visible galaxies near the edge of the observable universe ever actually exit the observable universe, or does their speed of recession merely approach the speed of light?

[Anonymous No.14667979]

>>14667947
>nowhere differentiable
Incorrect.
In fact, Lebesgue's space filling curve is a.e. differentiable.

[Anonymous No.14668078]

What are the assumptions for epsilon in the matrix form of linear regression?

[Anonymous No.14668423]

any way to estimate how much sulfuric acid / hydrogen peroxide i need to dissolve a given amount of organic material? asking for a friend.

[Anonymous No.14668441]

>>14668423
for rough estimate id say at least twice the volume of material you want to dissolve

[Anonymous No.14668790]

I've been working on repainting a bike lately and i noticed that sandpaper and microfiber towels like to stick together. is it working like a microscopic hooks and loops fastener or is there something else which causes it?

[Anonymous No.14668835]

>>14667978
>Do visible galaxies near the edge of the observable universe ever actually exit the observable universe
Yes because the rate at which the Universe expands accelerates. Although you won't see galaxies disappearing in your lifetime, in the far future the amount of visible galaxies will significantly decrease.

[Anonymous No.14669311]

Can someone that understands linear algebra well explain this to me? I'm at a loss as to where to even start.
A and B are both n×n matrices constructed into a bigger 2n×2n matrix as shown in the pic. You're supposed to show how that formula holds true but I don't know how to.

[Anonymous No.14669483]

>>14669311
>someone that understands linear algebra well
You're asking for a whole fucking lot.
>I'm at a loss as to where to even start.
Unsurprising because it's a gimmick problem.

We add the second column to the first (basic column operation, doesn't change the determinant) for [math]\pmatrix{A + B & B \\ B + A & A}[/math].
We then subtract the first row from the second (basic row operation, doesn't change the determinant) for [math]\pmatrix{A + B & B \\ 0 & A - B}[/math]
Result immediately follows.

[Anonymous No.14669783]

>>14669311
>>14669483
Wait a fucking second.
https://en.wikipedia.org/wiki/Block_matrix#Block_matrix_determinant
Did you ask this so you could edit the proof into wikipedia?

[Anonymous No.14669815]

>>14669483
Thanks
>>14669783
No, I'm slowly making my way through the optional exercises we have for LA at uni since I only have one last shot at completing the exam. I didn't realize it was on wikipedia as well

[Anonymous No.14669827]

>>14669815
>I didn't realize it was on wikipedia as well
It isn't, the result is there but it provides no proof or citation.

[Anonymous No.14670296]

If I want to move around a rigid frame with weightless rockets is the only thing that matters the position/angle of the rockets relative to the center of mass?

[Anonymous No.14670511]

>>14670296
moment of inertia also matters

[Anonymous No.14670548]

>>14670511
But not the path between the rocket and the center? The weird thing to me is that if you stack a bunch of rocket (ignore collision) it seems like it's just applying a multiple of the same force. But they all have different positions relative to the center of mass.

[Anonymous No.14670558]

>>14670548
>But not the path between the rocket and the center?
the path between a rocket and the center is a straight line, no?
>The weird thing to me is that if you stack a bunch of rocket (ignore collision) it seems like it's just applying a multiple of the same force. But they all have different positions relative to the center of mass.
youre gonna have to draw a picture or something for that one

[Anonymous No.14670653]

>>14670558
It seem like this should be applying 2 rockets of force to the endpoint either way but the relative angle of the second rocket to the center of gravity is different. Ignore slight offsets to the side. My real question is about magic rockets that don't interact with other matter except to push it.

[Anonymous No.14670894]

>>14670653
Moving a rocket parallel to its thrust direction doesn't change anything. The torque is r×F where r is the vector from the centre of gravity to some point on the thrust line, F is the force vector, and × is the cross product operator.

If P is some point on the thrust line, and D is the direction of the thrust line, then any other point on the thrust line can be expressed as Q=P+tD for some scalar t. If C is the centre of mass, r=P-C, let r'=Q-C = (P+tD)-C = r+tD. F=kD where k is the magnitude of the thrust.

r×F = r×kD = k(r×D)

r'×F = (r+tD)×kD = (r+tD)×kD = (r×kD)+(tD×kD) = k(r×D)+kt(D×D) = k(r×D) = r×F. For any D, D×D=0, i.e. the cross product of a vector with itself is always zero.

[Anonymous No.14671116]

>>14670894
Is the cross product the cosine of the angle vectors meet at, scaled by the magnitude of the vectors? If so it makes sense to me why that quantity doesn't change by moving the rocket parallel to its force.

Except in that case how is the portion of force that doesn't go into rotation determined? It seems like that should depend on something like the sine of the angle, at the place the force is being applied. But if the amount of rotation is independent of parallel position I can just move my rocket really far away, get the same rotation, and push my object. I assume that's a violation of conservation of energy, or something.

I don't think making the rocket non-magical can fix this, either. I can just say I'm taping it to the side of a giant lead disc, or something.

[Anonymous No.14671121]

Duuuude, what if we're living in another universe's big bang? And the heat death of this universe marks the beginning of the lepton epoch of the next universe? Imagine CCC but instead of taking place in sequence, all universes instead take place at once like a time fractal--each universe's entire lifespan is the first 1-10 seconds of the next universe, and said next universe runs that much slower--from our fleetingly miniscule perspective, anyway.
or, get this. what if we're just a bigger universe's quantum foam?
dude weed lmao

[Anonymous No.14671125]

>>14667894
>>14667918
>>14667929
>>14667930
oh i think maybe i confused the Reals for integers.

so are we finding the first integer value of k for which f(x) is smaller when the Real value x is greater than k?

so if k=4, then x could be 4.1 or 4.5 and f(x) would still be bigger at that point?

[Anonymous No.14671130]

Is there such a thing as having the condenser on a rotary evaporator too cold?

[Anonymous No.14671136]

>>14671125
the exact values of k and C dont matter, just that they exist. they dont have to be minimized, they dont have to be integers, you can pick whatever values you want if it makes the inequality easier to prove.

[Anonymous No.14671137]

>>14670894
>>14671116
Never mind, if I'm taping my rocket to a lead disc it seems intuitive that the quantity of non-rotational movement would stay the same.

Is there some nice formula for what fraction of a force is going into shifting the center of mass as opposed to rotation, though?

[Anonymous No.14671176]

>>14671136
but like for 2x^5 + x^3*logx

and 2x^5 + x^3*logx < 3x^5+x^5 = 4x^5. so C=4 and i initially thought k = 0 so x>0, since log(0) is undefined, but i was only thinking of x as an integer.

but k has to be 1 because if x only had to be bigger than 0, then (0.5)^4 is bigger than (0.5)^5, not smaller, so k=1

[Anonymous No.14671178]

>>14671176
oops nevermind i combined two problems into one, sorry

[Anonymous No.14671254]

>>14671176
i meant 3x^5 + (logx)^4

so (logx)^4 <= x^4 when x > 0, since 0 undefined

3x^5 + (logx)^4 <= 3x^5 + x^4

and x^4 is <= x^5 when x >= 1, since if x=0.5 then 0.5^4 is bigger than 0.5^5.

so now k has to be 1

so

3x^5 + (logx)^4 <= 3x^5 + x^4 <= 3x^5 + x^5 = 4x^5. when x >= 1. so C = 4 and k = 1.

[Anonymous No.14671310]

>>14669783
Based if true, Math Wikipedia needs fixing. All the linear algebra and differential geometry pages are mega-convoluted.

[Anonymous No.14671361]

>>14671116
> Is the cross product the cosine of the angle vectors meet at
sine. |a×b| = |a||b|sin(θ). The direction is perpendicular to both vectors. For a torque vector, the direction is the axis of rotation and the magnitude is the magnitude of the torque about that axis.

> Except in that case how is the portion of force that doesn't go into rotation determined?
It doesn't work like that. There is no "apportionment". The instantaneous linear acceleration of the object is the magnitude of the thrust divided by the mass. You can consider this as being the acceleration of the centre of mass or the net acceleration of all particles making up the object. It is unaffected by the position of the thrust.

If the thrust isn't aligned with the centre of mass, the object will experience rotational acceleration. The centre of mass will have the same linear acceleration regardless, points closer to the line of thrust will have greater linear acceleration, points on the opposite side to the CoM will have lesser (or negative) linear acceleration.

[Anonymous No.14671504]

>>14671361
>For a torque vector, the direction is the axis of rotation and the magnitude is the magnitude of the torque about that axis.
Is that the thing where they arbitrarily declared a positive vector means counterclockwise spin around it?

Here's a picture to show what I'm wondering about. Will this bar stay in the same location or shift?

>If the thrust isn't aligned with the centre of mass, the object will experience rotational acceleration. The centre of mass will have the same linear acceleration regardless
To me it sounds like you're saying that if I take a bar into space and whack it with a hammer from a certain direction, my bar will go flying in the same direction and speed independent of where I hit it. It will just be spinning at different rates. But isn't spinning a form of energy storage?

[Anonymous No.14671509]

>>14664222
hahahahaha

[Anonymous No.14671573]

>>14671504
For a right-handed coordinate system: if axb is pointing toward you, b is anticlockwise from a, if axb is pointing away from you, b is clockwise from a. If r is to the right and f is up, r×F points toward you. IOW, the torque is counter-clockwise when the vector points towards you. In 2D, r×F will always be perpendicular to the plane (i.e. x=y=0); positive Z is anticlockwise torque, negative is clockwise.

[Anonymous No.14671583]

>>14671504
>Here's a picture to show what I'm wondering about. Will this bar stay in the same location or shift?

If the forces are symmetric, there will be no acceleration, linear or rotational. If you removed the two right-most forces, there would be a clockwise rotational acceleration but still no linear acceleration.

[Anonymous No.14671676]

>>14661473
Does the X chromosome have more influence on offspring than the Y? My father is extremely smart but my mother is not and I came out unlike my father in terms of intelligence and more like my mother. If I were to copulate with a woman that had higher intelligence than my mother would it reverse regression to the mean?

[Anonymous No.14672305]

>>14661780
>muh golden ratio
You are ngmi

[Anonymous No.14672406]

Does a faraday cage need holes in it? Does it need electric current? Can I build a big box with styrofoam and aluminum foil layers?

[Anonymous No.14672721]

LaTeX newbie here, how do I make a subsubsection NOT break right after jumping a line?

[Anonymous No.14672772]

>>14671254
I'm not going to provide more help unless you learn to TEX, or provide an image with good handwriting

[Anonymous No.14672774]

>>14667698
Yo this kind of helped, I ended up making a different curve but it works out too

[Anonymous No.14672778]

Is the space of real sequences [math] \mathbb{R}^{\omega}[/math] equiped witht the (Tychonoff) product topology a countable union of compact sets? Is it Lindelöf? I assume the former implies the latter. I find this particuler topology hard to work with, can't construct an open covering that could provde a counterexample the open sets are rather counterintuitive and feel unfamiliar,

[Anonymous No.14673156]

If you move toward a target that is 1 light-year away at 99% the speed of light, you will get there in 99% of a year. But time dilation means that, from the perspective of the observer doing the moving, the trip toward the target takes much less than a year. So from their perspective, the star is moving toward them much faster than the speed of light. If there is no preferred perspective, does this not violate the cosmic speed limit?

[Anonymous No.14673270]

>>14673156
>So from their perspective, the star is moving toward them much faster than the speed of light
This is just not true

[Anonymous No.14673305]

>>14673270
but it must be
the trip's time is being reduced by more than 1%

[Anonymous No.14673416]

>>14672778
https://en.m.wikipedia.org/wiki/Tychonoff%27s_theorem

[Anonymous No.14673427]

>>14673156
>>14673305
The distance shrinks at relativistic speeds.

[Anonymous No.14673665]

>>14673416
I am familiar with Tychonoff Theorem plenty but all this proves is that the sequence space is not compact, which isn’t my question.

[Anonymous No.14673716]

>>14673665
is this loss?

[Anonymous No.14673803]

After moving to a new country, my body odor started to smell less strong which is a good thing, but I wonder what's the reason of it. I used to eat more processed food though in my home country, could that be the reason?

[Anonymous No.14674013]

Dear Dr. Anon,

I am trying to learn induction, but early on i am struggling because i can not make LHS = RHS even when i have correct LHS and correct RHS.

It involves some kind of mystical "reverse factoring" which i do not understand.

Please see pic related. Do i need to learn how to do this factoring, or can I just always rely on expanding both the LHS and RHS and seeing if they are equal?

Forever yours,
Anon-chan

[Anonymous No.14674311]

>>14674013
>can I just always rely on expanding both the LHS and RHS and seeing if they are equal?
I'm interpreting this as you expanding both sides out to the same intermediate expression to prove that all three are equivalent. This is fine logically, but it is better form to transform one of the expressions into the other.

As for the factoring, this is some very simple algebra I would highly suggest you try to understand. I can help you make sense of it if you want, but you should give it a good try by yourself first.

[Anonymous No.14674321]

>>14674013
Hello Master Anon,

It's good to hear from you. Middle school algebra can be tricky, so it's good that you're reaching out for help.

[eqn]
\left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^2 (k+1) = (k+1)^2 \left( \frac{k^2}{4} + k + 1 \right) \\
(k+1)^2 \left( \frac{k^2 + 4k + 4}{4} \right) = (k+1)^2 \frac{(k+2)^2}{2^2} = \left( \frac{(k+1)(k+2)}{2} \right)^2
[/eqn]

Hopefully this helps you in your quest to finally earn your GED.

Yours forever,
Me

[Anonymous No.14674649]

>>14672721
What do you mean? And why does your output look so weird (the spaces and the indent)?
Also, using \\ or \newline after begin document is typographically no bueno, so is using so many nested section.

[Anonymous No.14674791]

Apparently, certain models of space result in the cosmological event horizon emitting cosmological Hawking radiation. Wouldn't this imply that the non-observable universe "evaporates"?

[Anonymous No.14674808]

>>14674321
You're literally me.

[Anonymous No.14675005]

>>14673665
You can fill [math]\mathbb{R}^{\omega}[/math] with a countable number of copies of [math][0, 1]^{\omega}[/math], can't you?
All of which are compact.

[Anonymous No.14675190]

This may be completely unrelated to this board, but what the hell happened to "reddit spacing"? I swear it was called out every thread last time I browsed sci but I haven't been on 4channel in like 4 years

[Anonymous No.14675196]

>>14672721
\noindent iirc

[Anonymous No.14675217]

>>14661473
I posted >>14674613 in another thread but got nothing. So here it's again.
Get the Hamiltonian for the inclined plane and rolling disc with no friction in pic related.

I have come up with
[math]\mathcal{L}=\frac{M}{2}\dot{x}^2+\frac{MR^2}{4}\dot{\theta}^2+Mg(x-L)\sin(\alpha)[/math]


which leads me to:
[math]\mathcal{H}=\frac{3}{4M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

or, with another approach, to:
[math]\mathcal{H}=\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

so, which one is correct? I've omitted the details, but I'll post if asked to.

[Anonymous No.14675276]

>>14669311
watch 3blue1brown series on linear algebra, you will understand better what the determinant is

[Anonymous No.14675386]

>>14675217
>Get the Hamiltonian for the inclined plane and rolling disc with no friction in pic related.
With no slipping, you mean?
I don't wanna work through the calculations, were the two methods "reducing the Lagrangian to the coordinates [math]x, \dot{x}[/math] through the no-slipping condition and then calculating the Hamiltonian" and "calculating the multivariable Hamiltonian and then reducing to the coordinates [math]x, p_x[/math]"?
The first one works, the second one is tricky.

[Anonymous No.14675394]

Sup nerds got a real stupid question here. How would I best evaluate the following integral numerically?
[math] \int_0^{\infty} \frac{\sigma(\omega')d\omega'}{\omega'^2-\omega^2} [/math]
[math]\sigma[/math] is just some cross section I have a table for. Would I have to do some residue crap for the pole at [math]\omega[/math]? I'm not sure since apparently this relation came from a contour integral and I'm not sure if I would be going in circles, if that even makes sense. I feel like it shouldn't be that hard, right??

[Anonymous No.14675533]

>>14675386
>With no slipping, you mean?
yep, bad wording, sorry.
I did the first one with two different approaches and got those two different results. That's the problem. And seems I'm brainlet enough to not see what am I doing wrong and not to know how to do the multivariable Hamiltonian thing.

[Anonymous No.14675582]

>>14675533
>I did the first one with two different approaches and got those two different results. That's the problem.
You used [math]p = \dfrac{\partial L}{\partial \dot{q}}[/math] and [math]H = p \dot{q} - L[/math] and what was the other idea?

[Anonymous No.14675602]

Why is the speed of light 299792458 metres per second and not any other number? Something must define the upper limit to the speed of light, so what is it that stops it from going faster?

[Anonymous No.14675605]

>>14675582
Ok here are the details.

1st attempt.
[math]p_\theta=\frac{MR^2}{2}\dot{\theta}=\frac{MR^2}{2}\frac{\dot{x}}{R}[/math]

where

[math]\dot{x}=\frac{p_x}{M}[/math]
and therefore

[math]p_\theta=\frac{R}{2}p_x[/math]

then replacing in the expression for the Hamiltonian it gives:

[math]\mathcal{H}=\frac{3}{4M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

2nd attempt:

Manipulating the Lagrangian by applying $\dot{\theta}=\frac{\dot{x}}{R}$, it results in

[math]\mathcal{L}=\frac{3M}{4}\dot{x}^2+Mg(x-L)\sin(\alpha)[/math]
then
[math]p_x=\frac{\partial\mathcal{L}}{\partial\dot{x}}=\frac{3M}{2}\dot{x}[/math]
[math]\dot{x}=\frac{2}{3M}p_x[/math]

Then, the Hamiltonian will be given by

[math]\mathcal{H}=p_x\cdot\dot{x}-\mathcal{L}[/math]
[math]\mathcal{H}=p_x\cdot\frac{2}{3M}p_x-\frac{3M}{4}\Big(\frac{2}{3M}p_x\Big)^2-Mg(x-L)\sin(\alpha)[/math]
[math]\mathcal{H}=\frac{2}{3M}p_x^2-\frac{3M}{4}\Big(\frac{4}{9M^2}p_x^2\Big)-Mg(x-L)\sin(\alpha)[/math]
[math]\mathcal{H}=\frac{2}{3M}p_x^2-\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

[math]\mathcal{H}=\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

[Anonymous No.14675611]

>>14675605
>>14675582
Ok here are the details.

1st attempt.
[math]p_\theta=\frac{MR^2}{2}\dot{\theta}=\frac{MR^2}{2}\frac{\dot{x}}{R}[/math]

where

[math]\dot{x}=\frac{p_x}{M}[/math]
and therefore

[math]p_\theta=\frac{R}{2}p_x[/math]

then replacing in the expression for the Hamiltonian it gives:

[math]\mathcal{H}=\frac{3}{4M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

2nd attempt:

Manipulating the Lagrangian by applying [math]\dot{\theta}=\frac{\dot{x}}{R}[/math], it results in

[math]\mathcal{L}=\frac{3M}{4}\dot{x}^2+Mg(x-L)\sin(\alpha)[/math]
then
[math]p_x=\frac{\partial\mathcal{L}}{\partial\dot{x}}=\frac{3M}{2}\dot{x}[/math]
[math]\dot{x}=\frac{2}{3M}p_x[/math]

Then, the Hamiltonian will be given by

[math]\mathcal{H}=p_x\cdot\dot{x}-\mathcal{L}[/math]
[math]\mathcal{H}=p_x\cdot\frac{2}{3M}p_x-\frac{3M}{4}\Big(\frac{2}{3M}p_x\Big)^2-Mg(x-L)\sin(\alpha)[/math]
[math]\mathcal{H}=\frac{2}{3M}p_x^2-\frac{3M}{4}\Big(\frac{4}{9M^2}p_x^2\Big)-Mg(x-L)\sin(\alpha)[/math]
[math]\mathcal{H}=\frac{2}{3M}p_x^2-\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

[math]\mathcal{H}=\frac{1}{3M}p_x^2-Mg(x-L)\sin(\alpha)[/math]

[Anonymous No.14675621]

>>14674311
>>14674321
Dear Dr. Anons,

Your messages have received me well, I am much obliged. I thought they were doing that thing where you take things like k^2+2k+1 and turn it into (k+1)^2 which i dont know how to do for things other than this most basic one, especially with exponents higher than 2. I was scared this involved having the foresight to rewrite the fully expanded equation in some highly specific way that allowed it to be magically compacted into the RHS in just a couple steps, and that they were skipping steps.

I don't know why i struggled so much with this even having the answer, I thought I was mostly OK at manipulating equations, but anything with exponents higher than 2 or complex rewritings i dont know how to do.

I also thought i was supposed to cross multiply somewhere or multiply the whole thing by 4 and stuff.

Thank you for your kind assistance.

Thinking of you,
Anon-chan

[Anonymous No.14675625]

>>14675611
>>14675582
where
[math]\mathcal{L}=\frac{M}{2}\dot{x}^2+\frac{MR^2}{4}\dot{\theta}^2+Mg(x-L)\sin(\alpha)[/math]

and

[math]\mathcal{H}=\frac{p_x^2}{2M}+\frac{p_\theta^2}{MR^2}-Mg(x-L)\sin(\alpha)[/math]

[Anonymous No.14675651]

Gotta do good on this so please rate my answer.

a), the absic operation is runningSum > maxSum, input size should be measured by array length.
b) not sure, maybe [math] \sum_{i=0}^{n-1}\sum_{j=i}^{n-1} 1 [/math]? complete guess

[Anonymous No.14675739]

>>14675602
Fundamental constants are just that, fundamental. There is no reason they have the values they do, you can't calculate them from first principles, they are just how the universe is.

However if you believe string theory then the values could be anything but we happen to live in the universe with a particular random configuration of physical constants that we observe.

[Anonymous No.14675798]

>>14675611
>Manipulating the Lagrangian
The very moment you do this you change the definition of [math\p_x[/math] and your entire question of "why am I getting two different values" stops making sense.

Redo the second solution but reread the material on constrained Lagrangians and changes of parameters beforehand.

[Anonymous No.14675813]

Is it safe to keep Ammericium in a small glass bottle like this?

[Anonymous No.14675817]

>>14675798
>The very moment you do this you change the definition of [math]p_x[/math]
are you sure? I've been told the second solution I've got is the correct one. Now I'm more confused.
May I ask you for some reference material? I have some books already, but more won't hurt me.
Anyway, thanks anon.

[Anonymous No.14675906]

If P and Q are convex polygons with n and m points respectively, what is the minimum and maximum number of points that P union Q will have?

[Anonymous No.14675909]

If P and Q are convex polygons with n and m vertices respectively, what is the minimum and maximum number of vertices that convex hull of P union Q will have?

[Anonymous No.14675919]

>>14675909
The smallest should be three, I'm inclined to think the largest is (m+n)-1, but I'm not sure.

[Anonymous No.14675943]

>>14675651
your answer for a seems good as far as I can tell

your answer for b is a good start, but you didn't finish it

[Anonymous No.14675965]

>>14675813
yeah, it emits alpha particles

[Anonymous No.14675970]

Can someone show the math how airplanes use the earth's curvature to their advantage?
In pic related for example, the plane looks like it takes a longer path going way up north rather than the line made in orange. Please no bully, only completed grade 10 math, trying to learn again after 12 years.

[Anonymous No.14675975]

>>14675970
heres some reading material
https://en.wikipedia.org/wiki/Geodesic
https://en.wikipedia.org/wiki/Map_projection

[Anonymous No.14675981]

>>14675919
>>14675909
smallest is 3, largest is m+n? smallest is a triangle, largest would be the union of two half circle-like polygons (which would be m+n)?

[Anonymous No.14675983]

>>14675970
i think they avoid going over the water cause theres nowhere to land

[Anonymous No.14675991]

>>14675975
Thank you anon. Just those visuals helped my brain figure out what's going on.

>>14675983
Often times, flights from NYC to places like India and China, they fly right over the arctic ocean for a while, so I'm not sure if that's the reason.

[Anonymous No.14676073]

>>14675970
>earth
>curvature
you're on the wrong board kiddo

[Anonymous No.14676208]

Anyone know why microscope slide boxes open the wrong way instead of opening like a book which would be more natural?

[Anonymous No.14676582]

>>14676073
>>>/x/

[Anonymous No.14676935]

>>14675602
Also this value is entirely dependent on your choice of units, it could be 1 but those units wouldn’t make a whole lot of sense for trivial matters

[Anonymous No.14677365]

Let’s say Wanda is looking for a universe where her children are real but she’s dead. Now, let’s imagine that there is a 1% chance that her children exist in any universe. Now, she’s a pretty sturdy character, there’s little chance that her children could exist in a universe and she does not so let’s say there’s a one in a million chance that her children exists in a universe where she’s dead.

If let’s give her the benefit of the doubt and say her powers make it so she can scan a universe in 5 seconds to determine whether her children exist in a universe as orphans. How long would it take for her to find that universe with a reasonable chance? Say a reasonable chance is above 60%?

[Anonymous No.14677426]

>>14675970
A "great circle" is a circle formed by intersecting a sphere with a plane which pass through the centre of the sphere. It's the equivalent of a straight line but on the surface of a sphere. The shortest distance between two points on a sphere is always along a great circle. Note that lines of longitude (meridians) are great circles, lines of latitude aren't (their centres are along the axis, not at the centre of the earth).

Imagine you're near the north pole. You want to walk from 0°E,89°N to 180°E,89°N. A straight line on a lat-lon map would have you walk along the 89°N latitude line, i.e. a semicircle. The shortest distance is to walk straight over the pole; on a map, that path is an arc, first going north-east to the pole then south-east to the destination.

If you want to plot the curves, convert the start and and points from spherical coordinates (longitude, latitude, radius) to Cartesian coordinates (X,Y,Z), generate a bunch of samples by linearly interpolating between the end points, then convert the samples back to spherical coordinates (and ignore the radius).

[Anonymous No.14677535]

why do some anons on this board get angry when I tell them that intelligence has a molecular, not computational, basis?

[Anonymous No.14677694]

>>14675217
I'm gonna do the analysis in Example(i) of chapter 2.4 of Scheck's Mechanics verbatim.

We fix the coordinate system [math](x, \theta)[/math]. The resulting force subject to the constraints is [math]K = \left( \dfrac{2 mg\sin \alpha }{3}, - \dfrac{2 mg \sin \alpha }{3R} \right)[/math] (see https://www.youphysics.education/rigid-body/rotational-motion/rotational-motion-problems/rotational-motion-problem-4/).). We choose the generalized coordinates [math]q = x[/math] and compute [math]Q = \dfrac{\partial r}{\partial q} \cdot \left( \dfrac{2 mg \sin \alpha }{3}, - \dfrac{2 mg \sin \alpha }{3R} \right) = \dfrac{4 mg \sin \alpha }{3}[/math]
We choose [math]U(q) = - \dfrac{4 m \sin \alpha q}{3}[/math]. [math]T = \dfrac{m \dot{x}^2 }{2} + \dfrac{ m R^2 \dot{ \theta }^2 }{4} = \dfrac{3m \dot{q}^2 }{4}[/math].
The Lagrangian is [math]L(q, \dot{q}) = \dfrac{3m \dot{q}^2 }{4} + \dfrac{4 mg \sin \alpha q}{3}[/math]. [math]p = \dfrac{\partial L}{\partial \dot{q}} = \dfrac{3m \dot{q} }{2}[/math] and [math]\dot{q} = \dfrac{2p}{3m}[/math]
So [math]H = p \dot{q} - L = \dfrac{2p^2}{3m} - \dfrac{\dot{p}^2 }{3m} + \dfrac{4 mg \sin \alpha q}{3} = \dfrac{p^2}{3M} - \dfrac{4 m \sin \alpha q}{3}[/math]

[Anonymous No.14677737]

>>14677694
>[math]U(q) = - \dfrac{4 m \sin \alpha q}{3}[/math]
Forgot the g:
[math]U(q) = - \dfrac{4 mg \sin \alpha q}{3}[/math]

>[math]H = p \dot{q} - L = \dfrac{2p^2}{3m} - \dfrac{\dot{p}^2 }{3m} + \dfrac{4 mg \sin \alpha q}{3} = \dfrac{p^2}{3M} - \dfrac{4 m \sin \alpha q}{3}[/math]
Forgot the [math]g[/math]
[math]H = p \dot{q} - L = \dfrac{2p^2}{3m} - \dfrac{\dot{p}^2 }{3m} + \dfrac{4 mg \sin \alpha q}{3} = \dfrac{p^2}{3M} - \dfrac{4 mg \sin \alpha q}{3}[/math]

I'm not quite sure if this worked as is or if, since I'm treating [math]\theta[/math] as a regular coordinate, I need to use the regular coordinate energy formula, but whatever.

[Anonymous No.14677896]

>>14675217
Lagrangians can be constrained willy nilly because the action along a random path doesn't change.
Hamiltonians are trickier.
What you do in >>14675611 is first computing the multivariable Hamiltonian (which is why you have both [math]p_{\theta}[/math] and [math]p_x[/math]) and then trying to constrain it. Bad idea in general, you'd need to compute the entire vector field and orthogonally project it onto the constraint submanifold, and then integrate it for the hamiltonian.

[Anonymous No.14677899]

>>14677896
>What you do in >>14675611
*in the first attempt in.

[Anonymous No.14678052]

So does electricity need a closed path to flow or not?
where does the current go in lightning after the ground? does it go back to the clouds? or it stays in the ground? same with electrostatic discharge
also say single phase to ground fault, lets say a tree shorted a line to ground, does the current flow back to the generator? is the generator "grounded" in a literal sense, that is it's connected to earth?
or maybe a closed path isn't required and only a potential deference, or a gradient so to speak?

[Anonymous No.14678059]

>>14678052
>or maybe a closed path isn't required and only a potential deference, or a gradient so to speak?
this.
As long as there a potential difference current will flow. Current doesn't care from where it comes or where it goes. It's just a high concentration of charge (electrons or other charges particles) trying to set an equilibrium with a less concentrated charge distribution.

[Anonymous No.14678074]

>>14677896
>>14677899
are you the same anon at >>14677236 ?
so first solution is the correct one?
to be honest I did this problem a couple years ago and moved on. So I will have to give it another shot.

[Anonymous No.14678083]

Alright all you smart assholes, whats the best fan placement in a standard living room for maximum coolness

[Anonymous No.14678093]

I want to copy the principles of Concentrating Solar-Thermal Power but instead of directing the concentrated sunlight to a tower to simply boil water and turn a turbine I want to direct it towards a focal point on the ground sublimating any material in its path to create a borehole.
Is there any scientific reason this wouldn't work before I start saving up to create it? would the surrounding earth material just act like a giant heat-sink and dissipate the heat before the material under the focal point sublimates?

[Anonymous No.14678115]

>>14678074
>are you the same anon at >>14677236 ?
No.
>so first solution is the correct one?
No lmao, reread my post.

You can check pretty quickly that it's incorrect by using Hamilton's equations.
[math]\dot{x} = \dfrac{ \partial H }{ \partial p } = \dfrac{3}{2} p_x[/math].
But as you've posted in >>14675611 you have that [math]p_x = \dot{x} M[/math], which implies [math]\dot{x} = \dfrac{3M}{2} \dot{x}[/math].

[Anonymous No.14678229]

How do i do this step without expanding it all out?

[Anonymous No.14678237]

>>14678229
or did they just pull 1/4 to the outside of the bracket?

im having trouble with this way of rewriting equations to make them fit what the RHS looks like (induction proofs)

[Anonymous No.14678242]

>>14678229
[math]\dfrac{k}{4} + 1 = \dfrac{k + 4}{4}[/math] and then you pull the [math]4[/math] on the denominator out.

[Anonymous No.14678255]

>>14678059
Thanks, anon
Btw you just saved the janny from my wrath

>>14677529

[Anonymous No.14678300]

>>14678255
np. Just keep in mind charge need a reference to set the potential diff. Just as pic related analogy.

[Anonymous No.14678317]

>>14678300
This is excellent
Very much appreciated

[Anonymous No.14678523]

If the function [math] f [/math] is once continuously differentiably, then given what assumptions is
[eqn]
\int_0^{\infty} f'(x) \sqrt{x} dx
[/eqn]
finite? Is it enough if we have [math] f, f' \in L^2 [/math]?

[Anonymous No.14678553]

>>14678523
how many points do we get for solving your homework problem?

[Anonymous No.14678561]

>>14678553
It's the middle of the fucking summer, I'm neither a student nor a mathematician. I know it's a stupid question for non-brainlets, hence why I posted in this thread

[Anonymous No.14678705]

>>14678242
ty anon.

to prove 6 divides n^3-n with induction (on the k+1 part), i end up with:

k^3 + 3k^2 + 2k

can i add and subtract a k to do
k^3 + 3k^2 + 2k + k - k
and have
(k^3 -k) is divisible by 6 because of inductive hypothesis and

3k^2+3k = 3(k^2+k) is divisible by 6 because 6 is divisible by 3?

[Anonymous No.14678715]

>>14678705
or wait it must not be because 6 is divisible by 3, it must be that k^2 + k >= 6 when k > 0, and if k =0, 0 is divisible by 6 too

[Anonymous No.14678859]

>>14678705
To add another argument not using induction: we have that [math]n^3-n = n(n^2-1) = n(n-1)(n+1)[/math].

Reducing modulo [math]3[/math] gives [math]n(n-1)(n-2) = 0 \ [3][/math].
Thus [math]n^3-n[/math] is zero modulo 3.

The same applies modulo [math]2[/math], the expression becoming [math]n(n-1)^2 = 0\ [2][/math].
As such, [math]n^3-n[/math] is zero modulo 2.

Using the Chinese remainder theorem we get that [math]n^3-n[/math] is zero modulo 6.

[Anonymous No.14679024]

>>14678523
>enough...?
Idk, for instance, what about?
[math]f'=\frac{g(x)}{\sqrt{x}}[/math] with [math]g \in L^2[/math]

[Anonymous No.14679110]

>>14661473
I have to understand the gripes people have with 2nd order logic, and know whether it's criticisms are valid for a programming project I'm working on.

First of all, rumor I hear is that most mathematicians like 1st order and not 2nd order logic. The reasoning is that 1st order logic, although restricted in the kinds of things it can prove, has a complete set of relations that will prove everything provable. Because of Godel's incompleteness theorem, this alone is enough to show that 1st order logic isn't sufficiently advanced to do arithmetic.

What I hear, though, is that 2nd order logic is advanced enough to do arithmetic. The drawback is that you'll never get a complete set of inference rules with it. What does that mean, exactly? Does it mean that, when you use a 2nd order logic inference rule, it's a matter of 'opinion' whether it's valid or not, because there's no solid way to show that it is?

1st order inferences seem to me to be provable using mostly boolean algebra. Is 2nd order logic not the same way? Are the politics around 2nd order logic valid, or are they just politics?


And, further, isn't it true that whenever you use arithmetic, it comes down to 2nd order logic or some subset of it behind the scenes? Why would mathematicians against 2nd order be willing to use arithmetic?

[Anonymous No.14679160]

>>14678523
>Is it enough if
Don't think so.
Can't think of any non-trivial conditions either.
>>14679024
That's not sufficient.
>>14679110
>has a complete set of relations that will prove everything provable
Godel's completeness theorem.
>What does that mean, exactly? Does it mean that, when you use a 2nd order logic inference rule, it's a matter of 'opinion' whether it's valid or not, because there's no solid way to show that it is?
It means that things can be true or false without you being able to prove either way.

[Anonymous No.14679169]

>>14679160
>It means that things can be true or false without you being able to prove either way.
They can also be undecidable btw.

Undecidable isn't dumb stuff like "whether or not Remilia is the cutest 2hu is independent from the Peano axioms", it's stuff specifically in the language of the Peano axioms or whatever axiomatic system you're working with.

[Anonymous No.14679189]

>>14664222
The graph is for seeing how constant the apparent rate constants are. Nonlinear regression is probably better but this one can be done by hand on an exam.

First order integrated rate law can be ln(n0/n) = kt. But conversion x is 1 - (n/n0). So they've assumed it's first order and are plotting kt vs t. So the slopes are k. Ignoring T=140 where they need a sample at 5min (or there's something odd like a new mechanism killing the catalyst), the slopes are higher with higher T leading you to an activation energy. The confusing part is why their reaction goes backwards at the end: somebody decided log(1/(1-0.999999)) = 0 instead of something off the page.

I would answer the questions for 120C by reading k=2/20min off the graph, then 90% conversion is at 10 min* ln(1/(1-0.9)). Full conversion is just lots of 0.99 the exact number depends on instrumentation or economics or toxicology.

[Anonymous No.14679207]

[Anonymous No.14679305]

>>14678859
What does reducing do to turn n(n-1)(n+1) into n(n-1)(n-2)? are you plugging 3 into something?

[Anonymous No.14680303]

>>14679305
When working mod [math]3[/math], remember that [math]1 = 3-2 = -2 \ [3][/math].
When I say "reducing modulo [math]3[/math]", which might not be the proper way of say it, I mean looking at the projection
[math]\mathbf{Z}\rightarrow\mathbf{Z}/3\mathbf{Z}[/math]. This is just a fancy way of saying "Kill every multiples of [math]3[/math] from your expression".

[Anonymous No.14680744]

How come wearing clothes doesn't partially thermally isolate your balls from the environment and prevent your body from thermoregulating them correctly?

[Anonymous No.14680963]

by Gauss's lemma, if f = f(x) is a non-zero polynomial with integer coefficients that's reducible in Q[x], then f is also reducible in Z[x]. but can we say more? can we say that such f can in fact be written as a product of two *non-constant* polynomials in Z[x]?

[Anonymous No.14681025]

How do I find an explanation of what quantum mechanics is that would allow me to program a simulation of anything once I specify the values of the state?

[Anonymous No.14681037]

>>14681025
all you need is the schrodinger wave function.

[Anonymous No.14682200]

If I have a set [math] A[/math] of size [math] 15[/math], and a set [math] B[/math] of size 20, the number of ways of choosing 3 from the former and 2 from the latter is:
[eqn] \mathrm C^{15}_3 \times \mathrm C^{20}_2[/eqn]
or
[eqn] \frac{\mathrm C^{15}_3 \times \mathrm C^{20}_2}{2!}[/eqn]

[Anonymous No.14682282]

>>14682200
the first one

[Anonymous No.14682381]

>>14675943
>your answer for b is a good start, but you didn't finish it
I'm not really sure how to think about, what did I miss?

[Anonymous No.14682424]

Another similar question. In a), I think basic operation is the comparison if low > high, and input size is the length of the array (I feel like I'm missing something cuz this is too easy).

For b), I have no clue how to find the recurrence relation.

[Anonymous No.14682500]

>>14675943
Oh, I see. Was my summation fine? I didn't start the others because I wasn't sure if it was. I'll start thinking about a closed form and complexity, but could probably use a pointer.

[Anonymous No.14682516]

>>14675651
>the absic operation is runningSum > maxSum
I think random array access is the most expensive operation done per loop, but it's all O(1).
>not sure, maybe
Yeah that's pretty much it.
>>14682500
It might be easier for you to reason it out if you think of an explicit counter variable.

maxSum = -inf
operationCount = 0
for i = 0 to A.length - 1
]]runningSum = 0
]]for j = 0 to A.length - 1
]]]]runningSum = runningSum + A[j]
]]]]operationCount += 1
]]]]if runningSum > maxSum
]]]]]]maxSum = runningSum
return maxSum

What's operationCount at the end?

[Anonymous No.14682521]

>>14682516
>inb4 "why ] for indentation"
I'd have used > but then everything would be green.

[Anonymous No.14682566]

>>14682516
>I think random array access is the most expensive operation done per loop, but it's all O(1).
Where is random array access occurring there?

>Yeah that's pretty much it.
Cool, then I need the closed form and complexity. A quick check on wikipedia says the complexity is O(n).

>What's operationCount at the end?
well for each i it must run A[i] - i times right?

[Anonymous No.14682576]

>>14682566
>A[i]
I meant (A.length - 1) - i, I think.

[Anonymous No.14682591]

>>14682566
>Where is random array access occurring there?
A[j]
>A quick check on wikipedia says the complexity is O(n).
It ain't. You can do the problem in O(n) but
>well for each i it must run (A.length - 1) - i times right?
And that is [math]\sum_{i = 0}^{A.length - 1} A.length - 1 - i[/math] in maths, try to simplify it.

[Anonymous No.14682597]

>>14682591
>You can do the problem in O(n) but
But this solution isn't.

[Anonymous No.14682634]

>>14682597
>>14682591
By simplify you mean turn that summation into a closed form? If so, n(n+1)/2?

[Anonymous No.14682637]

>>14682634
which would of course imply the complexity is O(n^2). Does that all sound good?

[Anonymous No.14682648]

>>14682637
It sounds good.

[Anonymous No.14682650]

>>14682648
ty anon

:3

[Anonymous No.14682909]

>>14680963
Reread the definitions. Gauss's lemma is for primitive polynomials, those automatically can't be written as [math]p(x) = a q(x)[/math] with [math]a \notin \{ 1, -1 \}[/math] and [math]q \in \mathbb{Z}[x][/math]

[Anonymous No.14683123]

Is there any downside to HGH?
Does it make your dick bigger if you take it as a child/teen?
There's a soccer player (Lionel Messi) who took HGH as a kid and a teen and he's said to have a large penis.

[Anonymous No.14683304]

>>14680744
same raisin that wearing clothes in general doesnt prevent your body from thermoregulating
also remi has balls confirmed

[Anonymous No.14683390]

Can anyone explain tensors?

[Anonymous No.14683454]

>>14683390
Tensor is, roughly, a meme term that means "vector thingeridoo with a transformation rulerino."

[Anonymous No.14683739]

>>14682424
can someone coach my brain through this one too? I see the 'low > high' and 'low = high' are the operations that's being done on each call but no idea how to go from there setting up recurrance relation

[Anonymous No.14683839]

>>14682424
i not sure what "basic operation" means in this context, if you can show a definition from your lecture notes or something that would be helpful
b is a little tricky. the general idea behind the algorithm is that it keeps cutting the array in half until theres only one element. the function only returns a (real) value when low = high and it returns the value located at that index. lets say there are 16 elements in the array. this means that the function was called 16 times to count all the elements, but also 8 times from when it had to split the arrays of size 2, 4 times when it split the arrays of size 4, 2 times to split the arrays of size 8, and 1 more time right at the beginning, when it split the original array. so 31 times total, but the exact number isnt important, we just want to know how many MORE function calls there will be if we add one more element. if we add another element, then one of the arrays of 2 at the end will become an array of 3. it takes 3 function calls to count an array of 2 (split, count, count) but 5 calls to count an array of 3 (split, count, split, count, count). so theres two more steps. the recurrence is [math]a(n+1)=a(n) + 2[/math], which is [math]2n \in O(n)[/math].
you could also think of it like a binary tree, which obviously has roughly [math]2n[/math] nodes, where [math]n[/math] is the number of leaf nodes.
that was probably pretty confusing, let me know if you dont understand.

[Anonymous No.14683868]

>>14683839
thank you so much anon

[Anonymous No.14683881]

What will happen if we get a really bad geomagnetic storm like the Carrington Event again? Will our infrastructure be destroyed or will we be able to recover quickly? What's the plan?

[Anonymous No.14683884]

>>14683839
>The operation contributing most to the running time of the algorithm
What would it be for these two?:
>>14675651
>>14682424

Another anon said in the first one, 'random indexing' would be the most expnsive operation. I thought it was just the comparison of runningSum > maxSum. And what would it be in the second one? doing the line with mid = floor(high+low / 2)?

[Anonymous No.14683894]

>>14683454
No other replies, has this been considered an accepted answer?

[Anonymous No.14683953]

>>14683884
>>The operation contributing most to the running time of the algorithm
i thought that was probably what it was
>>14675651
lines 5 and 6 are called the most and equally frequently
>>14682424
im not 100% sure about this one. you could argue that lines 6 and 7 where it calls itself are the most computationally arduous, but that might not be allowed as an answer. if you "unravel" a single call to the function, line 1 gets called the most.
also, in C, line 5 is just mid = (high + low) >> 1, which doesnt have anything to do with the question but i thought it was neat

[Anonymous No.14683974]

>>14683894
in math, the question you ask shouldnt be "what is this thing?" but rather "what can i do with this thing?". if you look up the definition of a vector, for example, its something like "anything that adds and multiplies like a vector". so if youre asking what a tensor is, the most appropriate answer is "something that behaves like a tensor".
https://mathworld.wolfram.com/Vector.html
https://mathworld.wolfram.com/Tensor.html

[Anonymous No.14684011]

>>14683953
>im not 100% sure about this one. you could argue that lines 6 and 7 where it calls itself are the most computationally arduous, but that might not be allowed as an answer. if you "unravel" a single call to the function, line 1 gets called the most.
my gut is telling me it's line 5, is that wrong?

[Anonymous No.14684079]

>>14683974
>something that behaves like a tensor
what does a tensor behave like?

[Anonymous No.14684504]

>>14684079
A tensor is something that transforms in a particular way under coordinate change. It's that property that defines them.

[Anonymous No.14684605]

I have no idea what a field in physics is and at this point I'm too afraid to ask.

[Anonymous No.14684689]

>>14684605
A field is a ring in which every non-zero element has a multiplicative inverse.

[Anonymous No.14684763]

>>14684689
I mean in physics. Unless they use the same thing.
ie in Classical Field Theory or Quantum Field Theory.

[Anonymous No.14684807]

What this is

[Anonymous No.14684828]

>>14684605
"Something that has infinite degrees of freedom" is the best definition I can give you.

[Anonymous No.14685021]

I have been self-studying probability. Can someone tell me how to solve pic related?

For the first part I have been considering the two-state system (Roll is a 6, Roll is not a six) to get the probability that the nth score is 6 as
[eqn]p(n) = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ \frac{1}{5} & \frac{4}{5} \end{pmatrix}^{n-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix}
[/eqn]
for [math]n \geq 2[/math]. The Eigenvalues of the transiston matrix are 1 (since it's a stochastic matrix) and -1/5 (which is the diagonal minus 1). So the solution can be written in the shape
[eqn]p(n) = a + b \left( \frac{-1}{5} \right) ^n [/eqn]
Plugging in [math]p(1) = 1[/math] and [math]p(2) = 0[/math] gives
[eqn]p(n) = \frac{1}{6} - \frac{25}{6} \left( \frac{-1}{5} \right) ^n [/eqn]

The probability that the nth score is 1 can be gotten from that by using symmetry.
[eqn] \frac{1 - p(n)}{5} = \frac{1}{6} + \frac{5}{6} \left( \frac{-1}{5} \right) ^n [/eqn]

Now to the second half of this problem. If the result of the first die is [math]X_n [/math] and the second die is [math]Y_n[/math] then I got the relation that [math]Y_n[/math] has the same distribution as [math]X_n + n - 1[/math] if you consider both modulo 6.
So
[eqn]P(Y_n = 6) = P(X_n + n - 1 \equiv 6 \text{ (mod 6)})) = P(X_n \equiv 1 + 5n \text{ (mod 6)})) = \begin{cases} \frac{1}{6} - \frac{25}{6} \left( \frac{-1}{5} \right) ^n & \text{if } n \equiv 1 \text{ (mod 6)} \\
\frac{1}{6} + \frac{5}{6} \left( \frac{-1}{5} \right)^n & \text{else} \end{cases}
[/eqn]

Part 1/2

[Anonymous No.14685028]

>>14685021
On the other hand we should be able to calculate it as
[eqn]P(Y_n = 6) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \\ \frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \\ \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5 } \\ \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} \\ \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \end{pmatrix}^{n-1} \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
[/eqn]

Which gives different values according to Wolfram Alpha. So where is the error?

[Anonymous No.14685134]

>>14681037
It's not self explanatory enough for me to get everything

[Anonymous No.14685148]

>>14684807
leftovers from the entry rocket, probably the parachute

[Anonymous No.14685221]

>>14684504
What's the particular way?

[Anonymous No.14685242]

How do I solve this?
>doesn't give a single problem or example on unequal probabilities
>gives an exercise on it in the end

[Anonymous No.14685273]

>>14685242
Nevermind it is [math] 1 - e_k \vec p[/math], I guess.

[Anonymous No.14685425]

Where to learn (relearn) physics till high school (till 10th-11th grades) besides Khan Academy?

[Anonymous No.14685434]

what are the ramifications of https://hal.inria.fr/inria-00429965/document

[Anonymous No.14685511]

>>14685425
I like HC Verma. It is Indian but extremely comprehensive and serves a nice primer before undergrad. It has very good problems and fair amount of math.

[Anonymous No.14685555]

if metal can't reconnect after being cut due to oxidization and vice versa in vacuum... why gold bars don't fuse to each other since gold doesn't oxidize?

[Anonymous No.14685668]

>>14685555
iirc they do, you just have to hammer them together a bit

[Anonymous No.14685859]

>>14685668
interesting, Imma buy some and try it out myself when the price becomes less bullshit.

[Anonymous No.14685918]

>>14661473
Can someone explain the intuition behind Biot-Savart's law?
[math]dB = \dfrac{\mu_0}{4 \pi}\dfrac{I d\vec{l} \times \hat{r}}{|\vec{r}|^2}[/math]
I understand the constant part, but the part that irks me is the cross product.

Do we multiply the current with an infinitesimally small length dl (of the conductor) to scale down the current to fit a current within that small length? And why do we take the cross product of the unit vector r towards our point of interest? Is that because the magnetic field acting onto that point is only in the j hat direction?

[Anonymous No.14685957]

>>14685918
>Do we multiply the current with an infinitesimally small length dl (of the conductor) to scale down the current to fit a current within that small length?
yes. just a 'element of current'

>And why do we take the cross product of the unit vector r towards our point of interest?
magnetic field is perpendicular simultaneously to both the current flow 'line' and the radius to the point of interest.

[Anonymous No.14686237]

>>14685957
Man, I'm such a tard, never really understood the cross product but now I do. Does anyone else know why those two angles alpha are the same?
https://youtu.be/kb97TivvK50
What rule is behind it?

[Anonymous No.14686243]

>>14686237
See 31:28 ffs, didn't get the timestamp.

[Anonymous No.14686290]

>>14685957
And thanks by the way. I'm a faggot for not thanking you first anon. Have a good one.

[Anonymous No.14686296]

>>14686237
>>14686290
np anon, sometimes in case of doubt look for the very basics of what you're learning and you'll get some details you may forget with time.

[Anonymous No.14686342]

Why do power plants generate their electricity at million of volts if they have to step that power down immediately to transmission levels?

[Anonymous No.14686353]

>>14686342
>they have to step that power down
the what? they step it up afaik

[Anonymous No.14686355]

>>14686342
>>14686353
forgot pic related

[Anonymous No.14686369]

>>14686353
maybe i only mean the nuclear ones

[Anonymous No.14686392]

>>14686342
P = I * U
By keeping the voltage up you can transport the same amount of energy with a lower current.
The energy loss due to heat is proportional to the square of the current so you really want a low current.

[Anonymous No.14686399]

>>14686392
if i want a high voltage why do they step the voltage down

[Anonymous No.14686403]

>>14686399
Because touching a high voltage line immediately kills you.

[Anonymous No.14686620]

>>14686403
Strictly speaking, can't that also happen at a low voltage?

[Anonymous No.14686817]

>>14686620
It is a high current that fucks you up. If we assume we have an ohmic resistance (the human body has a pretty uniform resistance), then I = U/R. Increase the voltage and usually a higher current comes with it.

https://imperiaep.com/why-is-voltage-stepped-up-only-to-be-stepped-down/

Also an interesting read. I'm not that anon, just thought I'd share.

[Anonymous No.14687047]

What's going on in this image? These two specific galaxies? pairs of galaxies? look identical, but I don't see anything that looks like gravitational lensing. Is it just a coincidence?

[Anonymous No.14687050]

I've seen videos of people melting rocks and metal by using sunlight and a fresnel lens like this guy https://www.youtube.com/watch?v=ynUA8dY1btU
That's all done with a relatively small area to collect and concentrate the sunlight, if this were scaled up could you use it to replace an industrial forge, conventional digging tool, or even a steam boiler?

[Anonymous No.14687095]

So basically, thanks to time dilation and space contraction, for you as an observer in non inertial frame of reference, you can reach any point in space arbitrarily fast?

[Anonymous No.14687241]

>>14687095
yeah basically, as long as you can actually sustain that kind of propulsion.

[Anonymous No.14687249]

>>14687050
i assume the problem with scaling that up is space.
>>14687095
from your perspective yes. there are two problems aside from everyone you know being dead, however - hitting even extremely small, stationary things at that speed could make a serious dent in your craft, and don't forget before reaching your destination, you should slow down! and don't expect any sensors you have to work fast enough to compensate for tiny particles of gas out in space. the light they emit for detection may appear to travel away from your sensor at c even as you get close to c, but that doesn't stop the 20 ms delay from contact to detection to report, and in that time you've smashed into your destination (if you weren't a charred collection of scrap at that point)

[Anonymous No.14687339]

>>14684828
rip me I guess

[Anonymous No.14687374]

>>14686342
They don't. Nothing generates power at millions of volts. It's stepped up to a few hundred kV for transmission to reduce losses (higher voltage means less current which means thinner wires and less power lost in transmission).

[Anonymous No.14687501]

Does power factor correction essentially a resonant circuit? Say if you are lagging due to inductive load and add capacitors to correct the PF and if you get 1, haven't you just created resonance at the mains frequency?

[Anonymous No.14687654]

>>14687501
Yes. The point of PFC is to make the impedance purely real, which is what you get when an RLC circuit is at its resonant frequency. Usually, the load is inductive (motors), so you add a bank of capacitors to counteract.

[Anonymous No.14687673]

>>14687654
But isn't resonance bad in power lines as it could lead to current or voltage spikes? Even with a modest Q, 120v can become 3KV. I guess the series resonance is a voltage spike, but the parallel resonance is a current spike? Can't capacitive load resonate with the secondary winding's inductance and cause a huge current spike?

[Anonymous No.14687683]

How do these?

[Anonymous No.14687738]

>>14687683
the first one is [math]P(B_k)=P(C_k)(1-P(C_{k+1}))[/math], obviously.
a) could be rolling two different colored dice, not sure wtf Pebble World is.
b) is only independent when the area of the overlap of the rectangles is equal to the product of the two areas. this follows from the formula for independence it gives there.

[Anonymous No.14687768]

>>14687673
Not sure where you're getting this. Transformers don't behave like inductors unless you disconnect the other winding. Adding capacitance will tend to reduce voltage spikes. Current spikes don't really matter if they're transient, and the breakers will deal with larger spikes.

[Anonymous No.14687786]

>>14661473
Can science make my penis bigger and if not what good is it?

[Anonymous No.14687807]

I'm coming up quick on the ceiling of topics that I can learn with just a basic EE bachelor's worth of math education.

I need to teach myself proof-based math and analysis, especially since I want to learn about control theory, dynamical systems, chaos theory, cellular automata, optimization, variational calculus, and the like. Namely, I want to learn how to use them as tools to deriving new stuff. So, a really good working knowledge of all of them.

the problem is just can't fucking stand how pretentious proof-based analysis books are. All of the ones I've tried (and since forgotten) feel like they're just the authors jerking themselves off over how clever they are and how much obtuse, obfuscating terminology they can toss around.

Are there any proof-based analysis books that read like Strogatz's "Nonlinear Dynamics"? And, given the stuff I mentioned above, any other prerequisites I should bother learning about?

[Anonymous No.14687817]

>>14687786
Oh sure, you can turn me into an ugly woman but you can't make my penis bigger.

This is all the work of 'Big-Penis'

[Anonymous No.14687822]

>>14687817
Isn't it more cost effective to give me a big penis than turn me into a woman?

how long are we going to be beholden to 'Big-Penis'? Make with the penis growing technology already!

Also I want to drink soda and eat fried foods and not be fat. You suck science.

[Anonymous No.14688055]

>>14687738
Thanks.

[Anonymous No.14688440]

>>14687822
It's more efficient to just make all the holes smaller.

[Anonymous No.14688467]

>>14661473
Does anyone have sources on why those two 48 degree angles are equal?

[Anonymous No.14688470]

>>14688467
And is CED 46 degrees?

[Anonymous No.14688529]

Assuming 1000 coin tosses one after the other, what are the odds there are 7 tails in a row at any point?

[Anonymous No.14688719]

With n coin tosses the probability to get at lest k tails in a row at any point is
[eqn]\sum_{m=1}^{\lfloor \frac{n+1}{k+1} \rfloor} (-1)^{m+1}\Big[\frac{\binom{n-mk}{m}}{2^{m(k+1)}} + \frac{\binom{n-mk}{m-1}}{2^{m(k+1)-1}}\Big]
[/eqn]
Now plug in n = 1000 and k = 7 to get
[eqn]\frac{10519892902425252870246235126253986112794208512293257945206769575762018170169574684772197791099703041188563914193837170249809576996543186884653287255308936929641993587049326099743749256513492209486828040721912560021259617421487931163288807562360629002725162826529413195945445992793453139202566815297787}{10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376} \approx 98 \%[/eqn]

[Anonymous No.14688889]

>>14688719
Thanks, is this for at least 7 in a row? Is there a way to get the probability of exactly 7 or any other number in a row?

[Anonymous No.14688911]

>>14688467
>>14688470
Yeah there was a paper published, but it was on vixra

[Anonymous No.14689046]

heaters are "simple" because converting electrical energy into thermal is "simple
coolers (compressors in AC, fridge, freezer, etc.) are "complicated" because the "only way" to cool something via electricity is to extract heat and move it somewhere else (eg outside, back of the fridge, etc.)

why are cooling methods "complicated"? why is there not an electrical method that would extract heat (eg from air) and convert it into some other type of energy? and would I earn gorillion dollars if I discovered such a method?

[Anonymous No.14689102]

>>14689046
You can convert heat energy into electrical energy without issues as long as you have some heat differential somewhere.
In fact Coal, Gas and Nuclear Power Plants all work by first generating heat and then converting that heat into electicity.
Geothermal Power Plants convert heat directly.

[Anonymous No.14689110]

>>14689102
god dammit anon I knew that. how do I convert 40 celsius from my room via coal into electricity without the need of a window unit?

[Anonymous No.14689376]

>>14689046
>why is there not an electrical method that would extract heat (eg from air) and convert it into some other type of energy?
the second law of thermo states that you cant convert homogeneous heat into other forms of energy. sowwy anon.

[Anonymous No.14689710]

>>14688467
> Does anyone have sources on why those two 48 degree angles are equal?
Because the diagram says they are.

From the information given on the diagram, you can conclude that the triangles are congruent. Which means that AB and CD are parallel. You can use the law of sines to determine the length of any edge.

[Anonymous No.14689732]

>>14689046
It's not clear what you're asking. Thermoelectric (Peltier) coolers move heat without any moving parts. Similarly, thermocouples generate energy from a heat differential.

You can extract energy from a heat differential, or can use energy to create a heat differential. You can't extract energy from a uniform temperature, nor can you simply "delete" heat energy.

[Anonymous No.14691249]

New
>>14689399
>>14689399
>>14689399