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/sci/ - Science & Math

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11644245 No.11644245 [Reply] [Original] [archived.moe]

Previously >>11628445

>what is /sqt/ for
Questions regarding math and science, plus appropriate advice requests.
>where do I go for other SFW questions and requests?
>>>/wsr/ , >>>/g/sqt , >>>/diy/sqt , >>>/adv/ , etc.
libgen.is (warn me if the link breaks)
sci-hub (you'll have to google for a link, unfortunately)
>book recs?
>how do I post math symbols?
>a google search didn't return anything, is there anything else I should try before asking the question here?
>where do I look up if the question has already been asked here?
>how do I optimize an image losslessly?

Question asking tips and tricks:
>attach an image
>look up the Tex guide beforehand
>if you've made a mistake that doesn't actually affect the question, don't reply to yourself correcting it. Anons looking for people to help usually assume that questions with replies have already been answered, more so if it has two or three replies
>ask anonymously
>check the Latex with the Tex button on the posting box
>if someone replies to your question with a shitpost, ignore it

Good charts: https://imgur.com/a/kAiPAJx
Shitty charts: https://imgur.com/a/1Q1LIMk (Post any that I've missed.)
Verbitsky: https://imgur.com/a/QgEw4XN
Graphing: https://www.desmos.com/
Calc solver: https://www.wolframalpha.com/
Tables, properties, material selection:

>> No.11644347

Don't really need a proof, just curious. If [math] \frac{A}{B}\cong\frac{C}{B}[/math], does that imply that [math]A\cong C[/math]?

>> No.11644405
File: 42 KB, 1576x953, 1561550112606.png [View same] [iqdb] [saucenao] [google] [report]

what's the solution of 1 or 5?

>> No.11644419
File: 913 KB, 1412x1000, __patchouli_knowledge_rumia_and_koakuma_touhou_drawn_by_arnest__48ed0149560d45a775edd66e5d9b7eda.jpg [View same] [iqdb] [saucenao] [google] [report]

Has Yukariposter disappeared for longer than he usually does or is the quarantine goofing my sense of time?

Math questions:
>>11628616 , shoddily answered by >>11628711 and then continued by >>11628727

Probability and statistics:

Physics questions:

Chemistry questions:

Biology questions:

Advice requests:

/g/ questions:
>>11642896 (Most likely misclassified)

Stupid questions:

Selected posts that got no replies.

>> No.11644427

Skipped >>11644060 on accident.

>> No.11644594

>anon quoted my question before I reposted it

>> No.11644952

Somewhat stupid /sci/ newfag question:

How can I calculate pressure resistance on a hollow tube, and what data will I need? If I can use metric for area, that would be great, but I can totally work with Imperial too.

>> No.11645068
File: 12 KB, 764x138, AAAAAAAAAAAAAAAAAAA.png [View same] [iqdb] [saucenao] [google] [report]


>> No.11645120
File: 34 KB, 771x229, question.png [View same] [iqdb] [saucenao] [google] [report]

Can someone tell me if I'm correct in my understanding of how this person went from step (2) to (3)?
He essentially cancelled out the first two terms of the series expansion of cos(x) and then made a new series representation of where it was cut off?

>> No.11645125

Wouldn't it be [math] \sum_{n=0}^\infty\frac{(-11x)^n}{n!}[/math]?

>> No.11645131


>> No.11645135

Are there any postive whole number solutions for [math] n [/math] and [math] t [/math] for the equation:

[math] (2t-n)^2 = \frac{4}{3} (t^2+t) [/math]

[math] t [/math] can also have half integer solutions. I've only found [math] t = 1/2 [/math] and [math] n=0 [/math].

Also [math] n \in [0,4t] [/math]

>> No.11645145

Oh, there's also [math] n = 2 [/math] and [math] t = 1/2 [/math].

Are there any others?

>> No.11645153


>> No.11645158

is there any scientifically best way to study?

>> No.11645172


>> No.11645188

Yes but it might be difficult. You will probably have to do extra physics courses (grad schools typically outline the ones they expect incoming students to have), otherwise just do some research (preferably in physics department, EE should be fine), and do well on the physics GRE
It is a good field if you want to go into industry, even most phds i know will pick industry over academia for photonics. Masters degree should be enough if you just want a job though. Try to get into UArizona, Rochester, or UCF creol as they have great optics/photonics programs, otherwise you could try some with great EE programs in general like CUboulder, stanford, etc

>> No.11645194


>> No.11645206

cool cool ty

>> No.11645207
File: 197 KB, 595x437, 20200308_215340.jpg [View same] [iqdb] [saucenao] [google] [report]

What do you mean by pressure resistance? Do you mean like pressure gradient in a pipe? Then you use something called poiseuille's law if its laminar flow or else you rely on something called the Darcy-Weibach relation.
If you are talking about the hoop stresses in the metal of the pipe itself, then you can just look up the equation. Its like σ=gauge pressure×radius/thickness, or something like that.

>> No.11645489
File: 27 KB, 700x467, 1583878742628.png [View same] [iqdb] [saucenao] [google] [report]



>> No.11645962 [DELETED] 

Are there any ways to develop biological weapons to target a specific geographic region? What about simply capturing an air sample from that region and hoping there is something in it the biological agent can target? Would this work?

>> No.11645966


>> No.11645990

hello i am just a visitor to this board but i wanted to ask where i can find a decent free online IQ test, because i am worried i am pretty terribly slow, especially when it comes to grasping many new ideas or tasks. if anyone could help i would really appreciate it

>> No.11646000

Why do some compounds and elements have specific attributes? For example nitrogen and sulfur having a strong odor in certain compounds, carbohydrates being sweet, aromatic compounds being fragrant etc.?

>> No.11646005

fuck offfffffffffffffffff

>> No.11646026

nitrogen and sulfur are common products of decomposition, meaning they are a good sign of death/decay
Esters and aldehydes (aromatic compounds) go rancid easily, meaning they are a decent indicator of freshness
Carbs are high energy as fuck
>it's evolution, baby

>> No.11646091

If most of the volume of our body contains nothing, then why can we see our body/not see through it?

>> No.11646093

because light interacts with electrons, literally "quantum effects"

>> No.11646117

So im a hobbyist meddling with 3D graphics and while learning about rigging models I ran into quaternions like in a brick wall.
Ive watched the videos on youtube probably wasted a whole day on them and I still dont get it.
I understand the analogies but it doesnt click.

Now from experience I know that my entry point for learning this thing is bad so I come here asking is there a book that will show me how quaternions work,for computers at least?

>> No.11646145

Break it down into parts. What part about quaternions confuses you.
I think by now you must understand how to add, subtract, multiply them.

>> No.11646283
File: 154 KB, 1360x558, computer vision.jpg [View same] [iqdb] [saucenao] [google] [report]

Is digital image processing a good field to get into?

>> No.11646318
File: 315 KB, 920x518, fucking triangles.png [View same] [iqdb] [saucenao] [google] [report]

If A and D are each 88, then that means the red angle has to be 4, and the corresponding angle inside of B's triangle is 176. So how is B 64 degrees?

>> No.11646485

Can anyone explain why I hallucinate when I state at the mirror,my own reflection?
It's fun but I want to know WHY it happens

>> No.11646488

how do I know reality is real? the only argument I could think of is it "feels" real but this can also be said for dreams or hallucinations

what're some common arguments for us beingi n reality?

>> No.11646505

If I have a complex valued matrix (NxN), is there a way for me to approximate it with a sparser matrix by running PCA or something on it, taking the top n (n < N) principal components and zero padding?
Is there an easy way to do it in Python?

>> No.11646533
File: 315 KB, 920x518, 1588854375235.png [View same] [iqdb] [saucenao] [google] [report]

[math]y = -11x[/math], [math]e^y = \sum _{n=0}^{\infty} \frac{y^n}{n!}[/math] is the expansion at [math]x= y = 0[/math], and then you swap in the definition of y.
Second question in Advice.
>why not physics?
Basically, pic related tells you that [math]a_i = 180 - (b_i+b_{i+2})[/math], because of the angles of the triangle sum to 180 thing, where the indices are taken mod 5.
This lets you compute the sum of the [math]b_i[/math] in terms of the [math]a_i[/math], which is a classical result.
Then the rest is obvious.

>> No.11646647

t=0, n=0
t=1/2, n=0 or 2
t=3, n=2 or 10
If t can halve half-integer solutions, substitute t=s/2 where s is integer. You end up with 2s^2-6ns-2s+3n^2=0. You can use http://alpertron.com.ar/QUAD.HTM to find the general solution, which in this case starts with either s=0,n=0 or s=-2,n=-2, then any given solution can be iterated with either n'=5n-2s+2,s'=3n-s+1 or n'=-n+2s,s'=-3n+5s+1 to get additional solutions.

>> No.11646663
File: 142 KB, 920x518, triangles man.png [View same] [iqdb] [saucenao] [google] [report]

This is what's confusing me, though.
We know the angles of A and D, so we can find the angle of what I labeled as X (your a1).
Knowing X lets us find Y. It's a straight line so it adds up to 180.

The thing is, this means one of the angles of B's triangle is 176, meaning B plus the third angle have to add up to 4. But the answer's 64, so something's wrong with what I'm doing.

>> No.11646715

PCA-like operations will give you a lower-rank approximation, which isn't necessarily sparse with respect to the standard basis. If what you want is a matrix with fewer nonzero elements, why not just pick a threshold, and replace all elements with magnitude less than the threshold by zero?

>> No.11646741

>something's wrong with what I'm doing.
If you've made a mistake then I can't find it tbqh.
I'll try to see if I can find a website to actually construct the pentagram.

>> No.11646752

The problem is that you are inserting "each" into the expression. It's probably a bad translation from Japanese, but it all works if you take the phrase "angles A and D are x" as "the sum of angles A and D are x".

>> No.11646806

Yeah, that solves it. If you interpret "angles a and d are 88 degrees" as "a + d = 88" then everything checks out.

>> No.11646807

Why wouldn't a magnetic belt/chain running through an inductor work as a generator?

>> No.11646852

Oh that's a better idea, thanks!
And I know PCA gives a lower rank matrix, I was wondering whether I could take taht matrix, and zero-pad it till it's back to the old rank

>> No.11646907

does the z-transform of a signal always return a complex result?

>> No.11647051

Theoretically, if the coronavirus were developed in a lab, how long would that have taken? A couple weeks, months, years?

>> No.11647078

The Z transform of a sequence gives a function C->C. For a real sequence the coefficients in the Z transform are real, so applying the function to a real argument yields a real value. But it's more meaningful to consider arguments which are unit phasors, z=e^jω, as that gives the discrete-time Fourier transform.

>> No.11647121

Depends upon what sort of "development" you're talking about. Just taking it from wild animal populations and cultivating it? Selective breeding? Genetic modification? Basically, it would depend entirely upon the gap between what you start with and what you want.

If you just want to start an outbreak of some random coronavirus, you don't need a lab, just a country with minimal regulation of the meat trade and a population that will eat practically anything

>> No.11647128

Magnetic energy isn't static. A permanent magnetic field cannot store energy; the magnetic flux from the incoming belt would equalize the flux leaving the centroid of the inductor, thus you would get equal power.
An electro-magnet would require more power to overcome the resistance on its coil than what would be generated on the inductor.
This is why when you see youtube videos of people making 'perpetual motion' machines from "Back-EMF", it is a lie. There's an electric generator running it behind the scenes at a deficit of energy.

>> No.11647139

Does anyone know what the cloning laws are in China?
I was stoned with a friend and we were trying to imagine the craziest, most disturbing genetic experiments that could be going on in a lab somewhere.
I think you can only keep a synthetic zygote alive for 14 days in the US but I know China very relaxed about stem cell research. Can anyone give me any info?

>> No.11647188

This question is NOT stupid,but I can't post new thread because low memory

Are Sean carrolsls argument,and other studies,experiments and data, pointing to "no afterlife" correct and legit?
Is the Omega point a valid argument for human resurrection?

>> No.11647209

Am I understanding the photoelectric effect correctly...
Photons with higher frequencies have a higher chance of colliding with a particle in its path because they cover a lot more 3D space as they move
Is it really that simple, or is there quantum weirdness going on as well

>> No.11647232

Dunno what that other guy is talking about, but yes, basically. The only problem is that usually magnets arent passed through the solenoid, but over them.

>> No.11647303 [DELETED] 
File: 13 KB, 1834x124, parametric.png [View same] [iqdb] [saucenao] [google] [report]

Can anyone help me with this?
I've tried putting them together like cos^2(2t)+sin^2(2t) = 1 but that didn't work either. I've checked the solution manual and they didn't go over this problem.

>> No.11647326

How much math do I need to tackle Goldstein's classical mechanics book and Jackson's EM book?

I know analysis, algebra, ODEs (kinda), is that enough?

>> No.11647338
File: 19 KB, 948x142, parametric.png [View same] [iqdb] [saucenao] [google] [report]

Posted the wrong pic and I didn't even realize it.
Does anyone know how I can tackle a question like this? I don't know how deal with the function inside the sin/cos.

>> No.11647346

it's just a phase shift

>> No.11647352
File: 352 KB, 689x1020, __kawashiro_nitori_and_kagiyama_hina_touhou_drawn_by_mozukuzu_manukedori__fc9f8680855ad3e72d27afb6fdb7fbfa.png [View same] [iqdb] [saucenao] [google] [report]

Put [math]h(t) = \frac{7 \pi}{4} - t[/math].
Then, [math]\cos^2 h + \sin^2 h =1[/math].

>> No.11647356

What does phase shift mean?
Ooooh, got it.

Thank you both!

>> No.11647366

Make sure to double check the domain parts.
That is, [math]0 \leq t \leq \frac{ \pi}{2}[/math] implies [math]a \leq h(t) \leq b[/math] for some [math]a[/math] and [math]b[/math].

>> No.11647383

It's a circular arc, radius 1, moving clockwise from 7π/4 (SE) to 3π/4 (SW).

>> No.11647388

>[math]3 \pi /4[/math]
[math]7 \pi /4 - \pi /2 = 7 \pi /4 - 2 \pi /4 = 5 \pi /4[/math]

>> No.11647394

Does the domain change if I resubstitute it back? Thank you for the reminder.
Ohhh, thank you! I need to practice more with these!

>> No.11647480
File: 15 KB, 872x385, 2020-05-07 21.34.33.png [View same] [iqdb] [saucenao] [google] [report]

"what does this graph remind you of?"
send help

>> No.11647550

a wave

>> No.11647582

a trigonometric function

>> No.11647593

thanks, made me realize it's a fucked up sin/cos and found the error in my program

>> No.11647806
File: 1.84 MB, 2304x4608, IMG_20200507_141325.jpg [View same] [iqdb] [saucenao] [google] [report]

Did I fuck up in rearranging this equations? Not getting the value I'd expect

>> No.11647859

nope, it's fine

>> No.11647949
File: 28 KB, 707x786, any idea why i got this wrong.png [View same] [iqdb] [saucenao] [google] [report]

Any idea why I got this wrong?
I've tried replacing (t) with (y) as well but that didn't work either.
Sorry if my writing is illegible, I tried my best to write with a mouse. If there's something that you can't read please tell me!

>> No.11648026

Wait nevermind I'm fucking retarded

>> No.11648165

>Given a set A, what is the probability that a relation R: A A is reflexive?
[math]\frac{2^{|A|}}{2^{|AxA|}} = \frac{2^{|A|}}{2^{A^2}}[/math]
Reason being is that the sample space is 2^|AxA|, and there are |A| such reflexive elements in the relational set on A. I don't really see the intuiton behind forming the rest of the numerator though, I undertand there are 2^|A| ways to choose whether to include that element or not, but I don't see why this is our resultant numerator.

>B. Given a set A, what is the probability that a relation R: A A is symmetric?
Similar reasoning as above yields:

>C. Given a set A, what is the probability that a relation R: A A is reflexive and symmetric?
So for this we just take the product, which I give the result of below:
[math]\frac{2^{|A|} \cdot 2^{2|A|}}{(2^{2|A|})^2}[/math]

Do those all look right? Still struggling to see the intuition behind the numerators.

>> No.11648225

mdterm question so don't answer it, but isn't this kind of silly?:

Let A, B ⊆ N and |A| = |B| = |N|, where N is the set of natural numbers. Prove that |A U B| = |N| by constructing a bijective function, F: A U B N

It's a second semester discrete math class, we have not covered set density or anything like that. I'm inclined to think I should let A be evens and B odds, and show that the cardinality of the union is equal to the cardinality of N, but it's just a weird question imo. Alternatively, if they all just equal N (A=B=N), it's obviously true that the union of A and B is N, just like the union of N and N is N. I feel I just don't get what the question is really asking me.

>> No.11648231
File: 127 KB, 850x848, __rumia_touhou_drawn_by_goma_gomasamune__sample-03055e9c5d3f1f08a2553874a0344d02.jpg [View same] [iqdb] [saucenao] [google] [report]

We consider the diagonal relation [math]\Delta = \{ (x, x) : x \in A \} [/math].
A relation is reflexive if and only if it contains the diagonal, so the rest is just a short computation for [math]\frac{2^{|A|^2 - |A|} }{2^{|A|^2} }[/math], since the diagonal has [math]|A|[/math] elements, and we thus have [math]2^{|A|^2 - |A|} }[/math] reflexive relations.
>probability it's symmetric
Consider the set [math]B[/math] of subsets of [math]A[/math] having one or two elements.
It's natural to notice that each subset of [math]B[/math] corresponds to a symmetric relation. Actually computing this one is a pain.
>just the product
Is being symmetric independent of being reflexive tho?

I honestly can't follow any of your explanations.

>> No.11648247

>I'm inclined to think I should let A be evens and B odds
You can't do that because they aren't disjoint.

>> No.11648256

I don't think your probability of a random relation reflexive is accurate, it's equivelent to [math]1 - \frac{|A|}{2^{|A|^2}} [/math], which just seems really big. This SE answer goes over the question but I still don't get the reasoning for the numerators:

>Consider the set B of subsets of A having one or two elements.
>It's natural to notice that each subset of B corresponds to a symmetric relation. Actually computing this one is a pain.
This isn't necassarily true, is it? Consider the set (1,2),(2,3), this set is not symmetric, as it lacks (2,1) and (3,2), respectively, unlesss you meant something else here.

How do you know they aren't disjoint? Isnn't the proof super trivial if I just let A=B=N? The whole question is jsut weird

>> No.11648257

bump I wanna know this too since I'm taking classical mechanics next semester and I've only taken math courses

>> No.11648272

if you're taking intro level stuff it will be way below Goldstein, you should be fine if you know calculus, but reviewing tensor and vector calculus would probably be wise

>> No.11648273
File: 453 KB, 700x869, __remilia_scarlet_touhou_drawn_by_60mai__421429b4d24b871356d1d296f1ebcc64.png [View same] [iqdb] [saucenao] [google] [report]

>I don't think your probability of a random relation reflexive is accurate
See: https://en.wikipedia.org/wiki/Reflexive_relation#Number_of_reflexive_relations

>> No.11648274

Good resource to learn parametric equations/polar coordinates?
I'm so confused with it.

>> No.11648279

i'm so sorry i doubted you, so that SE guy is just straight up wrong? wtf?

>> No.11648284

it's a grad classical mechanics course. The undergrad uses Taylor and the grad one uses Goldstein. To put it into perspective, I've taken all of the core courses for a math degree and a few physics classes.

>> No.11648290
File: 343 KB, 900x1112, __remilia_scarlet_touhou_drawn_by_majamari__ec19b88a802e8f6734c0599b7d7ca1fa.jpg [View same] [iqdb] [saucenao] [google] [report]

[math]\frac{2^{|A|^2 - |A|}}{2^{|A|^2}} = 2^{-|A|}[/math]
I just left it like that so it would be easier to see the reasoning.

>> No.11648294 [DELETED] 

Which really makes me wonder on what level of brain rot am I, since your original formula simplifies to the same thing.

>> No.11648312
File: 119 KB, 1080x903, Screenshot_20200507-163036__01.jpg [View same] [iqdb] [saucenao] [google] [report]

Can anyone point me in the right direction to solving this problem? I can't figure out Re because I don't know RL, and without that I can't get the Rin(base) or Rin(tot). I know Ai is approximately equal to Ap, which is Ie/Iin, but I can't solve for either of those variable without Rin(tot) or Re.

>> No.11648321

Just watch Khan academy if you're confused about basic precalc/calc stuff like that, and then go back to your course textbook once you get it. Khan academy has its limits, but for the purposes of
>I'm stuck on a calc concept and I need help
it's pretty much the best tool available, short of paying a real tutor to put up with your brainletism.

>> No.11648325

oh, I misread the expression in your post originally (i read [math] \frac{2^{|A|} - A}{2^{|A|^2}} [/math]. Yes my formula does simplify to the same thing, but I still have trouble seeing the intuition behind either approach. So in your approach for example, why are you raising 2 to the power of [math]|A|^2 - |A|[/math]? The subtraction seems really counterintuitive to me, I don't see the intuition behind my approach either.

Anyway, thanks. The guy on SE made it sound like the same reasoning could be recycled for symmetric AND reflexive sets, but I think you're right about them not being independent, there's lots of overlap between the two.

>> No.11648331

(and yes I understnad |A|^2 is our total number of possible elements, and |A| is our total number of reflexive elements), I just don't get the substraction, I /kinda/ get the base of 2, since we have the option to include or exclude any given member we're looking at, I guess..?

>> No.11648362

"not disjoint" doesn't mean they're all equal. For an example, let A all numbers divisible by 6, and let B be all numbers divisible by 10. There are some elements in A but not in B, some elements in B but not in A, and some elements that are in both.
Hopefully you can see that in cases like this it's not completely trivial how to make a bijective function A U B -> N. That's the point of the question.

>> No.11648366

What have all of your experiences with undergraduate research been like? I'm just finishing up my freshman year and I've got a professor to take me on for the summer, but I'm a little worried that I'll be in over my head. Field is CS.

>> No.11648376
File: 173 KB, 456x532, __fujiwara_no_mokou_touhou_drawn_by_shangguan_feiying__05ab5dc84645fa62d7da2b145e774b14.jpg [View same] [iqdb] [saucenao] [google] [report]

>Yes my formula does simplify to the same thing
The original one? It doesn't.
[math]2^{|A|^2-|A|} \neq 2^{|A|}[/math]
The brain rot post was the actual brain rot.
>why are you raising 2 to the power of |A|2−|A|?
A reflexive relation [math]R[/math] breaks up into the disjoint [math]R = (R- \Delta) \cup \Delta = R' \cup \Delta[/math].
Then, [math]R' \subseteq [P(A)- \Delta][/math]. Since [math]P(A)- \Delta[/math] has [math]|A|^2-|A|[/math] elements, this gives [math]2^{|A|^2-|A|}[/math] choices of [math]R'[/math], each of which uniquely determines a reflexive relation.

>> No.11648383

>but I'm a little worried that I'll be in over my head
You are, but it's okay, that's the point. Undergrad research very rarely actually accomplishes anything, the point is mostly just to get you some experience doing research and working with your professors.
Profs don't expect much out of undergrads, and if you're a freshman they will be happy if you make it through the summer without breaking anything. The only way you will disappoint anybody is by making it obvious that you're not seriously trying to do a good job.

>> No.11648385
File: 483 KB, 650x773, __remilia_scarlet_touhou_drawn_by_60mai__36a90e9c14e16cbd2f80ae07c44a2468.png [View same] [iqdb] [saucenao] [google] [report]

>dude map the [math]i-th[/math] smallest element of [math]A \cup B[/math] to [math]i[/math] lmao
Proving this is a bijection is left as an exercise.

I don't think this is what your professor wants, tho, since it works for any unbounded subset of the naturals.

>> No.11648387

you really don't do anything desu. You simply don't know enough math to do anything. At the very least, however, they will get you to code something for them or solve/prove a very simple property that they are too lazy to prove.

>> No.11648389
File: 1.63 MB, 2304x2738, IMG_20200507_165158__01.jpg [View same] [iqdb] [saucenao] [google] [report]

Think I may have figured out Re but not very confident at this point since that formula for Vout wasn't in the textbook.

>> No.11648433

Well that's shit lol, can't have Re=1000ohm when it is RE of 1000ohm in parallel with RL, RL would be infinite. Maybe I need to go back and look for a rounding error

>> No.11648436

>simplify to the same thing
my original [math]\frac{2^{|A|}}2^{|A|^2}} = 2^{-|A|}, not sure if that's what we're talkng about anymore but that's the SE answer and what your formula simplifies to.

great explanation though, thank you!! I thnk I'm gonna take a break for a few minutes, hopefully I can figure out the symmetric probability and the probability of BOTH later, but they seem even more painful

>> No.11648826
File: 7 KB, 735x74, 095da5cbd6376d255ed9570a02146d5d.png [View same] [iqdb] [saucenao] [google] [report]

Anyone know why I got this one wrong?
The bounds are 0<thetha<2pi and my thing fits...
To calculate for 2, I did the pythagorean theorm and to get the pi/3, I did the tan^-1 (root3). However, when I typed it into wolfram alpha, the polar coordinates were (2, -2pi/3) which I didn't understand how they got the -2pi/3 part.

>> No.11648857

How does x=y become θ=pi/4 in polar?

>> No.11648861

>and my thing fits...
It doesn't fit. Plot (2,pi/3) on a grid and you'll see it's reflected into the opposite quadrant of your point.

You did the trig fine, the issue here is that tan(x) = tan(x+pi). So sometimes arctan will give you the wrong value by a shift of pi. The rule you'll have to follow is that if your point is below the x-axis you'll have to add pi to the angle to get it right.

Often polar angles are done from (-pi,pi) instead of (0,2pi), which is why Wolfram's giving you a negative answer here.
arctan(1) = pi/4

>> No.11648864

>arctan(1) = pi/4
oh im stupid

>> No.11648878
File: 119 KB, 491x484, 14fbbe195dfd103464d91983d99c4485.png [View same] [iqdb] [saucenao] [google] [report]

>you'll see it's reflected into the opposite quadrant of your point.
Can you elaborate a bit on this point? I plotted 2pi/3 onto the desmos polar coordinate graph and it's in Quadrant 1 which fits the domain of 0->2pi doesn't it? Sorry, I'm really bad at this.
>The rule you'll have to follow is that if your point is below the x-axis you'll have to add pi to the angle to get it right.
This makes a lot of sense, I changed it to 4pi/3 and got it right, I'll remember this for the future, thank you!

>> No.11648881 [DELETED] 
File: 294 KB, 984x798, 80e13b01a92be78533956ecf4e1234f3.png [View same] [iqdb] [saucenao] [google] [report]

Wait, I just graphed (2, arctan^-1(root3)) and yeah it was in quadrant 4. I completely understand what you mean by the under the x-axis part now.
Thank you so much kind anon!

>> No.11648882

>and it's in Quadrant 1
Exactly. But your _original_ rectangular point ( -1,sqrt(3) ) is not in quadrant 1. It's in quadrant 3. The polar and rectangular coordinates have to represent the same point in space.

>> No.11648891

Fuck, I sent the wrong picture but I get the point.
>The polar and rectangular coordinates have to represent the same point in space.
OOOOOOOO, I had completely missed this point in my lecture.
Thank you!

>> No.11649146

are there any sites where I can download a coursera textbook pdf instead of blowin 100$ on it?

>> No.11649247

Library Genesis?

>> No.11649251

I get ~37 Ω (12375/331 Ω).
1/Re = 1/1k+1/RL
Rb = 150*Re
1/Rin = 1/22k+1/27k+1/Rb
Rin=100*RL (20dB)
=> RL=0 or RL=12375/331

>> No.11649563

Thanks man, I'll have to play around with the formulas from the book and see if I can come up with a way to derive something that matches, you make it seem super easy for a 40 point problem so I guess the catch will be making it work with what the professor gave us formula wise. I don't think my solution will be nearly as elegant. I really appreciate you taking the time to solve it and share your solution, though. You're way better at recognizing what to do than I am.

>> No.11649585
File: 9 KB, 277x291, Screenshot_20200508_110320.jpg [View same] [iqdb] [saucenao] [google] [report]

Is there no answer to this or am I just retarded

>> No.11649604

There is at least one solution, and up to 3 solutions

>> No.11649608

right out of the bat, x=1 is an obvious solution.

>> No.11649628

do you know/recommend any good apps to organize notes from lectures?
i've always just wrote down everything in textbooks but i feel like that's a huge waste of time and money, i type waaay faster on keyboard but just writting it all down in notepad and then putting .txt in folders seems like a horrible way to do it, i don't know about any programs to do this but i can't really believe there's nothing like that

>> No.11649963 [DELETED] 

If I wanted to cut off the edges of a circle and turn it into a hexagon, and then calculate the new surface area, what would the formula or process for that be?

>> No.11650303

what would happen if the mean were to leave orbit? lets say it moves away at a speed decent for space bodies until its effects on earth or negligible
what would we observe on earth?

>> No.11650344
File: 924 KB, 877x1240, ae467c38e0e1ef6fbdc4c57f44f1120f87ac6cc2e6e8b2596ed3ae7fe1a3efd9.jpg [View same] [iqdb] [saucenao] [google] [report]

Would like to rephrase the /g/ birthday collision one.
I had no luck causing even a single collision with some made-up code using both MD5 and SHA1 hashing, so I can't see this solved with scripts.
So mathematically, what's the equation to solve this? I've got a pretty big range of numbers to test from 0000000 to 9999999 unhashed. Could it be related to XORing in any way?

>> No.11650389

I meant moon

>> No.11650607
File: 49 KB, 2584x170, Screenshot 2020-05-08 at 18.27.33.png [View same] [iqdb] [saucenao] [google] [report]

How do I prove this?

From what I understand div (r) = 3 so with gauss theorem, the answer would be that the second integral would be equal to three times the first. Why is this wrong?

>> No.11650616

Never mind I got it

>> No.11650656

Did you try at least 10^7 times to have a ~0.63 chance to hit?

>> No.11650672

thanks anon

>> No.11650747

The determinant of that matrix is x^3-3x+2=(x-1)^2(x+2), so the solutions are x=1 and x=-2. With x=1, all three rows are identical (rank 1, nullity 2, hence the duplicate root). With x=-2, the three rows sum to [0 0 0], i.e. any row is the negation of the sum of the other two (rank 2, nullity 1).

>> No.11650784

>There is a light bulb and a consciousness switching machine. Two people can enter the machine and switch their consciousness. Whenever two people switch their consciousness, the light bulb switches (on->off and off->on). You have access to as many people as you want. The light bulb starts off being off. Is it possible to using the machine in such a way that in the end, everyone's consciouness is back to where it started but the light bulb is on. If yes, give a method how to do it, and if not, explain why.
How would you explain the solution to this without any higher mathematics (digestible by a 10 year old child).

>> No.11650831

What's the question?

The probability of a collision when selecting n random values from a set of size d is given by
P(n,d)=1-\prod_{k=1}^{n-1}\left(1-{k \over d}\right)
This is just one minus the probability that all values are distinct. Note that for n>d, P(n,d)=1 (this results in a term with k=d => 1-k/d=0 => the product is zero).

For large d, this can be approximated with
P(n,d) \approx 1-e^{-\frac {n(n-1)} {2d}}

If you randomly select integers from the set [0,10^8), you need ~11775 values to have a 50% probability of a collision.

>> No.11650836

Hypothetically speaking, what would happen if we cut down every forest in the world to use for farmland?

>> No.11650850

No idea how you'd explain this to an average 10yo.

>> No.11650917
File: 118 KB, 462x454, __fujiwara_no_mokou_touhou_drawn_by_shangguan_feiying__4c2d507f8fb4dfac4cac95721545d7f4.jpg [View same] [iqdb] [saucenao] [google] [report]

Best I can do is digestible by a high schooler, since I can't find a simple proof of the parity thing outside of exploiting the [math]S_n \rightarrow GL(n)[/math] representation.

>> No.11650919

You can explain parity to a child using the inversion definition of the parity of a permutation of [1,2,...,n] being the number of pairs which are out of order.
So the point of the lightbulb problem is that if you stick numbers onto the heads of your people, swapping a pair of people changes the parity of the number of out-of-order pairs (which is a totally elementary counting argument). Then it follows that if you somehow swapped back to 0 out-of-order pairs you must've swapped an even number of times.

I would not say that kind of argument is digestible by a 10 year old but it's completely digestible by a 10 year old in a math club.

>> No.11650962

Can someone check if this is correct?

>Q: Let [math]\Gamma\subset \mathbb P^n[/math] be a finite subset of [math]d[/math] points, not all contained in a line. Prove that the polynomials defining the variety of those points can be chosen to be of degree [math]<d[/math].

My attempt:

We must have [math]d>2[/math] be assumption, so we have a line [math]L[/math] going through any two points in [math]\Gamma[/math] that does not go through a point [math]P[/math]. Induct on [math]d[/math]. Assume that [math]\Gamma[/math] has [math]d[/math] points and that it is true for any proper subset of [math]\Gamma[/math]. Let [math]\Gamma_P=\Gamma-\{P\}[/math] for some point in the set.

If the points of [math]\Gamma_P[/math] don't lie on a line, we apply the induction hypothesis to show that [math]\Gamma_P[/math] is generated by degree [math]d-2[/math] polynomials [math]F^\alpha[/math]. Let [math]M[/math] be a linear form vanishing at [math]P[/math]. Then the polynomials [math]F^\alpha M[/math] are of degree [math]d-1[/math] and vanish on [math]\Gamma[/math].

Now suppose [math]\Gamma_P[/math] lies on a line, and we can assume WLOG that the line is [math]L: X_0=X_1=...=X_{n-2}=0[/math] and the point is [math]P=[1:0:...:0][/math]. Let [math]\{F^\alpha(X_0,...,X_n)\}[/math] be the set of degree [math]d[/math] polynomials defining [math]\Gamma[/math]. We can perform the division algorithm to get [math]F^\alpha= X_0Q_0^\alpha(X_0,...,X_n)+X_1Q_1^\alpha(X_1,...,X_n)+...+X_{n-2}Q_{n-2}^\alpha(X_{n-2},X_{n_1},X_n)+K^\alpha(X_{n-1},X_n)[/math]. Notice that [math]K^\alpha[/math] vanishes for all the points of [math]\Gamma[/math], so in particular [math]F^\alpha- X_0Q_0^\alpha\equiv 0[/math] on [math]\Gamma[/math], and the polynomial [math]Q_0[/math] is of degree [math]d-1[/math]. Therefore we can write [math]\Gamma[/math] as the locus of [math]X_0[/math] and [math]Q_0^\alpha[/math].

Applying the method to the case [math]d=3[/math] yields the base case. QED

>> No.11651059

>Rin=100*RL (20dB)
So had a chance to play around with the numbers and this is what I'm missing, how can I derive this relationship? I understand it is true, but it isn't given by my text book. It is very helpful in this case, however.

>> No.11651098
File: 377 KB, 878x1228, __remilia_scarlet_touhou_drawn_by_puuakachan__2100d12ec2d99b7c388db0a53974e2d0.jpg [View same] [iqdb] [saucenao] [google] [report]

>then the polynomials [math]F^{\alpha} M[/math] are of degree [math]d-1[/math]
>and vanish on [math]\Gamma[/math]
Now here's the issue. While they do zero on [math]\Gamma[/math], they don't zero only on [math]\Gamma[/math]. Specifically, they zero on [math]\Gamma_P \cup N(F)[/math], so they don't define the gamma variety.

My algeo is weak as fuck tho, so this might be completely retarded.

>> No.11651120

Typo, it's [math]\Gamma_P \cup N(M)[/math]
Fuck, knew something was wrong.

>> No.11651123

How about this fix then: Let [math]M_\beta[/math] be a set of distinct hyperplanes whose intersection is [math]P[/math], each defined by their own linear form. Then the set of polynomials [math]F^\alpha M_\beta[/math] are still of degree [math]d-1[/math], and I'd assume that they can only coincide on [math]\Gamma[/math]

>> No.11651129

I made a mistake, instead of the product, I mean to write (as you'd write it), [math]N(F^\alpha,M_\beta)[/math]

>> No.11651145

I think that worked, yeah, but I could be wrong.
I'd choose hyperplanes that have empty intersection with Gamma P to be super sure.

>> No.11651194

Fuck I'm such a brainlet. That shit is like magic to me. Would appreciate if anyone could expand it to a pre calc 2 level as I'm a pleb ET major.

>> No.11651387

Prove [math]A \cup B = A \cap B \leftrightarrow A = B[/math]

Suppose [math] x \in A \land x \notin B[/math], then [math] x \in A \cup B[/math]. But [math]A \cup B=A \cap B[/math], so if [math] x \in A \rightarrow x \in A \cup B = A\cap B[/math], so [math] x \in B[/math].

A similar arguement can be made to show that [math] x \in B \rightarrow x \in A [/math]. Thus [math] \forall x | x \in A \leftrightarrow x \in B, A = B [/math].

Look good?

>> No.11651431

It's fine, but it's bad style to shoehorn in unnecessary proofs by contradiction like this. There is no point in supposing that x is not in B if you're going to never use that assumption and just going to go directly prove that x is in B. Including the assumption is pointless.

>> No.11651434

Yes, but you shouldnt prove by contradiction if you don't have to.

Suppose [math]x\in A[/math]. Then [math]x\in A\cup B [/math], hence [math]x\in A\cap B[/math], hence [math]x\in B[/math]. QED

>> No.11651448

thanks, and yeah I meant to edit the contradiction out as it was totoally unnecassary, sorry. It's how I started the proof yesterday and I realized I never needed it, just forgot to exclude it as I was mindlessly typing it up. I'm still having huge issues with this if anyone feels like dropping a hint: >>11648225.

>> No.11651465
File: 946 KB, 708x1000, __hinanawi_tenshi_yorigami_shion_and_yorigami_jo_on_touhou_drawn_by_katayama_kei__bdc9043d4ff2efe4ddd8bfcbb9ff7368.jpg [View same] [iqdb] [saucenao] [google] [report]

>you shouldnt prove by contradiction if you don't have to.
You shouldn't do [math]x \in [/math] argumentation either tbqh.
[math]A \subseteq A \cup B = A \cap B \subseteq B[/math], and similarly for [math]B \subseteq A[/math].

>> No.11651471

hold on don't give me any hints actually i think i get what the question is asking now

good point, just felt the question was so oddly trivial i wanted to show set equivelance by elements as it felt like a better answer, it was a midterm question and was extremely, no idea why it was on there

>> No.11651472

See >>11648385

>> No.11651476

i saw that, but it concludes with discouraging it as not something my prof wants.

>> No.11651482 [DELETED] 

can the answer be something like:
let A be the negative integers, let B be the positive ints. Let F be f(x)=|x|

is it that simple? I just feel i totally misunderstand the point of this proof

>> No.11651503

Define a function [math]F:A\cup B \to \mathbb N[/math] by defining it in [math]A-B[/math] and [math]B[/math]. You probably will have to consider separately the cases where [math]A-B[/math] is finite or infinite.

>> No.11651508

No. You don't get to choose A and B anon.

>> No.11651516

I see, that's my misunderstanding. So A and B are arbitrary infinite subsets of N (because cardinalities are equal), and the union of the two happen to be the domain of my bijective function mapping them to N.. Hmm.... Well now I understand how to think about it at least, thanks

>> No.11651539

ok, the best can come up with is a function that maps the smallest unmapped element in the union of A and B to the smallest unmapped element in N. So the smallest element in AuB is 1, the second smallest is 2, etc.. just gotta figure out how to properly notate this idea now

>> No.11651543

>literally >>11648385
See >>11651503
It's unironically probably what he expected.

>> No.11651555

yeah now that I understand the question better this feels very natural. I have
[math]x_i \in A \cup B \mid \forall x \in A \cup B, x_{i-1} \leq x_i \leq x_{i+1} \rightarrow f(x_i) = i[/math]

Is that ok? Not sure if there's some better notation available to express the idea.

>> No.11651564

nice digits me and maybe a better notation is
[math]x_i \in A \cup B \mid \forall x \in A \cup B, x < x_i < x \rightarrow f(x_i) = i[/math]


>> No.11651567

oh and obviously [math]x \neq x_i[/math] is supposed to be in there

>> No.11651574

final draft:

[math]a_i \in A \cup B \mid \forall x,y \in A \cup B \land x,y \neq a_i \rightarrow \exists x < a_i < y[/math]

So, in relation to the above we represent our function as [math]f(a_i) = i[/math]

sorry for tard posting

>> No.11651582

I’m going back to school for a masters in the fall. How do I decide what to focus on?
I’m looking at Data Science or Business IT. they’re both STEM programs at the University I’m looking at.
What would you pick /sci/?

>> No.11651594

dumb question, but in this arguement, [math]\Delta[/math] is being used to represent the reflexive _elements_ in the set of the reflexive relation, R, right?

>> No.11651601

can someone convince me why polynomial long division works or point me to a source which will show me why it works?
something simple and to the point preferably

>> No.11651625

also, what is P(A)? I only ask because that notation is often used for power set and probability, both of which are active themes in this question. It's obviously not the power set, as the cardinality of the power set is [math]2^{|A|}[/math], but I don't see what it would represent as a probability value either; R' is a subset of the probability of AxA - [math]\Delta[/math]?

just doesn't make sense to me presently, I know I can just sub P(A) for AxA but I wanna understand your reasoning

>> No.11651631

[math]R' \subseteq [P(A)- \Delta][/math]
not to nitpick, but you did mean [math]R' \subseteq [P(A-\Delta)][/math], right?

>> No.11651638

> how can I derive this relationship?
An emitter follower (common collector) has unity voltage gain; Vout=Vin (for the AC signal, ignoring DC bias). Power = V^2/R, so if the power dissipated in RL is 100x (+20dB) the input power, the input resistance must be 100x RL.

>> No.11651642

that question has nothing to do with math or science, sorry
for career advice maybe try >>>/adv/

>> No.11651649
File: 972 KB, 1000x1000, smile.png [View same] [iqdb] [saucenao] [google] [report]

[math]\Delta[/math] is defined in >>11648231
Essentially, it's the identity relation.
>what's P(A)
Power set.
Absolutely weird fucking typo, tho, should have been [math]A^2[/math]. Apologies.
No, that's correct as is (with P(A) substitued by A^2, as mentioned just now).

>> No.11651662

What are the flippin odds that a set a relation on set A is symmetric?

>my attempt
OK so the reflexive elements are the diagonal in the AxA matrix, so there's A reflexive elements. The symmetric elements are everything else, so there's S=AxA - A of them. The sample space of total possible subsets of AxA is [math]2^{|A|^2}[/math]. So the power set of AxA - S is [math]2^{|A|^2 - (A^2 - |A|} = 2^{|A|}. So the odds of a set being symmetric is [math]\frac{2^{|A|}}{2^{|A|^2}} = \frac{1}{2^{|A|}[/math]

>> No.11651670

>2^{|A|^2 - (A^2 - |A|} = 2^{|A|}. So the odds of a set being symmetric is [math]\frac{2^{|A|}}{2^{|A|^2}} = \frac{1}{2^{|A|}
[math]2^{|A|^2 - (A^2 - |A|} = 2^{|A|}[/math]. So the odds of a set being symmetric is [math]\frac{2^{|A|}}{2^{|A|^2}} = \frac{1}{2^{|A|}}[/math]

thanks, makes sense was just uber confusing given the context

>> No.11651673

So to solve for RL, how would I go about setting up the equation? I appreciate the help my man, but obviously my math isn't up to par. I have to set up the equation in terms of Rin, which is in terms of Rb? I've tried writing it out but plugging unknowns that many layers deep isn't something we've really done so far, and it has been almost a decade now since I took college algebra so any additional explanation on how to set this up would be most helpful.

>> No.11651676

Just try an example. Given polynomials A and B, you're looking for Q, R s.t. A=BQ+R, with the degree of R strictly less than that of B. That uniquely determines the leading term of Q; there is only one possible term which, when multiplied by B matches the leading term of A. So set Q=(q1+Q') => A=B(q1+Q')+R => A=Bq1+BQ'+R => A-Bq1=BQ'+R => A'=BQ'+R where A'=A-Bq1. Repeat until A' has degree less than B, meaning that Q'=0 and R=A'.

IOW, it's just like long division with numbers. What is the largest value (quotient) by which you can multiply the denominator without exceeding the numerator? With numbers, you find the largest leading (most significant) digit, then the next digit and so on. With polynomials, you find the largest leading (highest degree) term then the next and so on.

>> No.11651686

sorry, I don't know anything about sets
trying examples doesn't convince me 100% that we can always use it and it will always work and why
and saying "you can divide polynomials just like numbers" is a statement that requires validation for me

>> No.11651689

Data Science is mostly statistics, retard. Nobody at /adv/ will know about that.

>> No.11651690

statistics is not science or math
once again, try some other place for career advice, /sci/ is not for that

>> No.11651695
File: 111 KB, 850x822, uhoh.jpg [View same] [iqdb] [saucenao] [google] [report]

I feel like this is a really stupid question but can anyone tell me how to change this linear (homogenous?) recurrence relation
[math]\[x(0)=2, x(1)=3\ x_{n+1}=3x_{n}- 2x_{n-1}
[/math] for n > 0
so that it looks like
[math]x_{n} = a_{n-1}+ b_{n-2}[/math]
where a and b are some constants resulting from the above equation.
I've been trying to do it for the past 1 hour and I still can't do it.

>> No.11651699

Lol. You are a faggot

>> No.11651709

Scientifically, why do I get an erection whenever I play with my cat?

>> No.11651712

I'm not sure what you mean by [math]x_n = a_{n-1}+b_{n-2}[/math] . Are you just trying to solve for xn, like in the form [math]x_n = ar_1^n+br_2^n[/math] ?

>> No.11651719

I realize everything below is wrong, but I'm also not sure how else to approach the question either.

cleaner version of the same question ('odds that a set a relation on set A is symmetric?')
A symmetric relation R breaks up into the disjoint:
[math]R = (R-S)\cup S = R^s \cup S [/math].
[math]R^s \subseteq [P(AxA) - S]. [/math]
[math]P(AxA) - S [/math] has [math]|A|^2 - (|A|^2 - A) = |A|[/math] elements. This give [math]2^{|A|}[/math] choices of [math]R^s[/math], so the total probability is given by (where the sample space is the power set of (AxA)): [math]\frac{2^{|A|}}{2^{|A|^2}}[/math]

I get stuck because the odds of a set in AxA being symmetric depends on it's elements, like if (a, b) then it needs (b, a), for all a and b. It's not as simple as the reflexive question.

>> No.11651723

>and saying "you can divide polynomials just like numbers" is a statement that requires validation for me
Numbers are polynomials where you replace the x with a 10

>> No.11651737
File: 239 KB, 850x1000, uh.jpg [View same] [iqdb] [saucenao] [google] [report]

Yes, and in all the examples for solving those I could find online they start with a relation in the form of [math]x_n=a_1x_{n-1}+a_2x_{n-2}+...+a_kx_{n-k}[/math]
The fact that it's [math]x_n=a_1x_{n-1}+a_2x_{n+1}[/math] made me think that I have to change it so that it looks like the one above.
I'm really terrible at this

>> No.11651741

what about when you divide by (x - 33)? 33 exceeds 10 and should take the x's spot if that analogy was correct

>> No.11651755

I mean, you can shift the subscripts over one if you want and write [math]x_n = 3x_{n-1}-2x_{n-2}[/math], for n > 1.

>> No.11651758

Emitter resistance: 1k and RL in parallel
1/Re = 1/1+1/RL (1)
Base resistance (Ie=β.Ib => V/Re=β.V/Rb => Rb=β.Re; should actually be β+1, but this is close enough):
Rb = 150*Re (2)
Input resistance: 22k, 27k and amplifier in parallel:
1/Rin = 1/22+1/27+1/Rb (3)
Power gain:
Rin=100*RL (20dB) (4)

Reciprocal of (2) => 1/Rb=1/(150*Re) = (1/150)*(1/Re)
Substitute (1) => 1/Rb=(1/150)*(1/1k+1/RL)
Substitute into (3) => 1/Rin = 1/22+1/27+(1/150)*(1/1+1/RL)
= 1/22+1/27+1/150+1/(150*RL)
= (22*27+22*150+27*150)/(22*27*150)+1/(150*RL)
= 7944/89100+1/(150*RL)
= 662/7425+1/(150*RL)
Reciprocal of (4) => 1/Rin=1/(100*RL)
Equate the last two => 1/(100*RL) = 662/7425+1/(150*RL)
=>(1/RL)/100 = 662/7425+(1/RL)/150
=>3*(1/RL)/300 = 662/7425+2*(1/RL)/300
=>(1/RL)/300 = 662/7425
=>(1/RL) = 300*662/7425 = 2648/99
=> RL = 99/2648 (kΩ) = 99000/2648 (Ω) = 12375/331 = 37.4 Ω.

DESU, initially I just put the first four equations into Maxima and solved for RL. If I was doing this with a calculator I'd probably just use decimals rather than fractions.

>> No.11651762

I didn't say polynomials were numbers, I said numbers were polynomials.
Obviously polynomial division is a generalization, the point is that whatever reason you believe integer division works will work verbatim for polynomials.

>> No.11651782

>trying examples doesn't convince me 100% that we can always use it and it will always work and why
Then use your brain while you're working through the example.

It should be obvious that if deg(A)>=deg(B), there exists real c and integer n=deg(A)-deg(B) s.t. deg(A-c.x^n.B)<deg(A), i.e. the leading terms of A and c.x^n.B cancel. Thus you're now left with the problem of dividing a lower-degree polynomial. And that if deg(A)<deg(B) then A=BQ+R can only be satisfied with Q=0 and R=A. Recursive case, base case.

>> No.11651794
File: 608 KB, 637x719, hato.png [View same] [iqdb] [saucenao] [google] [report]

And that's what I initially did but I got a wrong answer, assumed that I'm doing something wrong and spend an hour looking for answers.
Your answer made me doublecheck it though, and apparently I just got the calculations wrong like the big retard I am.
Thanks for help

>> No.11651810

ignore these pls im getting it i think

>> No.11651820

if there are [math]\frac{n(n+1)}{2}[/math] such symmetrics sets in a relation on A, then the probability of a random set being symmetric is just the previous value divided by the sample space, right (the power set of (AxA))? Just seems way to easy after how hard it felt earlier.

>> No.11651821

Dayum, no joke, you're a better man than I. I hope that wasn't too much effort for you, didn't mean to have you go that far out of your way but I really really appreciate the effort and thorough explanation. Brain goals.

>> No.11651826

shit, I meant 'then the probability of a random set being symmetric is just TWO RAISED TO THE POWER OF the previous value divided by the sample space, right (the power set of (AxA))? Just seems way to easy after how hard it felt earlier.'

>> No.11651838

can I get a yay/nay on this >>11651574 as an answer to >>11648225

>> No.11651854

I gotta take an education class to get teaching competency in my country and I took sociology of education. Now they're making me write a 5k word essay on bullying. Should i TeX it? All the related works I could find are typesetted differently but I'm much more comfortable with TeX than office suits.

>> No.11651876

>but I'm much more comfortable with TeX than office suits.
You don't really need to be "comfortable" with Word to write an essay. You literally just open it up and start typing. Idiots who can barely operate Facebook write their school essays in Word.
It's not like you _can't_ write a text essay in TeX if you really want to for some reason, but that's not the correct tool for the job. It'd be easier for you to just use a word processor.

>> No.11651928

Literally the only reason to write an essay in TeX would be to show off. Imagine inconveniencing yourself and spending quadruple the time on formatting just to show off to normalfags.

>> No.11651971

Is every even number over 2 divisible by at least one uneven number?
I.e. if you wanted to test if a number is prime, could you just test if it is divisible by 2 or any uneven number between 3 and its squareroot?

>> No.11651984
File: 3 KB, 424x346, 4.png [View same] [iqdb] [saucenao] [google] [report]


>> No.11651995

Not sure if it's appropriate for this thread, but I think this board might provide a better response than /adv/. I got an offer both for the Uni of Edinburgh and St Andrews to read maths, and got rejected by all my other choices. Anybody here attended any of those two? Or would recommend one over the other?

>> No.11651996

>Is every even number over 2 divisible by at least one uneven number?

>> No.11652003

>Is every even number over 2 divisible by at least one uneven number?

>> No.11652018

Edinburgh is better regarded, but St.Andrews does a ton of applied research. Unless you're specifically really into applied math Edinburgh is probably a better choice.

>> No.11652031

Thanks this is sufficient.

>> No.11652222
File: 4 KB, 343x329, bruh.png [View same] [iqdb] [saucenao] [google] [report]

I have 3 white balls and 23 black balls. What is the possibility of getting no white balls if i get 5 randomly?
I think pic related but not sure cause if i get 25 ball it should be 100% aka 1.

>> No.11652226

sorry, 22 black balls, 25 in total. alll balls are the same equiprobable etc

>> No.11652252

[eqn]\frac{{22 \choose 5}}{{25 \choose 5}} = 57/115[/eqn]

>> No.11652254

which is incidentally the same number you'd get with your product, just an easier way to look at it

>> No.11652299


>> No.11652584
File: 252 KB, 1080x1372, Screenshot_20200508-223647__01.jpg [View same] [iqdb] [saucenao] [google] [report]

Why is the sign for Av1 inverted in this example? We use the same circuit with different values for a lab, but the professor didn't have us invert the gain. He reused the circuit for the exam and I'm comfortable calculating the gains based on the work we did in lab, but now I'm second guessing whether I should be inverting the gain? Supposedly the question is meant to be over chapter 6 but it also covers power efficiency which wasn't in our lab, so I do need to work ahead. Just wondering if I need to worry about that inversion since again, wasn't part of the lab.

>> No.11652588

Those are physics books, and not even quantum stuff at that. Analysis and algebra are irrelevant. You need to know how to solve/ approximate ODEs/PDEs and basic linear algebra ideas for Jackson. I don't recall Goldstein but I can't imagine needing much more, maybe some muiltilinear algebra (physicists' tensors and stuff).

>> No.11652594
File: 2.61 MB, 4608x2304, IMG_20200508_224322.jpg [View same] [iqdb] [saucenao] [google] [report]

What we used in lab.

>> No.11652642

>Why is the sign for Av1 inverted in this example?
Because it's an inverting stage; base voltage goes up => collector voltage goes down.

>> No.11652659

What's the deal with seleno cysteine? Why is it often not listed among amino acids?

>> No.11652673
File: 280 KB, 1080x1355, Screenshot_20200508-235157__01.jpg [View same] [iqdb] [saucenao] [google] [report]

I see. Any reason they chose not to mention that in the previous chapter? They talk about the Darlington voltage follower but gloss over the common-emitter section, trying to figure how this comes together with my lab work. Professor stopped lecturing after the midterm so definitely missed some details.

>> No.11652679
File: 286 KB, 1080x1361, Screenshot_20200508-235722__01.jpg [View same] [iqdb] [saucenao] [google] [report]

2nd page of the example. The circuit we used was from the chapter 7 example but with different values, but when we calculated the gains in lab we made no note of inversion

>> No.11652768

Why should 2-morphisms only go between morphism that share their source and target?
Is there a name for a structure where this is not true?

>> No.11652888
File: 112 KB, 229x263, 1576695930886.png [View same] [iqdb] [saucenao] [google] [report]

What are the chances that my organic chemistry professor expects me to independently figure out the mechanism for the Reformatsky reaction when we haven't even gotten to organometallics yet and the reaction isn't even in the textbook? I have a problem on an exam that I'm 99% positive that's what it is. Should I just do a Grignard-esque thing instead? It's supposed to be closed note and if I do the actual correct mechanism I'm sure he'll think I cheated.

>> No.11652892

I don't know why you would even consider using a word processor.

>> No.11653278
File: 596 KB, 2192x1472, retard.jpg [View same] [iqdb] [saucenao] [google] [report]

Guys, I'm retarded. How do I solve this? Will suck your dick if you help (no homo).

>> No.11653279


>> No.11653289

This is a popular hypothetical, there are lots of articles/videos online about it (look for "what if the moon disappeared" or something) but the two main side effects are usually that the tides of the ocean would pretty much disappear, since the gravity of the moon is why tides happen, and the earth's axis would tilt much more extremely than it does now (which would have severe effects on the climate in certain parts of the globe), since the moon stabilizes the axis of the earth

>> No.11653306

cos(a) * cos(b) = [cos(a+b) + cos(a-b)]/2

>> No.11653377
File: 460 KB, 3186x694, retard2.jpg [View same] [iqdb] [saucenao] [google] [report]

How about this one, mighty /sci/entist?

>> No.11653411

You don't gain anything from doing it that way; it's still a 2-morphism in the established sense.

>> No.11653428

>Should I just do a Grignard-esque thing instead?
yeah. it's just a simple oxidative insertion

>> No.11653493
File: 288 KB, 1080x1350, 7e496fa3337bf650c03489724548abf7.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.11653607

Can you learn to use your non dominant hand as you would use your dominant hand? I'm right handed but could I ever train my left hand enough as to both would feel the same and be very efficient or is that a pipe dream? does something prevent it? brain wiring?

>> No.11653667
File: 90 KB, 640x640, lionel-animals-to-follow-on-instagram-1568319926.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.11653677

You can build dexterity with your non-dominant hand but it's never going to be completely as natural as your dominant one.
You see this sort of thing happen when somebody e.g. breaks their dominant hand or something similar. So they have to live a month or two using exclusively their non-dominant hand for everything. They get better at it, but it's always a little awkward.

>> No.11653687

a little awkward? how? you see, what I want to get at is, is there any chance you could do something something that requires dexterity, fine motion such as drawing or painting with it?
how to take care of possible carpal tunnel?

>> No.11653702

>a little awkward? how?
At least for me, basically it's just that it was never an automatic process. I'm left-handed, and I don't have to focus on writing to write with my left hand. I can talk on the phone and just scribble something down without thinking about it.
When I had to learn to use my right for a while, I had to really focus on what I was writing in order to get it to turn out ok.

I'm sure it's possible to learn to draw with your non-dominant hand but I think it would take quite a long time.

>> No.11653724

To the math anons who can seemingly answer any math problem asked on here, how do you do it? Do you grind through textbooks all the time? How can I achieve your level of knowledge? I have a bachelor's in math and want to use my math skills to help anons as well

>> No.11653740

95% of the math asked on /sqt/ is pretty low level. If you have a Bachelor's and you can't help people with their calculus homework you fucked up hard.

>> No.11653754
File: 54 KB, 551x86, j.png [View same] [iqdb] [saucenao] [google] [report]

can someone explain where the j (i) in the numerator went?

>> No.11653856

Please respond

>> No.11653865

How can I solve this?

>The table below represents the books sold by a publishing company.

>Price | Number of books sold
>0 ⊢ 10 | 4000
>10 ⊢ 20 | 13500
>20 ⊢ 30 | 25600
>30 ⊢ 40 | 43240
>40 ⊢ 50 | 26800
>50 ⊢ 60 | 1750

>If the publishing company starts a sale with 25% of the lowest priced books, what is the highest price of a book that goes on sale?

>> No.11653886

the absolute value of
[math] | j·(5E-7)·\omega | = |j| · |5E-7| · |\omega| = |5E-7| · |\omega| [/math]
since the complex unit has norm 1.

>> No.11653890
File: 139 KB, 747x1059, robin.jpg [View same] [iqdb] [saucenao] [google] [report]

You need to be patient, hunny~! I'm assuming the 240 volts is rms. Let w=2pi*50, P=30 kW and Q=20 kVAr so that S=P+jQ.
[math] PF=\cos\theta=1\implies \theta=0,\ Q'=0 [/math]. So [math] S=P=30\text{ kW} [/math] and the reactive power due to the capacitor is [math] Q_C=Q-Q'=20\text{ kVAr} [/math]. Finally, [math] C=Q_c/\omega V_{rms}^2=20000/(2\pi\cdot50\cdot240^2)=1.11\text{ mF} [/math]
[math]PF=\cos\theta=.95\implies\theta=18.19^\circ ,\ Q'=P\tan\theta=9.86\text{ kVAr}[/math]. Now [math] Q_C=Q-Q'=10.14\text{ kVAr} [/math] and [math] C=Q_C/\omega V_{rms}^2=10139/(2\pi\cdot50\cdot240^2)=560\ \mu\text{F} [/math]
Obviously you need to check my work.
You're taking a magnitude. The magnitude of a the ratio of two complex numbers is the ratio of the magnitudes.
probably ecological if not social catastrophe. i like trees. this sound like a dark future. im not really sure what kinda response you were after.

>> No.11653923
File: 673 KB, 810x872, __izayoi_sakuya_and_remilia_scarlet_touhou_drawn_by_kirero__7761198488c3b5841838a0bb794f8d1f.png [View same] [iqdb] [saucenao] [google] [report]

>[math]2^{ \frac{n(n+1)}{2}}[/math]
Could you explain? I calculated [math]2^{n + \frac{n(n-1)}{2}}[/math] desu.
The trick is exposed here: >>11648231 , since [math]B[/math] has n choose 1 plus n choose 2 = [math]n + \frac{n(n-1)}{2}[/math] elements.
The part about division is correct.
[math]4000+13500+25600+43240+26800+1750 = 114890[/math].
A fourth of that is [math]28722.5[/math].
Eyeballing it, that goes up to the 25600 bracket, so between 20 and 30 dollars.

>> No.11653944

thank you, anon-kun~

>> No.11653957
File: 2.06 MB, 970x1200, 79468341_p0_master1200.png [View same] [iqdb] [saucenao] [google] [report]

its untested but i have a theory that playing 2hu fan games increases one's mathematical prowess exponentially

>> No.11653972

I guess it wouldn't change my answer since the professor wants it reported in dB and you can't calculate a log with a negative gain, maybe that's why he ommited the detail in lab.

>> No.11654045

>between 20 and 30 dollars
I got it down to that as well, but how can I be precise?

>> No.11654055
File: 63 KB, 750x422, 2267758_d55c.jpg [View same] [iqdb] [saucenao] [google] [report]

What algorithms are used to generate the tool-path so that the CNC machine mills the desired shape. pic related.Video related

>> No.11654073

I love you anon. You're the nicest person on 4chan.

>> No.11654092

You can't unless you make assumptions.
For example, the promotion has [math]28722.5-13500-4000=11222.5[/math] books in the [math]20-30[/math] tier.
Then [math]11222.5/25600 = 0.4384[/math], that is, it contains 43,84% of the books in the tier, so we can presume the books are equidistributed along the tier, so the largest price is 20+4.38=24.38 dollars.
But this is completely unjustified wild guessing, the best you can seriously do is between 20 and 30 dollars.

>> No.11654103

Beautiful. Thank you so much.

>> No.11654111

I ask questions on /sci/ because I'm too shy to ask my professors.

>> No.11654137

I'm being stupid

Two points determine a unique line in any projective space, in particular, [math]\mathbb P^3[/math]. Suppose the points are [math][1,0,0,0][/math] and [math][0,1,0,0][/math]. How would you write the line?

Starting with the equations of an arbitrary line [eqn]l:\begin{cases}aX+bY+cZ+dT=0\\eX+fY+gZ+hT=0\end{cases}[/eqn] substituting the points in we get immediately that the line is [eqn]l:\begin{cases}cZ+dT=0\\gZ+hT=0\end{cases}[/eqn] The fact that the planes actually define a unique line is expressed as the fact that [math]ch=dg[/math], and we can't have all coefficients being [math]0[/math]. But unless I have another piece of information, I can't say more on the coefficients, and hence cant determine the unique line

>> No.11654163
File: 3.49 MB, 2304x2418, IMG_20200509_130936__01.jpg [View same] [iqdb] [saucenao] [google] [report]

How did they calculate the VR1 and VR2? Is it just (VCC/2)-0.7V?

>> No.11654187

Where does Hydrostatic pressure act on a submerged body?

>> No.11654297

Have you tried computing the equations in affine space and homogenizing them?

>> No.11654303

are there any mathematicians who struggled with physics?
just looking for ways to make myself feel better..

>> No.11654307
File: 955 KB, 1000x1238, __remilia_scarlet_touhou_drawn_by_ainy77__17c3bd84967fdc0849b7fbde3e7c3bae.jpg [View same] [iqdb] [saucenao] [google] [report]

Herre you go.

>> No.11654309
File: 45 KB, 731x306, Screenshot_20200509-140347__01.jpg [View same] [iqdb] [saucenao] [google] [report]

Why do they call this the average power? Wouldn't average power be different, if this formula uses RMS wouldn't it be power out rms?

>> No.11654312

why'd my file get deleted? are you secretly a janny, 2hu-anon?

>> No.11654313

He deleted a bunch of mine too, tho.
Skipped some for no comprehensible reason.

>> No.11654315

converges in [-1,1], right?

>> No.11654324

Yes, because [math]\sum_{n=1}^{\infty} \frac{1}{n^2}[/math] converges and bounds it above term-wise.
Probably still converges a bit farther, tho.

>> No.11654357

Never mind, nope.
Just [-1, 1].

>> No.11654359

No it does not converge except in [-1,1]. Check out https://socratic.org/questions/what-is-the-interval-of-convergence-of-sum-x-2-n-n-2-1

>> No.11654362

The alternating series test shows it converges on [-1,0), and comparison test on the rest

>> No.11654486

can I get a yay/nay on this >>11651574 (You) as an answer to >>11648225 (You)

>> No.11654494

>Could you explain?
[math]n + \frac{n(n-1)}{2} = \frac{n(n+1)}{2} [/math]

>> No.11654502
File: 304 KB, 344x400, cbd6d235b35fef1464efc67859615219-imagegif.gif [View same] [iqdb] [saucenao] [google] [report]

Can I read Jackson EM instead of reading an undergrad EM book? I know real analysis, abstract algebra, etc. Probably need to brush up on some math methods like PDE stuff but should I do an undergrad book first? I fucking hate Griffiths.

>> No.11654528

Anime posting anon. Where are you???!

>> No.11654544

can you solve this equation without using an algorithm? i honestly forgot if there was a trick for equations like this

>> No.11654564

there's no exact solution for that particular one
best you can do is approximate with a computer

>> No.11654569

my ti says its 1.351993495854

>> No.11654590

You can then raise both sides to the seventh power like a retard to obtain a fourteenth degree polynomial.
In essence, no, it's numerics or nothing.

>> No.11654592

i see. is there any reason why not?
i did a quick excel approximation and it gave me 1.3519... too

>> No.11654594


>> No.11654596

My bad, only seventh.

>> No.11654603
File: 89 KB, 551x696, 1581212455815.jpg [View same] [iqdb] [saucenao] [google] [report]

based. checked.
>if this formula uses RMS wouldn't it be power out rms?
RMS power and average power are the same, that's the point. Instantaneous power is [math] p(t)=v\cdot i=V\cos(\omega t)\cdot I\cos(\omega t+\phi)=...=\frac{1}{2}VI\cos\phi +\frac{1}{2}VI\cos(2\omega t -\phi) [/math]. Now over a full period the average (real) power is [math] P=\frac{1}{T}\int_0^t p\text{ d}t=...=\frac{1}{2}VI\cos\phi=V_{rms}I_{rms}\cos\phi=\frac{1}{2}\mathfrak{Re}\{\mathbf{V}\mathbf{I}^*\} [/math] with bold being a phasor and * being conjugate.
you've really got to be patient
From White we've got for a planar surface:
>pic related
Where the Ixx and Ixy are moments and products of inertia of the planar lamina about the center of mass; hCG is the depth of the center of mass, theta is the angle from the horizontal; x and y axes originate from center of mass. For curved surfaces,
>The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the component
>The vertical component of pressure force on a curved surface equals in magnitude and direction the weight of the entire column of fluid, both liquid and atmosphere, above the curved surface.
Idk if center of pressure makes sense on a curved surface.

>> No.11654624

i often find 2hu-kun and furry-kun to be more helpful than my professors

>> No.11654657

So assuming a pure sine wave, to go from rms power to pk-pk power would be *2√2?

>> No.11654677

>is there any reason why not?
It's usually mathematically impossible to solve equations of degree 5 or higher in terms of roots and stuff. If you substitute u = x^(1/7) into your equation you get a polynomial of degree 7 that you'd have to solve, and that particular polynomial can't be solved.

>> No.11654692

>It's usually mathematically impossible to solve equations of degree 5 or higher in terms of roots and stuff
ive never heard of this before. what numeric category do these solutions fall into? theyre obviously algebraic but theyre different from something like [math]\sqrt{2}[/math]?

>> No.11654698

how decadent

>> No.11654703
File: 311 KB, 750x736, nea.t.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.11654718

The roots are still algebraic, but unlike something like [math]\frac{\sqrt{2+\sqrt{3}}}{2}[/math] it's provably not always possible to write them using only nth roots and arithmetic. You have to start introducing more complicated things to get a numeric expression for them.

>> No.11654726

Basically you need to be really careful with how you talk about AC power. Complex power is [math] S=\mathbf{V}_{\text{rms}}\mathbf{I}_\text{rms}^* [/math]. Active power or real power is the intuitive average power that you think about, and is the real part of S. If the phase between voltage and current is zero (this is really important), only then can you say that rms real power is half that of pk-pk real power. Its a factor of 2, not 2sqrt2.

>> No.11654728

so what is the name of the category for numbers that can be expressed with radicals?

>> No.11654735

I don't believe there's a specific name for them, other than "numbers expressible by radicals".

>> No.11654754
File: 322 KB, 460x690, Houjou.png [View same] [iqdb] [saucenao] [google] [report]

>numbers expressible by radicals
i always thought that all algebraic numbers were expressible by radicals
thanks for teaching me something

>> No.11654762 [DELETED] 

Element of the perfect closure, innit?

>> No.11654785

Related idea, but it's not the same thing. The "perfect closure" of Q would just be Q; Q is already a perfect field.
The idea of a perfect closure only does anything in characteristic p, and even then you build it by adjoining all pth-power roots, not all roots.

>> No.11654832

Hmm... Well my professor wants me to calculate the power in rms as well as peak to peak, but the textbook doesn't explain any of that. I can calculate the rms, would I really just double the value for peak to peak? I suppose it makes sense, I used the peak values to calculate the VCEQ and IC(sat). Thanks Anon! I know it is a gross oversimplification but the whole class is a gross oversimplification.

>> No.11654861

>I can calculate the rms, would I really just double the value for peak to peak
I think so? I'm not an EE and don't get everything they do, but from my understanding that's what he wants. yw~

>> No.11655023

doesnt converge uniformly on [math][1,\infty)[/math], right?

>> No.11655143
File: 357 KB, 1080x1875, Screenshot_20200509-192530__01.jpg [View same] [iqdb] [saucenao] [google] [report]

How would I go about find r'e by using fc1? I get as far as finding Rin, but doesn't seem useful since without beta I can't pull out r'e. Is there are way to solve for the beta?

>> No.11655168

Pretty sure it does

>> No.11655247

youre right, im retarded and i fucked the subtraction of [math]f_{n}(x)-f(x)[/math]

>> No.11655416 [DELETED] 
File: 167 KB, 1080x1344, Screenshot_20200509-215435__01.jpg [View same] [iqdb] [saucenao] [google] [report]

Can I just use this to calculate the gain? Something seems fishy.

>> No.11655539

is 35 too old to learn math?

>> No.11655738

No but it probably won't come intuitively

>> No.11655854

Is there any paper that discusses night cold shower to rheumatoid arthritis development? Whether there's no correlation or something?

>> No.11655883

How do I write 4a^3 + a^6 + 1 as a product?

>> No.11656071

i did a coursera course few months ago and it gave me matlab online license that should've expired on 27.04, today i went on matlab out of curiosity and i still can use it normally
have i won?

>> No.11656080

First you determine the roots of the polynomial.

>> No.11656122

I can't wrap my head around the Whitehead products signs. If [math]\alpha\in \pi_n(X), \beta\in \pi_m(X)[/math], why is [math][\alpha, \beta] = (-1)^{mn + m+ n}[\beta, \alpha][/math]? More precisely: where does the m+n come from?

>> No.11656148

In a bridge circuit, what does the current through the middle "bridging" resistor tend towards as the resistance in that resistor approaches 0? I know it increases intuitively but not what the limiting factor is

>> No.11656165
File: 237 KB, 1671x1600, 1585385099267.jpg [View same] [iqdb] [saucenao] [google] [report]

whats a good site or book to study some maths? like first year uni level.

>> No.11656205
File: 39 KB, 687x187, this one is from Whitehead's Elements of Homotopy theory.png [View same] [iqdb] [saucenao] [google] [report]

I recall someone asking the exact same before you.
Specifically, he was asking why didn't the old Whitehead paper and Fuchs-Fomenko have the same interchange theorem.
Anyhow, you just double check all the definitions and compute the degree of the switch mapping.

>> No.11656217


>> No.11656225

So I keep getting mixed messages about quantum entanglement. Can someone tell me where I have this wrong/what I am missing?

1) If two particles are entangled, their states become correlated such that if, say, the spin state of one of them is measured, then we can confidently say we know the spin state of the other, regardless of the distance that separates the two particles.

2) quantum entanglement CANNOT be utilized to transmit information faster than the speed of light.

3) measuring any property of a particle unavoidably and irreparably alters the state of that particle. In other words, if you measure the spin state of an electron, in so doing you also change the spin state of that electron.

(this is where I start to run into problems with understanding)

4) If 3 is true, than if we have a pair of entangled particles and we measure the spin state of one of them, would that not unavoidably and irreparably alter the spin state of the other in the process such that the new, altered state of the second particle has no correlation with whatever state it may have previously held? And if that is the case, how is it that we know that it ever was correlated with the state of the first particle if we can only measure one of them before altering the states of both?

>> No.11656241

holy shit I need to know as well.

>> No.11656282
File: 3 KB, 146x45, 1568869806196.png [View same] [iqdb] [saucenao] [google] [report]

I've been trying to solve this for 2 hours using various tricks but they're leading nowhere and I'm failing.
Can someone help me how the fuck do I solve this?

>> No.11656287

Nvm my figured it out myself.

>> No.11656291

That depends if you're willing to pay money or not.

>> No.11656294

>pay money
LOL faggot

>> No.11656298

Multiply the top and bottom by n^-12

>> No.11656313

If he's just after something free there's a million different free YouTube channels that can walk someone through 1st year uni math, and there countless text books that can be found on the net.

I assumed he was asking for more than just a list of a billion textbooks thrown at him.

>> No.11656315

I recall George Green being around that age.
[math]b = a^3[/math]
Remove the one because it adds nothing, raise the entire thing to the twelfth power and compute the leading coefficients of the numerator and denominator.

>> No.11656343
File: 40 KB, 700x500, 1560895929059.png [View same] [iqdb] [saucenao] [google] [report]

I got pic related, not sure what to do now
Am I retarded?
>Remove the one because it adds nothing, raise the entire thing to the twelfth power and compute the leading coefficients of the numerator and denominator.
That's gonna take me the entire day, it's supposed to be an easy high school problem.

>> No.11656352
File: 152 KB, 425x496, __fujiwara_no_mokou_touhou_drawn_by_shangguan_feiying__2aac5eb5f9f00cee6a50e49e9c86a27b.jpg [View same] [iqdb] [saucenao] [google] [report]

>that's gonna take me the entire day
You might be retarded, it takes thirty seconds.

>> No.11656363

Maybe, but that's still supposed to be a high school problem, the solution is supposed to be simple and should involve just manipulating the expressions.
I do want to see you compute it in 30 seconds though.

>> No.11656423

He's exaggerating; it doesn't take 30 seconds. Just by looking at it you should've known its [math]\sqrt[12]{3}[/math]. Fucking idiot.

>> No.11656430

I know it but I need to prove it, I can't just say "well duh the top highest power is 9/12 and bottom is 3/4 so they're the same blah blah".

>> No.11656442
File: 17 KB, 1161x325, fuck my teachers in their asshole.png [View same] [iqdb] [saucenao] [google] [report]

If i have lets say N balls and 3 holes, A; B and C. What are the cances of a balls landng in A, b in B and c in C if a+b+c=N?? i have been told the chances are pic related but it makes no fucking sense because if i do the process in different order it gives a different result. You can notice the mechanism i been told very easyly, so as you can realize it is WRONG

>> No.11656444

That's literally the high-school argument.

>> No.11656446

I know that if a N dimensional matrix has N distinct eigenvalues then the corresponding eigenvectors are a basis, is there a similar criterion for infinite dimensional operators?
My books finds for an operator eigenvalue 0, k and -k, and somehow it says that the corresponding eigenvectors are a complete set

>> No.11656461

Dude, it's a problem from a high school book and you're supposed to solve it simply by manipulating the expressions a little. It shouldn't take more than 5 manipulations, almost every single previous problem in the book had something to do with (a-b)(a+b)=a^2-b^2 and that was ir. If you can't see how it's solved then shut the fuck up, you're as clueless as me.
Try writing the high school argument you're suggesting on a high school test and let's see your score. Fucking PROVE it.

>> No.11656484

the a in that far right term should be a b

>> No.11656485

first there are N choose a ways for a balls to fall in A
second there are N-a choose b ways for b balls to fall in B, notice that the balls which can fall into B are the balls which didn't fall into A
and then the same for C, there are remaining N-a-b choose c ways for c balls to fall in C

your equation is close though, it's the same if you change N-a-c for N-b-c, you're just approaching it with starting with the hole B but the probability stays the same

>> No.11656488

Yeah sorry my point still stands. not the same
>but the probability stays the same
nope. try it out yourself with some values

>> No.11656494

uhm actually it is the same. I must have done some mistake before while calculating. now the method makes sense. Weird cause i tried it multiple times....

>> No.11656497

it is the same and must be the same, you're just calling the A hole B, the B hole C and the C hole A
what values are you trying?

>> No.11656500

what exactly do you mean by probability? both of those expression evaluate to integers

>> No.11656502

i fucked up a couple times while trying. now it is the same

>> No.11656505

yeah, you know, the probability it is that number divided by 3^N

>> No.11656510

Isn't that the multinomial distribution?

>> No.11656515

I'm not really sure if you're just bitching or actually don't know the theory.
[eqn]\lim_{n\rightarrow+\infty}\frac{\sqrt[12]{3n^9+3}+1}{\sqrt[4]{n^3+2}} = \sqrt[12]{\lim_{n\rightarrow+\infty}\frac{3n^9+3}{(n^3+2)^3}}+\lim_{n\rightarrow+\infty}\frac{1}{\sqrt[4]{n^3+2}} = \sqrt[12]{\lim_{n\rightarrow+\infty}\frac{3+\frac{3}{n^9}}{(1+\frac{2}{n^3})^3}} = \sqrt[12]{3}[/eqn]

>> No.11656549


>> No.11656645
File: 46 KB, 640x360, image.jpg [View same] [iqdb] [saucenao] [google] [report]

No problem. Now tell me what are you?

>> No.11657091

I have a function and a 3rd degree Taylor approximation of it. I'm asked to estimate the accuracy of it when x lies in a certain interval, but jesus christ I'm having a hard time figuring this out. What's the process look like for this?

>> No.11657108

is this as simlpe as plugging in both extremas of the interval and finding their difference?

>> No.11657132

one more question if y'all have the time, sorry I'm super fucking stressed rn:

I have a function and a 2nd degree taylor approximation for it, I'm asked to find the values for x in which the the error is less than a certain amount, how would you begin to structure a solution here? I have T_2 - F(x) < |error|, but I'm not really sure how to solve it for certain x yet.

>> No.11657158

When you're asked questions about the accuracy of the Taylor approximation you're generally supposed to maximize the Lagrange form of the remainder

>> No.11657223

The actual function I'm dealing with is [math]f(x)=(1+x^2)^{\frac{2}{3}}[/math], so it's a case of the Maclaurin of a binomial. The interval is [0.7, 1.3].

Am I wrong to think it's as simple, in this case, as just taking the maxmum value of [math](1+x^2)^{\frac{2}{3}} - T_3[/math] for x=0.7 and x=1.3?

I've read the wikipedia article but I just can't seem to apply it right now to get through this problem, or the other one. I get -50IQ when stressed out unfortunately (and my normal IQ is roughly room temp).

The other question I'm even more at a loss, I have T_2 expansion of e^x and I'm supposed to find x values for which the |error| is less than 0.001.

>> No.11657234

[math] f(x)=(1+x^2)^{\frac{2}{3}} [/math], so it's a case of the Maclaurin of a binomial. The interval is [0.7, 1.3]. Am I wrong to think it's as simple, in this case, as just taking the maxmum value of [math](1+x^2)^{\frac{2}{3}} - T_3 [/math]

>> No.11657339

sorry, not a maclaurin,it's evaluated at a=1.

>> No.11657346
File: 370 KB, 1080x2220, Screenshot_20200510-130434_Desmos.jpg [View same] [iqdb] [saucenao] [google] [report]

What is the answer to the summation of 3x-4, starting at x=3 and ending at x=7, and why does my text book say it's 115 and not 55?

>> No.11657384

Can someone walk me through this so I can check my work? Approximate [math]f(x)=(1+x^2)^{\frac{2}{3}}[/math] with [math]T_3[/math] about [math]a = 1[/math]. Use Taylor's Inequality to estimate the accuracy of the approximation [math]f(x) \approx T_3(x)[/math] when x lies in the interval [math] [0.7, 1.3] [/math].

The derivatives are really ugly for this, and my Taylor representation differs from Wolfram Alphas:

This shit is just not clicking for me, I also posted up here >>11657091, >>11657108 so you can ignore those posts now.

>> No.11657728

> If 3 is true, than if we have a pair of entangled particles and we measure the spin state of one of them, would that not unavoidably and irreparably alter the spin state of the other in the process such that the new, altered state of the second particle has no correlation with whatever state it may have previously held?
Yes (at least in the "change state upon collapse" interpretation) but I don't see where you have an issue with that.

>how is it that we know that it ever was correlated with the state of the first particle
By communicating after the fact.

Your mom, on her vacation in Germany, bakes a cherry cooky and a banana cooky. She tells you on a Skype call. She know you and your brother both like cherry. She loves you both, so she let luck decide who gets witch.
She puts both cookies in an envelope and gives them to the hotel manager. Next day she asks for them back, she now doesn't know which envelope holds which cooky. She sends one of them to you and one to your brother.
Some time later, your cooky arrives. You open the envolope and as you see the color of your cooky, you know what cooky your brother got.
You didn't know what you got before you opened the envelope, but you knew that as soon as you found out, you would know also what your bother got.

(The analogy is classical and doesn't match the QM situation insofar as the notion of probability in QM is backed into the ontology of it. In the QM situation, the state (flavor of your cooky) wouldn't even be thought of as determined before the envelope is opened. I.e. in the cooky situation, as soon as the envelope was sent, the outcome was already fixed, which it isn't in the QM case)

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