/sqt/ /qtddtot/ - stupid questions thread

[Anonymous No.15972871]

Previous thread: >>15951180

>what is /sqt/ for?
Questions regarding maths 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 search for proofs?
proofwiki.org
>where can I look up if the question has already been asked here?
warosu.org/sci
eientei.xyz/sci
>how do I optimize an image losslessly?
trimage.org
pnggauntlet.com
>how do I find the source of an image?
images.google.com
tineye.com
saucenao.com
iqdb.org

>where can I get:
>books?
libgen.rs
annas-archive.org
stitz-zeager.com
openstax.org
activecalculus.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
>online courses and lectures?
khanacademy.org
>charts?
imgur.com/a/pHfMGwE
imgur.com/a/ZZDVNk1
>tables, properties and material selection?
www.engineeringtoolbox.com
www.matweb.com
www.chemspider.com

Tips for asking questions here:
>attach an image (animal images are ideal, you can grab them from >>>/an/. Alternatively use anime from safebooru.donmai.us)
>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.15973047]

>>15972871
Please >>15964979 >>15968058

[Anonymous No.15973151]

>>15972774
Still need help on this. I keep doing practice problems but this one skill doesn't seem to get any better. Is it just shit spatial reasoning?

[Anonymous No.15973249]

>>15972476
But I still can't quite use the normal integration by parts because I'm dealing with functions [math]\mathbb{R^n} \rightarrow \mathbb{R}[/math]. If I could use Fubini's theorem then that could maybe work but the domain isn't necessarily a cartesian product so I can't see how to split it up.

[Anonymous No.15973465]

Information processing.
I realize why I fucking hate literary classes like history, philosophy, and english/writing, besides being very boring, and I will explain the exceptions.
It is because it is just word salad. It is nothing but barf and we're expected to remember every word in exact order. I'll prove it actually: suppose there are n words and m sentences. If each sentence has information that is necessary then we MUST remember all m sentences and n words in those sentences, thus we are demanded to remember n words in sequence. If I asked you to remember a sequence with no defining function for n terms you would struggle to remember it.
Therefore it is wrong and stupid because it simply cannot be the case to remember n words for p pages of q books. This is because if we are demanded to recite information it is then to demand the recitation of all such words.
So from a grade point of view only the sentences with informations to finite questions less than m are important. However again we cannot know this distribution, thus it is therefore not random and only a subset of m sentences matter meaning the information is both pedantic and misleading i.e. destructive to memory. If memory store x information with x < m*p*q then we are intentionally being abused and written information can thus be reduced. If it is not reduced then we cannot predict the number of ideas less than m to remember therefore we will remember then wrongly at a probabilistic distribution.
The exception are stories because stories are actually connected and meaningful, and we don't need to remember exactness to know a story, it can be reduced and optimized with little loss.

[Anonymous No.15973577]

if axb + bxc + cxa=0 how do I prove a,b,c are complanar

[Anonymous No.15973603]

>>15973577
Do the dot product of the LHS with any of the three vectors. The result is the determinant of the matrix whose columns are a,b and c. Since this determinant is zero, those vectors are linear dependent.

[Anonymous No.15973618]

>>15972871
suicide suicide suicide

[Anonymous No.15973631]

I'm reading through Roman's advanced linear algebra, and I got stuck.
Basically, [math] B [/math] is a set s.t. [math] \rvert B \rvert = \kappa [/math], [math] b\in B [/math] and [math] (F^B)_0 [/math] is the vector space of all functions with finite support from [math] B [/math] to a field [math] F [/math].
Recall that finite support of a function [math]
g:\mathcal{I}\to F [/math] is defined as [math] \mathrm{supp}(g) = \{i\in\mathcal{I}\,\rvert\, f(i)\neq0\} [/math].
Then he says that the functions [math] \delta_b [/math] defined as
[eqn]
\delta_b(x) =
\begin{cases}
1\quad x=b\\
0\quad x\neq b
\end{cases}
[/eqn]
form a basis of [math] (F^B)_0 [/math]. I get the linear independence part, but how is every function [math] f\in(F^B)_0 [/math] in the span of all [math] \delta_b [/math]'s?

[Anonymous No.15973636]

>>15973631
[eqn]f(x) = \sum_{i \in \text{supp}(f)} f(i) \delta_i (x)[/eqn]
Where is the problem?

[Anonymous No.15973644]

>>15973636
Oh, well, that makes sense. I'm not particularly intelligent.
Thanks anon.

[Anonymous No.15973883]

why does lim x->1+ sqrt(3 - 3x) = 0?

since x approaches from the right, that means 3 < 3x, so 3-3x should give a negative number, and you cant take the square toot of a negative nunber

[Anonymous No.15973889]

>>15973883
3-3x=3(1-x)

[Anonymous No.15973895]

>>15973889
ty anon but doesnt lim x->1+ mean x is equal to something like 1.000...01

so then it is sqrt(3-3.000...01) = sqrt(-000...01)?

[Anonymous No.15973902]

>>15973883
>you cant take the square toot of a negative nunber
Wrong, I can do this.

[Anonymous No.15973935]

>>15973895
first of all, 1.000...01 is not a number; this is the kind of nonsense the people who say 0.999... != 1 try to pull. second, that is not quite what lim means. lim x->1+ 1-x is obviously 0. if the limit of a square root of a negative gives you conniptions, consider that sqrt(1-x) = isqrt(x-1). what is i times 0?

[Anonymous No.15973938]

>>15972871
>>how do I optimize an image losslessly?
Why is this in the FAQ? Was it actually frequently asked?

[Anonymous No.15973944]

>>15973935
oh i didnt know that sqrt(1-x) = isqrt(x-1)

thank you anon this is very helpful

[Anonymous No.15973949]

>>15973938
The pasta is a few years old by now, and was based on /g/'s sqt blurb. I imagine the question comes from that.

[Anonymous No.15973969]

I am trying a online solver to figure out why the two limits in the image both arent "DNE"

they seem the same to me, i get 2/0 for the first by substituting x=2, and -3/0 for the second by substituting x=-1.

but the answers it gives for both are not expected. why are these answers not both DNE?

[Anonymous No.15973970]

>>15973969
sorry my screen snip program isnt working right

[Anonymous No.15974004]

>>15973969
>>15973970
DNE means a different sign infinity when you approach the limit from each side. That isn't true for the second equation since both signs are negative.

[Anonymous No.15974137]

Let [math]\mu = \frac{dx dy}{y^2}[/math] be the standard measure on the upper half plane [math]\mathbb{H} = \{x + iy \mid y > 0\}[/math]. Fix an open set [math]U \subset \mathbb{H}[/math] which is contained in [math]A \times (\delta, \infty)[/math] for some bounded set [math]A \subset \mathbb{R}[/math] and [math]\delta > 0[/math], and thus has finite measure.

Given [math]\epsilon > 0[/math], can we always find a finite collection of pairwise disjoint hyperbolic triangles [math]T_1 \dots T_n[/math] such that [math]\mu(\bigcup_{i=1}^n T_i \triangle U) < \epsilon[/math]?

[Anonymous No.15974269]

Can you calculate the determinant of a matrix by only adding the multiples of rows?

[Anonymous No.15974439]

>>15974269
No, but you can calculate it by adding, multiplying, and taking traces
https://en.wikipedia.org/wiki/Faddeev%E2%80%93LeVerrier_algorithm

[Anonymous No.15974639]

have i been psyopped into believing in freud from a psychological standpoint? on a neuroscience level is freud debunked? i know he modified his theories to appease the field but opinion seems to be that his theories ended up bastardized to accommodate other fields.

[Anonymous No.15975382]

>>15974269
A matrix whose diagonal elements are all 1 and non-diagonal elements are all 0 except for a single element has determinant 1, regardless of the value of the single non-zero off-diagonal element. The "elementary row operations" used to reduce a matrix to row-echelon form correspond to pre-multiplication by such matrices, so leave the determinant unchanged. The determinant of a diagonal matrix is the product of the elements on the main diagonal. IOW, you can calculate the determinant of a matrix by row-reduction to a diagonal matrix then finding the product of the diagonal elements.

[Anonymous No.15975466]

Using mesh current method there are two possible equations one can write here, given that [math]I_2=I_g=0,06A,E_1=12V,E_2=E_3=24V,R_1=300\Ohm,R_3=600\Ohm,R_4=600\Ohm[/math]:
[eqn]
900\Ohm I_1 + 600\Ohm \cdot 0,06A = 12V \\
600\Ohm I_1 + 1100\Ohm \cdot 0,06A = 0V
[/eqn]
But from the first one I get [math]I_1=-0,0267A[/math] and from the second one [math]I_1=-0,11A[/math]. Where am I going wrong?

[Anonymous No.15975473]

Using mesh current method there are two possible equations one can write here, given that [math]I_{II}=I_g=0,06A,E_1=12V,E_2=E_3=24V,R_1=300\Omega,R_3=600\Omega,R_4=500\Omega[/math]:
[eqn]
900\Omega I_I + 600\Omega \cdot 0,06A = 12V \\
600\Omega I_I + 1100\Omega \cdot 0,06A = 0V
[/eqn]
But from the first one I get [math]I_I=-0,0267A[/math] and from the second one [math]I_I=-0,11A[/math]. Where am I going wrong?

[Anonymous No.15975616]

>>15975473
You are retarded and need to review the chapter.
i2 =/= ig

[Anonymous No.15975620]

Let [math]\mathbf{0}, \mathbf{1}[/math] be the constant 0 and constant 1 sequences. Define [math]\tau: \{0,1\}^\mathbb{N} \to \{0,1\}^\mathbb{N}[/math] by [math]\tau (\overbrace{1 \dots 1}^{n-1 \textrm{ times}}, 0, x_{n+1}, x_{n+2} \dots) = (\overbrace{0 \dots 0}^{n-1 \textrm{ times}}, 1, x_{n+1}, x_{n+2} \dots)[/math] and [math]\tau(\mathbf{1}) = \mathbf{0}[/math]. Define also [math]\phi: \{0,1\}^\mathbb{N} \to \mathbb{Z}[/math] by [math]\phi(x) = \min \{ n \mid x_n = 0 \} - 2[/math]; we leave [math]\phi(\mathbf{1})[/math] undefined.

Is there an explicit formula for [math]\phi \circ \tau[/math] and [math]\phi \circ \tau^{-1}[/math]?

[Anonymous No.15975634]

>>15975616
It literally says so.

[Anonymous No.15975636]

>>15975616
>>15975634
In fact, the first equation gives the correct solution. I err somewhere in setting up the second one.

[Anonymous No.15976128]

>>15974137
I should mention that I know pretty much no hyperbolic / differential geometry or whatever specific topic your question comes from, so treat the following as just some comments.
If I'm correct, however, it seems the following are true:
There exists some [math]K>0[/math] such that [math]\mu(A\times[K,\infty))<\varepsilon[/math], and furthermore, [math]\mu[/math] restricted to the set [math]X:=A \times(\delta,K)[/math] is a finite measure which is absolutely continuous with respect to Lebesgue measure.
This latter fact means that for any [math]\varepsilon>0[/math] there exists some small [math]c>0[/math] such that for any measurable set [math]A[/math], [math]\lambda(A)<c[/math] implies [math]\mu(A)<\varepsilon[/math].
Since also [math]\lambda(X)<\infty[/math], you can find disjoint rectangles such that their symmetric difference with [math]U[/math] is less than [math]c[/math] (this result is pretty standard, I can prove it if you haven't seen it either), so that with absolute continuity you also have this approximation for mu.
(You can cut rectangles in half to create triangles (?) I'm not sure what a hyperbolic triangle is, precisely).

[Anonymous No.15976205]

can anyone tell me how in the green tick answer

[Anonymous No.15976207]

>>15976205
the part starting
a^4b^4=.... is true?
https://math.stackexchange.com/questions/1073870/for-an-ellipse-with-minor-radius-b-show-that-the-product-of-distances-from-th

[Anonymous No.15976538]

>>15972871
why do i get 2 different answers?

ln(2x^(2x)) = 2x*ln2x = 2x*(ln2 + lnx) = 2xln2 + 2xlnx

ln(2x^(2x)) = ln2 + lnx^(2x) = ln2 + 2xlnx

[Anonymous No.15976544]

>>15976538
>ln(2x^(2x)) = 2x*ln2x
There's your problem.
By doing this, you're acting as though the 2 is also being raised to (2x)

[Anonymous No.15976560]

>>15976544
why does this trick work anyways?

log(x^2) = 2logx

[Anonymous No.15976643]

>>15976560
It's not a trick, it's a property of logarithms: https://www.chilimath.com/lessons/advanced-algebra/proofs-of-logarithm-properties/

[Anonymous No.15976841]

How do I find a library that's subscribed to the particular journal I want? There's this one ACS Catalysis article I want that Sci-Hub doesn't have, and I emailed the author but they just sent me a manuscript(probably doesn't matter in the end but it's just ugly to me).

[Anonymous No.15976843]

>>15976841
Disregard this, just realized I could use Worldcat for articles too.

[Anonymous No.15977679]

Is there a textbook that specifically teaches you the algebra manipulations that are taken as "obvious" in higher level maths proofs?

i'm talking about the steps that are skipped and leave me sitting there for an hour trying to figure out how they got from one line to the next.

[Anonymous No.15977761]

>>15972871
How many rubs does it take for a piece of felt to lose all of its electrons when rubbing and discharging a charged rod?

[Anonymous No.15977985]

>>15972871
I need to do so chem stuff for my fucking math modele...
I want to solve the stoichiometric equation in pic rel, and find the values of [math] a,b,c,d[/math]. I am working under two hypothesis :
(i) Nitrogen is assumed not to change the equation state and, does not contribute to uniquely define the equation system.
(ii) b is assumed to be known.

Using (i), I write out the following linear system :
[math] \begin{align} 6 &= b+c \\
12+0.45b &= 1.79b+2d \\
6+2a&=0.57b+2c+d
\end{align}[/math]
which is pretty straightforward to solve using (ii).
Can one tell me if my formulation of the above system correct please?

[Anonymous No.15977989]

>>15977985
forgot to mention, first eq is for [math] C [/math], second is for [math] H [/math] and last for [math] O [/math].

[Anonymous No.15978029]

>>15976205
>>15976207
The equation [math]\epsilon[/math] at the start of the answer gives [math](b^2 x_0) x+(a^2 y_0)y -a^2b^2=0[/math] for any [math](x,y)[/math] on the tangent and for any [math](x_0,y_0)[/math] on the ellipse.
For the intersection point, take [math](x,y)=(x_0,y_0)[/math] since it is on both and if you re-arrange and multiply both sides by [math]a^2b^2[/math] you get the result.
>>15977679
I think basic mathematics by serge lang does this, though I've never read it.
>>15977985
Looks good to me.

[Anonymous No.15978033]

>>15975473
Your equations look right but you have two equations and only one unknown. Probably you either copied down something wrong or you have a retarded professor.

Note that since E_2 and E_3 cancel each other you would expect I_1 to be positive and I_g to be negative, so that's further evidence you just copied something down wrong

[Anonymous No.15979450]

bump

[Anonymous No.15979676]

>>15972871
WHY IS FLUORINE THE MOST ELECTRONEG ELEMENT
I have a quiz in an hour hepl me :(

[Anonymous No.15980008]

>>15972871
Hello /sci/, I'm stupid so I can't figure this out. I also tried using chatgpt but it couldn't help me, even though I tried explaining for like half an hour.
I want to make a RPG game, where my character starts with 120 HP at level 1.
For each level, my character gains 45 HP, which increases by additional 45 at every 5th level.
So at level 5 my character should have 290 HP, and at level 6 my character should have 380 HP.
However, I also want my character to have 44955 HP at level 100.
So I need an additional modifier to my HP formula, which increase exponentially (Chatgpt thought it could just add 14055 HP at level 1 and be done with it, that lazy nigger)
Please post a formula that helps me achieve my goal.

[Anonymous No.15980016]

>>15980008
>So at level 5 my character should have 290 HP
120+45+45+45+45= 300, not 290

It's no wonder chatgpt couldn't parse your nonsense

[Anonymous No.15980027]

How to draw this on the real line?

[Anonymous No.15980028]

>>15980008
You are not a real programmer unless you use a switch statement to solve this.

[Anonymous No.15980035]

>>15980016
Oh yeah you're right lol, like I said im stupid

[Anonymous No.15980039]

>>15978029
thanks.

[Anonymous No.15980170]

>>15980027
Anyone? Pls

[Anonymous No.15980312]

Why do laser photons copy the the properties of the stimulating photons? What makes it physically unfavorable for it to shoot off in some other direction?

[Anonymous No.15980375]

>>15980312
There isn't a simply answer to that unfortunately. To understand the why you would have to get into some complex and quite gnarly QED. You really just have to accept that when an incoming photon of the correct energy interacts with the excited state this results in decay process that is affected by the em-field of the original photon resulting in a new photon with the same quantum properties.

[Anonymous No.15980386]

>>15980375
Ah, that's too bad. I thought maybe it worked similarly to pair production and could be explained in those terms(phenomenon occurs the way it does because occurring any other way causes a simple physics violation).

[Anonymous No.15980398]

>>15980386
Think of it as more of a resonance than they are physically unable to do anything else. The likelihood at the precise, correct energy is approaching 1.

[Anonymous No.15980462]

>>15975634
This is strange, I think you are overfactoring Ii.
Redraw the circuit.

[Anonymous No.15980634]

If dark energy expands, it could tear the fabric of space-time
so what happens if it does?
I can't really ask what we'd find, what would be out there. If we tear a hole in the fabric of reality, we don't know what if any physics would exist outside of our universe, is there even an "outside" of our universe if you don't have any space or time outside of our reality?
Can someone make some sense of this to me? Can we know anything about what lies outside of our universe or is it fundamentally impossible to know? Even if dark energy tears a hole in space time, could we identify such a tear? Would it be physically or mathematically possible to detect and measure the nature of such a tear? Can we begin to speculate if it would have any properties, even if they're ones totally foreign to our universe?

[Anonymous No.15980655]

>>15980634
> it could tear the fabric of space-time
What pop-sci bullshit told you that? No such thing would happen. When space expands you just get more space. That's it, it is not some physical thing that can tear.

[Anonymous No.15980662]

>>15980655
Okay, so it would just be more empty space-time then filling it all in? You're saying there's no limit to how much reality can be stretched, and that space-time can be expanded infinitely by dark energy without breaking?
Not trying to be argumentative I'm just asking if I've understood you right.

But given that, if we assume a scenario where the rate of dark matter expansion increases which results in a "big rip" end to the universe, what happens beyond when subatomic particles are torn apart?
According the the theory the expansion would tear ever smaller structures apart, so what happens when it reaches the point where according the the math it should tear apart things smaller than the planck length? That's not possible, so does it just stop expanding at that point or do the physics break?

[Anonymous No.15980664]

>>15980634
>Can someone make some sense of this to me?
https://youtu.be/sDD_VcxmToE

[Anonymous No.15980667]

>>15980662
> space-time can be expanded infinitely
correct

> what happens beyond when subatomic particles are torn apart?
nothing, that's the end. the "big rip" conjecture is that at some point the rate of expansion becomes so great it pulls apart all atoms. at that point all you'll be left with is each particle in the universe becomes alone and separated by more and more light years of empty space.

[Anonymous No.15980669]

Thanks for answering my question guys :)

[Anonymous No.15980861]

>>15978033
>Your equations look right but you have two equations and only one unknown. Probably you either copied down something wrong or you have a retarded professor.
I just thought that I can use either mesh current to solve the problem. If only the first one will work, how do I know it's the one I should use?
>Note that since E_2 and E_3 cancel each other you would expect I_1 to be positive and I_g to be negative, so that's further evidence you just copied something down wrong
I don't see why [math]E_2[/math] and [math]E_3[/math] cancelling each other implies that [math]I_I[/math] and [math]I_g[/math] have to be opposites and how is that relevant.

[Anonymous No.15980866]

>>15978033
>Your equations look right but you have two equations and only one unknown. Probably you either copied down something wrong or you have a retarded professor.
I just thought that I can use either mesh current to solve the problem. If only the first one will work, how do I know it's the one I should use?
>Note that since E_2 and E_3 cancel each other you would expect I_1 to be positive and I_g to be negative, so that's further evidence you just copied something down wrong
I don't see why [math]E_2[/math] and [math]E_3[/math] cancelling each other implies that [math]I_I[/math] and [math]I_g[/math] have to have opposite signs and how is that relevant.

[Anonymous No.15981095]

Anyone knows a source that goes in depth about techniques of calculating Lebesgue measure for "weird" sets? Fichtenholz and other analysis books I've looked at don't seem to talk about it too much (probably since its a mostly pure math exercise).

[Anonymous No.15981577]

>>15980866
> If only the first one will work, how do I know it's the one I should use?
I'm saying this problem is set up incorrectly, either through your mistake copying it down or through an incompetent professor or TA. You are given too much initial data and it is inconsistent with two equations, as you saw. If you were not told the current I_g, you could use your two equations to solve for both I_I and I_g and everything would be consistent.

>how is that relevant
E_2 and E_3 canceling means you have only one battery in your circuit, E_1. Current flows from the positive terminal of the battery to the negative, so I_g should have the opposite sign. This is more evidence that the problem is set up incorrectly.

[Anonymous No.15981581]

>>15981095
What kind of sets do you consider to be weird?

[Anonymous No.15981771]

>>15980170

[Anonymous No.15981860]

>>15981577
Then I guess I'll just learn how to do it from YT and not from the book. Thnx.

[Anonymous No.15981899]

If I was an organism that lived on a 2d plane, is there any way for me to perform an experiment that would let me prove a 3rd dimension exists?

[Anonymous No.15981913]

>>15981899
Look up

[Anonymous No.15981927]

>>15981913

[Anonymous No.15981950]

>>15981927
https://www.youtube.com/watch?v=aS8S3ZxVWBw

[Anonymous No.15982050]

Just got called Dr. Anon in an email by a student for a class I'm TA'ing. I'm a PhD student and haven't completed my degree. What's the best way to correct students without making things awkward?

[Anonymous No.15982056]

>>15982050
Just ignore it, it's too trivial to even mention in text form without sounding akward. Next time you see him in real life and he uses Dr. again, just joke and say you are still doing PhD, not a doctor yet.

[Anonymous No.15982173]

>>15982050
It's post like this that make me realise you don't have to be smart to do a PhD.

[Anonymous No.15982360]

Whats the first step here? To pass the +26 to the right as negative and then solve the left?

[Anonymous No.15982362]

>>15982360
multiply both sides out
subtract like terms from both sides
etc.

[Anonymous No.15982399]

>>15982362
How?

[Anonymous No.15982411]

>>15982399
[math](x - 2)(x - 1) + 26 < (x + 4)(x + 5) \\
(x^2 -3x + 2) + 26 < x^2 + 9x + 20 \\
x^2 -3x + 28 < x^2 + 9x + 20 \\
x^2 -3x + 28 < x^2 + 9x + 20 \\
8 < 12x \implies \frac{8}{12} < x \implies x \gt \frac{2}{3}[/math]

[Anonymous No.15982423]

>>15982411
Thank you. Is there a reason why you passed everything to the left, like the +20 and the +9X?
What if I pass the +28 as -28 and get -8; and the -3X as +3X and get 6X. Would it be wrong?

[Anonymous No.15982425]

>>15982423
And get 12X*
I mean, is there an order? Or do I choose what to pass to the left and to pass to the right as convenient?

[Anonymous No.15982429]

>>15982423
No reason, answer would still be the same. Though negatives in equalities can be confusing.

[Anonymous No.15982498]

>>15973465
>If memory store x information with x < m*p*q then we are intentionally being abused and written information can thus be reduced

Looks like someone needs a few Grammar classes

[Anonymous No.15982803]

there are given the parametric equations of the lines l1 and l2 [math] \vec{r}=\vec{r}_{1}+t\vec{s}_{1} \text{ and }\vec{r}=\vec{r}_{2}+t\vec{s}_{2} [/math] what is the necessary and sufficient conditon for l1 and l2 to be intersecting lines(so have one intersection point)?
Can anyone answer this? I have an exam tomorrow

[Anonymous No.15983056]

Hello,

My professor wants us to do a logistic regression on a dependent variable that is not qualitative - it's an insurance dataset where the dependent variable is how much a company should charge. How am I supposed to do it?
Split the model into two different values of charges?

[Anonymous No.15983065]

>>15973465
All these word salad and barks per minute just to say that you suck at writing.

[Anonymous No.15983073]

I have a biology question
I was talking with my AI when it said that arteries contract and expand to help move blood along which is surprising since i never heard anything about this when i was taught about the cardiovascular system. It even went on to say that the arteries were able to selectively choose which areas needed more blood and expand/extract accordingly to get more blood to those areas
Is all this true?

[Anonymous No.15983087]

>>15983073
It's primarily a thing with capillaries in particular, but that's why frostbite will preferentially affect your fingers, for example: Your body will deliberately sacrifice them to keep more blood, and thus more warmth, in areas it figures to be more vital.

[Anonymous No.15983097]

>>15983087
Yeah that makes sense, it would be easier to move blood around on the small scales. I knew vessels did expand/contract to control the flow of blood but i thought it was more of a passive thing that just guided the blood, not like mini-hearts everywhere. Wild

[Anonymous No.15983157]

Consider the DEQ [eqn]\frac{dy}{dx} = -\frac{3y}{x}.[/eqn]This is a fairly simply problem right. A solution you'd immediately get when separating variables is [math]y(x) = -3x[/math]. Thing is I want to derive the non trivial-solution [eqn]\frac{1}{x^3}.[/eqn]Separation of variables doesn't work, as always leads to trivial solution. What can I do to compute the non-trivial one?

[Anonymous No.15983161]

>>15983157
[math]\frac{-3(-3x)}{x} = \frac{-9x}{x} = -9[/math]
[math]\frac{dy}{dx} -3x = -3[/math]
how are you getting your claimed solution?

[Anonymous No.15983163]

>>15983161
*9, not -9
I swear I'm only retarded sometimes, but my inability to multiply negative numbers does not change my point

[Anonymous No.15983171]

>>15983161
I did what they did here https://socratic.org/questions/what-is-a-solution-to-the-differential-equation-dy-dx-y-x.. Basically, I used the logarithm as the antiderivative on both the RHS and LHS.

[Anonymous No.15983177]

>>15983171
It's the proper method, at least, but you're not applying it correctly.
[math]\ln y = -3 \ln x + C[/math]
[math]y = e ^ { - 3 \ln x + C }[/math]
[math]y = e ^ { - 3 \ln x } e ^ { C }[/math]
[math]y = e ^ { \ln x^{ - 3} } C [/math], since e raised to an arbitrary constant is a (second) arbitrary constant
[math]y = \frac { 1 } { x ^ 3 } C[/math]

[Anonymous No.15983285]

>>15983177
Oh, now I see my mistake. Thank you.

[Anonymous No.15983685]

So it's my understanding that video cameras work by taking a series of pictures, hence frames per second. Is there such a thing as like a continuous video that is actually recording an actual single video instead of just a series of pictures? Is that possible? Does this question make sense?

[Anonymous No.15983735]

What am I doing wrong?
I'm getting only half of the solution. Where are the two 3 supposed to come from?

[Anonymous No.15983738]

>>15983735
Try plugging in 3 and seeing what value you get.

[Anonymous No.15983755]

>>15983738
I get the same result, no changes

[Anonymous No.15983764]

>>15983755
Let's rephrase that.
What value does your denominator take when x=3 exactly?

[Anonymous No.15983774]

>>15983735
I dont do math, but isnt it -5 since its -15 / 3

[Anonymous No.15983778]

>>15983764
1?
I dont need questions, I need answers dude

[Anonymous No.15983802]

>>15983778
3-3=1?

[Anonymous No.15983809]

>>15983802
0. Sorry, but where are you going with this?

[Anonymous No.15983814]

>>15983809
You're dividing by 0. Your function is undefined at that point, so naturally x=3 is not in the interval on which that relation is true.

[Anonymous No.15983818]

>>15983735
You screwed up the inequality on the right hand side. You implicitly assumed (3-x) is positive by not flipping the sign of the inequality and then you got x>5 which implies (x+3) is also positive. This can't be, since (x+3)/(3-x) is supposed to be negative.

So you must assume x>3 so that (3-x) is negative. This flips the sign of the inequality so that you get 3<x<5.

If you consider that you are also assuming (3-x) is positive on the other side, you see that you really have 9/5<x<3 too.

[Anonymous No.15983872]

>>15983818
Thanks
>So you must assume x>3 so that (3-x) is negative
Got it
>This can't be, since (x+3)/(3-x) is supposed to be negative.
Why?

[Anonymous No.15983911]

>>15983872
Because you suck at following the rules.

Stop smoking weed

[Anonymous No.15983932]

>>>15983872
But in (x+3)/(3-x) only one X is negative and its second, next to the three. All the rest is positive.

[Anonymous No.15983967]

>>15983872
>Why?
Because it is less than -4

[Anonymous No.15984055]

>>15983967
Thanks

[xxXSuiseisekiXxx No.15984228]

>>15974639
Yes. Psychology has no self evident results.
Like the meme of find x.
X is circled.
But that doesn't mean the author was being an smart arse. Could of had just circled it as part of an analytical thought process. Or even just to get the pen working.
They may say that it was a joke, but there's no certainty in self reporting.
Psychological theroy is more an insight to the academic mindset than anything.

[Anonymous No.15984443]

Can a non-interaction effect happen on a graph with a squared/sqrt plot?

(or just "Can non-interaction effect plots happen on graphs other than perfectly parallel lines?". I have a test for tomorrow.)

[Anonymous No.15984518]

Where do you lookup webassign homework answers nowadays? Back then it was Chegg and that shit was behind a paywall.

[Anonymous No.15984566]

>>15983685
The answer to all your questions is no.

[Anonymous No.15984964]

>>15972871
If there's evolution, then why aren't we evolved to have liquids of moderate ph variance not bother the inside of our noses?
Why must it burn when freshwater or vodka enters the nose?

[Anonymous No.15984974]

>>15984964
That's not really something that would help with survival, so there's no pressure in favour of it and it's not something we should explicitly expect to develop.
If anything, it seems like it'd be more of a hindrance from a survival standpoint. A pain response to water up the nose is a good first step in getting people to not drown themselves

[Anonymous No.15985078]

Psychiatry/Psychology question:
What mental disorder has people who get aroused with genital mutilation? Perform GM on other people not on themselves.

[Anonymous No.15985148]

Can anyone solve pic rel?

[Anonymous No.15985168]

How can I stop experiencing delusion of telepathy if even Peterson (after Jung?) believes that our skull and brain is not a boundary of a psyche? If my psyche stretches god-knows-where then it can reach as far as other psyches, right? And they fucking reveal their associated "thoughts" to me when I think about them as if they were separate agents. It's annoying and it makes me paranoid.
I can think in "the old way" but it takes my conscious effort and extra energy. I'm not gonna walk around and force myself to remember that I was supposed to "think in the old way" every time I want to think about someone else or about talking to them.

Does it sound ridiculous? Well, it's been my life in the past 8 years.

[Anonymous No.15986071]

>>15985168
If you cant make money off it, take your meds and go live with your mom for a while, crazy ass negro

[Anonymous No.15986156]

My professor used the following definition:
[math]
S = \{s \in \mathbb{Q}: s \geq \sqrt2\}
[/math]
Is it actually valid to do that? I was confused since square root of 2 is not in Q, can we really use a real number in the definition of a subset of Q?

[Anonymous No.15986164]

>>15986156
Sure, why not? Q is a subset of the real numbers and the s>= sqrt(2) is a subset of the real numbers and S is their intersection.

[Anonymous No.15986181]

>>15986164
I don't know, maybe it's my thinking from programming it felt like mixing data types. Also that set has no minimum, but that's not that weird either, since s > 0 also doesn't have a minimum.
You are right, that the set seems clearly defined, so I guess there really is no problem.

[Anonymous No.15986186]

>>15986181
If you multiply s by square root of 2 you a subset of Q that cant be defined by s, but is s nonetheless

[Anonymous No.15986479]

[math]V+U=V[/math] when [math]V[/math] is a vector space and [math]U[/math] a subspace. Is this true?

[Anonymous No.15986503]

>>15986479
Assuming that you mean [math]U[/math] is a subspace of [math]V[/math] specifically (it's pedantic, but it's importantly pedantic), then yes.
In that case every element in [math]U[/math] is also in [math]V[/math], so adding one to an element in [math]V[/math] is just adding two elements of [math]V[/math] together. Since [math]V[/math] is necessarily closed under addition, the result must also be a subspace of [math]V[/math], and since we can just use the zero vector as our element from [math]U[/math] we can get to any arbitrary vector in [math]V[/math], so our sum is [math]V[/math] itself.

[Anonymous No.15986511]

"Let [math]S\subset V[/math] be a subspace of [math]V[/math]. Then the map [math]f:\{T\,\rvert\,S\subset T\subset V\}\rightarrow \{U\,\rvert\,U\subset V/S\}; T\mapsto T/S[/math] is an order preserving (with respect to set inclusion) bijection."
I'm having a hard time proving the surjective part. Can you guys help me out?

[Anonymous No.15986520]

Anybody know how to only get real results from maxima? Keep getting complex numbers...

[Anonymous No.15986545]

>>15986520
use google you fuckwit.

[Anonymous No.15986551]

>>15986545
I looked and nothing worked, yes yes spoonfeed this and that sneed etc, this is the stupid questions thread, sorry that got you annoyed man I just want a few directions really.

[Anonymous No.15986821]

I thought a monomial had a coefficient (number), a variable and an exponent, so basically one 'term' like this

[math]2x^2[/math]

So why am I reading about multiplying and dividing monomials from this book?

[math]2a^2 * 8ab^5 -> 16a^3b^5[/math]

Isn't this a polynomial?

[Anonymous No.15986892]

>>15986821
monomial means a single term, not a single variable.

[Anonymous No.15986975]

>>15986156
You can always re-write it to use s^2>=2 instead.

[Anonymous No.15986976]

>>15986071
I was never prescribed anything. I was supposed to put my life together and stupid thoughts would "stop emerging". But it's hard to put life together with those thoughts in head. I mean, I spend a lot of time trying to figure out how to stop these thoughts instead of planning my life.

[Anonymous No.15987185]

In pic related, imagine R3 replaced with short circuit. How would you find equivalent resistance then?

[Anonymous No.15987193]

>>15986976
>I was never prescribed anything.
That was a mistake. You need professional help so get some.

[Anonymous No.15987197]

>>15987185
Anything in parallel with a short (0 resistance) is ignored
0 || R = 0×R / (0 + R) = 0

Of course, real wires have resistance, so this is only true in ideal circuit theory

[Anonymous No.15987238]

>>15987197
In my particular case I've reached a position like that after replacing sources with their impedences in order to find Thevanine's resistance. Does this fact change something about your statement? The circuit in my case would be like one in pic related with R3 replaced with a wire and A and B connected. My A would be between R5 and R6 and my B would be in the same position as in the pic ("my A and B" mark where load resistor was). In that case, you see I don't know which part of the circuit to ignore.

[Anonymous No.15987244]

is fake hair made out of keratin a thing yet? asking for a gf (female robot).

[Anonymous No.15987490]

>>15987193
It's a matter of a belief system. If you can't provide me a reliable (non-schizoid whatsoever) theory of the psyche, then I guess I'm stuck with mine. I prefer to fight the cause, not the symptoms. I'll be working on it.

[Anonymous No.15987498]

>>15987490
You belong on /x/ not /sci/

[Anonymous No.15987665]

>>15987498
And you belong in /s4s/ not in /sci/.

[Anonymous No.15987811]

>>15987490
The theory of the psyche is called psychology. There are lots of textbooks for it.

[Anonymous No.15987963]

do people actually read math books designed specifically for mathematicians? basically definition =>theorem =>proof =>corollary in loop without even one explanation of what's going on.
is this what makes a difference between an engineer and a mathematician?

[Anonymous No.15988328]

Is this the future for teaching children science?
https://www.youtube.com/shorts/6rwgGseOl0w
Low key I learned more from this short than my entire time in high school, I didn't take any bio in college because I wasn't a bio major, but still this might be effective.
I've read that manga that's supposed to teach you calculus and honestly it's fucking terrible, so for a long time I though education really could not be made to be interesting without an initial desire, but this proved me wrong.
My question is, how rich can I become from stealing this idea?

[Anonymous No.15988467]

As a zoomer how can I make sure that my offspring (gen sigma) will become as powerful as possible: physically and mentally
I know that all my great knowledge will help them, but how do I also pass on my impressions of the world so that they don't take their gifts for granted? How do I prepare them for the worse times of the world so that they can survive and improve them world?
You see with all the old and very evil generations dying before life extension is discovered there will be an enormous power vacuum during the sigma generation.
Zoomers are probably the most bifurcated generation, some are completely brain rotted hollows, and some are doomer-esque improovers.
Clearly though the next round of bad guys (evil) will be the millennials.
Obviously they will lead an early grave because of their bad health, but they will try to create a technological future for their children since their children are incompetent and will be the idiocracy.
So yeah obviously there's homeschools, not sending them to college, keeping them away from the techno-idiocracy, teaching them to look as machines as tool like hammers.
How else can we ensure that our children will become strong enough and able enough fix this pale world?
There is so much mental illness I doubt there will be a huge number of gen sigma to even be born from a western genesis.
That's why I call them sigmas, because they will break out from the thumb of oppression. They must be the generation of freedom and strength

[Anonymous No.15988478]

>>15988467
I wouldn't worry about it. You're clearly never going to have sex.

[Anonymous No.15988485]

>>15988478
Interesting. So this insult is implying that my existence in some way threatens you so the continuation of myself through breeding is also a threat to you, so you must state it by saying that will will not breed.
I wonder why the truth of the world accosts you.
Why do you live in fear and malice from fear?

[Anonymous No.15988510]

>>15987238
Here is my problem: I need to find Thevanine's resistance and Thevanine's voltage with R6 being the load resistor. If I remove R6 and then replace C with circuit break and E with short circuit I get this weird situation. It's easy to see that R1+R3+R4=15Ohm but I'm wondering what to do with this short circuit. Since the current certainly flows through R134 from A through R2||R5 to B, is the solution RT=R134+R2||R5?

[Anonymous No.15988546]

>>15983911
Why do you have this?

[Anonymous No.15988553]

>>15988510
Btw is Thevanine's voltage here 4,5333333333333333333333333V? Did I calculate this correctly?

[Anonymous No.15988713]

>>15987963
>>15988678
Yeah, I think so. It depends on what you mean by "without even one explanation of what's going on", but I assume you mean something like picrel.
In fact, there is plenty of structure in the theorems, and he also mentions (very briefly) what is going on and where he is going. I can think of no book with zero comments in between proofs.
I don't know about engineers.

[Anonymous No.15988727]

How are derivatives expressed in category theory? I have searched but there is no image of that and what are the best resources to learn that branch of mathematics?

[Anonymous No.15988787]

>>15988713
exactly that, my brain completely turns off when i try to understand what he's trying to say. i find proofs (mostly in analysis) extremely hard to understand but as soon as i find those explanation for laymen (i.e. engineers) i understand them and i'm also pretty good at applying them in physics, computer science, etc.
this leaves me with the feeling that i did not truly grasp the material even if it's not true and so i feel overwhelmed and demotivated to learn more.

[Anonymous No.15988804]

>>15988787
It probably has to do with experience, both with the material and reading terse books.
The picture is reasonably advanced, and I'd not be able to read it if I hadn't spent hundreds of hours on similar topics before, just to understand the terms he's using and what they mean.
Now that I do, it makes sense to me, the order in which he presents results and the relations they have to stuff I already understand.
As such, I appreciate the terseness and would be annoyed by very detailed digressions on 'basics' while reading.

I do think there's a big difference between, as you say, fully grasping material versus merely applying it in context.
I think if you just remember the statements of theorems and when to apply them, you start engaging with problems symbolically.
As in, this problem looks like <structure I've seen before> (literally, with certain symbols in certain places), and there I used <steps> to get <result>.
This line of thinking opposed to, say, understanding (visually or however) what a theorem/problem is saying and intuiting a good approach to a solution from that understanding.

If you want to get out of the mindset of merely reproducing stuff, there is no other way than the hard way, I'm afraid.
For me it probably took ~5 years before I stopped getting overwhelmed by proving (basic) theorems and such on my own.

[Anonymous No.15988847]

>>15988804
did you do it on your own or in university? i study computer science and all the math courses i took were all very practical. sometimes we would see some proofs in calc 2/3 but only during the lectures, the exam is just solving exercises.
the CS classes are very easy for me i feel like i'm wasting my years since i could've done this on my own (and i did).
i'm trying to catch up on my free time but it feels like i'm wasting even more time because i'm studying exams that i already passed a year ago but i did not really understand (linear algebra).

i know that my writing is really confusing but given that i'm also a serial procrastinator i feel like i'm wasting so much time i really wish i could turn my brain off and start everything from scratch

[Anonymous No.15988889]

Hello anons, I'm a complete retard when it comes to chemistry, its literally magic to me. I recently came across this channel where 18 year old dude built a pyrolysis reactor to melt plastic to get crude oil and natural gas like methane and butane out of it.
So where's the catch? why is this not more widespread? If some 18 year old can build this reactor in his backyard and melt 8 kilo plastic in 2 hours to get oil and gas back, how come its not industrialized

[Anonymous No.15989076]

>>15988889
550 ml of oil from 8 kg of plastic?
What happened to the other 7 kg?

[Anonymous No.15989201]

>>15988328
I dont think its very sustainable. For example, the nah i'd win thing would get old in a couple years at most, then you would have to redo the entire curriculum in the new meme format and hope it doesnt die in a week. It needs to be timeless comedy in the same sense as monty python and the holy grail
Im more of the opinion that its more effective to merge games and learning, like using a quick-time event calculus to create a super powerful potion or some bullshit.

[Anonymous No.15989225]

How do I get a girlfriend? Most of the other students I know don't have one and most of my professors are unmarried.

[Anonymous No.15989255]

>>15989225
Become a /lit/cel.

[Anonymous No.15989332]

>>15989255
I am a /lit/cel. I've read most of the well-memed books and even a lot of books that women wouldn't consider that off-putting.

[Anonymous No.15989803]

>>15988727
When you say "derivatives in category theory", unfortunately all I think of is zippers on ADTs

[Anonymous No.15990088]

I got a question. Can anyone take a stab at this? Finding some function of x that closely approximates it.

\[ \sum_{k=2}^{\infty} \frac{{(k-1)}^{(k-2)}{(x^k)}}{k!} \]

The only crummy idea I've got is to hopelessly turn it into an integral and use Integration by Parts. (I hope i used latex properly)

[Anonymous No.15990097]

I got a question. Can anyone take a stab at this? Finding some function of x that closely approximates it.

[math]\[ \sum_{k=2}^{\infty} \frac{{(k-1)}^{(k-2)}{(x^k)}}{k!} \][/math]

The only crummy idea I've got is to hopelessly turn it into an integral and use Integration by Parts.

[Anonymous No.15990105]

>>15990097

[Anonymous No.15990142]

>>15990105
Let [math]x = \frac{\pi}{2} [/math] then
[eqn]1 - \frac{\pi^2}{2^2 \cdot 2!} + \frac{\pi^4}{2^4 \cdot 4!} - \ldots = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = \cos(x) = 0[/eqn]

[Anonymous No.15990489]

>>15988727
For derivatives of function between manifolds you'd want to look into synthetic differential geometry and diffeological spaces and the tangent bundle functor

[Anonymous No.15990701]

If a,b,c,d are in arithmetic progression such that a<b<c<d, and ab= c+d -1

Then sum of all possible values of a is?

[Anonymous No.15990759]

>>15990701
c = 2b - a
d = 3b - 2a

ab = c + d - 1 = 5b - 3a - 1
(a - 5)(b + 3) = -16
There are infinite solution unless you restrict yourself to integers. Then you only have (a,b) = (-11,-2) and (a,b) = (-3,-1).

[Anonymous No.15990796]

is
(not A -> not B) and (A -> B) the same as A <=> B

[Anonymous No.15990798]

Is it pretty much guaranteed that all grammar (as in word declension) that human languages use is just fossilized word combinations? For example, the modern English equivalent would be if the past tense is properly formed like
>I walk ehd to the store
Where the word "ehd" serves to mark the past.
Which eventually became combined into
>I walked to the store
But obviously, this happened in some ancestor of Proto-Indo-European tens of thousands of years ago, since PIE already had a full grammar.
You can observe a modern phenomenon of e.g. "woulda", "shoulda", where the "have" is rolled into the word itself. If the "have" form eventually would disappear, English would have gained a new grammatical feature, instead of purely semantical/lexical.

[Anonymous No.15990909]

>>15988847
anon? i really need to calm down my paranoia

[Anonymous No.15990935]

>>15989332
Now put your knowledge into practice. You need to intersect the world with it as if making an "X."

[Anonymous No.15991007]

>>15988847
>>15990909 (?)
I did both. Studied something mathematical (a little bit) and took some courses from the math faculty, self-studying what I missed.
I wouldn't worry about being "the most efficient", and just study things you don't understand yet. You'll get there.

Also, stop procrastinating.

[Anonymous No.15991015]

>>15990796
Yes. B implies A is by contraposition the same as -A implies -B.

[Anonymous No.15991484]

>>15990935
Whoa. How do I do that?

[Anonymous No.15991495]

What is a^x

[Anonymous No.15991498]

>>15991495
Something raised to the power of something.

[Anonymous No.15991506]

>>15991495
You can just experiment with numerical values to confirm the rules.

[Anonymous No.15991666]

can anyone who studied human biology explain what this this?
it's like a pimple but kind of not? if I cut it will I die?

[Anonymous No.15991686]

>>15991666
That looks like a skin tag
https://en.m.wikipedia.org/wiki/Skin_tag

[Anonymous No.15991713]

Trying to prepare for a test and I can't for the life of me find and understand the answer to a question. Would prefer if somebody recommended relevant resources I can use to study instead of just answering, I tried to use Digital Communications from Proakis but I can't understand the way M-FSK systems are described.

For a M-FSK system to be coherent:
1. The period of the carrier wave must be an integer multiple of the symbol period.
2.The symbol rate to be an integer multiple of the frequency of the carrier wave.
3.The symbol rate to be at least twice the frequency of the carrier wave.
4. The symbol period must be an integer multiple of the period of the carrier wave.
Only one is correct, explain why.

[Anonymous No.15991725]

>>15991713
Going to go with 3 because it's the one that makes me think of aliasing

[Anonymous No.15991772]

If time is relative and everything in the universe is in motion doesn’t that mean that nothing we see is actually real?
The Earth is rotating around the sun and its relative position is always changing. At no time is the Earth a static object. Extrapolate this out to the end of the universe, what separates us from the end of the universe aside from our brains’ being limited to perceive things in three dimensions?

[Anonymous No.15991783]

>>15991772
> doesn’t that mean that nothing we see is actually real?
Define what you mean by 'see' and 'real'.

> what separates us from the end of the universe aside from our brains’ being limited to perceive things in three dimensions?
Translate this gibberish into English please.

[Anonymous No.15992258]

A question for the microbiologists:

I add 6-methoxy-2-benzoxazolinone into a soil microcosm (No nitrite, some nitrate). I get dose-independent response after adding it, reducing all the nitrate into unknown. WTF happened?

[Anonymous No.15992269]

My textbook says a point in spherical coordinates is given by [math]r \mathbf{e}_r[/math], where [math]r[/math] is the radial component and [math]\mathbf{e}_r[/math] the corresponding basis vector.

I don't understand this. Don't you need the two angles also to describe a point in 3d space? What do they mean by this?

[Anonymous No.15992386]

>>15992269
The two angles are absorbed into the definition of e_r. You can also define an e_theta and e_phi that form a complete basis. It's obviously not very useful for specifying the position of that point, but it is valid so see where they are going with it.

[Anonymous No.15992485]

I don't want this to sound political, but what's the point in reducing GHG emissions?
Given that it takes essentially forever on human-relevant timescales for natural sinks to draw down CO2 and that +4C of warming is pretty much baked in (consequences aside) by the end of the century given current concentrations and feedback loops, isn't it pointless to do so?
I'm all for protecting the natural environment, but I doubt it'll change anything in the near term.

[Anonymous No.15992510]

>>15992386
How could you e.g. get the point (1, 1, 1) by just using this basis? Under the assumption that it is a basis, it should be possible to somehow "get" this vector, right?

[Anonymous No.15992530]

>>15992510
You can just make that particular basis vector [math]<1,1,1>[/math]...
Again, it's kind of a useless definition, but there's nothing explicitly wrong about it.

[Anonymous No.15992536]

>>15992510
Yes, in principle no matter where e_r is you can express (1,1,1) in the basis, but you need to know where e_r is to actually do it concretely. So for the purpose of expressing a fixed point it seems useless, but this is not why the basis is being introduced in your textbook. Probably it has something to do with centripetal motion or something. Just read on.

[Anonymous No.15992565]

When I was 11yo I had my IQ tested and it came as 135. A few years later, don't remember how old I was, I tested again and it came as 107.
I did some of these online tests a few times and I remember getting 110 and 115.
I'm in the process of testing my IQ yet again now as an adult.
Is it possible to estimate the results of the test based on my previous two results, and the results of the online tests?

[Anonymous No.15992594]

Here I go again, another question pertaining to spherical coordinates. Exercise wants me to represent the Cartesian vector (0, 0, z)^T in terms of spherical coordinates. I don't get this at all. Isn't the z axis cut from the coordinate system? If so, how could you possibly find spherical coordinates for (0, 0, z)^T?

[Anonymous No.15992596]

Is there a super secret way to find ALL factors of a number that doesn't use "brute force"?

[Anonymous No.15992599]

>>15992565
IQ is a meme

[Anonymous No.15992609]

>>15992599
It is not.
Now, care to answer my question?

[Anonymous No.15992679]

>>15992596
There are algorithms that are better than pure brute force, current state of the art uses number field sieves. However they are only really beneficial when used on very large integers, e.g. > [math]10^{100}[/math] and they are still not "fast". Is it strongly suspected that no algorithm exists that can factor an integer in polynomial time, that it is an NP problem, but it has never been proven.

[Anonymous No.15992744]

Question about scalars. Physically, scalars are quantities that remain invariant under coordinate transformations, i.e. they are rank zero tensors. However, I'm hung up on the specifics of "coordinate transformation." Take speed, the magnitude of velocity, for example. We know speed is a scalar, which means that it should be invariant with respect to a coordinate transformation, but consider this example: In a rest Cartesian coordinate system, a particle is moving along the x-axis at with a velocity v (<< c). Clearly, the particle's speed must be v as measured from the rest coordinate system. Now, move to an identical coordinate system that is now moving along the x axis of the original coordinate system with a speed of v. This moving coordinate system now measures the speed of the particle to be zero, but speed is a scalar, so it should not change with respect to a coordinate transformation. So, when we say that scalars do not transform with a "change of coordinates," does that change of coordinates only include changes of coordinates that can be reached via an orthogonal transformation?

[Anonymous No.15992762]

>>15992744
Context matters. Speed is not a true scalar, it is a scalar quantity, it is not a rank-0 tensor. You proved that yourself, it is not invariant under a change of coordinates. Physicists tend to be loose with their mathematical terminology. Every rank-0 tensor is a scalar but not every 'physical scalar value' is a rank-0 tensor.

[Anonymous No.15992779]

>>15992762
>Every rank-0 tensor is a scalar but not every 'physical scalar value' is a rank-0 tensor.
This cleared it up for me, thank you!

[Anonymous No.15993189]

Brothers, what are your schedules and how much time do you dedicate to studying?

[Anonymous No.15993192]

>>15993189
as an addendum to my question, what productivity tools do you use? ex: Anki, etc

[Anonymous No.15993199]

>>15993189
>shonenkek looks 40
>moechad looks 17

[Anonymous No.15993450]

Question goes as follows: Calculate the mass of every salt in the table required to make a 250cm^3 solution whose concentration is c=2mol^dm^3.

I have a table of salts, first one is Iron (III) Chloride, then Copper (II) Sulfate, etc. Is my methodology correct? There were like only 2 examples I saw in what my professor gave me and in those two, n1=n2 were equaled so I'm guessing it's right?

[Anonymous No.15993480]

Consider the parametrization [math]\mathbf{r} = R \mathbf{e}_r + \phi \mathbf{e}_\phi + \theta \mathbf{e}_\theta [/math] in spherical coodinates. When I partially differentiate this parametrization wrt theta, do I get
>[math]\frac{\partial \mathbf{r}}{\partial \theta} = (1+R)\mathbf{e}_\theta - \theta \mathbf{e}_r [/math]
Looks wrong to me, but all I did was applying product and chain rule when differentiating the spherical basis vectors. So I don't see where my mistake is.

[Anonymous No.15993652]

>>15993480
It doesn't work that way. r= |r| e_r. Your units are even wrong for the other components. Stop posting questions about this without context. This basis is useless unless you do something like make r and the basis vectors time dependent.

[Anonymous No.15994097]

this question sounds easy but I probably just don't understand what it's asking
to clear up confusion
the delta symbol is symmetric difference
it's asking about cyclic groups and generators

[Anonymous No.15994103]

>>15975473
There is a server for electric circuits if you need more help. dPXSRu5Nvz

[Anonymous No.15994109]

>>15973938
>>15973949
It's a homage to optimizeanon.
He just randomly replied to people here with an optimized version of your image a couple years ago.
>>15973949
>was based on /g/'s sqt blurb
It wasn't tho.
I mean, at least not consciously.

[Anonymous No.15994289]

>>15994109
>It wasn't tho.
Oh my bad then. I said that because of >>11321297 and you adding it soon after. Guess I got exposed as a newfag.

[Anonymous No.15994315]

>>15994097
Since you know what all the symbols mean, you should just be able to compute the cyclic subgroup generated by A, right? Since any element is its own inverse this will have two elements.

[Anonymous No.15994357]

>>15994289
Don't worry, we're all newfags.
>>/sci/thread/11329061#p11330182

Sound logic on it being based on /g/'s sqt by the post you quoted.
But it's also easy to check up on how /g/'s sqt pasta looked at the time https://desuarchive.org/g/search/subject/sqt/start/20200115/end/20200120/
And it also still looks like that right now.

I'm pretty sure I wrote the whole thing ex nihilo by completely giving up on the old pasta's (here >>/sci/thread/10172376)) attempt at educating people who won't read it and making the entire thing purely dedicated to random autists who will actually read it.

[Anonymous No.15994392]

How do I do this problem? (Calc BC)
Show that x^2 grows at the same rate as cube root of (x^6+x^2)

[Anonymous No.15994403]

>>15994392
You have to show that the limit of their ratios approaches some non-zero constant.
[eqn]\lim_{x \to \infty} \frac{x^2}{(x^6 + x^2)^{1/3}}
= \lim_{x \to \infty} \frac{1}{(\frac{x^6 + x^2}{x^6})^{1/3}}
= \lim_{x \to \infty} \frac{1}{(1 + 1/{x^4})^{1/3}}
= \frac{1}{(1 + 0)^{1/3}}
= 1[/eqn]

[Anonymous No.15994478]

>>15994403
Ohh thank you!!

[Anonymous No.15994507]

>>15994315
Okay that's what I thought but it seemed too obvious and usually when that happens it's because I didn't read the question correctly

[Anonymous No.15994519]

How far away from a star do you need to be before it's light no longer illuminates you?

[Anonymous No.15994569]

>>15994519
First you have to define what you mean by "no longer illuminates you".

[Anonymous No.15994808]

>>15990142
Actually, it turned out that after taking the derivative with respect to X, that series changes into the Product Logarithm. So that series is the integral for the Product Logarithm, and it does have a closed form expression.

Thanks anyway for taking a shot at it.

[Anonymous No.15994851]

failed my semester 1 year 1 chem exams
do i drop out/switch course or do i push through and try to do well in semester 2?

[Anonymous No.15995091]

How can you express the product [eqn]\prod_{i = 1}^n (a - x_i)[/eqn]in terms of a sum. Is there such a thing as a binomial theorem for this?

[Anonymous No.15995094]

>>15995091
There is a way you can express this with symmetric polynomials https://en.wikipedia.org/wiki/Newton%27s_identities#Application_to_the_roots_of_a_polynomial

[Anonymous No.15995147]

>>15995094
looks interesting, thanks

[Anonymous No.15995615]

>>15987963
math books are to be written and purchased, not read

[Anonymous No.15995661]

Since the light is redshifted, it has lower frequency. A photon would then have less energy according to E = hf. Where does the energy go?

[Anonymous No.15995671]

>>15995661
It doesn't go anywhere.
Energy isn't necessarily invariant across frames of reference. (If you need convincing of this, consider the kinetic energy of an object in motion, and then of that same object in a frame of reference where it's stationary.)
What's happening is that it appears to have lower energy because you're moving away from it, but that doesn't change the energy of the photon itself. After all, if you were to start moving towards that object at such a speed that its light was no longer redshifted, you wouldn't actually be changing any properties of the photons it emitted, right? Just how they appear to you.

[Anonymous No.15995674]

>>15995661
Like velocity energy is relative, it depends on the reference frame. Energy is conserved within a reference frame but not between them. The photon is not losing any energy as it travels through space but different observers will each measure it to have a different energy (wavelength).

[Anonymous No.15995770]

Is differential geometry, linear algebra and abstract algebra enough of a math background to study quantum field theory?

[Anonymous No.15995834]

>>15995770
No. The math required would also need Complex Analysis and maybe some Group Theory. You also absolutely need the physics framework: Relativity (Special, not General), Quantum Mechanics to an advanced level, Statistical Mechanics, Lagrangian Mechanics (variational calculus). There's a reason QFT is seen as a hard subject.

[Anonymous No.15996054]

>>15995770
Yeah dawg just basic quantum mechanics alone demands strong familiarity with statistics, physics, linear algebra, and much more. QTField theory just cracked my head, was even more demanding, so I gave up.

Here’s a low-down of Quantum mechs so you can just skip it.

A particle’s position and its speed are fundamentally uncertain, yet still intertwined. Certainty of one, reduces certainty of the other. Waves explain all of that away; so, particles were actually condensed bundles of waves all along, not unlike the ripples darting along the surface of the waves you create while swimming.

But in congregation, the uncertainity relationship stops mattering because the average position and average speed win out, so the congregation stops behaving like wave.

The uncertainty and randomness isnt just for position and speed. It’s for everything, which causes other physical phenomenon such as electron orbits and electromagnetic repulsion.

Meanwhile, QFT doesnt explain physics, it comes UP with new physics, on top of reinventing everything. It’s way more bamboozaling than QM.

[Anonymous No.15996875]

i need to learn math from absolute 0, any book/tutorial that takes my hand until im ready to do precalculus?

[Anonymous No.15996910]

>>15995770
I'm seconding this guy's response >>15995834. Complex analysis is more important than differential geometry for learning basic QFT of scalar fields. For things like gauge theory or some more abstract quantum field theories you will need to know some things about Lie groups and algebras.

[Anonymous No.15996955]

>>15984228
i do agree but i do think there's something in concepts such as drive, sex, aggression, etc

when looking at things as units i believe there is more truth to some things than others

[Anonymous No.15998201]

bumppu~

[Anonymous No.15998292]

I'm trying to understand the interactions of photons with atoms as it relates to visible light. I'll explain my current understanding / thinking:

scenario: a white light, shines on a red object
1. the white light is a collection of photons of all visible frequencies (red, green, blue, etc)
2. the atoms in the red object are interested in absorbing all the frequencies except the red one
3. when a photon collides with one of the atoms and is absorbed it is then re-emitted in the opposite direction it arrived from (equivalent to passing right through). the atom is therefor 'transparent' to those colors
4. when a photon collides with one of the atoms and is reflected, the direction of reflection depends on where it collided with the atom, for this purpose the atom can be considered a sphere and you can just determine the normal of the 'impact' point and reflect across the normal to determine the reflection direction

does this sound okay as a model, or is it very wrong?

I've read that the whole photon absorption emission thing is actually a random distribution favoring some direction and it doesn't happen instantly, but the atom can't absorb energy forever and to conserve energy it must emit in the opposite direction right, so iI reasoned that it's the same as passing right through.

also the angle of impact wouldn't matter in this case as it would just emit along the same path offset on the opposite side of the atom, unless the emission always happens from the 'center' and is angled away to keep energy constant (in the case of the original collision being on the edge)?

[Anonymous No.15998347]

>>15998292
1. Yes

2. Pretty much right, but it doesn't necessarily work at the level of atoms. The atoms might be in some condensed matter structure and that can influence which frequencies are reflected

3. Sort of. There is actually a distinction between completely not interacting with the atom and passing right through (I think this is a better use of the word 'transparent'), and forward scattering where the photon is absorbed and reemitted in the same direction (which is what you described although you misused "opposite direction").

4. No. This is all quantum mechanical. If you have a definite frequency of light the photon is not well localized and it does not make sense to talk about an impact point. In any case the scattering is very much not like scattering of hard spheres. In a classical limit it is given by "Rayleigh scattering"

[Anonymous No.15998355]

Does anyone know good resources for antenna engineering? Preferably satellite-oriented.

[Anonymous No.15998369]

Is division best interpreted as the inverse of multiplication? Dividing 80 into 2 and 2/3 parts doesn’t make sense, but 30 * 2 and 2/3 makes 80, thus we get 30 for 80 divided into 2 and 2/3 parts.

[Anonymous No.15998408]

>>15998347
>There is actually a distinction between completely not interacting with the atom and passing right through
does the distinction matter when looking at it from my case where I'm just looking at how visible light would behave?

>4
so looking at this, I guess the frequency of the photon compared to the target atom's absorption ranges determines what angle it will reflect at?

[Anonymous No.15998412]

>>15998369
Yes.
In fact, some systems explicitly require that you define it that way.

[Anonymous No.15998633]

>want to take a copper supplement
>because i take a vitamin C supplement every day (powder), and i read taking plain ascorbic acid like that depletes copper because copper is required in its metabolism, and all natural vtamic C is usually found with copper, but when you take fake lab made vit C (ascorbic acid) it lacks the copper, so the body will leach copper from elsewhere in the body to process the vit C (dr berg said so, is it true?)
>not much copper in any multivitamin i see, only like 20%, while zinc is the same or higher
>taking zinc apparnetly further depletes copper, they are kinda antagonistic
>look for dedicated copper supplement
>can't even find one that's just copper
>the ones i can find come with even more zinc, which would further deplete my copper
>I DON'T WANT ZINC YOU CUNTS I GET ENOUGH OF THAT I JUST WANT PLAIN OLD COPPER BY ITSELF REEEEEEEEEEEEEEE
>...
>cleaning my desk drawer
>see old coins in there
>suddenly think
>what if i just put a penny in my glass of water every day and just keep refilling that with water and drinking from it but leave the penny in there, and some of the copper will leach out from the coin into the water im drinking
>get some copper, plus save a lot of money instead of buying another supplement, if i can even find it

So, would it work?

[Anonymous No.15998647]

why does this method not work for differentiating y=x^log(x)?

let u = log(x), then du/dx = 1/xln(10)

then y = x^u, so dy/dx = u*x^u-1 * du/dx

so dy/dx = log(x) * x^(log(x) - 1) * 1/xln(10)

I know the right way is i believe implicit differentiation, but i was wondering why this method doesnt work

[Anonymous No.15998719]

I had stupid thought at my warehouse job. If you take random two primal numbers for example 3 and 13, multiple them 3 x 13 is 39 then imagine you have 39 2d squares. If you try to assemble them into one square or rectangle the only way to do it is with sides of length of primal numbers you chose. Of course building a rectangle with a side of length 1 is forbidden. I kinda don't know why this rule happens. It seems to be very obvious.

[Anonymous No.15998733]

>>15998647
> y = x^u, so dy/dx = u*x^u-1 * du/dx
This part is wrong in many ways.

1) If you do a substitution you have to replace every occurrence of x with u. Since u = log(x) => x = 10^u
2) You did the differentiation dx/du, not du/dx.
3) The above doesn't matter since neither are correct because x and u and not interdependent. x = x(u) and u = u(x). That is why you have to do step 1).

[Anonymous No.15998779]

>>15998292
The problem you are having is that considering a single photon and a single atom cannot answer your last few questions. It does not explain them. Reflection is an effect caused by the coherent interaction of many photons and atoms in the material. When you combined the quantum probability amplitudes of all the particles involved you get the classical effect of reflection. It is why in most cases treating light as a wave is the best and easiest method when talking about such phenomena, it is the classic limit of complex quantum interactions.

[Anonymous No.15998780]

>>15998408
No I suppose the difference doesn't matter much for what you want to understand. But if it forward scatters it probably scatters in other directions too (actually there is a theorem called the optical theorem that makes this precise, but don't worry about it)

>4
The Rayleigh scattering formula, which describes the angular distribution of scattered light as a function of frequency, is something you can calculate in classical electromagnetism. If you want to do this with genuine photons you need to use quantum field theory and then you can calculate something called a "cross section" that describes the angles the photons bounce off and also the probability of interaction in the first place. It isn't an easy calculation (describing bound atoms is a little awkward in QFT) and at lowest order the answer just reduces to the classical Rayleigh scattering result anyway.

[Anonymous No.15998789]

>>15998733
thank you anon, i'm having a new problem:

i am trying to take the derivative of y = x^[(logx)^2 - 2]

and i am getting the correct answer for d/dx[ln(x^[(logx)^2 - 2])]

but not for d/dx(x^[(logx)^2 - 2]):

i start with: lny = ln(x^[(logx)^2 - 2])
=[(logx)^2 - 2]*ln(x)
=[(logx)^2]*ln(x) - 2*ln(x)

then i take the derivative of both sides and get
1/y dy/dx = d/dx([(logx)^2]*ln(x)) - d/dx[2*ln(x)]

and finally dy/dx = y*( d/dx([(logx)^2]*ln(x)) - d/dx[2*ln(x)] )

but once i multiply it by y it is no longer correct.

I think it has to do with d/dx([(logx)^2]*ln(x)) because this is a chain and product rule combined, and multiplying by y seems to not work.

but when i do this for a problem like y= x^(2x - x^2), then i get a correct answer for both dy/dx = y* d/dx(ln[x^(2x - x^2)]) and for d/dx(ln[x^(2x - x^2)]).

is there a problem with what i have done?

[Anonymous No.15998808]

>>15998789
>and finally dy/dx = y*( d/dx([(logx)^2]*ln(x)) - d/dx[2*ln(x)] )
What are you talking about, this is perfectly correct(?)

[Anonymous No.15998815]

>>15998789
You really should learn to use the [math] tag, that is a mess to try and read.

[Anonymous No.15998828]

>>15998808
>>15998815
sorry i meant i am trying to get the derivative of
y = x^[(logx)^2 - 2]
and i start by taking the ln of both sides to get
ln(y) = ln(x^[(logx)^2 - 2])
and i take the derivative of both sides

if i pretend that the question was actually
y = ln(x^[(logx)^2 - 2])
then when i check the derivative of this using mathmatica, my calculation up to this point matches the answer it gives

but when i check the derivative of
y = x^[(logx)^2 - 2]
on mathematica, my solution which i get by multiplying y by the derivative of
ln(x^[(logx)^2 - 2])
doesnt match the answer mathematica gives

so i don't get what is wrong since it should be that
1/y dy/dx = d/dx ( ln(x^[(logx)^2 - 2]) )
so then i should get
dy/dx = y * d/dx ( ln(x^[(logx)^2 - 2]) )

[Anonymous No.15998843]

I woke up today with a bunch of dark red spots on my forehad and they dont go awya when i press on them

theres more on the top of my scapl now like some kind of allerig rieaction

is there any could it be?

[Anonymous No.15998853]

>>15998828
Since you show neither the answer Mathmatica is giving, or the answer your get, or your calculations, what are you expecting us to say or do?

[Anonymous No.15998855]

OFFICIAL /sci/ SERVER

[Anonymous No.15998874]

Why do we call it a “root of unity” instead of just “root of 1”?

[Anonymous No.15998881]

>>15998828
> i should get
>dy/dx = y * d/dx ( ln(x^[(logx)^2 - 2]) )
That looks fine so far. There are only about 4 more steps after that
>simplify that ln
>multiply the result
>take the derivative
>multiply and simplify
Maybe you made an error in one of these? But your idea is right.

[Anonymous No.15998885]

>>15998828
Oh ok, I found your earlier post. Why are you writing things like
>(logx)^2]*ln(x)
? The log(x) in your problem is definitely supposed to mean ln(x). If you are reading log as “log base ten” or something, that could explain the different answer you are getting.

[Anonymous No.15998919]

>>15998885
> The log(x) in your problem is definitely supposed to mean ln(x)
That is triggering me too. If that anon is doing a math course you are right, however in a lot of sciences or engineering you will frequently see log used to mean log_10 since it comes up in the subject more often.

[Anonymous No.15998924]

>>15998874
probably history / tradition. "roots of unity" implies complex solutions are allowed. "root of one" means 1 only.

[Anonymous No.15998928]

What the fuck does M mean in this context? (J. Org. Chem 55, 6219-6223)

[Anonymous No.15999066]

So, tensors are basically vectors and not the other way around?

[Anonymous No.15999077]

>>15999066
tensors have Dimension and Rank.
a vector is rank-1.
https://youtu.be/bpG3gqDM80w

[Anonymous No.15999081]

>>15999077
I don't get it. The definition is very clear that tensors are elements of a vector space.
If it is an element of a vector space, it is a vector isn't it?
Also, when you say that
>a vector is rank-1
are you referring specifically to vectors in [math]\mathbb{R}^n[/math]? Or any kind of vector?

[Anonymous No.15999097]

>>15999066
yeah you can think of a tensor as just an n-dimensional grid of numbers. the retarded name “tensor” just comes from the fact that they were using that kind of grid to model tension in mechanics. I always wondered what the exact story was behind that decision
>mathematical physicist gets schizophrenia
>ngaaaaAAAAAA
>ALL vectors and matrices are a kind of TENSION!!!!
>MWAHAHAHAYAHA

[Anonymous No.15999114]

>>15999081
Think of the rank as the number of variables needed to index an element of the tensor.

[Anonymous No.15999117]

>>15999081
i'm not going to type out the video.
watch it or fuck off

[Anonymous No.15999149]

>>15999117
By the definition from the pic on my original post, a tensor is an element of a vector space. And from what I understand, if it is an element of a vector space, it is a vector.
From the video I get that a vector in [math]\mathbb{R}^n[/math] can be a tensor.
So a tensor is a vector (not necessarily in [math]\mathbb{R}^n[/math]) and a vector in [math]\mathbb{R}^n[/math] can be a tensor? Is that it?

[Anonymous No.15999154]

>>15999149
>is an element of a vector space
I mean, from that specific vector space described in the pic

[Anonymous No.15999176]

Is their book particular on continuity and differentiability of functions with rigorous problems ? Or any book which has a high emphasis on this topic ?

[Anonymous No.15999229]

>>15972871
Has anyone here studied Systems Theory? What domain did you come from before? Have you found work?
What does the future of Systems Theory look like academically and industrially?

[Anonymous No.15999317]

>>15999097
You are missing the point. The reason people use the name "tensor" has to do with the way it transforms under changes of coordinates. You could have a matrix of scalar numbers that don't transform under coordinates and that would not be considered a tensor (or if you want to be pedantic it could be a collection of rank 0 tensors)

[Anonymous No.15999331]

>>15999149
You are being confused by multiple meanings of "vector" in mathematics. A tensor is an object in a vector space in the sense of linear algebra (and in this sense it is a "vector"). The tensor product is a way to go from two vector spaces to a third vector space that it is in some sense the product of the two. This is what your book is trying to teach you.

When other people say a vector is particular rank of tensor, they mean "vector" in the sense of either differential geometry or physics.

[Anonymous No.15999351]

>>15999229
>Systems Theory
snake oil tier. avoid it 100%

[Anonymous No.15999540]

>>15999176
Bummpppoooo

[Anonymous No.16000349]

Where on the internet can a person buy generic prescription-only medicines (which said person does not have a prescription for)?

Asking for a friend :^)

[Anonymous No.16000373]

how far (rigorously) do i need to go to say the limit of x^(sqrt(log(x))) as x->inf is infinity?

x^(sqrt(log(inf)))
= x^(sqrt(inf))
= x^(inf)
= inf

is this too much?

[Anonymous No.16000387]

>>16000373
for this particular function I would just go with
>x^sqrt(log x) > x
whenever x > e (assuming log means natural logarithm, otherwise x > 10 or whatever). so the function goes to infinity by the squeeze theorem aka sandwich theorem

[Anonymous No.16000465]

>>16000349

the same place where you can buy illegal drugs using crypto: the darkweb

[Anonymous No.16000569]

>I have 2 parents
>I have 4 grandparents
>I have 8 great grandparents
>I have 2^n Nth-parents
>assuming that they all gave birth at age 50, an insane upper bound, that means there would be 80 generations between now and 2000 BCE
>that would imply there were 1.21E24 people alive in 2000 BCE
explain this. Of course there is some inbreeding and such, but even if every single person took 10 spots within the tree, it would still be an astronomically high number

[Anonymous No.16000586]

>>16000569
Probably by the time you get to n=5 or whatever it's not like anyone is screwing their sister for it to be "inbreeding" in you calculations. Historically, probably anyone within fucking distance would be related to you at that level. It won't be just a linear scaling you can take out.

[Anonymous No.16000626]

>>16000586
Even n=5 is high, people in the past lived in smaller, tightly-knit communities. Then there is also the fact of mortality rates and lower life expectancy. Half of all children never made it 15 years old so could never reproduce, and that only changed in the last few hundred years with modern medicine.

[Anonymous No.16000643]

>>16000569
Pedigree collapse

[Anonymous No.16000738]

>>16000569
bro what the fuck was I thinking, I feel unbelievably retarded for having had typed this

[Anonymous No.16000849]

>>16000569
I have seen estimates (based on those huge family-tree databases compiled from baptismal records etc) that typical people native to a given country (say, England) are all about 15th cousins in the worst case

[Anonymous No.16000857]

Not a question but time isn't real; no 4th dimension bullshit.. There is no past nor future. Those are just human concepts to explain verbally what happened before and what could come to be. We are always in the exact present no matter what. Niggers tongue my anus.

[Anonymous No.16000895]

>>15972871
under what conditions do
[math]a\otimes b = 0[/math] in [math]A\otimes_R A[/math]
imply
[math]a = 0[/math] or [math]b =0[/math]?

in my particular case I have that:
b has no annihilator
A is finitely generated, projective, faithful
and [math]a\otimes b = 0 [/math] and I want to show a = 0

[Anonymous No.16001078]

are the books on the former math list still recommended?
https://4chan-science.fandom.com/wiki/Math_Textbook_Recommendations

[Anonymous No.16001102]

>>15999077
Not a bad video, but I do think that it is a bit confusing to call a tensor either a matrix, like when he calls the 3x3 matrix a stress tensor. To fully represent the stress tensor, you need a basis in addition to the matrix, the matrix alone is not a tensor. Tensors themselves are independent of the chosen basis, you just need a basis to represent them in this way.

For the same reason one should be careful on what means by 'vector'. A 1xm and nx1 matrices can be called vectors, but they are not tensors either. They also require a basis to represent a rank-1 tensor.

The picrel shows two different representations of the same strain tensor (principal basis and basis with maximum shear)

[Anonymous No.16001393]

>>16000465
Where on the |)@r|< \/\/€|} ?

[Anonymous No.16001459]

Is it common for you STEM chads to be able to solve 2 digits X 2 digits multiplication in your heads? A have a bit of trouble with mental math and wanted to strengthen my math skills but Ive gotten into some self imposed paranoia that I would only start after learning how to multiply up to 5 digits by 5 digits in my head.

[Anonymous No.16001490]

>>16001459
Could probably do 2x2 in my head, but don't often bother since a calculator is easier. 3x3 would be a struggle and 5x5 is right out.

[Anonymous No.16001972]

>>16001459
see >>16001490
the reality of it is that mental mathematics may be a neat gimmick, but it has little significance. Your teachers lied when they said you wouldn't always have a calculator, and I really do suspect that the insistence on that is probably a huge part of why the general public despises the subject.
I mean, for fuck's sake, I wrote [math]\frac{1}{3} \times{2} = 1[/math] on one of my exams for diff eq and spent some 10 minutes trying to figure out why the math wasn't working out. Still got my degree, so...
as long as you understand the concepts, nobody of value is going to slight you for using a calculator even on 2 digit x 2 digit (unless it's something really obvious)

[Anonymous No.16001989]

>>16001972
Number is God's riddle. It is too bad you can't see the prison cell they broadcast around you, nor the key beside you. It's okay. You are guaranteed to roll better stats, eventually.

[Anonymous No.16002044]

My stupid question is in regards to archery.
Tl;dr the weight balance of an arrow has an effect on its trajectory "flatness" and its penetration upon impact, but I don't understand this relationship mathematically.
What I would like to understand better is how to put numbers to all this.
What maths do I need for this?
I get along well enough by having a "feel" for it, so it's not like my hunting will suffer from a lack of maths, and I'm not out to win any competitions, I would just like to to be able to play with it in paper.
I guess what I'm looking for is what do I need to be able to simulate the flight characteristics of arrows on paper?

[Anonymous No.16002103]

what do i do if a limit is too difficult to be solved with l'hopital rule?

like when taking the derivative just makes it even more complicated

[Anonymous No.16002126]

>>16002103
hard to say without knowing the specific function, but my immediate thought is the squeeze theorem

[Anonymous No.16002300]

If life originates from water, why isn't there an automatic assumption that most/all water in the known universe, on moons and planets, harbor some form of life even microscopic?
Like I know for a fact some of Jupiter's moons have water. Shouldn't that be pretty obvious something's living there?

[Anonymous No.16002309]

>>16002300
You also need some other things, like a nice atmosphere, a nice temperature, and some gooey aminos.

[Anonymous No.16002316]

>>16002309
With that blueprint, can we create life then?

[Anonymous No.16002317]

>>16002300
Because water by itself is not enough. You also need the right chemical and environmental conditions, combined with time and probably a whole lot of luck.

> Shouldn't that be pretty obvious something's living there?
Based upon what evidence? Our sample size is 1. So far we only know that abiogenesis happened once from our observations. What are the odds it has happened elsewhere? No one can definitively say.

[Anonymous No.16002320]

>>16001459
Mental multiplying two 2digit numbers is pretty easy, using the FOIL method.

Once while I was quite bored vacationing in Alaska, I practiced multiplying two 3digit numbers. It’s extremely slow, taking me ten minutes on average vs the ten seconds for two 2digits.

There is however some fancy method called phonetic code or something though, shown but not explained in this Ted-Ed video.

https://m.youtube.com/watch?v=e4PTvXtz4GM

[Anonymous No.16002339]

>>16002044
Objects in free fall will rotate around their center of mass. So weight distributions dictate how the arrow interacts with all that air passing by it.

The math you need is calculus, the math used for describing rates of change, such as “how impact changes if the mass changes.”

[Anonymous No.16002341]

>>16002316
People have tried that during the miller-urey experiment. They were able to create amino acids from various inorganic chemicals that were supposed to be present on earth.
>By introducing an experimental framework to test prebiotic chemistry, the Miller-Urey experiment paved the way for future origin of life research.[45] In 1961, Joan Oró produced milligrams of the nucleobase adenine from a concentrated solution of HCN and NH3 in water.[46] Oró found that several amino acids were also formed from HCN and ammonia under those conditions.[47] Experiments conducted later showed that the other RNA and DNA nucleobases could be obtained through simulated prebiotic chemistry with a reducing atmosphere.[48][49] Other researchers also began utilizing UV-photolysis in prebiotic schemes, as the UV flux would have been much higher on early Earth.[50] For example, UV-photolysis of water vapor with carbon monoxide was found to yield various alcohols, aldehydes, and organic acids.[

[Anonymous No.16002723]

>>16000373
This is not rigorous. By the same logic you could reason [eqn]\lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^n \\ = \left( 1 + \frac{1}{\infty} \right)^{\infty} \\ = \left( 1 + 0 \right)^{\infty} \\ = 1[/eqn]

[Anonymous No.16002782]

Having trouble with a specific polynomial exercise.

[math]
30am^2-40amn+16bmn-12bm^2
[/math]

The answer the book gives is [math]2m(3m-4n)(5a-2b)[/math] but I'm not getting it.

[math]
30am^2-40amn+16bmn-12bm^2
2m[15am-20an+8bn-6bm]
2m[5a(3m-4n)+2b(4n-3m)]
[/math]

One of the parentheses has the signs wrong. I thought maybe I missed a part in the book saying I could just multiply one of the groups by -1 but no... unless that's what I should do but it doesn't feel like that's correct if I multiply I should do it to the whole expression no?

[Anonymous No.16002783]

>>16002782
Not sure why it didn't skip lines there but you get it right?

[Anonymous No.16002797]

>>16002782
>>16002783
You don't need to multiply the whole expression by -1.
Notice that the sign on 2b is flipped from how the book has it written, too. That means that you can multiply both the 2b and the parenthetical expression you're multiplying it by by -1, which is equivalent to multiplying that entire term by 1.

[math]2b(4n-3m) = (-1)(-1)(2b)(4n-3m) = (-2b)(3m-4n)[/math]

[Anonymous No.16002852]

>>16000373
Pick any y you want no matter how large. I can find an x_0 such that x^(sqrt(log(x)) > y, for all x>x_0. That is basically what it means for the limit of a function to be infinity.

The proof is simple. As long as x>e, sqrt(log(x))>1

so x^(sqrt(log(x))>x, and so clearly I can just choose x_0 = y in the criterion above.

[Anonymous No.16002912]

Am I tripping or is this problem ridiculously annoying. First i try to integrate by parts u = x^2 dv = cos 3x

But then i have to use substitution to get v, and then we get anothet expression but have to integrate 1/3 * integral of 2xsin(3x), but we have to solve that by integrating by parts, but then we get another expression with 2 * [ -(1/3 x cos(3x)) - integral of (-(1/3) cos (3x)]

And so on. Is there some easier way to solve this? Or am i missing something? Because to me it seems extremely tedious and error prone. Looks like it just keeps expanding further and further.

[Anonymous No.16002917]

>>16002912
>Is there some easier way to solve this? Or am i missing something? Because to me it seems extremely tedious and error prone. Looks like it just keeps expanding further and further.
Nope. That's how you go about it.
Welcome to integration by parts with trig functions/exponentials.

Well, I guess you are missing the fact that it actually stops with that last expression you've written, if that's something you are indeed missing...

[Anonymous No.16002927]

>>16002917
What a massive headache. Here is what i got for final answer. Just kind of sucks because im doing online class and theres limited amount of space for any 1 problem. Idk any other way besides doing lot of mental calculations and erasing stuff as you go.

I wish they tested more on theory instead of how much tedium you can tolerate

[Anonymous No.16002938]

>>16002912
Well first take u=3x, and then find (1/27)\int u^2 cos u du, instead of worrying about a bunch of factors of 3 at each step.

u^2 cos u du
= d(u^2 sin u)- 2u sin u du
= d(u^2 sin u + 2u cos u) - 2 cos u du
= d((u^2-2) sin u + 2u cos u)

So the answer is
((u^2-2) sin u + 2u cos u)/27
where u = 3x

[Anonymous No.16003127]

My nephew got this Latin square today as homework. Both my brother and I are completely stumped.

I figured out that A3 must be 3 and B3 must be 1, and A4 and A5 are either 1 or 2.

This is more difficult for me than trying to write proof. Can anyone help?

[Anonymous No.16003224]

>>16003127
A4 and A5 have a 1 and 2 by 8
A3 and B3 have a 3 and 1 by 4. since A already has its 1, A3 is 3 and B3 is 1. You knew this much.

B5 and B6 are either 1 and 5 or 2 and 4 by 9. But you know B's 1 is taken, so it's 2 and 4. This also means that you know B1, B2, and B4 must be 3, 5, and 6.
5 tells you that B4 and C4 have 1 and 5 or 2 and 4. But we know B4 has to be one of 3, 5, or 6, so B4 must be 5 and C4 must be 1. So A5 is 1 and A4 is 2.
Furthermore, 3 tells you that B2 and C2 are 1/6, 2/5, or 3/4. From this list, B2 must be 3 or 6. But B2 being 6 would mean that C2 is 1, and C's 1 has already been used, so B2 is 3 and C2 is 4. B1 is 6, then.
10 says that C5 and C6 are 4/5 or 3/6, but they can't be the former, so they're the latter. This suggests that C3 and C1 are 2 and 5.
2 says that C1, D1, and E1 are some combination of 1, 2, and 3. Combining this with the previous, we know that C1 must be 2, so C3 is 5.
6 says that D5 and E5 have a 5 and 6, so that rules out C5 as 6. Thus C5 is 3 and C6 is 6. Because B1 and C6 are 6, that rules out A1 and A6, so A2 is 6 instead.
7 says that D6 and E6 are 3/5 or 2/6, but C6 is 6, so they have to be 3/5. That means that A6 can't be 5, so it must be 4 and A1 must be 5. Then B5 is 4 and B6 is 2.

We haven't even touched half of the squares yet.
By 11, D2 and D3 are 1/5 or 2/4. But 3 has both 1 and 5, so it has to be the latter, and C2 is already 4, so D2 is 2 and D3 is 4.
Also, since we know that D6/E6 is 3/5, F6 must be 1.
This forces D's 1 to be D1, and E's 1 to be E2, which in turn forces F2 to be 5.
2 also tells us that E1 is 3, so F1 is 4.
This forces E4 to be 4.
Since D5 and E5 are 5/6, F5 is 2, and that's sufficient to force the rest of the square.

[Anonymous No.16003243]

>>16003127
It's like a medium hard sudoku (one where you don't have to ever guess and check). I can give you the answer if you want. Here is a hint for the next certain cell:

You figured out B3 is 1, so you know B5 and B6 must contain 2 and 4 (rule 9). This implies B4 is 5 and C4 is 1 (rule 5).

[Anonymous No.16003249]

>>16003224
Oh well I wasn't gonna give him the whole thing, but there you go

[Anonymous No.16003425]

>>15998633
Pennies are made out of...?

[Anonymous No.16003757]

>>16002339
Thanks anon. Cheers.

[Anonymous No.16003852]

>>16002927
No, there does exist more speedier and more convenient way to do integration by parts, by making two columns. That youtuber BlackpenRedpen does it often (the columns are in blue on the right). I don’t know what the method is called.

[Anonymous No.16003916]

bump

[Anonymous No.16004099]

I might be retarded but I have been on this for two hours. I keep redoing it but I don't understand why A) isn't 1.133*10^-5

[Anonymous No.16004105]

>>16004099
Q' = 3at2 - 2bt + c
3*3.3*10^-6 t^2 -2*6.6*10^-6 t + 19.8*10^-6
When t is 0.5, ans is 1.5675*10^-5,
When 1 its 1.65*10^-5

[Anonymous No.16004107]

>>16004099
>>16004105
Might also be a sig fig issue

[Anonymous No.16004111]

>>16004105
I was supposed to be doing this using Q'(t)? I have just been doing Q(t) for two hours. Thank you anon.

[Anonymous No.16004125]

>>16004111
Rip anon. Current is the derivative of charge.

[Anonymous No.16004280]

>>16004276