/sqt/ /qtddtot/ - stupid questions thread

[Anonymous No.11586260]

Formerly >>11565779

>what is /sqt/ for
Questions relating to math and science, plus appropriate advice requests.
>where do I go for other SFW questions and requests?
>>>/wsr/ , >>>/g/sqt , >>>/diy/sqt , >>>/diy/ohm , >>>/adv/ , etc.
>books+articles in pdf
libgen.is (Warn me if the links break.)
>book recs?
https://sites.google.com/site/scienceandmathguide/
https://4chan-science.fandom.com/wiki//sci/_Wiki
https://i.imgur.com/MrGTBFB.png
>how do I post math symbols?
https://i.imgur.com/vPAp2YD.png
>a google search didn't return anything, is there anything else I should try before asking the question here?
https://scholar.google.com/
>where do I look up if the question has already been asked here?
>>/sci/
https://boards.fireden.net/sci/
>how do I optimize an image losslessly?
https://trimage.org/
https://pnggauntlet.com/

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

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

[Anonymous No.11586263]

How would I convert peak to peak wattage for an AB push-pull amp into RMS watts?

[Anonymous No.11586275]

Is there a periodicity to the wavelengths emitted by most obects/materials? We see approximately one octave of visible light, from 380 to 760ish nm, I'm wondering if that's because the emission ratios repeat so it'd be unecessary to see outside one octave

[Anonymous No.11586307]

Unanswered questions from the previous thread:

Math:
>>11567559
>>11567586
>>11573179 (I think it was answered in /mg/)
>>11574035
>>11577936
>>11578793
>>11585244
>>11586197 (Not quite unanswered, but it might actually be just notation I don't know about)
>>11586224

Physics questions:
>>11569817
>>11586249

Statistics and probability:
>>11566163
>>11574907
>>11579534

Chemistry:
>>11567346

Engineering questions:
>>11571018 (Technically not a question about engineering, but still a question an engineer would be able to solve)
>>11576045
>>11572115
>>11581704
>>11585783

Medicine:
>>11573450
>>11574840

/g/ questions:
>>11579518
>>11583079

Stupid questions:
>>11566181
>>11566477
>>11570382
>>11571341
>>11572491
>>11574708
>>11575493
>>11581861
>>11581932

Assorted non questions that received no replies:
>>11567165
>>11571340
>>11579622
>>11579989

Other requests for help:
>>11574322 (Pretty late now)

[Anonymous No.11586316]

Is there a periodicity to the wavelengths emitted by most obects/materials? We see approximately one octave of visible light, from 380 to 760ish nm, I'm wondering if that's because the emission ratios repeat so it'd be unecessary to see outside one octave

sorry for reposting but it got buried by the thing

[Anonymous No.11586328]

>>11586260
Is less than 1 the same as 1?

[Anonymous No.11586335]

Not a regular here so sorry if I do some stupid mistake in this post, but I'm curious about something and wondering if you guys could help.

A while back me and friends went to a museum that had a lot of small physics experiments for visitors to play with. Pendulums, static electricity, etc. One of them was two metal plates you'd put your hands in, and it'd measure... voltage? Don't recall the unit it was in.

My friends were struggling to get it going past halfway, even when combining efforts, but I immediately made it shoot past the scale. My friend concluded it's probably something to do with me being a fatso- more mass generating more electricity or something. It sounds believable enough for me, but I've another friend there who's only a few pounds lighter than me but proportionally was making a much much lower output, so I'm wondering if it might not be something else. Anyone know?

tldr- in a machine that measures your body's natural electricity, what makes some people measure higher than others?

[Anonymous No.11586349]

>>11586316
>>11586275
>>11586249
Please don't do this, anon. The answer to your question is no, btw.
>planck's law
>wien's displacement law
>>11581704
What text?
>>11586335
It was probably measuring your resistance/conductivity or something. If your hands were sweaty or if you made better or worse contact, the resistance fluctuates. Not to mention the absolute thickness and fat content of your hand.

[Anonymous No.11586351]

I can't seem to wrap my head around this.

[Anonymous No.11586381]

>>11574035
I'd try using cylindrical coordinates where the cylindrical axis is x. i.e.
[eqn]
R^2 = y^2 + z^2 \\
z = R \sin \theta \\
y = R \cos \theta [/eqn]
Then density becomes
[eqn]
\rho = x^2 + R^2 \sin^2 \theta [/eqn]
and the parabloid becomes
[eqn]
2(y^2 + z^2) - y^2 = 2R^2(1 - \cos^2 \theta) = 2R^2 \sin^2 \theta = 4x [/eqn]
Not sure if that helps, but it might simplify things

[Anonymous No.11586391]

Can someone explain fat tailed / heavy tailed / long tailed distributions? Is that one and the same thing? heavy vs long vs fat?

[Anonymous No.11586395]

>>11586381
Forgot to mention, the boundaries for \theta here are just 0 and 2[math]\pi[/math]. The boundries for R are 0 and
[eqn]
\frac{\sqrt{2x}}{|sin\theta|}
[/eqn]

[Anonymous No.11586412]

>>11586381
Easy to see, that integration over x from 0 to 2, remaining integral over dydz could be done with "spherical" coordinates(y=a*r*cos(\varphi), z=a*r*sin(\varphi), where a is a suitable constant).

[Anonymous No.11586459]

>>11586351
Well I can explain the second part to you right away: the kernel of any homomorphism is normal, and clearly eHe^{-1} = H, so that kernel must be contained in H as the intersection of H with other sets.
I'll think about the first part.

[Anonymous No.11586478]

>>11586351
Wild guess says that [math]g[/math] fixes [math]xH[/math] (or something similar) if and only if [math]g \in xHx^{-1}[/math]. The fixers of the entire thing is thus the intersection.
Fuck messing around with finicky coset shit, tho.

[Anonymous No.11586481]

>>11586351
>>11586459
The first part is just symbol-shuffling.
[eqn]\ker = \{g \in G: gxH = xH\} = \{g \in G: x^{-1}gxH = H\} = \{g \in G: x^{-1}gx \in H\} = \{g \in G: g \in xHx^{-1}\}[/eqn]
And since I should have been writing for all x in G that whole line, the last set is [math]\cap_{x \in G xHx^{-1}[/math]

[Anonymous No.11586535]

Hi, I'm learning linear algebra from Strang's intro book. I'm sure if he's trolling me or not, but my calculator doesn't like this and neither does numpy. Can someone explain what's going on. I'm a little slow. Thanks

[Anonymous No.11586545]

>>11586535
What are you not able to do?
The green circle isn't very explanatory.
If you're trying to do question 7, you just need to take the dot product between each pair of vectors using the
fact that
[math]
\mathbf{a} \cdot \mathbf{b} = | \mathbf{a} | |\mathbf{b} | \cos \theta
[/math]
where [math]\theta[/math] is the angle between the vectors.

[Anonymous No.11586549]

>>11586545
I meant
*you just need to take the dot product between each pair of vectors and use the fact that...
just in case that made it too unclear

[Anonymous No.11586564]

>>11586545
question 7d has me taking arccos of -5 / sqrt(10)*sqrt(5), which should equal theta but is showing up as NaN...?

[Anonymous No.11586585]

>>11586564
Sure you got your parentheses right?
If you just put in
- 5 / sqrt(10)*sqrt(5)
it would be parsed as
- ( 5 / sqrt(10)) * sqrt(5)
which would cause an error when you try to arccos is because it's not in the range [-1,1]

[Anonymous No.11586590]

>>11586564
>>11586585
or even better, simplify your shit first. You don't need to be using a calculator to do this; [math]\frac{-5}{\sqrt{10}*\sqrt{5}} = \frac{-1}{\sqrt{2}}[/math], and this is just 3pi/4.

[Anonymous No.11586593]

>>11586590
*the arccos of this is 3pi/4, rather

[Anonymous No.11586595]

>>11586349
thanks for replying. i checked out wiens and plancks laws but those deal with black body radiation for objects that are hot and are , well black bodies. im talking about typical visible objects, that normally reflect light but at certain wavelengths. im wondering something like this:

might an object that has peak reflection at N nm, such as a leaf, also have smaller or nearby peaks at N/2 or 2N nm, an octave above or below?

>>11586564
try phase shifting it

[Anonymous No.11586602]

>>11586590
desu senpai, I don't know the arccos of -1/sqrt(2) off the top of my head
I guess you can figure it out by considering a right-angled triangle with two sides of length 1, but eh it's easier to just type into a calculator and not think about it

[Anonymous No.11586607]

>>11586585
Ahh yes, thanks. That's in radians, so looks like the answer is 135 degrees in the end. Makes sense. Sweet, thanks again.

[Anonymous No.11586609]

>>11586602
>but eh it's easier to just type into a calculator and not think about it

[Anonymous No.11586620]

>>11586609
>spending time thinking about elementary school trigonometry when you could be working on whatever it is you're working on

ah yes, of course I'm just trying to prove that I'm not a dumb-dumb
but of course I am a dumb-dumb

[Anonymous No.11586644]

>>11586620
elementary students don't study trig

[Anonymous No.11586809]

In the extended reals, [math]\mathbb{R}\cup\{\infty\}[/math], if I define 'x' on the interval [math](0,\infty)[/math], does that include the negatives, or just the positives?

[Anonymous No.11586899]

>>11586260
In regards to computer architecture and branch prediction, was there any thought or research into executing both branches and discarding the incorrect results in a processor? I'm sure someone put some thought into it but I can't find anything about it. My immediate thoughts are that at 100% usage, the CPU performance would degrade due to essentially lengthening your program, but at lower usages, maybe 30%, you would received faster performance due to no mispredicted branch stalls. There would be increased hardware associated with computing both branches, but you would get rid of the prediction hardware.

[Anonymous No.11586903]

>>11586809
Are you talking about the extended reals in your post, or are you talking about the projective real line in your image? They are not the same thing. The extended reals have two distinct infinities, the projective line glues them together.
Either way the answer to your question is "just the positives" but for the projective line the reason is simply because the notation is defined that way; there aren't such things as intrinsic "intervals" in [math]\mathbb{RP}^1[/math] because there's no coherent way to compare elements to infinity.

[Anonymous No.11586939]

Is it safe to say that if we could realistically travel to a black hole (or into a black hole's orbit) then we would be fucked for being too close to that black hole?

>>11586351
is it just me or is this poorly redacted?

[Anonymous No.11587123]

hey anons, just trying to have some of my work approved before submission.this one's about finding a macluarin for f(x). I'm going to rewrite the terms of the Taylor series in this one tomorrow, but is it overall correct? the maclaurin looks accurate?

>1/2

[Anonymous No.11587125]

going to bed, big thanks for any help

[Anonymous No.11587148]

>>11587123
Ok, not asleep yet but wanted to say that I’m not sure if my T_n(x) thingys should be evaluated at a=0 (since that was given in the question) or if the thingys are just supposed to be treated as functions being differentiated with respect to x.. I’m inclined to think I royally fucked up T thingys and instead they should like the terms in the Maclaurin, correct?

[Anonymous No.11587151]

>>11587123
There are a lot of mistakes in this. Some of them are just bad typesetting, but the answer is wrong too.
>don't write [math]x-1^4[/math] when you mean [math](x-1)^4[/math]. You can't leave off the parentheses. Don't write [math]-1^n[/math] either.
>You forgot to put x terms in your T1 expansion.
>In your T1 expansion and in the final answer, you forgot to evaluate your derivatives at 0. Remember that the coefficients of the Maclaurin expansion are obtained from the derivatives __evaluated at zero__. It doesn't make any sense to have an x in the Maclaurin coefficient. Even for general Taylor series where you don't know that a = 0, it should be only a in the coefficient.
>You've flipped some signs. When n is odd, then (0-1)^n = -1. So, for example, your 4th-degree approximation should be [math]-2-6x-12x^2-20x^3-30x^4[/math]. In general, every coefficient will be negative.

You do have the derivatives right (once you fix the typesetting) and once you evaluate the coefficients of the bottom at 0, you'll have the correct general Maclaurin expansion too.

[Anonymous No.11587174]

Guys, so I have this Arrhenius kinetic with dimension of [1/s]. I want to convert it to surface reaction rate with dimension of either [mol/m2/s] or [kg/m2/s]. How do I do that?

[Anonymous No.11587207]

>>11587151
>>11587151
Agree about the parents and shit typesetting, fixing that first thing tomorrow. But wdym about
>You forgot to put x terms in your T1 expansion
? I’ve looked and looked but don’t see what’s wrong with it (but I do NOT understand this stuff yet, only started reading today).

>you forgot to evaluate your derivatives at 0
This too occurred to me after calling it a day. Initially I had done this for the Tn expansions, but I changed it to there more general form after being unsure. So my final line in the pic should be evaluated for x=0 then, right? Its that simple? The maclaurin series sigma notation still needs to be mulitiplied by x^n though right? I mean, if that’s true the whole things just zero, so I don’t think it’s true, but I don’t see why it wouldn’t be.

thanks a bunch anon

[Anonymous No.11587211]

if the sum of an and bn both diverge, can the sum of (an+bn) converge? It seems like this should be possible roight
For example if an = 1+1+1... and bn = -1-1-1...
Then (an+bn) converges to zero
Yes?
No?
Maybe so?

[Anonymous No.11587220]

>>11587211
Yes you fucking retard

[Anonymous No.11587231]

>>11587207
The Maclaurin series is [math]f(0)+f'(0)x+\frac{f''(0)}{2}x^2+...[/math].
What you've written for the final line is [math]f(x)+f'(x)x+\frac{f''(x)}{2}x^2+...[/math].
See the difference? The derivative out in front of x^n is not allowed to have an x in it. You need to evaluate only the derivatives at 0, but not the x^n pieces.

[Anonymous No.11587244]

>>11586595
See https://imgur.com/a/rHUhA6y

For an opaque surface (blackbody or not; graybody or not), [math] \text{radiosity}=\text{emissive power}+\text{reflected indecent radiation} \implies J=E+G_\text{ref} [/math]. Real spectral emissive power differs from a that of a blackbody by a factor called [math] \epsilon_\lambda [/math] (figure 1). Emissive power is then [eqn] E=\int_0^\infty \epsilon_\lambda(\lambda,T) E_{\lambda,b}(\lambda,T)\ \text{d}\lambda [/eqn] where [math] E_{\lambda,b} [/math] comes from Planck's law. Similarly, spectral reflectivity is largely found with experiment (figure 2). With spectral irradiation given, the reflected incident radiation is [eqn] G_\text{ref}=\int_0^\infty\rho(\lambda)G_\lambda(\lambda)\ \text{d}\lambda [/eqn] If you want spectral emissive and reflected power, just cover up the integral signs and imagine a subscript lambda in the LHS of the eqns above. So even though the spectral quantity vary with wavelength, periodic they are not.

[Anonymous No.11587249]

>>11587231
<3

Was my other pic >>11587125 just as fucked up? I still feel good about that one mostly

[Anonymous No.11587255]

>>11587249
No, that one looks correct.

[Anonymous No.11587261]

>>11587220
If I had 1+1+1+... = an and 1-1-1-... = bn
Then an+bn = 1-1+1-1+1-1...
Can this really have a sum. Is this not similar to inf-inf?

[Anonymous No.11587265]

>>11587261
Yes, that's fine. https://en.wikipedia.org/wiki/Riemann_series_theorem

[Anonymous No.11587476]

>>11587211
>>11587261
Clearly, yes, sum converges. This feature is at the heart of e.g. the Erdos discrepancy problem.

[Anonymous No.11587526]

Does anyone recognise this? Can you tell me what textbook its from

[Anonymous No.11587959]

>>11587261
It depends completely on how you sum the terms. It all changes if you sum certain terms before others.

[Anonymous No.11587985]

>>11587265
>there are actual people out there that call it "the Riemann series theorem"

[Anonymous No.11588073]

Have I gone wrong somewhere here? Steady State Approximation kinetics

[Anonymous No.11588376]

>>11586899
I found the answer to my question in
Wall, D. W. [1993]. Limits of Instruction-Level Parallelism

[Anonymous No.11588514]

Any ideas on where to go from here? I know I need to construct some neighborhood of [math]\{A\}[/math] in the quotient space that doesn't contain any of the [math]U_n[/math]s but I can't see how to make one from the original space.

[Anonymous No.11588520]

>>11587255
did the stuf f below 'delete this' look correct too? does that even need to be there if all I'm after is a third degree approximation?

[Anonymous No.11588591]

>>11588514
First of all, if we have an open set [math]X[/math] which contains [math]A[/math], then it's preimage at any copy [math]I_n[/math] contains a half-open interval up to some real number, that is, [math]\pi ^{-1} (X) \cap I_n \supseteq [0, a)_n[/math] for some [math]a>0[/math], by the definition of quotient topology, right?
So for whatever countable basis, you just need to construct an element which isn't in any basis element by making him stop halfway before this [math]a[/math] at some given coordinate.
So essentially, we have a real-valued function [math]f(X)[/math] such that [math]\pi ^{-1} (X_n) \cap I_n \supseteq [0, f (X_n) ~ )[/math]. Then if we have a supposed countable basis given by the [math]A_n[/math], we can construct [math]L \subset \prod I_n[/math] given by [math]L = \prod _n [0, f (A_n)/2 )[/math] whose projection thus doesn't contain any element of the supposed base.

My bad for any typos, possible mistakes and for the confusing notation, the products and the quotients also confuse me.

[Anonymous No.11588598]

how do i deal with constants when writing a differental equation in matrix form
for example
x' = 3t^2x - 10
x = x1
[x'] = [(3t^2 - 10/x1)] [x1]

is the 10/x1 okay? Like am I allowed to do that (put the 'x1' in the coefficient matrix)? This textbook covers no instances of it and web searches aren't really giving me what I want either

[Anonymous No.11588601]

>>11588591
>local idiot mixes up disjoint unions and products
I don't even know how do I do this stuff.

[Anonymous No.11588603]

>>11588598
ah nevermind I just checked the solutions (forgot I had that) and it just added another matrix at the end containing the constants

[Anonymous No.11588809]

Why is isopropyl alcohol so effective against insects? What does it do to them?

[Anonymous No.11588827]

>>11588591
I think I get what you're saying. Basically, for some basic neighborhood of the point [math]\{A\}[/math] in the quotient space, take its preimage and intersect it with one of the "rungs" of the interval ladder which makes up the original space. This must correspond to a subinterval that looks like [math][0,a)[/math]. Then create a new subinterval [math][0,\frac{a}{2})[/math].

Now, do this for every basic neighborhood. Then the union of all these new subintervals creates a new neighborhood containing the closed set [math]A[/math]. Then the projection of this neighborhood back to the quotient space is a neighborhood of the point [math]\{A\}[/math] which, by construction, does not contain any of the original basic neighborhoods.

[Anonymous No.11588858]

>>11588809
polar enough to creep into their airholes, volatile and dessicating, painful and mildly toxic
Remember, their airholes lead directly to their muscles and organs

[Anonymous No.11588867]

How far away are we from immortality? What about an extended lifespan?

My grandparents died today, and it got me thinking that this life is all I have, and once it's over it's fucking over. I won't remember it even if something like reincarnation exists.

It could just be that most people can't imagine a good future in our current society, but I honestly don't see why extending everyone's lifespan by 50 or so years wouldn't be possible in the near future.

Am I wrong? Is it impossible to delay old age? Just like it's impossible to stop cancer or heart disease or anything else from killing old people?

[Anonymous No.11588880]

>>11588601
>>11588591
Hey, good to have you back, don't leave us like that again ever

[Anonymous No.11588894]

>>11588867
>How far away are we from immortality?
Very far. Delaying or counteracting ageing is almost definitely possible to some extent, but it's a long way from being plausible in humans. It may happen in your lifetime. It might not. The stuff we've done is too primitive at this point, it's basically just mouse experiment-tier.
Probably the bigger issue than ageing itself is going to be all the shit like cancer and heart disease and strokes that we don't know how to cure or if they're fundamentally even curable. Simply pooping out due to old age one day is not a common way to go. Most people die _because of_ something that we can't currently stop.

[Anon No.11588895]

Can someone tell me the answer to this question.

[Anonymous No.11588918]

>>11588894
I see. I think a big part of the modern fear of old age has to do with a lot of people not being able to see a bright future in their lifetime or at all, as well as the natural fear that once you die you die. There's no going back everything you've done all your memories are gone.

I'd at least like to be able to live to the ripe old age of 150, but there's no telling how likely that is in our lifetime.

[Anonymous No.11588926]

>>11588827
Yeah, that's completely correct. I just mixed up a bunch of disjoint unions and products.
I'd still like to emphasize that for the n-th element of the "base" you pick out the n-th interval in the disjoint union when you're constructing the counter-example, since you seemed to skip this out and since the notation makes it hard to explain.
>>11588880
Hopefully my new IP doesn't get perma banned again.

[Anonymous No.11588945]

Would anyone be willing to help me through a series of linear questions? Willing to pay.

[Anonymous No.11588946]

>>11588926
>Hopefully my new IP doesn't get perma banned again.
Why would they perma ban you?

[Anonymous No.11588958]

can anyone how this inverse came about? Whenever I search about "inverse of a matrix of functions/solutions" I dont get any useful results

I get that the matrix of coefficients is [ 6 -3 2; -1 1 1; -5 1 1] and the inverse of THAT matrix is [ 0 1/4 -1/4; -1/5 4/5 -2/5; 1/5 9/20 3/20] and I suppose the LITERAL inverse (reciprocal) of [e^-t e^-2t e^3t] would be [e^t e^2t e^-3t] (I'm assuming that, please correct me if Im wrong. Idk how to take the inverse of a vector or a matrix of functions).
And since X = matrix of coefficients * [e^-t e^-2t e^3t] I assumed that its inverse would be the inverse of the coefficients * [e^t e^2t e^-3t] but apparently thats not the case according to pic related since matrix multiplication doesnt work that way
please help

[Anonymous No.11588986]

>>11588946
Dude fuck if I know.
>>11588958
Set [math]a = e^t[/math]. Then [math]a^2 = e^{2t}[/math], [math]a^3 = e^{3t}[/math], [math]a^{-1} = e^{-t}[/math] and [math]a^{-2}=e^{-2t}[/math].
Now pretend [math]a[/math] is a number and invert the matrix normally.
It's literally that stupid. It just looks funny because the e^t keeps changing powers.

[Anonymous No.11589008]

>>11588986
so I shouldnt have separated X into two matrices, but instead performed the inversion on the original X?

[Anonymous No.11589012]

>>11589008
Yeah.

[Anonymous No.11589026]

>>11589012
wow that seems like a total pain in the ass
I'll save that superpower for the final

[Anonymous No.11589040]

>>11589026
I mean, you can also give X as the multiplication of your matrix of coefficients and a matrix with the [math]a[/math]s on the diagonal and then use [math](AB)^{-1} = B^{-1} A^{-1}[/math].
But it needs to be a three by three matrix with the as, it can't just be a vector.

[Anonymous No.11589065]

>>11589040
I honestly have no idea what you meant by this (It's definitely not you, Im just an idiot) but I did realize that it's not actually a pain in the ass. I just replaced the last step in my inversion process with [a 0 0; 0 b 0; 0 0 c] and then just did the last few steps to convert them to 1s like you recommended in >>11588986
Yeah Im an idiot
Thanks for helping though

[Anonymous No.11589152]

Could someone explain the difference between definite and indefinite integrals briefly and in a simple manner? I have been doing indefinite integrals and just got introduced to definite integrals, I'm watching Prof. Leonard on YT.

My understanding is, indefinite integrals must have a constant C, because we're not sure what the area actually is, that's why they're called indefinite in the first place. They give you an area function of whatever function you integrated. So basically if you're doing indefinite integrals, you end up with a family of area functions, and until you set boundaries to that area function, you cannot determine C, aka pick one out.

The definite integral immediately gives you area? I'm not really sure about definite integrals since I've just been introduced to them, but I'm a bit confused about the whole C thingy. I understand a multiple (an infinite) number of functions can have an identical integral, so that's another reason we need C, because we don't know which function exactly it came from. But what determines C in the first place is the boundaries we set. Can someone elaborate on the constant please?

[Anonymous No.11589167]

How are people fighting the urge to cheat in online classes? I specifically requested that my teachers watch me via Zoom while I take exams, otherwise I'd be tempted. My one professor literally emailed us the exam and told us to get it back to him within 12 hours and "follow the honor system". The class average was a full 10 points higher than either of the previous two exams.

[Anonymous No.11589183]

>>11589167
That's autistic as fuck anon. You're literally setting yourself up for failure in life lol.

[Anonymous No.11589207]

>>11589183
Elaborate

[Anonymous No.11589224]

>>11589167
I'll be honest with you.
If I were a professor, at the worst I'd watch over stack exchange and quora so I could give my students incorrect answers as a joke.
But most professors I've known don't really care about undergrad students or the classes they have to give to undergrads. Sometimes they enjoy teaching, but what you do in tests isn't really a concern.

[Anonymous No.11589258]

>>11589152
Let's restrict ourselves to real-valued functions of a single real variable. Then the indefinite integral is really just the anti-derivative. That is, if the derivative of f() is f'(), then the indefinite integral of f'() is f(). The constant shows up because any function f()+C will have have the same derivative f'().
The definite integral actually computes areas. Consider the definite integral of f'() over the interval from a to b. If you consider this as a function of b, with a fixed, you can show (1) that it's derivative is f'(), and (2) that its value at b=a is 0. (1) Tells you that it is of the form f(b)+C, and (2) tells you that C = -f(a). So the value of the definite integral is f(b) - f(a).

[Anonymous No.11589342]

hi, I need a curious fact about basic 3 digit division for work, for example 117/3 if anyone know any. so far I have that 2520 its divisible by numbers 1-10

[Anonymous No.11589481]

How the fuck do I remove this connector?

[Anonymous No.11589499]

>>11589481
push the clip on the highlighted area and pull

may take some force

careful not to bend the PCB

[Anonymous No.11589568]

In chemistry or concerning drugs, what is the difference between the suffixes -crine and -line, e.g. quinacrine and quinoline?

[Anonymous No.11589589]

>>11589152
The terminology is kind of stupid. It should really be "antiderivative" or "primitive", and "integral", instead of "indefinite integral" and "definite integral."

[Anonymous No.11589594]

>>11588926
Then stop avataring you fucking idiot.

[Anonymous No.11589603]

>>11589594
>the audacity of this nigga

[Anonymous No.11589611]

I'm answering some questions on reaction rates. Some things that obviously increase reaction rate are temperature, surface area, and presence of a catalyst. In this particular reaction:

2NaCl(s) + AgNO3(aq)Na2NO3(aq) + AgCl2(s)

Would the reaction rate change if we used sodium chloride in the aqueous state rather than the solid state? My instinct is that it wouldn't make a difference. Apologies for no subscript numbers for the chemicals that need them.

[Anonymous No.11589682]

>>11589568
> In 1842, French chemist Charles Gerhardt obtained a compound by dry distilling quinine, strychnine, or cinchonine with potassium hydroxide;[4] he called the compound Chinoilin or Chinolein
so -line is a corruption of -olein, probably meaning "oil"
>http://www.chemspider.com/Chemical-Structure.8860.html?rid=b65fc4d1-a190-4ef4-9e04-bc8620ac0edf
the substructure of quinacrine is called acridine, so -crine is probably just some retarded drug marketing invention (aCRIdiNe), since words like "endocrine" are more familiar to normies

[Anonymous No.11589693]

>>11589594

[Anonymous No.11589723]

>>11589594
no you don't understand people on 4chan have to know which kawaii anime girl they're talking to
you can't just post anonymously on an anonymous imageboard

[Anonymous No.11589727]

Why isn't desmos graphing this?: https://www.desmos.com/calculator/lnv6se8pxf

[Anonymous No.11589736]

>>11589727
stop using square brackets

[Anonymous No.11589751]

oh man complimentary halloween gif good pic choice op even if it's weaboo-esque

[Anonymous No.11589847]

>>11589167
>>11589224
My prof said he would be monitoring 'forums' and chegg for cheating, but looking at my exam some questions are pretty straightforward. How would they prove you did cheat or not when the right answer will look similar across most students?

[Anonymous No.11590143]

OP THIS IS NOT FUCKING SCIENCE YOU FUCKING BASTARD. THIS IS ANIME WEEB SHIT BULLSHIT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I AM AN INTELLEGUTUAL!!! I WAN TSCIENCE

[Anonymous No.11590147]

>>11587244
thank you, so looks like there's just one peak, not particularly focused around visible. I looked at the spectral lines for wood too and there's some wibbling around a center but no periodicity outside that range

i wonder why visible light evolved in the range where it did and is approximately one octave. anyone have insight?

[Anonymous No.11590169]

is there any data on the vision ranges of animals that see infrared/uv, in terms of specific wavelengths? tried to find but cant

[Anonymous No.11590207]

>>11590147
>>11590169
nevermind guys i found the answer

[Anonymous No.11590212]

>>11586260
>Prove that if f is a linear functional on nxn matrices, f(AB)=f(BA) necessiates f is scalar multiple of tr(AB). (Use tr(AB)=tr(BA))
I am stucc. I ended up with f(AB)= sum of scalar(i) times AiBi= f(BA), not necessairly one scalar times the whole sum.
Did something went wrong here or can I proceed from here?
>let Eij be nxn matrix whose ij entry is 1, rest 0
>Eij are a basis of nxn matrix space
>f(AB)= f( sum on i sum on j of Eij times i,j entry of AB)
>f(AB)= sum on i(sum on j( f( Eij) times i,j entry of AB))
>f(Eij)= sum on l(sum on k( scalar(l,k)* f(l,k) (Eij))) = scalar(i,j)
>where f(l,k) = 1 if l=i and k=j, 0 otherwise
>so f(AB) = sum on i(sum on j( scalar(i,j) times i,j entry of AB))
>similiarily, f(BA) = sum on i(sum on j( scalar(i,j) times i,j entry of BA))
>for Every A and B, entry of AB = entry of BA if we compare entries on the diagonal
>so f(AB)= f(BA) implies that below is true
>sum on i, sum on j of scalar(i,j) * sum on k( A(i,k) B(k,j)) =sum on i, sum on j of scalar(i,j) * sum on k( B(i,k) A(k,j))
>and that can be true only if i=j, so we can keep only those scalars with i=j and omit one sum, rest is zero
>and here i am with
> f(AB)=sum on j of scalar(j) * sum on k( A(j,k) B(k,j)) =sum on j of scalar(j) * sum on k( B(j,k) A(k,j)) =f(BA) that is f(C)= sum on j( scalar(j)*C(j,j)) for arbitrary matrix C, where it is not necessairly a scalar multiple of trace of C.

[Anonymous No.11590231]

I want to learn probability. I even tried an online course but I still have difficulty knowing when to apply what. Could someone reccomend some books or resources on the subject?

[Anonymous No.11590277]

>>11588926
Thanks for the help. Pic related is my final draft of the proof

[Anonymous No.11590304]

>>11590212
You need to find 2 (very simple) matrices so that when you multiply them one way, f(result) is scalar(i), but when you multiply them the other way f(result) is scalar(j). That way you'll know scalar(i) and scalar(j) are equal.

[Anonymous No.11590322]

>>11590304
i thought it could be something like that, thanks. ill get on it when i wake up

[Anonymous No.11590359]

>>11586260
I often wake up at night drenched in cold sweat. I have done lots of tests and asked a lot of doctors but theu don't have a clue why this is happening. Is it that I eat too much at night? Is it a pinched nerve on the neck or back? No idea. Other than that I feel great. Some docbro outright told me "dude you're just weird"

[Anonymous No.11590373]

>>11589736
thanks I figured it out shortly after

now what do you know about plotting a series in gnuplot?

[Anonymous No.11590406]

>pi is equal to the circle's circumference divided by its radius
>pi is transcendental and it can't be expressed as a fraction

[Anonymous No.11590410]

>>11590406
pi is transcendental and it can't be expressed as ratio of integers***

[Anonymous No.11590952]

>>11590410
can it be expressed as ratio of rationals?

[Anonymous No.11591021]

>>11590212
>this monstrosity
The first thing is that, because of linearity, you just need to show that it's a multiple of the trace for the base. You don't need to actually decompose random A and B like that.
I get the impression the other anon was also thinking of this but forgot to actually mention it. I also get the impression you were thinking of this but didn't actually do it.
The second thing is, just to remind you, [math]E_{ij}E_{jk}=E_{ik}[/math].
>>11589751
Nice.
>>11590231
Cute Remilia.

[Anonymous No.11591069]

so first of all sorry for the shitty diagram.

I am studying Amperes law and was doing some exercises for myself to try and understand it more intuitively. Here I am trying to reason how
[math]\oint\vec{B}\cdot d\vec{s}=0[/math]
in this situation.

I have convinced myself by symmetry that all dot products of [math]\vec{B}[/math] and [math]d\vec{s}[/math] cancel except for at the points in the center of the paths BC and DA, at which the induced magnetic field by the current is parallel (for BC) or antiparallel (for DA) to the corresponding path element, and which there is no symmetrical elements for either. Furthermore it is obvious [math]||\vec{B}||[/math] at the point marked [math]y[/math] is greater than [math]||\vec{B}||[/math] at the point marked [math]x[/math], so I am not convinced they cancel out.

Do we simply ignore these points or approximate them as equal since they are infinitesimal pieces? I feel like I am missing something with my reasoning.

[Anonymous No.11591082]

>>11591069
I posted the wrong version of the diagram it seems. The marked point [math]x[/math] I referred to is in the center of segment BC and [math]y[/math] is the center of DA.

[Anonymous No.11591199]

>>11591021
continuing, if if f(AB)= F(BA) for every A and B and for every C f(C) = sum of cj* C(j,j), f(EijEji)=f(Eii)= ci = f(EjiEij)=f(Ejj)= cj, and this holds for every i and j, thus all the cj scalars, j 1 through n, are equal. So f(C)= c*trace(C)

[Anonymous No.11591261]

>>11591199
Bingo.
You can also use the same trick to clean up the earlier parts of the proof.
So, for [math]i \neq j[/math], we have [math]f(E_{ij}) = f(E_{ii}E_{ij}) = f(E_{ij}E_{ii})=f(0)=0=\lambda ~ trace(E_{ij})[/math] for any complex [math]\lambda[/math]. With the case [math]i=j[/math] you've just proved, this completes the proof across the whole base.

[Anonymous No.11591282]

It's been years since I've taken matrixes and I need someone to explain the solution for a simple m x 1 matrix. Why is pic related equal to 1? My first thought was you multiply each element with 1 and add them all up, which would then add up to 2.

[Anonymous No.11591290]

If I want to find the conditions where linear transformation T(A)=AB is isomorphic on B, is it enough to simply state that B needs to be invertible?

[Anonymous No.11591387]

>>11591290
Assume [math]B[/math] is invertible. Define [math]Z: M_n \rightarrow M_n[/math] by [math]Z(A)=AB^{-1}[/math].
Then [math]Z \circ T (A) = Z(AB)=(AB)B^{-1}=A[/math]. Since it's invertible, it's also an isomorphism.
Now, assume [math]B[/math] is not invertible. Then, [math]B^T[/math] is also not invertible, and [math]B^T v = 0[/math] for some collumn vector [math]v[/math] non-zero. Finally, [math](B^Tv)^T=v^T B = 0[/math].
So you can construct the matrix [math]C[/math] where every row is just [math]v^T[/math] for a quick and easy counterexample, since then [math]CB=0[/math].

[Anonymous No.11591399]

>>11591387
Thanks. I found that Ker(T) is 0 only when A=0 when B is invertible, thus injection. Then used the fact that the nullity of T=0, which means that rank of T = the dim of codomain.

I don't completely understand the second part, but your method is elegant.

[Anonymous No.11591428]

>>11591399
>I don't completely understand the second part, but your method is elegant.
It's not, it's really dumb. I'll give an "unskewed" version.
Set [math]T(A)=BA[/math]. If [math]B[/math] isn't invertible, there's some non-zero collumn vector [math]v[/math] such that [math]Bv=0[/math], where [math]0[/math] is the zero vector. Then, if we construct the matrix [math]C[/math] all of whose collumns are just [math]v[/math], then because of how matrix multiplication works, [math]BC=0[/math], where [math]0[/math] is the zero matrix. That is, matrix multiplication is always line times collumn. Any line times [math]v[/math] zeroes, but every collumn of [math]C[/math] equals [math]v[/math]. Then, [math]T(C)=0[/math].
The thing from earlier is the exact same, but with some transposes and more confusion because of the side from which B multiplies.

[Anonymous No.11591454]

>>11591282
>Why is pic related equal to 1
It isn't. Provide more context.

[Anonymous No.11591467]

I'm doing integrals, when doing definite integrals, the general statement is [math]\int^b_af(x)dx=F(b)-F(a)[/math]. This is saying that you're supposed to evaluate an integral from whatever arbitrary point to b minus integral evaluated from that same arbitrary point to a, and that difference would give the area you're looking for, ie a to b.

Now what I don't understand is this arbitrary point part: like if you're able to just choose some arbitrary starting point and evaluate the integral from that point to b, why couldn't you simply just choose that arbitrary point to be a in the first place, so instead of going around the problem, taking the difference of areas from some arbitrary fucking point to b and that same point to a, why not just make the arbitrary point a, evaluate at b and here's the area?

I just started studying this and I'm clearly missing something because this "workaround" makes no sense.

[Anonymous No.11591470]

Is there a difference in chance between one person rolling 1/3 to decide who to give an item to; compared to three people rolling 1/100 and the highest roll wins?

[Anonymous No.11591490]

>>11591467
maybe its just to do with the wording of whatever you are reading, but I am pretty sure what you are thinking in the second paragraph is what the actual case is. i.e. you evaluate it at an arbitrary point a and b and take the difference of these two evaluated expressions.

[Anonymous No.11591509]

how to interpret this equation to end up with a strain tensor?
[math]\mathbf{E} = \frac{1}{2} ( \mathbf{u} \mathbf{\nabla} + \mathbf{\nabla} \mathbf{u} )[/math]

del and u are vectors, but it is not a dot or cross product, the book has them next to each other, E is strain tensor which will just end up being 2x2 in this plane stress case

[Anonymous No.11591518]

>>11591454
It's for statistics, hypotheses testing. What throws me off is that later in the exercise I wrote r = 1 and drew a line to the C matrix like they're related in some way, but I can't remember why. Pic related, I have no clue why r = 1

[Anonymouss No.11591540]

>>11591470
The second method has to account for the possibility of ties.

[Anonymous No.11591549]

>>11591467
>Now what I don't understand is this arbitrary point part: like if you're able to just choose some arbitrary starting point and evaluate the integral from that point to b, why couldn't you simply just choose that arbitrary point to be a in the first place, so instead of going around the problem, taking the difference of areas from some arbitrary fucking point to b and that same point to a, why not just make the arbitrary point a, evaluate at b and here's the area?

how would you go about defining that anti-derivative?

[Anonymouss No.11591553]

>>11591467
The thing is, you don't explicitly choose the arbitrary starting point. Rather, you select an antiderivative of the function you are integrating, and end up with the starting point it uses.
For example, if you are integrating 2x, you should recognize this as the derivative of x^2, which has zero at x==0. The only way you can chose a different starting point is to subtract the value of x^2 at that point.

[Anonymous No.11591559]

>>11591509
Probably an outer product, that is, take it as the product of a column vector times a row vector. (Inner 'dot' product is as the product of a row vector times a column vector.)

[Anonymous No.11591564]

>>11591518
Relax, take a deep breath, and look at that sheet again. "r = 1" comes from the definition of the hypothesis you are testing.

[Anonymous No.11591590]

>>11591470
no

[Anonymous No.11591597]

>>11591470
100 isn't divisible by 3, there's no way to make the second one completely fair

[Anonymous No.11591605]

Alright /sci/, I have a question about psychiatry. I keep seeing videos in my youtube feed about about DID.. Now I was always told that DID or multiple personality disorder does NOT exist and it is fabricated by hacks and liars.

So is DID a real disorder, and a real diagnosis used by psychiatrists?

[Anonymous No.11591610]

>>11591597
>what the fuck did anon mean by this

[Anonymous No.11591619]

>>11588958
>>11589065
I suppose you might be confused between the inverse of a matrix and the inverse of a function.
To aid clarity, I will here write matrices and matrix-valued functions in bold, and real-valued variables normally.
If a matrix-valued function [math]\mathbf{X}[/math] maps the variable [math]t[/math] onto the matrix [math]\mathbf{X}(t)[/math], (i.e. [math] \mathbf{X}:t \to \mathbf{X}(t)[/math]) then inverse [math] \text{\emph{function}}[/math] is the function that maps the matrix [math]\mathbf{X}[/math] onto the variable [math]t[/m\th] (i.e. lets you know which t you would have to put into the function [math]\mathbf{X}[/math] to get the matrix [math]\mathbf{X}(t)[/math]) which would usually be notated [math]X^{-1}(\mathbf{\m}) : \mathbf{M} \to X^{-1}(\mathbf{M})[/math] i.e. the function [math]X^{-1}[/math] maps the matrix [math]\mathbf{X}(t)[/math] onto the real number [math]t[/math]. That function obeys the defining equation
[eqn] X^{-1} (\mathbf{X}(t)) = t[/eqn].
In contradistinction, the inverse of a matrix [math]\mathbf{X}[/math], which would also be notated as [math]\mathbf{X}^{-1}[/math] obeys the defining equation [math]\mathbf{X}^{-1} \mathbf{X} = \mathbf{X} \mathbf{X}^{-1} = \mathbf{I}[/math]. If that matrix depends on a real valued variable [math]t[/math] then that is just written as [math]\mathbf{X}^{-1}(t) \mathbf{X}(t) = \mathbf{X}(t) \mathbf{X}^{-1}(t) = \mathbf{I}[/math]. These two ideas are related but entirely distinct concepts.
It's useful to write the first equation i.e. the equation defining the inverse function as
[eqn] X^{-1} \cdot \mathbf{X} = f [/eqn]
where [eqn]f: x \to x[/eqn] is the identity function, i.e. the function that maps its input to its output without changing it.

Basically here you're taking the inverse of a matrix which happens to depend on a real-valued variable, not taking the functional inverse of that function. In this case, you just invert it as you would any other matrix.

[Anonymous No.11591623]

>>11591559
yep I think that's it thanks, I got the arithmetic to work out
it wasn't clear for me because I thought [math]\mathbf{\nabla} \mathbf{u}[/math] and [math]\mathbf{u} \mathbf{\nabla}[/math] were different because u in front of del wouldn't be the same as del operating on u, but I guess in this case that doesn't hold up, maybe that's only for dot products

[Anonymous No.11591631]

>>11591470
You might have ties with three people rolling, depending on how you resolve ties it can be either different or the same.

[Anonymous No.11591634]

>>11591605
>is DID a real disorder,
no
>[is it] a real diagnosis used by psychiatrists
unfortunately

[Anonymous No.11591637]

>>11591605
None of these things are real.

[Anonymous No.11591646]

>>11591634
>>11591637
Thanks anons. Glad my sense of bullshit is still working.

[Anonymous No.11591812]

>>11591553
So what you're saying is that the starting point is usually (always?) 0, unless tampered with?

[Anonymous No.11591832]

how do you guys manage to study complex subjects for so long?
I want to work through a textbook but I can only complete like 3 pages a day and I really have to force myself to do it, at that pace it'd take me a year to work through a single textbook or like 2 months for a single chapter
is that normal or not? I've tried stimulants but they only seem to help a little and I get extremely antisocial on them, I start getting into heated arguments with people, telling them they're stupid, get overall angry and irritable, etc

[Anonymous No.11591837]

>>11591812
>>11591553
Also, you do seem knowledgeable, so I'll shoot another question, taking an integral of some function gives an area function of it. But integrating is summing up an infinite number of rectangles under a graph of the function of which we're trying to find the area of. So basically, the conclusion is that summing up an infinite number of rectangles (unless we evaluate their heights to get an actual value) gives us a new function, which is the area function? Sorry if this sounds autistic, I'm just trying to connect what I've learned so far.

So basically [math]\int f(x)dx=F(x)+c[/math]

[Anonymous No.11591846]

>>11591832
Find a good source of information that is also fun and interesting. If a textbook is boring, you'll hate it.

[Anonymous No.11591870]

>>11591846
but that's the thing about complex subjects, it's usually very hard to find a fun and entertaining source
all the textbooks are dull and boring and unengaging because the subjects are difficult and there's no easy way to explain them
at least that's my experience

[Anonymous No.11591891]

>>11591870
>it's usually very hard to find a fun and entertaining source
tru
>textbooks are dull and boring and unengaging because the subjects are difficult and there's no easy way to explain them
fals

[Anonymous No.11591903]

>>11591891
>fals
and that's what I'm asking about, why is it true for me but false for others?
I just can't enjoy it but some people can for some reason

[Anonymous No.11591942]

>>11591903
Well I don't know how complex of a subject we're talking about here, but you should still be able to find a textbook or source of knowledge that suits you. I hate textbooks for one, so I try and find a good complete series of lectures. Try MIT or any other open course website, you might be able to find something that suits you. You don't necessarily have to cram a textbook. That's my opinion though

[Anonymous No.11591993]

>>11591903
>why is it true for me but false for others?
it has nothing to do with you. Good teaching is difficult. It's a soft science and an art.
The people writing the books are doing it for money, vanity, or out of a misguided belief that textbooks are anything more than a bloated shitty reference.
Professors and lecturers understand this; this is why their class notes boil the 1000 page textbook into a 60 page powerpoint. State schools, mid-size schools, and open courseware schools like MIT should all have publicly available lecture slides. Use those.

[Anonymous No.11592016]

>>11591837
I'm not sure I follow what you are asking. But strictly speaking, it isn't summing any particular infinite set of rectangles, but rather taking the limit of sums of rectangles as they represent an ever-finer refinement of the area. I only raise this point because people are throwing confusion about infinite sequences, series, and sums, in other threads.

[Anonymous No.11592031]

>>11591993
thank you anon... I was starting to doubt myself

[Anonymous No.11592060]

I want to study math in uni one year from now. As someone who barely remembers math from high school, is it possible to get good enough for uni math in a year if I study diligently every day? I plan on using this boards wiki, is that optimal? Any other advice is welcome

[Anonymous No.11592076]

>>11592060
Professor Leonard, start at the bottom. yw

[Anonymous No.11592084]

>>11592060
You'll be fine if you just do a little every day. I agree with the other anon, professor leonard and khan academy are good resources. Don't worry so much about going super intense with the stuff you've learned before, it will come back to you quickly once you build momentum.

[Anonymous No.11592252]

I'm trying to find the basis of the intersection between two spans. I'm trying to follow this example we did in class and found the dim of U+V so I know the dim of U intersect V is 3. I have this row reduced 4x7 matrix with 4 leading 1's and 3 columns with two non-zero elements.

Where do I go from here? I don't understand how my prof went from this matrix to the basis.

[Anonymous No.11592273]

Why is it for series solutions to ODEs about a regular singular point with an indicial root of degree two (the indicial roots are equal) a second solution can be derived from the first solution (y sub one) via

[math]y_2 = y_1 + \frac{\partial y_1}{\partial m}[/math]

?

I know there's another way to do this using integrating factors, but I want to know why this way works.

[Anonymous No.11592288]

>>11592273

m represents the indicial root of the series solution

[Anonymous No.11592490]

wut is goin on here?

[Anonymous No.11592591]

Why has /mg/ been so bad lately?
>>11592490
The Maclaurin series of the product is the product of the Maclaurin series.

[Anonymous No.11592682]

Why do some people speak so FUCKING LOUDLY.

[Anonymous No.11592709]

>>11592591
Insecure animeniggers who need constant validation aren't receiving any from the outside world because right now because of isolation, so they look for it here by shitting up threads with 250 posts of blogs and circlejerking

[Anonymous No.11592782]

>>11592709
That's definitely a good part of it, but I feel like there's also something else. The political discussion last thread was absolutely horrendous, for example.

[Anonymous No.11592842]

>>11592591
no like with the tables and stuff., wtf is going on with that actual algebra?

[Anonymous No.11592860]

>>11592842
I apologize for it looking schizophrenic.

[Anonymous No.11592874]

>>11592490
seriously though i do not understand the format here, what is happening between "polynomials: " and "thus"?

[Anonymous No.11592887]

>>11592860
oh, so distribute the x (the first term of the macluarin expansion of sin(x)) first, then add like terms, then multiply by -1/6(x^3), then add like terms, then.... ? is that the process?

sorry i just don't fucking get it, it's laid out super wierd

[Anonymous No.11592897]

>>11592887
Yes, that's entirely correct.
I did all those red scribbles earlier, but it might be easier to grasp like this.

[Anonymous No.11592899]

Is the "Dirac delta" a valid probability distribution? Say I'm drawing random samples from [math][0,1][/math] but I choose the distribution to be a deterministic one, e.g. draw [math]0.5[/math] with probability [math]1[/math]. I suspect that, since [math]\{ 0.5 \}[/math] has measure 0, something about this is not formally correct.
But if one were to do this in practice, you could say approximate the "desired" delta distribution with a Gaussian with width less than floating point precision [math]\varepsilon[/math]. Analytically, the first moment would be exactly as desired, and the second moment would be [math]< \varepsilon[/math]. So rigorous results about the distribution in expectation would be valid. And in practice, well even if you wanted to simulate drawing samples from the obscenely small tails, you couldn't tell the difference because your "analytic" distribution has those parameters well below machine precision. So you quite literally HAVE to define your distribution function as just a deterministic value.

[Anonymous No.11592901]

>>11592682
it is, canonically, the best way to spread irrational influence
and geT PEOPLE HYPED!!!! OMG!!!

[Anonymous No.11593016]

>>11592899
>I suspect that, since {0.5} has measure 0, something about this is not formally correct.
I think you are confusing Lebesgue measure and probability measure. The probability distribution doesn't care that your space happens to be special and you know how to define a different measure on it. The fact that {0.5} is a null set in some different measure doesn't matter.

[Anonymous No.11593018]

Let f(x) = x when x is rational, and f(x) = -x when x is irrational on [0,1]. I want to prove that f is not integrable. A proof on stack exchange says sup{f(x) in [a_i-1, a_i]} = a_i but at the same time inf{f(x) in [a_i-1, a_i]} = -a_i
How can this be the case if a_i is either rational or irrational?

[Anonymous No.11593291]

does anyone know of any waveform generators? i want to produce an audio clip of something like
>y=sin(10x)sin(400x)

[Anonymous No.11593386]

>>11593018
Doesn't matter whether a_i is rational or not, all that matters is there are both rationals and irrationals arbitrarily close to it.

[Anonymous No.11593833]

I have a very basic grasp of what EEG is, and an awareness of more intrusive methods of implanting electrodes on or in the brain for sensing brain electrical differentials. Is there an established field of study of inducing electrical differential near, on, or in the brain to create desired effects? I'm imaging something along the lines of sampling numerous hours of raw data for X action, Y emotion, or Z thought, and then trying to recreate X Y or Z by sending that common derived signal back into the brain. Is there any legitimacy and research regarding this?

[Anonymous No.11594050]

do we need to wetten starches for them to break down or will simply heating them do the trick, e.g., can i simply roast unboiled beans and make them soft and safe to eat, or is water somehow necessary?

[Anonymous No.11594061]

>>11593018
It might not be rigorously correct but what matters is that for any constructed interval the supremum and infimum never converge; the supremum is always strictly greater than the infimum so it does not satisfy the definition of integrability

[Anonymous No.11594309]

Brainlet here,
[math]\frac{dy}{dx}|_{x=0}=0[/math]
What does it mean?

[Anonymous No.11594325]

how the fuck do i NOT forget some shit that i can't really learn by repetition?
i just completed a machine learning course on coursera but i just realized that i already forgot how to implement some shit that i've did like a month ago on the course
what can i do to actually memorize all that shit? i feel like in another month or two i won't be able to do a single thing that i've did on that course, that'd be an immense waste of time

[Anonymous No.11594350]

i don't really get it, with overfitting and regularization is there ever a point in decreasing the amount of features?

[Anonymous No.11594363]

>>11590952
no
>>11591069
>current into page
>[math] \int B\cdot\text{d}s=0 [/math]
One of these is false. Circulation of B is proportional to the current enclosed.
>>11591509
>>11591623
literally just treat [math] \mathbf{u}\nabla [/math] as [math] (\nabla\mathbf{u})^T [/math]
>>11592709
imagine watching anime
>>11594309
The slope of the function y(x) at x=0
>>11593291
[math] \sin(10 t)\sin(400 t)=-.5\cos 410t+.5\cos 390t [/math]
use this https://download.cnet.com/Dual-Function-Generator/3000-2169_4-75717132.html and superimpose the two signals

[Anonymous No.11594406]

>>11594325
not using machine learning every day => you will eventually forget and it's inevitable. BUT it's okay, you can't hold every single thing in your head. the important thing is that when the time comes and you need to use ML, it's true that you will need to revise, but it will take you only a small fraction of the time you needed the first time. definitely not a waste of time.

[Anonymous No.11594460]

>>11594363
>One of these is false. Circulation of B is proportional to the current enclosed.
the current into the page (marked as the circled x on the left side) is outside of the rectangular amperian loop ABCD. The enclosed current of the loop is thus zero, hence ∮B⃗ ⋅ds⃗ = 0 must hold, which is what I was trying to reason.

[Anonymous No.11594461]

>>/wsr/809597
>>/wsr/809620
>2
Draw a free body diagram and make sure to pick proper coordinates. Recall Hooke's law for springs. You should get the equation of motion for the mass: [math] F(t)-kx=m\ddot{x}\implies 4\ddot{x}+9x=\sin(3t/2) [/math]. Write down the characteristic equation and use the method of undetermined coefficients to get [eqn] x(t)=C_1\cos(3t/2)+C_2\sin(3t/2)-(1/12)t\cos(3t/2) [/eqn] Solve for the constants with initial conditions, C1=0, C2=1/18.
>3
It's just a separable ODE. Write down [eqn] \int2\text{ d}x=\int\frac{3\text{ d}y}{(1+y)(1-2y)}=\int\frac{1}{1+y}+\frac{2}{1-2y}\text{ d}y [/eqn] and compute.
>4
Part i is very easy. Part ii involves an integration factor. Recall that for [math] u'+f(x)u=g(x) [/math] you can let [math] \mu=\exp\int f(x)\text{ d}x [/math] then [math] u=\int\mu g(x)\text{ d}x/\mu [/math] and you finally end up with [math] u=2x+Cx^2 [/math]. You can easily get y now.

[Anonymous No.11594473]

>>11594460
Oops. Yeah, you're right about the magnitude of B at x and y, but consider than the integrals along the top and bottom of the rectangle are both not zero. In fact, they are both negative and have same magnitude. These two combine with the difference of the integrals along BC and DA to get zero.
>>>/wsr/809597
>>>/wsr/809620
>2
Draw a free body diagram and make sure to pick proper coordinates. Recall Hooke's law for springs. You should get the equation of motion for the mass: [math] F(t)-kx=m\ddot{x}\implies 4\ddot{x}+9x=\sin(3t/2) [/math]. Write down the characteristic equation and use the method of undetermined coefficients to get [eqn] x(t)=C_1\cos(3t/2)+C_2\sin(3t/2)-(1/12)t\cos(3t/2) [/eqn] Solve for the constants with initial conditions, C1=0, C2=1/18.
>3
It's just a separable ODE. Write down [eqn] \int2\text{ d}x=\int\frac{3\text{ d}y}{(1+y)(1-2y)}=\int\frac{1}{1+y}+\frac{2}{1-2y}\text{ d}y [/eqn] and compute.
>4
Part i is very easy. Part ii involves an integration factor. Recall that for [math] u'+f(x)u=g(x) [/math] you can let [math] \mu=\exp\int f(x)\text{ d}x [/math] then [math] u=\int\mu g(x)\text{ d}x/\mu [/math] and you finally end up with [math] u=2x+Cx^2 [/math]. You can easily get y now.

[Anonymous No.11594567]

My calc 4 professor mentioned that the Riemann integral is "insensitive to measure-zero changes to the function".

Clearly this choice of words is not precise, because the indicator function [math]\chi _{\mathbb Q \cap [0,1]}[/math] differs from the zero function only on a countable set, whereas only the latter is Riemann-integrable.

I would like to make this statement accurate. My guess is the following: Let [math]f,g[/math] be a.e.-equal, and suppose they're *both* Riemann-integrable; then, their integrals coincide. Is this correct? If so, could we relax any of the assumptions I just stated? Also, could you hint how to prove this (for someone with no knowledge in measure theory and Lebesgue integration)?

[Anonymous No.11594581]

My calc 4 professor mentioned that the Riemann integral is "insensitive to measure-zero changes to the function".

Clearly this choice of words is not precise, because the indicator function [math]\chi _{\mathbb Q \cap [0,1]}[/math] differs from the zero function only on a countable set, whereas only the latter is Riemann-integrable.

I would like to make this statement accurate. My guess is the following: Let [math]f,g[/math] be a.e.-equal, and suppose they're *both* Riemann-integrable; then, their integrals coincide. (Obviously equivalent to: if [math]h[/math] vanishes a.e. and is Riemann integrable on some box [math]Q \subset \mathbb R^n[/math], then [math]\int_Q h(x)\mathrm d x = 0[/math].)

Is this correct? If so, could we relax any of the assumptions I just stated? Also, could you hint how to prove this (for someone with no knowledge in measure theory and Lebesgue integration)?

[Anonymous No.11594671]

>>11593018
Any interval [math][a_{i-1}, a_i][/math] with [math]a_{i-1}<a_i[/math] contains both a rational and an irrational number. Then, the infimum of [math]f[/math] on the interval is at least smaller than [math]-1[/math] (which implies it is in fact -1), and the supremum is at least larger than [math]1[/math] (and thus is 1, yada yada).
>>11594581
Assume [math]f[/math] and [math]g[/math] are Riemann integrable, and their difference is a.e. zero. Then [math]\int _a ^b f(x) ~ dx - \int _a ^b g(x) dx = \int _a ^b [f(x)-g(x)] ~ dx[/math]. Since [math]f-g[/math] is a.e. zero, it's Lebesgue integrable even if [math]f[/math] and [math]g[/math] aren't ( source for the sake of it: http://mathonline.wikidot.com/lebesgue-integrability-of-functions-equalling-0-a-e-on-gener ), so we can use [math]\int_a ^b [f(x)-g(x)] ~ dx - \int _{[a, b]} [f(x)-g(x)] d \mu = 0[/math], and then finally [math]\int _a ^b f(x) ~ dx = \int _a ^b g(x) dx[/math].
I'm not sure if I did anything illegal or not, but I think it works as is.

[Anonymous No.11594677]

>>11594671
Typo, should be [math]\int_a ^b [f(x)-g(x)] ~ dx = \int _{[a, b]} [f(x)-g(x)] d \mu = 0[/math] .
The argument works because when both integrals exist they coincide.

[Anonymous No.11594734]

>>11594671
Thanks anon, I'm afraid I don't know anything about Lebesgue integration (that [math]\mathrm d \mu[/math] symbol means nothing to me) but at least it's good to know that my guess was right. If you have any idea on how to prove this using only Riemann integrals that'd be swell

[Anonymous No.11594769]

>>11594734
Essentially, use the Riemann integral instead of the Darboux integral.
More specifically, for any partition [math]\delta[/math] of the interval of integration, we can always pick points in the intervals where [math]f[/math] and [math]g[/math] coincide. As the partition is arbitrarily defined, all the sums are always equal, so they converge to the same number.

[Anonymous No.11594803]

>>11591564
Ah, damn, I've mislead myself. Thanks for the advice and help, appreciate it!

[Anonymous No.11594816]

Why do darker plastics resist UV radiation better than lighter colours?
I've been told this by several companies and I've seen it myself.

[Anonymous No.11594823]

What's an intuitive explanation of signal energy and signal power? Feel free to use analogies and explain like I'm 5

[Anonymous No.11594876]

>>11594769
Got it. Thanks a bunch

[Anonymous No.11595017]

so I know this is going to make you guys mad.... but I have an online math exam today and I was wondering if theres a website where I can pay someone to answer math questions live. Exam is timed so I can't just wait for an answer

[Anonymous No.11595041]

Am i being fucking dense? You can't solve the bottom one without some kind of estimation surely?

[Anonymous No.11595049]

>>11594823
Power is energy/second. You know that voltage is energy/charge by definition, so power must be like (energy/charge)*(charge/second)=voltage*current. But, the current through a component with some impedance is proportional to the voltage over it. This means power is proportional to voltage^2 which is also proportional to current^2 (where these are peak to peak voltages/currents/power). If you care about average power, then you must consider average voltage and current. If you integrate the square of a sine function over its period and take the square root of the mean, you get the average. It's not to hard to show that (average of peak-to-peak value)=(peak-to-peak)/sqrt(2). So RMS power for a sinusoidal signal is (average voltage)*(average current)=(peak-to-peak voltage)(peak-to-peak current)/2

[Anonymous No.11595057]

>>11595041
[math]e^{0.5y}=a[/math].
You now have a polynomial in [math]a[/math].
>>11595017
>so I know this is going to make you guys mad
Kek.

[Anonymous No.11595093]

>>11595057
That leaves me with a quartic function...

That isn't reasonably solvable by pen and peper is it? (unless there's a cheap way to do it)

[Anonymous No.11595105]

>>11595093
You *can* solve a quartic exactly, it just sucks. Like, ten pages of calculations.

[Anonymous No.11595113]

>>11595105
Shove it into wolfram alpha.

[Anonymous No.11595143]

>>11595049
I'm a CS student reading a DFT course so I have no idea about circuit and electrical appliances. I'm more curious of an intuitive explanation and the motivation behind explaining a DFTd NASDAQ graph or any periodic sequence with fucking electrical voltage of all things.

[Anonymous No.11595146]

>>11595143
>DFTd NASDAQ graph
wat

[Anonymouse No.11595168]

>>11595093
The question is, what would you accept as a solution? Iterative methods for solving this would converge pretty quickly, but never be exact, while "exact" methods leave you with a mess with root signs all over the place.

[Anonymous No.11595204]

can anyone explain how this eigenvector was found because whatever Im trying isn't giving me a result

[Anonymous No.11595247]

>>11595204
Multiply the Matrix by the z vector and you get two equations which equal 0.

[Anonymous No.11595259]

>>11595247
I mean yeah thats one of the things Ive tried and that hasnt worked
Ive also tried row reduction

[Anonymous No.11595274]

nevermind I realized z1 was taken as a free variable 2s

[Anonymous No.11595275]

>>11595259
Where is it not working for you? You will get
z1-iz1 + 2z2 = 0
and
-z1 - z2 - iz2 = 0

[Anonymous No.11595279]

Math or CS?

[Anonymous No.11595562]

>>11595279
if you're equally motived/indifferent toward either field, CS for sure.

[Anonymous No.11595721]

>>11595259
>Ive also tried row reduction
You're trying to find the kernel (null space) of (A-rI), so you want to use column operations to get a zero column.
[eqn]
\begin{pmatrix}1-i & 2 \\ -1 & -1-i \\ \hline 1 & 0 \\ 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix}0 & 4 \\ 0 & -2-2i \\ \hline 2 & 0 \\ i-1 & 2 \end{pmatrix}
[\eqn]

[Anonymous No.11595725]

>>11595259
>Ive also tried row reduction
You're trying to find the kernel (null space) of (A-rI), so you want to use column operations to get a zero column.
[eqn]
\begin{pmatrix}1-i & 2 \\ -1 & -1-i \\ \hline 1 & 0 \\ 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix}0 & 4 \\ 0 & -2-2i \\ \hline 2 & 0 \\ i-1 & 2 \end{pmatrix}
[/eqn]

[Anonymous No.11595741]

What really is the Lorentz group? By that I mean the group of symmetries that physicists refer to when they say "blah is Lorentz invariant." Are they referring to the whole of [math]O(3,1)[/math]? Or is it [math]SO(3,1)[/math]? [math]O^{+}(3,1)[/math]? [math]SO^{+}(3,1)[/math]
I understand the differences between these objects, I just don't know which one is the relevant one for physics.

[Anonymous No.11596225]

I'm interested in going to college for neuroscience and psychology. I'm still on the fence about it, since I don't want to get into the medical field and I'm only interested in R&D and production. Is it a solid field with good pay and interesting stuff?

[Anonymous No.11596258]

I need some feedback in order to make sure I'm not going full retard here: is it obvious that the map induced by an exact functor between abelian categories is a well-defined function between their Grothendieck groups (definition at the bottom, just in case)?

Even though it's easy to prove, I'd say it isn't completely straightforward, since not every element in [math]G_0(\frak{A})[/math] can be represented by the class of an object in [math]\frak{A}[/math], but I might be not seeing it clearly, and am just overcomplicating stuff. My brain is actually melting and I don't trust myself anymore. Send help.

*Given an abelian category [math]\frak{A}[/math], let [math]\mathscr{F}[/math] be the free abelian group generated by the isomorphism clases of objects in [math]\frak{A}[/math], and take the subgroup [math]\mathscr{R}[/math] generated by the elements [math]A-B+C[/math] for each short exact sequence [math]0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0[/math] in [math]\frak{A}[/math]. The Grothendieck group of [math]\frak{A}[/math] is [math]G_0(\frak{A})=\mathscr{F}/\mathscr{R}[/math].

[Anonymous No.11596318]

>>11596258
>since not every element in G0(A)G0(A) can be represented by the class of an object in A
The generators can, though.
It's exactly like the usual way you construct a morphism of groups G->H given via a presentation; you send all the generators somewhere with a set map, and if this map takes relations in G to relations in H then it extends uniquely to a morphism in the obvious way.
In this case there's really nothing to even check since the exactness assumption is exactly that the relations (S.E.S.) are preserved.

[Anonymous No.11596320]

>>11595741
I'm personally used to seeing [math]O(3, 1)[/math] as Lorentz and [math]SO^+(3, 1)[/math] as proper Lorentz.
>>11596258
Right, so if we call the two categories [math]A[/math] and [math]B[/math], there *is* an induced map [math]F^*: \mathscr{F}_A \rightarrow \mathscr{F}_B[/math] between the free groups. This is because isomorphic objects have isomorphic images under functors. Also, there's a projection map [math]\pi: \mathscr{F}_B \rightarrow \mathscr{F}_B / \mathscr{R}_B[/math].
The composition gives a map [math]\pi \circ F^* : \mathscr{F}_A \rightarrow \mathscr{F}_B / \mathscr{R}_B[/math].
To check if it's defined on [math]\mathscr{F}_A / \mathscr{R}_A[/math], it's necessary and sufficient that [math]ker \pi \circ F^* \supseteq \mathscr{R}_A[/math]. For that, it's sufficient to check that the generators are in the kernel. And then that follows because of the exactness.
I think it works.

[Anonymous No.11596329]

>>11596320
*proper orthocronous Lorentz.

[Anonymous No.11596379]

Is there a mechanism that albeit inefficiently or too slowly particularly with age, clears out accumulated prions, or is it just the case that any amount will be there forever? If mis-folded prions are no longer replicated or added to the total population of prions, will they still be there in some years?

[Anonymous No.11596505]

>>11586260
Hi, is it possible to have the contrary to a panic attack? Like a bliss attack?? All of the sudden I felt so incredibly good and relaxed today, it was crazy and new to me. I need an explanation please.

[Anonymous No.11596523]

>>11596505
well seemingly it is

[Anonymous No.11596530]

>>11596505
>i'm in my manic phase

[Anonymous No.11596535]

>>11596530
No bro. It was chill.

>>11596523
Yeah. I need to know a little more about it.

[Anonymous No.11596560]

Can someone explain to me why is it valid this jordan block for complex eigenvalues? I can't demonstrate it for some reason.

[Anonymous No.11596747]

>>11596505
In all likelihood you just decided to feel "euphoric" like that retarded atheist bastard, rather than that you that you were actually attacked by some cocaine

[Anonymous No.11596789]

a nibba needs help....

[Anonymous No.11596805]

>>11596789
>just differentiate both sides bro

[Anonymous No.11596809]

>>11596805
can you show me how ,,,

[Anonymous No.11596835]

>>11596805
>>11596809
I got
f(x) = -5x^3
but not sure what to do after

[Anonymous No.11596840]

>>11596835
Substitute, and solve for a.

[Anonymous No.11596876]

>>11596840
ahhhhhhhhhhhhhhhh
i jsut cant do it

[Anonymous No.11596887]

if a fair coin is tossed 8 times, how do i find out the odds that it will land heads exactly 5 times?

I know the sample is 2^8, but I don't know to figure out the event space.

[Anonymous No.11596931]

>>11596887
is it just 2^5/2^8?

[Anonymous No.11596966]

>>11596887
Binomial distribution: [math]P(H = 5) = {8 \choose 5}(\frac{1}{2})^{5}(\frac{1}{2})^{8-5}[/math]
You select 5 out of the 8 coins to be heads, and the other three are tails. The 5 head coins each have a 1/2 chance of being heads, and the other (8-5) tail coins have a 1/2 chance of being tails.
Exact answer is 7/32.

[Anonymous No.11597175]

>>11596966
i knew combinations and binomials had somehting to do with it, i just couldn't find any explicit statements and couldn't pencil out myself, thanks!

[Anonymous No.11597238]

did i do this right?

[Anonymous No.11597268]

>>11597238
nvm i found it on chegg.
P(-$15)=1/1

[Anonymous No.11597488]

There is a theorem that for a Noetherian integral domain R, if every minimal (nonzero) prime ideal of R is principal then R is a UFD.
Could anyone give me an example for why the Noetherian assumption is necessary? It is used in the proof, but I can't think of a ring ugly enough to show the assumption is actually required and not just convenient.

[Anonymous No.11597635]

>Glassing - When an area is bombed to a degree that nothing remains. Based off of the idea that when a nuke goes off in the desert, the sand turns into a glowing sheet of irradiated glass.

>Based off of the idea that when a nuke goes off in the desert, the sand turns into a glowing sheet of irradiated glass.
Does this actually happen?

[Anonymous No.11598074]

If I have a sinusoid and I add a line of a known slope to it, is there a quick way of calculating the resultant curve's minimum (circled in the pic)?

[Anonymous No.11598078]

>>11598074
Quick as in without traversing the curve in a for loop.

[Anonymous No.11598093]

>>11598074
Yeah it's called the derivative

[Anonymous No.11598114]

>>11598093
I was doing some more thinking and tinkering in excel and it occurred to me that if the cos is the rate of change then it must be something to do with acos, I think. But I've got to find the angle which gives the maximum slope of the cos minus the line so I use acos to get the angle assuming the line is zero slope. The steeper the line the closer to 0 deg the steepest point on (cos minus line) is - so I need to feed acos the maximum steepless minis the line slope or something.

This is as far as my thinking can take me.

[Anonymous No.11598128]

>>11598074
Solve
[eqn]
f'(x) + a = 0
[/eqn]
for x

[Anonymous No.11598130]

>>11598114
I mean acos (0) will give me the angle at which the sine changes fastest, so 90. I get that.

When I add the line into it, I'm still looking for an acos but of what? Do I subtract the line from the cos and get the acos of that? Doesn't seem to work, looking at my excel testing.

[Anonymous No.11598131]

>>11597488
Have you tried taking the integers and just adding all the n-th roots of 2?

[Anonymous No.11598134]

>>11598114
the slope of sin(x) at any given point x is cos(x). The slope of cos(x) at any given point x is -sin(x).
If you want to find a minimum of sin(x), solve cos(x) = 0. If you want to find the minimum of sin(x) + ax then solve cos(x) + a = 0. You also have to check that you haven't got a maximum instead of a minimum.

Honestly you should probably read an introductory calculus book's chapter on differential calculus.

[Anonymous No.11598143]

>>11598134
I appreciate what you're saying but this is for a computer program and I'm aware that it may involve calculus but I'm more of a coder than mathematician and I can see the general nature of my problem and have reasoned it out as much as I can.

However, what I'm looking for is something that can take a known slope and, given a sine, spit out a minimum in one "step". Is there a function that can do that? It's for calculating a bounding box for efficient redraw on my program.

Does such a function exist?

[Anonymous No.11598148]

>>11598143
Honestly my brain can't make it to this final step, this is as far as my maths knowledge takes me.

[Anonymous No.11598157]

>>11598143
Assuming your sinusoid is of the form [math]a \sin bx[/math], and that the line is [math]cx+d[/math], we obtain [math]\frac{d}{dx}[a \sin bx +cx + d]= ab \cos bx + c[/math]. Equalling it to zero, we get [math]\cos bx=-\frac{c}{ab}[/math], and then [math]x = \arccos (- \frac{c}{ab})/b [/math]
Could have made a typo somewhere, tho.

[Anonymous No.11598194]

>>11598157
a = amplitude, b = frequency and x is phase angle?

[Anonymous No.11598201]

>>11598194
I think I get you - dx (delta x) is essentially the sample rate - what fraction of 2pi per step. And cx is line's value at that angle.

But what is d?

[Anonymous No.11598214]

>>11598201
cx + d is the line's value at that phase bx. d is the value of the line at 0 angle. If the line always passes through the origin you can safely set d to 0 and remove it from your equations. dx is not the sample rate, b is frequency, i.e. how many full cycles per unit x.

[Anonymous No.11598225]

>>11598157
So for the unit case (ignoring frequency too, instead considering it just as an angle) doesn't this resolve down to simply arccos (-c)?

[Anonymous No.11598233]

>>11598214
Ah it's an offset, I see. Also if this >>11598225
is true then this is precisely what I've been looking for - thanks so much!

[Anonymous No.11598254]

What's the term for an equation like [math]\frac{1}{\frac{1}{x}+\frac{1}{a}}[/math], where x is clearly first order (wrt. polynomials), but is convoluted with some variable a?

[Anonymous No.11598286]

fluid mechanics problem:

Someone please help and explain this to me in simple terms. i need to know and im so tired of trying to find an adequate explanation i can live with.
What causes the signs before the area terms? How can i tell when they're negative and not? I've thought it was simply dictated by the flow of the waterstream, but that doesn't work apperantly.

Thanks

[Anonymous No.11598288]

>>11598233
I'd be careful with it to be honest. I'm fairly certain that the equation gives a maximum, not a minimum, unless the sign of c is negative.

[Anonymous No.11598345]

>>11598286
I figured it out yay

[Anonymous No.11598432]

should you take days off while learning?

[Anonymous No.11598523]

>>11596535
https://en.wikipedia.org/wiki/Mania

[Anonymous No.11598524]

>>11598288
It also occurred to me that what if the slope is outside +-1 (i.e. the range output by the sine and cosine functions)?

I just need a one-shot way of telling my program the lowest point on the curve without me having to traverse it step by step. If acos (-c) gives me the angle at which the minimum occurs then getting the actual minimum value is cos (acos (-c)) - cx? That doesn't seem to make sense because the cos and acos cancel out, leaving -cx?

Is the answer to my question really that simple?

[Anonymous No.11598534]

>>11598524
No, wait, sin (acos (-c)) - cx, is it?

[Anonymous No.11598561]

>>11598524
>the lowest point on the curve
the lowest point on the curve is [math]-\infty[/math] as long as the coefficient of the linear term is non-zero. I guess you must be looking for a [math]\textit{local}[/math] minimum, but there are many local minimums (see pic related, vertical line is at x=arccos(-c)) so you have to think about what range you need to be in.
It's most certainly possible to find the minimum within a certain range, but you'll have to tell us what the range is.

[Anonymous No.11598577]

>>11598524
At this point, I'd recommend just testing stuff to check if the signs are correct.
Try summing or subtracting pi/2 to the arccos to check what works.
>>11598561
You might notice from OP's original picture that he just wants the first local minima after zero.

[Anonymous No.11598586]

>>11598577
fair
in which case the first local minima after x=0 is at [math]2\pi - \arccos(-c)[/math] and its value is [math]-\sqrt{1-c^2} +c(2\pi - \arccos(-c))[/math] assuming I've not made a mistake

[Anonymous No.11598598]

>>11598254
"Adding in parallel"

[Anonymous No.11598601]

>>11598561
I just want to know the minimum value per cycle. When the slope of the line is zero then this is -1 constantly (let's say the amplitude is 1 for now and I'll deal with scaling it later, so the unit case).

Adding the line means the minimum changes, and so does the angle at which this occurs.

I just wanna know the minimum per cycle.In the first cycle it's in the red circle in >>11598074, then the blue curve repeats itself from there, where there will be another minimum for this cycle. This continues for as long as you like.

For every cycle of this sloped sine wave, I wish to know in advance what the physical value will be at the trough.

But I only need to know it for the first cycle then I can use multiples of the line slope to work it out for every subsequent cycle.

[Anonymous No.11598612]

>>11598601
I see what you mean. It's [math]-\sqrt{1-c^2} +c(2\pi - \arccos(-c))[/math] for c in range -1,1. For c outside that ranges it's something else

[Anonymous No.11598621]

>>11598612
For c outside that range it's always on the corners of the cycles. Just graph it.

[Anonymous No.11599376]

>ee homework
>have to find the resonance frequency of circuits
i seriously, unironically, truly, cannot do a single fucking one of these
im going to end it, i fucking hate fractions, i fucking hate 6 fractions nested in each other like a russian doll that wants nothing more than to see you fail, i fucking want to die

[Anonymous No.11599884]

What method was used here?

[Anonymous No.11599947]

How come a lightbulb in the 2700 K range can give a blueish light when according to a black body radiation curve it should be yellowish?

[Anonymous No.11600089]

>>11599376
Did you have a question?
>>11599884
This requires context
>>11599947
[math] 2700\approx2898 [/math] so maximum spectral emission occurs at ~1 micrometer. This is infrared light. But because of the shape of the Boltzmann distribution, something like ~25% of power emitted is at smaller wavelengths (so visible light including blue) (table 12.2 in Bergman).

[Anonymous No.11600091]

>>11600089
>This requires context
It's a linear system.

[Anonymous No.11600110]

wtf did I do wrong? if n=2, my probability is negative

[Anonymous No.11600111]

>>11600089
>Did you have a question?
no, i was just seeing red and the only thing i could think to do was shitpost
ive calmed down now but im still retarded
heres a question: im probably going to be living NEET mode this summer and i want to relearn all of circuits up to this point. do you recommend any book(s) i could use to study from the very basics all the way up to pic related (and possibly beyond)? i feel like leaving circuits 2 feeling like i havent learned a thing is setting myself up for failure.

[Anonymous No.11600122]

>>11600110
Hint: 4/4 =/= 4

[Anonymous No.11600124]

>>11600122
proof?

[Anonymous No.11600144]

Explain???

[Anonymous No.11600206]

>>11600111
Checked. Also, Sadiku covers all that I think. If you've got a working understanding of those in the pic, just move on to signals or system dynamics or something.

[Anonymous No.11600212]

>>11600206
i dont have a working understanding of any of that, that was my point. i need to go back to the basics and work my way up

[Anonymous No.11600279]

What is a good flashy data science topic with a real world application?
I need to make a presentation

[Anonymous No.11600548]

What can I factor out of
x/(2-x)^2
to end up with 1/(1-?x)^2

[Anonymous No.11600575]

>>11600548
[eqn]\frac{x}{(2-x)^2}[/eqn]
[eqn]\frac{1}{(1-?x)^2}[/eqn]
lern 2 tex

[Anonymous No.11600577]

Hello. I'm trying to learn functions.
How do I approach a problem like this exactly?

[Anonymous No.11600609]

>>11600577
A decent place to start is always writing out exactly what the problem statement means: remainder zero when divided by (2x+1) and by (4x-1) means that [math]f(x) = (2x+1)q(x), f(x) = (4x-1)p(x)[/math] for some polynomials p and q.

Now you should see that for each of those two equations, there's exactly one very nice value to evaluate them at. For example, you can look at x = -1/2; this turns the first equation into f(-1/2) = 0., and that gives you some info about f. Do that again for the other factor, and you have enough info to get the coefficients.

[Anonymous No.11600621]

>>11590207
what is the answer?

[Anonymous No.11600632]

>>11600577
>>11600609
math is so cool

[Anonymous No.11600752]

>>11600609
Thanks for the help.

I would have never guessed to do what you're doing now. I don't have a deep understanding of functions at all. I understand that when I evaluate the second of your equations at 1/4 it will equal zero. I'm not sure where these polynomials p and q are coming from and I'm not sure how to get the coeffs from this information. I appreciate the help - sorry for being so new.

[Anonymous No.11600824]

>>11600752
Well, you're not expected to just immediately see what to do the first time you try it. But it's not at all a guess, either. Especially the first step; all I did was rewrite exactly what the problem says, but in symbols instead of words. Writing down what you know is always a good place to start.

If you don't understand what the p and q are, it's probably because you don't really understand polynomial division and polynomial remainders clearly, so you should look back at that section.

Anyway, if you have f(-1/2) = 0, that means if you stick (-1/2) into the formula for f, you will get 0. That looks like [math]m(-\frac{1}{2})^4+n(-\frac{1}{2})^3 +68(-\frac{1}{2})^2-(-\frac{1}{2})-6 = 0[/math]. If you simplify this, which is kind of gross, I think you end up with [math]-m+2n = 184[/math].

Now you can't quite solve that yet, because you have only one equation but 2 unknowns.
But if you do the same thing again with 1/4, then you'll get a second equation, and you'll have two equations and two unknowns. Hopefully you know how to deal with two equations and two unknowns.

This is a fairly involved problem for a basic functions course (as you can see), so it's not unexpected that you don't see what to do if you're not comfortable with the material yet.

[Anonymous No.11600871]

What are some good physics textbooks that cover this material and are relatively accessible? It’s for a sophomore-level physics course at my school, i.e. it comes right after Mechanics and E&M. I want to minor in physics, and I think it might be good to get a head start over the summer.

[Anonymous No.11600905]

>>11600871
im finishing that course rn (its called Waves & Heat at my uni) and we use pic related. those particular topics are spread out between volumes 2 and 3.

[Anonymous No.11600909]

>>11600824
Thank you for this very, very helpful explanation. The course I'm doing doesn't use a text and is very poorly organized. Does this look correct to you?

[Anonymous No.11600913]

>>11600905
Interesting, thanks! What were your thoughts on the course?

[Anonymous No.11600923]

>>11600909
Looks clean as far as I can see.

[Anonymous No.11600927]

>>11600909
you can always check by dividing by the given polynomials
also
>solving systems by hand
laughs in engie
>>11600913
my professor was a very good lecturer so i thought the course was a breeze (98 average for exams going into the final)
entropy and lens were probably my weakest areas, but everyones different

[Anonymous No.11600935]

>>11600923
I'm still not entirely sure what the function of p and q are though. I can do polynomial division. I feel like I'm on the cusp of understanding this strategy, just not quite there yet...

[Anonymous No.11600945]

>>11600935
it doesnt matter what p and q are, the motivation behind writing the function as [math]f(x)=(2x+1)q(x)[/math] was to show that [math]f(-\frac{1}{2})=0[/math]

[Anonymous No.11600954]

>>11600945
This is true, but you still need to justify that it's possible to write f that way in the first place.

>>11600935
They're just from the definition of polynomial division.
When you do polynomial division of f by (2x-1), you get two things out; a quotient p, and a remainder r. If you remember, or look up, how division works, that means that f = (2x-1)p+r. But r is zero, by assumptions of the problem. So f = (2x-1)p.

[Anonymous No.11600955]

>>11600945
I think I get that, yes. Thanks a lot again for the guidance. I'm upgrading highschool marks to try and get into engineering. I have a long way to go, as you can see...

[Anonymous No.11600967]

>>11600954
>you still need to justify that it's possible to write f that way in the first place
>>11600955
to elaborate a bit more, when the question states that "[math]f(x)[/math] is divisible by [math]2x+1[/math]", think about it saying instead that "[math]2x+1[/math] is a factor of [math]f(x)[/math]"
you can then take advantage of this, because its basically giving you a zero of the function for free (remember in grade school when to find the zeros of a function, you had to factor it?), and you use those zeros (theres two from the two given polynomials) to create the system

[Anonymous No.11600982]

>>11600967
This is becoming a lot clearer now. Thanks.
Is it normal to feel so discouraged learning math? I seriously need to practice this.

[Anonymous No.11600995]

>>11600982
If you don't feel like a retard 95% of the time while doing math you should probably call up Princeton and arrange an interview.

[Anonymous No.11601001]

>>11600995
I think with the help of you fine folks, I'm going to feel like a retard 100% of the time

[Anonymous No.11601008]

>>11600982
i know that feel, getting demoralized while trying to learn really sucks
its important to remember that the more effort you put into trying to learn things on a fundamental level, the easier future material will be (well, it wont be easy, but it will be possible instead of impossible)
>>11601001
also remember that its all a matter of perspective, im not very good at math at all and i tutor for dif eq, you should see how utterly hopeless some of the retards in there are, but they have to be at least semi-competent because theyre there in the first place

[Anonymous No.11601029]

>>11601008
I think/hope being addicted to the work might go a long way. Since corona I've had a lot of free time and I'm watching/ reading gilbert strang's linalg and professor Leonord's calc I lectures non-stop and doing lots of trig and algebra. When I don't understand something I get very angry and I tend to push all other commitments away and grind. I'm probably jumping around topics a bit too much, but at least it's a good time. I can only imagine how much being forced to do math when you hate it would suck. It really is a lovely feeling when something clicks.

[Anonymous No.11601039]

>>11601029
>When I don't understand something I get very angry
lol heres me earlier >>11599376
when i find out how to stop throwing a tantrum when i dont understand something ill let you know

[Anonymous No.11601061]

>>11601039
>>11599376

> ee homework

I'm a retard with an English degree failing at getting a crystal oscillator to work in a breadboard...

[Anonymous No.11601063]

>>11601061
well i dont even have a breadboard, and i dont know what the fuck a crystal oscillator is

[Anonymous No.11601095]

NEW

question, pic related. can you help me finish the proof?

[Anonymous No.11601096]

>>11601063
It's a crystal that oscillates. It's used to clock instruction cycles in your computer. Man, I fuckin want to be in EE so bad... I went down the wrong path.

[Anonymous No.11601110]

>>11601095
The next time you see the person who is insisting you do this by induction, please kick them in the balls for me

[Anonymous No.11601118]

>>11601096
stem is for autists, you should be more proud of studying the arts
its cool that you have a passion for EE, tho, shits really cool, i remember at the end of digital logic when i had to build a full adder and display it on a 7-segment, i thought it was the pinnacle of human achievement

[Anonymous No.11601122]

>>11601110
yeah I don't really the need for induction either, but i think i've almost laid it out. I know that through successive multiplications of primes, we can present any arbitrary prime factorization, I just don't know how to wrap this up into induction language with n+1 and frens

[Anonymous No.11601153]

>>11601118
English programs are cancerous now. It's no longer safe to speak your mind in humanities departments. I remember at my university there were dedicated hotlines for males looking to drop out. They were posted everywhere, even in the department's own buildings. It's much more effective and enjoyable to just read and write on one's own or with a small group over beers.

[Anonymous No.11601162]

>>11601153
>there were dedicated hotlines for males looking to drop out
haha what the fuck?

[Anonymous No.11601172]

>>11601162
The dropout rate for male undergrads was approaching some ridiculously high percentage (and likely still is) - it's not a good look for the university.

[Anonymous No.11601174]

find a recurrence relation to determine how many bit strings of length end in 111, while containing no other sequence of three 1's anywhere else in the string.

a_3 = 111, so 1
a_4 = 0111, so 1
etc

[Anonymous No.11601188]

>>11601174
[eqn]a_6=7\\a_n=2a_{n-1}-1[/eqn]

[Anonymous No.11601193]

>>11601188
oh my b
[eqn]a_6=4[/eqn]

[Anonymous No.11601196]

>>11601193
yeah i figured it out already, thx though, it's a_n = a_n-3 + a_n-2 + a_n-1

same question different phrasing here: https://math.stackexchange.com/questions/2814625/find-a-recurrence-relation-that-gives-a-formula-for-the-number-of-arrangements-o

[Anonymous No.11601197]

>>11601193
wanna take a gander at this one: >>11601095
?

[Anonymous No.11601567]

>>11598612
Perhaps outside +-1 is when the curve stops being sine-like with peaks and troughs and always has its minumum at 0deg, sort of like a sigmoid curve.

[Anonymous No.11601787]

>>11601095
We induct on [math]n[/math].
If [math]n=0[/math], [math]N=1[/math], and we're done.
Assuming that the statement is valid for [math]k[/math], we prove it for [math]k+1[/math].
First, we split [math]N= \prod_{i=1}^{n+1} p_i^{a_i}=p_{n+1}^{a_{n+1}} \prod_{i=1}^n p_i^{a_i}[/math]. For any divisor of [math]N[/math], we can split it into the multiplication of a divisor of [math]p_{n+1}^{a_{n+1}}[/math] and another of [math]\prod_{i=1}^n p_i^{a_i}[/math] (because they're coprime). Also, multiplying any divisor of the first by a divisor of the latter gives a divisor of the multiplication. This bijection between the divisors of [math]N[/math] and the product of the two sets of divisors essentially concludes the proof.
>>11601122
Eh, the usual way you'd do it is just considering that each exponent of a divisor is an integer between [math]0[/math] and [math]a_i[/math] and then conclude by the number of possible words.

[Anonymous No.11601797]

>>11598612
This works brilliantly so far - thanks! I now need to know the equivalent for finding the maximum when the line slopes downward. Is it the same because I'm having trouble working the negative case in my code.

[Anonymous No.11601935]

>>11601797
>finding the maximum when the line slopes downward
minimum, I mean.

[Anonymous No.11601955]

Which algebra course should i do next year; Rings and modules, Groups and representation or Galois theory. Mainly want to know which is the most beautiful according to 4chan

I've already done a basic course on abstract algebra covering groups up to quotient groups and actions, Rings up to quotient rings and euclidean domains.

[Anonymous No.11602005]

>>11601797
The derivative of sin(x)+cx is cos(x)+c. This is equal to zero at extrema => cos(x)+c=0 => x=2πn±acos(-c) for integer n. The x=2πn+acos(-c) values are the maxima, x=2πn-acos(-c) are the minima. Plugging these values into the original equation gives
c(2πn+acos(-c))+√(1-c^2) for the maxima
c(2πn-acos(-c))-√(1-c^2) for the minima
If |c|>1 then there are no real solutions; the slope is never zero.

[Anonymous No.11602494]

>>11601196
the simplest way to phrase it is "number of different bit strings without three consecutive 1's", then just slap 0111 at the end of the string
its easy to see that without the three 1's restriction, the relation would be [math]a_n=2a_{n-1}[/math]
this takes all the cases of a string thats one bit smaller, and adds a 0 or 1 to the start
however, if we add a 1 to a string that already had two 1's at the beginning, thats an invalid string, so we subtract one
[eqn]a_1=2\\a_n=2a_{n-1}-1[/eqn]
where n is the number of bits in the string (not including the 0111 at the end)

[Anonymous No.11602503]

>>11602494
hmm, actually that invalid string reasoning only works if n >= 3, so we need two initial conditions
[eqn]a_1=2\\a_2=4\\a_n=2a_{n-1}-1[/eqn]