/sqt/ /qtddtot/ - stupid questions thread

[Anonymous No.11028644]

Previously >>11006433
I happen to be an expert on this subject edition.

[Anonymous No.11028700]

Unanswered questions from the previous thread.
In loving memory of Yukariposter, 19XX - 2019.

/g/ questions:
>>11006500
>>11015426

Chemistry questions:
>>11007963
>>11008891

Physics questions:
>>11008737 (I think Yukari had answered.)
>>11012336
>>11013447
>>11018076
>>11018323
>>11026474

Math questions:
>>11009254 (I think he asked it again here >>11012002 , but that's answered.)
>>11012849 followed up by >>11012852
>>11015878 (Roman's, or you can just pick it up from a diffy topology text.)
>>11017558
>>11017998 (Yukarihomo also answered this.)
>>11022741 (Yakumoposter.)

Stupid questions:
>>11006789 (Nah)
>>11006999
>>11008115
>>11008146
>>11016506
>>11021837
>>11023238
>>11023663
>>11024072
>>11024706 (Nah)
>>11024771 (Yes, it's a linear operator. Deja vu. Maybe someone else answered this and also got banned? I definitely don't remember a Yukari.jpg)
>>11025085 (Implying.)
>>11025881
>>11027054
>>11028462 (No.)

[Anonymous No.11028707]

why is wolfram doing this

[Anonymous No.11028710]

>In loving memory of Yukariposter, 19XX - 2019.
What did they mean by this?

[Anonymous No.11028712]

>>11028707
try using a different letter

[Anonymous No.11028715]

>>11028712
no

[Anonymous No.11028718]

>>11028707
Put an asterisk between 3d and x

[Anonymous No.11028720]

>>11028707
>dx
You're asking for it.
>>11028710
A bunch of his posts vanished, so he was probably banned.

[Anonymous No.11028739]

Does compressing an electrically conductive material generally make it more conductive?

[Anonymous No.11028768]

How is spooky action at a distance an action?
If i had a machine producing pairs of marbles- one always black, the other always white - and took one of the marbles with me somewhere not knowing it's color beforehand, did it react with the other marble when i discovered it's color?

[Anonymous No.11028811]

do the electric charges from the polar water molecules and salt ions in a completely saturated solution balance out each other?

[Anonymous No.11028918]

Does anyone know a simple, intuitive proof to show that [math] S^1 \times S^2 \neq S^3 [/math]? Something geometric/visual would be best. I need a simpler way to show someone why we really do need the Hopf fibration, because quite honestly I don't really understand the geometry of it myself.

[Anonymous No.11028946]

>>11028918
>why is [math]S^1 \times S^2 \neq S^3[/math]
Because [math]S^1 \times S^1 \neq S^2[/math].
If you'll insist on a formal proof, [math]S^3 = S^2 \wedge S^1 \neq S^2 \times S^1[/math]

[Anonymous No.11028959]

>>11028768
The classical analogy breaks down, so that's not really a good way to think about it.

It really depends on your interpretation of QM. In the coping harder interpretation, projective dynamics *literally* happens, and when someone measures their (say) qubit of an entangled state (say, a Bell state), it causes the entire state to *literally* evolve from [math]| 00 \rangle + |11 \rangle \mapsto | 00 \rangle[/math] (I'm not bothering with normalization).

[Anonymous No.11028981]

>>11028946
You can also argue from [math]\pi _n (X_1 \times X_2) = \pi _n (X_1) \times \pi _n(X_2)[/math]

[Anonymous No.11029006]

>>11028946
>>11028981
Ah I see. I tried using the torus argument but wasn't sure why that discluded higher dimensional spheres from "magically" working. I'll read up about the smash product though, I like that approach.

[Anonymous No.11029114]

>>11028700
>>11028462 (No.)
ty anon, i wont give up

[Anonymous No.11029130]

>>11028644
Is yukariposter really banned or on hiatus?

[Anonymous No.11029246]

im applying for grad school in december but my professor ive been doing undergrad research with has already offered me to stay around and turn it into a masters thesis with funding, it should still be fine to ask him for letters of recommendation to other schools? the program here isnt that bad but the ones im applying too are top schools and so i dont even plan to apply here. he is pretty old so i think he will be fine, just dont want him to be salty and try to sabotage my shit

[Anonymous No.11029267]

>>11029246
have you tried talking to him about it and telling him how you feel?

[Anonymous No.11029270]

>>11028768
>not knowing it's color beforehand
a coherent quantum marble wouldn't have a definite color beforehand. it is mathematically defined as a complex linear combination of black and white. The concept of superposition has no good translation into everyday English.

[Anonymous No.11029316]

>>11029267
nah then i would have to talk to him

[Anonymous No.11029345]

>>11029316
presumably you have to talk to him to get the rec letters

[Anonymous No.11029479]

I wanted to calculate escape velocity. I retardedly only calculated the energy to move against gravity until reaching the end of the atmosphere, apparently not realizing that the Moon is not inside the atmosphere. My answer was ~1/10 of what it should have been. The correct upper limit of integration is infinity.
Can an object really be "at rest" if it is subjected to gravities from every chunk of mass in the universe? Is Ceres accelerating me right now? Is it like electrostatic force where every charged particle in the world is interacting on every other charged particle in the world?

[Anonymous No.11029491]

>>11029479

[Anonymous No.11029566]

>>11029479
>Is Ceres accelerating me right now? Is it like electrostatic force where every charged particle in the world is interacting on every other charged particle in the world?
Brainlet answer: yes, but since gravity follows an inverse square law and the gravitational constant is so small, low-mass and distant objects have almost no effect compared to stars and planets.
The escape velocity from Earth is the velocity a ballistic object needs to never return on a stable orbit around the Earth, assuming the Earth is the only massive body in the universe. The escape velocity from the entire solar system, for example, is higher than this, because it takes into account all of the mass in the solar system. So you could escape Earth with the Earth escape velocity, but you would still end up orbiting the Sun at some larger-radius orbit than the Earth does. The sun contains like 99% of the mass of the solar system.

[Anonymous No.11029670]

Could someone explain what Zorn's lemma tries to say? The way it's worded it seems like it guarantees the existence of a maximal element (and not just a supremum) of at least a single chain in the partially ordered set. All chains have upper bounds by hypothesis, I guess that isn't really enough to guarantee that all or at least some of them have a least upper bound. But when it's said that the set has a maximal element, does it actually just mean that at least some chain (that can't be strictly contained into another) has a maximum?

[Anonymous No.11029712]

Do you guys find any subjects interesting to read about, but dislike doing actual work with them? I'm usually like that with Physics.

[Anonymous No.11029715]

I know I'm incorrect, but I can't explain why. Can you help me understand why I'm wrong using the wording I'm using?

Flip two coins, at least one is heads, what are the chances at least one is tails?

The obvious way to do this is to draw out all four possibilities, eliminate the TT, and conclude that 2/3 have a tails.

However when I think about it like this I can't figure out what I'm doing wrong: You flip two coins at least one is heads, what are the chances that there is a tails?

Okay so either left going is heads:
HH
HT
or right going is heads:
TH
HH

So it's 2/4? I know this is wrong, but can someone explain why I'm being so retarded?

[Anonymous No.11029851]

>>11029670
You start out with a chain. Obviously, either it's supremum is a maximal element, or it isn't. Assume it isn't.
Then, I choose (emphasis on the word) an element that's higher than it. This one is either maximal, or not. So I just keep choosing.
Of course, this makes obvious sense for a countable set, but if it isn't, then all I need to do is "choose" an uncountable amount of times.
Once I've chosen all of these, I have a totally ordered set, I take the supremum, and it's a maximal element. If it isn't, I choose harder.
>>11029715
2/3 is correct.
Maybe the issue is you pretending HH and HH are two separate cases when, you know, they aren't.

[Anonymous No.11029876]

>>11029712
Most stuff in applied maths.
Differential equations.
Half of real analysis.
Riemannian geometry.

[Anonymous No.11029982]

>>11029851
Okay thanks, there are a couple things I still don't get though.
>Once I've chosen all of these, I have a totally ordered set, I take the supremum, and it's a maximal element
How did we get there? A chain is by definition a totally ordered subset of a partially ordered set, right? So it was totally ordered from the start, as a subset. And in case you meant the whole set is totally ordered, how do we get to totally order a partially ordered set by picking elements from one chain?

And in general what prevents me from applying the same reasoning for, say, the usual order < in [math] \mathbb{R} [/math], and "proving" it has a maximum for every subset (which doesn't happen for example in the cas of open sets). Or is this something like the well ordering theorem where the order isn't arbitrary?

[Anonymous No.11029993]

why do functions and their inverses typically have very different derivatives and antiderivatives, despite being essentially the same function graphed with respect to different axes

for example, e^x and ln(x). e^x is of course its own derivative and antiderivative. However, the antiderivative of ln(x) is xln(x)-x. the ln(x) term is still in the function, but it's still not its own antiderivative.

Likewise, the derivative of ln(x) is 1/x, a hyperbola. Where does this hyperbola go for e^x? Why doesn't e^x have a hyperbola in its derivative?

[Anonymous No.11030027]

>>11029982
>And in case you meant the whole set is totally ordered, how do we get to totally order a partially ordered set by picking elements from one chain?
We don't totally order the set, we just gouge out a "maximal" totally ordered subset by always following upwards.
>what prevents you from doing the same to R
R is already totally ordered, it just doesn't have any supremum.

[Anonymous No.11030033]

>>11029993
>essentially the same function
>same
>different axes
>different

[Anonymous No.11030038]

>>11030033
a function is not the axis it is graphed on

[Anonymous No.11030062]

Hi, ESL here, I have a question, well it might be a grammar problem.
So this question “The quotient of a number and -5 has a result of 2. What is the number?”
So here, i’m not sure who is divided by who, and i gave the answer n=2.5, which was apparently wrong and n was -10.

My question is does the original question imply who is the divident and who is the divisor, or was it a poorly worded question.

[Anonymous No.11030072]

>>11030027
>R is already totally ordered, it just doesn't have any supremum.
I'm sorry, by the same I meant the process of picking an element, and if it's not the maximum then going higher indefinitely. It would seem that doing that I would eventually get to a maximum of for example, the set (a,b). Obviously this interval (under the order <) doesn't have a maximum, but it would seem that repeating the process I could just always go higher until I eventually can't, and I have a "maximum". Specifically from:
>Once I've chosen all of these, I have a totally ordered set, I take the supremum, and it's a maximal element. If it isn't, I choose harder.
I guess the question would be what prevents me to choose harder in (a,b) until I get a "maximum" too?
>we just gouge out a "maximal" totally ordered subset
Yeah I think I got it wrong from the start. So the maximal element I'm trying to find is not a maximal element from the chain, but a maximal chain from the set of totally ordered chains, then? Like, a chain that contains the initial chain I had and has the same total order?

[Anonymous No.11030086]

>>11030062
in english, unless otherwise specified, the first number mentioned is the dividend, and the second number is the divisor

so, "a number" would be the dividend, and also the variable to be solved for.

[Anonymous No.11030094]

>>11030072
>I would get a maximum
No, remember the statement of the theorem? You know that chains have suprema, and you derive from that the existence of a maximum element. The proof is based on constructing a chain that's so big that it's supremum needs to be a maximal element, but if chains don't have suprema to begin with the entire thing goes instantly into the trash.
>a chain that contains the initial chain I had and has the same total order
Yes. It needs to be "maximal" since if it weren't, it's suprema wouldn't necessarily be a maximal element.

[Anonymous No.11030142]

Why can't I stop jacking off to humiliation porn

[Anonymous No.11030175]

>>11030094
I think I'm starting to get it now. I'll think about it some more, but thanks a lot.

[Anonymous No.11030289]

>>11028918
>Does anyone know a simple, intuitive proof
You could say the entire field of algebraic topology exists because of how intractable this is.

[Anonymous No.11030493]

>>11028700
>>11008891
>Is there anything I should be keeping in mind while self learning chemistry?
Once you've taken your intro courses, the only chemistry books worth reading are preparitive guides. Preparative organic, preparative inorganic, preparitive organometallic, etc. Hell, even comp chem books should have you running running simulations irl.
Chemistry is a massive, nebulous field and the only way you'll learn anything useful is with hands-on experience. If you can't do it yourself, look up demos on yt

[Anonymous No.11030496]

Is there a simple, elegant way of representing 'i' in binary like two's compliment? Like, if an 8-bit complex integer had assigned bits {i, -64, 32, 16, 8, 4, 2, 1}

[Anonymous No.11030564]

>>11028700
>>>11022741 # (Yakumoposter.)
What did xe mean by this? I was answered by the author of the paper itself.

[Anonymous No.11030602]

>>11029993
Consider the graph [math]y = f(x)[/math] of some function as a curve in [math]\mathbb{R}^2[/math]. A tangent vector to the graph at a point [math](x,y)[/math] has coordinates [math](f'(x),1)[/math] (you can check this directly be looking at the equation for tangent line passing through the point). And also conversely if you have a vector [math](K,1)[/math] which is tangent to the graph, then [math]K[/math] is the derivative at that point. You can rescale to [math](1,\tfrac{1}{f'(x)})[/math] and it follows, by swapping the axes, that [math]\tfrac{1}{f'(x)}[/math] must be the derivative of the inverse function at [math](x,y)[/math]. So far everything looks symmetric, but when you're talking about the inverse function, you naturally want things to depend on [math]y[/math] and not on [math]x[/math], so you need to plug in [math]x = x(y)[/math]. That's where the symmetry breaks.

Example: [math]y = e^x[/math] and the inverse is [math]x = \ln y[/math]. Derivative of [math]y(x)[/math] at [math](x_0,y_0)[/math] is [math]e^{x_0}[/math], therefore derivative of [math]x(y)[/math] at that point is [math]\tfrac{1}{e^{x_0}}[/math]. However [math]e^{x_0} = y(x_0) = y_0[/math] and so the derivative expressed with dependence on [math]y[/math] becomes [math]\tfrac{1}{y_0}[/math].

[Anonymous No.11030605]

>>11028644
god, i really want to kill remi with kisses and headpats

[Anonymous No.11030628]

>>11028918
honestly, the fundamental group is your best shot. you can easily convince them that S3 is simply connected if you regard it as the compactification of R3. S1 x S2 can even be visualized as a solid torus, but with an identification at the boundary. it's not that hard to see that the identification doesn't make the longitudal loop contractible.

[Anonymous No.11030632]

>>11030602
it should be [math](1,f'(x))[/math], [math](1,K)[/math] and [math](\tfrac{1}{f'(x)},1)[/math], sorry

[Anonymous No.11030657]

[math] f_n(x) := \frac{x^n}{n+x^n} [/math] on [0, infinity). Prove that it does or doesn't converge uniformly.

What I did was found that [math]f_n(n^\frac{1}{n}) = 1/2[/math], therefore it can't converge uniformly. Is this valid?

[Anonymous No.11030776]

>During the preview of a movie, 2000 tickets were sold with a profit of $22800. The ticket cost $15 for adults and $9 for children. How many adults and children watched the preview?

[Anonymous No.11030794]

>>11030776
a+c = 2000
15a + 9c = 22800
Then you get c=2000-a, so 15a+9c = 15a + 9*2000 - 9a = 6a + 18000 from which you get 6a = 4800, and so a = 800. Now c = 2000 - a = 2000 - 800 = 1200, and you get that there were 800 adults and 1200 kids.

[Anonymous No.11030837]

Can you multiply a vector component-wise?
i.e.
v3 = v1 * v2
v3 = {v1.x * v2.x, v1.y * v2.y}

Is that defined?

[Anonymous No.11030840]

>>11030657
>Prove that it does or doesn't converge uniformly.
Here's a nonstandard proof: each f_n(x) is continuous and you know that the limit of uniformly convergent functions is continuous. But limit of f_n is not continuous, therefore they don't converge uniformly.

[Anonymous No.11030842]

>>11030657
>What I did was found that fn(n1n)=1/2, therefore it can't converge uniformly
How does that follow? The way you can show it directly is notice that there's a discontinuity at x=1. So take a small epsilon and note that for large enough n's, you can get |f_n(1+epsilon) - f_n(1-epsilon)| > 1/2 no matter what epsilon you choose

[Anonymous No.11030865]

>>11030837
Depends on the vector space. For example, you can do that for pairs of real numbers (a, b) and (c, d) and get (ac, bd), but you can also multiply those vectors like (ac - bd, ad + bc), if you instead think of them as complex numbers. Both are vectors in R^2 over R, but the multiplication of vectors isn't unique. That is why you may have been told you can't do it (uniquely) for general vector spaces.

[Anonymous No.11030870]

>>11030837
You can define it like that, but it's pretty useless for the following reason: it's highly basis dependent. Say you have a 2 dimensional space and 2 lin independent vectors. In one basis they can be represented like (1,0), (1,1) and their product is (1,0), but in another you can choose them to be basis, so that they become (1,0), (0,1), and then their product is the zero vector. That shouldn't happen for nice operations of vectors, and that's why nobody uses it.

[Anonymous No.11030878]

>>11028644
Are you set for life once you choose your undergrad? Realistically is it possible that i went to economics or engineering once after graduating?
t. biomajor

[Anonymous No.11031052]

>>11030837
yes (you just defined it) but it's useless

[Anonymous No.11031217]

What did I do wrong? Pls no bully

[Anonymous No.11031221]

>>11031217
A^3 is incorrect

[Anonymous No.11031234]

>>11031221
I wrote 9 instead of 19

[Anonymous No.11031244]

How do you find the direction of a 3D vector?
For example, you want to turn a character in a game toward a specific point. How do you know where he's pointing? What can you use to tell that?

no-one has been able to answer this question.

[Anonymous No.11031252]

>>11031244
How do you quantify the direction? There are many ways to do that.

[Anonymous No.11031254]

>>11029670
Sorry, I think the guy who answered you didn't really fix the problem with your understanding. You're wrong about the conclusion. Zorn's lemma says that if you have a partially ordered set with the property that if you take any chain in the set, there's some element bigger than anything in the chain, THEN (we're back to talking about the whole poset now and not an individual chain) there is a maximal element for the whole poset. Here, maximal doesn't mean it's bigger than everything, maximal means that there's nothing bigger than it.
So now for the other dude's intuition: Take an element. This is a chain, so there's an element which dominates the chain. Add it to the chain (the proof deals with the nitty gritty here, but this is just the idea). Do this over and over: clearly, you can keep finding bigger and bigger things on top. But will you get a maximal element for the whole set? Maybe not. But you're chipping away each time.
By choice you can declare the existence of a choice function which picks each element bigger than the previous ones. And now just pick until you can't pick any more. Cardinality doesn't matter when you have a function. You may have heard of transfinite induction, that's a nice way to think of it. Once you can't pick anymore, there's nothing in the partially ordered set which is bigger than your last element. So it's maximal.
Why can't you use this on R? Well, there are chains which dont have a cap. For example, {1, 2, 3,...} or just R itself. Instead, Zorn tends to be very useful for things like sets of sets where you can take arbitrary unions or intersections to get maximal elements.

[Anonymous No.11031255]

>>11031252
I don't know; it seemed simple in 2D but I don't know how it's represented in 3D.

[Anonymous No.11031257]

>>11031244
You calculate the dot product of the vector you are interested with one of your (unit) basis vectors. The direction of the vector of interest is the inverse cosine of the dot product over the magnitude.
[math]\mathbf{v}\cdot\mathbf{\hat{e}}_i=v\cos\theta[/math] and theta is the angle between the vector and the basis vector.

[Anonymous No.11031269]

>>11030175
Nah, sorry, that dude still doesn't know what he'a talking about. Please look at
>>11031254 . And the proof doesn't work for R or a unit interval because you eventually get to a chain where you can no longer pick anything bigger than it. Remember, we're not saying "lol pick a sequence," eventually the idea is that you pick out a set of cardinality strictly higher than your original set and that's a contradiction, so you terminated earlier. For R the instant you hit a countable sequence which goes off to infinity you're toast, there's no more dominating elements.
You can't think of Zorn's lemma as a finite sequence. It just doesn't work.
>>11030094
You're not getting to the point. Zorn's lemma is trying to find a maximal element for THE WHOLE POSET. In the process, as a side effect, it creates a maximal chain from the set of chains ordered by supremum. Well, of fucking course it does! But you went right into the proof when the dude doesn't even know what he's looking for.

[Anonymous No.11031288]

can someone explain how this limit is equal to -1?
I got two different limits when I split it up so I thought ot wouldnt exist

[Anonymous No.11031291]

>>11031288
forgot the pic like a potato

[Anonymous No.11031361]

>>11031291
why are you setting x and y to zero? You should be setting them to 1, since the point is (1,1)

[Anonymous No.11031368]

>>11031361
still not the same :(

[Anonymous No.11031372]

>>11031368
you can't put in x=0, you're dividing by 0, retard

[Anonymous No.11031373]

>>11031291
Consider the change of variable to polar coordinates centered at x,y = 1,1.

[eqn] x = 1 + r\cos\theta [/eqn]
and
[eqn] y = 1 + r\sin\theta [/eqn].

Then taking the limit as r tends to 0 will be the same as taking the limit of you're function as x,y, tend to 1,1. If your resulting function still depends on theta after taking the limit, the the limit does not exist.

[Anonymous No.11031380]

>>11031373
>If your resulting function still depends on theta after taking the limit, the the limit does not exist
In this case, the limit exists though, so what this tells you is that the limit exists from every direction because you get - rcos(theta) + r^2 sintheta cos theta/rcos(theta) which tends to -1 independent of theta as r -> 0

[Anonymous No.11031391]

>>11031255
> I don't know how it's represented in 3D.
Euler angles (yaw/pitch/roll), axis-and-angle, unit quaternion, 3x3 matrix. There are probably other possibilities, but those four are the main ways that are used to represent an orientation.

If the object is at P and you want it to face Q then the local forward axis (which we'll call Z) is just Q-P. But you can rotate around Z by any angle and it will still be facing Q, so you need to choose which way is "up". Assuming that you want the local vertical axis (Y) to be in a plane containing both Q-P and the global vertical axis (V), you'd calculate the local horizontal axis (X) as X=Z×V (where × is the cross product); then the local vertical axis (Y) as Y=X×Z. That gives you a 3x3 orthogonal matrix (all axes are perpendicular) which can be converted to one of the other (more compact) representations if desired.

[Anonymous No.11031406]

>>11031372
where did I use x=0?

[Anonymous No.11031408]

>>11031406
Meant x=1. When you substitute x=1, the denominator becomes zero. You can't do that. The function is not defined whenever x=1

[Anonymous No.11031422]

>>11031408
so how exactly should I evaluate the top limit? x=y would be the same as evaluating x at 1 right?

[Anonymous No.11031446]

>>11031422
To evaluate it, you first need to prove that it exists. Fortunately that's easy, you just need to prove that no matter which direction you approach it, it's always the same. Use the polar parametrization to show that. Once you've done that, approaching it from any direction will give you the correct result. Including the direction t-> (t,t) (which is the same as setting x=y and taking x to 1)

[Anonymous No.11031472]

Why dafuq do physishits always assume plane wave solutions? For the love of gawd I cannot find any physishit who actually solves Klein-Gordon or Dirac generally. They always resort to the plane wave bullshit.

[Anonymous No.11031490]

>>11031472
Although I agree with your notion of physishits, I must admit I got a lot of different cancers from your post.

[Anonymous No.11031713]

>>11031254
So it's meant that there is (at least) a single element contained in the poset such that no element in any chain is bigger than it? Like for example, if I were to take the partial order of inclusion in the power set of R then the maximal element would be R itself? I guess that example is kinda trivial since R contains all of its subsets. But something I think I've been confused with is if the upper bound has to be an element of the poset or not. I've usually just seen stated that every chain has to be bounded above, but that does mean that that bound is included in the poset as well? Or can I take for example P(R)\R with set inclusion order, so every set is bounded above by R but R is not included in the set?
>>11031269
>doesn't work for R or a unit interval because you eventually get to a chain where you can no longer pick anything bigger than it
This is the last part that's bugging me I think. If I took an open unit interval (a,a+1) (so every chain has an upper bound, albeit not in the interval), it's obviously totally ordered and thus partially ordered with the < order. I think this makes sense if what I asked above is true, cause then yeah if I took a subinterval like (a+0.5,a+1) then it doesn't have an upper bound within the original interval. Is this what is meant by "every chain has an upper bound", then?

[Anonymous No.11031807]

>>11031291
Factor the numerator.
xy-y-2x+2 = (x-1)(y-2)

[Anonymous No.11031844]

>>11028644
How do I become smart? I want to become something but Im a fucking idiot.

[Anonymous No.11031901]

>>11031844
prove [math]a^2 + b^2 = c^2[/math] without looking anything up

right now, do it. I'm waiting

[Anonymous No.11031920]

>>11031901
Im not sure I can, I cant even remember if i can √ on both the √(a + b) or just √a or √b, if that even is the correct way to go, like I said, im not smart.

[Anonymous No.11031925]

>>11031920
I'm going to give you a hint, do it geometrically. You need to be drawing triangles on paper

[Anonymous No.11031942]

>>11031925
I failed you, i googled it after i wrote the last thing, I thought it look familiar, still cant prove it tho... I could solve it if there was some number tho..

[Anonymous No.11031965]

>>11031942
try proving
[math] 1^2 +2^2+\cdots+n^2=\frac{1}{6}(n+1)(2n+1) [/math] for all positive integers [math]n[/math]

[Anonymous No.11031971]

>>11031965
Thats impossible.

[Anonymous No.11031974]

>>11031942
>>11031965
If you need a hint first look at [math] 1+2+ /cdots+n=\frac{1}{2}(n+1) [/math] and how to prove it *cough* mathematical induction

[Anonymous No.11031976]

>>11031974
Litteraly dont have the slightest clue

[Anonymous No.11031991]

>>11031976
Then go read the clues he gave you dude, being "smart" doesn't mean that the knowledge magically shows up in your head, if you don't learn shit and you don't do anything to at least try to learn shit then you won't know shit. If you don't get it the first time you read it then it's fine, even most people who are in the field don't when they are just starting. Just write down what you understand and check some examples, or whatever it is that you find helps you most.

[Anonymous No.11032021]

>>11031942
you build a square out of triangles like this

remember that when things are multiplied you are forming a square with them, that's literally the definition of multiplication. So this does indeed follow logically if you think about it.

[Anonymous No.11032045]

is it correct to say a NxN+1 matrix is always linearly dependent?

[Anonymous No.11032052]

>>11032045
if by "linearly dependent" you mean it has linarly dependent columns, then yes

[Anonymous No.11032063]

>>11032052
is there something else i could be referring to? legitimate question

[Anonymous No.11032083]

>>11032063
lol

[Anonymous No.11032086]

>>11032083
please im not trying to be an asshole i am getting ass raped in linear algebra rn

[Anonymous No.11032090]

>>11032086
No you smacked his ass down, it was hilarious

[Anonymous No.11032095]

>>11032090
this is serious!

[Anonymous No.11032118]

>>11031965
Dont laugh to hard but...
(1+2+⋯+k)2+(k+1)=(k+1)+1)(2k+2)+1) / 6
Is this anywhere close to being on the right track?

[Anonymous No.11032119]

>>11032118
that 2 on the left is suposed to be square thing, but it didnt work apperantly...

[Anonymous No.11032126]

>>11032063
the rows could be linearly independent.

[Anonymous No.11032151]

>>11032118
This is not an easy problem. It's kind of easy once you know how of course.

If you figure this out I will be very impressed

[Anonymous No.11032154]

>>11032118
Yes this is very good. In a formal proof you would also show the algebraic work you used to obtain that result. But also show it works for [math] n=1 [/math]. So what you did you showed that if it is true for a certain [math] n [/math] it must now also be true for [math] n+1 [/math] so if it is true for [math] n=1 [/math] it must now also be true for [math] n=2,n=3,n=4,... [/math] etc.

[Anonymous No.11032155]

>>11032118
I love how after the equals sign there's 2 opening parentheses and 4 closing parentheses.
That's genuinly impressive.

I think at this point you've done enough to prove you're a brainlet (which you already admitted).
So I don't know why people are still giving you problems. Maybe they just want a laugh.

[Anonymous No.11032181]

>>11031965
just looking at the equation, it reminds me of a square pyramid of balls stacked n balls high
i wonder if that’s a hint

[Anonymous No.11032210]

>>11032181
well the more pure way of solving it is geometric of course, so you can give it a try

[Anonymous No.11032216]

>>11032155
Auch, well atleast I tried, eh?

[Anonymous No.11032224]

>>11031965
Answer. Do not read this if you don't want to be spoiled

[eqn]\sum_{k=1}^{n} (k+1)^3 - k^3 = (2^3 - 1^3) + (3^3 -2^3) + (4^3 -3^3) +\text{......}+((n+1)^3 -n^3)[/eqn]
[eqn]\sum_{k=1}^{n} \xcancel{k^3} +3k^2 +3k +1 - \xcancel{k^3}= (\cancel{2^3} - 1^3) + (\bcancel{3^3} -\cancel{2^3}) + (\cancel{4^3} - \bcancel{3^3}) +\text{......}+((n+1)^3 -\cancel{n^3})[/eqn]
[eqn]3\sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n}k + \sum_{k=1}^{n}1 = (n+1)^3 - 1[/eqn]

and since we know that
[eqn]\sum_{k=1}^{n}1 = n[/eqn]
and
[eqn]\sum_{k=1}^{n}n = \frac{n(n+1)}{2}[/eqn]

we know that
[eqn]3\sum_{k=1}^{n}k^2 = n^3 + 3n^2 + 3n - \frac{3}{2}(n^2+n) - n[/eqn]
[eqn]\sum_{k=1}^{n}k^2 = \frac{n}{3}(n^2 + \frac{3}{2}n + \frac{1}{2})[/eqn]
[eqn]\sum_{k=1}^{n}k^2 = \frac{1}{6}n(n + 1)(2n + 1)[\eqn]

This is called a "telescoping sum". It would be insanely difficult to think of this yourself

[Anonymous No.11032237]

>>11032224
another beautiful telescoping sum (kinda) is the proof that the harmonic series is divergent

[Anonymous No.11032238]

>>11032237
Deleted it because the latex is all fucked up, god dammit.

Anyway you get the idea

[Anonymous No.11032240]

>>11032224
That's a fun way proving it my dude, too bad about the Tex though

[Anonymous No.11032260]

what is grad school like in good universities? in my shitty state university it's not much different from undergrad, the same lecture/problem solving classes and crap

[Anonymous No.11032303]

>>11032260
My grad school is basically just me doing stuff on my own and reporting to my supervisor once a week.

[Anonymous No.11032308]

>>11031965
that's not even true...

[Anonymous No.11032320]

>>11032308
Yeah it's actually 1/6 n (n+1)(2n+1)

[Anonymous No.11032330]

>>11032320
>11032320
fug I know, sorry mang

[Anonymous No.11032345]

>>11031965
I thought about it for a bit, and here's my proof, assuming we fix the expression to say 1/6 n (n+1)(2n+1)
1. Let S_n be the sum. For a sequence a_n let d(a_k)_n = a_(n+1)- a_n.
Then we easily see that d^3 S_n is a constant sequence
From this it follows that S_n = p(n) for a 3rd degree polynomial p. This is because we can construct it: start with an arbitrary polynomial a_3 x^3 + a_2 x^2 + a_1 x + a_0 and by applying d (which is a linear operator) you can uniquely set the coefficients.
2. Now that it's established that S_n is a cubic in n, it suffices to show that your polynomial matches S_n in 4 places (because a cubic that has 4 roots is identically zero). This is easy for small values of n.
QED
More general result shown: sum of the first n k'th powers is a (k+1)th degree polynomial in n.

[Anonymous No.11032347]

how do you prove that the set of points on a line is set-theoretically isomorphic to the set of real numbers?

[Anonymous No.11032352]

>>11032347
>set-theoretically isomorphic
The expression you're looking for is "has the same cardinality".
And that depends on what exactly you mean by a line. If you mean a set of points in R^2 satsifying the equations ax+by=c with not both a and b zero, you can easily do that by explicitly constructing a bijection.

[Anonymous No.11032355]

>>11028700
>>>11015878
Roman's advanced linear algebra book is maybe the best linear algebra textbook i've ever seen; it covers multilinear algebra and tensors and all that. I also like Cooperstein's book by the same title, which has to its advantage that 1) about half the exercises have solutions in the back and 2) Cooperstein's book is interspersed with some of his own commentary. Problem with Cooperstein is that he pursues a pretty nontraditional approach in some cases, (such as defining the characteristic polynomial as the product of the invariant factors and defining the determinant and trace of a matrix as being certain coefficients in the characteristic polynomial), but if you're going to study the whole book it won't matter because he eventually circles back to the traditional approach.
Btw, I would if you haven't learned Jordan normal form and all the machinery which goes into it (the theory of the minimal and characteristic polynomials, T-invariant subspaces, etc) then I would recommend learning that stuff first. Its not a prerequisite, but its something that can be learned with only an undergraduate understanding of linear algebra. For that material, chapter 4 of Cooperstein is one of the best expositions I've ever seen.

[Anonymous No.11032376]

>>11032347
Say the line is in R^n. Project the line onto R^(n-1), then project the result onto R^(n-2)...until finally you project the result onto R.
In R^3 this looks like projecting the line onto the xy plane, then projecting that result onto the x-axis

[Anonymous No.11032380]

>>11032352
by a line i mean a geometric object with infinite length and zero width and height, like a line in synthetic euclidian geometry. obviously, the line from analytic geometry, by definition, has the same cardinality as the set of real numbers, but i was wandering whether it was somehow possible to show that it indeed is the line from classical synthetic geometry... i don't know whether this question even makes sense

[Anonymous No.11032386]

>>11032380
>line from classical synthetic geometry
And what is that?
If you mean euclidean geometry as in actual euclid or hilbert, it's not based on set theory so it doesn't even make sense to talk about cardinalities.

[Anonymous No.11032422]

>>11032380
>>11032386
I thought about your question a bit more and came to the conclusion that if you could find a model for these axioms of synthetic geometry, just like R^2 is a model, but in which cardinality of lines is greater than R, then that would answer the question in the negative. This may sound a bit autistic, but it's what I came up with: take R^2 and consider (R^2)^(2^R) (the set of functions from 2^R to R). It has a natural structure as a R^(2^R) vector space (where R^(2^R) has the structure of a direct product of 2^R copies of R, and you can do geometry on it. Points in this space correspond to 2^R points in R^2. Essentially you work on many many tabs, just like in a browser. I think all of the axioms are also satisfied in this space, yet lines have a much much higher cardinality than R.

[Anonymous No.11032446]

>>11032422
Oh, much simpler proof! Take the field of constructable numbers, which is a countable field. It can act as a model of euclidean geometry. Then every line is also countable.

[Anonymous No.11032447]

how do you divide a polynomial with two variables by a polynomial

[Anonymous No.11032474]

Can we, the same way Gödel showed that there doesn't exist a demonstration to every theorem, show that there doesn't exist a "computational solution" to every problem?

If we show that, we would have shown that general AI (an AI that solves all problem) is impossible. And thus make shut up "singularity" faggots.

What do you think?

[Anonymous No.11032505]

>>11032474
>general AI
>an AI that solves all problem
just no

[Anonymous No.11032509]

>>11032474
Why is every faggot that spews shit about the incompleteness theorem such a fucking retard? And a frogposter too, just to double down.

[Anonymous No.11032524]

>>11032447
you don't, unless the polynomial on top factors perfectly with the one of bottom.

[Anonymous No.11032528]

>>11032474
>What do you think?
I think that you are a fucking retard and have no idea what you're talking about.

[Anonymous No.11032529]

>>11032447
Best reference for this is the book Ideals varieties and algorithms by Cox Little O'shea or something liek that. There are different algorithms.

iirc, one of them goes something like: do polynomial division like normal imagining the other is constant. Then when you can divide no more, continue with the other variable. iirc, this doesn't give a unique solution, but can be shown to be unique up to an element of the Gröbner basis (this im pulling out of my ass, it's been a while).

[Anonymous No.11032544]

>>11032447
Lads, I have a real analysis test tomorrow.
Give me last minute recommendations.

[Anonymous No.11032566]

>>11032544
Go study for it.

[Anonymous No.11032584]

if i transform/multiply/whatever a 2d vector by a 2x3 matrix, do i just “ignore” the last column of the matrix and do the calculation normally?

[Anonymous No.11032596]

>>11032584
You can only multiply if the dimensions are right. This is one of the rules of matrix algebra

2x3 3x4 = 2x4

[Anonymous No.11032619]

>>11032596
well then wtf do i put for this problem?

[Anonymous No.11032634]

>>11032619
I don't know the context of the problem. But if you made "h" a row vector you could do it by doing hxa

or you could do it the other way, turn A on it's head and do Axh.

But yeah other than that you can't do it

[Anonymous No.11032642]

>>11032634
exact wording is “with T(x) defined as T(x) = Ax, find a vector x whose image under T is b”

[Anonymous No.11032652]

>>11032642
so, do that

[Anonymous No.11032653]

>>11032642
So they're asking you what vector you run through A to get b as a result. Sort of a "matrix division".

If it were a square matrix I would just do an inverse, but you can't invert a non-square matrix.
So I think in this case you literally just have to use algebra.

Here's what you should do. 1x3 vector like this [x, y, z], then multiply it through A. Then use algebra to solve for x, y, and z.
There might be more than one solution, but remember it's just asking for "a vector" so you can pick any old solution you want.

[Anonymous No.11032654]

>>11032566
I meant book recommendations.

[Anonymous No.11032660]

>>11032642
excuse me, that should be a 3x1 vector, and you should multiply A through it
my bad.

Also, since we know there are mutiple solutions, I wonder if you can eliminate one of the columns in A and turn it into a square matrix.
Maybe some anon can fill me in here, it's just a thought

[Anonymous No.11032677]

>>11032619
about the idea I had here: >>11032653

Yes, I believe you can actually do row reductions on A and use that to eliminate one of the columns. Then you can turn A into a square matrix and invert. That would be sort of a "matrix math" way to solve the problem, that's how a programmer would solve it.

But in this case I think they literally just want you to use algebra.
So do this:
A x [x, y ,z] = [-2,-2]
and then use algebra to solve for x, y, and z.
And since there are only two equations, but three variables, you will get more than one answer. But that's ok, because they're just asking for "a vector", so you can pick any answer you want

[Anonymous No.11032683]

How do you stop being sexually attracted to certain things? They all generate the same dopaminergic rush, so why can't you just slap the reflex on and off anything you want so you can be sexually attracted to like eyelids and owl bacon?

[Anonymous No.11032692]

>>11032683
>How do you stop being sexually attracted to certain things?
gimme your email, ill let you know as soon as i find out

[Anonymous No.11032702]

this parametric graph looks a lot like a down-translated Lambert W function

is this a coincidence?

[Anonymous No.11032705]

>>11032677
ah fuck, i see
a 3d vector through a 2x3 matrix is a 2d vector, i’m retarded and didn’t realize that
so how would i prove (or “justify”) that the vector isn’t unique? cause they’re asking that too

[Anonymous No.11032711]

>>11032683
https://en.m.wikipedia.org/wiki/Paraphilia

[Anonymous No.11032713]

>>11032692
I'[e-mail]

[Anonymous No.11032716]

>>11032713
>hotmale
sensible chuckle

[Anonymous No.11032718]

>>11032705
If there is more than one answer.

So for that specific problem, you will generate two equations like this:
1x - 5y - 7z = -2
-3x + 7y + 5z = -2

Two equations, three unknowns. Thus you will get an infinite number of solutions.

Remember, each equation has three variables. You can think of each variable as describing a three dimensional plane in space.

What does the intersection of two three planes look like? A line.
That's why there isn't a single answer, but an entire spectrum of answers all along the line.

[Anonymous No.11032723]

>>11032705
>>11032718
*each equation as describing a three dimensional plane in space

*The intersection between two planes looks like a line

[Anonymous No.11032725]

>>11032718
alright, i see
tyvm

[Anonymous No.11032726]

>>11032702
there's nothing on wikipedia about a parametric form of the W function but boy howdy it looks pretty close, does anyone know anything about this

[Anonymous No.11032774]

Does anyone have any tips for getting better at drawing curves/surfaces?
I don't have a problem with the initial part of using math to find properties of the curve, but my sketches look like something a gorilla drew with his nondominant hand. I can't draw anything more complicated than a simple polyhedron in R3 without it becoming a totally unintelligible mess.

[Anonymous No.11032814]

>>11031713
>>11031713
Yes, the bounds certainly must be included in the poset. You keep saying this "maximal element is not smaller than any element of any chain" thing which is stupid because any element on its own is a chain. A maximal element is just an element which is not bigger than any element of the poset.
You can't just look outside the poset and imagine those things exist.
We call a poset "inductively ordered" when it has the property that "every chain in the poset has an upper bound in the poset." P(R) is inductively ordered, and has the obvious maximal element R. P(R)\R is not inductively ordered as shown by the chain {(-n, n)} for n in N - there's no set bigger than all of these besides R. But P(R)\R still has maximal elements, namely R\x for any fixed x.
What Zorn's lemma tells you is that any inductiely ordered poset contains an element which is maximal over the whole poset.
>unit interval thing
Yeah, this becomes stupid if you allow any random set. Just take the union of the whole poset then, if they're sets, and this is maximal. You've got to stay in your little domain.

[Anonymous No.11032961]

Lads, how do i improve on my proof writing in math? I try to do tons of practice problems but i always struggle with figuring out proof strategies for problems. Any literature or suggestions on this topic would be appreciated. Thanks.

[Anonymous No.11032976]

>>11032814
Thanks, I'm starting to see it now. Just a few more questions
>You keep saying this "maximal element is not smaller than any element of any chain" thing which is stupid because any element on its own is a chain
I don't see how this is a problem? If I were to consider chains where the maximal element is not comparable to the elements of the chain, then the maximal element certainly wouldn't be smaller (since it can't be compared to begin with through the partial order). If I were to take a chain where the elements are comparable to the maximal element, then it is bigger than all of them by definition so it's also not smaller than any of them. And if I were to take the singleton chain of the maximal element, well it can't be smaller than itself right? Don't know if I'm missing a case or if it's a syntactic thing. The reason I'm saying it like this is because from the little I know about posets, in order to say an element is (not) bigger/smaller, we have to do it with respect of some chain of elements, since of course all elements may not be comparable. Once I get more comfortable or get a more correct notion I'll probably reword it in a more simple way.

[Anonymous No.11032982]

>>11028644
Okay, I know something like xenon hexafluoride would never exist in nature, but explain why the fuck we can't just force it under laboratory conditions?

[Anonymous No.11032988]

>>11032355
>>11028700
Thanks, I'll check out Roman first. I'm still not at a grad level but I like the topic so hopefully this goes well.

[Anonymous No.11033002]

Haven't seen physics since high school so sorry if this is a stupid question.
I have this conception which might be wrong and I must have heard somewhere that at an atomic scale a particular set of rules, quantum physics, are used; while at a more common macro scale old newtonian physics, or classical mechanics, are used. This is because newtonian physics don't work, or can't predict things accurately at that scale.
My question is: can quantum physics work at a macro scale? If so, why are classical mechanics used? Is it because they are simple and get the job done? Or do you actually need them to make sense of physics at this scale?

[Anonymous No.11033184]

How would you solve this?

[Anonymous No.11033200]

>>11033184
cramers rule

[Anonymous No.11033247]

>>11028644
retard here, how does inertia actually "work".
if i have moment of inertia I and i apply some torque f(T) then how do i calculate instantaneous acceleration?
should i just do alpha=(f(T)/I ) to get acceleration at any given time?
and if i have several "masses" in "series" (connected by massless gears) should i just "cascade" them ?

should just treat it like a damper?

[Anonymous No.11033279]

>>11033200
>cramers rule
God I fucking hate linear algebra, Damm my multivar teacher who couldn't teach it for shit which left me in the dust

[Hoaxfag No.11033291]

>>11033002
Quantum mechanics is a hoax.
It will never make sense to you because it's gibberish.
That is why no one can answer any of your questions, it's because they are just pretending to understand so they don't feel stupid ;)
Classic hubris.

I'll give you an example
Pretend this is an exam
Q1, provide evidence for the theory of quantum mechanics.

There is no evidence.

[Anonymous No.11033360]

professor never went over legendre polynomials and it isnt in the book so i googled the answer

[Anonymous No.11033368]

>>11033291
hey HoaxFag

>>11033002
it really all comes down to the mathematics of waves. at the macroscale, the frequency of the wavefunction is so high and the interactions with other matter is so high, that quantum phenomenon are negligible. think of a stationary propeller and one rotating at the speed of light, where it would be like a wall

[Anonymous No.11033383]

anyone have that /sci/ math learning pic? im just an EE/physics fag and only did up to diff eq and linear algebra but want to go beyond and also want to rebuild my mathematics base for the future as the foundation is crumbling at this point

[Anonymous No.11033476]

if i am not 100% sold on doing a phd should i just apply for an MS first and then see if i like research before going for the phd? only did one "research" project during undergrad because had no clue i wanted to go for more than a bachelors until senior year

[Hoaxfag No.11033485]

>>11033368
Hey man, what's doin?

Got any evidence yet for your 100 year old autism festival?

Oh hey, you know how wave particle duality is the cornerstone for your insane theory about how we are multimensional hyperdimensional entities experiencing themselves as subjective hallucinations,

Can you provide any EVIDENCE for this?

Other than your own personal subjective and biased experience?

Or am I supposed to just take your word for it?

[Anonymous No.11033508]

>>11033485
>evidence
http://web.mit.edu/course/5/5.73/oldwww/Fall04/notes/Experimental_Evidence_for_Quantum_Mechanics.pdf

[Anonymous No.11033522]

>>11033485
electrons callapse into a theory of simulation aka superpostition when observed with direction

[Anonymous No.11033531]

>>11033184
gaussian elimination

[Anonymous No.11033540]

Explain to a brainlet why this is unacceptable

[Anonymous No.11033545]

>>11033540
It's unacceptable because it's meaningless. The "+..." doesn't mean anything, and the infinity is meaningless in this. You have to define what you mean so that what you wrote has actual meaning, then we can discuss what's wrong with it.

[Anonymous No.11033560]

>>11033184
Gaussian elimination.

>>11032619
> well then wtf do i put for this problem?
What's the problem? Solve Ax=b for x? That gives you a 1-D family of solutions: [3,1,0]+t*[-3,-2,1]

The "correct" way to arrive at that solution depends upon what you've just been taught. Typically, you'd reduce the system to column echelon form to identify the null space (which gives you the t*[-3,-2,1] part) and solve the non-null portion to obtain one solution.

A simpler way is to set z=0 to get 2 equations in 2 unknowns and solve that to get [3,1,0], then set z=1 and solve to get [0,-1,1], subtract the two to get [-3,-2,1].

[Anonymous No.11033563]

>>11032528
>>11032509
>>11032505
It's not called stupid question thread for nothing.

My idea doesn't even ring a bell to you ? Does it evoke stuff that already exists and that I could check up to be less uncultured?

[Anonymous No.11033573]

>>11033563
You have no idea what you're talking about. Stupid questions are questions where you ask something simple, not where you pretend to understand something you don't even have superficial understanding of.
You have no idea what the incompleteness theorems say, you don't know what general AI means, and judging how dumb your post is, you probably also don't know what singularity means.
Try being more humble and stop pretending to know more than you do, because you come across as a massive faggot.

[Anonymous No.11033658]

>>11033573
>ooga booga i will tell he stupid so i be lookin smart
All of your assumptions are false. Are you one of those butthurt singularity faggots?

[Anonymous No.11033663]

what if your body made testosterone to STOp your body getting too horny?

[Anonymous No.11033703]

>>11033658
You have to be over the age of 18 to post on 4chan.

[Anonymous No.11033722]

l[math]\lim_{x\to3} 1+ \frac{1}{x} (2b+3 + \frac{5b+15}{x-3})[/math]

For this limit to exist, I have to end up with the last fraction being [math]\frac{0}{0}[/math]. Help me.

[Anonymous No.11033754]

>>11033722
b=-3

[Anonymous No.11033810]

>>11033754
Yes but if you follow through with that you end up with [math]\frac{x^2-6x+9}{x(x-3)}[/math] so you get [math]\frac{0}{0}[/math] for the limit

[Anonymous No.11033827]

>>11033810
x^2 - 6x + 9 = (x-3)^2

[Anonymous No.11033833]

>>11033827
Yup, thanks for that. Sometimes the obvious shit gets me.

[Anonymous No.11033885]

would it be possible to change genders with CRISPR ?

[brainlet No.11033902]

Not looking for an answer here, just want to understand what they're asking and how to start.

[Anonymous No.11033909]

>>11033902
rate of change = derivative
average = integrate and divide by the length of the interval.

[Anonymous No.11033919]

>>11032988
>still not at grad level
Lad, I was reading GTM books on my freshman year. The only difference is that they expect you to grasp the intuition by yourself instead of spoonfeeding it to you, plus skipping proofs. The first is actually a good thing, since constructing intuitions based on the formal concepts by yourself typically leads to stronger foundations, and the second one is ambiguous.
>>11029006
The advantage of the smash product argument is that it's extremely obvious that the smash product and the categorical product never coincide.
The advantage of the homotopy group argument is that it's three seconds to show that S^3 is simply connected.
>>11029130
A bunch of his posts vanished quite a bit later than he could delete them, so I give it a 90% chance of him having been banned for avatarfagging.
The tragedy of being unable to attach your waifu to your every post.
>>11030142
Who knows?
>>11030564
Yakumoposter did reply to your post, tho.
And please don't call me xe.
>>11032774
>drawing surfaces
Give up right now. You either can do it or you can't.
>drawing curves
Step one:
>draw the axis
Step two:
>plug some points in and mark them out
Step three:
>find any singularities and mark them out
Step four:
>smoothly connect the dots
>>11032961
Practice.
Also, euclidean geometry and linear algebra are the ideal starting points for proof writing, but if you struggle with those two, try set theory.
>>11033360
Okay.
>>11033383
Pic related?

[Anonymous No.11033929]

>>11033919
>Yakumoposter did reply to your post, tho.
Ah. Now I understand who this Yakumo poster is. A completely useless answer, as his answers to my questions tend to be. Well, at least he tries to help, and I appreciate that.

[Anonymous No.11033990]

>>11033902
classic use of tai's method

[brainlet No.11034007]

>>11033909
thanks, i eventually figured it out.

[Anonymous No.11034016]

>>11033919
that should be good, thanks

[Anonymous No.11034081]

>>11034007
Don't be too hard on yourself anon

[brainlet No.11034082]

>>11033990
is tai's method (x+h)-x))/h?

[Anonymous No.11034086]

>>11034082
"Tai's metod" is a jocular way to say integrate. The joke is that a guy named Tai published a biology paper in which he claimed to discover trapezoid rule for integration. The paper was published and cited many times. The author himself called the method "Tai's method"

[Anonymous No.11034125]

I saw a news article some years ago that I can't seem to find anymore. It was about a method of making solar panels out of any semiconductor using thin graphene strips with a current running through it to amplify or create the photovoltaic effect. They needed a power source to "turn on" but once the system was bootstrapped it supplied it's own current and the efficiency was something like 10-14%.

Can any of you point me towards it or the paper it was based on?

[Anonymous No.11034131]

citizen kane is overrated and problematic

[Anonymous No.11034152]

How can I use linear algebra irl? How complex of a system can you model with it?

[Anonymous No.11034217]

>>11033485
https://medienportal.univie.ac.at/presse/aktuelle-pressemeldungen/detailansicht/artikel/2000-atoms-in-two-places-at-once/

[Anonymous No.11034352]

How can I prove this by induction? I did the base case at n = 1, but after that I'm stuck

[Anonymous No.11034393]

>>11034352
Divide both sums by n.
Apply quick logic.

[Anonymous No.11034443]

>>11031472
Because plane waves are easy to work with. You can then create physically relevant wave packets via Fourier methods.

[Anonymous No.11034449]

>>11034352
Easy. Just use the fact that x + 1/x >= 2 for all x. Pair inverses and use this inequality n+1 times

[Anonymous No.11034503]

In my calc 1 class it was mentioned that e^(log(x)) = x for all x. I don't see how this is true for x<0 since log of a negative number does not exist

[Anonymous No.11034511]

>>11034503
For all x means for all positive x.
Obviously it doesn't work for x<=0 because you can only get positive numbers by exponentiation.

[Anonymous No.11034640]

If I have two uniform, oppositely charged clouds with the same radius and have them overlap, what would the resulting electrical field look like?
I'm guessing it'd look something like a 'Pure Dipole' in pic related, but I have no idea. How would I go about finding the new electrical field for the overlapping areas? I'm really stumped here.

[Anonymous No.11034643]

>>11034640
Stupid me, forgot to mention that they never fully overlap, they only approach eachother

[Anonymous No.11034692]

How am I supposed to interpret the notation here?

[Anonymous No.11034720]

>>11034511
R(i)'s ur path

>>11034692
parts per billion
or
between 57 and 67% and between 87 and 97%

[Anonymous No.11034738]

>>11034720

Wouldn't 62 +/- 1 be between 61 - 63%?

[Anonymous No.11034743]

>>11034738
yeah my b, misread it

[Anonymous No.11034758]

>>11034743
Doesn't it also show, that less workers died using higher concentrations? This makes no sense.

[Anonymous No.11035130]

How do you know when to use the product rule and when to just expand everything? I know that they want you to do difference of squares on number 9, but why can't you expand number 3 as well? I can see that expanding number 9 leaves you with just a polynomial, but you also have a polynomial if you just distribute e^x in number 3 don't you?

[Anonymous No.11035144]

>>11034503
> I don't see how this is true for x<0 since log of a negative number does not exist
The logarithm of a negative number is complex, with imaginary part π.
log(-x)=log(-1*x)=log(-1)+log(x)=log(x)+πi.
e^log(-x)=e^(log(x)+πi)=e^log(x)*e^πi=x*-1=-x.

[Anonymous No.11035162]

>>11035130
its really depends on the question, 3 and 9 can be differentiated in many different ways some are easier when you expand, others are easier when you do not, for 3, you could just do product rule for e^x and 3x^2 - 5x which would be e^x(3x^2-5x) + e^x(6x -5) if you dsitributed the e^x you would have to do product rule twice which is less efficient. The only reason classes like calc 1 and 2 tell you to do something in a certain method is so you are aware of these methods when you approach more complex questions, in higher level math classes you can differentiate and integrate however you want, all professors care about is if you get the right answer

[Anonymous No.11035180]

>>11035130
> How do you know when to use the product rule and when to just expand everything?
Whichever is simpler.

> why can't you expand number 3 as well?
You can. But it's marginally less effort to apply the product rule first than to expand first.
d/dx((3*x^2-5*x)*e^x)
= (d/dx(3*x^2-5*x))*e^x+(3*x^2-5*x)*(d/dx e^x)
= (6*x-5)*e^x+(3*x^2-5*x)*e^x
= (3*x^2+x-5)*e^x
vs
d/dx((3*x^2-5*x)*e^x)
= d/dx(3*x^2*e^x-5*x*e^x)
= (d/dx 3*x^2*e^x)-(d/dx 5*x*e^x)
= ((d/dx 3*x^2)*e^x+3*x^2*(d/dx e^x))-((d/dx 5*x)*e^x)+5*x*(d/dx e^x))
= (6*x*e^x+3*x^2*e^x)-(5*e^x+5*x*e^x)
= (3*x^2+x-5)*e^x

[Anonymous No.11035216]

cab one of you guys help me with this? I dont just want the answer tho, if you could guide me however it would be much appreciated

[Anonymous No.11035226]

>>11035162
>>11035180
ok thanks

[Anonymous No.11035239]

>>11035216
TIP: That shit's an inner product son.

[Anonymous No.11035284]

>>11035216
>ghostery and abp
cringe and unbased. use ublock origin, privacy badger and decentraleyes.

[Anonymous No.11035305]

>>11028644
How do you rank organic molecules based on their acidity and how basic they are? Anywhere I can read about this or watch a YT video?

Very urgent. Need this for my extractions lab

[Anonymous No.11035316]

>>11028918
I thought this was a stupid question thread

[Anonymous No.11035586]

When using the product rule on the (1+e^x)^2 it appears that this person is setting this up to then factor out a (1+e^x) from all terms. Can you factor things before applying the differentiation to the (1+e^x)^2?

[Anonymous No.11035857]

>>11035586
pleb bump

[Anonymous No.11035865]

>>11035586
I don't get what you are saying. If you mean the part that's highlighted, they are just applying chain rule and the (1+e^x) is just a factor after the differentiation, they are not factoring anything before finishing the differentiation step.

[Anonymous No.11035874]

>>11028700
Yep, I was banned for posting yaoi of Yachie on /jp/.
>>11028918
Think of it like this: if [math]S^1 \times S^2 \simeq S^3[/math], then they are also homotopic over [math]S^2[/math] and the fibre bundle [math]0\rightarrow S^1 \rightarrow S^3 \rightarrow S^2 \rightarrow 0[/math] is weakly homotopically equivalent to the trivial bundle [math]0\rightarrow S^1\rightarrow S^2 \times S^1 \rightarrow S^2 \rightarrow 0[/math]. Of course this would imply that all sections [math]S^2 \rightarrow S^3[/math] are trivializable and so the long exact sequence in homotopy gives [math]\pi_3(S^2) = 0[/math], which contradicts the fact that [math]\pi_3(S^2)=\mathbb{Z}[/math] with the generator given by the Hopf map.
>>11030837
https://en.wikipedia.org/wiki/Hadamard_product_(matrices)
>>11031472
Anon, I can't stop you from embarrassing yourself in the future but please:
1. understand what Born-von Karmen boundary conditions are, and
2. study spectral theorems for bounded elliptic compact operators on the n-torii
before making a post like this next time.
>>11033002
From an effective perspective, [math]\hbar[/math] serves as an energy scale the separates the physics of the classical from the quantum, similar to how the Planck mass [math]M_P[/math] separates the physics of the Newtonian from the Einsteinian GR. The main tenet of effective field theory is that the physics at different scales decouple, and that every theory we devise is an effective theory of another.
>>11035216
Denote by [math]P[/math] the projection onto an [math]n[/math]-dimensional subspace of [math]C^1([a,b])[/math]. Your goal is to show that [math]A[/math] is invertible iff it's an automorphism on [math]\operatorname{im}P[/math].

[Anonymous No.11035880]

>>11028644
are niggers a more retarded race?

[Anonymous No.11035959]

>>11035865
What I was referring to is how he got to the highlighted part because we haven't learned the chain rule yet. I used a differentiation calculator and it says that (1+e^x)^2 becomes 2e^x(e^x+1). The yellow part appears to be the same, but he puts the 2 and the e^x on different sides for some reason.

I don't even know how you differentiate that because I don't think we have learned it yet.

[Anonymous No.11035998]

>>11035959
You should check your notes again, it'd be pretty strange if you have all the differentiation rules except for the chain rule. Anyways, the main gist of it for real functions is that [math] f(g(x))' = f'(g(x))g'(x) [/math]. It's for composition of functions, which is what's happening in your case, you can take g(x) as (1+e^x) and f(x) as x^2, so of course f(g(x)) = (1+e^x)^2, and using the formula above you get the f'(g(x)) as 2(1+e^x) (basically you just differentiate the square while the inner part remains the same) and g'(x) is obviously just e^x
You get the same result with the product rule which apparently is what the instructions want you to do, by taking it as (1+e^x)*(1+e^x), but it just makes it unnecessarily longer. Since both factors are the same, you can notice that both resulting addends will have (1+e^x), as you said earlier. In which case, yes, you could see it as skipping the step up expanding the whole sum and you just factor (1+e^x) from the start (you are not factoring before the differentiation, you are just differentiating first but fast-forwarding the process to get to the end result, which you then factor). Ultimately both results are the same, but most people would use the chain rule in that case just for simplicity's sake. Even the solution you posted also seems to be using it, which is kinda funny but well, that's what's going on.

[Anonymous No.11036006]

How do I get smart? serious question

[Anonymous No.11036013]

>>11036006
Think of your mind as a muscle, think of getting smart as going to a gym. Train your mind, let your mind rest, eat and sleep well. Alcohol is bad for gains.

[Anonymous No.11036057]

>>11035998
Ok thanks. The chain rule is it's own section and it comes after power rule chapter, addition/subtraction, quotient/product, and the trigonometric functions. Not sure why this problem is even here.

[Anonymous No.11036076]

>>11036006
Learn Russian and mathematics.

[Anonymous No.11036078]

Prove that 10^(3n+1) (for any integer n) cannot be written as a sum of the cubes of two integers.

[Anonymous No.11036184]

How come cos(x)x^2 doesn't equal cosx^3 ?

[Anonymous No.11036248]

>>11036184
Because [math]x\in\mathbb{R}[/math] but [math]-1 \le \cos{y}\le 1[/math]. Now try x=2.

[Anonymous No.11036264]

>>11036076
>Russian
You mean Greek, right?

[Anonymous No.11036326]

>>11036248
x=0 works too

[Anonymous No.11036345]

>>11036326
True. Good point. It's been a while since I've used numbers as anything but indices, so "tricks" like that have been forgotten.

[Anonymous No.11036356]

>>11035305
There are pKa charts for common groups. Many of the molecules in your pic are testing your understanding of how enol tautomerization works.
Those are some dubious structures btw

[Anonymous No.11036453]

Is there any subset of R that includes the irrationals that can be counted?

[Anonymous No.11036477]

How do I prove by induction that any natural number x greater or equal than 2 can be written as 2y+3z, where y and z are nonnegative integers? these "x can be written as" proofs really confuse me

[Anonymous No.11036634]

>>11036477
decompose into 2 cases:
x is odd and x is even

[Anonymous No.11036653]

>>11036477
x even -> y = x/2 and z=0
x odd -> y = (x-3)/2 and z = 1

[Anonymous No.11036666]

>The Lewis-diagram for element X is (see pic) Denote the three simplest structures this element can make with Chlorine, and which of the three are most likely to form.

Am I fucking retarded? Why is this so hard to solve

[Anonymous No.11036718]

If I have to find the supremum and infimum of a set that's only one element, like {a}, are they equal to each other? Or do they not exist?

[Anonymous No.11036770]

Determine all prime numbers p and q for which the number pq is defective.
I know the number pq is called a semiprime and that σ(pq)<2pq to be called deficient. How do I continue?

[Anonymous No.11036776]

>>11036718
Yeah, inf {a} = sup {a} = a, iirc for any partial order.
Doesn't quite work otherwise.
>>11036770
Hmm, I sure wonder what divisors does a semiprime have. Couldn't be that many. Definitely.

[Anonymous No.11036802]

>>11036776
>what divisors does a semiprime have
1, the two primes which made the number, and itself?
Do I solve something like 1+p+q<pq? Since that should make it a deficient number

[Anonymous No.11036834]

I think I'm too retarded for CIP priority rules
Why is this not 2R, 3S?
For the number 2 carbon, the CH2OH group is first on the priority list, then the number 3 carbon group, then the hydroxide and finally the hydrogen.
Similarly for the number 3 carbon, the CHO group is first, then comes the number 2 carbon group, then the hydroxide and the hydrogen.
What the fuck am I missing here?

[Anonymous No.11036845]

>>11036834
I forgot to mention the number 3 carbon group needs to "turn" in order for CIP to apply but even then I don't get why the molecule is 2S, 3R.
Why does HO have more priority than CH2OH (which is what I'm assuming is the reason for the #2 carbon to be S), and likewise, why does the #3's hydroxide group have priority over CHO?
Is it because it's implied that with CHO and CH2OH, the carbon atom is the first to bond, which means it has less priority than the oxygen in the hydroxide groups?
So if instead of CHO it had been OHC, and OCH3 instead of CH2OH, that would've changed the entire CIP layout?
Sorry for the retarded questions I just don't get it

[Anonymous No.11036856]

>>11036802
Yes, something like that.
You might want to assume that the smaller prime factor is larger than say, 5, and prove the opposite inequality.

[Anonymous No.11036857]

>>11036856
Thank you anon

[Anonymous No.11036949]

>>11036453
No. R is uncountable, Q is countable, thus R-Q (the irrationals) is uncountable and so is any superset of it.

[Anonymous No.11036983]

>>11036834
>>11036845
Never mind, I think I got it:
For the carbon #3, the rotation goes clockwise because CHO means it's an aldehyde (it would've been an alcohol had it been COH, so the order does matter), so there's a double bond meaning the oxygen is counted twice, therefore it has higher priority than anything else linked to the asymmetric carbon except OH.
For the #2 I was just being a retard since it's obvious OH has higher priority due to being directly linked to the asymmetric carbon, whereas CH2OH is linked through one carbon first, and then the oxygen, therefore it loses.

[Anonymous No.11037063]

Fr is the only torque on pic related, right?

[Anonymous No.11037129]

can someone point out what im doing wrong? im supposed to compute the determinate in two different ways but im getting different results

[Anonymous No.11037144]

>>11037129
3(3-3) should be 3(9-1)
3(4+9) should be 3(4-9)

so the determinant is 0 in both cases

[Anonymous No.11037147]

>>11037144
sheit, what tf am i doing
tyvm

[Anonymous No.11037148]

>>11037063
yes

[Anonymous No.11037223]

Im helping friend but havent done this shit in a while pls respond

if f(x) = x +tanx, what is (f^-1)'(x)?

.

[Anonymous No.11037431]

>>11030496
{-8i, 4i, 2i, i, -8, 4, 2, 1}

[Anonymous No.11037476]

How can I find the limit of a recurrence sequence? Specifically x_(n+1)=(2+x_n)^(1/2), where x_1 = 2^(1/2)

[Anonymous No.11037479]

>>11037476
[math]\lim_n x_{n+1} = \lim_n x_n = x[/math]

[Anonymous No.11037534]

>>11032702
Well if you say for the parametric that x=t^t then ln(x) = t*ln(t).

with the y part: y=ln(t) which means t=e^y.

plugging this t value into the x formula gives us ln(x) = ye^y

taking the Lambert W function of both sides gives us y=W(ln(x)). This parametric gives us the lambert W of ln(x), hence why it looks similar to the regular Lambert

[Anonymous No.11037749]

is there any money in being a TeX monkey or does basically everyone do their own thing?

[Anonymous No.11037918]

If fn converges uniformly to f and gn converges pointwise to g (both with domain (0,1)) and fn+gn converges uniformly to f+g, then gn converges uniformly to g.

Is this even true?

[Anonymous No.11037924]

>>11037918
Yes.
TIP: (f_n+g_n)-f_n.

[Anonymous No.11037938]

Please tell me how the hell this works.
"Answer" in next post

[Anonymous No.11037943]

>>11037938
This is the answer. It doesnt check out at all if you put in the numbers. Please help

[Anonymous No.11038105]

Okay somehow I have been invited to an ivy league school. Let's say I even have the money to go via scholarships and stuff. What the fuck man. I didn't know getting a near 4.0 in community college for the two semesters I went would warrant this. What do I do? Do I go? What's even a place like that well, like? What am I even doing? What is this? Help.

[Anonymous No.11038287]

>>11037749
Most people working on biology, medicine and other natural sciences (and even some in experimental physics) can't TeX for shit, so they hire someone to do it for them, so they can then submit that to a journal. If you enjoy looking at a horrible jumble of tables and equations written in M$ Word then yes, you can make an easy buck off those geezers

[Anonymous No.11038513]

if you are differentiating using the quotient rule, can you always rewrite it as multiplication by rewriting with negative exponents? If I have x/cosx can I rewrite as xcosx^-1, or does that mess things up since the ^-1 usually means the arccos?

[Anonymous No.11038586]

>>11031965
Brainlet Anon chiming in here. Is pic related even remotely correct?

[Anonymous No.11038601]

How do i do d)
I don't have any idea how to begin, i did all the others.

[Anonymous No.11038651]

>>11038513
It works fine, if it didn't then you would have counterexamples that could break either the quotient or the product rule, just by taking the multiplicative inverse of a function (which is still a function, at least when it has non zero values)
If you are gonna do that then you have to learn the difference between an inverse function and the multiplicative inverse. 1/cos(x) is NOT arccos(x), and even if you used the ^-1 notation, (cos(x))^-1 and cos^-1(x) are two different things.

[Anonymous No.11038673]

>>11038586
Just the counterexample would have been enough, but yeah that's fine. If you wanna prove something that's true instead try >>11032320
Also in the substition, you take (2(4+1)) instead of (2*4+1). The formula says (2k+1), not (2(k+1)). It doesn't change your result cause the statement is still false, but if you wanna try proving the statement above then keep that in mind.

[Anonymous No.11038741]

>>11038673
I revised the problem with the true proof. Thanks, anyway.

[Anonymous No.11038756]

>>11038601
Is this from a engineering program or are you a mathematics undergraduate?

[Anonymous No.11038767]

>>11038756
Its a mathematical foundations class from a graduate level engineering course

[Anonymous No.11038908]

If I have a finite group [math]G[/math], and then I have [math]G'[/math] be an extension of [math]G[/math] corresponding to [math]\text{id}\in H^2 (G; H_2 (G; \mathbb{Z} ))[/math], what is this [math]G'[/math] supposed to be like?

[Anonymous No.11038910]

>>11038741
Uhh we did this in class 9 if I remember correctly

[Anonymous No.11038913]

>>11038601
Reminding you that [math]x^T y=<x, y>[/math]
The reduced formula gives immediately that the diagonal is everywhere zero.
The full formula shows that A_{ij}=-A_{ji}, which is the skew symmetry condition.

[Anonymous No.11039080]

Suppose I've got a price is right wheel with an outcome with probability p and an outcome with probability q and then a bunch of other outcomes that I don't care about. If I spin it a bunch I'll get A of the first outcomes and B of the second outcome and some other stuff that isn't important to me. I'm interested in the probability of getting exactly a of the first outcome and b of the second in n spins.

Can I treat A and B as binomial random variables and do this:
[eqn]
\begin{align}
P(A=a \cap B=b) &= P(A=a)P(B=b|A=a) \\
&= \binom{n}{a}p^a (1-p)^{n-a} \times \binom{n-a}{b}q^b (1-q)^{n-a-b}
\end{align}
[/eqn]

Or am I overcounting?

[Anonymous No.11039181]

I'm not sure if I'm doing this right

"Calculate the mass of the acetic acid with a concentration of 9% that can be obtained from 600g of 80% 1,1,1 tricloroethane"

C2H3Cl3 + 3H2O -> C2H4O

80% = 480 grams of pure C2H3Cl3
M(C2H3Cl3) = 133.5g/mol
M(C2H4O) = 44g/mol

480/133.5 = x/44 => x = 126.84g of 9% acetic acid solution => 11.41g of concentrated acetic acid

Am I retarded?

[Anonymous No.11039233]

>>11039080
pretty sure you should replace [math] q [/math] with [math] \frac{q}{1-p} [/math]

[Anonymous No.11039256]

turbo brainlet here
why was schrodinger's cat ever a problem?
how could you expect the system to not decohere immediately?
or was the question more about what the cat perceives if we somehow could keep it in a coherent state?

[Anonymous No.11039313]

How do I get [math]x(x-1)^8[/math] from [math]x(1-x)^8[/math] ?

[Anonymous No.11039329]

>>11039233
I think I don't because the [math]P(A=a)[/math] part is already ""checking for"" there being no more of the first result in the remaining [math]n-a[/math] observations. I think I've convinced myself I'm right but I might have also just invented a new wrong interpretation of probability.

[Anonymous No.11039357]

>>11039313
x(1-x)^8 = x(-(x-1))^8 = x(x-1)^8
because if n is even, then (-a)^n = a^n

[Anonymous No.11039405]

>>11039357
cheers m8, I've just spent nearly 2 hours on an integral over this issue lmao

[Anonymous No.11039458]

>>11039329
No your formula is not correct. Let's just look at an example:
Let's say the wheel is equally divided in 4 sections "A", "B", "C", and "D".
And we want to know the probability of getting exactly one "A" and exactly one "B" in two spins.
So the variables are: n=2, a=1, b=1, p=1/4, q=1/4.

Let's list all the possible outcomes of the two spins (these outcomes all have equal probability):
AA, AB, AC, AD, BA, BB, BC, BD, CA, CB, CC, CD, DA, DB, DC, DD
As you can see 2 of the 16 outcomes have exactly one "A" and exactly one "B". So the probability is 2/16 = 1/8.

But if you plug the variables into your formula, you get 3/32, which is wrong.
While if you plug the variables into my formula, you get 1/8, which is correct.

.

>the P(A=a) part is already ""checking for"" there being no more of the first result in the remaining n−a observations
that's precisely the problem
the second part of your formula is calculating as if you haven't already checked for this, while my formula corrects for this

[Anonymous No.11039486]

Why the fuck do I randomly wake up with moderate stomach pain that causes intense anxiety/panic/depression until I inevitablely vomit up some yellow bile, sometimes I can just shit to make it go away, otherwise I just pace around until it goes away???

This has been going on for years.

[Anonymous No.11039546]

>>11039486
And now I'm perfectly fine and heading back to sleep

[Anonymous No.11039650]

>>11028644
I'm just curious Stupid Question Thread guy. What is your IQ? You seem to know something about everything.

[Anonymous No.11039663]

What is the standard procedure for finding if a sequence is bounded or limited? My professor sometimes just finds un+1-un and then concludes from there, sometimes takes the limit of the sequence and sometimes finds the first term, but I'm not sure what to apply

[Anonymous No.11039666]

>>11039650
He's a moron. Don't even start.

[Anonymous No.11039675]

>>11039650
>he thinks I've ever taken an IQ test
>>11039666
Based.
>>11039666
>how to find if a sequence is bounded
There's really no universal method, but the two main tricks are calculating the limit (if it has one it's bounded) and extending it to a function on the entire real line (like in the harmonic series trick) and showing it's bounded.
You can also wildly guess a bound and check if it works, or bound it above or below by bounded sequences.

[Anonymous No.11039679]

>>11039675
Third part was for >>11039663

[Anonymous No.11039746]

>>11038908
It's called the universal central extension.
Any other central extension is obtained by quotienting some of H_2(G) out.

[Anonymous No.11039757]

Am I seeing things or did epsilon one just become negative for no reason?

[Anonymous No.11039953]

>>11039746
Cheers, m808!

[Anonymous No.11040400]

Geobros, any tips about classifying rocks according to the QAPF diagram?
Identifying orthoclase is easy as fuck, but distinguishing between plagioclase and quartz is kinda difficult.

[Anonymous No.11040401]

if an irrational number has infinitely many digits, does that mean that if you could convert the number into binary and from that into text, every combination of letters and every possible thing you could say in any language would be in there somewhere?

[Anonymous No.11040408]

what are some good textbooks on homological algebra?

[Anonymous No.11040410]

>>11040401
No, that's a common misconception.
Consider the irrational number (base 10)
0.10100100010000100001.....
It's irrational because its expansion is not periodic. Yet clearly the decimal expansion will never contain the sequence 456. Same if you consider the same digits but base 2: the number is also irrational but it will never contain the sequence 1111.

[Anonymous No.11040415]

>>11040408
Weibel.

[Anonymous No.11040421]

>>11040415
thanks. was thinking about choosing between maclane, cartan and weibel

[Anonymous No.11040426]

>>11040410
>0.10100100010000100001.....
what if it's something like 0.5207384275489836471164381083648419472...where all numbers can happen?

[Anonymous No.11040439]

>>11028644
How do I say a set has no duplicates in predicate logic? I know sets usually don't have duplicates but this is for a BON diagram of some software

ive got ∀x∀y: x, y ∈ SET: x ≠ y

[Anonymous No.11040450]

>>11040426
You're dumb. Did anyone ever tell you that?

[Anonymous No.11040457]

>>11040439
You're dumb too. Sets can't have duplicates because it doesn't even make sense to talk about duplicates. A set either has an element or it doesn't.
If you want a "set" where you can be able to talk about whether or not it has duplicates, you need multisets.

[Anonymous No.11040458]

>>11040450
>enter stupid question zone
"you're dumb!"

very insightful comment lmao

[Anonymous No.11040461]

>>11040458
What are you supposed to say when you answer the question and he asks the same question again?

[Anonymous No.11040467]

>>11040457
it's an invariant to make sure the collection is a set, that's what the predicate is for.

[Anonymous No.11040484]

>>11040450
if I weren't dumb I wouldn't be asking dumb questions don't you think

[Anonymous No.11040511]

can someone please walk me through how you integrate x/(x+c)^2 with respect to x? im in calculus 3 rn and my enginnering prof just reamed the whole class for not being able to solve it after like 10 goddamn minutes, i was really trying, i tried splitting it up into a product and using chain rule but that didnt fucking work aaaaaaa

[Anonymous No.11040514]

>>11040511
>chain rule
product rule*

[Anonymous No.11040530]

>>11040511
>integrate
take the derivative*

[Anonymous No.11040634]

I just beat a puzzle challenge and was entered into a random draw for winning

[Anonymous No.11040641]

>>11040426
>0.10100100010000100001.....
> what if it's something like 0.5207384275489836471164381083648419472...where all numbers can happen?
Did you not notice that the example
>0.10100100010000100001.....
contains both zeros and ones?
A number being irrational simply means that its expansion (in any base) doesn't end in a fixed sequence which repeats infinitely. It doesn't preclude any other constraints. Maybe it consists entirely of pairs of digits (e.g. 0.11223344...) which would mean that 123 never occurs. Or maybe it never contains pairs of digits. Or some other constraint which would preclude certain sequences from ever occurring.

Numbers where every digit sequence of a given length is equiprobable are called normal numbers. While many irrational numbers are believed to be normal (including e, pi and sqrt(2)), and it can be proven that almost all irrational numbers are normal (i.e. non-normal numbers are a sparse subset of the irrationals), there are very few irrational numbers which have actually been proven to be normal.

[Anonymous No.11040752]

>>11040511
>>11040514
>>11040530
x/(x+c)^2
take multiplicative inverse if you want to use the product rule, doesn't really matter but I'll do it
x*(x+c)^-2
product rule, you'll have to use chain rule in the right
technically you can use product again but fuck that it's too unnecessarily long
1*(x+c)^-2+x*-2*(x+c)^-3
and just simplify it a little bit
(x+c)^-2-2x(x+c)^-3
((x+c)-2x)/(x+c)^3
(c-x)/(x+c)^3

[Anonymous No.11040770]

>>11040752
tyvm