New SQT thread

[Anonymous No.8424626]

I'll start. How in the sweet baby Jesus do you answer my next post

[Anonymous No.8424629]

How 2 do q1-3 (op here)

[Anonymous No.8424654]

>>8424629
>not using the already existing SQT thread
>posting your image sideways

[Anonymous No.8424721]

>>8424654

Pls

[Anonymous No.8424740]

Bump

[Anonymous No.8425070]

How do you take the gradient for this if you know [math]x \in \bf{R}^2[/math]?

Do you just replace x with [x1, x2] and do all the matrix multiplications? Or can you do it straight away?

[Anonymous No.8425073]

>>8424626
>SQT thread
>Stupid Question Thread thread
GTFO out of here, NEWFAG!

[Anonymous No.8425088]

How can we get rid of the racist White Man's science and replace it with POC science?

https://www.youtube.com/watch?v=C9SiRNibD14

[Anonymous No.8425167]

Why is the force not simply -2kx i?

[Anonymous No.8425185]

>>8425167
because the springs are pulling in different directions, so the vertical pull of the two springs cancels out and only the horizontal pull is left. So you use pythagorians theorem to convert the side length from the hypotenuse/other side and you get that funky square root thing. Too lazy to read and actually deduce

[Anonymous No.8425258]

>>8424629
Use suvat equations and newtons second law equation

[Anonymous No.8425628]

I know what the EM tensor is and how raising/lowering indices work

[math]F^{\alpha \beta} = \eta^{\alpha \gamma}\eta^{\beta \delta}F_{\gamma \delta}
[/math]

But what is

[math]F^{\alpha}\! _{\beta}[/math] ?

We use it to find to find the Stress-Energy tensor, but I don't see how I would write it out like I can with the others in matrix form.

t. new at tensor calculus

[Anonymous No.8425718]

>>8425628
Got any good books on tensor calculus?

[Anonymous No.8425726]

I am pretty confused

how do you calculate slope of a vector in 3d?

for every one unit you move along a line, how many units do you move upwards (in units on the Z axis), I mean to say?

The main idea is to calculate where you'll be after moving one unit forward I think, which I think I could work out how to do. Then just take the difference in Z units and that's your slope

[Anonymous No.8425735]

>>8425070
can someone give me a pointer

[Anonymous No.8425738]

>>8425735
sure

0xbff5a3f6

[Anonymous No.8425742]

>>8425738
thnx

[Anonymous No.8425744]

>>8424629
M1?

[Anonymous No.8425760]

>>8424629

[Anonymous No.8425790]

>>8425735
yo why doesn't anyone told me. I already found it using the stupid method with x1 and x2 but I swear there must be something more clever

[Anonymous No.8425793]

>>8425790
>yo why doesn't anyone told me.
because you should learn to do your own homework autist

[Anonymous No.8425800]

>>8425793
>implying you even know the answer

i've already done it i just wanna learn better methods. also i meant s/told/tell in the previous post

[Anonymous No.8425802]

>>8425800
what do you mean by the 'stupid method'?

[Anonymous No.8425804]

I somehow can solve complex equations with no issue, however babby fundamentals elude me.

Anyway let's say we have a survey for income with options being
1. 0 - $24,999
2. $25,000 - $50,000
And so on, would this be an interval or ratio scale?

[Anonymous No.8425807]

>>8425802
I replaced [math]x[/math] with [math](x_{1}, x_{2})^{T}[/math] and did all the matrix multiplications and solved for x (I'm given a, b and B numerically)

[Anonymous No.8425823]

>>8425807
whats stupid about that?

[Anonymous No.8425828]

>>8425823
there must be some tricks like (x-b)^T B (x-b) looks like some standard form that should have some easy derivation without multiplying all matrices? (x and b are vectors)

[Anonymous No.8425834]

>>8425828
learn to use google dummy, if you can't even use google you're going nowhere in school or life

[Anonymous No.8425837]

>>8425834
just show me a link if you found something

[Anonymous No.8425839]

>>8425837

[Anonymous No.8425844]

>>8425839
:)

[Anonymous No.8425882]

>>8425718
No, this is taught as a side piece in a course in electrodynamics I'm taking.

[Anonymous No.8425924]

I'm taking babby's first analysis class, and i'm trying to go back and get an intuitive grip on some of the more important definitions and theorems.

In particular, i'm trying to get a better grasp on the definition of an accumulation point of a set. The definition i have is: Let S be a subset of R. Then s in R is an accumulation point of S iff for all epsilon > 0 there exists t in S such that t != s and |t-s| < eplsilon.

I think i have a good intuitive picture of this: we have a subset S of the real line, and some point s is an accumulation point if no matter the epsilon we can make an interval (deleted neighborhood) around s that contains no gaps. I guess my question is, what about the definition guarantees there are we have the right epsilon so that there are no gaps or missing points in this interval? Would we need a greatest epsilon for this, or some type of thresh hold? or does the universal quantifier before epsilon ensure that whatever one we start with, we are sure than any epsilon_i smaller will also be satisfied?

Also, regarding the existential quantifier before the t in the definition, how can we be sure there doesn't simply exist only one t that satisfies the condition and thus there is/are hole(s) in our interval? or is that why we have the universal quantifier for epsilon, to guarantee an interval for any interval between any s and points s + epsilon or s - epsilon (we can always make epsilon smaller).

Also, the actual defintion in my book says there must exist at least one t. If there were only one t, mustn't we be at the point immediately adjacent to s (on either side)? are there ever instances where we deal with this?

[Anonymous No.8425937]

>>8425924
mate how about you put that gibberish into plain english? and accumulation point of a set in a topological space is a fairly simple concept.

[Anonymous No.8425950]

>>8425937
Youre right, i should have reread it before i posted it. perhaps i will come back after trying to dissect the definition a little more carefully and you correct any misunderstandings? Perhaps I should look into the definition in terms of a topological space as well (though i don't have any topology under my belt, sadly)

[Anonymous No.8425955]

>>8425950
stick with R for now mate

just think about where your problem is arising. figure out exactly which part of the definition doesn't make sense? then the answer is trivial.

[Anonymous No.8425961]

>>8425924
Your problem with understanding is that your "good intuitive picture" is completely wrong.

An accumulation point, in simple English, means that some points of S cluster extremely closely around that point. So closely that no matter how small of a neighborhood you choose there's always one inside it.

There's no guarantee at all that neighborhood is "gapless"; consider the set S = {1/n | n is natural number}. 0 is a cluster point. You can't take any interval around 0 with "no gaps" because you only have rationals in S.

[Anonymous No.8426085]

/k/ here. We pretty frequently talk about disaster preparedness, and it got me thinking. What do I do about my prescription meds?

I'm on a proton pump inhibitor for acid reflux, and a beta blocker for a hereditary heart defect. Obviously I can just stockpile a lot, but that's only a temporary solution. Are these things I could make at home?
Or am I doomed to have heartburn until I die of a stroke?

[Anonymous No.8426166]

>>8425955
thanks, yeah sometimes i get too carried away and start studying things beyond the scope of what i should

>>8425961
thank you for the correction. i now see i was trying to force my preconceived notions onto the mathematical definition

[Anonymous No.8426177]

>>8425726
bump

even a purely theoretical answer would be helpful

[Anonymous No.8426225]

>>8426177

Use direction cosines.
https://en.wikipedia.org/wiki/Direction_cosine

[Anonymous No.8426240]

Is CS just a meme? What are /sci/'s top picks?

[Anonymous No.8426252]

>>8426225
oh neat thank you

I also came up with an alternate solution which is a lot more roundabout, but I noticed it was easy to calculate an X and Y line from the top-down. Then you just plug in any two points on that line into a function which calculates its intersection with your plane (like pic related). Then you have to is compare your origin coords to that and do some more trig to get the rise/run... Would probably be a lot slower though

[Anonymous No.8426280]

>>8426240
shitposting

[Anonymous No.8426359]

>>8426252
>using floats
don't

[Anonymous No.8426370]

>>8426359
pretty massive difference between floating points and doubles concerning memory management

If this was scientific it would be a bigger deal but I seriously doubt anyone will notice I'm .0000002 off

[Anonymous No.8426372]

>>8426370
what the fuck are you even doing that you can't spare some KB's of memory for more accurate results?

[Anonymous No.8426389]

>>8426372
it's just needless

this is 3d collision detection though to answer your question, and this is calculating the player's physical interaction with a mesh (speed has to be slower the steeper the slope is, which means you need to take a plane, a direction, and a XYZ coordinate and return the slope of the mesh directly under and in front of the player)

[Anonymous No.8426410]

>>8426240
lol, who put chemistry up there?

[Anonymous No.8426484]

I need to prove this lemma.

I believe I have to use the number theory theorem of [math]d\mid m [/math] and [math]d\mid n[/math], then [math]d\mid m + n[/math].

I don't know how to start developing it though. Could you guide me a little?

[Anonymous No.8426496]

>>8426484
lcm(5,2)=lcm(5,7)?

[Anonymous No.8426593]

How do you calculate the original Kinetic Energy here?

[Anonymous No.8426596]

>>8426593
here is pic, pls help

[Anonymous No.8426611]

I need help with this one. I've been trying to clear "s" to prove what they're asking me to do. Any idea?

[Anonymous No.8426627]

>>8426496
Come again? That's an example, sure. It doesn't help me generalize the lemma though.

[Anonymous No.8426632]

>>8426593
>>8426596
anyone??

:(

[Anonymous No.8426667]

>>8426627
lcm(5,2)=10
lcm(5,7)=35

[Anonymous No.8426689]

>>8426667
pls see
>>8426632

[Anonymous No.8426696]

>>8426667
So you meant it as a counterexample?

That's how the exercise is written letter for letter. I skipped mentioning a and b are positive integers but I don't think it changes much in this case.

Also, I'm not sure if the order changes the product when it comes to lcm. lcm(2,5)=/=lcm(5,2) if we consider the guideline of this exercise, right? Because in the former a stays as 2, in the other a stays at 5. Is there a rule on lcm that it has to be ordered starting from the smallest? I'm pulling shit out my butthole at this point but I'm out of ideas so I don't know what else to do.

[Anonymous No.8426699]

20.19?

[Anonymous No.8426700]

>>8426696
No the lcm doesn't depend on the order. The exercise is wrong as written; I'm guessing they meant gcd instead of lcm.

[Anonymous No.8426706]

>>8426700
Ok let's assume they did fuck it up, how can I develop the demostration using the theorem I mentioned before?

[Anonymous No.8426717]

>>8426706

Well you said that if d|a and d|b then d|a+b

So any common divisor of a and b is also a common divisor of a and a+b

Therefore gcd(a,b)<=gcd(a, a+b)

(why? we showed that any common divisor of a and b is a common divisor of a and a+b. In particular gcd(a,b)|a,b, so it is a divisor of a and a+b. Any divisor of a and a+b cannot be larger than the *greatest* common divisor of a and a+b)

Now how would you show that gcd(a,a+b)<=gcd(a,b)?

[Anonymous No.8426728]

>>8424626
Hey, why does MIT OCW physics have this little video lectures? That's pretty damn sad :(

[Anonymous No.8426729]

>>8426728
because MIT physics lecturers are known pedophiles whose videos need to be removed to preserve MIT's stellar reputation

[Anonymous No.8426784]

>>8426729
Oh shit. wtf, I had no idea that you aren't actually joking.
https://www.quora.com/What-is-the-reaction-to-MIT-removing-Walter-Lewins-lectures-from-OCW

[Anonymous No.8426934]

I have a question about a wheatstone bridge like pic related. Assuming it's balanced, how could you express the voltage at B and D in terms of the cell voltage? I would normally just use the voltage divider equation but the fact that the current is split at A is confusing me, because you don't know how much current is flowing through B and D

[Anonymous No.8426938]

>>8426934
shit forgot to upload pic

[Anonymous No.8426962]

>>8426934
I assume by cell voltage you mean supply and not Vg. If you assume it is balanced, you know Rx, and D = B. Vg can be treated as open. Given this information, the system can be treated as two resistors in parallel for determining the current distribution from A (current divider), if you find that entirely necessary. I don't think you need to know that to answer the question, given that it is balanced.

[Anonymous No.8426971]

>>8426962
>given that it is balanced
so to find the voltage between D and C I can use the normal voltage divider equation? I.e Vo = Vi(R2/(R1+R2))? I know this is really basic but i can't quite wrap my head around how the full voltage of the cell would be going through both D and B, I thought it would have had to be divided at A

[Anonymous No.8426986]

>>8426971
>so to find the voltage between D and C I can use the normal voltage divider equation? I.e Vo = Vi(R2/(R1+R2))

Correct.
Vo = Vi(R2/(R1+R2)) = vi(Rx/(Rx+R3))

>I know this is really basic but i can't quite wrap my head around how the full voltage of the cell would be going through both D and B, I thought it would have had to be divided at A

This part I don't quite understand. The branches are parallel, so they both drop the full supply voltage, A - C.

The current however is divided, meaning unless Rx+R3 = R1+R2, Id != Ib. However, D and B are not a function of branch current, only total voltage drop and the ratio of the resistors.

[Anonymous No.8426989]

>>8426986
Correction: In wheatstone bridges, Vo = Vd - Vb. If balanced, Vo = 0.

Vd = Vb = Vss(R2/(R1+R2)) = Vss(Rx/(Rx+R3)) would be better. Vss being supply (or cell voltage).

[Anonymous No.8426996]

>>8426986
thank you so much for this post you just clarified all the little bits that have been fucking with me.
Thank you anon

[Anonymous No.8427123]

when finding the roots of x^4 - 16 it has complex and real roots right? how do I analyticalally know that it has both?

When I make y = x^2 and do y^2 - 8^2 = 0 then (y - 8)^2 then (x^2 - 8)^2 I get the real roots but when I do the quadratic formula I get imaginary roots.

[Anonymous No.8427125]

>>8427123
x^4 - 64
not x^2 - 16

[Anonymous No.8427156]

>>8425804
Are these logged values? Then you would have log ratios being expressed
>Everyone thinks about log-dollars as a natural unit

[Anonymous No.8427164]

>>8426085
You're gonna die and it will probably be both painful and embarrassing as you kick the bucket during the revolution not bc of some battle-sustained injury but because your heart gives out.
>Feel sorry for whoever is next to you when you shit yourself immediately after death.

[Anonymous No.8427412]

How do I git gud at real analysis proofs/problem solving? I know I should just "do problems" but the problems I look at in my book (not Rudin) seem complex and I don't know how to even get started on a lot of them.

[Anonymous No.8427441]

>>8427412
Honestly, it's not hard at all...

[Anonymous No.8427449]

>>8427123
>y^2 - 8^2 = 0 then (y - 8)^2
no

[Anonymous No.8427568]

Is MyMathLab retarded or am I missing something here?

[Anonymous No.8427569]

>>8427568
Nevermind, was suppose to be in radians even though the fucking shits have been using degree for the rest of the questions

[Anonymous No.8427780]

>>8425828
> there must be some tricks like (x-b)^T B (x-b) looks like some standard form

x^T.x is just |x|^2.

Set B=R^T.R, then
x^T.B.x = x^T.(R^T.R).x
= (x^T.R^T).(R.x)
= (R.x)^T.(R.x)

I.e. it's |y|^2 where y=R.x.

If R is orthonormal, then |y|=|x|.

Note that B is typically symmetric; making it so with B'=(B+B^T)/2 doesn't change the equation. Setting B=R^T.R guarantees that B is symmetric. Any symmetric matrix can be factored in that way, although if any of the diagonal elements of B are negative, R will be complex.

[Anonymous No.8427785]

>>8427123
> when finding the roots of x^4 - 16 it has complex and real roots right? how do I analyticalally know that it has both?
Real and imaginary (not complex) roots.

> When I make y = x^2 and do y^2 - 8^2 = 0
Should be:
y^2-4^2 = 0
=> y=±4
=> x^2=±4
if x^2 = 4 => x=±2
if x^2 = -4 => x=±2i

[Anonymous No.8427793]

>>8427412
Real analysis is the first course where most people really understand how important definitions are in mathematics. It's also one of the reasons a lot of people struggle with it.

The problems likely seem complex because there's a lot of funky jargon you just learned, and they may seem jumbled together arbitrarily.

Every time you read a problem you should be constantly keeping in mind what the primitive definitions of each word are. If you do this it should be easier to see what the problem is _really_ asking you to do and most problems will become much simpler.

[Anonymous No.8428004]

[eqn]\lim_{x\to0} \frac{xy^{5}}{x^{2}+y^{6}}[/eqn]

[Anonymous No.8428011]

>>8428004
As [/math](x,y) \to (0,0)[/math]
Wrote the limit wrong, in any case, how do I find this, through the pathing method right?

[Anonymous No.8428012]

>>8428011
[math](x,y)\to(0,0)[/math]

[Anonymous No.8428052]

>>8428012
No limit unless you're working with complex x and y in the Riemann sphere or if they're reals in a specific quadrant or if the direction of the convergence is specified. Or maybe other things too, but with just what's given, there is no limit.

[Anonymous No.8428053]

>>8428052
Alright, thanks, that's all I really needed

[Anonymous No.8428086]

How the hell do space companies know their rockets can reach orbit before they launch?

I've been playing a lot of Kerbal Space Program recently and been practising some orbital mechanics but I don't get that shit.

I'm comfortable with all the manuevers once up in orbit but I ave no idea how to approach the maths for things like:

>Will this vehicle make it into orbit
>How long will it take between ignition and orbit
>Timing so that the vehicle rendezvous with a satellite after reaching orbit

[Anonymous No.8428095]

>>8428086
rocket thrust
fuel depletion rate
distance to travel
load
gravity
all these need to be considered

[Anonymous No.8428098]

>>8428095
It's easy to think about what things will have an effect but it's something more to combine those in the right way

[Anonymous No.8428178]

>>8428098
Engine tests give you the relationship between thrust and fuel consumption. Aerodynamic theory will give you a fairly accurate figure for drag as a function of altitude and speed. You know almost exactly how much the rocket itself will weigh from the design.

Those tell you how much mass your rocket can put into a given orbit to within sufficient accuracy to know whether it's worth actually building the thing. Once it's built, you have an accurate weight measurement, and test flights will give you even more accurate figures for drag, thrust, and fuel consumption.

From there, working out how much payload you can get to any given orbit is a relatively straightforward calculation. It doesn't have to be perfectly accurate, but the maximum payload is dictated by the worst-case error, so any inaccuracies eat into your effective payload capacity.

[Anonymous No.8428317]

How do you draw your nu so it doesn't look exactly like a v

[Anonymous No.8428490]

Can one get a job with a computer information systems degree? If so, what jobs do they entail? Are they third fiddle to the Computer Science and engineering degrees?

[Anonymous No.8428546]

>>8424626
Here is a hw problem that I am unsure of.

You have a 100kg cart A moving to the right at 8m/s toward a 200kg cart B at rest that is connected to a spring. the two carts move together after impact. (I assume this implies perfectly inelastic collision.) Neglect friction. Spring constant is 3000 N/m

Find the velocities after impact, spring deflection (what is meant by this?), and energy lost during the impact.

pic is my attempt but I am unsure about the energy loss part. Thanks

[Anonymous No.8428547]

>>8425070
>Calling the differential of a vector field gradient
Stop

[Anonymous No.8428549]

Anons! I e-mailed with a lab and they want me to come in for an interview!

Does anyone have any idea on what I should expect? Any tips on preparing?

[Anonymous No.8428693]

>>8425628
You need a metric tensor for each index change, hence if you're only changing one index, you're going to only need one metric tensor.

[math]F_{\gamma}\!^{\beta} = g_{\gamma \alpha} F^{\alpha \beta}[/math]
[math]F^{\gamma}\!_{\beta} = g^{\gamma \alpha} F_{\alpha \beta}[/math]

For minkowski space [math]g_{\alpha \beta} = \eta_{\alpha \beta}[/math], also [math]\eta_{\alpha \beta} = \eta^{\alpha \beta}[/math]
I will be taking [math]\eta_{\alpha \beta}[/math] with signature (- + + +), I will also be taking c = 1.

[math]F_1\!^{\,0} = \eta_{11} F^{10} = -E_x[/math] (since [math] \eta_{\alpha \beta} = 0[/math] for [math]\alpha \neq \beta[/math])

[math]F^1\!_0 = \eta^{11} F_{10} = E_x[/math]

[math]F_0\!^1= \eta_{00} F^{01} = -E_x[/math]
[math]F^0\!_1 = \eta^{00} F_{01} = (-1)(-E_x) = E_x[/math]

So if [math]F^{\alpha \beta} = \{ \{ 0, E_x, E_y, E_z \} , \{-E_x, 0 , B_z, -B_y \}, \{-E_y, -B_z, 0 , B_x \},\{-E_z, B_y, -B_x, 0 \} \}[/math] in Mathematica's notation for matrices (sorry, don't know how to write matrices on 4chan's latex (pic related, for easy reference).

[math]F_{\alpha \beta} = \{ \{ 0, -E_x, -E_y, -E_z \} , \{E_x, 0 , B_z, -B_y \}, \{E_y, -B_z, 0 , B_x \},\{E_z, B_y, -B_x, 0 \} \}[/math]

[math]F^\alpha\!_\beta = \{ \{ 0, E_x, E_y, E_z \} , \{E_x, 0 , B_z, -B_y \}, \{E_y, -B_z, 0 , B_x \},\{E_z, B_y, -B_x, 0 \} \} [/math]

[math]F_\alpha\!^\beta = \{ \{ 0, -E_x, -E_y, -E_z \} , \{-E_x, 0 , B_z, -B_y \}, \{-E_y, -B_z, 0 , B_x \},\{-E_z, B_y, -B_x, 0 \} \} [/math]

[Anonymous No.8428705]

[Anonymous No.8428748]

>>8426611
Shouldn't it be [math](\partial_x U)^2 - (\partial_y U)^2 = (\partial_r U) - 1/r^2 (\partial_s U)^2[/math]?

[Anonymous No.8428761]

>>8426611
Shouldn't it be [math](\partial_x U)^2 - (\partial_y U)^2 = (\partial_r U)^2 - 1/r^2 (\partial_s U)^2[/math]?

[Anonymous No.8428833]

>>8428549
anyone?????

[Anonymous No.8428868]

a) and b) were rather easy, but I'm having some difficulty with c). Any insights?

[Anonymous No.8428871]

>>8428868
do it by induction

[Anonymous No.8428891]

>>8424626
>5kg "particle" of mass

that's hilarious.

[Anonymous No.8429188]

How much time would it take to learn physics from zero to college-level? (Petroleum industry major)

[Anonymous No.8429208]

>>8429188
do you mean starting college, finishing an engineering BS, a physics BS, a physics MS, just the core classes from a physics BS... what do you mean?

[Anonymous No.8429221]

>>8429188
I'm currently thinking about switching my major to oil-industry related stuff.
You need to take a physics exam for that in our country. From zero is because we had a very bad physics teacher at our school and I was a lazy cunt in general

[Anonymous No.8429223]

>>8429221
meant for >>8429208 of course

[Anonymous No.8429357]

>>8428693
Thank you, excellent answer!

[Anonymous No.8429410]

How to prove that
a) If 'a' if positive and 'b' is negative, then 'ab' is negative
b) If 'a' if negative and 'b' is negative, then 'ab' is positive
Using only pic's rules?

[Anonymous No.8429415]

>>8429410
Example how to prove, but it doesnt help me much

[Anonymous No.8429460]

>>8429410
Isn't the first part literally a variation of POS 1?
Just say according to POS 1 if a nd b are positive then ab is positive.
But given that b is negative and a is positive,an is negative.

[Anonymous No.8429462]

>>8429460
>an is negative
Sorry I meant ab

[Anonymous No.8429527]

>>8425738
I did not c that coming m(2^3)

[Anonymous No.8429559]

I'm so bad at factoring. like x^2 - x - 6 = 0 took me too long to factor no joke. is there an analytical method to find the factored form? How do you know if it's factorable?

[Anonymous No.8429562]

>>8429559
Anything beside the product sum thingy?

[Anonymous No.8429617]

>>8429559
Quadratic formula: (-b±sqrt(b2-4ac))/2a

There are formulae for cubic and quartic, but they're a fair bit more complex (and may require the use of complex numbers even if the roots are real).

[Anonymous No.8429746]

>>8429460
Yeah, a is easy-enough, but that about 'b'? How could it be positive, if 2 numbers are not positive as in pos1?

[Anonymous No.8429756]

>>8429460
I'm either not seeing something, or your reasoning is flawed. POS 1 never says anything about a and b that are not both positive. By saying "given that b is negative and a is positive, ab is negative" you are just restating the theorem you're trying to prove.

>>8429410
Are these the only rules we're allowed to use? No axioms for addition/multiplication?

[Anonymous No.8429773]

>>8426596
Not entirely sure what the question is asking; since the table is frictionless, that means no kinetic energy is lost during the collision.

But, we can still calculate the original kinetic energy. Since B is initially at rest, all energy in the system comes from puck A. Kinetic energy is equal to 1/2*mass*velocity^2. Find the kinetic energies of each puck after the collision (since velocities and mass are given) then add them together to find the kinetic energy of puck A before the collision. Should be 0.0792 J, which is the same answer as given by the yahoo patron, but this method is a lot simpler.

[Anonymous No.8429808]

>>8429756
> No axioms for addition/multiplication?
well, its allowed, but not something like "we know that neg*neg=pos because for example -1*-1=1, 'coz of multiplication rules"

[Anonymous No.8429822]

>>8429808
I think this video might have something to do with the conversation you guys are having

https://www.youtube.com/watch?v=ij-EK-MZv2Q

[Anonymous No.8429843]

Breaking Bad S4E12 "End Times"
Walt is seen making a homemade bomb out of cold packs
I'm curious if that's even possible, with most of the series having plausible references to chemicals and such.

[Anonymous No.8429849]

What would be the best way of starting this? I dont know how to make the power of the X's into something that makes sense and will lead me to the answer? Or else i'm a retard.

[Anonymous No.8429859]

Why not suck the air out of the room in power plants to lower the boiling point of water so it needs less energy to make boil?

Would maintaining less than one atmosphere use more energy and it be a wash?

[Anonymous No.8429863]

>>8429843
Yep, it's possible. Instant cold-packs contain ammonium nitrate. That show did a good job of keeping things quasi-realistic.

[Anonymous No.8429868]

>>8429863
thanks
I did some quick searching and couldn't find any components of typical cold packs that were explosive
I would assume that there would be a easier way to get ammonium nitrate

[Anonymous No.8429876]

>>8429808

A) a > 0 and -b > 0, hence a(-b) > 0. Now use the distributive law to prove that ab is the additive inverse of a(-b), and you're done.
B) is similar

[Anonymous No.8429956]

I'm defeated. Is there something wrong with my expansion?

[math] -\frac{1}{10}log|\frac{1}{100}\bf{I} + \bf{x}\bf{x}^T| [/math]

Outer product of the 2 vectors

[math]-\frac{1}{10}log|\frac{1}{100}\bf{I} + \begin{bmatrix} x_{1}^2 & x_{1}x_{2}\\x_{1}x_{2} & x_{2}^2\end{bmatrix}|[/math]

Add the identity matrix

[math]-\frac{1}{10}log|\begin{bmatrix} x_{1}^2 + \frac{1}{100} & x_{1}x_{2}\\x_{1}x_{2} & x_{2}^2 + \frac{1}{100} \end{bmatrix}|[/math]

Calculate the determinant

[math]-\frac{1}{10}log(\frac{x_{1}^2}{100} + \frac{x_{2}^2}{100} + \frac{1}{10000})[/math]

Did I mess up anything up to here?

[Anonymous No.8430068]

I'm at Cal Poly Pomona changing my engineering major to business major. Chances of getting a girlfriend that will love me and support me?

[Anonymous No.8430099]

Hello internet friends! I am having trouble solving the Fourier Transform of exp(-abs(x)).

[Anonymous No.8430111]

>>8429956
looks fine

[Anonymous No.8430113]

>>8430099
what have you done so far? its a straightforward integral

[Anonymous No.8430116]

Probability question: What does evidence [math]P(x)[/math] tell you in Bayes' theorem?

What does it mean when P(x) is higher/lower?

[Anonymous No.8430119]

Hi, short question on GR:
Whenever an object doesn't expereince any force, it goes in geodesics. Does it also mean it doesn't change its speed? Or not exactly?

Thanks

[Anonymous No.8430122]

>>8430119
Oh shit I'm a brainlet. Of course it does change its speed.

Ok, then I correct myself:
When the object goes in closed orbit around say earth does it change its speed?

[Anonymous No.8430127]

>>8430113
I have pretty much done nothing. I was thinking that I should maybe rewrite the imaginary part using Eulers identity? I’m having problem cooping with the absolute value.

[Anonymous No.8430129]

>>8430122
You don't have to know a single bit about GR to know that in elliptical obits the speed of the satellite changes

Only in a perfectly circular orbit does a satellite's speed remain constant

http://www.braeunig.us/space/

[Anonymous No.8430136]

>>8430127
>I’m having problem cooping with the absolute value.
its the same thing you do with any absolute value in an integral, just split it into two cases, one where |x|=x and one where |x|=-x

[Anonymous No.8430139]

>>8430129
Yeah, you're right, thanks, and thanks for the link.

[Anonymous No.8430141]

>>8430136
Thanks! its been a long day at uni seems like im forgetting the basics. Anyways i figured out the solution.

[Anonymous No.8430149]

>>8430116
What is [math]x[/math]?

[Anonymous No.8430159]

>>8430149
x is the data, and [math]\theta[/math] are the parameters of the model

[math]P(\bf{\theta}|x) = \frac{P(x|\bf{\theta})P(\bf{\theta})}{P(\bf{x})}[/math]

[Anonymous No.8430202]

I would like to learn about economics and how things like stock work. Pretty much how our societies economy work and interacts with each other.

I would just like to be knowledgeable enough to hold my ground in conversations

pls respond

[Anonymous No.8430234]

>>8430116
Evidence=average likelihood of the data x (where the average is taken with respect to the prior). So in a sense it's your "average uncertainty" of the result x, with the average being taken over all possible states of your model.

[Anonymous No.8430244]

>>8429849
>Auto
pls

[Anonymous No.8430871]

How the fuck do i solve this?
I know it's fundamental theorem of calculus but I can't really wrap my head around it and I can't find any examples either.

[Anonymous No.8430966]

>>8430871

Don't confuse yourself. You can evaluate it exactly how you think you would. That is, Integrate the function, plug in the bounds, then take the derivative with respect to that variable.

The only thing that's different here is that the theorem lets you cut out a bit of the work by thinking about what you're actually doing.

The term that was evaluated at t^2 is obviously going to zero when you take the derivative with respect to x, and so you're just taking the derivative of your function evaluated at the relevant bound.

(d/dx)F(x) = x'f(x) = f(x)

(d/dx) F(x^2) = 2x f(x^2)

(d/dx)F(x^3) = 3x^2 f(x^3)

...

In your case,

(d/dx)F(x^3) = 3x^2 f(x^3)

f(x) = sin(x)/x

and so f(x^3) = sin(x^3)/x^3 and then the final answer is just

3x^2 [sin(x^3)/x^3]

and for the second one, there is a negative. That's because if you think of just carrying out the entire operation. You'd have F(x^3) - F(t^2) and when you take the derivative with respect to t you're just left with d/dt(-F(t^2) )

d/dt(-F(t^2)) = -2t f(t^2) = -2t sin(t^2)/t^2

[Anonymous No.8430978]

>>8430202
Thomas Sowell's book "Basic Economics " is what you're looking for.

[Anonymous No.8430987]

>>8430202
A level economics. I recommend studying the singapore-cambridge syllabus. The international syllabus is crap

[Anonymous No.8431002]

>>8430966
Thank you so much senpai, I really needed some closure on this. I'm pretty much ready for my exam now. <3

[Anonymous No.8431007]

>>8424626

hypothetically:

If I had access to 100% accurate and complete pi what could I do with it

[Anonymous No.8431164]

>>8431007
Nothing important, you can already count the size of the entire universe down to a single hydrogen atom using just 16 digits.

[Anonymous No.8431214]

Doesn't Ohms Law negate the concept of voltage being "electric potential."

Taking v=ir, at i=0, v=0
therefore, there should be no voltage, correct?
Yet, voltage is called potential energy because even without a current of charge carriers, there remains an attractive force. So, then is ohm's law valid only at i>0?

Explain, geeks.

[Anonymous No.8431905]

How good of a combination is a major in psychology and cognitive behavioral neuroscience? Is it good enough to get me into grad school? Get me into behavioral research? Get a decent job in the field(s)?

[Anonymous No.8431922]

>>8431007
could finally determine whether it's a normal number or not

[Anonymous No.8431943]

>>8429849
No stats experts on /sci baka ill try help you bro give me a fewvminutes

[Anonymous No.8431947]

>>8431214
First of all voltage isn't electric potential, it's the difference in potential between 2 points.

Apart from that you're sort of right, Ohm's law is an approximation that is valid only for conductors. If you connect a conductor to a voltage there will always be current, if you connect something like rubber there won't be or it'll be immeasurably small but no one cares about that anyway.

[Anonymous No.8431975]

>>8430129
Where this pic is from? Is it a simulator?

[Anonymous No.8432152]

Why is f(4,4)=8 the maximum when f(16,1)=17 seems to meet the requirements?

In the Lagrangian I do get an implicit restriction/result that x=y, but why does it have to be like that and why is f(16,1) invalid?

[Anonymous No.8432180]

How to improve my mathematical foundations to get to an olympiad math level? I want get better at problem solving.

[Anonymous No.8432190]

>>8432152
Not sure how wolfram alpha does this but that function shouldn't have a maximum.

Put x=16/y, then f(x)=x+16/x which goes to infinity as x goes to zero

[Anonymous No.8432215]

>>8432152
f(4,4)=8 is a local minimum. f(-4,-4)=-8 is a local maximum.

The other two are wrong.

[Anonymous No.8432238]

>>8432190
Yeah, I know it has no global max but this is a constrained max problem.

>>8432215
You're right, f(4,4) is a minimum not a maximum. I just don't see why f(16,1) can't be a local maximum.

[Anonymous No.8432249]

>>8432152
I think it's simply because the criteria of the method of Lagrange multipliers is not met (you're not optimizing over a closed, bounded region), so it is not guaranteed to yield correct results.

[Anonymous No.8432270]

>>8432238
>>8432215
Actually, you're wrong, f(4,4)=8 is a local maximum just checked my results.

>>8432249
Might be this, I think my understanding of local maximums is lacking a bit.

[Anonymous No.8432271]

>>8432238
> this is a constrained max problem.
No it isn't. xy=16 reduces it from 2-dimensional to 1-dimensional, but the domain of x is the reals.

> I just don't see why f(16,1) can't be a local maximum.
Because f(16+d,16/(16+d)) increases as d increases.

A local maximum is a point where a change in the argument in any direction results in the value decreasing. That isn't the case at (16,1).

[Anonymous No.8432287]

>>8432270
> Actually, you're wrong, f(4,4)=8 is a local maximum just checked my results.
No, you're wrong, it's a minimum.
f(4+d,16/(4+d))>f(4,4) for all d<4.

I.e. all "nearby" points satisfying xy=16 have f(x,y)>f(4,4), i.e. (4,4) is a local *minimum*.

[Anonymous No.8432340]

Why are vector spaces required to be defined on fields?
The axioms only seem to require a commutative ring

[Anonymous No.8432362]

>>8432340
Modules are the analogous structure over rings.

[Anonymous No.8432439]

I don't this ii. and iii. are solutions but I'm having trouble putting that into words and justifying it

[Anonymous No.8432490]

Hey, i would greatly appreciate advice on this.
The time unit "second" is somewhat defined by the frequency of the nuclear cesium clock ( 9192631770 Hz) i am supposed to find out how many years it would take the time to be off by 1 second (not paying attention to the earth's rotation slowing down and so on). I also have given an accuracity of this clocks frequency by [math]10^-5[/math].

Do i just have to divide the frequency by the error accuracy and as a result i get the oscillations (not periods) until the 1 second error comes along. Followed by multiplying with 60×60×24×365.25 to get the years.

Is this valid? I've only been given the frequency and the error accuracy so i guess this might be the only way?
Thanks in advance

[Anonymous No.8432515]

how do i fight procrastination?

[Anonymous No.8432600]

>>8432515
Set aside a particular time of day to study or do whatever you have to do.
Find a way to make whatever you have to do easier; pave the way.

[Anonymous No.8432605]

I've got a problem /sci/ I'm a mere mortal man who needs to take breaks in order to be productive and I also hold the knowledge that short, regular breaks are better for you than long, infrequent ones. The problem for me is however that all of my hobbies are mentally straining - shit like reading books. So what should a guy like me do when he wants a break?

[Anonymous No.8432615]

>>8432605
jack off

[Anonymous No.8432629]

>>8432615
That also makes you tired. It's also hard to stop once you get into it - your breaks will be too long. I also wouldn't want to work while being distracted by the fact that my dick is still oozing out cum.

[Anonymous No.8432648]

>>8432629
go for a short walk

[Anonymous No.8432658]

>>8432340
results are nicer over fields

for example the idea of dimension of a vector space doesn't translate well to rank of a module, as there's some modules where R^n isomorphic to R^m doesn't imply m=n like in the vector space case

[Anonymous No.8432690]

Does anyone know how to specify where contours should start/end and what the spacing between the contours is in MatLab?

[Anonymous No.8432809]

What can I possibly hope to do with a Bachelor's in physics?

I am struggling through the master now but realise I'm too stupid for this, at least definitely for a Ph.D.

Should I just end myself?

[Anonymous No.8433425]

>>8433320
>>8433320
>>8433320
>>8433320
>>8433320

[Anonymous No.8433438]

>>8432605
Reading books isn't that mentally straining, unless you're reading fucking Ulysses in your breaks. If you want to read during your breaks, just pick something lighter to read. Alternatively, just take short walks.

[Anonymous No.8433441]

>>8433425
>It's not homework because I already submitted the homework
I hope the mods do their job and ban you.

[Anonymous No.8433461]

>>8433441
>Having a question about a process is the same thing as getting /sci/ to do my homework
ok

[Anonymous No.8433464]

>>8433461
>Starting a thread about your homework is considered a homework thread, which is explicitly against the rules because it shits up the board
Wow, who knew?

[Anonymous No.8433493]

>>8433464
The rule is to not ask /sci/ to do your homework, I'm just referencing homework WHICH I ALREADY HAVE THE ANSWERS FOR. I don't know what's so hard for you to understand about this

[Anonymous No.8433805]

>>8433464
You'd really have to be "special" to think that homework is what makes this board shit.

[Anonymous No.8434001]

>>8433438
>. If you want to read during your breaks, just pick something lighter to read.
Such as?

I think I might genuinely not understand what a 'light read' is.

[Anonymous No.8434039]

Is archeology science?

[Anonymous No.8434090]

Is it worth getting a PhD in Chem or Physics if I want to go into research? How much money can a researcher get?

[Anonymous No.8434297]

Can anybody help with 5a) on here? I've tried substituting tx_0 and x_0 into the expression at the top but I can't seem to get them to be equal.

[Anonymous No.8434360]

I'm a physics major looking to go to applied math grad school next fall.

I've taken the cal sequence, diffeq, pde's, complex analysis, and fourier analysis. What should I start studying in order to not get shit on my first semester?

[Anonymous No.8434397]

What's the hell are [math]\phi_m(x)[\math]?

[math]P_m(x)[/math] are Laguerre polynomials.

[Anonymous No.8434424]

>>8434360
real analysis

[Anonymous No.8434425]

I got a D in Chemistry in my Secondary school final exams.

Would I be able for a Chemistry course in college? I enjoy studying Chemistry, and I would need a high final GPA in order to do the entrance exam for Medical School here in Ireland.

What do you guys suggest? Any advice?

[Anonymous No.8434433]

Having proven that E is irrotational how do i then use the gradient theorem?

[Anonymous No.8434456]

>>8434397
probably the m-th cyclotomic polynomial

[Anonymous No.8434477]

Let's say we have a vector going from 0,0 in the direction of PI/3

If I were to go to x=1 what would the y be provided that point is on the vector?

[Anonymous No.8434487]

>>8434433
Find a scalar field [math]f[/math] such that [math]\nabla f = E[/math], then the solution would be [math]f(2,0)-f(0,1)[/math].

[Anonymous No.8434567]

>>8434487
Would it not be -∇f=E ?

[Anonymous No.8434588]

>>8434487
Then the answer would be [math]f(0,1)-f (2,0)[/math].

[Anonymous No.8434596]

>>8434588
Ahh ok thanks. The notes i have seem to be confused on the issue

[Anonymous No.8434600]

So is [math]H_1(X, pt)=H_1(X)[/math] in the Eilenberg-Steenrod axioms alone or not?

I just look at this exact sequence:
[math]H_n(pt)\overset{i_*}{\longrightarrow\,} H_n(X) \overset{j_*}{\longrightarrow\,} H_n(X, pt) \overset{\partial}{\longrightarrow\,} H_{n-1}(pt)[/math]
If I substitude [math]n=1[/math] I get
[math]0\overset{i_*}{\longrightarrow\,} H_1(X) \overset{j_*}{\longrightarrow\,} H_1(X, pt) \overset{\partial}{\longrightarrow\,} H_0(pt)[/math]
..Eh? How do I get isomorphism here? And do I?

Excision axiom doesn't seem to help to

[Anonymous No.8434606]

>>8434600
*too

[Anonymous No.8434615]

I'm able to brute force my way through math/physics problems reasonably well, but when exams roll around I have trouble doing things in the allotted time, even when I know exactly how to work through the given problems.

Anyone have tips for being faster at solving stuff?

[Anonymous No.8434690]

I can work out that the matrix
0 0 -1
0 1 0
1 0 0
Represents rotating the xyz unit vectors so that the the whole plot is rotated by 90deg around the y axis but just thinking about where the Cartesian unit vectors would land.

I want the same transformation but in polar coordinates, I'm finding it tough thinking about where each polar unit vector would end up but I've just had a thought, will the same matrix work the same to describe the same transformation regardless of the unit vectors I choose?

[Anonymous No.8434734]

>>8434600
Ah, I've figured out it myself.
[math]H_n(pt)\overset{i_*}{\longrightarrow\,} H_n(X) \overset{j_*}{\longrightarrow\,} H_n(X, pt) \overset{\partial}{\longrightarrow\,} H_{n-1}(pt)\overset{i_*}{\longrightarrow\,} H_{n-1}(X)[/math]

[math]0\overset{i_*}{\longrightarrow\,} H_1(X) \overset{j_*}{\longrightarrow\,} H_1(X, pt) \overset{\partial}{\longrightarrow\,} H_0(pt)\overset{i_*}{\longrightarrow\,} H_{0}(X)[/math]
Since the sequence is exact if we prove that [math]\ker i_* = 0[/math] we prove that [math] \partial = 0[/math] and hence [math]j_*[/math] is an isomorphism.

Now we have a sequence [math]pt\overset{i}{\longrightarrow\,} X \overset{\pi}{\longrightarrow\,} pt[/math] (composition is id) by functoriality we have [math]H_0(pt)\overset{i_*}{\longrightarrow\,} H_0(X) \overset{\pi_*}{\longrightarrow\,} H_0(pt)[/math] (composition is id). Since composition is id [math]\ker i_*[/math] is essentially zero.

[Anonymous No.8434961]

is material science the stem equal to a business degree aka to broad of an approach?

would chemistry be a better degree if you are trying to get into research?

[Anonymous No.8434968]

Can you have two standard form polynomials multiply and result in a monomial?

No terms like (x-x) and no zeroes ala (0x^2) and such

[Anonymous No.8434989]

>>8434968
The order of the product of two polynomials of order n and m is n+m.

So

p(X) = a·X
q(X) = b

p(X)·q(X) = a·b·X

is the only solution.

What can happen, however, is that the product of two polynomials has fewer terms than the two previous ones, but the first example is already of non-trivially high order.

[Anonymous No.8434997]

I'm supposed to solve this in terms of variables. However whenever I solve it using a matrix, I get all 0s. Somehow I keep getting linearly dependent equations. Help pls

[Anonymous No.8434998]

>>8434961
>to, not too
cringe

if you want to get into research you have to be dedicated, so choose the one you think you'll have more passion for.
how well you do will have a far greater influence on whether or not you get into research than the number of research positions available

[Anonymous No.8435002]

>>8424626
Why is this thread still alive

[Anonymous No.8435004]

>>8434690
Not in the sense you mean. This same matrix in polar coordinates would "rotate" 90 degrees around the [math]\theta[/math] axis, which would mean something completely different.

[Anonymous No.8435019]

>>8434968
No.

If you perform the binomial expansion of
(a[n]*x^n + ... + a[1]*x + a[0]) * (b[n]*x^n + ... + b[1]*x + b[0])
you end up with
(a[n]*b[n]*x^(2*n) + ... + a[0]*b[0])

1. The intermediate terms are sums of several a[i]*b[j] products, but the first and last terms only contain a single product.

2. The only way that such a term can be zero is if either a[i]=0 or b[i]=0.

3. Making such a substitution then reduces either the second or second-to-last term from a sum of products to a single product.

4. GOTO 2.

The result is that you end up constraining all coefficients except for one to be zero. IOW, the product of two polynomials is a monominal if and only if both polynomials are monomials.

Note that you /can/ multiply polynomials and get the sum of a monomial and a constant. E.g.
(x^4-sqrt(2)*x^2+1)*(x^4+sqrt(2)*x^2+1) = x^8+1

[Anonymous No.8435044]

>>8434989
>>8435019
Thanks lads

[Anonymous No.8435064]

>>8434297
I'll try and do it in 1D cause I think it's easier that way and the idea is the same.

Set [math]x=tu[/math] then [math] \frac{d}{dx}=\frac{du}{dx} \frac{d}{du}=\frac{1}{t}\frac{d}{du} [/math].
Plug this into the first equation and you get [math] tu\ \frac{1}{t} \frac{df(tu)}{du}=0 [/math] so [math] u\ \frac{df(tu)}{du}=0 [/math]

From that [math] x \frac{df(tx)}{dx}=x \frac{df(x)}{dx}=0 [/math]
Derivatives are linear so this is the same as [math] x\frac{d}{dx}(f(tx)- f(x))=0 [/math] and [math]f(tx_0)=f(x_0)[/math] up to a constant

[Anonymous No.8435074]

>>8434997
I1.R1 + I3.R3 = V
-I1.R1 + I4.R4 - I2.R2 = 0
-I3.R3 + I2.R2 + I5.R5 = 0
IV + I1 + I4 = 0
-I1 + I2 + I3 = 0
-I3 - I5 - IV = 0
-I2 - I4 + I5 = 0

You can delete any one of the last 4 equations (KCL) to get an invertible 6x6 matrix. Multiply that by the column vector [V 0 0 0 0 0] to get the 6 currents.

Currents are positive top-to-bottom or left-to-right.

More generally: to get the currents for a tree with N nodes and E edges:

First choose any spanning tree (i.e. any graph where all nodes are connected with no loops). Such a tree will have N nodes and N-1 edges. The corresponding cotree (the set of edges not in the spanning tree) will have E-N+1 edges. Every edge in the cotree has a corresponding loop, to which you apply KVL.

In this case, the graph has 4 nodes and 6 edges, the tree has 3 edges, the cotree has 3 edges, so you get 3 loops.

Applying KCL at each node gives you N equations. One of them is redundant, so delete any one to get N-1 equations.

So in total you have (E-N+1)+(N-1) = E equations in E unknowns (the current for each edge).

[Anonymous No.8435092]

>>8425070
>>8428547
what this guy means is DIVERGENCE AND CURL
[math] \nabla • \vec f [/math] vs [math] \nabla \times \vec f[/math]

[Anonymous No.8435107]

>>8435074
Thank you so much

[Anonymous No.8435167]

>>8435074
Say for the sake of example you have values for the resistances (in ohms), say
R1=1
R2=2
R3=2
R4=5
R5=4
And the voltage from the battery is 10V.
What would the matrix look like?

[Anonymous No.8435206]

Is it true that all level sets have empty interior?

[Anonymous No.8435222]

>>8435206
To be more specific, a level set embeded in the space where the graph of the funcion is located.

[Anonymous No.8435236]

>>8435167
The matrix would be:
[ 1 0 2 0 0 0]
[-1 -2 0 5 0 0]
[ 0 2 -2 0 4 0]
[ 1 0 0 1 0 1]
[-1 1 1 0 0 0]
[ 0 0 -1 0 -1 -1]
[ 0 -1 0 -1 1 0]

Delete any one of the last four rows to get an invertible square matrix; note that the last 4 rows sum to a zero row, i.e. they aren't linearly independent.

Also note that the battery voltage doesn't appear in the matrix, but on the right-hand side.

Regardless of which row you remove, the inverse is
[ 48 -8 10 40 -36 40]
[ 6 -12 15 60 78 60]
[ 42 4 -5 -20 18 -20]
[ 12 20 8 32 24 32]
[ 18 8 23 -40 -30 -40]
[-60 -12 -18 60 12 -72]
divided by 132 (the determinant of the original matrix is +/- 132, the sign depending upon which row is removed).

This is multiplied by [10 0 0 0 0 0]^T, so only the left-hand column is relevant. This gives:
[480 60 420 120 180 -600]^T / 132
= [40 5 35 10 15 -50]^T / 11

Thus, the solution is:
I1=40/11, I2=5/11 I3=35/11 I4=10/11 I5=15/11 IV=-50/11

[Anonymous No.8435260]

How do I git gud at measure theory? Got a Real 1 midterm tomorrow covering measure spaces and sigma algebras and I'm pretty meh at the subject.

>>8435222
I'm not sure exactly what you mean but I'll take a stab at it.

A level set of a function from n dimensional space to R has at most dimension n; if you're looking at the graph in n+1 dimensional space, then since the level set has at most dimension n, it can't be open.

[Anonymous No.8435285]

>>8435004
Thanks

[Anonymous No.8435292]

If a metal has enough tensile force applied to it that it enters plastic deformation, but after removing the force and reapplying it it regains elastic deformation, what does this mean is happening on a material level? Does it mean crystalline structures are reforming or something? Pic related is what I mean

[Anonymous No.8435397]

Why are PDFs still around when we have DJVU?

[Anonymous No.8435430]

>>8435292
It means the individual atoms are relaxing to their lowest possible energy positions in terms of LJ potential between stress applications. Bond lengths returning to normal as much as they can given the dislocations that have been introduced. Some fraction of the dislocations should also annihilate during the unloading phase.

Just a WAG.

[Anonymous No.8435536]

>>8435397
Jews

[Anonymous No.8435621]

whose dick do i have to suck to learn what this means and how to compute it
[eqn]\int_\Omega X(\omega)\ \text{d} P(\omega)[/eqn]

[Anonymous No.8435648]

i started python on codeacademy last night
why did it loop through the dictionary in this order?

as an aside, codeacademy seems to be pretty shit
i think if they fixed all the bugs then it would be good for an elementary school kid to learn coding, but it's pretty annoyingly slow while not covering much, and it fucks up all the time

[Anonymous No.8435664]

>>8435621
Do you have a specific example in mind?

[Anonymous No.8435742]

>>8435648
Python dictionaries are unordered, you have no guarentee on how it will be iterated.

[Anonymous No.8435998]

Is there a rule or theorem that lets me say a/b/c/ = a/b*c

[Anonymous No.8436022]

I'm back with Eilenberg-Steenrod axioms.
How do I prove by them alone that [math]H_1(\Sigma X)=H_0(X)[/math]? Although I've immediately solved the general case but I've spent hours trying this case, with no luck.

[Anonymous No.8436030]

>>8435621
See this https://people.math.ethz.ch/~kowalski/measure-integral.pdf

[Anonymous No.8436044]

>>8436022
Halp

[Anonymous No.8436122]

So you have a bag of black and white balls with a known generative process (prior). And you sampled N balls and W were white. How do you compute the probability of the next ball being white?

[Anonymous No.8436168]

>>8428547
What? It's a vector of all partial derivatives. Why is not the gradient?

[Anonymous No.8436184]

i live in nevada and im not sure if i should vote yes on question 3. basically its

>yes
make energy market open and ban monoplies

or

>no
keep it the way it is (keep the monopoly intact)

basically both arguments in the ballet are convincing and both seem to cherrypick data to support their claims. its essentially comes down to "open market means lower prices, here are the stats vs no it fucking doesnt, here are the stats"

[Anonymous No.8436206]

>>8436122
The probability of the next ball being white is the summation over K of the probability of the bag containing K while balls multiplied by the probability of the next ball being white given that the bag contains K white balls.

The second part is easy, the first isn't (and was answered a couple of threads back; not sure if that was you).

Note that the probability of the bag containing K white balls depends upon the generative process and the number of white balls obtained previously.

[Anonymous No.8436509]

I'm trying to tab out a song but I have problems hearing certain parts.

Can I decompose an mp3 file to sines and get the notes at certain times from these sines?

[Anonymous No.8436511]

>>8436184
When were monopolies ever a good idea?

[Anonymous No.8436512]

How do I learn bayesian statistics to understand posterior predictive distribution and stuff like this

[Anonymous No.8436516]

>>8436184
read the reports and find out why the stats contradict each other

[Anonymous No.8436548]

>>8436511
The main problem with that line of reasoning is that utilities tend to naturally form monopolies. So the choice is often between a regulated /de jure/ monopoly or an unregulated /de facto/ monopoly.

[Anonymous No.8436554]

Does

1. [math]P(X \geq x)$ = 1 - $P(X < x)[/math]

or

2. [math]P(X \geq x)$ = 1 - $P(X \leq x)[/math]

If 1 is true and 2 is false, how do I get the value of 2?

[Anonymous No.8436557]

>>8436554
ignore dollar signs

[Anonymous No.8436643]

How do you calculate the posterior predictive distribution using MAP plug-in?

[Anonymous No.8436800]

What's the purpose of accumulation points in analysis? why do we define things like the derivative with them, as opposed to simply requiring the domain to contain an interval or neighborhood around the point(s) of interest?

[Anonymous No.8436815]

>>8436557
Lol you use piazza don't you

[Anonymous No.8436838]

Let G be a finite group. Show that in the disjoint cycle form of the right regular representation Tg(x) = xg of G, each cycle has length |g|

What is this even asking? We're on the isomorphisms of groups chapter and I haven't seen "right regular representation" mentioned yet. .

[Anonymous No.8436860]

How do I show the first part of this question? Plotted both functions in Wolfram and I'm just confused

[Anonymous No.8436871]

>>8436815
I type my math into mathjax beforehand to verify it before posting. Mathjax uses $

[Anonymous No.8436911]

>>8436554
1 is true
2 should be P(X >=x)=1-P(x>X)

[Anonymous No.8436921]

>>8436838
not 100% sure but for example if G=S_3 and g=(123) then
Tg(e)=(123)e=(123)
Tg(12)=(123)(12)=(13)
Tg(13)=(123)(13)=(23)
Tg(23)=(123)(23)=(12)
Tg(123)=(123)(123)=(132)
Tg(132)=(123)(132)=e

so the disjoint cycle form would be [(12)(13)(23)][e(123)(132)] which agrees with the idea that each cycle should have length |g|=3

[Anonymous No.8436923]

>>8436921
looks like i did left regular representation by mistake but you should hopefully get the idea

[Anonymous No.8436943]

haploidy/monoploidy is having one set of chromosomes

diploidy is having two sets of chromosomes

polyploidy is having more than two sets of chromosomes

Is there a word for having more than one set of chromosomes? i.e. encompassing both diploidy and polyploidy but not haploidy/monoploidy

[Anonymous No.8436973]

Every night I lay in bed in total darkness and silence, at a very comfortable temperature, with no TV or cellphone in the room. Yet I don't feel sleepy until way over 1 a.m. and sometimes it just doesn't happen.

Physchiatrists have prescribed me meds that only work the first couple nights I take them, then it's back to insomnia land. I work out during the day, eat healthy meals, I stick to a routine and take a heavy dose of valerian root and passiflora for good measure. Yet when bed time comes my brain just won't shut the fuck up and sleep.

What do /sci/?

[Anonymous No.8436978]

>>8436973
Don't sleep and be productive for 20+ hours a day.

[Anonymous No.8437007]

>>8436978
I want to do the uberman sleep schedule so bad but every time I set the alarm for a nap I just end up wasting 30 minutes trying to sleep and failing because my hypothalamus is a dick.

Could there be an alarm app for the phone that starts counting the moment you fall asleep instead of when you hit 'start'?

[Anonymous No.8437037]

Can someone help me find the equivalent capacitance of this setup? I don't know the method to finding the equivalent capacitance in a parallel and series circuit

[Anonymous No.8437048]

>>8437037
It's the opposite of resistance. Capacitors in parallel add (C=C1+C2), capacitors in series add reciprocals (1/C=1/C1+1/C2).

[Anonymous No.8437057]

>>8437037
use the parallel method to find the equivalent capacitor for 2C and 3C (denote as CE). then use the series method to find the equivalent capacitor of C, CE, and 6C

[Anonymous No.8437061]

>>8437057
forgot my source
http://farside.ph.utexas.edu/teaching/302l/lectures/node46.html

[Anonymous No.8437064]

>>8437048
How do I know which capacitors to add in series and to add in parallel?

I'm new to this stuff so I haven't had time to understand the rules very well

[Anonymous No.8437124]

>>8437064
Look at the picture.
2C and 3C are parallel.
C, the 2C|3C pair, and 6C are in series.

[Anonymous No.8437183]

>>8436943
I'm not sure I understand your question... polyploidy means having more than one set in general

[Anonymous No.8437199]

>>8437124
So I can just add C, the 2C&3C pair, and the 6C in series and that's my answer?

[Anonymous No.8437286]

What method do I use to find the limit of

(1+1/x^2)^x^2

to be e?

[Anonymous No.8437313]

>>8437286
(1 + 1/x^x)^(x^x)

[Anonymous No.8437321]

>>8437313
After looking around a lot I found a way that is basically subbing the 1/x^2 as a constant and I'm able to do that. I'm not sure what you're trying to say but I would like to know, so please explain a bit more.

[Anonymous No.8437334]

>>8436548
>utilities tend to naturally form monopolies
Uhh, source? Governments often auction off utilities in a way that artificially creates monopolies, if that's what you mean.

[Anonymous No.8437337]

>>8436973
Have you tried cutting out all screen time within 2 hours of going to bed?

[Anonymous No.8437397]

>>8427569
classic

[Anonymous No.8437520]

>>8437334
The cost of supplying electricity (or gas, water, telecoms) to one house in a neighbourhood is a significant fraction of the cost of supplying all of them. IOW, the relationship between market share and cost is highly non-linear, while revenue is more linear. So profitability falls of quickly as market share drops, meaning that the economics strongly favour incumbents over new entrants.

And that's before you factor in the monopoly bonus, i.e. the fact that you can get away with charging more if you don't have any competition, which means that anti-trust enforcement needs to be rigorous.

And if you do manage to create a competitive market, it's a safe bet that suppliers will focus on cream-skimming the densely-populated areas and leave rural areas swinging in the wind. A monopoly is usually required to provide universal service.

[Anonymous No.8437629]

Any suggestions on how to approach this problem? I was thinking graphically.

[Anonymous No.8437643]

>>8437629
Im starting to see some patterns. Looks like every 5 units up there is a new value of 1, 1mod5. I suspect there is pqttern at 3, 18mod15 as well

[Anonymous No.8437700]

Can someone help me?
1. Done
2. 3. 4. help ;_;
I have Morton E. Gurtin book "An Introduction to Continuum Mechanics"
Do you know a good book from genesis library?

[Anonymous No.8437861]

>>8437700
Should be able to get the gist for from this webpage:

http://www.continuummechanics.org/principalstress.html

But why use the inverse? Seems like an unnecessary step to calculating principle stresses.

[Anonymous No.8437909]

>>8437861
Thanks buddy, but I need the trace of inverse tensor too.
>But why use the inverse?
Teacher want to do all the course like "linear algebra" (mathematical proof, logic, show, prove, establish, demonstrate, etc). He hates engineers ;_;

[Anonymous No.8437924]

I want to see how the recurrence relation is gotten, already made a thread but no one responds I already spent hours on this im tired etc. end my suffering

[Anonymous No.8437967]

Chemistry noob is back

I'm trying to calculate moles of a substance corresponding to moles of another substance from a titration. Is my calculation correct?

Substance B: 0.2 mol/Litre (titrant)
Substance A: what I'm finding

0.2 mol/L * 15ml of titrant = 0.003 moles of titrant

So then I should do a ratio between moles of Substance B and Substance A?
Supposing Substance B is 3 and Substance A is 4:

0.003 * 3/4 = 0.00225 moles

Therefore, 0.00225 moles of Substance A

[Anonymous No.8437982]

>>8437909
Yeah. I can't think of any matrix properties off the top of my head that would simplify it. You could invert the stress tensor by hand using the adjugate method. Or you could google something like "inverse of a 3x3".

After that, it looks like a few lines of algebra to get from a to b.

[Anonymous No.8437990]

>>8437924

Answer is B

g(n) = g(n/2) +n^2

apply an unknown transform given by T*

T[g*(n)] = T[g*(n/2)] +T[n^2]

n^2 can be dropped and so you have

T[g*(n)] = T[g*(n/2)]

no longer a function of n and so you generalize it to

T[g*(u)] = T[g*(u/2)]

and drop the transform notation

G(u) = G(u/2) (in this new space)

expand the relation

g(u, u1, u2, u3, ....) = g(u, u1, u2, u3, .... / 2)

Set the sum of ui = m

u + u1 + u2 + u3 + .... = m

G(m) = G(m/2)

map backward

G(m) ---> T[g*(n)] you want n = 2^m as the problem says and so

G(m^2) = T[g*(n))

Pull the exponent out of the transform function and put it in terms of n

2G(m^n^x) --> T(g*(n))

The term we had dropped at the start was was n^2

and so

2G(m^n^2)) reduces to 4G(m) by the same rule

4G(m) ---> T[g*(n)],

reversing the transform corresponds to three operations on n in m space

4G(m) ----> 3g(n)

G(m) and g(n) have domain equivalence and so
you can set them equal but you gain an extra term in G because you lost information

g(n) = 4/3 G(m+c)

g(n) and G(m+n) have the relation that

That when g(n) = 0, G(m+n) = 0, but G(m+n) = 0 if m+n = 0. That is G(0) = 0

g(1) = G(m+1) = 0 ==> m = -1

g(n) = 4/3 G(c-1)

well c is then just going to be that constant dropped and so pulling c-1 out of G,

G(c-1)-->(c-1)G(1) = (c-1)

g(n) = 4/3 (n^2-1)

[Anonymous No.8438118]

>>8436871
there's a tex button in the corner of the reply box you dummy

[Anonymous No.8438121]

>>8436815
what universities use piazza?

[Anonymous No.8438133]

>>8438121
my uni uses pizza as incentive for good grades :^3

[Anonymous No.8438138]

>>8438133
how? what uni?

[Anonymous No.8438242]

[eqn]\lim_{x\to 0} xsin(\frac{1}{x})[/eqn]

[eqn]\lim_{x\to 0} \frac{sin(\frac{1}{x})}{\frac{1}{x}}[/eqn]

Use L'hopital's rule as the limit yields to [math]\frac{\infty}{\infty}[/math]

[eqn]\lim_{x\to 0} \frac{(\frac{-1}{x^2})sin(\frac{1}{x})}{\frac{-1}{x^2}}[/eqn]

[Anonymous No.8438244]

>>8438242
Woops that last sin should be a cosine, in any case the [math]\frac{-1}{x^2}[/math] cancels out with the one below and you end up getting infinity

[Anonymous No.8438246]

>>8438242
>>8438244

This is retarded.

sin(1/x) is bounded.

The limit is 0.

[Anonymous No.8438252]

>>8438246
Whoops again, I haven't done basic limits in a long time so excuse my retardation

Anyways, the original poster was conflicted on whether to use the sandwich theorem or using the basic limit [math]\lim_{x\to 0} \frac{sinx}{x} = 1[/math]

[Anonymous No.8438256]

Help me understand my homework.
Which of these links of R xR -> R, (x,y) -> x*y are associative

x*y:=x-y
How the fuck do you possibly show the proof?
How do you start it?

[Anonymous No.8438318]

Anyone know if silica gel desiccant only work in an environment where damp air flow and pass through it along the way, or if it can suck the remaining moisture from food in a ziplock bag squeezed from most of the air?

[Anonymous No.8438352]

>>8438318
It's highly hygroscopic, so it wil attract water molecules in the container. First it will dry up whatever air there is, this will induce evaporation of water in the food because of the concentration delta, and this will continue until the silica gel is saturated or there is no more humidity to adsorb.

[Anonymous No.8438385]

>>8438352
That's perfect! A million thanks, anon!

[Anonymous No.8438418]

how many ways are there to sit 7 girls and 5 guys in a circular table so that no guy is next to other guy?
i got 7!5!/12 but i think this is wrong

[Anonymous No.8438420]

>>8438418
i meant (7!5!) / 12

[Anonymous No.8438448]

>>8438256
do you know what associative means?

[Anonymous No.8438476]

>>8438448
Yeah, more or less I am solving it.
Like
x*y*z = x+y+z+xyz
=(x+y)+z+xyz
=x+(y+z)+xyz
=x + y+(z+xyz)
=x+(y+z+xyz)
=(x+y+z)+xyz
=(x+y+z+xyz)
?

[Anonymous No.8438486]

>>8432439
ii is a solution (x-4t)^2
You can always just check since you've got the equation they have to satisfy [math]\frac{\partial^2{f}}{\partial{x^2}} = \frac{1}{v^2}\frac{\partial^2{f}}{\partial{t^2}} [/math]
Where v is constant or the wave's speed here.

[Anonymous No.8438495]

>>8438476
i have no idea what you're doing here

you need to show whether (x*y)*z=x*(y*z)

in your case (x*y)*z=(x-y)*z=x-y-z
and
x*(y*z)=x*(y-z)=x-(y-z)=x-y+z
so not associative

[Anonymous No.8438534]

>>8438418
Define "ways".

If you only care about whether a given seat is occupied by a boy or a girl (so swapping 2 boys or 2 girls is considered the same "way"), then it's 36.

Fill the first 10 seats with alternating B,G:
B,G,B,G,B,G,B,G,B,G

That leaves 2 spare girls. There are six distinct places to put the first one: before the first boy, or adjacent to an existing girl (it doesn't matter which side).

And there are also six distinct places to put the second one. If the first girl goes on the left, you have:
G,B,G,B,G,B,G,B,G,B,G
and it's just a choice of which of 6 girls to put the 7th next to (again, it doesn't matter which side).
Otherwise, you have the original layout
B,G,B,G,B,G,B,G,B,G
where one of the G is actually 2 girls. Again, you can place on the left or next to any girl (and again the side doesn't matter, nor does placing one girl between two others).

If you care about which boy/girl goes in each seat, then it's 36*5!*7! (i.e. pick one of 36 "templates" then permute the boys then permute the girls).

[Anonymous No.8438558]

Given these 2 sequences, why does the definition work for the lower one also? I mean if you have a monotone sequence of course for any given n>N you'll have it closing onto the convergence, but the other?

[Anonymous No.8438579]

>Without using the divergence theorem, calculate the flux of the vector field [math]\sin((\pi x/2)),y)[/math] across the square with vertices at [math](-1, -1), (1, -1), (1, 1)[/math] and [math](-1, 1)[/math].

How would I even start this? Really just at a complete loss

[Anonymous No.8438582]

>>8438579
Oh, I'm stupid and there should obviously be another bracket before sin and only 1 directly after it

[Anonymous No.8438615]

>>8438558
> why does the definition work for the lower one also?
What do you mean "why does it work"? That's the definition of convergence. If that statement holds, the sequence converges, otherwise it doesn't.

For any convergent sequence X, you can define
L[n] = min {X[i] : i >= n}
U[n] = max {X[i] : i >= n}

Clearly:
L[n] <= X[n] for all n (i.e. L is a lower bound for X).
U[n] >= X[n] for all n (i.e. U is an upper bound for X).
i>j => L[i]>=L[j] (i.e. L is monotonic increasing)
i>j => U[i]<=U[j] (i.e. U is monotonic decreasing)
lim(L)=lim(U)=lim(X)

If X itself is monotonic increasing, then L[n]=X[n] and U[n]=lim(X). If X itself is monotonic decreasing, then U[n]=X[n] and L[n]=lim(X).

[Anonymous No.8438641]

>>8438579
>>8438582
Haven't done these things for years but shouldn't you use Stokes'?

[Anonymous No.8438645]

>>8438641
We're supposed to do it directly from the definition of flux, so not in this case. We haven't actually covered Stokes' theorem yet anyway- just Green's

[Anonymous No.8438660]

I am in Community College(yes, CC). I am taking both Pre-calculus and trigonometry at the same time as separate classes. I find Trig so much more difficult. Why is this? I also use MyMathLab as my primary source of teaching also.

I wasn't a fan of Geometry when I was in MS/HS either. I forgot which grade I took it in. Could this be connected in someway?

I will list something I don't understand for instance:

2sin^2(theta)-sin(theta)-1=0

Okay. I get this is a quadratic equation and I should factor.

(2sin(theta)+1)(sin(theta)-1)=0

I apply the zero-product property for each factor getting:
sin(theta)=(-1/2) and sin(theta)=1

I am supposed to solve in the interval [0,2pi]

Why are the solutions theta=(7pi/6),(11pi/6), and (pi/2) for the second factor?

[Anonymous No.8438678]

>>8438579
>>8438582
>>8438645

Flux = [math]\iint_{S}dS\, \bar{F}(x,y,z)\cdot \bar{n}[/math]

Your surface lies on the x-y-plane, meaning it's normal vector everywhere is (0,0,1)

Since your vector field has no z-component the product between the field and the normal vector is zero everywhere, and the flux is zero.

This is just a guess.

[Anonymous No.8438739]

What's the property that says [math]x^Ty = y^Tx[/math] with [math]x,y \in \boldsymbol{R}^n[/math]

[Anonymous No.8438769]

>>8438739
In general:
(x.y)^T = y^T.x^T
=> (x^T.y)^T = y^T.(x^T)^T = y^T.x
=> x^T.y = ((x^T.y)^T)^T = (y^T.x)^T

If x and y are column vectors, then x^T.y and y^T.x are scalars, and a scalar is its own transpose.

[Anonymous No.8438833]

Is it better to do a math BS and pick CS courses then to do a CS BS? I want into either machine learning or embedded systems.

I realize that there is a difference in difficulty. I might be a delusional brainlet but I just can't see how anything can be impossible to learn.

What do? Should I just try some opencourseware to see if I can hack a math degree?

[Anonymous No.8439295]

>>8438833
So you want to do a bs in math and then one in cs? Why not just do both at once?