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# /sci/ - Science & Math

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File: 101 KB, 500x707, __hinanawi_tenshi_touhou_drawn_by_d_i__e455f35b8a6576f1a437842361a7244d.jpg [View same] [iqdb] [saucenao] [google] [report]

Previously >>10527130
>>10554314 edition.

 >> Anonymous Sat Apr 13 11:14:27 2019 No.10554331 >>10554321Thanks I feel anxious when there's no /sqt/
 >> Anonymous Sat Apr 13 11:16:16 2019 No.10554335 If I printed enough counterfeit money could I hypothetically devalue the supply enough to basically wipe away my countrie's debt?
 >> Anonymous Sat Apr 13 11:17:22 2019 No.10554337 >>10554335no because the debt in measured in other countries currencies.Double inflation = double debt
 >> Anonymous Sat Apr 13 11:29:24 2019 No.10554367 File: 77 KB, 911x662, FY2018-preliminary-to-whom-does-the-US-government-owe-money.png [View same] [iqdb] [saucenao] [google] [report] >>10554337Depends on the country and the debt - sometimes countries will issue debt that is to be repayed in their own currency, sometimes (especially if they need the debt to buy foreign goods) they will issue debt that is to be payed back in a foreign currency. The US debt is all owed in US dollars, for example, while Argentina has a bunch of bonds that have to be payed back in USD, since foreign lenders didn't trust them not to inflate their local currency.So in theory, with enough counterfeiting you could inflate away a countries debt, or counterfeit the foreign currency you need to pay the money back in. However, actually faking enough money to be noticeable at the scale of even a small nation would be extremely hard to do without being caught.
 >> Anonymous Sat Apr 13 11:47:52 2019 No.10554405 Why is Tenshi so hot?
 >> Anonymous Sat Apr 13 12:37:18 2019 No.10554517 If you were to iterate a powerset operation on a finite set to it's "limit" at infinity, would you end up with a set of cardinality equal to the natural numbers, or the highest ordinal?
 >> Anonymous Sat Apr 13 12:48:56 2019 No.10554571 >>10554517Which one is the highest ordinal?
 >> Anonymous Sat Apr 13 12:50:49 2019 No.10554578 >>10554517Pick up a category theory text lad.
 >> Anonymous Sat Apr 13 12:51:45 2019 No.10554581 >>10554571Sorry, what I mean is aleph-omega. As in, would the iterative application yield aleph-naught, or would it "reach" aleph-naught and then continue to increase in cardinality?
 >> Anonymous Sat Apr 13 12:54:19 2019 No.10554597 >>10554578Do you have any recommendations, or would really any general text help for this type of problem? The wiki doesn't seem to have anything on the topic.
 >> Anonymous Sat Apr 13 12:57:11 2019 No.10554605 >>10554597Handbook of categorical algebra.You won't use much, just until the definition of categorical limits and colimits.
 >> Anonymous Sat Apr 13 13:00:40 2019 No.10554612 >>10554605Thanks fren.
 >> Anonymous Sat Apr 13 13:40:08 2019 No.10554729 Is this the place to ask for help with my Biomechanics homeworkM
 >> Anonymous Sat Apr 13 14:24:32 2019 No.10554849 What is the good way to prove statements involving factorials. For example, I have to prove[eqn]\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}[/eqn]but I don't know how.Pls help I'm new to this stuff.
 >> Anonymous Sat Apr 13 14:35:42 2019 No.10554870 >>10554729yes
 >> Anonymous Sat Apr 13 14:46:10 2019 No.10554892 >>10554849There are n red balls and 1 blue ball. The number of ways to choose k of them is n+1 choose k.The number of ways to choose k of them if black must be chosen is n choose k-1.The number of ways to choose k of them if black is not chosen is n choose k.Either black is chosen or not, so n+1 choose k equals n choose k-1 plus n choose k.
 >> Anonymous Sat Apr 13 14:49:54 2019 No.10554899 >>10554849for these types of statements you're usually required to give either a technical "algebraic" proof or a more verbose "combinatorial" proof. because i dont fucks with the algebra, i'll just say that you have to come up with a story for both sides of the equation are counting and show that they describe the same amount.in your example, the left hand side is counting the number of ways to choose k students in a class of n+1 people. on the right hand side, you're getting a similar story, take student #n+1: either he was chosen, in which case you need to pick k-1 more students from the remaining n, or he wasn't and you have to choose k out of the remaining n. go it mate?
 >> Anonymous Sat Apr 13 15:07:21 2019 No.10554936 >>10554849Induction
 >> Anonymous Sat Apr 13 15:11:41 2019 No.10554943 File: 20 KB, 460x100, Annotation 2019-04-13 190905.png [View same] [iqdb] [saucenao] [google] [report] can I learn all this and get ready for the final exam in 14 days?
 >> Anonymous Sat Apr 13 15:18:23 2019 No.10554960 >>10554892>>10554899Well I don't really understand meaning of any of that.I'm not doing combinatorics, I'm learning calculus and none of that combinatorics stuff is even mentioned. Actually, this is an exercise after chapter on mathematical induction, where factorial function is an example of inductively defined function. Again, none of that stuff you guys are talking about is even mentioned so I'm probably expected to give algebraic proof (since you don't need to know anything about combinatorics to do that).But it just seems so hard to do it algebraically. If you can, please give me a hint on how to do this albebraically. If I can't figure it out in either case I'll just read the solution in the book itself.>>10554936Still not sure how to do this inductively either, but anyway it is explicitly stated that no induction is required for that proof.
 >> Anonymous Sat Apr 13 15:20:02 2019 No.10554963 >>10554943Stop shitposting and study.
 >> Anonymous Sat Apr 13 15:20:27 2019 No.10554964 >>10554943How about you start studying and find out?
 >> Anonymous Sat Apr 13 15:21:48 2019 No.10554971 >>10554960Option one: write out the extensive expression for both sides, and apply algebraic manipulation until you get 1=1. Then pay attention to what exactly it is that you did and apply it to the original formula to get the equality you want.Option two: attempt induction on n, k and n+k.
 >> Anonymous Sat Apr 13 15:22:19 2019 No.10554974 >>10554960Write out the definition of the right-hand side using factorials. Find a nice common denominator to add them, and see what happens
 >> Anonymous Sat Apr 13 15:25:14 2019 No.10554980 File: 13 KB, 657x527, R14kkDj.png [View same] [iqdb] [saucenao] [google] [report] >>10554963>>10554964you're right, thank you friends
 >> Anonymous Sat Apr 13 15:41:39 2019 No.10555032   Can anyone give hints on this? topology$Let Y be connected, and assume a homeomorphism B : Y \rightarrow Y exists such that B \circ B(y) = y. for all y \in Y Prove that, for every continuous function g :Y \rightarrow \mathbb{R} there exists y \in Y such that g(y) = g(B(y))$.This chapter had to do with generalization of intermediate value theorem in topology and I'm stuck on how that relates to this.
 >> Anonymous Sat Apr 13 15:50:01 2019 No.10555055 >>10554974Hmm I did as you have said but I got[eqn]\frac{n!(k(n-k)+1)}{k!(n-k)!}[/eqn]I can't continue... I wonder if I made a mistake or I simply don't see what to do next. If you want I'll write out all the steps I applied.>>10554971>Option two: attempt induction on n, k and n+k.How would I actually do that? Until now I used induction on expressions with only one variable i.e. $1+\dots+n=\frac{n(n+1)}{2}$.
 >> Anonymous Sat Apr 13 15:51:03 2019 No.10555061 >>10555055I don't know why that didn't format correctly.
 >> Anonymous Sat Apr 13 15:52:19 2019 No.10555066 >>10555055>i.e.It always happens...
 >> Anonymous Sat Apr 13 15:52:20 2019 No.10555067 >>10555055You can still use induction on one variable while treating the rest as parameters.So here you can use induction over n and treat k as just a number (or 0,<=k<=n but it will work either way)
 >> Anonymous Sat Apr 13 15:54:22 2019 No.10555075 >>10555055No, I meant attempt on n, and if it doesn't work try k, and if that doesn't work either try n+k.
 >> Anonymous Sat Apr 13 16:17:08 2019 No.10555133 >>10555067>>10555075Ok, I'll try induction.But also please someone point out where is a mistake (I guess there is one) in my derivation.[eqn]\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{(k-1)!(n-k-1)!}+\frac{n!}{k!(n-k)!}[/eqn][eqn]\frac{n!k!(n-k)!}{(k-1)!(n-k-1)!k!(n-k)!}+\frac{n!(k-1)!(n-k-1)!}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn][eqn]\frac{n!k!(n-k)!+n!(k-1)!(n-k-1)!}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn][eqn]\frac{n!(k!(n-k)!+(k-1)!(n-k-1)!)}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn][eqn]\frac{n!(k(n-k)!+(n-k-1)!)(k-1)!}{k!(n-k)!(k-1)!(n-k-1)!}[/eqn][eqn]\frac{n!(k(n-k)!+(n-k-1)!)}{k!(n-k)!(n-k-1)!}[/eqn][eqn]\frac{n!(k(n-k)+1)(n-k-1)!}{k!(n-k)!(n-k-1)!}[/eqn][eqn]\frac{n!(k(n-k)+1)}{k!(n-k)!}[/eqn]There you have it.
 >> Anonymous Sat Apr 13 16:22:39 2019 No.10555139 >>10554870Sweet.On a gravity of 1.62m/s^2, I'm throwing a ball with the v0 of 25m/s, at an angle of 42degrees from the X axis.-How do I calculate what's the highest it's gonna get?-How do I calculate how long will it take it to fall back to the ground?
 >> Anonymous Sat Apr 13 16:27:29 2019 No.10555148 How would punching someone on the Moon or another low but non-zero gravity environment be different than on Earth?
 >> Anonymous Sat Apr 13 16:35:11 2019 No.10555169   >>10554849Just put in the definition and put everything on the same denominator. It's a two minute exercise.
 >> Anonymous Sat Apr 13 16:38:47 2019 No.10555173 >>10555148Do you mean in terms of damage inflicted, or are you wondering if you can punch somebody into orbit?In terms of the former, there is no difference, because the energy you transfer via the mass of your arm will be the same, provided your stance allows for a stable platform. Concerning the latter, if you vector the punch correctly and the target is sufficiently low mass, and the planet is like the little prince's planet, it is possible.
 >> Anonymous Sat Apr 13 16:45:12 2019 No.10555186 >>10555173Would you be able to plant your feet well enough to throw a punch? Lower normal force means lower friction, so I was guessing you'd probably slip or at least not be able to throw your weight behind the punch as easily.
 >> Anonymous Sat Apr 13 16:46:42 2019 No.10555195 >>10554849$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$so$\sum_{k=0}^{n+1} \binom{n+1}{k} x^k = (1+x)^{n+1} = (x+1) \cdot (1+x)^n = \sum_{k=0}^n \binom{n}{k} x\cdot x^k + \sum_{k=0}^n \binom{n}{k} x^k = \sum_{k=0}^n \left( \binom{n}{k-1} + \binom{n}{k} \right) x^k + x^{n+1}$
 >> Anonymous Sat Apr 13 16:46:57 2019 No.10555197 $\sum_{k=0}^{n}(-1)^k \binom{n}{k} (n^2-kn)!$ How do I simplify this?
 >> Anonymous Sat Apr 13 16:54:24 2019 No.10555211 >>10555133$\binom{n}{k-1} = \frac{n!}{(k-1)!(n-k+1)!}$
 >> Anonymous Sat Apr 13 16:57:03 2019 No.10555217 >>10555197Unless that's already the expansion of a known function, I don't think you'll have much luck reducing it.But it leads me to another questions - are there some general results on how $\dfrac { (n^a)! } { (n!)^a },$say for a=2.behaves. From experimenting with Mathematica, it appears that even the log of that still grows at least polynomialy.
 >> Anonymous Sat Apr 13 17:00:29 2019 No.10555223 >>10554321If a 70kg body loses around ~1°c of temperature every hour after it dies and 1 calorie is needed to raise 1 gram of water 1°c then I figure:70,000 *24h: 1680kc a dayAm I accounting for everything here? Is this right?Do we really use most of our energy just keeping us warm?
 >> Anonymous Sat Apr 13 17:03:24 2019 No.10555229 >>10555197My gut instinct says stuff cancels out. Try opening it up.
 >> Anonymous Sat Apr 13 17:08:57 2019 No.10555242 >>10555186Well I wasn't exactly picturing a paved planetoid. If there is gravel, dirt or any terrain you can push against, that would be enough.
 >> Anonymous Sat Apr 13 17:28:50 2019 No.10555300 >>10555211Yeah... fuck.
 >> Anonymous Sat Apr 13 17:47:33 2019 No.10555345 >>10555139Okay, I managed to do the first one (28.5m), but still struggling with the second. What do?
 >> Anonymous Sat Apr 13 17:49:32 2019 No.10555350 >>10555223the energy isnt being used to keep us warm. The energy being used is what produces the warmth. there is a very distinct difference.
 >> Para !EemC8p8CHE Sat Apr 13 17:49:58 2019 No.10555351 >>10554321Sort of stumbled across an interesting idea. We have the placebo effect, cognitive bias and eye witness testimony all pointed towards a single potential effect. Belief has a profound effect on the way we interpret information. Could this be an in-built system rooted within the brain? Could hallucinations and other things entirely of the imagination also be effected to a large degree? Is this already covered in psychology to a the degree it should be? It's all very interesting and I want to know your opinion.
 >> Anonymous Sat Apr 13 17:50:16 2019 No.10555352 why do schools / unis arrange curriculum so that 3+ subjects or so are taken concurrently? are there enough cognitive benefits to merit this being the de facto standard? for example, see:the a converse would be arranging curriculum as a series of individual classes, like:>intro programming in haskell -> applications in haskell -> calculus I - III w/ applications in haskell -> linear algebra w/ applications haskell each course of course would likely be about 4x shorter, since it's the only course you';d be taking at that time. could someone sight the cognitive benefits of the standard approach, where all the abvoe classes would basically be taken at once instead?
 >> Anonymous Sat Apr 13 17:51:26 2019 No.10555356 >>10555351my opinion is that you shouldn't have a trip, especially so if you're an idiot with nothing of substance to say
 >> Anonymous Sat Apr 13 17:52:36 2019 No.10555358 >>10555351My pet hypothesis is that prior beliefs/biases activate some kind of active gain mechanism in the brain to provide positive feedback to stimuli that are in-line with the prior.t. EE with no background in cognitive psychology, so this is very unlikely to be a whole truth.
 >> Anonymous Sat Apr 13 17:53:29 2019 No.10555361 >>10555345If y(t) is the altitude, solve y(t)=0
 >> Anonymous Sat Apr 13 17:54:28 2019 No.10555365 >>10555350No way, you are saying it's waste heat? From what?I thought for sure we only kept warm to keep our muscles ready at any time and situation unlike reptiles
 >> Para !EemC8p8CHE Sat Apr 13 17:54:49 2019 No.10555366 >>10555356I'm asking at least one important question we don't have an answer to. Do beliefs effect hallucinations? The rest is less important and well documented.
 >> Anonymous Sat Apr 13 18:04:59 2019 No.10555378 >>10555366>Do beliefs effect hallucinations?fucking DUH, retard. you think everyone envisioning jesus is pure chance? kill urself
 >> Para !EemC8p8CHE Sat Apr 13 18:10:55 2019 No.10555386 >>10555378>Being this butthurt over some letters that appear on your screenIt's like telling your enemies exactly what you don't like. You fucking sensitive little faggot.
 >> Anonymous Sat Apr 13 18:24:34 2019 No.10555411   File: 6 KB, 205x246, brainlet.png [View same] [iqdb] [saucenao] [google] [report] How do I solve cos(10°)/sin(80°) without a calculator? According to my book the answer is 1 but I don't even know how to get there.
 >> Anonymous Sat Apr 13 18:30:23 2019 No.10555421 File: 517 KB, 650x390, Circle_cos_sin.gif [View same] [iqdb] [saucenao] [google] [report] >>10555411not very rigorous but: since the cosine measures the X axis and the sine the Y axis you can tell that they are the same here since the sine is 10° away from being at full height and the cosine is 10° away from being fully to the right.
 >> Anonymous Sat Apr 13 18:32:10 2019 No.10555422 >>10555411cos(10) = cos(90 - 10) = use the angle addition property
 >> Anonymous Sat Apr 13 18:33:37 2019 No.10555425 >>10555411cos(10°)=cos(90°-80°)=cos(90°)cos(80°)+sin(90°)sin(80°)=sin(80°)
 >> Anonymous Sat Apr 13 18:40:04 2019 No.10555436   >>10555421Not quite what I was looking for, but pretty interesting all the same.>>10555422>>10555425That's what I was thinking initially, but I don't know how to find cos(80°) without a calculator. If it were something like cos(15°) then I could just use the unit circle for cos(45°- 30°).
 >> Anonymous Sat Apr 13 18:41:32 2019 No.10555441 >>10555436cos(90°)=0 so it doesn't matter, that leaves you with cos(10°)=sin(90°)sin(80°) hence the result.
 >> Anonymous Sat Apr 13 18:42:04 2019 No.10555445 >>10555386>calling me sensitive while clearly getting butthurtlmao, you absolute idiot
 >> Anonymous Sat Apr 13 18:45:39 2019 No.10555449   >>10555441Oh, okay. I realize I'm going full brainlet here, but then how do you know what sin(80°) is?
 >> Anonymous Sat Apr 13 18:47:40 2019 No.10555454 >>10555449We don't, cos 10=1 sin 80 is just the equality you wanted to prove.
 >> Anonymous Sat Apr 13 18:49:45 2019 No.10555458   >>10555454Oh jesus christ now I see, thanks anon.
 >> Anonymous Sat Apr 13 18:55:20 2019 No.10555470 File: 196 KB, 1478x796, file.png [View same] [iqdb] [saucenao] [google] [report] does the author here just switch from the L^2 norm to the square of the L^2 norm without telling us? he did something similar earlier in the section. not sure if i'm just failing to understand the simplification properly
 >> Anonymous Sat Apr 13 18:57:31 2019 No.10555476 File: 45 KB, 1524x238, file.png [View same] [iqdb] [saucenao] [google] [report] >>10555470sorry, here's the referenced 2.67
 >> Anonymous Sat Apr 13 18:58:20 2019 No.10555477 >>10555470You're referring to the sub and superscript of 2 in eq. 2.69 right?
 >> Anonymous Sat Apr 13 19:00:05 2019 No.10555480 >>10555477yes. the subscript is denoting L^2 norm. the sqrt in the upper equation is expressing the Frobenius normd is a 1 dimensional matrix (big D isnt)
 >> Anonymous Sat Apr 13 20:40:56 2019 No.10555742 I need to find the degree of polarization of light that is reflected off a surface given the indices of refraction and incident angle. I've got the Fresnel coefficients and read into the Stokes vector but I'm not sure how to get from one to the other for an unpolarized source.
 >> Anonymous Sat Apr 13 21:05:08 2019 No.10555785 >>10555742Pretty sure you can just assume the incident unpolarized wave is an equal proportion of p and s polarisation and then use the Fresnel coefficients to determine the resulting proportion
 >> Anonymous Sat Apr 13 21:27:41 2019 No.10555830 File: 1.18 MB, 2452x3000, 1533497190923.jpg [View same] [iqdb] [saucenao] [google] [report] Please recommend books about the history of science. Wasn't there one by a nobel laureate?
 >> Anonymous Sat Apr 13 21:33:12 2019 No.10555839 >>10555785What proportion exactly?
 >> Anonymous Sat Apr 13 23:09:47 2019 No.10556035 is inertial mass the same as gravitational mass?specifically, does an object with high relative momentum "pull" more strongly gravitationally on an object in relative rest?
 >> Anonymous Sat Apr 13 23:39:15 2019 No.10556102 How do audio spectrograms work? fourier transforms don't work instantaneously
 >> Anonymous Sun Apr 14 00:16:43 2019 No.10556170 >>10556102not if you're using the fastest Fourier transform in the west
 >> Anonymous Sun Apr 14 00:40:38 2019 No.10556213 >>10555830Modern Physics by Tipler and Llewellyn
 >> Anonymous Sun Apr 14 00:57:46 2019 No.10556237 >>10555470Minimizing the L2 norm is the same as minimizing the square of the L2 norm. Usually, dealing with squares is simpler because they're easier to differentiate.
 >> Anonymous Sun Apr 14 02:00:49 2019 No.10556346 how were the summation properties/formulas discovered, just trial and error? or is there some method like $\sum_1^n = n(n+1)/2$
 >> Anonymous Sun Apr 14 02:03:03 2019 No.10556351 >>10556346the formula is guessed and then proved using induction
 >> Anonymous Sun Apr 14 02:06:48 2019 No.10556353 File: 128 KB, 548x575, are_you_fucking_serious (2).jpg [View same] [iqdb] [saucenao] [google] [report] >>10556102>something that plots a function of time>Fourier transform
 >> Anonymous Sun Apr 14 02:47:35 2019 No.10556420 Can someone post that one meme of a trench that bases the depth of a math branch by difficulty/complexity?
 >> Anonymous Sun Apr 14 03:08:05 2019 No.10556446 >>10556420God I hate that meme. Just google mathematics trench
 >> Anonymous Sun Apr 14 03:18:17 2019 No.10556468 >>10556446kek, can't believe that works
 >> Anonymous Sun Apr 14 03:56:40 2019 No.10556524 File: 44 KB, 800x450, brainlettttt.jpg [View same] [iqdb] [saucenao] [google] [report] If I know the current in a loop, and there's only 1 battery in the loop, how do I know how much of that current comes from the battery, and how much of it comes from the other loop(s)?
 >> Anonymous Sun Apr 14 03:58:11 2019 No.10556526 What transport equations did the scientists at the Manhattan project use to determine the critical mass of the Fat Man core?
 >> Anonymous Sun Apr 14 03:59:13 2019 No.10556527 Does a light path get manipulated by gravity itself (the influence of a mass) or by the curvature of space-time caused by gravity?
 >> Anonymous Sun Apr 14 05:48:10 2019 No.10556691 I have a graph where I plot 3 lines with some new data for two of the lines each day. The remaining line will be the absolute sum of the two other lines. Does anyone know what area of math I could look at if I wanted to find the optimal combinations of data for the highest absolute sum at, for example, 30 days from now?
 >> Anonymous Sun Apr 14 07:29:36 2019 No.10556860 >>10556346> like $\sum_1^n = n(n+1)/2$That one is quite trivial:S(n)=1+2+...+(n-1)+nS(n)+S(n) =1+2+...+(n-1)+n +n+(n-1)+...+2+1= (n+1)+(n+1)+...+(n+1)+(n+1)= n(n+1)=> S(n)=n(n+1)/2This can be extended to polynomials of any degree. If f(n) is a degree n polynomial, the difference f(n+1)-f(n) is a degree n-1 polynomial. So for the sum of terms given by a degree n polynomial, calculate the difference for a generic degree n+1 polynomial, equate coefficients, and solve the resulting system of linear equations.A common approach is to find some transformation such that "shifts" the series, mapping each transformed term to the preceding or successive term. E.g. for the geometric seriesS(n)=a+ar+ar^2+...ar^(n-1)rS(n)=ar+ar^2+ar^3+...ar^nrS(n)=S(n)-a+ar^n=> (1-r)S(n)=(1-r^n)a=> S(n)=a(1-r^n)/(1-r)Also: difference equations are usually strongly analogous to differential equations. So if you can solve the differential equation, the solution of the difference equation usually has a similar form. E.g. dy/dx=k^x => y=A.k^x+B.Telescoping is a common technique. If you can write f(n) as g(n)-(g-1) for some g(), then the sum is (g(1)-g(0))+(g(2)-g(1))+(g(3)-g(2))+...+(g(n)-(n-1)) = g(n)-g(0) as all of the intermediate terms cancel out (each positive matches an identical negative in the successive term, except for the first and last terms).
 >> Anonymous Sun Apr 14 09:43:35 2019 No.10557093 >>10556102Short-time Fourier transform (STFT).Only a small "window" of the signal is transformed (fraction of a second), and the window moves over the signal (usually by a distance that is shorter than the window length, so the windows overlap).The frequency spectrum for that small snippet is manipulated and transformed back - the signal gets synthesized by adding the snippets back together continuously (overlap-add-method).
 >> Anonymous Sun Apr 14 09:46:26 2019 No.10557100 why is this?$\|A\|_F = \sqrt{ \Tr{AA^T} }$
 >> Anonymous Sun Apr 14 09:47:28 2019 No.10557104 >>10557100that should be the trace operator... guess we don't support that
 >> Anonymous Sun Apr 14 09:56:38 2019 No.10557128 >>10556446>he works on a field near the topHeh, nothing personal kid.
 >> Anonymous Sun Apr 14 10:03:49 2019 No.10557153 >>10556526they have a grad student stand in the desert and clap two plutonium half-spheres of successively larger sizes together til something interesting happened
 >> Anonymous Sun Apr 14 10:08:56 2019 No.10557164 >>10557104>>10557100oh nm i get it now
 >> Anonymous Sun Apr 14 11:04:14 2019 No.10557300 >>10556353if you don't know what i'm talking about, take small brain somewhere else
 >> Anonymous Sun Apr 14 11:14:29 2019 No.10557318 File: 5 KB, 204x247, download.jpg [View same] [iqdb] [saucenao] [google] [report] So I'm kinda lost.Covering magnetic fields, and one of the first formulas I get is for flux.$flux = \int_c BdS,$(B is magnetic induction, S is the area of the contour, might be translating these somewhat wrong) Okay, seems reasonable enough, but next chapter it says the law of conservation is:$\int_c BdS = 0,$Does this mean that the flux is always equal to zero? I mean, I know it doesn't, but what am I missing here?
 >> Anonymous Sun Apr 14 11:18:55 2019 No.10557331 >>10557318>hey guys, the book is saying that the amount of energy in the system is constant, but also that objects can have different amounts of energy through time wtf?
 >> Anonymous Sun Apr 14 11:33:49 2019 No.10557343 >>10557331Oh, the second fomula just implies that it's constant?How come it isn't just::$flux = \int_c BdS = 0$My college taught us literally 0 things about integrals before shoving them onto us.
 >> Anonymous Sun Apr 14 12:17:57 2019 No.10557441 How do I show the identity$exp[\hat{A}+\hat{B}] = exp[\hat{A}] + \int_0^1 dz exp[z(\hat{A}+\hat{B})] \hat{B} exp[-z\hat{A}] exp[\hat{A}]$?>>10557318I think the key is what you are integrating overFlux is stuff moving through an area (dS)>One way to better understand the concept of flux in electromagnetism is by comparing it to a butterfly net. The amount of air moving through the net at any given instant in time is the flux. If the wind speed is high, then the flux through the net is large. If the net is made bigger, then the flux is larger even though the wind speed is the same. For the most air to move through the net, the opening of the net must be facing the direction the wind is blowing. If the net is parallel to the wind, then no wind will be moving through the net. The simplest way to think of flux is "how much air goes through the net", where the air is a velocity field and the net is the boundary of an imaginary surface.I don't get what you mean with the law of conservation sentences>>10557343Well look at it this wayWe know that energy is conserved alwaysNow consider you take a big rock in your arms and walk up a stair and place down the rock and walk down againYou obviously expended energy, we could write down integrals for that, dW = F.dsWhat matters is that your energy is loweredBut yet, the total energy is conserved! How does that work?!It works because we look only at the system that is you, in which energy is not conservedBut in the system that is the universe, energy is conservedWe don't even have to go that big, we can just look at the you-rock systemYou lost energy but the rock gained energy, your energyAssuming perfect everything the rock gained all of the energy that you expendedSo energy is not conserved in the you system, nor in the rock system, but it is in the you-rock systemI don't know what you mean with your flux = 0 but I hope this makes it a bit clearer
 >> Anonymous Sun Apr 14 12:20:43 2019 No.10557452 File: 1.23 MB, 495x341, green lantern is fucking stupid.gif [View same] [iqdb] [saucenao] [google] [report] Did someone ever figure out the nerve circuitry responsible for feeling pleasure during orgasm? When will be finally able to simulate and stimulate without burning the skin on our dicks?
 >> Anonymous Sun Apr 14 12:37:52 2019 No.10557480 >>10557452Yes, it's called dopamine.We also have substances that give us that, they're called drugs.
 >> Anonymous Sun Apr 14 12:43:43 2019 No.10557489 >>10557318>>10557343For an open surface:$\Phi_B = \iint_S {\mathbf B} \cdot d{\mathbf S}$For a closed surface:$\oint_S {\mathbf B} \cdot d{\mathbf S} = 0$or (can't remember if oiint works here)$\oiint_S {\mathbf B} \cdot d{\mathbf S} = 0$IOW, for a closed surface, the net flux is zero; any flux which enters the region bounded by the surface must leave it. There's no such thing as a magnetic monopole; flux goes around in loops with no start or end points.
 >> Anonymous Sun Apr 14 12:44:21 2019 No.10557490 >>10557480I was thinking more of electrodes in your shaft.
 >> Anonymous Sun Apr 14 12:46:41 2019 No.10557493 >>10557490Literally no reason to go put shit near your shaft when you can pop a pill and get the same result, no?
 >> Anonymous Sun Apr 14 13:29:51 2019 No.10557591 >>10557343No, the second formula implies that you should read the actual text and see what it is you're integrating over instead of just noting down the formulas.
 >> Anonymous Sun Apr 14 13:48:20 2019 No.10557641 >>10557591Yeah, I understood the concepts, it was just the formulas messing with me.Mostly it was just me noticing the difference between $\oint_s$ and $\int_s$.>>10557489>>10557441I think I get it now, thanks!
 >> Anonymous Sun Apr 14 15:54:12 2019 No.10558016 >>10557480Or you could go next level and take a bunch and drugs *and then* fuck for exponential returns.
 >> Anonymous Sun Apr 14 16:20:00 2019 No.10558082   I'm stumped lads, how do I get this into rectangular or polar form for the purpose of graphing? I typed it into cymath and it just told me to fuck off. I'm fully aware I'm retarded btw.r = 1 + cos θ
 >> Anonymous Sun Apr 14 16:24:19 2019 No.10558090 >>10558082$( cos \theta + cos^2 \theta , ~ sin \theta + sin \theta cos \theta)$
 >> Anonymous Sun Apr 14 17:18:06 2019 No.10558268 How the fuck do I calculate charge-to-mass ratio given initial velocity, displacement, time, and E? I don't think it's possible to use kinematics to get mass, so that makes me think I need to use E (units are N/C), but I can't figure out how. Any help would be appreciated.
 >> Anonymous Sun Apr 14 19:17:22 2019 No.10558531 >>10558268Calculate acceleration from displacement, velocity and time (a=v^2/2s). F=ma, F=qE => ma=qE => q/m = a/E.
 >> Anonymous Sun Apr 14 20:40:26 2019 No.10558719 File: 13 KB, 692x92, file.png [View same] [iqdb] [saucenao] [google] [report] how do i take the gradient of:$f(\boldsymbol x) = \frac{1}{2} \| \boldsymbol A \boldsymbol x - \boldsymbol b \|_2^2$?i know it's supposed to look like pic related but i'm unclear on how to get to it. i know that we can rewrite as$f(\boldsymbol x) = \frac{1}{2} (\boldsymbol A \boldsymbol x - \boldsymbol b)^T \cdot (\boldsymbol A \boldsymbol x - \boldsymbol b)$not sure how to get any closer though... when i work the multiplication out it doesn't seem like i'm on the right track.
 >> Anonymous Sun Apr 14 20:45:16 2019 No.10558732   >>10558719the \cdot shouldn't be there; omitting it, it's equivalent to dot product
 >> Anonymous Sun Apr 14 20:48:49 2019 No.10558737 Idiot here. I need to calculate some probabilites that are adding onto each other like this>1 in 6 chance to get the item you want>1 in 13 for it to be the color you want>1 in 20 for it to have something else you want>1 in 50 and so on on.Each of these are sequential. If step 1 didn't happen, step 2 is irrelevant because step 1 didn't occur. How do I calculate what the likely hood is to complete each of these steps? Obviously, the first is 1 in 6, but how do I find out the second or third and so on?
 >> Anonymous Sun Apr 14 20:49:51 2019 No.10558738 >>10558719Expand $||Ax-b||=\sqrt{tr(Ax-b)^T(Ax-b)}$
 >> Anonymous Sun Apr 14 20:50:36 2019 No.10558739 >>10558719It's just the chain rule, you should look up Matrix Calculus for how to extend a lot of your single variable understanding of calculus to derivatives-by-vectors, which the gradient is (derivative of a scalar function by a vector x)
 >> Anonymous Sun Apr 14 20:55:32 2019 No.10558748 >>10558739i know that.. i'm just not sure clear on the algebraic simplification to get to the point prior the derivative actually being taken
 >> Anonymous Sun Apr 14 21:02:31 2019 No.10558758 >>10558748But that's the thing, it's already as simple as it needs to be for application of a vector derivative. Expanding it out is more of a hindrance than a help. A straightforward application of the derivative gives you (1/2) * 2 * A'(Ax - b) immediately
 >> Anonymous Sun Apr 14 21:04:42 2019 No.10558764 >>10558758ah - ok thanks for pointing that out
 >> Anonymous Sun Apr 14 21:07:14 2019 No.10558770 >>10558764np, the only real trick that Matrix calculus adds on top of vector or single variable calc is keeping the transposes and ordering correct, which can change whether you're considering your vectors to be rows or columns. Otherwise most of the rules work with the same intuition as single variable
 >> Anonymous Sun Apr 14 21:28:21 2019 No.10558811 What's the point of adding so many different collection types when they all serve the same purpose. For instance, queues, stacks, and lists all store multiple objects. Most are just implementations of lists anyways so why add these extra collections.
 >> Anonymous Sun Apr 14 21:43:17 2019 No.10558849 >>10558811they have different interfaces and so internally can be implemented differently. therefore certain operations will be more performant with one type than another
 >> Anonymous Mon Apr 15 06:57:00 2019 No.10559789 File: 69 KB, 1260x444, Screen Shot 2019-04-15 at 3.53.34 PM.png [View same] [iqdb] [saucenao] [google] [report] herps me
 >> Anonymous Mon Apr 15 09:01:20 2019 No.10559981   >>10559789I'm pretty sure there's a section in Concrete Mathematics that goes into detail on how to construct these sequences using that example.
 >> Anonymous Mon Apr 15 09:45:28 2019 No.10560071 >>10559789This is quite straightforward once you dispel any assumptions you may have about what constitutes a "polygon". Given any ordered set of points, a polygon can be formed by joining adjacent points. The resulting polygon won't necessarily be convex and may self-intersect.Also, the question should probably have stated that no pair of lines are parallel, i.e. each pair of lines has a point of intersection.There n! ways to choose an ordering of n lines, each of which forms a polygon. But a cyclic permutation (of which there are n) yields the same polygon, as does reversing the order. So there are n!/n/2 = (n-1)!/2 distinct polygons.
 >> Anonymous Mon Apr 15 09:58:34 2019 No.10560098 >>10554321Are there any notations for "complex numbers" that avoid $i$?
 >> Anonymous Mon Apr 15 10:00:42 2019 No.10560103 >>10560098$\sqrt{-1}$
 >> Anonymous Mon Apr 15 10:06:12 2019 No.10560118 >>10560103I mean along the lines of tuples or vector spaces.
 >> Anonymous Mon Apr 15 10:09:30 2019 No.10560124 >>10560118(a, b) for the number a + ib is sometimes used but the multiplication isn't really obvious when you write it like that.
 >> Anonymous Mon Apr 15 10:10:41 2019 No.10560127 >>10560118>I mean along the lines of tuples or vector spaces.We give $\mathbb{R}^2$ the operations $(a, ~ b)+(c, ~ d)=(a+c, ~ b+d)$ and $(a, ~ b)*(c, ~ d)=(ac-bd, ~ bc+ad)$.
 >> Anonymous Mon Apr 15 10:10:44 2019 No.10560128 Does anything actually ever fall into a black hole from the perspective of an outside observer?When a star collapses into a black hole, what happens to the mass that was at the center of the star when the black hole formed? Does if just get pushed out by the event horizon? It can't pass the event horizon because nothing can do that from the perspective of an outside observer right?
 >> Anonymous Mon Apr 15 10:15:27 2019 No.10560137 >>10558811Because each one describes certain problems better. You can use a list for almost anything, but if it makes sense to use a queue, then using one will make your code easier to read an more maintainable.
 >> Anonymous Mon Apr 15 11:01:22 2019 No.10560229 >>10558737Please respond, I'm too dumb for this.
 >> Anonymous Mon Apr 15 11:23:26 2019 No.10560274 >>10560229Ignore the order. Effectively, all conditions have to be met, so the required probability operation AND, which means that the required arithmetic operation is multiplication. Think of the fractions as a bunch of unlikely outcomes. If you multiply them, then the result is something more unlikely. For example if you multiply a coinflip by another coinflip and want two heads in a row, then the probability of that happening is 1/2 * 1/2 = 1/4. The reason for this is that the number of possible outcomes increases to 4:head, headhead, tailstails, headtails, tails
 >> Anonymous Mon Apr 15 11:25:02 2019 No.10560278 >>10558531I tried that, but either it's wrong or I fucked up somewhere. I have these givens:v0=2.32m/s, d=.0698m, t=0.2sobject is travelling upwards, so:g=9.8m/s^2and E=3.6*10^3 N/CI can find Vf, and then acceleration, but even doing that I still get the wrong answer.
 >> Anonymous Mon Apr 15 11:47:20 2019 No.10560345 >>10560274So 1/6 and 1/13 would be 1/78, right? for all three it would be 1/6 x 1/13 x 1/20 = 1/1560? Alright, thank you very much. That helped.
 >> Anonymous Mon Apr 15 12:13:06 2019 No.10560413 >>10560345I believe so, unless the probabilities are somehow dependent ( https://www.khanacademy.org/math/ap-statistics/probability-ap/probability-multiplication-rule/a/general-multiplication-rule )That doesn't seem to be the case for the first two probabilities in your question. They're independent because choosing an item doesn't remove a color from existence.
 >> Anonymous Mon Apr 15 12:18:58 2019 No.10560427 >>10560413No I don't think so. The only thing linking is that the first chance must have occurred before moving on to the second, but the first thing doesn't change or influence anything in the second run. Again. thank you very much.
 >> Anonymous Mon Apr 15 13:32:26 2019 No.10560559 >>10560278> I have these givens:> v0=2.32m/s, d=.0698m, t=0.2sd and t are the distance and time until the object comes to rest? If that's the case, d=v0*t+(1/2)*a*t^2 => a = 2*(d-t*v0)/t^2 = -19.71 m/s^2Otherwise, state the exact problem.For constant acceleration, you have distance, initial velocity, final velocity, time and acceleration. Given any four you can find the fifth.> object is travelling upwards, so:> g=9.8m/s^2If the object is charged and in an electric field, then gravity won't be the only force.q/m = a/E assumes that the acceleration is due to the electric field alone. If you have gravity in the mix, then the actual acceleration experienced by the object will be the sum (or difference) of acceleration due to gravity and acceleration due to the field.
 >> Anonymous Mon Apr 15 14:54:57 2019 No.10560774 >>10560559The problem reads:>In a region where there is a uniform electric field that is upward and has magnitude 3.60×10^4 N/C , a small object is projected upward with an initial speed of 2.32 m/s . The object travels upward a distance of 6.98 cm in 0.200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)? Assume g = 9.80 m/s2, and ignore air resistance.I found that a=-19.71m/s^2 using that kinematics equation, but after that I don't know what to do. a/E doesn't give me the right answer using 19.71 (positive or negative) for a and 3.6*10^4 for E.
 >> Anonymous Mon Apr 15 16:41:40 2019 No.10561043 >>10560774> The problem reads:Okay, so -19.71 m/s^2 is the total acceleration. -9.81 m/s^2 of that is gravitational, leaving -19.71-(-9.81)=-9.9 m/s^2 for the electric field, which gives q/m=a/E=-9.9/3.6e4=-2.75e-4 C/kg.PS:> Given any four you can find the fifthCorrection:Given any three you can find the other two. The mean speed is 0.0698m/0.2s = 0.349 m/s = (v0+v1)/2 => v1=2*0.349-2.32 = -1.622. v1-v0=-1.622-2.32=-3.942, a=(v1-v0)/t=-3.942/0.2=-19.71.
 >> Anonymous Mon Apr 15 18:48:45 2019 No.10561332 >physics professor's teaching style never clicked with me>textbook sucks>only kind of half understanding what I'm doing at any given time>would you look at that, finals are in two weeksare there any resources anyone can recommend so that I can fill in the gaps in my Calc-Physics I course knowledge?I get lost in the equations not knowing how the fuck they figured how to use what and where they're pulling what they're using out of their ass, and something just isn't clicking
 >> Anonymous Mon Apr 15 19:07:02 2019 No.10561372 >>10561043Wow, I'm stupid. Thanks for your help. I didn't think you were supposed to combine the acceleration values.
 >> Anonymous Mon Apr 15 19:12:28 2019 No.10561385 >>10561332For physics, if you're having trouble with figuring out which equation to use, write your givens out and then find the equation that has the givens you have, and what you're trying to find. In Phys 1, you're probably doing mostly kinematics, force (including gravity and friction), centripetal force, and maybe torque. If you can give an example of a problem, I might be able to help more. Khan Academy is a great source though if you want get a better understanding.
 >> Anonymous Mon Apr 15 19:17:13 2019 No.10561397 File: 74 KB, 567x368, 2019-04-15-191446_567x368_scrot.png [View same] [iqdb] [saucenao] [google] [report] >>10561385I don't have any problems I need to solve right now, but here's one we were assigned recently to give you an idea.I can write out all of the givens but no equations ever just pop out at me, and then the way the textbook manipulates the equations is extremely foreign to me and the straightforward/elegant ways my calculus textbook uses.
 >> Anonymous Mon Apr 15 19:20:59 2019 No.10561402 File: 7 KB, 696x88, 4help_15.png [View same] [iqdb] [saucenao] [google] [report] Little stumped on this one.
 >> Anonymous Mon Apr 15 19:35:30 2019 No.10561427 >>10561402course?
 >> Anonymous Mon Apr 15 19:38:58 2019 No.10561435 >>10561402I dunno lad, what's the definition of average again?
 >> Anonymous Mon Apr 15 19:39:50 2019 No.10561436 >>105614351/b-a integral from a to b of f(x) dx
 >> Anonymous Mon Apr 15 19:42:45 2019 No.10561440 >>10561427calc 1
 >> Anonymous Mon Apr 15 19:44:14 2019 No.10561442 >>10561436Yeah, and the integral of $4x^{-2}$ is $-8/x$. Evaluated at 1 we have -8, at c it's -8/c. So it's [-8/c-8]/(1-c)=1.
 >> Anonymous Mon Apr 15 19:45:17 2019 No.10561445 >>10561442>that entire formulaI meant [-8/c+8]/(c-1)=1.
 >> Anonymous Mon Apr 15 19:51:08 2019 No.10561461 >>10561442>>10561445the answer isn't 1, I already tried that
 >> Anonymous Mon Apr 15 19:54:28 2019 No.10561467 File: 18 KB, 617x168, 2019-04-15-195328_617x168_scrot.png [View same] [iqdb] [saucenao] [google] [report] >>10561461are they looking for something like this?I wouldn't think so since that's Calc II for me, but maybe your course introduced this concept in Calc I
 >> Anonymous Mon Apr 15 20:00:11 2019 No.10561481 >>10561397If you're given a formula sheet (which I'm guessing you are), don't worry too much about being able to just know what equation to use. As long as you can piece together what you have, and figure out what you need, you should be able to figure out what equation to use. If you can categorize the problem, it will make it easier to do this though. "Angular speed" should be a pretty big hint, and a "uniform rectangular rod" might also be a hint at moment of inertia. In my high school physics classes, my teacher sometimes had us solve problems by simplifying equations until we got down to what we needed to find rather than giving us values for variables and having us find an actual value. I found that to help a lot because you get a better understanding for the relationship between some variables.
 >> Anonymous Mon Apr 15 20:02:46 2019 No.10561484 >>10561467okay I randomly guessed and it said the answer 4 is correct, could someone explain this to me?
 >> Anonymous Mon Apr 15 20:08:00 2019 No.10561490 >>10561484the average of a function over an interval (a,b) is 1/(b-a) times the integral over (a,b)so set that equal to one and solve for c$\displaystyle 1= \frac{1}{c-1} \int_{1}^{c} 4 / x^2 dx= \frac{4}{c-1} (-1/c + 1) = \frac{4}{c-1} \frac{c-1}{c} \\1 = 4/c\\c = 4$
 >> Anonymous Mon Apr 15 20:24:01 2019 No.10561530 >>10561490it all makes sense now
 >> Anonymous Mon Apr 15 20:31:02 2019 No.10561556 >>10561481We're not given a formula sheet which is a big part of the problemMr professor pretty much just writes down solutions from his lecture notes on the board without explaining them (and gets them wrong anyway half the time) so I'm having a real difficult time getting any intuition like you're describing in your second paragraph
 >> Anonymous Tue Apr 16 02:20:54 2019 No.10562243 -1=i*i=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1why not
 >> Anonymous Tue Apr 16 07:15:36 2019 No.10562591 How do I go about calculating the following: "A graph has 30 vertices, 63 edges and all vertices have the degree 3, 4, or 5. It is known that there is one vertex with the degree 3. How many vertices have degree 4 and how many have degree 5?"
 >> Anonymous Tue Apr 16 09:10:59 2019 No.10562787 >>10562591How about you write down the equations forX = summation (number of vertices of degree D) times DN = number of verticesX will be twice the amount of edges in the graph because it counts every edge twice. Then you'll have a system of equations with two variables, because X is 1*3 + a*4 + b*5, N is 30 is 1+a+b
 >> Anonymous Tue Apr 16 09:49:24 2019 No.10562847 What to do if I read the solution but still don't get it?
 >> Anonymous Tue Apr 16 09:52:31 2019 No.10562858 >>10562243>sqrt(-1)*sqrt(-1)=sqrt(-1*-1)Incorrect
 >> Anonymous Tue Apr 16 09:59:08 2019 No.10562870 >>10562847check what specific step in the solution you do not understand and start from there.
 >> Anonymous Tue Apr 16 10:30:33 2019 No.10562910 File: 57 KB, 645x729, pqafkb6d9ba01.jpg [View same] [iqdb] [saucenao] [google] [report] >>10562870Heh I still don't get it...
 >> Anonymous Tue Apr 16 10:32:26 2019 No.10562913 >>10562910Reread the definitions and try to picture them in your head.
 >> Anonymous Tue Apr 16 10:35:36 2019 No.10562920 >>10562910Then ask about that specific step
 >> Anonymous Tue Apr 16 10:57:04 2019 No.10562963 >>10562913>>10562920Ok, I have to prove[eqn]\sum^l_{k=0}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}[/eqn]I'm also given a hint to apply binomial theorem to $(1+x)^n(1+x)^m$.So I did. I more or less figured out by myself that[eqn]\sum^n_{k=0}\binom{n}{k}x^k\times\sum^m_{j=0}\binom{m}{j}x^j=\sum^{n+m}_{l=0}\binom{n+m}{l}x^l[/eqn]But then comes the part that I don't understand. The author just says that from this formula it is obvious that coefficent of $x^l$ on the left-hand side is $\sum^l_{k=0}\binom{n}{k}\binom{m}{l-k}$ and then it completes the proof.
 >> Anonymous Tue Apr 16 10:59:29 2019 No.10562971 >>10562963>then it completes the proofNo, it's better to say "with that it completes the proof".
 >> Anonymous Tue Apr 16 11:23:13 2019 No.10563033 Am I fucked for getting into grad school without doing research in undergrad?
 >> Anonymous Tue Apr 16 13:18:15 2019 No.10563311 >>10562963It is obvious. Try it with n = 3 and m = 4 for example.You can also prove it by counting number of ways for choosing $l$ elements from two disjunct sets $A \text{ and } B$ of sizes $n \text{ and } m$ respectively, in two different ways. One is obvious (RHS), and the other one is obtained by finding number of ways to choose $l$ elements by taking $k$ elements from $A$ and $l-k$ elements from $B$ and then summing from $k = 0 \text{ to} k = n$ (LHS).
 >> Anonymous Tue Apr 16 13:28:59 2019 No.10563333 File: 7 KB, 284x168, intersect.png [View same] [iqdb] [saucenao] [google] [report] I want to be able to calculate where lines(vectors?), spheres, cubes etc intersect in 3D space but dont even know where to start. what do? anyone have some good learning material for that stuff?
 >> Anonymous Tue Apr 16 13:32:26 2019 No.10563341 >>10563333Do you know analytic geometry in plane?
 >> Anonymous Tue Apr 16 13:36:49 2019 No.10563357 >>10563341my knowledge of maths doesnt go much further than the basic trig I learned in school unfortunately
 >> Anonymous Tue Apr 16 13:39:25 2019 No.10563366 >>10563341ok i shouldve looked that up before replying, while i've never heard of it with that name, quite a lot of that stuff does look familiar. so yea, i somewhat do know analytic geometry i guess
 >> Anonymous Tue Apr 16 13:47:18 2019 No.10563399 For ξ and η elements of a Hilbert space, what does ξ|η stand for?
 >> Anonymous Tue Apr 16 13:55:49 2019 No.10563425 >>10563366Okay. Even if you don't know calculus, any beginner text on vector calculus typically covers how to describe line and shapes and planes in the 3D space in the first chapters. Check that out. Basic lin alg covers that too, but it's more abstract.
 >> Anonymous Tue Apr 16 14:08:01 2019 No.10563466 >>10563311Desu it is more clear when you phrase it like that.But even though I see why it is true when I look at the original equation I still don't understand what I need to do to the second one to obtain the first one. In other words, how do I algebraically manipulate the second equality to obtain the first one?>summing from k=0 tok=n (LHS)Did you mean "to $l$"?
 >> Anonymous Tue Apr 16 14:13:51 2019 No.10563478 >>10563333nice quads bro
 >> Anonymous Tue Apr 16 14:21:08 2019 No.10563508 >>10563425>>10563478thanks, shouldnt be too hard to learn that stuff eh
 >> Anonymous Tue Apr 16 14:21:23 2019 No.10563511 I don't understand this question, any help would be appreciated:>For the system equations in problem 1, determine what the value of y(t) will be after a long time (i.e as t approaches infinity) if the initial conditions are i. y(0)=0, and ii y.(0)=0.1Problem 1 simply lists equations, here is an example>1. a. y'=1-y^2Am I trying to find c1 and c2 after taking the anti-derivative, or should I be finding a limit or some combination of both? Perhaps I'm completely off base. This is for a Linear Engineering Systems class, so it's Linear Algebra and differential equations.
 >> Anonymous Tue Apr 16 14:23:18 2019 No.10563520 >>10563466I meant from $k = 0 \text{ to } k = n$, but since $a \choose b = 0 \text{ for } b > a$ they are the same.
 >> Anonymous Tue Apr 16 14:25:03 2019 No.10563526 >>10563520${{a} \choose {b} } = 0 \text{ for } b > a$
 >> Anonymous Tue Apr 16 14:28:03 2019 No.10563536 >>10561556Oh, that is strange. I've never not been given a formula sheet, even on the AP test and in college I get one. My best advice in that case is to try working through a few problems (even ones you've already done) and doing them how I previously described; don't use values, just variables. When I was in physics in high school, we usually had 10-20 problems due online biweekly through WebAssign, and when you're doing that many problems, you become so used to using the equations you start to memorize them without really trying. I suppose if you're really feeling SOL you could put the equations on a note card and try to look at them while your teacher isn't looking, or if you have a calculator you can take notes in, store them in that.
 >> Anonymous Tue Apr 16 14:52:21 2019 No.10563591 A solution to Snake is to come up with a route that passes through every square on the map exactly once and ends where it started. It's easy to create such a route for any grid with at least one even side, but it seems impossible for any grid with two odd sides (except 1x1). How do you prove this?
 >> Anonymous Tue Apr 16 14:58:04 2019 No.10563606 >>10563511You're not supposed to use integration. The point is to describe the behaviour of the system, not solve it explicitly, since a given system of differential equations will often have a solution that can't be expressed in terms of standard functions. Post all the equations.
 >> Anonymous Tue Apr 16 15:05:15 2019 No.10563630 So I just learned about the equivalence of norms in euclidean spaces, and I know the formal definition + the gist of the proof. But I feel like I don't quite grasp the usefulness of that property yet. The name "equivalence" makes it seem like a very strong relation, as if norms inherited several properties because of this relation. But from the definition of equivalence I can't see why.So how powerful is the equivalence of norms? What properties are shared between equivalent norms, that I should know of?
 >> Anonymous Tue Apr 16 15:20:32 2019 No.10563793 >>10563606So, I should be looking at the set of equations, rather than each one? If I was to find the equilibrium points, would I do it individually, or for the entire set?Here is the list of equations:a. y' = 1-y^2b. y' = y^2-1c. y' = y(y-1)(y+1)d. y' = (y-1)^2e. y' = (y+1)(y-1)^2f. y' = (y^2+1)(y-1)g. y' = cos(y)
 >> Anonymous Tue Apr 16 15:32:16 2019 No.10563824 >>10563511The formula for y'(t) tells you how the value of y changes over time.When t=0, the rate of change of y is 1-0^2 = 1, so y will get bigger. Once y has increased to, say, 0.5, the rate of change is 1-0.5^2 = 0.75, so it's still increasing but more slowly. When y=1 the rate of change is 1-1^2 = 0, so the value of y remains stationary at 1.You want to plot a vector diagram. As there's only one variable, this will be a one-dimensional plot, i.e. a straight line. Find all the stationary points (for the equation you gave, the points where 1-y^2 = y) and mark them. Then work out how the value of y behaves inbetween the stationary points (you'll find it points towards one of the stationary points), and mark this with arrows. By looking at your diagram you'll be able to see where y will end up if it starts at 0 and 0.1.You should get stationary points at y=1 and y=-1. 1 will be a stable point and -1 will be an unstable point, so:any initial value greater than -1 will converge to 1if the initial value is -1 exactly it will stay at -1any initial value less than -1 will tend towards negative infinity
 >> Anonymous Tue Apr 16 15:36:38 2019 No.10563841
 >> Anonymous Tue Apr 16 15:45:04 2019 No.10563871 >>10563793Nah, do each one individually. Sometimes you might get multiple variables per set, example:x' = y-xy' = 2x-yIn which case you'd need a 2D plot and probably graphing software to make it easier. But all yours are single-variable, they're separate questions. Find the equilibrium points individually.
 >> Anonymous Tue Apr 16 15:46:07 2019 No.10563872 >>10563824> (for the equation you gave, the points where 1-y^2 = y)This is wrong, it should be the points where 1-y^2=0.
 >> Anonymous Tue Apr 16 16:07:19 2019 No.10563910 File: 304 KB, 1488x1737, 20190416_210400.jpg [View same] [iqdb] [saucenao] [google] [report] >>10563793Seems like it doesn't matter if you start at 0 or 0.1, you always end up at the same point regardless. I would've expected at least one difference. Probably a good idea to do them yourself to make sure.
 >> Anonymous Tue Apr 16 17:36:34 2019 No.10564120 Okay so I feel retarded but I'm having trouble with my calc III homework. basically I have to find the line integral of the gradient of $f(x,y,z) = xy^3z^2$ with the curve $r(t) = < e^{tcos\left(t^2+1\right)},ln(t^2+1), \frac{1}{\sqrt{t^2+1}}>$. I found the gradient of f to be  and then plugged in r(t) to give me [eqn]\left(f\:\circ \:r\right)(t) = <\frac{ln^3\left(t^2+1\right)}{t^2+1},\:\frac{3e^{tcos\left(t^2+1\right)}ln^2\left(t^2+1\right)}{t^2+1},\:\frac{2e^{tcos\left(t^2+1\right)}ln^3\left(t^2+1\right)}{\sqrt{t^2+1}}>[/eqn]. I then had to take the derivative of r(t) with respect to t which gave me: [eqn]<\mathrm{e}^{t\cos \left(t^2+1\right)}\left(\cos \left(t^2+1\right)-2t^2\sin \left(t^2+1\right)\right),\:\frac{2t}{t^2+1},-\frac{t}{\left(t^2+1\right)^{\frac{3}{2}}}>[/eqn]. I use this to find the dot product and take the integral on the bounds provided to me and get: [eqn]\int _0^1\:\frac{ln^3\left(t^2+1\right)\mathrm{e}^{t\cos \:\:\left(t^2+1\right)}\left(\cos \:\:\left(t^2+1\right)-2t^2\sin \:\:\left(t^2+1\right)\right)}{t^2+1}+\:\frac{6te^{tcos\left(t^2+1\right)}ln^2\left(t^2+1\right)}{\left(t^2+1\right)^2}+\:\frac{-2te^{tcos\left(t^2+1\right)}ln^3\left(t^2+1\right)}{\left(t^2+1\right)^2}dt[/eqn] and there is no fucking way that shit is even remotely possible. Should I just try to integrate this or am I doing something horrendously wrong? sorry if the formatting is fucked btw, this is my first time using it.
 >> Anonymous Tue Apr 16 17:46:16 2019 No.10564138 >>10564120>line integral of the gradient
 >> Anonymous Tue Apr 16 17:55:05 2019 No.10564158 >>10564138>just looked back at my textbook>fundemental theorum of line integrals[eqn]\int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}} = f\left( {\vec r\left( b \right)} \right) - f\left( {\vec r\left( a \right)} \right)[/eqn]>this could have been so much easier for metime to an hero
 >> Anonymous Tue Apr 16 17:55:40 2019 No.10564159 File: 76 KB, 500x672, 65763543354354.jpg [View same] [iqdb] [saucenao] [google] [report] If I have two objects and I reduce the distance between them by half each second, how come there is a moment when they touch? Does that means the second before they touch they are at a distance that can not be divided by 2? And if so, what is that distance?
 >> Anonymous Tue Apr 16 18:06:00 2019 No.10564186 >>10564138thank you anon for pointing out how retarded I was being, it was actually really easy when I stopped small-braining.
 >> Anonymous Tue Apr 16 18:17:04 2019 No.10564212 >>10564159>If I have two objects and I reduce the distance between them by half each second, how come there is a moment when they touch?If we are talking about classic rigid bodies, this isn't true. There is no "smallest distance"
 >> Anonymous Tue Apr 16 18:29:28 2019 No.10564242 >>10563871>>10563910Ok, so that all makes sense, but I'm still wondering if I should be considering all the equations at once, or individually like>>10563606 said. I would think individually, since it seems to make more sense with the work we've done so far in class. However, this is my first experience with linear algebra/dif eq, so I'm not sure how to interpret "system equations". >>10563824Ok, so I'm assuming I'm looking at the equations individually then? I found equilibrium points for each equation, and now I'm working on making sense of plugging in t=0 and getting either y=0 or y=.1. When I first read it, I assumed it to be an initial value problem, but seeing the responses I got, it seems I misunderstood the problem. If it's not IVP, then it would appear to be a limit, but that doesn't seem right. I understand what you mean in your last 3 statements, I'm just confused as to why the problems is asking about plugging in t=0 and getting either y=0 or y=.1. I really appreciate everyone's help, it's definitely giving me a better understanding of this.
 >> Anonymous Tue Apr 16 19:49:57 2019 No.10564506 File: 78 KB, 1270x402, file.png [View same] [iqdb] [saucenao] [google] [report] why are the equal signs far apart on the left and close together on the right?i'd like the right hand side to look like the lefthow do?
 >> Anonymous Tue Apr 16 19:57:24 2019 No.10564517 >>10564492That's because of the chain rule.
 >> Anonymous Tue Apr 16 20:01:34 2019 No.10564526   >>10564506Attempt.P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}$P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}$
 >> Anonymous Tue Apr 16 20:03:56 2019 No.10564536 >>10564506Attempt.P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}[eqn]P(y = y \mid x = x) = \frac{ P(y = y, x = x)}{P(x = x)}[/eqn]
 >> Anonymous Tue Apr 16 20:04:51 2019 No.10564538 File: 86 KB, 1226x424, file.png [View same] [iqdb] [saucenao] [google] [report] >>10564536renders properly on 4chins but not on my local
 >> Anonymous Tue Apr 16 20:05:42 2019 No.10564540 File: 57 KB, 482x549, 1493405349916.jpg [View same] [iqdb] [saucenao] [google] [report] How the fuck do I generate a square wave with a period of 1s in Matlab?
 >> Anonymous Tue Apr 16 20:08:08 2019 No.10564548 >>10564538I don't know man. Does /g/ have a qtddtot?
 >> Anonymous Tue Apr 16 21:34:39 2019 No.10564751 File: 8 KB, 884x469, vecTriangle.png [View same] [iqdb] [saucenao] [google] [report] Is it possible to make a triangle with the values on the right?How can it be explained with math?
 >> Anonymous Tue Apr 16 21:42:53 2019 No.10564781 >>10564751Sure.$y=90$$a=\sqrt{c^2-b^2}$$b=\sqrt{c^2-a^2}$$c=\sqrt{a^2+b^2}$$x=h$
 >> Anonymous Tue Apr 16 21:43:55 2019 No.10564786 >>10564781>x=hForgot to erase that.
 >> Anonymous Tue Apr 16 21:47:54 2019 No.10564800 >>10564751No, since the largest side is always opposite to the largest angle, and b is opposite to y.>>10564781Setting up values doesn't show it actually exists.
 >> Anonymous Tue Apr 16 21:54:18 2019 No.10564826 >>10564786And your angles[eqn]\sin(x) = \frac{a}{c}[/eqn]$h = 90-x$This would've been in the first post but I had to look up laTEX for sine (and now I feel retarded).
 >> Anonymous Tue Apr 16 22:02:32 2019 No.10564848 >>10564800>the largest side is always opposite to the largest angleHow can this be explained with math?
 >> Anonymous Tue Apr 16 22:08:34 2019 No.10564870 >>10564800>Setting up values doesn't show it actually exists.But in this case it does. All I described was a right triangle where one side is inbetween the other side and the hypotenuse in length. The sum of all angles in a Euclidean triangle is 180, so y has to be 90, or the triangle non-Euclid. FUCK!I just realized that doesn't work because y can't be a right angle if c is the hypotenuse. I mean, assuming we're only using positive values.
 >> Anonymous Tue Apr 16 22:08:39 2019 No.10564871 >>10564848Apply the sine law to the biggest>90 and biggest<90 cases.
 >> Anonymous Tue Apr 16 22:21:43 2019 No.10564918 >>10564871Thank you
 >> Anonymous Wed Apr 17 02:35:38 2019 No.10565495 >>10564212Please explain it to me as if I was dumb, because I'm afraid that's the case and I really want to get this.
 >> Anonymous Wed Apr 17 02:44:47 2019 No.10565511 >>10560098Complex numbers can be represented by particular real 2-by-2 matriceshttps://en.wikipedia.org/wiki/Matrix_(mathematics)#Applications
 >> Anonymous Wed Apr 17 02:46:22 2019 No.10565515
 >> Anonymous Wed Apr 17 02:47:58 2019 No.10565518
 >> Anonymous Wed Apr 17 06:00:53 2019 No.10565821 File: 437 KB, 1920x1080, 1554456584839.jpg [View same] [iqdb] [saucenao] [google] [report] >>10565515>>10565518I'm afraid these do not explain this event at all. Both tell me that theoretically the distance can always be smaller, ok, makes sense, but in practice there is a moment in time when both objects touch, and a time when they do not. A time when the distance between is 0 and a time when it is not. Maybe at a near atomic level there is a moment when 2 objects are so close that if let go they slam together and end up touching, idk, I'm pulling this theory totally off my ass. But could that be an explanation?
 >> Anonymous Wed Apr 17 06:52:04 2019 No.10565878 >>10564540you uninstall and run python instead, like a sane person.
 >> Anonymous Wed Apr 17 07:45:24 2019 No.10565955 File: 350 KB, 1200x1000, 1550617206182.jpg [View same] [iqdb] [saucenao] [google] [report] >>10554321How do I learn enough to pass a calc 3 exam in 1 day?
 >> Anonymous Wed Apr 17 07:55:26 2019 No.10565972 >>10565821https://en.wikipedia.org/wiki/Half-lifeNot directly related to distance, but still.
 >> Anonymous Wed Apr 17 08:19:17 2019 No.10566010 >>10556035yes they are same and it has been experimentally proven momentum is not equal to mass but there is another thing called relativistic mass,though i am not sure if it affects gravitational mass.
 >> Anonymous Wed Apr 17 09:35:43 2019 No.10566163 >>10564242>Ok, so that all makes sense, but I'm still wondering if I should be considering all the equations at once, or individuallyEach equation is a separate question. Solve each one individually. Period.>so I'm not sure how to interpret "system equations"In your case, each 'system' only has one equation. If this doesn't make sense to you, just don't think about it. 100% you need to solve each equation individually.>I found equilibrium points for each equation, and now I'm working on making sense of plugging in t=0 and getting either y=0 or y=.1What? No, you're not supposed to try and "get" 0 or 0.1, those are the values you start with. Look at the graphs on >>10563910, that's the approach you need to take. Imagine you dropped a ball at 0 and it moved according to the arrows (which represent whether y'(t) is positive or negative for that value of y). If the arrow points left, the ball rolls left. If the ball reaches a stationary point, it stops there. And so on. Wherever the ball ends up (either at a stationary point or rolling off into infinity) is your answer. Do this for each equation, finding out where the balls ends up if it starts at 0, then if it starts at 0.1.For systems with multiple equations, you get more complicated answers, like the ball might be caught in a loop. But this can't happen with one-equation systems, which is all you're dealing with right now.t is just a time variable, it parametrises the equation. y(t) means "the value of variable y at time t". If y(0)=0.1, that means y's initial value is 0.1. What you want to find is where y ends up.
 >> Anonymous Wed Apr 17 09:39:43 2019 No.10566172 [eqn]e(x)=\sum_{j=0}^{\infty}\frac{x^j}{j!}[/eqn]Starting from this definition only, how do I prove that e(0)=1?
 >> Anonymous Wed Apr 17 09:41:09 2019 No.10566175 >>10566172Just substitute x = 0 lol.
 >> Anonymous Wed Apr 17 09:52:28 2019 No.10566200 >>10565821>I'm afraidgo see your shrink then
 >> Anonymous Wed Apr 17 09:53:45 2019 No.10566201 >>10565955I'm struggling to reach that .75 boys. Wish me luck.
 >> Anonymous Wed Apr 17 09:53:47 2019 No.10566202 >>10565821>I'm pulling this theory totally off my ass.yes, that's the explanation?
 >> Anonymous Wed Apr 17 10:05:38 2019 No.10566222 >>10566172lmao just substitute x by 0 and remember 0^0=1
 >> Anonymous Wed Apr 17 10:09:22 2019 No.10566232 >>10565821.5+.25+.125+.0625+...=1 Right?Provided the sum converges, it turns out that summing an infinite number of terms takes a finite amount of time and results in a finite distance. That's why the objects touch. Zeno didn't get that, hence the "paradox" (it isnt a paradox).
 >> Anonymous Wed Apr 17 11:08:47 2019 No.10566362 >>10566175>notices your undefined term>0^0 what's this?
 >> Anonymous Wed Apr 17 11:52:15 2019 No.10566471 File: 73 KB, 1200x1200, 814266.0.jpg [View same] [iqdb] [saucenao] [google] [report] I have a knife that I use for woodcarving and cutting modeling clay- when it gets cold large chunks of the stuff are hard to work. Will dragging a knife through a soft but dense material dull it?Surprisingly hard to google
 >> Anonymous Wed Apr 17 12:45:27 2019 No.10566589   File: 4 KB, 153x123, 1489686137131.jpg [View same] [iqdb] [saucenao] [google] [report] amirite?$\forall a\in A on a topology T if a \in B \subseteq A, B\cap C=\{a\} and B,C are open, then a is not limit point of B.\\Proof. Assume for contradiction that a is a limit point of B. Then every open set containing a must contain some point z\in B, but the only points that B and C share is a.\bot\\\square$
 >> Anonymous Wed Apr 17 12:48:49 2019 No.10566596 File: 55 KB, 581x525, fx7un7vpmgh21.png [View same] [iqdb] [saucenao] [google] [report] $\forall a\in A$ on a topology $T$ if $a \in B \subseteq A, B\cap C=\{a\}$ and $B,C$ are open, then $a$ is not limit point of $B$.Proof. Assume for contradiction that $a$is a limit point of $B$. Then every open set containing$a$must contain some point $z\in B$, but the only points that $B$ and $C$share is $a$.$\bot$$\square$
 >> Anonymous Wed Apr 17 12:58:10 2019 No.10566626 >>10566596YOU WHATTOPOLOGIES ARE CLOSED UNDER FINITE INTERSECTION LAD{a} IS FUCKING OPEN, IT ISN'T A LIMIT POINT OF ANYTHING
 >> Anonymous Wed Apr 17 13:05:31 2019 No.10566643 >>10564242If the system converges to a stable equilibrium, then y'=0 at that point. So substitute y'=0 and solve. If there are multiple solutions, the initial conditions will determine which one it converges to.
 >> Anonymous Wed Apr 17 13:10:50 2019 No.10566659 Is differential equations useful for higher math courses? Everything so far has been cookbook work that only works for specific problems, and apparently that's fine for sciences, but if you come across those "unsolvable" problems in other studies I don't feel like they would just drop the problem.
 >> Anonymous Wed Apr 17 13:31:59 2019 No.10566702 >>10566202No Anon, that's the disclaimer.
 >> Anonymous Wed Apr 17 13:52:07 2019 No.10566754 >>10566626Right, but is my proof good? It's just a stupid question. That's acknowledged. Why you are in a stupid questions thread acting surprised is a different question.
 >> Anonymous Wed Apr 17 13:54:51 2019 No.10566762 >>10566626Wrong. The topology T could contain nothing but {X,O,{a},{c,d},{a,c,d},{b,c,d,e}} and b would be a limit point of a subset {a,b,c)
 >> Anonymous Wed Apr 17 14:23:41 2019 No.10566869 >>10566762>implying {b} is open in your exampleViper.>>10566754No, it's horrendous. Still correct, but for the wrong reasons.
 >> Anonymous Wed Apr 17 15:08:39 2019 No.10567011 Why does Europa have so much water? If all the moons around Jupiter formed from leftovers of the same blob of matter, why wouldn't the water be more evenly distributed amongst them? Is it possible the more volcanically active ones DID have water originally, but outgassed and ejected them all due to tidal forces (because closer to Jupiter)? Or is it just "one of the moons got a fuckload of water for some reason"? If so, what the hell is the reason?
 >> Anonymous Wed Apr 17 20:36:13 2019 No.10567911 File: 86 KB, 1540x370, file.png [View same] [iqdb] [saucenao] [google] [report] did the author make an error transcribing this formula? seems incorrect to me
 >> Anonymous Wed Apr 17 20:44:20 2019 No.10567937 >>10567911>turns a function of n variables into a function of one variable times a function of n variablesI hate stats so much.But no, the formula isn't incorrect. It's just Bayes's theorem.
 >> Anonymous Wed Apr 17 20:53:27 2019 No.10567958 File: 42 KB, 1138x346, file.png [View same] [iqdb] [saucenao] [google] [report] >>10567937ok so please help me understand how the examples map to the definition. see pic related. let's take the first line. ostensibly, the first part of the first line -- P(a | b,c) -- maps to the second part of the right hand side: a maps to x^i, so b and c must map to x^1 and x^2 (i.e., x^i-1). clear enough. but the first part of the right hand side is P(x^1)... the second part of the first line of pic related is P(b,c), so two values, rather than one. we have an impedance mismatch here afaictwhat am i missing?
 >> Anonymous Wed Apr 17 20:56:30 2019 No.10567965 >>10567958It's literally "the probability of a, b and c is the probability of a happening given that b and c happened, times the probability of b and c happening."The second line is the usual Bayes, and the third line is just swapping the values in.
 >> Anonymous Wed Apr 17 21:01:33 2019 No.10567972 >>10567965i hear you and i understand the intuition. i'm just struggling to understand the notation. what specifically does x^1 represent?
 >> Anonymous Wed Apr 17 21:03:20 2019 No.10567976 >>10567972The event x^(1). Formally, x^(1) is a subset of the set of events that we consider as "x^(1) happened" and also call x^(1).
 >> Anonymous Wed Apr 17 21:04:19 2019 No.10567979 >>10567976but in the example it's "a", right?so why does nothing in the example reflect P(a)?
 >> Anonymous Wed Apr 17 21:10:41 2019 No.10567997 >>10567979Same same. a, x^(1), heads, you can call the event whatever.It doesn't reflect P(a) because it doesn't need to. There is no "immanent order of events", so we can consider it a and b happening simultaneously to be the same as a happening and then b happening or b happening and then a happening. Consider something like a is "a dice falling on an even number" and b is "a dice falling on two".Both events happen simultaneously, but we nonetheless know that if it falls on two, then it must fall on an even number, or that if it fell on an even number, it has a 1/3 chance of having fallen on 2.Notation wise, this reads P(b)P(a|b)=P(a)P(b|a);
 >> Anonymous Wed Apr 17 21:15:45 2019 No.10568006 >>10567997i apologize if i'm being dense so bear with me.the whole point of the example is to show how a joint probability of many variables can be decomposed into conditional distributions over only 1 variableP(x^1) is the one variableP(a,b,c) = P(a|b,c)*P(b,c) ostensibly is an example of the decomposition being appliedwhere is the one variable in from the decomposition
 >> Anonymous Wed Apr 17 21:27:47 2019 No.10568034 File: 28 KB, 1044x124, file.png [View same] [iqdb] [saucenao] [google] [report] >>10567997>>10568006here is how wikipedia defines the chain rule.i hope you can see why to me it seems like the book didn't define it properlyafaict the author is being inconsistent with notation and should not have referred to P(x^1)
 >> Anonymous Wed Apr 17 21:30:58 2019 No.10568045 >>10568034Anon, the comma and the cup stand for the same thing.>>10568006Yes.
 >> Anonymous Wed Apr 17 21:33:09 2019 No.10568052 >>10568045>Anon, the comma and the cup stand for the same thing.i know....?
 >> Anonymous Wed Apr 17 21:38:15 2019 No.10568062 am i being retarded about this? i feel like i havent gotten a straight answer to the question
 >> Anonymous Wed Apr 17 21:54:25 2019 No.10568099 fuck sorry
 >> Anonymous Wed Apr 17 22:39:39 2019 No.10568211 File: 6 KB, 676x139, nword.png [View same] [iqdb] [saucenao] [google] [report] okay, a little confused here
 >> Anonymous Wed Apr 17 22:54:13 2019 No.10568238 >>10568211literally just divide the interval into two simpler subintervals
 >> Anonymous Wed Apr 17 23:08:23 2019 No.10568264 How do we get people (especially young people) re-invested into society and healthy social structures?
 >> Anonymous Wed Apr 17 23:19:30 2019 No.10568292 >>10568211int f(x) from 0 to 1 + int f(tx) from 1 to 2
 >> Anonymous Wed Apr 17 23:20:20 2019 No.10568295 >>10568264by getting rid of weak fathers
 >> Anonymous Wed Apr 17 23:27:52 2019 No.10568320 >>10568295Weak fathers is just another symptom i think. I grew up without a father (mine died when i was under 10 due to heroin overdose) and despite having only a mom I turned out better than plenty of others so that's not a magic bullet in either case
 >> Anonymous Wed Apr 17 23:29:34 2019 No.10568326 >>10568320>Weak fathers is just another symptom i thinkweak fathers are a symptom of previous weak fathers
 >> Anonymous Wed Apr 17 23:32:21 2019 No.10568333 >>10568326Honestly, fair enough. It seems unfair to put the onus onto the guys but i suppose we are supposed to be the tough ones. We're the ones that are supposed to crack down on our children and make them be good and nice to the people around them, one way or another. The real question is how do we solve that.
 >> Anonymous Thu Apr 18 00:28:15 2019 No.10568439 Two things ive been thinking about. Harnessing energy from the earths magnetic field, and mass planting trees ti decrease carbon pollution
 >> Anonymous Thu Apr 18 01:18:42 2019 No.10568496 >>10568439>Harnessing energy from the earths magnetic fielddoable but ultimately not as worthwhile as solar or fusion (if doable).>mass planting trees ti decrease carbon pollutionWe can just suck out co2 of the air and sequester it into solid forms, its not even that costly anymore. The problem is methane is a more potent greenhouse gas. Artificial leaves will also probably start being using in architecture fairly soon when they are able to be readily produced as theyll be easier to maintain than natural plants
 >> Anonymous Thu Apr 18 06:09:36 2019 No.10568803 If you have the equation of a circle, how would you the find the values for c in y=c-x if you know that the straight line touches the circle at one point?
 >> Anonymous Thu Apr 18 06:49:36 2019 No.10568857 >>10568803I assume you mean a circle around the origin.If you want the line to be tangent to the circle and have slope -1, there's obviously two possibilities for that point: one is at $\vartheta = {\pi \over 4}$, the other at $\vartheta = {5\pi \over 4}$.In cartesian coordinates, solve for ${r \over \sqrt{2}} = c - {r \over \sqrt{2}}$
 >> Anonymous Thu Apr 18 09:42:18 2019 No.10569246 >>10568496ok nice, i had no idea i was shooting in the dark. loland im not sure where your getting your source. http://www.cleanerandgreener.org/resources/air-pollution.html
 >> Anonymous Thu Apr 18 10:31:49 2019 No.10569358 >>10568803you can set the two equations equal to each other in terms of c
 >> Anonymous Thu Apr 18 13:46:11 2019 No.10569821 >>10554849Write out the right side. Get it to a common denominator. Add up. Done
 >> Anonymous Thu Apr 18 13:59:45 2019 No.10569847   File: 3 KB, 436x91, 267.gif [View same] [iqdb] [saucenao] [google] [report] How do you go from step 1 to step 2?
 >> Anonymous Thu Apr 18 14:06:12 2019 No.10569854 >>10567911>>10567958He just split up the probability from left to right on top and from right to left on the bottom. This is possible because P(A,B) = P(B,A) and P(C)*P(D) = P(D)*P(C)
 >> Anonymous Thu Apr 18 14:14:20 2019 No.10569863 >>10569358I know that this is the answer somehow but I keep fucking it up. I just end up spending 20 lines of algebra to prove c=y+x or something dumb.The circle is $x^2 +y^2 -4x-2y-3=0$ and through graphing the values for c are 7 and -1 but how I arrive at those values are missing me
 >> Anonymous Thu Apr 18 15:32:38 2019 No.10570061 >>10569863Substituting y=c-x into the equation for the circle gives you a quadratic in x:2x^2-2(c+1)x+c^2-2c-3=0Applying the quadratic formula gives youx = ((c+1) ± sqrt(-c^2+6c+7))/2The discriminant is -c^2+6c+7. The equation has two real solutions if the discriminant is positive, one solution if it's zero, and no real solutions if it's negative. The line is tangent to the circle when the equation has a single solution, i.e. when the discriminant is zero:-c^2+6c+7=0=> -(c-7)(c+1)=0=> c=7 or c=-1
 >> Anonymous Thu Apr 18 15:40:40 2019 No.10570080 >>10568264Probably get rid of social media and celebrity culture and replace with STEM role models and people with moral fiber in the limelight.
 >> Anonymous Thu Apr 18 16:06:11 2019 No.10570149 >>10570061Thank you very much. I tried a variation of this but I think I got confused when it came to the values for the discriminant.
 >> Anonymous Thu Apr 18 16:39:07 2019 No.10570236 If stuff falling towards an event horizon, in the reference frame of a remote observer, appears to slow down and stop, doesn't that mean the apparent matter flux across the EH is zero (in said reference frame)? Does that mean BHs cannot be observed to grow?
 >> Anonymous Thu Apr 18 18:12:20 2019 No.10570455 Am I supposed to write it as "10ml" or "10 ml"? I'm assuming the latter, but I see the former a lot and don't know if I'm actually the one in the wrong.
 >> Anonymous Thu Apr 18 18:33:55 2019 No.10570507 >>10570455There should be a space between the number and the units.
 >> Anonymous Thu Apr 18 21:11:17 2019 No.10570822   File: 21 KB, 767x251, really now.png [View same] [iqdb] [saucenao] [google] [report] OpineThis is wrong?
 >> Anonymous Thu Apr 18 22:52:06 2019 No.10571015 What notation do I use if I want to get min(two non-numerical values) by supplying a function to map the values to numbers? e.g.:min(F(x), F(y)), if F(x) < F(y) then it would give x and not F(x)Of course I could just say "x if F(x) < F(y) otherwise y" .
 >> Anonymous Thu Apr 18 23:31:39 2019 No.10571089 >>10555351This is kinda covered. Humans have incredibly strong pattern recognition. This pattern recognition is a larger part of our conscious minds than most people suspect. As such, doing things that fit the pattern of something can produce an illusory effect since our own perceptions don't play as big a role in our thoughts. Sorry if this isn't understandable I'm a tired drunk esl
 >> Anonymous Thu Apr 18 23:35:59 2019 No.10571098 >>10555366>>10555378>>10571089Also yes culture and beliefs are more or less the source of where many hallucinatory images come from. Likewise, environment, culture, and expectations have a huge effect on drugs like LCD, alcohol, and others.
 >> Anonymous Fri Apr 19 05:33:02 2019 No.10571608 Submitting this one for stupidest question of the year please be gentleWhen an equation has π in it how many decimals of π are you supposed to use.
 >> Anonymous Fri Apr 19 05:56:00 2019 No.10571641 >>10571608Use the pi function on your calculate but by hand anyone using more than 3 is a try hard.
 >> Anonymous Fri Apr 19 05:59:08 2019 No.10571647 How much of basic lab work in any given field could be done with people who never studied the subjects?Like can you ask someone who has never studied molecular biology to do Western blotting after giving him of her a training?Just wondering how much "science" stuff is accessible to most yet locked behind a wall.
 >> Anonymous Fri Apr 19 06:01:45 2019 No.10571651 >>10571641>using decimals while calculating by handWho does this? Just use 22/7 or 355/113
 >> Anonymous Fri Apr 19 06:12:02 2019 No.10571670 >>10571608Depends, if the formula has Pi, you can use less than if the formula has Pi^10 in it.
 >> Anonymous Fri Apr 19 09:01:46 2019 No.10571952 Is just working through the recommended textbooks enough to be able to solve Putnam problems? Currently working though Spivak and a lot of Putnam problems still give me alot of trouble. I know they’re not the point of math but it’d still be nice to be able to solve them.
 >> Anonymous Fri Apr 19 09:45:23 2019 No.10572023 >>10571608It depends on the precison you want in your result.If you're an engineer, 3.14 covers you. If you're a mathematician, you might as well use literally no digits. An astronomer would use a whole lot more.
 >> Anonymous Fri Apr 19 11:30:35 2019 No.10572251 Why does matter settle into a flat ring around a planet instead of a full sphere?
 >> Anonymous Fri Apr 19 11:32:21 2019 No.10572257 >>10572251> Be earth> Have a moon that's a full sphere
 >> Anonymous Fri Apr 19 11:36:20 2019 No.10572264 >>10572257You're funny anon
 >> Anonymous Fri Apr 19 11:36:38 2019 No.10572266 >>10571608As many as the most precise number is in that formua. A circle with radius 1.1 m doesn't have a circumference of 6.9115038378975451246178154432149063452337726786252328 m. 1.1 m implies a precision of 0.05 m, so stay at that precision and just write 6.9 m
 >> Anonymous Fri Apr 19 12:48:51 2019 No.10572425 Taylor Polynomials. Obviously they're very useful and I already know of their applications. Obviously, they work.Yet, what I'm having trouble seeing/building intuition for is, if you necessarily have to have an explicit function to take its derivative (as far as my up-to-calc II knowledge goes, this is probably what's limiting me), how then do you use a taylor polynomial to model a function without an explicit definition?
 >> Anonymous Fri Apr 19 13:18:44 2019 No.10572496 >>10572425What exactly do you mean? Implicit functions?Remember that you don't need an algebraic expression inside the Taylor expansion. The nth derivatives _at a particular point_ are used. For finding these, even numerical methods might suffice depending on the application.
 >> Anonymous Fri Apr 19 13:57:50 2019 No.10572606 File: 191 KB, 253x280, firefox_iEYypnb2Ru.png [View same] [iqdb] [saucenao] [google] [report] I'm in a draw of 30 people. Two people will be picked.So first it'd be 1/30, then 1/29.How do I work out the overall probability of being picked in this draw?It's not simply 1/15 either, is it?
 >> Anonymous Fri Apr 19 14:10:24 2019 No.10572639 >>10572606$\left( \matrix{ 30 \\ 2 } \right)$ possible draws. 29 of them have you in it.
 >> Anonymous Fri Apr 19 14:43:18 2019 No.10572728 >>10572606They pick one person among thirty, and then pick the second among 29, giving 29x30. Since we don't care about order, that's 29x15 possible choices. 29 include you, so that makes 1/15 chance of you being picked.
 >> Anonymous Fri Apr 19 14:48:07 2019 No.10572747 >>10572606It actually is. Don't know how to tex it, but 29 / >>10572639 is 1/15. Alternative way to calculate:1/30 + 29/30 * 1/29 = 1/15[getting picked on the first pick] + [not gettin picked on the first pick and getting picked on the second one]
 >> Anonymous Fri Apr 19 16:20:50 2019 No.10572986 File: 1.13 MB, 535x301, tumblr.gif [View same] [iqdb] [saucenao] [google] [report] Take a prime natural NSubtract a prime natural X, where X < N If N-X < X then add N to the difference and subtract X. If you do this for N steps, you'll end up at 0.I'm a brainlet and I don't understand this. I know it's because of modular arithmetic, but why does it always seem to come up to N steps with primes?
 >> Anonymous Fri Apr 19 16:36:14 2019 No.10573028 >>10572986https://en.wikipedia.org/wiki/Field_(mathematics)Read the introduction and then the classical definition.https://en.wikipedia.org/wiki/Characteristic_(algebra)Read the introduction and then the case for fields.
 >> Anonymous Fri Apr 19 16:39:33 2019 No.10573037 >>10573028That was kinder than I was expecting, and very helpful.Thanks much anon
 >> Anonymous Fri Apr 19 17:52:47 2019 No.10573201 File: 7 KB, 661x131, bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb.png [View same] [iqdb] [saucenao] [google] [report] how do I evaluate this if there isn't an elementary way to integrate sqrt(sin(x))
 >> Anonymous Fri Apr 19 18:31:58 2019 No.10573282   >>10572425If a function $f(x)$ exists, explicit or not. All the taylor series asks for are the functions derivatives, $f^{(n)} (x)$. Some times, that's all you would know about a function: its value at a particular x, and the first and second derivative....sometimes more.Let say you have a graph of a function in a spread sheet. You know the value of the function at particular points. You also know an approximation of the derivative of a function at particular points.....and so on. You can use this to reconstruct an approximate explicit function.Taylor expansions are also useful to solve for difficult limits. Because a function behaves almost idiotically to it's partial taylor expansion in the neighborhood of it's limit point of interest.
 >> Anonymous Fri Apr 19 18:34:37 2019 No.10573295 >>10572425If a function f(x) exists, explicit or not. All the taylor series asks for are the functions derivatives, f(n)(x). Some times, that's all you would know about a function: its value at a particular x, and the first and second derivative....sometimes more.Let say you have a graph of a function in a spread sheet. You know the value of the function at particular points. You also know an approximation of the derivative of a function at particular points.....and so on. You can use this to reconstruct an approximate explicit function.Taylor expansions are also useful to solve for difficult limits. Because a function behaves identically to it's taylor expansion in the neighborhood of it's limit point of interest.
 >> Anonymous Fri Apr 19 18:36:38 2019 No.10573298 >>10573201You integrate the square. Every slice of your integral is actually a disk around the x-axis with radius y(x) and thickness dx. To get the volume, you calculate y^2 * dx. To sum all disks up, you integrate.
 >> Anonymous Fri Apr 19 18:38:56 2019 No.10573305 >>10573201You don't have to integrate sqrt(sin(x)). The square root actually makes the problem easier.
 >> Anonymous Fri Apr 19 18:41:09 2019 No.10573314 >>10573295Alright, now it's starting to make more sense. I'm sure once I get to diffeq it'll make a lot more sense.>>10573298>>10573305Thank you, it's been 4 months since we've covered it, and I'd completely forgotten the methods to find solids of rotation.
 >> Anonymous Fri Apr 19 18:43:06 2019 No.10573318 >>10573314You either remember the formula $\integral \left(y(x)\right)^2 \mathrm{d}x$ or you remember what you're trying to achieve, which is basically the thing with the disks from >>10573298
 >> Anonymous Fri Apr 19 19:00:00 2019 No.10573361 How do I calculate local sidereal time
 >> Anonymous Fri Apr 19 19:00:26 2019 No.10573364 >>10573314>I'm sure once I get to diffeq it'll make a lot more sense.You don't necessarily deal with the Taylor series in diffeq. You deal with the more general power series. Search for "Method of Frobenius" https://en.wikipedia.org/wiki/Frobenius_methodYou can solve for some differential equations using this method.
 >> Anonymous Fri Apr 19 19:40:21 2019 No.10573471 File: 3 KB, 262x42, helpimdumb.png [View same] [iqdb] [saucenao] [google] [report] So I need prove that pic related's a equivalence relation but I also need to list the equivalence classes the relation has, but if x^2=y^2 then x=y, right? so why do I need to list every equivalence class if that's the case?
 >> Anonymous Fri Apr 19 20:06:28 2019 No.10573526 >>10573471>but if x^2=y^2 then x=y, right?youre forgetting negatives
 >> Anonymous Fri Apr 19 20:07:53 2019 No.10573531 is campbell biology a good textbook to teach myself freshman biology?
 >> Anonymous Fri Apr 19 20:08:29 2019 No.10573534 >>10573471Gee, I wonder what are the equivalence classes. Surely that must be a problem for someone on Euler's level.
 >> Anonymous Sat Apr 20 12:04:20 2019 No.10574882 File: 48 KB, 420x375, 1350587791929.jpg [View same] [iqdb] [saucenao] [google] [report] Why does mathematics use 0 instead of +0 and -0?We have no problem thinking of infinity as a number that's constantly expanding, so why isn't zero treated as a number that's constantly collapsing into itself? Why isn't it given a positive or negative sign like other values?And I understand that the direct answer is "The math checks out", but just because an answer works in mathematics doesn't mean it has to be 100% correct right? Neutral zero being a substitute for positive zero and negative zero might work in certain cases, but in others, such as divided by zero, it does not. Zero might be such a small number that you can essentially say that it is worth nothing, but is that actually correct? Or just a hotfix solution?
 >> Anonymous Sat Apr 20 13:11:25 2019 No.10575013 >>10574882Maybe you'd like to read this essay by GZA, the genius? There's a lot of sci in it:>The importance of the “O” it’s shape and what it means to me and some of the things it reminds me of by GZA, “genius” >first of all the “O” represents a cipher the numeral 0 of a thing or person of no importance, [...]>The lens too a camera, binoculars, or telescope are all round they are like eyes which are also round enabling us to rotate or move it any direction. [...] >When dealing with mathematics or numbers the “O” represents 0 or nothing but when you put it next to any number it multiplies that number by 10 but when you remove the “O” the number is devised by 10, it’s the only number that has no value unless it stands next too another number.>Quote “together we stand, separated we fall”.[...]> The “O” or circular shape plays an important role in our lives, imagine if tires weren’t round the bicycle, the car, roller skates, the skateboard, the baby stroller, or even horse carriage probably wouldn’t exist, if so then travelling would be extremely difficult and uncomfortable.>The steering wheel in most motor vehicles has a circular pattern similar to the earth which is a clock in its self thats how we got 24 hours in a day. Even in the world of music the circular shape plays an important role from the vinyl records too the turn tables we play them on they are all round, from the compact disc’s that stores our music the headphones we use to listen to the songs, the drums that holds the beat too the woofers in the speakers are round, VHS, Cassettes tapes and film also play in circular motion.
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