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Jun 7 2018 03:05pm
Quote (Hanako @ Jun 7 2018 03:01pm)
boring as fuck

for some reasons the image is rotated, but right click -> open image in a new tab does the trick

http://i65.tinypic.com/28k6iqp.jpg



What surprised me the most is that someone has really made it :D
Calculus is pretty funny (sometimes), but this integral is indeed a bit exhausting.
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Jun 8 2018 02:29am
Guys. Another one here:

Maria goes to Walmart. She buys 23 Watermelons, 4 onions, 3 bags of chips, 14 salmons and a glass of pickles. The Watermelons cost twice as much as the chips and 1 salmon is 2/3 of the price of the onions. These are equal to the price of the chips. She pays 54.76$.

Whats the colour of her leggins?

This post was edited by TheBabyPixel on Jun 8 2018 02:31am
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Jun 8 2018 05:16am
Quote (feanur @ 7 Jun 2018 21:16)
Here's a little one :

Suppose you have a square garden of side L, and you decide to plant 2 trees at random (meaning X-coordinate and Y-coordinate are both evenly chosen between 0 and L).

What is the average distance between the trees ?


1/15 L (2 + sqrt(2) + 5 arcsinh(1))

This is interesting but convolving is boring



Quote (TheBabyPixel @ 8 Jun 2018 10:29)
Guys. Another one here:

Maria goes to Walmart. She buys 23 Watermelons, 4 onions, 3 bags of chips, 14 salmons and a glass of pickles. The Watermelons cost twice as much as the chips and 1 salmon is 2/3 of the price of the onions. These are equal to the price of the chips. She pays 54.76$.

Whats the colour of her leggins?


Probably gold or bright green because only a n would buy so many watermelons and ns are tacky as fuck

This post was edited by Hanako on Jun 8 2018 05:29am
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Jun 9 2018 08:12am
Quote (Hanako @ Jun 8 2018 12:16pm)
1/15 L (2 + sqrt(2) + 5 arcsinh(1))

This is interesting but convolving is boring

http://i68.tinypic.com/wahbv9.jpg

(...)


Congratulations !

I can see that electronic methods have gone very far ... That's impressive.

Now, what about a paper proof ?
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Jun 9 2018 03:05pm
Quote (Hanako @ Jun 7 2018 02:22pm)
Your existence is incorrect, I can't really be arsed to find where it went in the wrong though (probably right at birth, even though you're now going to pretend you were trolling)
Now listen here, I'm always right, premed brainlet. As I said it depends on your definition of "intersect" for the green line but both of the lines I found definitely share the point B with the original polynomial curve and are tangent at a point (a,f(a)), which you did not bother to define so I could pick any a.
(and assuming Y is the plot line associated to y, which you did not define either)
(and assuming the tangent of f(x) at A is the tangent of f at A, else it is meaningless)
According to variations they are the only ones with these properties

http://i68.tinypic.com/f2owoi.png



and you're 25? damn, med students really are bainlets. The meme is actually true. I hope you fail because having doctors this slow being surgeons and shit is really scary.
There is nothing to know to solve this problem (the only prerequisites are extreme boredom and patience to try and understand your ill-stated problem which was supposed to be entertaining)


Mate, this isn't the kind of stuff I use -ever-. I browsed through it for 6 weeks for the MCAT and nothing remotely related to maths for the 5 years before that. I am human, I forget things. You would too.

If 13 is one solution you can find the other by factoring (x-13) out of x^3 - 11x^2 - 13 - p*x + 13p. Quotient is x^2 + 2x + (1-p). I think you'll find that p and q are both 0.
/e But I admit your answer looks pretty damn correct now that I bother to actually look at it lol, but I would define an intersection as "f(x) = g(x) /\ f ' (x) =/= g ' (x)", that is to say, their slopes differ. If they would not differ, it would not be an intersection but a tangent. But that's my interpretation.

This post was edited by Forg0tten on Jun 9 2018 03:09pm
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Jun 11 2018 12:46pm
Quote (Forg0tten @ 9 Jun 2018 23:05)
Mate, this isn't the kind of stuff I use -ever-. I browsed through it for 6 weeks for the MCAT and nothing remotely related to maths for the 5 years before that. I am human, I forget things. You would too.

If 13 is one solution you can find the other by factoring (x-13) out of x^3 - 11x^2 - 13 - p*x + 13p. Quotient is x^2 + 2x + (1-p). I think you'll find that p and q are both 0.
/e But I admit your answer looks pretty damn correct now that I bother to actually look at it lol, but I would define an intersection as "f(x) = g(x) /\ f ' (x) =/= g ' (x)", that is to say, their slopes differ. If they would not differ, it would not be an intersection but a tangent. But that's my interpretation.


So the curve associated to the cube function does not intersect the x-axis?

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Jun 11 2018 02:27pm
Okay, you've got a valid point, my description is flawed :) Is this also true for the negative p+q value you found?
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Jun 11 2018 03:41pm
Quote (Forg0tten @ 11 Jun 2018 22:27)
Okay, you've got a valid point, my description is flawed :) Is this also true for the negative p+q value you found?


If an intersection is simply a shared point, then there are indeed two solutions, (p,q) = (0,0) and (p,q) = (196,-2548), i.e. p + q = 0 and p + q = -2352, because a tangent does share a point with the curve it is tangent to.
Also, I think it's weird that they specifically ask for p+q instead of (p,q)... unless there is a trick which allows for the determination of p+q directly
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Jun 12 2018 02:57am
Yeah, I've heard quite a few ways of solving this particular exercise so far.

This one's way out of my league, but an acquaintance of mine sometimes drop these for me to try and interest me back into the world of maths I was once into :p
Maybe you'll make more sense of it and can appreciate it.
Copied from his wall:



As a self-learner in the math and sciences, I found this research rather interesting. The ‘Lyapunov exponent’ characterizes the rate of separation of infinitesimally close trajectories Z(t) and Z₀(t) of a dynamical system in phase space, such that |δZ(t)| ≈ exp(λt) · |δZ₀ | where λ is the Lyapunov exponent. (See graphic). In other words, it describes when a system diverges into pure chaos. You can have a spectrum of Lyapunov exponents in a dynamical system, with the largest Lyapunov exponent determining limits on system predictability. In fact, the inverse of the largest exponent represents the ‘Lyapunov time’ until chaos wins. Mathematically, the ‘Kuramoto-Sivashinsky equation’ describes the dynamics under investigation in this particular research. However, an extension neural network computation framework called ‘Reservoir Computing’ is able to model this nonlinear system without any use of the Kuramoto-Sivashinsky equation, but merely based on its internal reservoir map and randomly recurrent connected structure. While this is remarkable, I’m reminded of a recent observation by Google’s A.I. researcher, Ali Rahimi, that machine learning has become like a form of alchemy. By that, he meant, without any in depth understanding of how benchmarks are achieved by these ‘black box’ processes, is this necessarily a good thing. More to the point, if there’s little ‘interpretability’ beyond the limits of our own computational abilities, do we continue to trust results from black box methods. Obviously, the hope is that such interpretability improves significantly, going forward in A.I. research. Since large scale brain oscillatory dynamics are also nonlinear, understanding those processes, along with, the artificial computational frameworks to emulate them, might possibly one day be integral to emulating human-like cognition in machines. Very interesting research!
Reference to article: https://www.quantamagazine.org/machine-learnings-amazing-ability-to-predict-chaos-20180418/
Address of attached image to this post: https://scontent-amt2-1.xx.fbcdn.net/v/t1.0-9/33869811_10212022125294024_2810992804593926144_n.jpg?_nc_cat=0&oh=a9200433fc218be1a226f9512db78da4&oe=5BA7671E
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