[QUOTE=Swebonny;42644271]Just had my discrete mathematics midterms a few hours ago. Its result statistics is kinda worrying. Last year 40 % failed that exam and the rest got A 3%, B 6%, C 13%, D 16%, E 20%.
All I can think is in sets now.[/QUOTE]
I passed! Wooo.
[QUOTE=JohnnyMo1;42845658]Hey Joey, you're the only person I know of on FP who could potentially answer this.
I was talking to my professor in office hours and he claimed without proof that an isometry from a smooth manifold M to itself can't push a vector V at p forward into to W at p if the geodesic generated by V is closed and the geodesic generated by W is open, or vice versa. Is that true, and more importantly, why? I don't need a proof, just the intuition.[/QUOTE]
I haven't thought about it too hard, so I could be talking rubbish, but it seems that the stated direction is fairly clear - an isometry maps geodesics to geodesics, specifically if V is pushed forward to W, then the geodesic generated by V is mapped to the geodesic generated by W.
If the first is closed, then it can't be mapped smoothly into one which is open (since a loop can't be mapped [i]smoothly[/i] onto a line)
The converse is less obvious, and if it wasn't the same point it wouldn't be true (e.g. an isometry of a plane onto a cylinder). I haven't really thought about it hard enough to think whether it's obviously true (it's always obvious once you've thought about it enough), but I can't think of a counterexample... It seems pretty plausible at least, since isometries are in some sense a map showing the spaces are 'the same' (but only locally...)
It's certainly not always injective, since there is a non-injective isometry of a cylinder onto itself which doubles the angle of every point, but perhaps this is an oddity of zero curvature. Not sure.
I'm also thinking about all my examples as 2D manifolds, so more interesting stuff could happen at higher dimensions that I haven't even imagined. It's been too long (over 6 months) since I did this stuff properly!
[QUOTE=Joey90;42855352]I'm also thinking about all my examples as 2D manifolds, so more interesting stuff could happen at higher dimensions that I haven't even imagined.[/QUOTE]
Higher dimensions are fucked up. Lookin' at you, R^4!
Nerds.
[sp]I want to be like you.[/sp]
Joey90 are you a physics student?
[QUOTE=Swebonny;42855698]Joey90 are you a physics student?[/QUOTE]
Not quite, I was (up until June this year) a maths student. Which does mean I could pretty much just do theoretical physics if I wanted, but I generally steered towards the purer stuff.
I found a book on Galois theory by I.A. Stewart. Is it any good? It seems pretty interesting already....
To the maths students in here: Did any of you stop going to lectures after a few weeks in your first year? I've been to maybe 2 or 3 in the last 5 or 6 weeks
My first year? No. Honestly, I don't really ever skip math lectures if I can help it, except that I'm taking advanced calculus this semester and I've already taken real analysis which is basically the same thing only much harder and twice as long. I skip a bit of that.
I skip physics lectures all the time, though, but it's an awful habit. Don't do it. Just get in the habit of going even if you don't want to.
Haven't skipped a single lecture yet, but I've just started Freshman year. They actually ask questions in many of my lectures that we answer with little multiple-choice remotes, so they know if you skip and deduct points.
[editline]14th November 2013[/editline]
Well, not really just started, finals are in a month.
Ew, that sucks. As much as I think attendance is a good thing, I don't ever think it ought to be mandatory. I'm paying for this education dammit, I'll learn how I please!
The only reason I'm going to university is to learn math. I feel like I'd be wasting my life if I didn't go to lectures.
My campus is really small too, if you don't show up they know you're not there. My number theory class has eleven students.
I went on the first few Discrete Math lectures, then I skipped the rest of them. I just felt that the teacher was reading straight from the book. I generally just skip thing I know I'm sure about.
[QUOTE=sambooo;42861043]To the maths students in here: Did any of you stop going to lectures after a few weeks in your first year? I've been to maybe 2 or 3 in the last 5 or 6 weeks[/QUOTE]
Definitely not - after the first week or so if you missed a lecture then you were basically stuffed until you'd managed to get someone else's notes and go through them yourself.
There were a tiny handful of people who didn't go to any lectures and tried to learn everything via books, but they generally struggled (unless they'd done a similar course at a different university or something).
My courses potentially were at an unusually rapid pace though [sp]at Cambridge[/sp]
Mayer-Vietoris is pretty neat. Made calculating the homology groups of the figure eight feel pretty easy.
[QUOTE=sambooo;42861043]To the maths students in here: Did any of you stop going to lectures after a few weeks in your first year? I've been to maybe 2 or 3 in the last 5 or 6 weeks[/QUOTE]
My lectures are online, so I can't miss them. History and English are a different story, but I've only missed a couple of those due to being in the hospital and dealing with issues at home. I've never understood the point of skipping, unless you absolutely know the material already. I'm at a small college, but I still pay around $15 per lecture on average, and I'm going to get my money's worth!
[QUOTE=Joey90;42863147]Definitely not - after the first week or so if you missed a lecture then you were basically stuffed until you'd managed to get someone else's notes and go through them yourself.
There were a tiny handful of people who didn't go to any lectures and tried to learn everything via books, but they generally struggled (unless they'd done a similar course at a different university or something).
My courses potentially were at an unusually rapid pace though [sp]at Cambridge[/sp][/QUOTE]
I'm only at Nottingham and it's way different for me here. We have reading lists but almost no one ever has anything from them in first year since you don't need them. I have 3 other maths friends living in my block and we've all stopped attending since the maths is easy enough that we can catch up in next to no time. Started revising for an exam 4 hours before it started today and got all but one question right.
I'm hoping I'll be more motivated to go into lectures when the material stops being slow and trivial but I'll probably prioritise Presidential stuff over it anyway.
Recently my maths teacher found it extremely odd that I could mentally solve something like 59*11 and 49*49 fairly quickly yet had a lot of trouble doing 7*8. The 7 and 8 times table are hard for me to get along with for some reason. Does anyone find themselves sometimes getting affected by even the basics?
Hey guys, I'm finishing up my assignment and I'm kinda stuck on the very last question. I took a picture of the question for you to check out.
[img_thumb]http://img199.imageshack.us/img199/884/fwb0.jpg[/img_thumb]
Now, my professor is the kinda guy where he doesn't really give you any hints in the classroom, it's kind of a fend for yourself until you go see him in his office, in which he still doesn't do that great of a job explaining it.
I think it has something to do with finding Area of Polar Coordinates, but I'm not quite sure where to start. I couldn't find anything in my notes of what we did in class that pertains to this type of problem.
[QUOTE=GreenDolphin;42895551]Recently my maths teacher found it extremely odd that I could mentally solve something like 59*11 and 49*49 fairly quickly yet had a lot of trouble doing 7*8. The 7 and 8 times table are hard for me to get along with for some reason. Does anyone find themselves sometimes getting affected by even the basics?[/QUOTE]
64-8?
[QUOTE=Falubii;42896767]64-8?[/QUOTE]
That would also refer to finding 8*8 embarrassingly enough for me.
[QUOTE=Zareox7;42896338]Hey guys, I'm finishing up my assignment and I'm kinda stuck on the very last question. I took a picture of the question for you to check out.
[img_thumb]http://img199.imageshack.us/img199/884/fwb0.jpg[/img_thumb]
Now, my professor is the kinda guy where he doesn't really give you any hints in the classroom, it's kind of a fend for yourself until you go see him in his office, in which he still doesn't do that great of a job explaining it.
I think it has something to do with finding Area of Polar Coordinates, but I'm not quite sure where to start. I couldn't find anything in my notes of what we did in class that pertains to this type of problem.[/QUOTE]
The rope is half the circumference of the circle so r = 2πx/2 = πx where x is the radius of the silo. Now take the area of r (2πr²) and subtract the area of the silo (2πx²)
Thus 2πr²-2πx²
Since r=πx, and x=r/π replacing x to be in terms of r; 2πr²-(2πr²/π)
=2πr²-2r²
=2r²(π-1)
(Somebody correct me if I'm wrong, but that's just a general overview of how I'd approach that question.)
That would mean the rope could go through the silo.
[QUOTE=Falubii;42897462]That would mean the rope could go through the silo.[/QUOTE]
[QUOTE=mastoner20;42897245]The rope is half the circumference of the circle so r = 2πx/2 = πx where x is the radius of the silo. Now take the area of r (2πr²) and [B]subtract the area of the silo[/B] (2πx²)
[/QUOTE]
[QUOTE=mastoner20;42897245]The rope is half the circumference of the circle so r = 2πx/2 = πx where x is the radius of the silo. Now take the area of r (2πr²) and subtract the area of the silo (2πx²)
Thus 2πr²-2πx²
Since r=πx, and x=r/π replacing x to be in terms of r; 2πr²-(2πr²/π)
=2πr²-2r²
=2r²(π-1)
(Somebody correct me if I'm wrong, but that's just a general overview of how I'd approach that question.)[/QUOTE]
As Falubii says, you're just treating the area the cow traces out as a semicircle, which it certainly isn't, it looks something like a Cardioid (although I don't think it actually is one). Also relabeling something they give in the question and then using the same name for something else is very confusing (I'm talking about r).
-snip-
I've actually got a cleverer method. Coming soon (maybe)
First split the problem in half across the middle, then break it up again:
[img]https://dl.dropboxusercontent.com/u/4081470/silol.png[/img]
The green section is easy - quarter of a circle radius [;\pi r;] has area
[;\frac{1}{4} r^2 \pi^3;]
The blue section we need to do an integral. After careful consideration, the best thing to integrate over is actually the piece length of rope [i]not[/i] pressed against the silo, marked [;x;] below:
[img]https://dl.dropboxusercontent.com/u/4081470/silol2.png[/img]
Set alpha to be the angle round from the centre. Now consider 'unwinding' by [;dx;], this corresponds to a change in angle of d alpha. Conveniently, since [;x;] and [;\alpha;] are directly proportional, [;dx = r d \alpha;]
We want to calculate the yellow shaded area, and we do this by approximating it as a sector of a circle. In which case it has area
[;\approx \frac{1}{2}x^2 d \alpha = \frac{1}{2r}x^2 dx;]
Now we can integrate over [;x;] from [;0;] to [;\pi r;]:
[;\frac{1}{2 r} \int_0^{\pi r} x^2 dx;]
[;=\frac{1}{6 r}\left[ x^3 \right]_0^{\pi r};]
[;=\frac{1}{6}r^2 \pi^3;]
adding the two together gets you
[;\frac{5}{12}r^2 \pi^3;]
then double it (since we've only done half) to get
[;\frac{5}{6}r^2 \pi^3;]
[QUOTE=mastoner20;42897609]stuff[/QUOTE]
This doesn't account for that. Subtracting the area of the silo only means the cow can't intersect it, but not the rope.
You know that the cow can walk around the silo, and at the point opposite the tether, the cow will be touching the silo since the rope only goes halfway around. However, if the area available for grazing is a circle with a radius pi times that of the silo, then he should be able to be quite a distance from the silo when he reaches the opposite side of it. Something is wrong here - in actuality you lose some rope as you move around the silo.
The solution (probably, I have no idea) might require integrating some polar function.
[QUOTE=Joey90;42905394]First split the problem in half across the middle, then break it up again:
The green section is easy - quarter of a circle radius pi r = (r^2 pi^3)/4
The blue section we need to do an integral. After careful consideration, the best thing to integrate over is actually the piece length of rope [i]not[/i] pressed against the silo, marked x below:
Set alpha to be the angle round from the centre. Now consider 'unwinding' by dx, this corresponds to a change in angle of d alpha. Conveniently, since x and alpha are directly proportional, dx = r d alpha.
We want to calculate the yellow shaded area, and we do this by approximating it as a sector of a circle. In which case it has area x(x+dx)d alpha/2 = x(x+dx)dx/2r which approximates to x^2/2r dx.
Now we can integrate over x from 0 to pi r:
1/2r integral 0 to (pi r) x^2 dx
= (r^2 pi^3)/6
adding the two together gets you 5(r^2 pi^3)/12
then double it (since we've only done half) to get 5(r^2 pi^3)/6[/QUOTE]
This problem could be greatly simplified if this oppressive farmer allowed his cows to be free range.
Edit:
Also, carets are lame. What are the chances this forum will ever get superscripts/subscripts?
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