What if it's in R[sup]n[/sup] but all the disks must have a nonzero radius?
[QUOTE=Krinkels;39948399]What if it's in R[sup]n[/sup] but all the disks must have a nonzero radius?[/QUOTE]
This:
[QUOTE=Joey90;39947467]It's [i]not[/i] possible in R[sup]n[/sup] - a union of more than one open discs is not connected (just take one disc and the union of all the others) whereas a larger disc clearly is connected, thus they can't make up a larger disc.[/QUOTE]
[QUOTE=Jellyman;39908983]Yeah, that has to be the best part about learning math for me, it feels so satisfying once you get something down.[/QUOTE]
This is why I intend on doing Physics + Maths at university now as opposed to sociology/psychology/politics.
It actually feels satisfying to do the work.
[QUOTE=JohnnyMo1;39949619]This:[/QUOTE]
Couldn't they be closed?
[QUOTE=Krinkels;39951586]Couldn't they be closed?[/QUOTE]
That's an interesting question, and I'm not actually sure (at 2am in the morning) what the answer is.
Certainly in R it's possible - a disc is then just an interval, and you can certainly join closed intervals with disjoint interior to get a bigger closed interval.
You could rephrase it as 'Can a disjoint union of open discs be a [i]dense[/i] set in a larger disc.' The answer to which I think is [i]almost certainly[/i] yes - Usually via some kind of construction where you place a disc at all points with rational coordinates. I can't think of an actual proof at the moment, but I may do in the morning.
It's obvious, of course, that such a construction would have discs of arbitrarily small size, but they would be nonzero.
[QUOTE=Krinkels;39951586]Couldn't they be closed?[/QUOTE]
oh yeah wasn't paying attention and thought you were referring to open balls in R[sup]n[/sup]
[editline]17th March 2013[/editline]
[QUOTE=Joey90;39952469]You could rephrase it as 'Can a disjoint union of open discs be a [i]dense[/i] set in a larger disc.' The answer to which I think is [i]almost certainly[/i] yes - Usually via some kind of construction where you place a disc at all points with rational coordinates. I can't think of an actual proof at the moment, but I may do in the morning.[/QUOTE]
I don't see how they could be disjoint by that construction. If there is a disk at a point with rational coordinates (e.g. in R[sup]1[/sup] for simplicity's sake), for any finite radius the disk must contain another rational point. So if there is a disk at every point with rational coordinates, the discs must intersect.
[QUOTE=JohnnyMo1;39952932]
I don't see how they could be disjoint by that construction. If there is a disk at a point with rational coordinates (e.g. in R[sup]1[/sup] for simplicity's sake), for any finite radius the disk must contain another rational point. So if there is a disk at every point with rational coordinates, the discs must intersect.[/QUOTE]
Yeah, I guess I don't mean [i]all[/i] rational points. How about something like this:
Choose an open subspace X of R[sup]n[/sup] (in this case we are considering a large open disc)
The rational points are countable, so order them x[sub]1[/sub] x[sub]2[/sub] etc.
Now at each step choose the least n such that x[sub]n[/sub] is [i]not[/i] covered by [i]the closure of[/i] existing discs, and put an open disc at that point of radius small enough that it doesn't intersect any others.
It's always possible to do this since the complement of the closures of the existing discs (in X) is an open set, so any point in it has an open disc around it (still in X).
Thus you get a countable set of disjoint open discs, where the union of the closures cover every rational point in X, so must intersect X at a dense set, but this means it must equal the closure of X (i.e. they equal the large closed disc).
What do you mean by discs being "connected"?
They are not the disjoint union of two open sets.
Why do trigonometric functions exist, and how can I come to stop hating them?
To describe the sides of right triangles as well as the coordinates of points on a circle. Are you in pre-calc right now? In pre-calc I had to memorize a ton of values and prove trig identities. After you get through that you'll only need to know the values on the unit circle. And most frequently just where they equal -1, 0, and 1.
[QUOTE=credesniper;39972729]Why do trigonometric functions exist, and how can I come to stop hating them?[/QUOTE]
Watchu talkin bout willis trig functions are great they have nice simple derivatives and integrals
[img]http://i.imgur.com/2HX1bwB.png[/img]
how should I start by proving this by Combinatorial proof?
Welp.
Rewriting n choose k in terms of factorials and using induction might work.
E: Oh wait that's not combinatorial sorry
[QUOTE=Krinkels;39974502]Rewriting n choose k in terms of factorials and using induction might work.[/QUOTE]
This. If there's a thing you have to prove for every n then you will have to use induction 99% of the time.
[QUOTE=Mr_Razzums;39974293][img]http://i.imgur.com/2HX1bwB.png[/img]
how should I start by proving this by Combinatorial proof?
Welp.[/QUOTE]
If you want to do a proof by induction you will need to work out the same series with just a k in it (not k[sup]2[/sup]) so if you've done that, then it's ok (or you can also prove that series by induction, if you can guess what the answer needs to be...).
A snazzy (but slightly non-obvious) way to do it is by taking the binomial expansion for (1+x)[sup]n[/sup], differentiating once and twice, then letting x=1 in each of them.
A strictly combinatorial way of doing it would be something like this:
Take a set of n objects, then k[sup]2[/sup](n choose k) is the number of subsets of size k with 2 (possibly coinciding) distinguished points.
Thus the sum of these is the number of possible subsets (of any size) with 2 distinguished points.
To count these, you look at the two possibilities:
If the distinguished points are distinct, pick those first in n(n-1) ways, then you just have to choose a subset from the n-2 remaining points, of which there are 2[sup]n-2[/sup]
If the distinguished points are the same, pick that point in n ways, then choose a subset from the remaining n-1 points, of which there are 2[sup]n-1[/sup]
Hence in total there are n(n-1)2[sup]n-2[/sup]+n2[sup]n-1[/sup] such subsets, as required.
To come up with this, I looked at k[sup]2[/sup](n choose k) to try and think what it could be counting. The n choose k obviously suggests taking a subset of k objects from n, but then what could have k[sup]2[/sup] possibilities on this set? Answer: pick one of the k objects, then pick one again.
[QUOTE=JohnnyMo1;39973302]Watchu talkin bout willis trig functions are great they have nice simple derivatives and integrals[/QUOTE]
They're super easy providing you got a decent education on identities and all that. I was not taught trig identities and characteristics very well so I have to look up/Wolfram Alpha every problem with a trig identity that isn't super simple like cos or sin.
Speaking of that, how would you go about finding something like the 5000th derivative of cos? I mean, it's cyclic so it'd either be -sin, -cos, or sin but is there some way you can set it up using notation?
Well, it cycles every 4, so you'd do 5000 mod 4 = 0 which is just cos.
As for notation you could write it like (d[sup]5000[/sup]y / dx[sup]5000[/sup]) or, assuming f(x) = cos(x), f[sup](5000)[/sup](x)
[QUOTE=Joey90;39975574]If you want to do a proof by induction you will need to work out the same series with just a k in it (not k[sup]2[/sup]) so if you've done that, then it's ok (or you can also prove that series by induction, if you can guess what the answer needs to be...).
A snazzy (but slightly non-obvious) way to do it is by taking the binomial expansion for (1+x)[sup]n[/sup], differentiating once and twice, then letting x=1 in each of them.
A strictly combinatorial way of doing it would be something like this:
Take a set of n objects, then k[sup]2[/sup](n choose k) is the number of subsets of size k with 2 (possibly coinciding) distinguished points.
Thus the sum of these is the number of possible subsets (of any size) with 2 distinguished points.
To count these, you look at the two possibilities:
If the distinguished points are distinct, pick those first in n(n-1) ways, then you just have to choose a subset from the n-2 remaining points, of which there are 2[sup]n-2[/sup]
If the distinguished points are the same, pick that point in n ways, then choose a subset from the remaining n-1 points, of which there are 2[sup]n-1[/sup]
Hence in total there are n(n-1)2[sup]n-2[/sup]+n2[sup]n-1[/sup] such subsets, as required.
To come up with this, I looked at k[sup]2[/sup](n choose k) to try and think what it could be counting. The n choose k obviously suggests taking a subset of k objects from n, but then what could have k[sup]2[/sup] possibilities on this set? Answer: pick one of the k objects, then pick one again.[/QUOTE]
Alright! After reading your post about 10 times, I think I'm starting to understand.
Can you elaborate a bit more here:
[QUOTE]
If the distinguished points are distinct, pick those first in n(n-1) ways, then you just have to choose a subset from the n-2 remaining points, of which there are 2n-2
If the distinguished points are the same, pick that point in n ways, then choose a subset from the remaining n-1 points, of which there are 2n-1
[/QUOTE]
I'm just a little confused on how you came up with those figures. Sorry, I'm very new to counting.
Thank you!
[QUOTE=account;39975657]Well, it cycles every 4, so you'd do 5000 mod 4 = 0 which is just cos.
As for notation you could write it like (d[sup]5000[/sup]y / dx[sup]5000[/sup]) or, assuming f(x) = cos(x), f[sup](5000)[/sup](x)[/QUOTE]
sin[sup](n)[/sup](x) = sin(x + nπ/2) is clearly the superior way to do it.
shit i am very out of practice with my integrals and its starting to bite me in the ass in physics
once I started higher-level math courses, where everything is proof based and explicit calculation of integrals is almost never done anymore, I started to lose my ability to do them and now quantum mechanics is like 90% ugly integrals and it sucks
Flux! I was trying to remember that term earlier. Somehow I managed to forget it.
Man it's been a while.
Yeah that's not necessarily flux but it could be.
[QUOTE=Mr_Razzums;39975824]Alright! After reading your post about 10 times, I think I'm starting to understand.
Can you elaborate a bit more here:
[quote]
If the distinguished points are distinct, pick those first in n(n-1) ways, then you just have to choose a subset from the n-2 remaining points, of which there are 2n-2
If the distinguished points are the same, pick that point in n ways, then choose a subset from the remaining n-1 points, of which there are 2n-1[/quote]
I'm just a little confused on how you came up with those figures. Sorry, I'm very new to counting.
Thank you![/QUOTE]
So you're looking for subsets which have 2 distinguished points, but to count them in a different way, we pick the distinguished points [i]first[/i] and then the rest of the subset.
If they're different points, there are n choices for the first point, which leaves n-1 choices for the second. Then to choose our subset we take those two points along with some subset of the remaining n-2 points, of which there are 2[sup]n-2[/sup]
e.g. suppose you have n=5, call your n points A, B, C, D, E
First you pick two distinguished points, say A and B. Order [i]does[/i] matter, so lets label them with a dash and an asterisk: A', B*, C, D, E. There are 5(5-1) = 20 arrangements for the dash and asterisk (on different letters)
Now to pick a subset with these points, you need to choose a subset from the unmarked points - in this case from C, D, E... If we choose for example {C, D} then we get in total {A', B*, C, D} and there were 2[sup]5-2[/sup] = 8 possible subsets to choose from.
(The number of subsets of a set of size k is 2[sup]k[/sup], since you choose for each element whether it's in the subset or not, so get 2*2*...*2 [k times] possibilities)
Overall then, there are 20*8 = 160 possible subsets containing two distinct distinguished points.
For the other case, when the two points are the same, then you just do the same thing, but treat it as one distinguished point A'*, B, C, D, E. For which there are 5 choices etc.
I hope that helps, it's hard to explain much more clearly without being in person to wave my arms a lot!
That helps a ton. Thanks man
Easy math for you experts here. What I basically have to do is create a rational function that meets the given criteria.
[quote]Horizontal Asymptote y=2
Vertical Asymptote x=1
Hole (-1,3)[/quote]
For this one, I can't figure out how to make a hole at (-1,3). I can only figure out how to make it at x=-1.
[quote]Vertical Asymptote x=3
Oblique Asymptote y=2x-3[/quote]
For this one, I'm somewhat clueless. I ended up multiplying a vertical asymptote (x-3) by 2x-3 in order to find the numerator, but I don't think I can do that since it ends up creating a hole at x=3 instead of a vertical asymptote.
[editline]20th March 2013[/editline]
Ok I think after asking around and sacrificing a few children to the math gods, I was able to figure it out.
For the first one I got f(x)=2(x^2-x-2)/(x^2-1). For the second one I got f(x)=(2x^2 - 9x + 10)/(x - 3)
[QUOTE=titopei;39984036]For this one, I can't figure out how to make a hole at (-1,3). I can only figure out how to make it at [/QUOTE]
(x+1)(x-3)/(x+1)(x-3)
I figured it out, but doesn't that just make holes at x=-1,3?
Ahh, I misread.
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