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Comments:
<0> # Reduce [ x + 10 > 7 - y]
<0> % Reduce [ x + 10 > 7 - y]
<0> % Reduce[x+10>7-y]
<0> I ****.
<1> K^Holtz : http://mathworld.wolfram.com/Multichoose.html
<2> bluefoxicy: mbot cant hear you whenit'sa not on irc, funnily enough
<3> mbot's a jerk that way
<4> implicit functions ftl!
<5> hi
<6> 'ftl' ?
<2> i believe the young folk use that to mean "for the lose"
<5> I have been spending hours trying to prove a binomial theorem using induction theorem :(
<3> true.
<5> it is sum(k=0 , n ) (n k) = s^n
<4> _llll_: I hope that even if i make it to age 60, i will use these :p
<0> _llll_: oh
<7> tubbie: where s=2
<7> tubbie: hint: look at (1+1)^n
<5> yea, it should be 2
<8> % 1/I
<8> Eh.
<7> no mbot
<1> I/-1
<7> -I
<3> K^Holtz: 4^8
<8> Thanks.
<9> ski: http://www.computerhope.com/jargon/f/ftw.htm
<9> Shorthand for For The Win, FTW is often using in gaming and gaming related definitions to indicate a user's win. The complete opposite to FTW would be FTL, which is short for For The Loss.
<6> mhm
<6> (indicate how ?)
<5> int-e: the problem is I have to prove the sum of (n k) = 2^n, not the binomial theorem itself :\
<1> Use the binomial theorem to prove it.
<3> it's a line from that wretched tom bergeron whoopi goldberg game show
<7> tubbie: can you use the binomial theorem?
<3> hollywood squares
<5> int-e: it was not given during cl***es
<7> tubbie: hmm. how about (n k) + (n (k+1)) = ((n+1) (k+1)) ?
<5> I have proven that one a few minutes ago
<7> it follows from that by induction.
<7> if combinatoric arguments are allowed, remember that 2^n is the number of subsets of an n-element set.
<5> I'm only allowed to use induction theoreom
<3> you don't need the binomial theorem to prove that \sum C(n,k) = 2^n by induction
<5> now I have sum (N+1, k = 0 )(N+1 k)
<3> (you wouldn't even quote induction if using the BT, anyway)
<5> equal to 1 + sum(k=0 , N) C(N+1,k)
<3> yes
<3> you're going to want to peel off the C(N+1,0) and C(N+1,N+1) terms, apply pascal's rule to each of the remaining terms, and then regroup what's left and split the 1's you took off between them appropriately.
<4> memories of the horror of discrete math !
<4> I lost vital internal organs in that course
<7> why horror?
<10> <3 combinatorics
<7> this is all pretty.
<4> Its fun, sometimes, but the exams, here atleast, are insanely hard and not fun :p
<4> like the "beauty" of generating functions
<4> All i remember is that I was terrorized
<11> First homework in my algorithms cl*** we were told generating functions existed, and then given problems
<5> HiLander: why should I peel off C(N+1,0)?
<11> Decided that if there wasn't going to be any discussion in cl*** on topics on tests, it was a good idea to drop it =p
<7> tubbie: to avoid having to use C(N,-1)
<5> oh, yea. indeed
<5> then I got this. 1 + sum(k=1, N) C(N+1,k)
<5> then I got this. 1 + sum(k=1, N) C(N+1,k) + 1
<3> so far, so dangerous.
<3> keep going.
<7> Hmm, what's the precedence of the sum notation?
<5> C(N+1,k)= C(N,k)+C(N,k-1)
<3> int-e: ?
<7> How should I read Sum(k=1, n) a_i + b_i ?
<7> hmm. the index should be i.
<3> if b's subscript is the index of summation, i'd ***ume it was in the sum
<3> otherwise, i think it's generally ambiguous.
<12> given {1,2,...,n}, can I compute how many subsets there are of k items?
<5> I got this so far. 2 + sum (k=1,N)C(N,k) + sum(k=1,N)C(N,k-1)
<3> tubbie: good.
<5> is 1 + sum (k=0,N)C(N,k) + sum(k=1,N)C(N,k-1)
<3> myrkraverk: that depends on how clever you are
<5> is 1 + 2^N + sum(k=1,N)C(N,k-1)
<12> HiLander: I just want to use a computer to do it for me ;)
<5> HiLander: I dunno what to do with the last part :(
<7> change the variable (k) to make it easier.
<5> l = k -1?
<7> yes
<12> HiLander: someone told me to use sterling, is that true?
<5> but what would k=0 become then?
<7> k=0 isn't in the sum
<3> myrkraverk: for subsets of size k? not so much.
<7> it starts at 1
<3> that's the important part.
<5> then it would become sum (l = 0,N)C(N,l)
<12> HiLander: not so much? (I don't really use sterling, or not so many subsets?)
<3> myrkraverk: there are several ways to justify it, but the number of subsets of size k of an n-set is C(n,k)
<7> tubbie: be careful there. you only changed the lower index.
<12> HiLander: what is C ?
<7> tubbie: sorry, limit
<5> because l = k -1 => k = l + 1, since k = 1, then l = 0?
<3> myrkraverk: it stands for "choose", that notation is read "n choose k" (it's hard to express the real notation on irc)
<3> myrkaverk: C(n,k) = n!/(k! * (n-k)!)
<12> HiLander: ak
<3> myrkraverk: are you familiar with n! being the number of ways of ordering a set with n elements?
<12> HiLander: you mean that stuff that looks like 1 x 2 matrix?
<5> but then I end up with 1 + 2^N + 2^N :\
<12> HiLander: I was, I think I've forgotten this stuff ;)
<3> myrkraverk: it usually looks like a 2 x 1 matrix (rows first), but yes. like that. they're called binomial coefficients.
<7> tubbie: yes, the 0 is correct. but somehow the sum has now n+1 terms instead of the n you started with. what's wrong?
<12> HiLander: ok (I was just (re) learning about matrices today, and I'm not too good at recalling which is usually mentioned first, rows or columns ;)
<5> that's the induction. I plugged N+1
<3> myrkraverk: the rough idea is you pick a k-subset of an n-set by ordering all of the elements (n! ways) and choosing the first k: you get the same k-set no matter how those k are ordered (k! ways) or how the remaining elements are ordered ( (n-k)! ways)
<7> tubbie: you had 1 + 2^N + sum(k=1,N)C(N,k-1). you changed that to 1 + 2^n + sum(l=0,N)C(N,l) which isn't correct.
<3> myrkraverk: so the number of distinct k-sets has to be n!/(k! * (n-k)!), since each k-set shows up as the first k elements in k! *(n-k)! such orderings.
<5> the step from k to l is wrong?
<5> I meant 2^N
<7> tubbie: remember that l=k-1. so what range should l have when the rango of k is 1 to N?
<7> . o O ( o instead of e is a weird typo )
<7> tubbie: no, the step that I quoted is wrong. the step from there to 1+2^N+2^N is correct again, but based on a wrong formula.
<13> i need help please
<7> kermn: we only do math.
<13> the number of cups of coffee sold in a cafeteria during lunch, Is it discrete variable or continous?
<13> i guess it's discrete correct?
<7> yes.
<5> 1 + 2^N + sum(k=1,N)C(N,k-1). is correct?
<13> thanks int-e :)
<7> tubbie: yes
<12> HiLander: ok, is that what is called (or may be called) binomial is CASes?
<13> int-e, do you know why it's not continnous?
<7> because it's discrete. it's a whole number.
<13> because the number of coffee sold might be different
<5> the range for l is then l=0 to N-1 ?
<7> they don't sell half cups, quarter cups, pi cups, etc.
<7> tubbie: yes
<13> ohhhh :)
<13> very clear now, thanks int-e
<7> the 1 gets changed to 1-1 = 0 and the N to N-1.
<4> The world of natural numbers is fasniating :o
<5> so the last part is then sum(l=0,N-1)C(N,l)
<7> yes. and what could the 1 be? :)
<5> that is equal to sum(l=0,N)C(N,l) - C(N,N)
<7> kermn: be careful, it's not about integers or not. it's about having only a finite number of values in a finite range. So if they'd sell quarter cups and half cups (don't know how, maybe they'd freeze them first), but no other fractional quantities, the value would still be discrete.
<5> this last part is equal to -1
<5> int-e: ?
<7> tubbie: yes, you can do it that way.
<5> and I get 2^N + 2^N = 2 *2^N= 2^N+1
<7> 2^(N+1) please. yes.
<7> I wouldn't know what to answer. It's dense, but countable and not complete, so I'd say it's neither discrete nor continuous. Are these terms rigerously defined?
<5> int-e: thank you for the help :) really appreciate it. :)
<5> I have spent like 5 hours on this one. :S
<14> int-e - interesting question. I don't think there's a formal definition for bare sets as such, but one is implied by distinguishing the cardinality of the reals as 'having the property of the continuum'. If we wanted a definition, I'd say then that uncountabiilty is a necessary condition for a set to be 'continuous'; a discrete set is one that is at most countably infinite.
<15> Would you say that an integer that has exactly two positive factors, the integer itself and 1 is a prime number?
<1> Yes.
<3> not only would i say that
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