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Comments:
<0> can anyone explain to me what k^<insert random fractionary number> means ?
<1> like k^(2/3)?
<0> yup
<2> wotdu: k^(n/m) = m-th root of k^n
<2> example k^(1/2) is just the sqrt root
<0> somiaj, k^pi
<2> k^(1/3) cube root
<0> what about now? :)
<1> it's the limit as a rational number approaches pi
<3> and now somiaj's explanation falls apart :P
<2> wotdu: k^pi would be the limit of rational exponents
<1> mathpwn!
<4> Or if you prefer, e^(pi*log(k))
<2> thneed: nope, for an irrational number, just take the limit of rational exponents that approach the irrational number
<0> i see...
<3> Ok, I guess that works.
<0> it's kind of weird : /
<0> thanks guys
<2> wotdu: basically square roots, cube roots etc are all power functions for non integer powers, and enxtending the idea to irrationals isn't that diffcult, though caulculating out 3^pi by hand could be a pita
<1> pffft, just write it as in infinite product!
<0> another thing
<0> why is k^0 == 1 ?
<1> only nonzero k, don't forget
<0> y if k != 0
<1> 0^0 is just undefined... 0^x is zero for any nonzero x
<2> wotdu: why? that depends on what you consider evidence, for most cases k^0 is defined to be 1, for all k!=0
<5> CHodapp: you can make 0^0 be undefined, or 1, or whatever you want. this isn't religion, it's math.
<2> wotdu: if you look at set theory and defining powers based power sets of sets, you can get a why, or you can just look at how say 2^x behavies as x -> 0, in that case you'll see it gets close to 1
<5> CHodapp: define 0^0=17, prove that (a*17)^b is divisible by 17 for any a,b integers b>=0
<5> oops, define x^0=17 for all x too
<0> thanks guys
<0> brb
<4> Also there's the relation x^n * x^m = x^(n+m), so if you apply that to x^1*x^0 you get x^1, and so x^0 = 1 if x != 0
<1> TRWBW: every reference I'm seeing says that the only 1 and undefined make much sense.
<5> somiaj: there is no evidence, this isn't empiracle. you can define 0^0="hot apple pie" if you want. you can define a^b=a/b, why not? or how about a^b is the longest arithmetical expression of b or more symbols that evaluates to a? or a^b=a^(b-1)
<1> TRWBW: go look up "Axiom" and then come back.
<5> CHodapp: that's my point, nezzbit
<6> Uh I dunno you Chodapp, but TRWBW is a pretty knowledgable guy.
<7> for every axiom there exists anti-axiom such that axiom + anti-axiom = null-axiom.
<1> Then are we talking about "How it's generally defined" or "The fact that it's an axiom and is unprovable which means people can freely rant about it without really getting anywhere"?
<5> CHodapp: how its generally defined is a valid topic, but it's not math. lots of math is actually about defining standard things weird ways and seeing what happens.
<1> Is wotdu talking about such an area of math?
<8> pkrumins: null-axiom?
<5> sure you can define 0^0=1, you can define 0/0=1, 1/0=inf, if you are formal and cover all the cases. they all lead to different things.
<8> Is that the axiom that doesn't say anything at all?
<8> The one that's consistent with all consistent theories?
<1> In all your judgement, does it appear wotdu is aiming for information in those areas of math?
<5> 0^0=1 makes some things work, it breaks other things, like a^b=e^(b*log(a))
<7> a jokes
<7> it was a joke
<7> yes probably consistent with all consistent theories.
<9> oh, hmmm, my internets are dying
<10> can anyone give me an example of an equation that gives a singular solution?
<11> adx2: by singular you mean one and only one?
<10> xerox: i don't know what it means, that's why i want to see an example so that i can get it
<2> adx2: http://en.wikipedia.org/wiki/Singular_solution
<11> maybe this can be helpful http://en.wikipedia.org/wiki/Mathematical_singularity
<5> adx: it's a song from the musical "A Chorus Line". "One, singular solution, ..."
<11> oh, nevermind.
<10> is it that it has infinite solutions?
<10> like 0x=0 ?
<10> how is a system called that has no solutions?
<4> Inconsistent?
<12> windows?
<10> Kasadkad: and how is called a system that has infinite solutions?
<4> I dunno
<5> underdetermined
<10> thanks
<4> Ah right
<10> which is more efficient, gaussian elimination or gauss-jordan ?
<4> Gaussian elimination
<5> i'm suprised, i thought they were both N^3
<4> Hmm, well
<4> Doesn't Gauss-Jordan elimination take it to row echelon form and then to reduced row echelon form?
<5> i thought the difference was just whether you eliminated above the diagonal and generated the full inverse, or whether you just do an LU decompositiion (with some P's in there for pivoting).
<4> Yeah, but doesn't that mean that Gauss-Jordan goes through all the steps of Gaussian elimination but then also some others to eliminate above the diagonal?
<13> hello?
<13> anyone here?>
<2> joshabts: /who, lots of people here, if ya have a question best to just ask it and then there is a possibily someone will respond
<13> i have a trig question
<13> i need to evaluate the following problem using right triangles
<13> tan^-1(2*cos(pi/3))
<2> well cos(pi/3) is a known from the 30/60/90 triangle, so start there
<13> so i draw my 30/60/90
<13> and the cos pi/3 is 1/2
<2> so then you want to find tan^-1 (2*1/2) = tan^-1(1)
<2> and what triangle has a tangent of 1?
<13> isnt tangent opposite over adjacent?
<13> so inverse tangent would be adjacent over opposite?
<2> its opposite over adjacent
<5> joshabts: no that would be cotangent = 1/tangent
<2> so the inverse tanget of 1/1 is a triangle who has equal sized legs
<2> so its the 45/45/90 triangle
<13> yeah 1-1-2 O-A-H
<2> the inverse tanget gives you the angle (in radians)
<13> so it would end up being 45 which would be pi/4?
<2> correct
<5> jum, 1-1/sqrt(2)
<5> 1-1-sqrt(2)
<5> 1-1-2 isn't a right triangle
<2> oh yea, it is 1-1-sqrt(2), but the answer to the question is pi/4
<5> sure
<13> ok so im trying to refresh myself on this, i haven't touched this stuff in over a year and a half
<13> so my first step is to find the cos(pi/3)
<13> which is 1/2
<5> yes
<13> so then i have the tan^-1(2*1/2) which is the tan^-1(1)
<5> yes
<5> and yes
<13> at this point i can draw a right triangle with a side of 1 and a hypotenuse of 2?
<5> god no
<13> ergh
<13> can someone enlighten me?
<5> you want a triangle whose opposite and adjacent sides are the same, so that their ratio is 1
<13> why does their ration have to be 1?
<13> *ratio
<5> because you want tangent to be 1, duh
<5> tangent is the ratio of ...
<14> hi
<15> http://karlscalculus.org/calc6_1.html
<16> Title: Karl's Calculus Tutor - 6.2 The Farther We Go, The Faster We Get There
<15> I wish some teacher explained the ln & e thingy to me that way before.
<17> sets with countably many elements have measure zero?
<18> Lebesgue measure zero.
<19> thanks
<19> the set of all functions from the naturals to the naturals is uncountable?
<18> The set {0,1}^N of all functions f: N -> {0,1} is uncountable.
<20> That is to say, yes
<21> ekiM: consider decimal expansions of real numbers
<18> Or simply Cantor's theorem.
<21> oh, you said {0,1}.. on my font it looks like (0,1)
<21> so i was having trouble seeing your point
<20> (0, 1) doesn't make a lot of sense :/
<21> (0,1) = open interval {x : 0 < x < 1}
<21> i still can't tell the brackets apart visually, have to rely on contextual information
<21> need to change this damn font
<22> try helvetica or something.
<20> Oh, true
<20> And I suppose that's less applicable
<19> I see
<23> "Prove that a finite domain is a division ring". What is meant by finite domain exactly?
<4> A finite ring without zero-divisors
<23> Kasadkad: so basically a ring with integral domain?
<4> Well, integral domains are also commutative
<23> i don't think I understand zero divisors, according to the wikipedia definition
<23> it's an element a in R such that ab = 0 for some b not 0 in R
<23> but a does not have to be non zero?
<4> Yes
<4> Er
<4> Sorry, misread
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