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<Arafangion> astrolabe_: That's suspiciously easy!
<saba> x^x, x->0+ is 1
<ods15> nature create 0, E, and Pi...
<zero0> then it became illogical to have 1 = 0
<ods15> saba: yeah that's what i told him
<zero0> so natural selection kicks in to create -1
<astrolabe_> Arafangion: It seems like a good guess though.
<zero0> 1 - 1 = 0
<zero0> now we have 0, 1, -3
<zero0> err 0, 1, -1
<zero0> :)
<Arafangion> astrolabe_: Is there any way to verify this?
<ods15> zero0: the only things nature invented is E, Pi, and 0. the rest is out imagination
<zero0> and then through natural selective of lowest state, all primes were created
<zero0> after primes, all other numbers are created
<zero0> therefore we have the real
<ods15> natural or real? :)
<zero0> natural first
<zero0> then real
<ods15> i'd think first rational though
<zero0> that's how nature created number
<zero0> yah rational
<zero0> but u get the gist
<zero0> i'm going to create a poster
<astrolabe_> Arafngion, So p lives in <1,x,x^2,x^3>. Say p = a+bx+cx^2+dx^3. Then the image of p is (p(1),p(2),p(3),p(4))
<Arafangion> astrolabe_: The nearest example I have in my book is: T(p(x)) = p(2x - 1), but that is very different.
<ods15> saba: hey, help me out, are there more prime numbers than there are natural numbers?...
<zero0> just need to iron out a few details
<ods15> cause, for ex., there's more numbers between 0 to 1 than there are natural numbers
<astrolabe_> Arafangion@ That map is linear.
<zero0> oh my god my number is looking crazy haha :)
<zero0> i need to use my system for my big calculation
<saba> i'm an applied mathematician, whatever the hell that means :) ask someone else
<zero0> gotta go
<zero0> see yah!
<astrolabe_> ods15: There are the same number of each.
<ods15> astrolabe_: aye
<ods15> so i figured
<Arafangion> astrolabe_: So, what is p(1), p(2), p(3), p(4)?
<astrolabe_> ods15: What 'the same number' means is that you can pair them up.
<ods15> ?
<ods15> i know little of infinity but i know there are different "levels" for it, like, 2*inf is not bigger than inf, but 2^inf IS bigger...
<saba> aleph-null stuff.
<Arafangion> ods15: Infinity is a concept, not a number.
<astrolabe_> Arafangion: (p(1),p(2),p(3),p(4)) means the vector in R^4, or wherever with first coordinate p(1) etc.
<ods15> aye
<Arafangion> astrolabe_: I don't actually _have_ p.
<Arafangion> astrolabe_: I do not have a "Let p(x) = 1 + x + x^2" or anything.
<astrolabe_> Arafangion: Do you need it?
<Arafangion> ods15: So, 2*infinity is NOT greater than infinity.
<Arafangion> astrolabe_: I'm just confused.
<astrolabe_> Arafangion: What are you trying to do?
<Arafangion> astrolabe_: I am trying to transform a vector from one basis, to another.
<ods15> Arafangion: thats what i said...
<Arafangion> astrolabe_: From a parametric basis, using a provided tranformation (That's the original question), to an elementary basis.
<astrolabe_> ods15: It would be more accurate to say 2*Aleph_0 =Aleph_0
<Arafangion> astrolabe_: Well, suppose I have the transofrmation: T(p) = [p(1), p(2), p(3), p(4)], how do I transform the vector [1, 3, 5, 6]?
<Arafangion> astrolabe_: Given a parametric source basis (1, x, x^2, x^3), and an elementary target basis?
<astrolabe_> Arafangion: Ah, there were commas in the expression for T(p)?
<Arafangion> astrolabe_: No, it's actually a vertical vector.
<Arafangion> astrolabe_: With the p's rendered in bold.
<astrolabe_> Arafangion: ok, better and better.
<Arafangion> Harder or easier?
<astrolabe_> Is [1,3,5,6] a column too?
<Arafangion> Essentially, I'm really confused as to what p(2) means.
<Arafangion> I'm guessing that p is a vector.
<astrolabe_> The more information, the easier it gets.
<Arafangion> astrolabe_: I'll /msg you the question.
<Arafangion> astrolabe_: YOu'll notice that I did not ask for help on the exact question, but a modification. I do not want to be "given" the answer - I want to be able to understand it and thus answer the real question myself.
<astrolabe_> Arafangion: Good, I'll see what I can do :)
<Arafangion> Thanks, because I'm stumped as to the _meaning_ of it.
<astrolabe_> So, you need to understand why what they give is a linear transformation, or even a map from P_3 -> R^4?
<Arafangion> astrolabe_: Well, I know _what_ a linear transformatino is, and I can handle almost every linear tranformation except for that one.
<Arafangion> astrolabe_: Because that one doesn't give a definition for p.
<Arafangion> astrolabe_: ie, I can't figure out what's supposed to happen if I shove a vector into the tranformation, let alone calculate the general matrix of the transformation.
<astrolabe_> Well, you know p is in P_3, so you know what it looks like
<Arafangion> astrolabe_: Well, yes, but I don't know if it's just 2, or 1+x+3242x^3
<astrolabe_> Arafangion: The formula is for any element of P_3
<astrolabe_> You need to say which element of P_3 you put in before you know what you get out of the transformation.
<Arafangion> astrolabe_: So, suppose T(p) = [p(1) p(2) p(3) p(4)], for a general vector [a b c d], the result is actually a 4x4 matrix ((a b c d) (2a 2b 2c 2d ... )
<Arafangion> ?
<astrolabe_> Arafangion: No, first, convert [a,b,c,d] into a polynomial (and stop missing out commas!)
<astrolabe_> Arafangion: Where [a,b,c,d] is in P_3
<Arafangion> astrolabe_: The syntax I am taught is that spaces separate elements in a vector/matrix, while comas separate entire vectors in a set.
<astrolabe_> Arafangion: Sorry, better stick with that then.
<Arafangion> This is a wierd transformation!
<astrolabe_> As a clue to my question, the notation [a b c d] in a vector space means a*e_1 + b*e_2 + c*e_3 + d*e_4, where the e_i are the basis elements.
<Arafangion> I understand that now, first expand, then transform.
<Arafangion> Hmm.
<astrolabe_> Arafangion: So if [a b c d] is in P3, what is it as a polynomial?
<Arafangion> astrolabe_: a + bx + cx^2 + dx^3
<astrolabe_> Arafangion: yay!
<astrolabe_> Arafangion: So now if that is p, what is p(3)?
<Arafangion> a + b(3) + c(3)^2 + d(3)^3
<astrolabe_> Arafangion: Right, a +3b+9c+27d
<Arafangion> astrolabe_: So, given that I'm dealing with elementary vectors, how is this significant?
<Arafangion> astrolabe_: Say, [1 0 0 0] for the first vector, the result, regardless of wether I have to do p(0) or p(1231), it's still [1 0 0 0]?
<astrolabe_> Arafangion: So now the biggie. T(p) is a vector: a vertical sequence of 4 numbers. What are they in terms of a,b,c and d.
<astrolabe_> Arafangion: Yes, if [1 0 0 0] = p as an element of P3 then p(z) = 1 for all z.
<Arafangion> [(a + bx + cx^2 + dx^3) (a + bx + cx^2 + dx^3) (a + bx + cx^2 + dx^3) (a + bx + cx^2 + dx^3)]
<Arafangion> So, how does one do the matrix of transofrmation, then?
<astrolabe_> Arafangion: that isn't what I meant.
<Arafangion> Did you mean for [p(1) p(2) p(3) p(4)]?
<astrolabe_> The top element of T(p) is p(-3) right? What is that in terms of a,b,c and d?
<Arafangion> [(a + b + c + d) (a + b2 + b4 + b8) (a + 3 ... ]
<astrolabe_> You could do p(3), you should be able to get p(-3).
<Arafangion> Or in the case of the top element being p(-3), it's (a - 3b + 9c - 18d)
<astrolabe_> If 18 = 27, that's exactly right ;)
<Arafangion> :)
<Arafangion> So, the top element of the matrix of transofrmation is (1 -3 9 -27) ?
<Arafangion> That sorta makes sense...
<astrolabe_> Arafangion: no, it's not, sorry.
<astrolabe_> Carry on along my path
<astrolabe_> You have worked out the top element of T(p), what is the second to top element?
<Arafangion> a - b + c - d
<astrolabe_> Yeah! So what is T(p)?
<Arafangion> [(a - 3b + 9c - 27d) (a - b + c - d) (a + b + c + d) (a + 3b + 9c + 27d)
<Arafangion> ]
<astrolabe_> Excellent.
<astrolabe_> So, you know how to go from an arbitrarty element in [a b c d] in P_3 to its image in R^4
<Arafangion> astrolabe_: If I select the basis vectors, (1 0 0 0), (0 1 0 0) and so forth, I can do a general matrix of transormation?
<astrolabe_> Arafangion: Right, you use their images as columns of the matrix right?
<Arafangion> Yes.
<Arafangion> I think I see why the P is in bold now.
<Arafangion> It's really [a b c d]^T[1 t t^2 t^3]
<Arafangion> Thank you _heaps_ for the help :)
<astrolabe_> Arafangion: It's really [a b c d]*[1 t t^2 t^3]^T, but you have the right idea.
<astrolabe_> It's cause it's a vector.
<astrolabe_> You are welcome. Good luck with the next bit!
<Arafangion> I've got the next bit figured out, thanks :)
<Arafangion> I hate it when it's the _key_ thing that I'm stuck on in the ***ignment!
<astrolabe_> Yeah, it'll seem obvious next week too.
<Arafangion> All my focus is for system software :(
<Arafangion> I'm _killing_ it.
<Arafangion> Heck, the lecturer wants me to write some papers on it.
<Arafangion> But I just have to make sure I p*** this one just the same :(
<delta> Hi.
<Safrole> How would you solve tanh^2(g*t/Vt) = 2 for t
<Safrole> How would you solve tanh^2(g*t/tau) = 2 for t
<Safrole> is there some identity for the tanh^2?
<ichor> Safrole, could you not take the square root of both sides?
<bobwhoops> I thought tanh was restricted to > 0 and < 1
<Safrole> yeah ichor
<ichor> bobwhoops, for complex numbers too?
<Safrole> I didn't think about that...
<bobwhoops> Oh, I thought you were keeping it in the reals
<Safrole> for some reason when I see tanh
<Safrole> I think differently.
<Arafangion> astrolabe_: Maybe it's because it's 2am and I cant' think, but my change of base matrixes _are_ the new bases? That right? (THe source is elementary)
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