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Comments:
<0> roxlu: well im not sure what the problem IS
<1> Oke I'll try to make it clear
<0> you see....im not a programmer....
<1> I'm making a new post
<1> Actually it is like this: http://rafb.net/paste/results/itsi2b40.html
<1> I'n fact I've got only 3 rows, but lets say I've got "virtually" 6 rows like on the square
<1> Now, I need to recalculate (row,col): (3,0) to (6,0)
<2> Well, there are several ways of renumbering them.
<1> JabberWalkie: you know what I mean?
<1> oke
<0> mmm'
<0> so there are relations between all these entrys?
<3> roxlu, lets say the top is (i, j), the bottom is (x ,y)
<1> yes
<2> Which ones are you moving ?
<1> Hmm what do you mean with "moving" ?
<2> Well, you're rearranging them, so they get moved to accomplish that.
<1> oh oke, well the 'real layout' is the top one
<2> Which items move to give you the bottom one ?
<1> columns 6 to 11
<1> and those rows.. so actually the last two 'blocks' of 3x3
<2> OK, if the column is greater than or equal to 6, reduce it by 6 and increase the row by 3.
<2> Otherwise, do nothing.
<3> hold on i got a one liner
<1> Olathe: I need to do the other way around
<2> Alright.
<2> if (row >= 3) { row = row - 3; column = column + 6; }
<1> wow
<1> thanx
<2> No problem.
<3> (i, j) = (x % 3,
<3> darn, cant get y without and if
<3> *an
<3> (i, j) = (x % 3, (x > 2)?y:y+6)
<3> oops
<3> (i, j) = (x % 3, (x < 2)?y:y+6)
<3> (i, j) = (x % 3, 6 * (x / 3) + y ) ... theres the one liner
<4> Hi! Can anyone determine a from the following equation: sin(a)+b*cos(a)=c, when b and c are given??
<3> yeesh
<4> CoffeeBuzz: :-) ???????????
<2> % Solve[Sin[a] + b Cos[a] == c, a]
<5> Olathe: {{a -> -ArcCos[(c - c/(1 + b^2) - Sqrt[b^2 + b^4 - b^2*c^2]/(1 + b^2))/b]}, {a -> ArcCos[(c - c/(1 + b^2) - Sqrt[b^2 + b^4 - b^2*c^2]/(1 + b^2))/b]}, {a -> -ArcCos[(c - c/(1 + b^2) + Sqrt[b^
<5> 2 + b^4 - b^2*c^2]/(1 + b^2))/b]}, {a -> ArcCos[(c - c/(1 + b^2) + Sqrt[b^2 + b^4 - b^2*c^2]/(1 + b^2))/b]}}
<6> veery nice :)
<2> % Solve[Sin[a] + b Cos[a] == c, a] //FullSimplify
<6> quite elegant, BTW
<5> Olathe: {{a -> -ArcCos[(b^2*c - Sqrt[b^2*(1 + b^2 - c^2)])/(b + b^3)]}, {a -> ArcCos[(b^2*c - Sqrt[b^2*(1 + b^2 - c^2)])/(b + b^3)]}, {a -> -ArcCos[(b^2*c + Sqrt[b^2*(1 + b^2 - c^2)])/(b + b^3)]}, {
<5> a -> ArcCos[(b^2*c + Sqrt[b^2*(1 + b^2 - c^2)])/(b + b^3)]}}
<2> A little shorter.
<6> now I would like to see all the elaboration...
<4> Olathe: cool!!!
<2> I don't think mbot can show its work (and it probably uses methods easy for computers rather than humans, anyway).
<3> half way there lol
<7> in polar coordinates, which comes first? the distance or the angle?
<2> I've usually seen the distance written first.
<8> yes
<9> if m is the numeric day of the month (an integer in the range [1,31]), and w is the numeric day of the week (an integer in the range [0,6]), how would you go about creating a formula for s, the numeric day of the week on which the first of the month falls?
<10> hi all
<9> m and w refer to today
<7> Olathe: have you even seen the angle come first?
<8> no
<11> let S_1 be the number of the week day on the first day of the first month. Then the number of the week day on the first day of the second month is S_2 = (S + 31) mod 7. Similarly S_3 = (S_2 + 28) (mod 7) (but watch for leap years), and so on.
<9> but you don't know S_1
<9> you know today's weekday number and dayofthemonth number, and you want to calculate the weekday number of the first day of the current month
<9> it's not hard to do, but i'm not sure if i have the simplest expression
<9> i have s = (w - (m mod 7) + 8) mod 7
<11> That's just (w - m + 1) mod 7.
<9> ***uming modulus works on negative numbers
<11> Why wouldn't it?
<9> well, i know it does theoretically, but haven't tried it in javascript before :)
<9> hmm. -37 mod 7 gives me a -2
<2> Change '+ 1' to '+ 36'.
<2> Then 36 - m is always positive.
<11> % Mod[-37, 7]
<9> nice
<5> Catfive: 5
<9> well, javascript isn't really a mathematically-oriented language, i guess i shouldn't be surprised
<12> javascript ****s in every way you look at it :P
<7> affine transformation!
<13> heya
<14> % Solve[x==Sqrt[a x^2+b x+c]+Sqrt[d x^2+e x+f]+Sqrt[g x^2+h x+i],x]
<5> |Steve|: "Time limit exceeded for computation."
<14> weak
<15> you keep squaring, and geta polynomial, but it wont be soluble
<14> Yeah.
<14> I was just wondering if there was a closed form solution.
<15> there's no closed fomr solution for even a degree 5 polynomial
<14> Right.
<14> But this wasn't a general degree 5 polynomial.
<15> hmm, what degree do you get actually? i was ***uming 6, but maybe only 4
<14> I have no idea. I wasn't _that_ interested. Someone just mentioned it in #math on efnet.
<14> 6 would be my guess
<15> first you get <quadratic>= <quadratic> +sqrt(<x^2>)
<15> so i guess it's justa quartic int he end
<14> % Simplify[(Sqrt[a x^2+b x+c]+Sqrt[d x^2+e x+f]+Sqrt[g x^2+h x+i])^2]
<5> |Steve|: (Sqrt[c + x*(b + a*x)] + Sqrt[f + x*(e + d*x)] + Sqrt[i + x*(h + g*x)])^2
<14> % Expand[(Sqrt[a x^2+b x+c]+Sqrt[d x^2+e x+f]+Sqrt[g x^2+h x+i])^2]
<5> |Steve|: c + f + i + b*x + e*x + h*x + a*x^2 + d*x^2 + g*x^2 + 2*Sqrt[c + b*x + a*x^2]*Sqrt[f + e*x + d*x^2] + 2*Sqrt[c + b*x + a*x^2]*Sqrt[i + h*x + g*x^2] + 2*Sqrt[f + e*x + d*x^2]*Sqrt[i + h*x +
<5> g*x^2]
<14> heh, Yeah, I really don't think I care _that_ much.
<15> :)
<14> % Simplify[Expand[(Sqrt[a x^2+b x+c]+Sqrt[d x^2+e x+f]+Sqrt[g x^2+h x+i])^4]]
<5> |Steve|: c^2 + f^2 + 6*f*i + i^2 + 6*b*f*x + 2*e*f*x + 6*f*h*x + 6*b*i*x + 6*e*i*x + 2*h*i*x + b^2*x^2 + 6*b*e*x^2 + e^2*x^2 + 6*a*f*x^2 + 2*d*f*x^2 + 6*f*g*x^2 + 6*b*h*x^2 + 6*e*h*x^2 + h^2*x^2 + 6*
<5> a*i*x^2 + 6*d*i*x^2 + 2*g*i*x^2 + 2*a*b*x^3 + 6*b*d*x^3 + 6*a*e*x^3 + 2*d*e*x^3 + 6*b*g*x^3 + 6*e*g*x^3 + 6*a*h*x^3 + 6*d*h*x^3 + 2*g*h*x^3 + a^2*x^4 + 6*a*d*x^4 + d^2*x^4 + 6*a*g*x^4 + 6*d*g*x^4 +
<5> g^2*x^4 + 4*f*Sqrt[c + x*(b + a*x)]*Sqrt[f + x*(e + d*x)] + 12*i*Sqrt[c + x*(b + a*x)]*Sqrt[f + x*(e + d*x)] + 4*b*x*Sqrt[c + x*(b + a*x)]*Sqrt[f + x*(e + d*x)] + 4*e*x*Sqrt[c + x*(b + a*x)]*Sqrt[f +
<5> x*(e + d*x)] + 12*h*x*Sqrt[c + x*(b + a*x)]*Sqrt[f + x*(e + d*x)] + 4*a*x^2*Sqrt[c + x*(b + a*x)]*Sqrt[f + x*(e + d*x)] + 4*d*x^2*Sqrt[c + x*(b + a*x)]*Sqrt[f + x*(e + d*x)] + 12*g*x^2*Sqrt[c + x*(b
<5> + a*x)]*Sqrt[f + x*(e + d*x)] + 12*f*Sqrt[c + x*(b + a*x)]*Sqrt[i + x*(h + g*x)] + 4*i*Sqrt[c + x*(b + a*x)]*Sqrt[i + x*(h + g*x)] + 4*b*x*Sqrt[c + x*(b + a*x)]*Sqrt[i + x*(h + g*x)] + 12*e*x*Sqrt[c
<5> [5 @more lines]
<15> nice
<14> wow
<14> I'm quite done.
<14> It's good to see that mbot won't spam too much.
<16> Is this allowed in math: http://www.irclogs.net/freenode/math/ ?
<15> hmm it's a bit rude to do that withotu asking permission
<15> so whoever is doing that, please dont be antisocial
<2> (21:33.35) * Joins: ^betul^ (n=anjuna@ip-152-234.sn2.eutelia.it) | (21:36.04) (JabberWalkie) haha | (compare with http://www.irclogs.net/freenode/math/2006/07/01/)
<16> Olathe, It seems him.
<16> +it is
<15> that 'anjuna' seems to ring a bell, i tihnk someone complained about this before
<15> so.. do people care? or is this a good thing? or for all i know everyone already decided this was a good idea :)
<11> it's an abuse of irc and extremely poor netiquette
<15> i tend to agree
<16> Well, at least he's making money out of the logs.
<7> can anyone show me the formula for projective transformation in the plane?
<17> what is God?
<18> god is epsilon < 0
<17> are you sure?
<2> Yes.
<17> Olathe: how?
<2> It's part of category theory.
<17> Olathe: so what is God?
<2> Mulder answered that.
<17> I don't know category theory
<2> I don't either, unfortunately.
<19> Does anyone besides MacLane really know cat theory?
<15> of course
<15> and maclane is no longer with us
<17> why is maclane no longer with us?
<19> Oh, that's right >_<
<19> I'd forgotten :(
<15> well he died
<17> ****
<17> how?
<19> Old age, I thought.
<17> really?
<15> afaik
<17> and IRCer dying of old age
<17> this is a new concept to me
<17> pitty it had to be MacLane that introduced it to me
<20> hmm
<17> is anybody here an academic?
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