TRWBW | citrus: not sure what you mean by that. here, if you have triangle ABC with the angle at point B, so vectors V1=A-B and V2=C-B, and N1=V1/|V1| (normalized V1) same with N2=V2/|V2|, then the bisector vector is N1+N2. citrus: that works in 2 dimensions or 3 dimensions (or technically any number of dimensions) |

citrus | ah thnks |

TRWBW | citrus: if you want to be fancy, since any multiple of a bisection vector is just as good, you could also use V1*|V2|+V2*|V1|, which is the same thing multiplied by |V1|*|V2|. citrus: that is you can get the same length by multiplying as easily as by dividing |

citrus | ok sounds good |

TRWBW | citrus: i just put that there so you understand there isn't a "unique" bisection vector, it's really just a direction, so any multiple works just as well. |

citrus | ok i'm writing code, so multiplication is faster than devision but on today's hardware even that isn't an issue |

eugman | heh, find a way to use the power of bitshifting. |

bsmntbombdood | given n points how do you find the nth degree polynomial that goes through them all? |

TRWBW | bsmntbombdood: easiest way conceptually is probably vandermonde matrix |

bsmntbombdood | er, given n+1 points |

TRWBW | bsmntbombdood: first off it's n-1 degree polynomial. think of the matrix M with m_ij = (n_i)^j where n_i is the i'th number 0..(n-1) and j is an exponent 0..n-1 bsmntbombdood: you with me so far? |

citrus | practialy there is Newtonian and Talor series |

bsmntbombdood | yes |

citrus | http://en.wikipedia.org/wiki/Newton_polynomial 2cents :P |

TRWBW | bsmntbombdood: okay, now think of the coefficients of a polynomial p(x)=c_0+x*c_1+x^2*c_2...x^(k-1)*c_(k-1) as a vector C={c_0, c_1, ..., c_(k-1)}^t |

bsmntbombdood | what's t? |

TRWBW | bsmntbombdood: then A=M*C, work it you, you'll see A={p(n_0), p(n_1), ... , p(n_(k-1))}^t. oops, i mixed up n's and k's there. bsmntbombdood: transpose. i just meant it's a column vector, but i wrote it as a row vector transposed because i didn't want to have to type in... c_0 c_1 c_2 c_3 ... |

bsmntbombdood | ok |

TRWBW | c_(k-1) |

anubiss | is x^2 monotonous or is ist x^3 ? and then whats the other word for 'non-monotonous' i got another fun question |

TRWBW | bsmntbombdood: anyways, you get the idea? you use the points to generate a matrix, then your problem because a linear system A=M*C, where M comes from the points, A is the values you want the polynomial to take, and C is the coefficients you want to figure out |

anubiss | one integral is area under a surface, whats a double, triple integral ? |

TRWBW | bsmntbombdood: so it's just use gaussian elimination, or invert the matrix, or any other kind of method for solving a linear system |

bsmntbombdood | oh |

eugman | How many 5 digit combinations like 01223 is needed to satisfy the two conditions: The sum of the digits for each combonation is a multiple of 4(counting 0), and for each digit there is an equal number of 0's, 1's, 2's and 3's over the total combinations. |