Cin | what does this dot mean? 211 what does ab mean? |

TRWBW3 | _shellcode_: say you got v1=a*u1+b*u2, v2=c*u1+b*u2. and you also have u3=e*v1+f*v2. okay, then u3-u3=e*v1+f*v2-u3=e*(a*u1+b*u2)+f*(c*u1+b*u2)-u3=0. that's a LD on u1,u2,u3. |

_shellcode_ | Because u3 depends on u1 and u2? |

TRWBW3 | _shellcode_: um, an LD is a linear combination of vectors with at least one non-zero coefficient that sums to zero. _shellcode_: so because it fits the definition of LD |

_shellcode_ | Well, I know what an LD is but I don't see how proving that u3-u3=0 is an LD |

TRWBW3 | _shellcode_: it isn't until you plug in the formula for u3 in terms of v1 and v2, that you get in the first step _shellcode_: copy down everything i've said and just think about it until you see it. it's all there. good luck. |

_shellcode_ | OK. Thank you. |

TRWBW3 | http://www.microwaves101.com/encyclopedia/images/Switches/treeofwoe.jpg |

Cin | what does ab mean? what is that dot? is it showing up? |

nexact | it's a multiplication. but I may be wrong... |

Cin | do you know what the symbol is called? perhaps I could search for it |

TRWBW3 | _shellcode_: okay, one last hint. say you got u1=3*v1 and u2=2*v1. how can you show u1 and u2 are LD? well you can use v1=u1/3, then u2=2*v1=2*(u1/3)=(2/3)*u1. then use 0=u2-u2=(2/3)*u1-u2, same idea, just with 3 vectors instead of 2. |

nexact | Cin, You may recall using a "x" as the multiplication symbol in arithmetic. This is generally avoided in algebra because x is the most common variable used in expressions and equations. It would be extremely confusing if x were used for both a variable and a multiplication sign. |

JIMJONESBALLIN | But of course, x is used for the cross product |

Cin | nexact, ah, thanks! indeed, in school we wrote the letter x in italics to differentiate |

rhodopsine | given y = sin 2pi(x - 1/2), the phase shift is 1/2, and it follows that for the general form y = A sin(Bx + C) = sin B(x + C/B) and that the phase shift, if C/B < 0, |C/B| units to the right, otherwise C/B > 0 then C/B units to the left given this, how do they end up with 1/2 as the phase shift? (2*pi) * -1/2 would be -pi which would be pi as the phase shift, no? |

nvictor | % D[Sqrt[3-5x],x] |

mbot | nvictor: -5/(2*Sqrt[3 - 5*x]) |

nvictor | % Limit[Sqrt[x-5a]-Sqrt[x],x->3] |

mbot | nvictor: -Sqrt[3] + Sqrt[3 - 5*a] |

nvictor | anybody online? |

JIMJONESBALLIN | yes |

nvictor | I can't find this limit :( h->0 of (f[x+h]-f[h])/h where f[x] = Sqrt[3-5x] I've been trying for one hour now... I have tried to change unknown but it doesn't work either |

JIMJONESBALLIN | do they want you to compute it from the definition there? |

nvictor | yes |

JIMJONESBALLIN | or can you just use the rules for derivatives? |

nvictor | well, we are not allowed to use derivatives for the moment |

JIMJONESBALLIN | yes, which? ok well... f(x) = (3-5x)^(1/2) |

nvictor | :) |

JIMJONESBALLIN | so f'(x) = -5/2(3-5x)^(-1/2) Thats what you should get at the end. |