## #math - Sun 25 Feb 2007 between 00:00 and 01:26

### NY Lost Funds

 Cin what does this dot mean? 211what 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 leftgiven 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 + 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.