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

Cinwhat 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.
Cinwhat does ab mean? what is that dot? is it showing up?
nexactit's a multiplication.
but I may be wrong...
Cindo 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.
nexactCin, 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.
JIMJONESBALLINBut of course, x is used for the cross product
Cinnexact, ah, thanks! indeed, in school we wrote the letter x in italics to differentiate
rhodopsinegiven 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]
mbotnvictor: -5/(2*Sqrt[3 - 5*x])
nvictor% Limit[Sqrt[x-5a]-Sqrt[x],x->3]
mbotnvictor: -Sqrt[3] + Sqrt[3 - 5*a]
nvictoranybody online?
nvictorI 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
JIMJONESBALLINdo they want you to compute it from the definition there?
JIMJONESBALLINor can you just use the rules for derivatives?
nvictorwell, we are not allowed to use derivatives for the moment
ok well...
f(x) = (3-5x)^(1/2)
JIMJONESBALLINso f'(x) = -5/2(3-5x)^(-1/2)
Thats what you should get at the end.

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