#math - Fri 9 Mar 2007 between 17:39 and 20:02

NY Lost Funds



deltaihope, ok. a, A, d and D are arbitrary?
ihopeYes, arbitrary real numbers.
deltaihope, did you try something like p(X) = bX^2 + (a-d)X - c, etc?
ihopeWhat would that do?
deltaihope, the discriminant of this trinom is the first term. Being < 0 meaning there's no real root.
ihope, same for the second. So, the third one is the discriminant of p + q.
ihopeThe third is the discriminant of p+q?
(b+B)^2 - 4(a+A)(c+C) = b^2 - 4ac + B^2 - 4AC, isn't that?
delta? :)
(p + q)(X) = (b+B)X^2 + (a - d + A - D)X - (c + C).
ihopeOh yeah.
deltaSo, if bB > 0, you're done.
Investigate others cases maybe.
ihopex^2 + (a+d)x + ad-bc is the polynomial I got the stuff from.
At least, I think it is.
Also, the existence of a nice definition of e^x is nice, no?
mike8901can someone guide me to the solution of this linear algebra question? The question is: If the vector (a,b) is a multiple of the vector(c,d) show that the vector(a,c) is a multiple of (b,d). The hint given(a very vague one) is that if A=[a,b;c,d] then AB=I and BA=I. However, I thought the matrix wouldn't be invertable.
*wouldn't have an inverse
seb--mike8901: what do you mean 'multiple of the vector' ?
mike8901: (a,b) = k (c,d) for some integer k?
mike8901(a,b)=s*(c,d) where s is a scalar multiplier
FatalErrormike8901, consider [a,b] = [c,d] and since [c,d] is a multiple of [a,b] we can rewrite it k[a,b] = [ka, kb]
mike8901right
seb--ok
mike8901FatalError: I got that far :P
so [ka,kb]=[c,d]
then ka=c kb=d
FatalErrorwell then you have the vectors [a, ka] = j*[b,kb] -> a*[1,k] = j*b*[1,k]
seb--yes..i see it now too
mike8901hmm
seb--mike8901: so are you happy?
mike8901not really :/
[a, ka] = j*[b,kb] how did you get there?
seb--mike8901: he worked backward from your goal
mike8901: do that
mike8901: ASSUME what you are trying to prove and massage it into something obvious
doyley|ukDCC SEND MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
analpearhi
fruitbagHey
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