ohexploitable hey Joined Dec 31, 2009 Messages 4,741 Location sarajevo Gender Female HSC 2015 Uni Grad 2017 Jan 15, 2011 #2
ohexploitable hey Joined Dec 31, 2009 Messages 4,741 Location sarajevo Gender Female HSC 2015 Uni Grad 2017 Jan 15, 2011 #3 Last edited: Jan 15, 2011
NewiJapper Active Member Joined Jul 19, 2009 Messages 1,010 Location Newcastle Gender Male HSC 2010 Jan 15, 2011 #4 Wow! How the hell do you manage to do that doing General Maths!?
NewiJapper Active Member Joined Jul 19, 2009 Messages 1,010 Location Newcastle Gender Male HSC 2010 Jan 15, 2011 #5 ...................lol
ohexploitable hey Joined Dec 31, 2009 Messages 4,741 Location sarajevo Gender Female HSC 2015 Uni Grad 2017 Jan 15, 2011 #6 NewiJapper said: Wow! How the hell do you manage to do that doing General Maths!? Click to expand... We do that stuff in General
NewiJapper said: Wow! How the hell do you manage to do that doing General Maths!? Click to expand... We do that stuff in General
B Bored Of Fail 2 Banned Joined Jan 13, 2011 Messages 354 Gender Male HSC N/A Jan 15, 2011 #7 cant you do the first part, showing it instead of stating it, you are a mad noob ohexploitable, that parts easy
cant you do the first part, showing it instead of stating it, you are a mad noob ohexploitable, that parts easy
D deterministic Member Joined Jul 23, 2010 Messages 423 Gender Male HSC 2009 Jan 15, 2011 #8 see below.... Last edited: Jan 15, 2011
D deterministic Member Joined Jul 23, 2010 Messages 423 Gender Male HSC 2009 Jan 15, 2011 #9 (a) Let P(x)=px^4+qx^3+rx^2+sx+t where p, q, r,s, t are real P(a)=0 implies pa^4+qa^3+ra^2+sa+t = 0 Show P(conj(a))=0 Sub into P(x), use the fact conj(a^n)=conj(a)^n, pconj(a)=conj(pa) if p is real. try it yourself
(a) Let P(x)=px^4+qx^3+rx^2+sx+t where p, q, r,s, t are real P(a)=0 implies pa^4+qa^3+ra^2+sa+t = 0 Show P(conj(a))=0 Sub into P(x), use the fact conj(a^n)=conj(a)^n, pconj(a)=conj(pa) if p is real. try it yourself