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oh math

i topped 4u :)

Re: MX2 Integration Marathon i subbed u = x-x^2 in, solved the quadratic for x (take positvie root, its the same either way i think), and cancelled difference of two sqaures and solved it throguh. got different answer -1/3(1-2u)^3/2 + (1-2u)^1/2 it might be the same expressiont hough

Re: HSC 2013 4U Marathon oh yhy, i wrote the wrong thing online

Re: HSC 2013 4U Marathon im not quite sure if this is right?? i got the expression I = k^n n!/(m+1)(m+n+1)(m+2n+1)...(m+kn+1) plus if you sub in m = 1, n= 1, k = 100, the inequality doesn' work. i dunno, i might be worng about this

i would like to go to the 4u, if its not already filled. thanks

Re: HSC 2013 4U Marathon seemed a bit on the easy side for putnam surely???

Re: HSC 2013 4U Marathon Is it (-1)^n-1*n!*k^n???

Re: HSC 2013 4U Marathon oh same solution as sys thatll teach me to scroll down

Re: HSC 2013 4U Marathon i might have a quicker way and easier to find, looking at x^6(y^-2 - x^-2) > y^6(y^-2 - x^-2) *just assume the greater than sign is greater than or equal to, this inequality can be easily shown for x>y and y>x to hold true, then all you have to do is move the...

Re: HSC 2013 4U Marathon oh actually that wouldn't work would it, ok then

Re: HSC 2013 4U Marathon just tell me its not P(|x|), before i go any further with this lol

Re: HSC 2013 4U Marathon oh right yeh, n-1 for the second part. your method is pretty neat too. The one jsut above it, though, i have a solution for, but it was pretty long, and involved the pascal addition theorem and induction and looking at n = even and n= odd case, though it eventually...

Re: HSC 2013 4U Marathon thought of something abit dodgy for this one, looking at (1-x)^n/(1-x) = (1-x)^n(which is true) for |x|<1, if we equate x^n coefficients the result comes out, by eqauting 1/1-x as 1+ x + x^2 + x^3... how did you guys do it????

Re: HSC 2013 4U Marathon tan-1(n/n+1) - tan-1(n-1/n) = tan-1(1/2j^2)

actually you don't need a lot of algebra if you do it the simple way, right down your equation of motion tanOx - x^2sec^Og/2v^2 + h = 0 and then implicitly differentiate that. You get the condition that tanO = V^2/gx, which if you sub that in, kills it. its much easier than fiddling around...