Recent content by mathing

I know very little about history and am interested to understand the basics of what modern history. What would you history buffs recommend so that I can get a decent overview? Thanks!

Re: HSC 2014 4U Marathon You got it!

Re: HSC 2014 4U Marathon - Advanced Level He's on fire today.

Re: HSC 2014 4U Marathon - Advanced Level Wonderful!

Re: HSC 2014 4U Marathon If you want to punish yourself have a go at this inequality: (lx + my + nz)^2 \le (l^2 + m^2 + n^2)(x^2 + y^2 + z^2)

1a) Look at the graph of it and the unit rectangles formed, then take the integrals 1b) As above 2) Prove then use a^3 + b^3 >= a^2 + ab^2 3) Looks like a typo, no d on LHS 4) apply (x + y + z) >= 3(xyz)^(1/3) with x -> 1 + x etc. we get (3 + x + y + z) >= 6 so we have x + y + z >= 3 Now...

Not sure if there is an easier way but you can think about how \theta changes with time and then you want to make sure the difference of these two functions is an integer multiple of 2 \pi. v_1 = 2 \pi r_1/t so we can solve for t and create \theta_1(t) which at that time is 2 \pi and so on.

Re: HSC 2014 4U Marathon - Advanced Level Please tell us the solution!

Re: HSC 2014 4U Marathon - Advanced Level Not sure if this is what you are looking for but we can apply the tan addition formula to get \lim_{x \to 0} \frac{\ln(\tan(x) + \tan(\pi/4)) - \ln(1 - \tan(x)\tan(pi/4))}{\tan(x)} \\ $Since $ \tan $ is continuous at$ x=0 $ we have:$ \\ \lim_{u = 0}...

Re: HSC 2014 4U Marathon - Advanced Level Nice Sean! Please post the alternative solution to continuing f for x < 0.

Re: HSC 2014 4U Marathon - Advanced Level I see. In the original question it says the 2nd derivative is continuous, so I wasn't sure what the assumptions were for the 2D case. Is the 2nd derivative existing still an assumption for the 1D case?

Re: HSC 2014 4U Marathon - Advanced Level In the 2D case do we assume all the right hand directional derivatives agree? If that's the case wouldn't it work.

Re: HSC 2014 4U Marathon - Advanced Level For the more complicated reflection, can we look at f'(0+) and f(0) then shift our origin up f(0) units on the y axis and apply a rotation by f'(0+) so it's now 0. With the derivative 0 in these co-ordinates can we now do the negative reflection...

Re: HSC 2014 4U Marathon - Advanced Level I think I see the extending way. We have the differential equation y''(0) = f''(0+) and y'(0) = f'(0+) and y(0) = f(0) so we can solve for some quadratic.

Re: HSC 2014 4U Marathon - Advanced Level Good counter examples :) I think we only have to worry above the 2nd derivative changing signs if the first derivative is zero at x = 0. Can we take the union of the two approaches? So if f'(x+) = 0 mirror it around x = 0 otherwise take the negative...