Search results

Re: HSC 2014 4U Marathon Another opportunity to illustrate the power of Jensen's inequality! $As $f(x):=1/\sqrt{1-x}$ is convex on $(0,1)$, we have:\\ \\ $LHS=\sum_j f(x_j) \geq n f\left(\frac{1}{n}\sum_j x_j\right)=nf(1/n)\\ \\=\frac{\sqrt{n}}{\sqrt{n-1}}\geq \frac{\sum_j...

Re: HSC 2014 4U Marathon Haha yep, right idea just a bit of clutter.

Re: HSC 2014 4U Marathon $If $a<b$ are the two rationals, then $a+\frac{1}{\sqrt{2}}(b-a)$ is an irrational between them.\\ \\ It is easy to check that it lies between $a$ and $b$, and if it were rational, we could deduce that $\sqrt{2}$ was rational, which is absurd. $

Re: HSC 2014 4U Marathon Using that the sum, product, and composition of continuous functions are continuous. (And that the exponential function a^p is continuous). Actually "proving" these things is sort of overkill, and comes pretty easily from the epsilon-delta definitions of limits and...

Re: HSC 2014 4U Marathon Nice! Yep, to get to rationals from here: If a < b are rationals, we can choose a common denominator and write a = p/r b = q/r where p < q. We have M_p =< M_q, let x_k=y_k^{1/r} (and note that this is a bijection on the non-negative reals.) Convert every 'x' in...

Yep, the problem is when you "visualise" that upper rectangles would suffice, I assume you have a nice and continuous function f in mind. If f is not continuous, there is no reason to be sure that upper Riemann sums and lower Riemann sums converge to the same fixed value (which we then call the...

Re: HSC 2014 4U Marathon Well you are just dealing with a single variable problem atm (in trying to see what happens when p->0 while the xj are fixed). Think of what sort of tools you have for looking at limits and manipulate to bring the expression you are looking at to a form that is more...

Re: HSC 2014 4U Marathon Could work. I did it a little differently, but the thing I used would have to be proven separately in an MX2 level solution.

Re: HSC 2014 4U Marathon Whoops, my bad, I forgot to include a factor inside the brackets. You need to average the power sum before taking the p-th root. Edited appropriately.

Re: HSC 2014 4U Marathon $Let $x_1,x_2,\ldots,x_n$ be positive real numbers. Let \\ \\$M_p:=\left(\frac{1}{n}\sum_{j=1}^n x_j^p\right)^{1/p}$\\ \\ for positive real $p$.\\ \\ Prove that $M_p\leq M_q$ whenever $p\leq q$.\\ \\ Prove that $\lim_{p\rightarrow 0^+} M_p = \left(\prod_{j=1}^n...

Get back into exercise. Get decent at chess again. Get published. Learn a few more songs (guitar).

Not that many careers will require you to explicitly be able to use say calculus or trigonometry. But, as others have mentioned, the same is true for every other subject in the HSC. It is the SKILLS rather than the knowledge that will be of use. The skills you get from essays transfer to skills...

Re: HSC 2014 4U Marathon Nope, there will always be four knights on dark squares and four knights on light squares (as this is the case to start with, and each move changes the colour of each knights square).

I wouldn't call it a "dumbo move". I essentially did the same thing with Chemistry and it worked out fine (kept doing 12 units to appease my parents but never went to chem and didn't really do anything for it). At the end of the day, you know your capabilities so it's your call, but I don't...

Re: HSC 2014 4U Marathon You would of course have to check the b=0 case separately, but it is obvious this won't give us any issues thanks to symmetry. Regardless, here is a third way :p. By Cauchy-Schwartz twice: (a+b)^4=((a+b)^2)^2\leq 4(a^2+b^2)^2\leq 8(a^4+b^4).

Re: HSC 2014 4U Marathon I can't really imagine such a thing existing, but I might well be wrong of course. I guess "easier" is pretty subjective.

Re: HSC 2014 4U Marathon Oh right cool, yeah those triangles are exactly the ones in my construction. Just numbers vs a picture.

Re: HSC 2014 4U Marathon Haha I know, I normally don't answer these for a while so other people have a chance. Felt like spoiling the fun today though. Am interested to see your first solution, did you do it nonconstructively or something?

Re: HSC 2014 4U Marathon T_n+T_{n-1}=\frac{n(n+1)}{2}+\frac{n(n-1)}{2}=n^2\\ \\$If $n^2=2T_k=k(k+1)$, we have:$\\ \\k<n<k+1,\\ \\$a clear contradiction.$

I agree that it is quite an intuitive fact that the circle is optimal in the isoperimetric sense, but the point is to find a mathematical way to PROVE this. The whole point of proof is that it gives us absolute certainty. I could easily list about 50 counterintuitive facts in mathematics. When...