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Be more careful please. It does hold for n=4. Let the quadrilateral be ABCD, and the diagonals intersect at K. Then AK+BK > AB BK+CK > BC CK+DK > CD DK+AK > DA Adding, we get 2(AK+BK+CK+DK) > AB+BC+CD+DA so with d=sum of diagonals, p=perimeter, 2d > p and p/4 < d/2, hence it is true for...

No.That's not pathetic at all. Your question (except for the n>=5 bit, which really should be n>=4) is equivalent to the first part of the 1984 IMO Q5: Let d be the sum of the lengths of all the diagonals of a plane convex polygon with n > 3 vertices. Let p be its perimeter. Prove that n - 3...

I put up my solutions to the 2005 HSC paper at http://www4.tpgi.com.au/nanahcub/2005hscsol.pdf

Let n be a positive integer > 1 and &sigma;(n)=sum of positive divisors of n and h(n)=n-th harmonic number=1/1+1/2+...+1/n and f(n)=h(n)+e^h(n)ln(h(n)). Then f(n)>&sigma;(n) This is equivalent to the Riemann Hypothesis and is easy to verify on the calculator for small values of n...

There are some Newton's method approximation questions on cubics and quartics in past HSC papers, and although it's not required in these questions, nevertheless it is possible to solve them exactly using the cubic and quartic formulae: Solve for x&epsilon;R: 1...

No need. It is already at the 2 unit resources page http://www.boredofstudies.org/view.php?course=1

Good. Thanks for that. So &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;))=4.000000303.... > 4 (by calculator) and therefore &pi;⁴+&pi;⁵ < e⁶. But if the idea is to avoid calculators altogether, won't you need to prove &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;)) is...

I tried it a similar way to brett86 and got &pi; < 3.14159266 & 2.718281828 < e. So &pi;⁴+&pi;⁵ < 3.14159266⁴+3.14159266⁵ = 403.4287797363725532846657585016733756588576 (exactly) <...

I disagree. "e=2.7... . Therefore e < 3" is a perfectly valid proof.

I just think it's sad that no_arg still thinks &pi;⁴+&pi;⁵=e⁶ exactly. It doesn't. &pi;⁴+&pi;⁵=403.42877... and e⁶=403.42879... Hence &pi;⁴+&pi;⁵&ne;e⁶ It's quite simple, really.

I was trying to edit a post in the <a href="http://community.boredofstudies.org/showthread.php?t=81019">pi and e thread</a> when McLake closed it. The - should have been a + in my last post in that thread at the bottom. So here it is again. Here's another proof using only integers...

Here's another proof using only integers. e*10⁹ > 2718281828 and &pi;*10⁸ < 314159266 Hence (e⁶-&pi;⁴-&pi;⁵).10⁵⁴ > 2718281828⁶-314159266⁴*10²²-314159266⁵*10¹⁴...

Unfortunately, they are both offline at the moment, so we'll have to wait till they come back later.

Exactly! And the reference to Castellanos is Castellanos, D. "The Ubiquitous Pi. Part I." Math. Mag. 61, 67-98, 1988. no_arg has no publication refuting Castellanos's claim, and if he ever submits such a thing for publication, it would get rejected!

Well, you're the one claiming &pi;⁴+&pi;⁵=e⁶ exactly, and your invective does not constitute proof. I've now proved &pi;⁴+&pi;⁵&ne;e⁶ two ways, and several others on this thread have also proved it. You're still claiming...

How about this. &pi; < 3.14159266 & 2.718281828 < e. So &pi;⁴+&pi;⁵ < 3.14159266⁴+3.14159266⁵ = 403.4287797363725532846657585016733756588576 (exactly) < 403.428793083965001476126676589903866851868865301397704704 =...

1989 four unit HSC Question 8(b)(i)(&alpha;) "The difference between a real number r and the greatest integer less than or equal to r is called the fractional part of r, F(r). Thus F(3.45) = 0.45. Note that for all real numbers r, 0 < F(r) < 1. Let a = 2136 log₁₀2. Given that...

Yeah. Anyway at least we know that &pi;⁴+&pi;⁵ is not equal to e⁶! no_arg said they are equal, so I guess you could just deduce from this that they aren't! Really? Well how about this. If f(x)=sin(x), then f''(x)=-sin(x), so f''(0)=0, and f'''(x)=-cos(x)...

It is known that they all lie in the critical strip. The Riemann Hypothesis says that all the nontrivial zeros of &zeta;(s) lie on the critical line, Re(s)=1/2. If that's what you meant to say, then theoretically, yes. But you're a braver man than me if you try to do it by hand. Not even...