The Proof is Trivial!

This is favourite of mine. A lot of you will know it, so please don't give the answer away if you've seen it before

Problem 43 **/***

Evaluate $\displaystyle \lim_{m \rightarrow \infty} (\lim_{n \rightarrow \infty} \cos^{2n}(m!\pi x))$

(To do it rigorously you obviously need analysis, but don't bother if you've done no Analysis)

My guess is that it's 1 if x is rational and 0 otherwise.

Indeed! Can you explain why?

Problem 42*/**

Show that $\displaystyle\sum_{r=0}^{88} \dfrac{1}{\cos r \cdot \cos (r+1)} = \dfrac{\cos 1}{\sin^{2} 1}$

(angles measured in degrees)

Solution 42:

$\dfrac{1}{(cosr)(cos(r+1))} = \dfrac{A}{cosr} + \dfrac{B}{cos(r+1)}$

$Acos(r+1)+Bcos(r) = 1$

Let r = 90:

$Acos91 = -Asin1=1 \Rightarrow A = -cosec1$

Let r = 89:

$Bcos89 = Bsin1 = 1 \Rightarrow B = cosec1$

As such, we are summing:

$f(r) = \dfrac{cosec1}{cos(r+1)} - \dfrac{cosec1}{cosr}$

In this sum, all terms except for two will cancel out:

$S = \dfrac{1}{cos89sin1}-\dfrac{1}{sin1}$

$= \dfrac{1-sin1}{sin^21}$

We know that:

$cosec1cosr-cosec1cos(r+1)=1$

Let r = 0:

$1 - cos1 = sin1$

So cos1 = 1 - sin1, so our sum is $\dfrac{cos1}{sin^21}$, as required.

See that's what I did except.

cos(1)=0.9998476952
1-sin(1)=0.9825475936.

So I got confused an gave up. I checked the first bit out numerically too and that was dodgy too.

I know. I don't like it; and wouldn't have submitted the solution if it weren't for the partial fractions bit saying that it's apparently true.

The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter?

I think the problem is with the partial fractions bit. You're finding A and B only for the cases where r= 90 and 89. It isn't an identity.

The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter?

Problem 41***

A (countably) infinite class of Cambridge students who believe in the Axiom of Choice are taking their finals for the Mathematical Tripos. At the start of the exam, the examiner places either a white or a black hat on each student at random. There is only one question in the exam: "What colour is your hat?" If only a finite number of students answers incorrectly then everyone becomes a Wrangler (very good), otherwise the Wooden Spoon is given to the whole class (very bad). Everyone can see the hats of everyone else besides their own, and since it's exam conditions they are not allowed to communicate. Students may not remove their hats or try to look at it in any way.

What strategy could the students devise together so to avoid failure and the impending doom of unemployment? Being clever clogs from Cambridge, memorising infinitely large amounts of data is no problem to the students.

Note: A student suggested that each person just guess at random. However this has already been ruled out by the cleverer students who realised that if they were really unlucky, infinite of them could guess wrong.

I've got it. It you plot lateral binary graph associated with the symbolic mathematical solutions, you can reduced the problem down to simple SNF's. And by using Dalek's contraction algorithm the answer is: the physical quantity of dimensions m to the ml to the lq to the qt
to the ttheta to the x has planck quantity root g minus m plus l minus 2q
plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q.

This is what I was thinking so I'm guessing it's wrong!

You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confused

You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confused

This is why I said I found the order of the limits confusing. In the case m<q, the first limit is 0, but then in the second limit we have m->∞. I would then say that for m<q, the limit is 0, but you told me that it was 1. Hmm...

I dunno. We'll probably have to find out what Felix was after; as the solution isn't an agreeable one.