The Proof is Trivial!

Really? How did they used to say it? And how did they distinguish between a request to prove something and a statement of fact?

You know how in textbooks these days have a set of exercises? Imagine those, except any question which asked you to prove something would just look like a statement of fact. I think it was to save space and paper which these days isn't as a big a deal

I did a sneaky check on wolfram before I posted to check whether 'a' gave three distinct solutions... it did!

One thing that bothers me about the question is that it asks to find the sum of the values, when there seems to be only one value.

Nope. As the expression, $a_{n}=m\sqrt{n}$ is not correct, the subsequent equalities in your solution are also incorrect. And finally, your conclusion is not true.

Really, I do not know. After the first exercise the authors said: In subsequent propositions, we shall omit the words "Prove that" and "Show that", and thus stating all problems as theorems.

Let me suggest another lovely functional equation. It seems as though you are the only one who enjoys doing functional eqs.

Problem 99**

Find all continuous functions $f(x)$, defined for all $x>1$, which satisfy $f(xy)=xf(y)+yf(x)$, for all $x,y >1$.

I got 99 problems but my bitch ain't one

Problem 99*/**

Rigorously prove that the area of a rectangle of side lengths $p$ and $q$ is $pq$ (for real numbers $p, q$).

a) Surely it depends upon the units used?

b) Does using integration count? Or is that circular reasoning?

Problem 99**

Find all continuous functions $f(x)$, defined for all $x>1$, which satisfy $f(xy)=xf(y)+yf(x)$, for all $x,y >1$.

Yes. You will need to define the concept of area in your proof.

That would form a circular argument I do believe. Do note that in an elementary proof of this type, using calculus is a little bit mad (and one may argue that in order to be rigorous you would have to include a proof, not only of the fundamental theorem of calculus, but of this operation in a 2 dimension plane being equivalent to area when considering positive values.)

Btw, by rigorous I don't mean for it to be of a university level of rigour, but just explanative justification.

What makes you think that? :P

Oh hmm... But if you were to suppose it tended to a finite limit why is it incorrect to suppose that $\frac{a_{n}}{\sqrt{n}} \to m$ as $n \to \infty$? (and then proceed in construction a contradiction)

If you don't know why a convention is used, why use it?

And I like functional equations too, I'm just nowhere near as good at them. The only widely applicable way of solving them I know is trying special values, like $x=y,\ x=1,\ x=0$ etc.

This problem looks a lot like product rule, although I'm not sure if that has any significance
If you take the case $x=y$, then $f(x^2)=2xf(x)$.

If $f(x)=ax+c$, then $f(xy)=axy+c=2axy+(x+y)c$. As both x and y are positive integers greater than one, the only possible solution is if both constants are 0. Therefore a possible solution is $f(x)=0$
This is about as far as I've gotten

The best I can do at this hour of the night:
$(*)f(xy)=f(xy') \iff \dfrac{f(x)}{x}=-\dfrac{f(y)-f(y')}{y-y'}$
let $y'=y+\epsilon$ and let epsilon tend to zero giving:
$\dfrac{f(x)}{x}=-\dfrac{df(y)}{dy}$. Continuity guarantees existence.
let y=x:
$\dfrac{f(x)}{x}=-\dfrac{df(x)}{dx}\Rightarrow f(x)=-Ax$
but $f(x^2)=2xf(x) \Rightarrow A=0$
so f identically zero is only option.
I suspect there's a flaw as calculus is very often fickle.

Well. If we take $\displaystyle \lim_{n\to \infty} \frac{a_{n}}{\sqrt{n}}=m$, from the recurrent relation, we obtain $\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{\sqrt{n}}=\lim_{n\to \infty} \frac{a_{n}}{\sqrt{n}}+\lim_{n\to \infty} \frac{1}{a_{n}\sqrt{n}}$, which is $m=m$.
I doubt that you can obtain a contradiction for this part.

There are hundreds of techniques when it comes to functional equations. It is annoying that there are no books on functional eqs

Why is it not valid to omit the limit notation if accompanied with the assumption that n is very large? (sorry if these are stupid questions)

I've never really tried functional equations questions. For me these methods seem non-rigourous (not justifying that all solutions are found). Similar to curve sketching

There are two possibilities:
You write either: $\displaystyle \lim_{n \to \infty} \frac{a_{n}}{\sqrt{n}}=m$, or: for each positive number $\epsilon$ there exists $N$ such that for all $n > N$, $\displaystyle |\frac{a_{n}}{\sqrt{n}}-m|<\epsilon$.
This is the concept, you have $\displaystyle \frac{a_{n}}{\sqrt{n}} \to m$ as $n \to \infty$, not $\displaystyle \frac{a_{n}}{\sqrt{n}}=m$ for all sufficiently large $n$; in other words, for all sufficiently large $n$, $\displaystyle \frac{a_{n}}{\sqrt{n}}$ is arbitrarily close to $m$ but it is not equal to $m$.

I shall post my solution to problem 97 in order to demonstrate rigorous solution to functional equation. Although, I would say that Lord of the Flies' solution is perfectly justified.

Solution 100

Right! I have finally found the golden nugget!

Solution 90 (part 2)

We require that there are 3 distinct real roots. We must conclude, based on the fact that the polynomial is of degree 4 (i.e. cannot be reduced to a polynomial of a lower degree) and that all imaginary roots come in pairs (i.e. theory of conjugate pairing), that there are 4 real roots. From this the deduction has been made that there exists exactly one repeated root. This is correct.

What isn't correct is that we have assumed that, in the quadratic factorisation found, the repeated roots must be that of those in the same quadratic. This is incorrect! (I believe siklos addresses issues like this in two different questions in A.P.I.M.)

One way to address this is to find all other ways in which to factories the polynomial. Allow me to propose another strategy. Starting from the factorisation found in your solution, we deduce two possibilities either the roots are shared between one of the polynomials (leading to $a=\frac{23+4\sqrt{19}}{100}$ only) or one root from one is equal to one root from another. There is only one such root for which this could occur and the other roots must be equal and opposite (there are a number of strategies that could be used but we only need to find one so observation is sufficient). A keen eye will spot that both quadratics have the root x=0 when 1-a=0 (i.e. a=1).

Substituting this in yields $x^2(x+3)(x-3)=0$ and hence, for a=1, the quadratic has roots x=0, x=3 and x=-3. Therefore $\sum a = \frac{123+4\sqrt{19}}{100}$. QED.

Ok, I will add the last part to my post too. That site brilliant.org is fantastic! When you complete a whole week's solution can you have a go at the next level or do I have to wait until next week now?

Okay awesome. I hope people besides those who post are doing/trying some of these problems. It's such an awesome problem-solution collection that has been created We've completely gone into olympiad territory now though. Would be really nice to have some more like problem 66 don't you think? I was so chuffed with that one! It's so nice you the irrationality of pi is a consequence of the convergence of the maclaurin expansion for e

Unfortunately not! I managed to complete all the number theory and geo/combinatorics last week on level 4 and so found out that way

My advice to you if you have just started on the site is to try and work out what level you want to be at and make sure you don't rush their "initial test" (take time and get some paper!) What I did was to fire through the first 3 questions on each test and then pretty much guessed the 4th on each one. This resulted in me being placed in level 3 for both olympiad sets (though working your way up is good training). I am such that you would be suited to starting on level 4 or 5 so make sure you do well on the test!

Also note that the format/style of the questions are not similar to british olympiad questions but are very similar to the US's.

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May I ask what cyc on the bottom of a sigma means?