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Yeah that works deterministic, some sort of squeeze theorem is needed to make the argument rigorous though. (We cannot write things like f'(a) before we know that the defining limit exists.) edit: mathman, you haven't really found a contradiction...in your working "K" is not necessarily...

And how does that imply that f is constant? (Also you have made the assumption that f is differentiable everywhere, this follows from the given inequality though.)

$Nice question!\\ Suppose first that $g$ is non-decreasing.\\ This immediately gives us: $f(x+y)-(1+y)f(x)\geq e^xg(x)(e^y-1-y)\geq 0.\\$(Where the last inequality follows from using basic calculus to show that the bracketed term is non-negative for non-negative $y$).\\ To prove the converse...

Tiny signs mistake in your first line of working but otherwise perfect :)

$Let $f$ be a function with domain the real line. Show that:\\ $|f(x)-f(y)|\leq |x-y|$ for all $x,y\in\mathbb{R}$\\does \emph{not} imply that $f$ is constant, whilst \\$f(x)-f(y)\leq (x-y)^2$ for all $x,y\in\mathbb{R}$ \\does. $

$By summing the geometric series we get: \\$\sum_{k=0}^n \cos((2k+1)\theta)=\frac{1}{2}\sum_{k=0}^n z^{2k+1}+z^{-2k-1}=\frac{z^{2n+2}-z^{-2n-2}}{2(z-z^{-1})}=\frac{\sin(2(n+1)\theta)}{2\sin(\theta)}.

Yep, so actually not that hard a question apart from being fairly unintuitive. You can replace the property of "birth on Thursday" by an arbitrary property. If the property is rare, your probability will be close to 1/2, if the property is common, your probability will be close to 1/3 :).

you have my kudos :) try the question including the Thursday part though, bonus point if you are able to explain why that makes a difference :P

Ignoring the "Thursday" part for now (which in my opinion brings up the most counter-intuitive aspect of the question). The critical reason why this simpler question has answer 1/3 and not 1/2 is that we are not stipulating order at any stage. So asking "What is the probability of the other...

;) Its not as silly as it sounds MrBrightside.

Try this one: A couple has two children, one of whom is a boy who was born on a Thursday. What is the probability that the other child is also a boy? (This isn't a lateral thinking question). Edit: Although admittedly with more precise wording, the problem messes with the mind less. It is...

\pm $ means plus OR minus, not plus and minus, the reason you get a polynomial that works in the end is entirely because of the squaring step. Try it for the example I suggested and you will see that $ P\left(\frac{x+\sqrt{x^2-4}}{2}\right)\neq 0$ at $x=5/2.

My point is that you only end up with an equation that is satisfied at (a+1/a, b+1/b, c+1/c) after the step where you square both sides to eliminate the square root sign. It is not obvious that this method is valid without use of the +- sign. If you try the same method for say...

$Although it leads you to the correct answer for this particular polynomial, simply substituting $ \frac{x+\sqrt{x^2-4}}{2}$ into your polynomial is a little dodgy since: $\\y=x+x^{-1}$ does \emph{not} necessarily imply that $ x=\frac{y+\sqrt{y^2-4}}{2}.\\$ So you run into problems for...

Not a clue, personally I would do the extra work to be safe...the BOS is kind of unpredictable in their expectations of rigour.

$Not quite, the uniqueness of $ p_k $ and $ q_k $ indeed implies that $ p_{k+1} $ and $ q_{k+1} $ are uniquely determined as defined using your recursive definition below.\\ However what is to say that there does not exist a different pair: $ (p'_{k+1},q'_{k+1}) $ which also satisfy the property...

BT is more about group theory yes, but the necessity of AC for constructing nonmeasurable sets feels more like set theory. The group theory comes into play when making the statement about moving the finite number of pieces of the sphere to reform two spheres, it has to do with the group of...

Intuition i guess, the best hope of evaluating series like these is either by relating them to a series you already can evaluate or using a telescoping sum. Recurrence relations are usually a pretty good indicator that the latter will be applicable.

$hups question:\\ From the recurrence relation given: $\frac{1}{1+u_n}=\frac{u_n}{u_{n+1}}=\frac{1}{u_n}-\frac{1}{u_{n+1}}.\\$Hence by telescoping: $\sum_{k=1}^n \frac{1}{1+u_k}=\frac{1}{u_1}-\frac{1}{u_{n+1}}.\\$It is easy to see that $1/u_n\rightarrow 0$ so we can conclude that...

Hey yeah, and simply replacing ZF with another set theory in order to avoid invoking AC feels like cheating. Its pretty easy to get bogged down in these sort of things haha...set theorists are crazy. I'm studying these highly symmetric functions on hyperbolic space sometimes known as...