I get it now. Thanks, TheDude.
Incidentally, are you all (apart from Ricky) only assuming that x is positive? Can't x be negative as well? After all, if you multiply a number four times, you do get a positive number. Ricky's method may allow for negative numbers but I don't really know what you can do with the method.
From all the numbers I've tried, I'm inclined to believe that there is no negative solution. I'm just wondering if you can definitely prove that no negative solutions exist.
We can prove thatis increasing by contradiction.
Suppse it is not. Then there existin I such that .
Alsois either or .
Thus we have eitheror . This contradicts the fact that both f and g are increasing on I.
Thereforeis increasing on I.
You can prove thatis also increasing on the interval I by the same method.
1. A function :R->R has a PROPER RELATIVE MAXIMUM VALUE at c if there exists d>0 such that f(x)<f(c) for all x that satisfy 0<|x-c|<d. Prove that the set of points at which has a proper relative maximum value is countable.
This is one of those questions which are intuitively obvious but present such a challenge to prove. I can only suggest using the metric-space properties of the real numbers. Try proving that the set of points at which has a proper relative maximum value is nowhere dense in R. Countability should follow from the fact that all nowhere-dense subsets of R are countable. (Are they? I think so anyway. )
In theory, the betting game could go on forever, but in practice no game can go on forever. You can't possibly spend an eternity in the casino. Besides, you'll miss your bus.
A more practical phrasing of the question might be, say, what is the probability that you have £50 within 50 bets? Assuming you only have time for 50 bets before your bus arrives.
I *think* this result can be generalised so that instead of f' and f'', you have f[sup](m)[/sup] and f[sup](n)[/sup], where m and n are respectively odd and even.
You'd also need f to be differentiable max(m,n) times rather than just twice.
Haven't worked out a solid proof yet though.
Intuitively, f', f'' > 0 means the function is convex and strictly increasing; such a curve should intuitively go to infinity as x goes to infinity. I just wanted a concrete analytical proof of what was intuitive to me.
In fact f' and f'' don't have to be positive on the whole real line, they just have to be positive on the interval [a,∞) for some real number a. This can even be easily shown - just replace 0 by a in the your proof above.