Recent content by elliotician

So that definition would give the 'average rate of change of P(x)' i.e. change in profit/change in quantity For average profit which is the same as any average (arithmetic mean) i.e. total/quantity

Yes but what about sinx?

Where did you get your definition of average profit? Average profit = Total Profit/Quantity

OK well I am working with Probability and random processes By Geoffrey Grimmett, David Stirzaker, the google books preview covers the section i am dealing with (1.1 and 1.2 right at the start). This uses the definitions given in the exercise...

For 3 we need the union of any two intervals to be in G For the sigma field property we want all possible unions of all possible intervals to be in GPoint 3 basically requires [0,x) be in G for x<=1, which we do have. In order for that sigma to hold we would need the interval (0,1) to be in the...

Thanks yes! OK so we say A=[ai,bi)U[a(i+1),b(i+1)U...U[ar,br) Ac = omega/A Then 3 is now easy A=[ai,bi)U[a(i+1),b(i+1)U...U[ar,br) B=[aj,bj)U[a(j+1),b(j+1)U...U[as,bs) so AUB = [ai,bi)U[a(i+1),b(i+1)U...U[ar,br)U[aj,bj)U[a(j+1),b(j+1)U...U[as,bs) AUB=[ak,bk)U[a(k+1),b(k+1)U...U[at,bt)...

Homework Statement Let \Omega = [0,1) Let G be the collection of all subsets of \Omega of the form [a1,b1),\cup[a2,b2),\cup...\cup[ar,br) For r any non-negative integer and 0<=a1 and a1 <=b1 <= a2 ... Show that G is a field Show that G is not a \sigma-field Homework...

Do you have a link or any follow up for this, would be interested to have a look :bugeye:

You can move through time. To the extent that if you travel fast enough, you could return to find the Earth extinct, or perhaps even the entire universe. Travelling at light speed you would experience zero time and so presumably you can travel to the end of the universe! I don't study physics...

I meant n>=5 :)

So to conclude By the standard comparison test: since (1-1/n)^3 >= 1/2 for n>=2 and n\sqrt{1+n^{-7}+2n^{-8} <= 2n we must have \frac{(1-1/n)^{3}}{n\sqrt{1+n^{-7}+2n^{-8}}}\geq\frac{1/2}{2n}=\frac{1}{4n}\] More simply by the limit comparison test...

Ah no i made a mistake! (again!) Limit comparison works, tends to 1 when i divide by 1/n. Thanks!

No i meant for the comparison test, however if i use 1/n in the limit comparison i get an/bn tending to 0

I do know the limit comparison however i just get the sequence tending to 0 rather than a finite L. So doesn't seem to be of use Comparison test should work for 1/4n, for n>=2 \frac{(1-1/n)^{3}}{n\sqrt{1+n^{-7}+2n^{-8}}}\geq\frac{1}{4n}\] So by comparison test the series diverges?

I'm not thinking clearly at the moment, need a break, but I think 1/4n would work though for n>=3.