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
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 G
Point 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...
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...
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...
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...
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 think the arbitrary term in my series is less than 1/n except for very small values of n below zero, so 1/n isn't suitable.
With some manipulation we can express the nth term as as:
\frac{(1-1/n)^3}{n\sqrt{1+n^{-7}+2n^{-8}}}
Homework Statement
Determine whether the following series converges:
\sum \frac{(n-1)^3}{\sqrt{n^8+n+2}}
Homework Equations
Definition of convergence:
Let \sum a_{n} be a series.
If the sequence of (sn) partial sums converges to L (finite). Then we say the series converges to L or has...
If he says yes in the first question you know he is lying so its on at 9.
If he says yes in the second question you know he is telling the truth so its on at 10.
If he says no in either case your at a dead-end, you would have to keep asking both questions till you got a yes. May be a better way...
Is it true that you are lying and the football is on at 9?
If he is telling the truth he will say no since he is not lying as the statement is always false when he is telling the truth.
If he is lying then he will say yes if its on at 9 and no for 10.
Using the and logical operator means he...
I'm guessing you have probably gone wrong somewhere unfortunately i don't know where. But the solution to your integral:
http://www5a.wolframalpha.com/Calculate/MSP/MSP8641961aa713bih43db00000ebbd4gc0a51fd4i?MSPStoreType=image/gif&s=15 [Broken]
Thanks both,
I think then if i let gu(x,v) = f(x, v)
Then i will integrate as before to get
\partial /\partial u [g(u,v)- g(a,v)]= g_{u}(u,v)
so following on we can say if
gu(x,v) = f(x, v)
then
gu(u,v) = f(u, v)
(since x is the dummy variable)
Would this be correct...
Homework Statement
Show
\partial /\partial u \int_{a}^{u} f(x,v) dx = f(u,v)
Homework Equations
The Attempt at a Solution
Basically i understand that we hold all other variables constant, and i understand that we will get our answers as a function of u and v. But to show that we have...