Does anyone else fist pump when they get an exercise question right when doing homework? :embarassed:

so it turns out my solution isn't quite limiting enough. i'm going to keep working on it but ultimately i'm looking for a finite upper bound on the number of powers of 2 in the sequence 2^(2^k)+k+n for some arbitrary integer n as k goes to infinity. does anyone have any insight as to whether or not this exists or not and possible approaches to the problem?

Can you state precisely the theorem you'd like to prove? Maybe I'm being dumb, but I don't understand.

i am trying to find whether or not there is a finite upper bound independent of n for the numbers of powers of two in the sequence 2^(2^k)+k+n. n is an arbitrary integer and k goes from 0 to infinity.

if n=0 the sequence is 2^(2^k)+k=[2,5,18,259,...]. if n=3 the sequence is [5,8,21,262,...].

so what i'm trying to do is say if i'm given an arbitrary one of these sequences then i can give a numerical upper bound on how many numbers of the form 2^j exist in the sequence.

for instance for n=0 i know there is 1 power of 2 exactly (2^1) because after k=0 by my result from my previous post after k=0 i can't have another power of two every again.

for n=-3 i have found 2 powers of two, names for k=1 i get 2^(2^1)+1-3=2 and k=3 2^(2^3)+3-3=256.

not sure if this is clearer, this is one of those things that i feel is a lot easier to explain with pen and paper than through text.

What does this mean?

the solution i got limits the number of powers of 2 for a given n values is n. what it turns out i need is an upperbound that is independent of n. i think it is helpful to know that for any n there are finitely many powers of 2 but what it turns out i need is something like a numerical upperbound for any n independent of n.

I've managed to prove to myself (not rigorously) the bound is 2...

i got a similar feeling, since you can get potentially get one at k=0 and k=-n (n has the be negative here) and i can't think of another way to generate them. the problem is i haven't found a concrete way to show that this is the upper bound.

Well, how about this, it's still not quite rigorous but it gives you a starting point:

The sequence goes: {2+n, 5+n, 18+n, 259+n, 65540+n, ...}

For each index, there's a finite amount of n's that satisfy (I'm not considering positive n's, they're easy to deal with).

2+n: {0, -1}
5+n: {-1, -3, -4}
18+n: {-2, -10, -14, -16, -17}
259+n: {-3, -131, -195, -227, -243, -251, -255, -257, -258}
...

And an interesting pattern takes hold. For each index k, the n's that satisfy must live in this set:
A_k U B_k, where

A_k={-k}
B_k=[-2^(2^k) - k + 1 , 2^(2^k-1) - 2^(2^k) -k]

on one end of B_k we have the n that makes the sequence component = 2^0, and on the other, the n that makes the sequence component = 2^(2^k-1)

And as it turns out, all the A_k sets are disjoint, as are all the B_k. So each n can live in at most one A_k, and in at most one B_k type set.

So the bound is 2.

I have no idea where to go with this problem: Use Ito's Formula to prove that

X(t) = (W(t) + t) exp(-W(t) - 0.5t) is a martingale.

Assume that W(t) is standard brownian motion.

I can't really figure out why Ito's Formula is helpful here.

limey you could turn this whole world on its ass with your unparalleled comedic genius but instead you waste your time with mathematics and this gay ass forum

suit yourself

*awkward silence*

We have X(t) = f(W(t),t), where f(x,t) = (x + t)exp(-x - 0.5t). By Ito's rule,

Show that the dt integrals vanish.

Strictly speaking, doesn't that show it's a local martingale?

apparently my problem can be disregarded, my advisor says he has proved that for any intersection like the one i'm trying to find a bound for there will be some sequences that have n elements.

lol, yeah i came to the same conclusion as you. i usually meet with him thursday afternoon so i figure he'll show me the proof then and i can relay it as best i am able then.

If you have nothing of substance to contribute and/or refuse to post in a civil manner then please go elsewhere. Thanks.

Useful suggestion:http://forumserver.twoplustwo.com/62/bbv4life/

-Zeno

Hey guys first post in the maths forum.

It's basic stuff that i'm unsure whether or not the answer is correct.

Given the equation f(x) = x/(1+x^2)

Find:

1 - f '(x)
2 - f "(x)
3 - the stationary points

so, here is what i have f '(x) = (1-x^2) / (x^4 + 2x^2 + 1)

f "(x) = (2x^3 - 6x) / (x^6 + 3x^4 + 3x^2 + 1)

I think that both of these are correct. The part I'm struggling with is the stationary points for some reason.

Stationary point dy/dx = 0

f '(x) = (1-x^2) / (x^4 + 2x^2 + 1)

which simplifies to:

x^2(3+x^2) = 0 (i think i got this part correct).

therefore x=0 and x= +/- sqrt3

Sub the values of x back into the original equation to get the values for y.

y = sqrt3/(1+3) (x = +sqrt3)
= 0.43
The point is (sqrt3, 0.43)

y = -sqrt3/ (1+3)
= -0.43
The point is (-sqrt3, -0.43)

the line intercepts the axes at (0,0)


Wasn't sure if i was doing this correctly.

One more Qu. 2p2.

This one is on implicit differentiation:

5x^2 + 8xy +5y^2 = 45

I came up with:

10x + 8xy' + 10y = 0

using the product rule on 8xy'

10x + 8xy' + 8y + 10y = 0

simplifying to:

y' = (-18y-10x) / 8x


Still trying to wrap my head around implicit diff. How did i go with this one guys?

I think I figured this out. I just have to show that dX(t) = v(t)dW(t), where v(t) is bounded and adapted (which it is in the above problem). I can then just string a couple of the theorems in our book together, and show that it is a martingale.

I figured out how to approach most of the Ito Formula problems. I have just one last problem on this set, and I can't seem to do it because it involves a Poisson Process. I have to show that

M(t) = exp(a*N(t) - lambda(exp(a)-1)t)

is a martingale, given that N(t) is a Poisson Process, with parameter lambda, and a is a constant. I'm having trouble with this because I'm not entirely sure how to differentiate the function. Is N(t) just exp(-lambda)lambda^t/(Gamma(t))?

The boundedness is important. You can check that if X(t) = |B(t)|^{-1}, where B(t) is a 3-dimensional Brownian motion started away from 0, then differentiating using Ito's formula gives a process with no drift term (so it's a local martingale), but not a true martingale.

You should not be differentiating anything. Write

E(M_t|F_s) = E(exp(aN_t-\lambda t(\exp(a)-1))|F_s)
= E(exp(a(N_t-N_s)+aN_s-\lambda t(\exp(a)-1))
= exp(aN_s-\lambda t(\exp(a)-1)) Eexp(a(N_t-N_s))

So you have to compute Eexp(a(N_t-N_s)). Use the fact that N_t-N_s ~ Exp(\lambda(t-s)) and it should be straightforward enough.

True. We can complete the proof by showing that

for all t. This would follow, of course, if the integrand were bounded. But it is not, in this problem, because W is not bounded below.

I think it should be 10x + 8(y + xy') + 5(yy' + y'y) = 0.

Thanks for a help guys! I feel really dumb asking, but hopefully I'll get a better grasp of it soon.