so you're saying:
2(Pc) + X(Pc) = X
=Pc(2+X) = X (to which you divide both sides by (2+X))
=Pc = X/(2+X)

thanks!

Depends on what you're trying to use them for. If you want to assess the effect a diploma has on odds of unemployment and income, you should run a normal and a logistic regression where you control for other common determinants of these things. Otherwise, the effect you think is due to diplomas is really caused by a lot of things.

Diplomas and the like are usually taken to decrease the odds of unemployment and increase income via two channels: Signal value, a diploma signals something positive to employers, and actually making you more productive by developing your skills etc. If it's the former you're interested in, you want to include productivity related controls. If it's the latter you need panel data to take into account selection bias (more productive indivuals might be more likely to choose to get a diploma, leading you to overstimate the producitivy increase due to diplomas if you don't take it into account)

Solve for A:



I can't figure out where I'm going wrong, but the homework software tells me my answer is incorrect. The problem specifies, "Use the sum-to-product formula to find the exact value," but I don't see why I can't solve it using the following method? My answer is incorrect so, obviously there's a problem somewhere.



I'll work with this in terms of alpha and beta for the next few steps...





So...



Convert it to radians:



So far, so good? Now I just use the half-angle identity:




This is in the first quadrant, so I'll take the positive square root. Also, I'll find the common denominator in the numerator:





So, A = 2 - sqrt(3)? The homework software disagrees.

Thank you.

Hey, I need some help on some Quantum Chemistry problems. I have an exam on friday but the problems due tomorrow. Shoot me a pm. The problems (4) are on molecular orbitals and energy states it appears. Will gladly compensate for any help.

Thanks!

The book says the paraboloid is positively oriented and the disk is negatively How is the 2nd surface negatively oriented? I'm having trouble understanding the concept and the only way I can figure out out how to determine orientation is the z component of the normal, but I am not sure what it is in this case as z = 0 and any way I can think of to find the cross product will get a balanced term for z in terms of x and y such that is could be positive or negative depending on where we are in the disk...

Why don't you post them?

Disregard. The problem actually said, cos(255 degrees) - cos(195 degrees), rather than cos(225 degrees) - cos(195 degrees). Sorry for the spam. Don't ban me please!

That doesn't seem like the right figure to use, because your non-diploma workers are *part* of the group of 'average Americans'. Surely you want to compare non-diploma workers with diploma workers?


Similarly here, you talk about "the other group", but it isn't a separate group. People without diplomas are *part of* the group of all Americans.

I think you want to work out how much more likely someone with a diploma is to be employed than someone without a diploma, and how much more someone with a diploma earns than someone without a diploma. If you know what fraction of Americans have diplomas, those calculations would be easy to do using the data you gave above.

Consider a heuristic which in each step minimizes a cost function C(n,W), alternating with respect to W (continuous) and n (discrete). I am asked to show that this heuristic converges. Now what I have shown already is that C(n,W) is decreasing. Moreover, this cost function is bounded by some lowerbound proved somewhere in the book, but for the sake of argument it may as well be bounded by 0. Hence C converges.

How do I now show that the heuristic converges to a particular solution of n and W? I was thinking about the set of solutions induced by the level sets of C, which is a decreasing sequence, ie, S(t) is a subset of S(t-1) for all iterations t, where S(t) = { (n,W) | C(n,w) =< C(t)}. However, S(\infty) is not necesarily a set with cardinality 1, so this approach seems to be a dead end.

Is it possible in general that such a heuristic finds an infinitismal improvement each iteration, yielding a change in n, yielding a small change in W, which then again implies a discrete change in n? If this is the case, can we still say the procedure converges?
This is about the heuristic of Doll and Whybark (1973) if anyone is interested, but I don't think the specifics matter here.

How is F dot dS defined? It's F.n dS. Which way does "n" point on the disk? Which way on the paraboloid? The usual choice for a closed surface is that "positively oriented" means that the normal points outward from the region contained by the surface.

Since I don't see anything in the question about orientation, if I were integrating over this, I'd assume that the normals from the disk bounding S below point downward(!), and those on the paraboloid point outward.

Are you allowed to use the divergence theorem?

http://en.wikipedia.org/wiki/Divergence_theorem

The normal should be pointing out always in your closed object defined by the disk and paraboloid. You will have to do a lot more work if you explicitly evaluate the surface integral rather than the divergence volume integral. Are you studying introductory differential geometry?

Otherwise just do the surface integration like here;

http://en.wikipedia.org/wiki/Surface_integral

I think I have solved it. I have shown that S(1) is a bounded finite set. If we then assume the heuristic does not cycle, there must be a time where a solution is considered for the second time. Then the solution must also have been considered in the previous iteration, as else the solutions in the iterations in between must have had the same objective value, which implies that the heuristic does cycle. Hence the heuristic has converged at this point.

Density of Q in R proof...

WTS: For every two real numbers a and b, with a < b (and a and b both positive), there exists a rational number r satisfying a < r < b.

Proof: We need to find m and n in N s.t. a < m/n < b. So we choose an n large enough so that 1/n < b-a. Now we just need to find an m small enough so that a < m/n < b. Here's where I get confused:

My book says "multiplying our inequality (a < m/n < b) by n yields na < m < nb. But we're setting out to show that there exist m, n in N such that a < m/n < b...so how can we use this in our proof?

Obviously I'm missing something but would appreciate the help.

Thanks guys (Wyman),
Mariogs

1/n < b-a

So 1 < nb-na; hence there is an integer between them. Call it m. \qed

I'm confused, we're trying to show there's a rational between any two reals. How does 1 < nb - na show us this...?

Sorry just a bit lost on this one.

work backward from the end of the proof and see that you're done if you show what I showed...

Think a bit harder on this.

I'm working on a project for an econ class and I'm looking at flat taxes and growth of a states GSP. To get flatness I'm using the difference between a states top and bottom marginal rate. I also plan to have a dummy variable for whether or not the year was one where a regression took place. I'm struggling to set up the regression though. I want to do it over many years, with all 50 states.

Is this the best way to set it up, GSP being the dependent variable and flatness/recession being the two independents?

Ok yeah, the way the book does it is really counterintuitive to me (still seems like they're using assuming the conclusion as a premise).

So we're just saying, "ok, pick n big enough s.t. 1/n < b-a". Then 1/n is an increment smaller than b-a, and because of this, there must be some m that makes m/n fall in between b-a. That argument makes sense to me, is that what you mean?

I think the book is trying to say that proving a < m/n < b is the same as proving na < m < nb, which is true.
And this last inequality follows from Wyman's hint

yup, I just multiplied it up so that we were saying "If x < y and |x-y| > 1, then there is an integer M such that x < M < y."

And actually you should prove this. Yes it's obvious, but lots of "obvious" things are false...

If there is at least 1 integer n such than a<n<b then that integer is the rational number in question.

If there are no integers between a,b we can deduce than x=b-a<1.
If we consider an integer m such that m>1/(b-a)=1/x and we then multiply a,b with m we will get A=a*m,B=b*m and A-B=m*(b-a)>1/x*(b-a)=1/x*x=1

So A-B>1 which means there is an integer n such that A<n<B. Basically
the rational n/1=n is between A,B so clearly

m*a<n<m*b and a<n/m<b

So the rational n/m is the required rational between a,b.

The fact that if B-A>1 implies an integer is between A,B is trivial.

If that were not true then the next integer lower than A and the next integer higher than B would have a distance larger or equal than A-B>1 and you would have 2 integers A1,B1 that differ by more than 1 without another integer between them existing which is false because A1+1 is such integer clearly.

By the way if A itself is an integer we do not need the above A1,B1, we can immediately consider A+1 as a number between A,B that is an integer. (same logic on the reverse if B was an integer).


Also if a-b=1 then either both are integers or an integer n exists between a,b. So either n or (a+b)/2 is the required rational.

Trying to figure out at what frequency my opponents need to respond to my steals to make it neutral EV preflop. So made the formula, EV=B(1-S) + SP, where B =steal size, S=steal success frequency and P=preflop pot size(playing in a cash game that has antes, hence why P isn't constant)

So I want to isolate S, how do I do this? I get stuck on this spot:

Make EV=0

0=B-BS+SP

?? Help?

Memory has returned, figured it out :s

I think we just had a question like this on the last page. Factor out S.

BS - PS = B
S(B-P) = B
S = B/(B-P)