Hi was wondering if anyone can help me with the last part of this? The notes I have basically state a few things and say hence: and then gives the result with no steps and I don't really know how to start. The textbook uses ladder operators in a slightly different context.

For (v) I obtained:

|1 -1>
|1 1 >
|0 0 >
|1 0 >

For (vi):

s1 m1 s2 m2 J M
|1/2 1/2 > | 1/2 1/2> = | 1 1 >
|1/2 -1/2 > | 1/2 -1/2> = | 1 -1 >

Any help at all even a slight hint would be appreciated.

Do you know about Clebsch–Gordan coefficients? Basically, it wants you to show that
|1,0> = A|1/2, -1/2>|1/2, 1/2> + B |1/2, 1/2>|1/2, -1/2>

Where you find A and B from using the ladder operation on either |1, 1> or |1, -1>

I tried that and got the same coefficient for each.

Namely SQRT(2)Hbar.

So,

|1 0> = SQRT(2)Hbar | 1/2 - 1/2 > |1/2 - 1/2> + SQRT(2)Hbar |1/2 1/2> |1/2 1/2>

Which is wrong:

I am having a mindblank and can't apply this correctly?:

J+-=J1+- + J2+-

Its 2:20AM basic quantum mechanics is owning me right now. I forgot how to apply the product rule a minute ago. FML.

The m values are supposed to add up to the total M. Remember the definition of the raising or lowering operator is. Also, the coefficients should be the same.

Ah I think I have it now thanks. I seperated the two terms I need to apply the raising/lowering operator to into the small j's and found THEIR coefficients. Then equated that with the coefficients I had found previously.

Ended up with:

| 1 0 > = 1/SQRT(2) | 1/2 1/2 > | 1/2 -1/2> + 1/SQRT(2) |1/2 -1/2 > | 1/2 1/2 >

That looks correct. Congrats!

Given the following difference equation:
2y(n+1)-3y(n+1)+y(n) = 2
Find the general solution.

So I can find some solutions to this kind of questions on line, however I am stuck on something written in the answers.

Homogeneous part:
2y(n+1)-3y(n+1)+y(n) =0
(2L^2 -3L +1)y(n) = 0
with characteristic equation
2x^2 -3x +1 = 2(x-1)(x-0.5) =0
Hence x = 1 or x = 0.5
(L-1) 2 =0 (ofcourse 2 lagged one period is still 2, so this makes sense)
Solution of original equation satisfies:
(2L^2 -3L +1)(L-1)y(n)=(L-1)2 = 0
Characteristic equation:
2(x-1)(x-1)(x-0.5) =0
Sols x=1 (multiplicity 2), x = 0.5

Now this is all clear, however then she concludes that
Hence every solution has the form
y(n) = C1 + C2n + C3 (1/2)^n

Why is this? I get that C1 = C1*1^n, but where does the C2n come from?
I cannot seem to find it on line, do not really know what to look for.

Say I have two players that are choosing effort levels x and y in [0,1] that results in a level of a good say x+y. The utilities are given by

u1(x,y) = (x+y)*exp(-x)
u2(x,y) = (x+y)*exp(-y)

Part A:

Now, if these two players choose effort levels simultaneously, what is the pure Nash eq (does it even exist)? I took the respective derivatives and ended up with

x* = 1-y
y* = 1-x

Is this the Nash eq?

Part B:

Suppose player x chooses first and then y chooses his effort level. What about now? Well player y's optimal choice is y* = 1-x (for any given level of player x) but x anticipates this and will obviously choose x = 0 but then finally when player y has to make a decision he will also choose y = 0. So the PSNE is (x,y) = (0,0), did I get this right?

Part C:

How do I write up a continous game in normal form (the game in part B). Also I have to show that this game has infinitely many PSNE.

Anyone?

Part A :

Yes, any (x,y) with x+y=1 is a NE.

Differentiating to get the max is good. Not sure if you're aware of this but finding the points s.t. df/dx=0 isn't enough, you also need to check that the max can't be on the boundary (in this case it's easy since the derivative is > 0 for x<1-y and <0 for x>1-y).

Part B :

Yes x=0 but then player 2's optimal choice is still 1-x=1, so the NE is (0,1).

Thanks! One more question regarding game theory (Cournot model):

Say we have k firms in a market where each of the firms will choose a quantity to produce [0,100]. Let the total quantity be defined as Q = q1 + ... + qk where each of them have the same payout

p(q1,...,qk) = (100-Q)*qi

if Q<= 100

It is easy to find the Nash equilibrium for 2 firms using best responses etc. but I want to find a nash equlibrium (pure) for each k (in a market with k firms and not only 2). I guess a start is to write each payoff and take the derivative w.r.t. each firm, right?

p1 = (100-Q)*q1
.
.
.
pk = (100-Q)*qk

then partial derivatives of each one and then I get the best choice for each firm. Then what would one do?

This is the same thing...given find maximising the payoff for player i (best response).

Then find all the k-uplets such that for each i, is a best response to . This gives you k equations.

I haven't done the calculations but I doubt it's harder than your previous game.

Thanks, I solved it in Matlab very easily.

If you have n = 2 then you get a 2x2 equation system with 2 unknowns, n unknowns leads to a nxn eq system with n unknowns etc. I found that the total quantity Q = q1 + .... + qn converges to 100 as n => inf which seems about right.

Anyone got time to help me with series maclaurin and taylor etc pm me

trying to help a buddy with a homework problem:

If you have a pyramid of 36 pennies, how many "moves" does it take to to turn the pyramid upside down...For example, if you have a pyramid consisting of three pennies, you simply move the top penny on the opposite side of the two pennies; thus the solution would be 1 "move" (to turn the pyramid upside down)

I actually did this with real pennies and found the following:

3 pennies = 1 move
6 pennies = 2 moves
10 pennies = 3 moves
15 pennies = 5 moves
21 pennies = 7 moves
28 pennies = 9 moves
36 pennies = 12 moves

I thought the solution was y=4x-x where x = the number of moves and y = the total amount of pennies however the formula doesn't work for the 10 penny pyramid and the 28 penny pyramid.

Any help would be greatly appreciated...

Thanks

[n/3]

where [.] is the greatest integer function

For this one I would let Qi = Q-qi, ie the sum of all the OTHER producers, which is independant of qi. Then pi = (100-Qi-qi)*qi, which is maxed at qi=50-Qi/2. This leads to a NE where qi = 100/(k+1) for all i=1,\dots,k.

Can anyone tell me what kind of fallacy this is?

I hate the idea that people are taking advantage of the welfare system. Therefore, we should abolish it.


Thanks

So, this is a homework problem for me. I can't figure out the answer. Can someone break one of them down for me and help me figure out how the hell to do it? I have a quiz tomorrow and am worried!

Problem 1

6+11+16+21....+61
The sum of the sequence is (...)


Problem 2

Use the formula S= n squared to find the sum of 1+3+5+....+459 (hint, to find n add 1 to last term and divide by 2)


Problem 3
Use the formula S= n squared to find the sum of 1+3+5+....+857 (hint, to find n add 1 to last term and divide by 2)

Faulkner, you need to find some rule of the series.

Such as problem 1:
Consider the series of terms (5n+1).

fallacy of necessarily.

(sry no pc atm google it

I have a proof/answer that I'm not quite sure about, so I was wondering if someone here could read it over and check that it's ok.

Given two smooth planar autonomous flows

x'=f(x) and x'=g(x)

If f(x)(dotproduct)g(x)=0, then show that if f contains a limit cycle, then g contains a fixed point. (Note that the bolding indicates vector functions, though I'll get lazy from this from now on, because I need to be quick).

Answer:

f(x)(dotproduct)g(x)=f1g1 + f2g2 =0

Limit cycle happens when r'=0, ie that

0 = x*f1+y*f2

x*f1=-y*f2

Fixed points occur when g1=g2=0

f1g1+f2g2=0
f1g1-(x/y)*f1g2=0
f1(g1-(x/y)*g2)=0

As f1 cannot equal zero, this has a solution when g1=g2=0, ie when g contains a fixed point.

Also has a solution when yg1=xg2, but, well, that doesn't have any meaning I guess.

Ok, so the proof was typed up fairly roughly, and it wasn't much good to begin with, but was it acceptable/correct, and have I missed anything?

The question also askks why this only works for planar flows, which I'm assuming is to do with problems with limit cycles in R^3, but since that part of the question isn't actually worth any marks, I'm not really fussed.

Any help would be appreciated, thanks

Do you understand the method used to sum 1+2+...+n?

2(1+2+...+2)=
(1+2+...+n)
+(n+(n-1)+...+1)
(n+1)+(n+1)+...+(n+1)
=n(n+1)

Just do the same thing with these sums.

Forgot to mention this, but x=(x,y), which is some lazy labelling, but what can you do.

Hi everyone.

I assume this is the right place to post this -- making a thread seems a dumb.

I am several years removed since completing my last college course. At uni, I did pretty well in organic chemistry. I remember next to nothing. I'm going back to school in the summer and am going to spend the next few months re teaching myself this stuff so I can progress through the premed pre reqs.

I'm not connected at all to the academic community anymore, and am posting this in pursuit of study guides/text recommendations/anything that will help me along the way. I no longer have my book and will be heading to a library near my house for this, but am uncertain if I'll be able to comprehend 100's of pages of this stuff. I remember the book below being just okay. I learned a lot in lecture/study groups.

thx a ton.

this doesn't look completely kosher to me, but I am not a dynamical systems guy, so some of the terminology might be escaping me. If you're still interested, in this, PM me, and I'll have a quick back and forth with you about definitions, etc.