f is even

Thanks for catching that...

Think of 4 digit number abcd as point (a,b,c,d) in 4 dim space.

I haven't done too much thinking in spaces >3 dimensions, but here is how I attacked it:

Let abcd be a 4 digit number, with a,b,c,d each one digit. We want to find all a+b+c+d = 13, st a,b,c,d are inetgers between 0 and 9, inclusive.

So the total number of integers for this is C(16,13).
Then, I went around and subtracted off the cases where one of the numbers was 2 digits. Is this roughly equivalent to your method?

I suck at visualising things but I think the idea is that if you plot a,b,c,d such that a+b+c+d = 13 you get the surface of a 4-d tetrahedron with vertices at (13,0,0,0) etc, then when you remove 'impossible solutions' where one of a,b,c,d > 9, this removes a little bit at each corner.

Yes.

Yes. But it really is a 3d tetrahedron with 4 vertices. (All real solutions to a+b+c+d = 13 form a 3D space).

The Chebyshev metric doesn't have a derivative in Rⁿ, does it?

Let f be a continuous function on the reals. Assume that f(r)=0 for all rationals r. Show that f(x)=0 for every real x.

This is totally obvious but i dont know how to write the proof. i think it would be something with a constant function.

something like

|f(x)-0|<e and |0-0|<delta

f continuous at x0 means that for every epsilon > 0, there exists a delta > 0, such that for |x-x0| < delta, |f(x)-f(x0)| < epsilon.

Now, let x0 be any irrational such that f(x0) =/= 0. Let epsilon = f(x0)/2. Then there exists a delta > 0, such that |x-x0| < delta, |f(x)-f(x0)| < epsilon. i.e. on the interval (x0-delta,x0+delta), all function values are nonzero. But this is absurd, since the rationals are dense.

OK one more,
Let g and f be continuous functions on R. assume that f(r) = g(r) for every rational r. show that f(x) = g(x). for every real x.

f continuous means that for every epsilon > 0, there exists a delta > 0, such that for |x-a| < delta, |f(x)-f(a)| < epsilon.

from here im not sure

You need to think harder about this one, and use what we just showed.

I really stink at this analysis stuff!

Ok so what do we know? We know that f and g are continuous functions and they meett the epsilon delta proofs definitions. from the earlier question we proved that f(x)=0 for every real x

The question tells us that f(r)=g(r)

I also know that in the previous question we proved this by contradiction. I beleive what we did was to say that:

Asumme all values are non zero, we did this because we wanted to contradict they were all zero. So we showed that our assumption was wrong thereby confirming our original statement.

Can you please give me another idea to ponder reagrding the second question, such as other info im missing?

The point is that if x is the limit of the (Cauchy) sequence x_1, x_2, .. then, by continuity of f, f(x) is the limit of f(x_1), f(x_2), ..., which, when each x_i is rational, is the sequence g(x_1), g(x_2), .., which, by continuity of g, has the limit g(x). So f(x) = g(x) because sequences have unique limits. Every real number is the limit of a sequence of rationals

Hopefully you didn't see what I originally wrote.

How can we use the first question to help us answer the 2nd?

Hmmm, if only we had some function that was 0 on all the rationals...

Afternoon all, I was hoping someone might be able to give me some pointers on where to go with this problem.

"3. (a) Use eigenfunction expansion to nd the bounded solution of

on -1 =< x =< 1 [Hint: write the equation in self-adjoint form to determine which
expansion functions to use.]
(b) What can you say about the solution of

with u again bounded on [-1,1]?"

Probably the easiest way is for me to outline what I've done, and then (hopefully) someone can either show me where I've gone wrong, or give me some pointers on where to go next.

For 3) a) I first converted the function to self adjoint form through the process:



Using an integrating factor of

To give



Which at first looked like it had a simmilar form to Legendre or Chebyshev, but now I'm not sure. It's defiently a singular Sterm-Liouville system though, rather than a periodic or Regular SL system, as they go to infinity at the end points, which I'm pretty sure satisfies the singular SL system conditions.

But it's at this point I got pretty damn stuck.

I think the next step is to go and do

[img]http://latex.codecogs.com/gif.latex?L(u)+\lambda\rho*u=0[/img]



Which actually is starting to look like it's taking the form of a Hermite DE, but I'm not really sure, especially since hermites exist on

So....any help possible?

Much obliged.

If you didn't get it by now with Wyman's hint you really are stuck, so here's another "hint".

If f(r) = g(r), then f(r)-g(r)=0.

Game theory paper, stuck on one question, dont know how to tackle it, my thoughts below.


The activity of firm F has a positive externality. If it goes bankrupt, then
there will be a loss to society of V . If the firm is approaching bankruptcy,
the government might be able to bail it out at cost D. But before learning
whether the firm will get into difficulties, the government, G, can commit to
being hard-nosed, and not bailing the firm out.1 The advantage to such a
commitment would presumably be that it would give the firm an incentive
to try harder not to get into difficulties. The steps of the game are:
1. G decides whether to publicly promise not to bail out failing firms,
P 2 {0, 1}.
2. F sees G’s decision, and decides whether to put in low or high effort
into keeping itself afloat, E 2 {L,H}.
3. Nature chooses whether the firm gets into difficulties or not, with probability
(pie)L if E = L and probability (pie)H if E = H. Assume that
0 < (Pie)H < (Pie)L < 1.
4. If the firm gets into difficulties, G decides whether to bail it out by
paying $D.
Payoffs are: V to G and W − E to F if the firm doesn’t get into difficulties, V − D − PT to G and W − E to F if the firm gets bailed out, and 0 to G and −E to F if the firm gets into difficulties and is not bailed
out.

The interpretation of the parameter T is the political cost to G of being
seen to break a promise. Assume that V − D − T < 0 < V − D and
0 < L < H < W.



QUESTION
Under what condition(s) will the firm choose high effort in step 2 if
P = 1?



ANSWER
In all my text books I cant seem to understand the the actually want me to answer, are they want a written answer or do they want a calcuated answer.
I think it could be something like.--> They will choose H on the condition that G is rational and will not bail them out, as pH > pL then there needs to be a substantial value of ..........
If (pie sumbol)L > (pie symbol)H, I cant see how they can choose H given that the payoffs are equal...what conditions is it?!

Thanks for your help

Stuck on this problem:

Find the result of using the following operator (x = multiply)

(1/r^2 x d/dr x r^2 x d/dr) + (2/r)

To operate on: Ae^(-br)

What must the values of A and b to be to make this function an eigenfunction of the operator? Normalize the eigenfunction to determine A.

Of(r) = ((1/r^2 x d/dr x r^2 x d/dr) + (2/r)) Ae^(-br)

After the first step I get:

= (1/r^2 x d/dr x r^2)x(-Abe^(-br)) + (2Ae^(-br))/r

Then I do r^2 on (-Abe^(-br)) to get:

(1/r^2 x d/dr)x(-Abr^2e^(-br)) + (2Ae^(-br))/r

Taking the derivative (-Abr^2e^(-br)) and using the product rule I get:

(1/r^2) ( Abr^2e^(-br) - 2Abre^(-br) ) + 2Ae^(-br)/r

Finally I multiply (1/r^2) to ( Abr^2e^(-br) - 2Abre^(-br) ) and cancel out r's.

After the operator I get:

Abe^(-br) - (2Abe^(-br))/r + (2Ae^(-br))/r

I have no idea if this wrong or right. (99% chance its wrong ldo) From there I am supposed to normalize by taking the absolute value of the square of the function after operation then find the values of A and b to make it an eigen function.

Any help?

I believe you are missing the bolded part, but that this is otherwise correct. For this kind of thing it's probably worth learning to use the LaTeX codecogs.

Do you know what eigenfunctions and normalization mean? Once you've gotten this far you've done the bulk of the work. Don't worry about normalizing yet; what has to be true of A and b given this form for you to have an eigenfunction?

I am so f-******ed. I cannot figure out the simplest problem...

Consider a gas-phase N2 molecule. Its RMS speed is given by

Vrms = (v^2)^(1/2) = (3kT/m)^(1/2)

What speed would the gas molecule have if it had the same energy as a photon of the following wavelengths of IR, visible, and X-Ray electromagnetic radiation?

lambda = 1.00 x 10^4 nm

I have no idea how we are supposed to use the root mean square equation above to help solve the problem. What I thought was since it has the same energy as a photon, we can use the relationship of E = hv to find the energy associated with the wavelength of light. E = hv = hc/lambda. We have hc and lambda so we can solve for the energy. From the energy what would I do to solve for the speed of the gas molecule? Would I use E = 1/2 mv^2? Or how do I use the given equation in the problem to solve for speed?

The E I get for solving E = hc/lambda is 1.9878E-20J. How do I proceed from here?

Well I haven't taken a formal math class that actually taught eigenfunctions (taking this for Q. Chem), so I am learning this from the start. From what the prof has taught us so far eigenfunctions are functions when operated on give the original function + some constant. And in this case I believe the eigenfunction would be (b - (2b^2)/r + (2/r)), that look correct? Normalizing in Q. Chem gives the probability density for a certain wavefunction...or in otherwords when you integrate the abs. value of the square of the function it will be equal to 1.

lol, im a scrub at this stuff...

neverheard of LaTeX...probably will google it after I finish my HW.

Given this form, A and b must not be 0? I am completely lost on this part. I am not sure how to think about it.

Not quite. If I have an operator A acting on a function \Psi, I call \Psi an eigenfunction if the following is satisfied:



where a is some scalar called the eigenvalue of A. So what you need to do is figure out for what choice of b the result of the operator looks like some eigenvalue E times the original function.

EDIT: It might be easier if you combine terms:




jason's sticky at the top of the forum explains how to use latex.codecogs.com to produce LaTeX images, but that won't help you much until you learn how to use LaTeX. However, for the amount of stuff you're likely to want to do, you can learn it in 20 minutes. Greek letters are things like \psi or \Psi (for uppercase), subscripts are _{text here}, superscripts are ^{text here}, nice looking fractions are \frac{numerator}{denominator} and that's about 95% of what you're going to be likely to use at first.

haha, okay I thought LaTeX codecog was something else related to eigenfunctions. I'll take a look at it before I post a question again, thanks.

If b = 0 I get the function to become (2/r) times the original function Ae^-br. Is this closer to what I am trying to figure out? So eigenvalues can never be (5 + 2/r) or of that nature? Are they always some scalar value multiplied by the original function? Or can it be 2 scalar values like (4 + 2r).

If I have (2/r) multiplied by my constant if b = 0 , then must A = r?

Here is another example of what I am trying to say:

ae^-3x + be^-3x with the operator d^2/dx^2

Taking the 2nd derivative would be 9ae^-3x + 9i^2be^-3ix. So I get the original function multiplied by (9 + 9i^2). Is this not an eigenfunction then?

Anyway, thanks for the help so far! I hope you are on in an hour or so...gotta go to biochemistry class, already late lol.

I guess 9 + 9i^2 would be an eigenfunction with an eigenvalue of 9? What I'm trying to ask is eigenvalues must always be multiplied to the original functions and cannot be like (3x + 4y) where the original function is xy.