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Old 10-15-2009, 04:27 PM   #201
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Re: The Official Math/Physics/Whatever Homework questions thread

I'm having a tough time figuring out how to be more clear so I'll just give a couple examples:

e^(ax) is an eigenfunction of the operator d / dx, with the eigenvalue a.

e^(ax) + e^(bx) with a not equal to b is not an eigenfunction of the operator d / dx. To see this, take the derivative and get ae^(ax) + be^(bx). Any constant you multiply the original function by must leave the same coefficient on both terms, but since we've stipulated that a does not equal b, that can't possibly be the case for our function, and so that sum is not an eigenfunction of the operator d / dx.

sin(ax) is not an eigenfunction of the operator d / dx because we get a*cos(ax) as a result.

sin(ax) is an eigenfunction of the operator d^2 / dx^2 with eigenvalue -a^2.



And so on.
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Old 10-15-2009, 04:44 PM   #202
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Re: The Official Math/Physics/Whatever Homework questions thread

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I'm having a tough time figuring out how to be more clear so I'll just give a couple examples:

e^(ax) is an eigenfunction of the operator d / dx, with the eigenvalue a.

e^(ax) + e^(bx) with a not equal to b is not an eigenfunction of the operator d / dx. To see this, take the derivative and get ae^(ax) + be^(bx). Any constant you multiply the original function by must leave the same coefficient on both terms, but since we've stipulated that a does not equal b, that can't possibly be the case for our function, and so that sum is not an eigenfunction of the operator d / dx.

sin(ax) is not an eigenfunction of the operator d / dx because we get a*cos(ax) as a result.

sin(ax) is an eigenfunction of the operator d^2 / dx^2 with eigenvalue -a^2.



And so on.
Okay, that makes sense then. Thanks
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Old 10-15-2009, 05:01 PM   #203
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Re: The Official Math/Physics/Whatever Homework questions thread

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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.
So with that information you posted above. Can you give me some hint on how to proceed with this? Can't really come up with an A and b to get to the original function.
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Old 10-15-2009, 05:32 PM   #204
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Re: The Official Math/Physics/Whatever Homework questions thread

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So with that information you posted above. Can you give me some hint on how to proceed with this? Can't really come up with an A and b to get to the original function.
What is the original function?

Also, remember that the eigenvalue must be a scalar but it doesn't need to be 1.
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Old 10-15-2009, 05:50 PM   #205
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Re: The Official Math/Physics/Whatever Homework questions thread

So after looking at it for a bit...if b = 1 and A = any integer you will have an eigenfunction.
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Old 10-15-2009, 05:59 PM   #206
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Re: The Official Math/Physics/Whatever Homework questions thread

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So after looking at it for a bit...if b = 1 and A = any integer you will have an eigenfunction.
Yes.

Once you know b, you can go about normalizing - since as you point out for any A it's an eigenfunction, you can pick whatever A you want. Normalizing means picking A such that the integral of the square of the function Ae^(-br)(since your function is real) over r = 0 to infinity is 1. This is because (as you pointed out earlier) normalizing is supposed to be capturing the idea that the square of this function is a probability density. In this particular case, you are solving for what turns out to be (in some system of units) the radial wavefunction of a 1s electron in a hydrogen atom.
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Old 10-15-2009, 06:12 PM   #207
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Re: The Official Math/Physics/Whatever Homework questions thread

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Yes.

Once you know b, you can go about normalizing - since as you point out for any A it's an eigenfunction, you can pick whatever A you want. Normalizing means picking A such that the integral of the square of the function Ae^(-br)(since your function is real) over r = 0 to infinity is 1. This is because (as you pointed out earlier) normalizing is supposed to be capturing the idea that the square of this function is a probability density. In this particular case, you are solving for what turns out to be (in some system of units) the radial wavefunction of a 1s electron in a hydrogen atom.
Cool man, thanks a lot for the help!
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Old 10-15-2009, 06:13 PM   #208
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Re: The Official Math/Physics/Whatever Homework questions thread

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I am so f-retarded. 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?
This one got lost in the array of posts from my last question. Does anyone know how to proceed from here?

We aren't given any temperature and I am not sure if we are supposed to be using that root mean square velocity equation at all. We have E = J, we have N2 where we can get a mass and we can get a velocity using (mv^2)/2. But it seems like we should be using the Vrms equation.
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Old 10-15-2009, 06:23 PM   #209
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Re: The Official Math/Physics/Whatever Homework questions thread

Came up with the idea of just using that mass of a diatomic nitrogen atom and using KE = (mv^2)/2 and solving for v and not using the Vrms equation.
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Old 10-15-2009, 07:01 PM   #210
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Re: The Official Math/Physics/Whatever Homework questions thread

We also have the Debroglie equation where lambda = h / mv. Would that equation be the correct one to use instead of the method I used?
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Old 10-17-2009, 04:39 PM   #211
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Re: The Official Math/Physics/Whatever Homework questions thread

A statistical question which I thought about:

You know the height of all the people in population A.
You know that the average height of the people in population A is 1.500 meters.

Inside population A, you have 10000 people who eat at least one kiwi a day. Let's call this sub-population A(k). You know the height of all the people in A(k). The average height of A(k) is 1.540 meters.

Inside population A, you have 25 people who eat at least one orange a day. Let's call this sub-population A(o). You know the height of all the people in A(o). The average height of A(o) is 1.543 meters.


Now, you have a population B, whose heights are known and whose average is also 1.500 meters. If you were to take the sub-population B(k) or B(o) in order to maximize the average height, which one would you choose?

Intuitively, you would tend to say that you should take B(o), but I don't think this is necessarily correct. ( I'll explain further if you think it should obviously be B(o) )

Thanks a lot.
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Old 10-18-2009, 07:53 PM   #212
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Re: The Official Math/Physics/Whatever Homework questions thread

Can anyone here help me out with a little simple measure theory for my Stochastics class? I don't think this is very hard but I've never had a class on it before and its not covered well in my book.

Given a queue-length Markov Chain where p is the probability of an arrival and q is the probability of a service completion at any discrete time. Show,



is an invariant measure for the MC, where



The next part is more stochs, but bonus points if you want to give it a shot.

Prove that the chain is ergodic iff p<q and in this case the stationary distribution is

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Old 10-18-2009, 09:04 PM   #213
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Re: The Official Math/Physics/Whatever Homework questions thread

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Can anyone here help me out with a little simple measure theory for my Stochastics class? I don't think this is very hard but I've never had a class on it before and its not covered well in my book.

Given a queue-length Markov Chain where p is the probability of an arrival and q is the probability of a service completion at any discrete time. Show,



is an invariant measure for the MC, where



The next part is more stochs, but bonus points if you want to give it a shot.

Prove that the chain is ergodic iff p<q and in this case the stationary distribution is

OK I think I actually kinda understand now, an invariant measure just needs to satisfy M=MP.

I get my transition probabilities as:

P(0,1)=p(1-q)
P(0,0)=1-p(1-q)
P(i,i+1)=p(1-q)
P(i,i-1)=q(1-p)
P(i,i)=pq + (1-p)(1-q)

In which case:



...and with some algebra the general case turns out to work as well. Not that bad really, I'll try to tackle the rest of the problem and see how it goes.
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Old 10-21-2009, 01:46 PM   #214
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Re: The Official Math/Physics/Whatever Homework questions thread

are the two groups Z_2 X Z_12 and Z_4 X Z_6 isomorphic?

now can I show isomorphism for two groups namely:

Z_2 X Z_12 and Z_4 X Z_6

buy showing the two groups are merely a a form of

Z_2 X Z_2 X Z_2 X Z_3
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Old 10-21-2009, 01:50 PM   #215
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Re: The Official Math/Physics/Whatever Homework questions thread

How many elements of order 4 does each group have? Isomorphisms preserve the order of elements, so if these are not equal, the groups cannot be isomorphic.
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Old 10-21-2009, 01:59 PM   #216
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Re: The Official Math/Physics/Whatever Homework questions thread

when you ask how many elements of order 4 does each group have. Does the following have 4 elements?

Z_2 X Z_2 X Z_2 X Z_3

or 24 elements?

In the original question each group has 24 elements.

and under the operation we have different ordered pairs.

Z_2XZ_12=(0,0),(1,1),(0,2).........

Z_4 X Z_6=(0,0),(1,1),(2,2).....

So im not sure what is meant by elements of order 4?

Im thinking that if we made a table there would be 144 entries and then we would have to check for isomorphism, but breaking them down in to the product of primes, I wouldnt have to go through that agony
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Old 10-21-2009, 02:06 PM   #217
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Re: The Official Math/Physics/Whatever Homework questions thread

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when you ask how many elements of order 4 does each group have. Does the following have 4 elements?

Z_2 X Z_2 X Z_2 X Z_3

or 24 elements?

In the original question each group has 24 elements.

and under the operation we have different ordered pairs.

Z_2XZ_12=(0,0),(1,1),(0,2).........

Z_4 X Z_6=(0,0),(1,1),(2,2).....

So im not sure what is meant by elements of order 4?

Im thinking that if we made a table there would be 144 entries and then we would have to check for isomorphism, but breaking them down in to the product of primes, I wouldnt have to go through that agony
2 things:

1) It looks like you don't understand the group operation in Z/mZ x Z/nZ, so make sure you know how to add elements [e.g. (1,5) + (1,8) in Z/2Z x Z/12Z ]. What is the identity in that group? Then what does it mean for an element to have order 4?

2) Z/4Z is NOT isomorphic to Z/2Z x Z/2Z, even though they both have 4 elements. Every element in Z/2Z x Z/2Z has order at most 2, and Z/4Z has 2 elements of order 4.
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Old 10-21-2009, 02:16 PM   #218
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Re: The Official Math/Physics/Whatever Homework questions thread

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Originally Posted by Wyman View Post
2 things:

1) It looks like you don't understand the group operation in Z/mZ x Z/nZ, so make sure you know how to add elements [e.g. (1,5) + (1,8) in Z/2Z x Z/12Z ]. What is the identity in that group? Then what does it mean for an element to have order 4?

2) Z/4Z is NOT isomorphic to Z/2Z x Z/2Z, even though they both have 4 elements. Every element in Z/2Z x Z/2Z has order at most 2, and Z/4Z has 2 elements of order 4.
1. identity (0,0), (1,5)+(1,8)=(0,1) order 4 would mean that 4 operations bring us back to the identity.

2.

Z/4Z= {0,1,2,3}, I dont see the 2 elements of order 4 here

Z/2Z x Z/2Z = {(0,0),(1,1)}
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Old 10-21-2009, 02:23 PM   #219
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Re: The Official Math/Physics/Whatever Homework questions thread

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1. identity (0,0), (1,5)+(1,8)=(0,1) order 4 would mean that 4 operations bring us back to the identity.

2.

Z/4Z= {0,1,2,3}, I dont see the 2 elements of order 4 here

Z/2Z x Z/2Z = {(0,0),(1,1)}
1. x has order 4 if x + x + x + x = identity.

2. What is the order of 0? 1? 2? 3? in Z/4Z

3. Z/2Z x Zx2Z has 4 elements, not 2.
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Old 10-21-2009, 02:37 PM   #220
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Re: The Official Math/Physics/Whatever Homework questions thread

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1. x has order 4 if x + x + x + x = identity.

2. What is the order of 0? 1? 2? 3? in Z/4Z

3. Z/2Z x Zx2Z has 4 elements, not 2.
2.

0= {0}=order1
1= {0,1,2,3}=4 elements
2={0,2} two elements
3={0,3,2,1}= 4elements

3.
{(0,0),(0,2),(2,0),(2,2)}

So let me ask, with something like Z/2Z X Z/2Z. Do i work the set out first for each individual then take the ordered operation? So number 3 above would be the correct way to compute it. Or should I start with the identity, in this case (0,0) i hope, and add the generator (1,1) then perform the given mod op? So i was thinking the way to do Z_2 X Z_3:

Take the identity and add the generator

{(0,0),(1,1),(0,2),(1,0),(0,1),(1,2),}

And in all honestly Im not sure what Z/4Z, It looks like its just another way to write Z mod 4. which is what I read. But something more complex like:

(3,1)+<(1,1)> in (z_4Xz_4)/<(1,1)>

i have no clue what this says

Last edited by smcdonn2; 10-21-2009 at 02:46 PM.
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Old 10-21-2009, 02:43 PM   #221
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Re: The Official Math/Physics/Whatever Homework questions thread

Are you asking what addition is in the product G*H of the (additive) groups G and H? If so, it's pointwise: (g, h) + (g', h') = (g+g', h+h').
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Old 10-21-2009, 03:13 PM   #222
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Re: The Official Math/Physics/Whatever Homework questions thread

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2.

0= {0}=order1
1= {0,1,2,3}=4 elements
2={0,2} two elements
3={0,3,2,1}= 4elements

3.
{(0,0),(0,2),(2,0),(2,2)}

So let me ask, with something like Z/2Z X Z/2Z. Do i work the set out first for each individual then take the ordered operation? So number 3 above would be the correct way to compute it. Or should I start with the identity, in this case (0,0) i hope, and add the generator (1,1) then perform the given mod op? So i was thinking the way to do Z_2 X Z_3:

Take the identity and add the generator

{(0,0),(1,1),(0,2),(1,0),(0,1),(1,2),}

And in all honestly Im not sure what Z/4Z, It looks like its just another way to write Z mod 4. which is what I read. But something more complex like:

(3,1)+<(1,1)> in (z_4Xz_4)/<(1,1)>

i have no clue what this says
First of all, mathematics is a very precise thing. You should read your statements before you send them, so that your question is clear. This also gets you in the habit of writing well, and whoever reads your proofs will appreciate that.

Z/4Z is the factor group (or quotient group, depending on the language your book uses) that you're calling "Z mod 4". I would call it "Z mod 4Z", and I write it Z/4Z instead of Z_4, because (at least for p prime), Z_p means something else to a lot of mathematicians.

It is the set of cosets of the subgroup 4Z in Z. There are 4 such cosets:
0+4Z, 1+4Z, 2+4Z, and 3+4Z. This forms a group under the operation
(a+4Z) + (b+4Z) = (a+b)+4Z.
We often omit the (implied) 4Z when talking about this group and list the elements as 0,1,2,3.

To your questions above:
Z/2Z x Z/2Z has 4 elements: {(0,0), (0,1), (1,0), (1,1)}.

Your calculation that (1,1) has order 6 in Z/2Z x Z/3Z is correct.

Quote:
(3,1)+<(1,1)> in (z_4Xz_4)/<(1,1)>
First, <(1,1)> is a (normal) subgroup of Z/4Z x Z/4Z. (Z/4Z x Z/4Z)/<(1,1)> is the set of cosets of <(1,1)>.

<(1,1)> = {(0,0), (1,1), (2,2), (3,3)}
(3,1)+<(1,1)> = {(3,1), (0,2), (1,3), (2,0)}
Notice that (3,1)+<(1,1)> = (2,0)+<(1,1)>.
(3,1) and (2,0) are different choices of coset representatives for this coset.

In other words, (3,1) and (2,0) are the same element in (Z/4Z x Z/4Z) / <1,1> in just the same way as 1, 5, and 9 are the same element in Z/4Z.

Lastly, 2 elements represent the same coset if and only if their difference is in the normal subgroup (in this case, notice (3,1)-(2,0) = (1,1), which is in <(1,1)>).

edit: You can now see why I write Z/4Z. This group is Z/<4>, and <4> = 4Z.

Maybe this is helpful and will spark some questions.
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Old 10-21-2009, 03:35 PM   #223
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Re: The Official Math/Physics/Whatever Homework questions thread

How many ways can you arranged the letters in the word aaabbbccc so that you do not have three of the same letters adjacent to each other?

Is there an easier way than my method?

Spoiler:
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Old 10-21-2009, 04:15 PM   #224
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Re: The Official Math/Physics/Whatever Homework questions thread

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Originally Posted by Wyman View Post
2 things:

1) It looks like you don't understand the group operation in Z/mZ x Z/nZ, so make sure you know how to add elements [e.g. (1,5) + (1,8) in Z/2Z x Z/12Z ]. What is the identity in that group? Then what does it mean for an element to have order 4?

2) Z/4Z is NOT isomorphic to Z/2Z x Z/2Z, even though they both have 4 elements. Every element in Z/2Z x Z/2Z has order at most 2, and Z/4Z has 2 elements of order 4.
Ok so back to the original question, These two groups are not isomorphic becuse their order is not the same right?

Z_2 X Z_12 = has order 12

Z_4 X Z_6 = has order 6

is this right?
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Old 10-21-2009, 04:23 PM   #225
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Re: The Official Math/Physics/Whatever Homework questions thread

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Ok so back to the original question, These two groups are not isomorphic becuse their order is not the same right?

Z_2 X Z_12 = has order 12

Z_4 X Z_6 = has order 6

is this right?
No.

The order of a group is its size, i.e. its number of elements. Both of these groups are order 24.

The order of an element in a group is the size of the subgroup generated by that element.
e.g. in Z, 1 has infinite order, since <1> is infinite.
e.g. in Z/4Z, 1 has order 4, 2 has order 2, 3 has order 4, and 0 has order 1.

My question to you is: How many elements of order 4 are in Z/4Z x Z/6Z, and how many are in Z/2Z x Z/12Z? List them.
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