Two Plus Two Poker Forums The Official Math/Physics/Whatever Homework questions thread
 Register FAQ Search Today's Posts Mark Forums Read Video Directory TwoPlusTwo.com

 Notices

 Science, Math, and Philosophy Discussions regarding science, math, and/or philosophy.

10-08-2009, 02:18 PM   #176
Wyman
Carpal \'Tunnel

Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 12,023
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by solsek I am having trouble with this last problem in my homework set. The problem is Integrate: e^(-4x^2)dx from -infinity to +infinity. I tried looking up identities for integrating e^(-ax^2) but only managed to find one possible hopeful hint. I found that integrating e^(-ax^2) from 0 to +infinity has the solution of (pi/4a)^(1/2). So I just split the integral from -infinity to 0 and from 0 to +infinity (I get (pi/16)^(1/2) for the integral from 0 to +infinity). I am not sure how to evaluate the -infinity to 0 portion. Is there a better way to solve it?
f is even

10-08-2009, 02:20 PM   #177
solsek
Carpal \'Tunnel

Join Date: Feb 2008
Posts: 6,003
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Wyman f is even
Thanks for catching that...

10-08-2009, 02:22 PM   #178
thylacine
Pooh-Bah

Join Date: Jul 2003
Posts: 4,029
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Myrmidon7328 I did the math out, and I got to 480, but I don't quite understand the 4C(6,3) part. I had a much longer way to find a sum that adds up to 4C(6,3). I also don't understand the tetrahedron part. Would it be possible to explain this? Thanks!
Think of 4 digit number abcd as point (a,b,c,d) in 4 dim space.

10-08-2009, 05:09 PM   #179
Myrmidon7328
old hand

Join Date: Dec 2007
Posts: 1,960
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by thylacine 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?

 10-08-2009, 05:21 PM #180 Pyromantha veteran   Join Date: Dec 2007 Posts: 2,227 Re: The Official Math/Physics/Whatever Homework questions thread 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.
10-08-2009, 09:09 PM   #181
thylacine
Pooh-Bah

Join Date: Jul 2003
Posts: 4,029
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Myrmidon7328 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?
Yes.

Quote:
 Originally Posted by Pyromantha 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. But it really is a 3d tetrahedron with 4 vertices. (All real solutions to a+b+c+d = 13 form a 3D space).

 10-14-2009, 02:08 PM #182 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,244 Re: The Official Math/Physics/Whatever Homework questions thread The Chebyshev metric doesn't have a derivative in Rn, does it?
 10-14-2009, 02:53 PM #183 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread 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|
10-14-2009, 03:00 PM   #184
Wyman
Carpal \'Tunnel

Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 12,023
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 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|
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.

 10-14-2009, 03:24 PM #185 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread 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
 10-14-2009, 03:26 PM #186 Wyman Carpal \'Tunnel     Join Date: Mar 2007 Location: Redoubling with gusto Posts: 12,023 Re: The Official Math/Physics/Whatever Homework questions thread You need to think harder about this one, and use what we just showed.
 10-14-2009, 04:41 PM #187 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread 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?
 10-14-2009, 04:47 PM #188 lastcardcharlie Carpal \'Tunnel     Join Date: Aug 2006 Location: QED, I think Posts: 7,527 Re: The Official Math/Physics/Whatever Homework questions thread 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
10-14-2009, 05:16 PM   #189
Wyman
Carpal \'Tunnel

Join Date: Mar 2007
Location: Redoubling with gusto
Posts: 12,023
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 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?
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...

 10-14-2009, 10:17 PM #190 Skillgannon journeyman   Join Date: Jan 2009 Posts: 283 Re: The Official Math/Physics/Whatever Homework questions thread 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 $img=http://latex.codecogs.com/gif.latex?(1-x^{2})u''-xu'+2u+=+4x^{3}(1+4x)$ 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 $img=http://latex.codecogs.com/gif.latex?(1-x^{2})u''-xu'+4u+=+x^{4}$ 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: $img=http://latex.codecogs.com/gif.latex?u''-\frac{x}{(1-x^{2})}u'+\frac{2}{(1-x^{2})}u+=+0$ Using an integrating factor of $img=http://latex.codecogs.com/gif.latex?\int+e^{\frac{x}{(1-x^{2})}}+=+\frac{1}{(1-x^{2})^{0.5}}$ To give $img=http://latex.codecogs.com/gif.latex?\frac{d}{dx}(\frac{1}{(1-x^{2})^{0.5}}u')+++\frac{2}{(1-x^{2})^{1.5}}u=0$ 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] $img=http://latex.codecogs.com/gif.latex?\frac{d}{dx}[\frac{1}{(1-x^{2})^{0.5}}u']+++(\lambda\rho+++\frac{2}{(1-x^{2})^{1.5}})u=0$ 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 $img=http://latex.codecogs.com/gif.latex?\infty+=<+x+=<+\infty$ So....any help possible? Much obliged.
10-15-2009, 06:54 AM   #191
Pyromantha
veteran

Join Date: Dec 2007
Posts: 2,227
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 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)
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.

 10-15-2009, 08:19 AM #192 Jub. Pooh-Bah     Join Date: Oct 2007 Location: New Zealand Posts: 3,902 Re: The Official Math/Physics/Whatever Homework questions thread 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
 10-15-2009, 02:27 PM #193 solsek Carpal \'Tunnel     Join Date: Feb 2008 Posts: 6,003 Re: The Official Math/Physics/Whatever Homework questions thread 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?
10-15-2009, 02:37 PM   #194
gumpzilla
Carpal \'Tunnel

Join Date: Feb 2005
Posts: 13,973
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by solsek Taking the derivative (-Abr^2e^(-br)) and using the product rule I get: (1/r^2) ( Ab^2r^2e^(-br) - 2Abre^(-br) ) + 2Ae^(-br)/r
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?

 10-15-2009, 02:38 PM #195 solsek Carpal \'Tunnel     Join Date: Feb 2008 Posts: 6,003 Re: The Official Math/Physics/Whatever Homework questions thread 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?
10-15-2009, 02:44 PM   #196
solsek
Carpal \'Tunnel

Join Date: Feb 2008
Posts: 6,003
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by gumpzilla 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?
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.

 10-15-2009, 02:49 PM #197 solsek Carpal \'Tunnel     Join Date: Feb 2008 Posts: 6,003 Re: The Official Math/Physics/Whatever Homework questions thread 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.
10-15-2009, 02:51 PM   #198
gumpzilla
Carpal \'Tunnel

Join Date: Feb 2005
Posts: 13,973
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by solsek From what the prof has taught us so far eigenfunctions are functions when operated on give the original function + some constant.
Not quite. If I have an operator A acting on a function \Psi, I call \Psi an eigenfunction if the following is satisfied:

$img=http://latex.codecogs.com/gif.latex?A+\Psi+=+a+\Psi$

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:

$img=http://latex.codecogs.com/gif.latex?Ab^2+e^{-br}+++2A+(1+-+b)+\frac{e^{-br}}{r}$

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.

 10-15-2009, 02:53 PM #199 solsek Carpal \'Tunnel     Join Date: Feb 2008 Posts: 6,003 Re: The Official Math/Physics/Whatever Homework questions thread 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. Last edited by solsek; 10-15-2009 at 03:01 PM.
 10-15-2009, 03:21 PM #200 solsek Carpal \'Tunnel     Join Date: Feb 2008 Posts: 6,003 Re: The Official Math/Physics/Whatever Homework questions thread 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.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Links to Popular Forums     News, Views, and Gossip     Beginners Questions     Marketplace & Staking     Casino & Cardroom Poker     Internet Poker     NL Strategy Forums     Poker Goals & Challenges     Las Vegas Lifestyle     Sporting Events     Politics     Other Other Topics Two Plus Two     About the Forums     Two Plus Two Magazine Forum     The Two Plus Two Bonus Program     Two Plus Two Pokercast     The Best of Two Plus Two Marketplace & Staking     Commercial Marketplace     General Marketplace     Staking - Offering Stakes     Staking         Staking - Offering Stakes         Staking - Seeking Stakes         Staking - Selling Shares - Online         Staking - Selling Shares - Live         Staking Rails         Transaction Feedback & Disputes     Transaction Feedback & Disputes Coaching & Training     Coaching Advice     Cash Game Poker Coach Listings     Tournament/SNG Poker Coach Listings Poker News & Discussion     News, Views, and Gossip     Poker Goals & Challenges     Poker Beats, Brags, and Variance     That's What She Said!     Poker Legislation & PPA Discussion hosted by Rich Muny     Twitch - Watch and Discuss Live Online Poker     Televised Poker     Two Plus Two Videos General Poker Strategy     Beginners Questions     Books and Publications     Poker Tells/Behavior, hosted by: Zachary Elwood     Poker Theory     Psychology No Limit Hold'em Strategy     Medium-High Stakes PL/NL     Micro-Small Stakes PL/NL     Medium-High Stakes Full Ring     Micro-Small Stakes Full Ring     Heads Up NL     Live Low-stakes NL Limit Texas Hold'em Strategy     Mid-High Stakes Limit     Micro-Small Stakes Limit Tournament Poker Strategy     STT Strategy     Heads Up SNG and Spin and Gos     Mid-High Stakes MTT     Small Stakes MTT     MTT Community Other Poker Strategy     High Stakes PL Omaha     Small Stakes PL Omaha     Omaha/8     Stud     Draw and Other Poker Live Poker     Casino & Cardroom Poker         Venues & Communities         Regional Communities     Venues & Communities     Tournament Events         WPT.com     Home Poker     Cash Strategy     Tournament Strategy Internet Poker     Internet Poker         Winning Poker Network         nj.partypoker.com         Global Poker     Commercial Software     Software         Commercial Software         Free Software General Gambling     Backgammon Forum hosted by Bill Robertie.     Probability     Sports Betting     Other Gambling Games 2+2 Communities     Other Other Topics         OOTV         Game of Thrones     The Lounge: Discussion+Review     EDF     Las Vegas Lifestyle     BBV4Life         omg omg omg     House of Blogs Sports and Games     Sporting Events         Single-Team Season Threads         Fantasy Sports     Fantasy Sports         Sporting Events     Wrestling     Golf     Chess and Other Board Games     Video Games         League of Legends         Hearthstone     Puzzles and Other Games Other Topics     Politics     History     Business, Finance, and Investing     Science, Math, and Philosophy     Religion, God, and Theology     Travel     Health and Fitness     Laughs or Links!     Computer Technical Help     Programming International Forums     Deutsch         BBV [German]     Français     Two Plus Two en Espańol

All times are GMT -4. The time now is 09:00 AM.

 Contact Us - Two Plus Two Publishing LLC - Privacy Statement - Top