mooeypoo Posted February 26, 2011 Share Posted February 26, 2011 Hey guys, I'm practicing for a thermo test later this week. We have this question in the book: Zipper Problem: A zipper has N links; each links has a state in which it is closed with energy 0 or open with energy [math]\varepsilon[/math]. We require, however, that the zipper can only unzip from the left end, and that the link number s can only open if all links to the left (1,2,...s-1) are already open. (a) Show that the partition function can be summed in the form: [math]Z=\frac{1-\exp{(\frac{-(N+1)\varepsilon}{\tau})}}{1-exp{(\frac{\varepsilon}{\tau})}}[/math] Okay, so I approaced the problem by defining the possible energies as a sum of epsilon(n) where it goes from 0, 1, 2, 3 etc. Hence, my partition equation: [math]Z=\sum \exp{(\frac{-n\varepsilon}{\tau})} = \sum (\exp{(\frac{-\varepsilon}{\tau})} )^n[/math] I have been trying to manipulate this further forever. I know it should look like something starting with 1 + exp(..) since with n=0, the exponent will be =1, but I couldn't see how to transform it. Finally, I resorted to the answer sheet, and it says that the above is right, and therefore it's obviously [math]Z=\frac{1-\exp{(\frac{-(N+1)\varepsilon}{\tau})}}{1-exp{(\frac{\varepsilon}{\tau})}}[/math] Obviously? What am I missing? How did they get from the ^n to that? meh. Help! ~mooey p.s I see in another source that they rewrote the summation as: [math]\sum_{s=0}^{N} x^s = \frac{1-x^{N+1}}{1-x}[/math] where [math]x=\exp{(\frac{-\varepsilon}{\tau})}[/math] Great... that looks like an expansion of a power series, and I guess it makes sense if you have the instinct to translate it like the above. Assuming I got stuck with this for an hour, is there any other way to do this, or is this simple a "remember your power series expansion" problem... ? Link to comment Share on other sites More sharing options...

DrRocket Posted February 27, 2011 Share Posted February 27, 2011 Hey guys, I'm practicing for a thermo test later this week. We have this question in the book: Zipper Problem: A zipper has N links; each links has a state in which it is closed with energy 0 or open with energy [math]\varepsilon[/math]. We require, however, that the zipper can only unzip from the left end, and that the link number s can only open if all links to the left (1,2,...s-1) are already open. (a) Show that the partition function can be summed in the form: [math]Z=\frac{1-\exp{(\frac{-(N+1)\varepsilon}{\tau})}}{1-exp{(\frac{\varepsilon}{\tau})}}[/math] Okay, so I approaced the problem by defining the possible energies as a sum of epsilon(n) where it goes from 0, 1, 2, 3 etc. Hence, my partition equation: [math]Z=\sum \exp{(\frac{-n\varepsilon}{\tau})} = \sum (\exp{(\frac{-\varepsilon}{\tau})} )^n[/math] I have been trying to manipulate this further forever. I know it should look like something starting with 1 + exp(..) since with n=0, the exponent will be =1, but I couldn't see how to transform it. Finally, I resorted to the answer sheet, and it says that the above is right, and therefore it's obviously [math]Z=\frac{1-\exp{(\frac{-(N+1)\varepsilon}{\tau})}}{1-exp{(\frac{\varepsilon}{\tau})}}[/math] Obviously? What am I missing? How did they get from the ^n to that? meh. Help! ~mooey p.s I see in another source that they rewrote the summation as: [math]\sum_{s=0}^{N} x^s = \frac{1-x^{N+1}}{1-x}[/math] where [math]x=\exp{(\frac{-\varepsilon}{\tau})}[/math] Great... that looks like an expansion of a power series, and I guess it makes sense if you have the instinct to translate it like the above. Assuming I got stuck with this for an hour, is there any other way to do this, or is this simple a "remember your power series expansion" problem... ? [math]\sum_{s=0}^{N} x^s = 1 +x \sum_{s=0}^{N} x^s - x^{N+1}[/math] [math] (1-x) \sum_{s=0}^{N} x^s = 1-x^{N+1}[/math] [math]\sum_{s=0}^{N} x^s = \dfrac {1-x^{N+1}}{1-x}[/math] 1 Link to comment Share on other sites More sharing options...

mooeypoo Posted February 27, 2011 Author Share Posted February 27, 2011 Thanks! Sorry to be really slow here, but I'm not sure I get how you got to this step: [math]\sum_{s=0}^{N} x^s = 1 +x \sum_{s=0}^{N} x^s - x^{N+1}[/math] Shouldn't the sum in the right hand be s=1 to N now? I'm just trying to understand the steps. The 1+ came from n=0, right? (Sorry, this is obviously "low level" series manipulation, but sadly, I've never really taken any real math classes in those, so it's really a lot of hard work and guessing and tryingt o figure it out on my own here. Sorry if the questions sound stupid.) And thanks for your time! ~mooey Link to comment Share on other sites More sharing options...

DrRocket Posted February 27, 2011 Share Posted February 27, 2011 Thanks! Sorry to be really slow here, but I'm not sure I get how you got to this step: Shouldn't the sum in the right hand be s=1 to N now? No, the "x" in front takes care of that, but adds an "N+1" term that is subtracted out in the last term. I'm just trying to understand the steps. The 1+ came from n=0, right? right 1 Link to comment Share on other sites More sharing options...

timo Posted February 27, 2011 Share Posted February 27, 2011 I think your book assumed that you realize the term is the geometric series. Note that it should be [math]Z=\frac{1-e^{-(N+1)\varepsilon/\tau}}{1-e^{ -\varepsilon / \tau}}[/math], not [math]Z=\frac{1-e^{-(N+1)\varepsilon/\tau}}{1-e^{ +\varepsilon / \tau}}[/math] (i.e. that you forgot the minus sign in the denominator). Link to comment Share on other sites More sharing options...

mooeypoo Posted February 27, 2011 Author Share Posted February 27, 2011 Okay, I see it now, though there's no chance in hell I could've seen it earlier without really.. knowing this trick... But it helps knowing at least why it's like that, so thanks Dr Rocket! And timo, you're right, oops, I forgot the minus - but only online. It came out okay on paper. LaTeX is annoying. In any case, thanks a lot for helping.. I was very frustrating over the fact that it seems I got the answer but not quite the answer, and.. I had no clue why. I guess I should go over a bit of series expansions.... If you have any suggestions on what to start from, that would be great. Otherwise, I guess I'm going google hunting. Thanks! ~mooey Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now