Sutherland Potential Second Virial

[Xittenn]

Could anyone explain how this

 

3 * integral 1 to infinty ( rho^2 ( 1 - exp [ 1/(x * rho^6) ] ) ) d rho

 

becomes this

 

3 * Sum from n=1 to infinity of ( x^-n / n!(6n-3) )

 

I'm a litle foggy when it comes to integral evaluation through series expansion

 

I do understand that exp [ 1/x ] gives the series expansion 1 + x/2! + x^2/3!...

 

how did they get the (6n-3) term

[Xittenn]

e^x.........................ooops

 

I know expand and integrate term by term, it's just not happenin'.................................

 

mathematicas quick and easy error function just won't cut it this time.......................

 

not that I've ever used mathematica except for this time............................guess I don't make amateur mathematician....................

 

the following is a link to a pdf I've been examining to help me understand the steps taken. The final results match the book I've been studying Joseph Noggles Physical Chemistry. I can get pretty close to the final answer now but I still have an extra term 1/rho^(mn-3) m=6 and rho is x in the doc.

 

http://personal.rhul.ac.uk/UHAP/027/PH4211/PH4211_files/sutherland.pdf

 

another thing I've been having a problem with is the subsuming of the leading 1 in 1 + 3*integral(x^2 * exp[1/x*rho]) d rho to be the first term of the expansion how is this allowed???????

 

excuse my mess just trying to iron this one out...................

 

wrong subsuming still not sure how they did subsume the term they had in the document??? Now I'm not sure what happened to the 1 when this 1 + 3*integral(x^2 * exp[1/x*rho]) d rho was expanded and integrated.......it probably explains where 1/rho^mn-3 goes...........I seem to also be caught between two different placements of signing convention theirs and the books!

Edited September 21, 2008 by buttacup
multiple post merged

[Bignose]

1) Can you please use LaTeX to write out the step where you are confused? Using TeX, you can write the math equation pretty much exactly like the paper you linked to.

 

2) Can you please ask a clearer question? I am unsure even exactly what you are asking. (This is at least a little bit related to 1) above.) That is, exactly what step are you confused with? Or what term?

[Xittenn]

:embarass:It took two weeks for me to figure out exactly what I was doing wrong but I finally solved the equation step by step last night. I'm actually quite ashamed to say what I forgot to do so I just won't..............I hope nobody tries hard enough to figure it out, it's kind of embarasing. Thanks for taking the time to at least peak at the question though. I know my question was hard to understand, I was hoping someone with a good understanding of physical chemistry would just look at it and know. The question itself was actually pretty easy, I just haven't been doing much math lately. Sometimes I forget to do the simplest things..............As for Tex I use public access library computers so I don't think I can. When I really feel it's necessary I use ASCII special character codes to write it out but even that's pretty tough to read. Sorry at any rate if there is a next time I'll take the time to write the question out more clearly.

 

Thanks again for at least looking it over!

[Bignose]

You don't need any special software to post in TeX, you just have to use the [ math] and [/ math] tags (without the spaces)

 

For example. Which is easier to read: 3 * integral 1 to infinty ( rho^2 ( 1 - exp [ 1/(x * rho^6) ] ) ) d rho

 

or

 

[math]3\int^{\infty}_{1} (\rho^2( 1-e^{\frac{1}{x \rho^6}}))d\rho [/math]?

 

See http://www.scienceforums.net/forum/showthread.php?t=4236&highlight=latex for a good intro

[Xittenn]

hey thanks I'll look into it...............I wish someone had told me this ages ago!