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Help de-cyphering brevity in a publication

by the.drizzle
Tags: brevity, decyphering, publication
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Nov8-12, 11:03 PM
P: 10

I'll keep this brief for the moment as I think this may be a bit on the involved side, and I don't want to bore people if I can avoid it.

Basically, I found the article I have attached as jones1971.pdf and it relates directly to some work I am trying to do at the moment, and it would be very useful for me to code the author's algorithm in some language--I'm using python. To this end, I've been mostly successful and as he was so kind as to publish tables of results, I've been able to verify the computational accuracy of my code.

However, I am stuck on the last step! The problem I am having is that I cannot figure how he has taken the steps that he has on pp. 621, right column, where he has done:

If we substitute the expansion

p_n = p_{n0} + \frac{1}{P_e^2}p_{n1} + \frac{1}{P_e^4}p_{n2}+\cdots

in the expression

F\left(\eta\right) = \displaystyle\sum_{r=0}^\infty \frac{\left(-1\right)^r}{P_e^{2r}} F_r\left(\eta\right)

and equate to zero, we obtain

F_0\left(p_{n0}\right) =0

The short version of my problem is that I simply cannot figure out the mechanics of the substitution he has alluded to, nor what he has equated to zero. I need to know this in order to structure my code correctly, otherwise it will simply be garbage in --> garbage out, which I'd prefer to avoid if at all possible.

I realize there are quite a few variables here, and this problem as I have presented it probably makes no sense--hence why I have attached the source document.

Thanks in advance for any help, I'm getting close to being out of ideas.
Attached Files
File Type: pdf jones1971.pdf (364.5 KB, 10 views)
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Nov9-12, 10:52 PM
Sci Advisor
HW Helper
P: 9,816
It is being substituted for η. pn is defined to be a root of F(η), so the expression evaluates to zero. I think you then have to consider lim Pe→∞ to get the F0(pn0) = 0 result.
Nov11-12, 04:59 PM
P: 10
Yup, that and expanding F as a Taylor series about p_n seems to do it. Perfect, thanks!

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