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Power Series Expanded, Arfken 5.7.16

  • Thread starter citra
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Homework Statement


The behavior of a neutron losing energy by colliding elastically with nuclei of mass A is described by a parameter ξ1,
ξ1 = 1 + [itex]\frac{(A-1)^2}{2A}[/itex]*ln[itex]\frac{A-1}{A+1}[/itex]

An approximation, good for large A, is
ξ2= [itex]\frac{2}{A+2/3}[/itex]

Expand ξ1 and ξ2 in powers of A−1. Show that ξ2 agrees with ξ1 through (A−1)2. Find the difference in the coefficients of the (A−1)3 term.

Homework Equations



Taylor Expansion, see wiki page

ln(1-x)= x+ [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+[itex]\frac{x^n}{n}[/itex]

ln(1+x)= x- [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+(-1)^(n+1)[itex]\frac{(x^n)}{n}[/itex]

The Attempt at a Solution



Alright so since it is in powers of A−1 I decided to substitute x=A−1. I get:

ξ1= 1+ [itex]\frac{x((1/x)-1)^2}{2}[/itex]* ln(1-x)-ln(1+x)

I use the expansions from above and substitute.
Now I do not know how to expand the entire thing from here.
 
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Answers and Replies

  • #2
vela
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Homework Statement


The behavior of a neutron losing energy by colliding elastically with nuclei of mass A is described by a parameter ξ1,
ξ1 = 1 + [itex]\frac{(A-1)^2}{2A}[/itex]*ln[itex]\frac{A-1}{A+1}[/itex]

An approximation, good for large A, is
ξ2= [itex]\frac{2}{A+2/3}[/itex]

Expand ξ1 and ξ2 in powers of A−1. Show that ξ2 agrees with ξ1 through (A−1)2. Find the difference in the coefficients of the (A−1)3 term.

Homework Equations



Taylor Expansion, see wiki page

ln(1-x)= x+ [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+[itex]\frac{x^n}{n}[/itex]

ln(1+x)= x- [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+(-1)^(n+1)[itex]\frac{(x^n)}{n}[/itex]
Suggestion: Use LaTeX for the entire expression instead of for just bits and pieces.

You should clean up your math a bit. Those expressions above aren't true because the series on the righthand sides have an infinite number of terms. They don't terminate with the nth term.

The Attempt at a Solution



Alright so since it is in powers of A−1 I decided to substitute x=A−1. I get:

ξ1= 1+ [itex]\frac{x((1/x)-1)^2}{2}[/itex]* ln(1-x)-ln(1+x)

I use the expansions from above and substitute.
Now I do not know how to expand the entire thing from here.
Again, what you wrote is incorrect. Use parentheses (or square brackets, etc.)!
[tex]\xi_1 = 1+\frac{x}{2}\left(\frac{1}{x}-1\right)^2[\log(1-x)-\log(1+x)][/tex]You just need to substitute the series in, combine terms, and then multiply everything out and simplify. In other words, do a bunch of algebra. Note that since you're only looking for the first few terms, those are the only ones you really need to worry about. Anything that results in higher powers of x you can ignore.
 

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